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COURSE
MATHEMATICS.
IN TWO V LUMES.
COMPOSED E USE OP
THE ROYAL MILI'lARY ACADEMY.
CHARLES HUTTON, LL.D. F.R.S.
FORMERLY PROFESSOR OF MATHEMATICS IN THAT
INSTITUTION.
CONTINUED AND AMENDED BY
OLINTHUS GREGORY, LL.D. F.R.A.S.
VOL. I.
TWELFTH EDITION,
WITH CONSIDERABLE ALTERATIONS AND ADDITIONS,
BY
THOMAS STEPHENS DA VIES, F.R.S. L. & E. F.S.A.
ROYAL MILITARY ACADEMY.
LONDON:
LONGMAN, ORME, & CO.; J. M. RICHARDSON; J. 0. F. & J. RIVINOTON; HAMILTON
& CO.; WHITTAKER & CO.; DUNCAN & MALCOLM; 8IMPKIN, MARSHALL. * CCi
J. SOUTER; COWIE & CO.; SMITH, ELDER, & CO.; ALLEN ft CO.; HARVEV ft
DARTON; HOULSTON & STONEMAN; H. WASHBOURNE ; L. A. LEWIS; C. DOLMAN ;
AND G. ROUTLEDGE. EDINBURGH : STIRLING ft CO.
1841.
xJ .1
LONDON:
GILBERT AND RIVINOTON, rUINTEUS,
ST. John's square.
THE
EDITOR^S PREFACE.
The circulation of nearly thirty tliousand copies of Hutton's Course, suflB-
ciently attests the estimation in which it has been held by mathematical
teachers and students throughout the countrj'.
" Long experience in all seminaries of learning," says the author in his
original preface, "has shown that such a work was very much wanted, and
would prove a great and general benefit ; as for want of it, recourse has
always been obliged to be had to a number of other books by different
authors ; selecting a part from one and a part from another, as seemed most
suitable to the purpose in hand, and rejecting the other parts — a practice
which occasioned much expense and trouble, in procuring and using such
a number of odd volumes of various forms and modes of composition ;
besides wanting the benefits of uniformity and reference which are found in
a regular series of composition."
Dr. Mutton's Course of Mathematics was greatly in advance, as to the
manner of treating the subjects contained in it, of all works which had then
appeared ; and for many years the author continued to improve the succes-
sive editions, as new discoveries were made or new methods invented. At
the close of his long, laborious, and useful life, he committed the work to
the care of Dr. Gregory, who, by continual additions and modifications,
endeavoured to assimilate it to the growing spirit of inquiry produced by
a long period of general peace. Into the last edition, however, greater
changes were introduced than had been made in the work since its first
composition ; and he did me the honor, soon after my appointment to
this Institution, to request me to make the greater part of those altera-
tions, under his editorship. It was, however, a matter of deep regret to
both of us, that owing to the haste with which the work was urged through
the press, adequate time was not given to complete our contemplated
improvements. The same cause, also, gave rise to a great number of
errata. I am noiw, however, not without a hope, that the present edition,
whilst as free from errata as any mathematical work extant, will be found
a2
iV THE editor's PREFACE.
to justify the views under which the alterations were commenced, and to
give it that preference as a text-book for mathematical instruction which
the original work so long enjoyed.
The state of the health of my lamented friend and coadjutor not allowing
him to give the attention essential to the editorship of the work, has com-
mitted it wholly to my care, to carry out our joint views to the best of my
ability. It was not, however, without some reluctance and much anxiety,
that I undertook it : and for more than twelve months the present volume
has been the unremitting object of my entire labour. Even yet, I am
obliged to defer a few of our contemplated improvements for a future
edition.
In the arithmetic very little alteration has been made, except a few
occasional notes ; and in the early part of the algebra comparatively few
essential alterations have been made from the last edition. In the multi-
plication and division I have given prominence to the use of detached
coefficients and the synthetic method of division. An elementary investiga-
tion of the latter process is annexed, as that of Mr. Homer is not easily
understood except by students whose progress is considerably more ad-
vanced : but a still simpler and more direct one is given amongst the
" Additions" at the end of the volume, and which I discovered since sheet
K, (p. 128), was printed off. To the simpler operations of algebra, where
the reason of the step is not apparent at once, investigations are annexed,
to secure to the student a complete understanding of the logic of his pro-
cesses.
In the chapters on simple and quadratic equations, the introductory
remarks and suggestions, as well as the examples chosen for illustrating
the methods by actual working, have been generally exchanged for others
better adapted to show the true character of the operations. In the
quadratics, the Hindu method of completing the square is enforced, as
being generally superior, in respect of facility, to the Italian or common
one.
The chapter on the general resolution of numerical equations has been
wholly recomposed ; and I hope it will be found free from those logical
defects which are so liable to insinuate themselves into abbreviated treatises
on subjects involving so many distinct principles as this does. The theory
of equations, is, however, carried no further than is requisite for numerical
solution : though to this extent, great pains have been taken to render it
logically complete. Legitimate proofs, on elementary principles, are given
of the criteria of De Gua and Budan, for detecting the imaginarj' roots of
an equation ; and as brief a form of investigating Sturm's criterion as I
could devise, has also been added. Though I am as fully impressed as any
THE EDITOR S PREFACE. ^
one can be of the great beauty and importance of Sturm's theorwn, I have
been led, I confess, to introduce it here more in accordance with the dictum
of the mathematical public, than from my own conviction of it* practical
utility in reference to numerical solution — at leaat till 800)6 method 1cm
operose and practically embarrassing than is yet known, shall be dincnvcied.
of forming his successive auxiliary functions subsequent to the first deri-
vative. Should such a method, in any way analogous to Ilorner'ii proccM
for transformation, ever be invented, Sturm's theorem will become prac-
tically useful : — but not till then *.
Upon Homer's method of continuous approximation to the roots of equa-
tions, I have dwelt at sufficient length to render it easy of comprehension.
As the first attempt ever made to compose an elementary treatise on this
subject was made by myself in the previous edition of this work, my
attention was naturally directed to it subsequently with sufficient precision
to enable me to separate the essential and the useful part of that coni|)o-
sition, from the parts which I found superfluous, and make such additions
as experience might suggest during my professional use of the volume.
The chapters on indeterminate coefficients, piling of balls, the binomial
and exponential theorems, and on logarithms, it will be seen are all written
anew, and with especial reference to the order in which the subjects natu-
rally present themselves in a systematic course of study. The same may
be said of the chapters on series and finite differences.
The early part of the geometry is unaltered, though in a future edition I
propose to remodel it entirely. The doctrine of ratio is put altogether in a
new and, I persuade myself, a perfectly logical form ; and the theorems
depending on ratio are changed in their manner of demonstration, to be in
accordance with the same principles. A few theorems of great practical
value are added. The chapters on the geometry of planes and solids have
also, for the most part, been modified and rewritten.
The practical geometry has been entirely recomposed, and in especial
reference to the circumstances under which the problems themselves occur
in practice. A number of constructions of this kind, which arc believed
to be new, and are adapted to peculiar exigences, have been intro<luccd.
• This page was in type before I met with the elegant and instrurtire .yfatkrmatical IHaerlm-
/tons of Professor Young. In that volume, an important improvement i» ni»<lc in •iniplifvinf
the numerical process, and especially the initial step of it. .V few more »U( h iniprorrin*nU
would entirely remove the objections which, in a practical point of view, are urged >{;mimt
that important and beautiful process.
Independently of the discussion of the different point* connerted with Stunn't crilprion. lJ>«
volume of Mr. Young deserves the attention of the mailiematiral Mudcnt, bcjond any oik«r
■work that I could suggest to him : as he will be at once led to study the logic of the procriw*
involved in elementarj- mathematics, with more precision than in «ny other pn<d»ic«ion •"■
which I am acquainted.
VI THE EDITOR S PREFACE.
The chapter on practical geometry in the field contains a series of pro-
blems of great importance to the military profession, to engineers and sur-
veyors, and which form the substance of a course of lectures just delivered
at the Royal Artillery Institution.
In the plane trigonometry nothing besides the examples for exercise, of
the last edition, remains in this. To give every thing essential to elemen-
tary trigonometry investigated in a direct and simple manner, and en-
tirely to exclude all matters of mere scientific curiosity, has been my guid-
ing principle in the composition of these chapters. Trigonometry, there-
fore, instead of forming two separate treatises in two successive volumes,
is now brought entirely into the first ; and the examples that are changed
in place have been marked by a quotation of the places in which they
previously stood, for the convenience of those who wish to make reference
to any works founded on the preceding edition. The few additional ex-
amples here given, will ofier no difiiculty to those who have fully mastered
those of that edition.
In the mensuration the investigations are, for the sake of continuity,
classed difierently, but the problems have the same order as before ; and a
few easy additional examples are added, as the paucity of these in the
former editions has often been a subject of complaint. In the artificers'
work and land surveying no changes are made in this edition, but a com-
plete revision of them will be given hereafter, — and, but for practical
obstacles, would have been given now, as no parts of the entire course
require it more than these.
The figures in this edition are nearly all newly-cut, and every attention
has been paid to the arrangement of each page, both for convenience of
reading and reference, and of losing no space that could possibly be filled
up with useful matter. Much of the phraseology, and the entire notation,
of former editions has been modernised, and an attempt has been made to
render it, with the exceptions already specified, consistent and systematic
throughout.
T. S. Davies.
lioycd MU'dnrif Ac^ulemy, Wtxjiicteh,
lOlh FSruarif, 1841.
CONTENTS.
The alterations made in this edition have caused tlic paging of the work to lie ahrrr<i ; and om
in some works references are made to tlie last edition, it lias been rnnfidercd advantngcouk to
give, besides the pages of this edition, the corresponding pages of the prtcfding one : ilic fint
column refers to this, and tlie second to the preceding edition. The few articles brought from
the second volume are marked with an asterisk.
ARITHMETIC.
PAOE.S
Notation and numeration 3 3
Addition 7 7
Subtraction 8 8
Multiplication 9 9
Division 13 13
Reduction 18 17
Tables of weights and measures 18 18
Comparison of Frpnch and English... 2.5 23
Compound addition 27 25
Table of Officers' pay 29 27
Compound subtraction 30 28
multiplication 32 30
division 34 32
Rule of three 3(5 34
Compound proportion 40 37
Vulvar fractions 41 39
Reduction 42 40
Addition .50 4fi
Subtraction 51 47
Multiplication 51 48
Division 52 49
Rule of three 52 49
Decimals, or decimal fractions 54 49
Addition 54 50
Subtraction 55 50
* Multiplication .55 51
Division 57 53
Reduction .59 54
Ruleof three 62 57
Duodecimals 63 58
Involution and EvoltUion 65 60
Extraction of the square root... 66 62
cube root ... 69 64
any root 73 69
Progressions and ratios 75 79
Arithmetical progi'ession 76 80
Geometrical progression 79 82
Harmonical progression 82 85
Single fellowship 83 86
Double fellowship 84 88
Simple interest 85 89
Compound interest 88 91
Allegation, medial 89 92
, alternate 90 93
Position, single 93 96
, double 93 97
Practical questions 96 100
ALGEBRA.
Introductory chapter 100 104
Definitions, notation, and exer-
cises 102 106
Addition 112 115
Subtraction 115 119
Multiplication 117 121
• , by detached coef-
ficients 120 123
Div
-, by detached coefficients ... 127
-, syntliclic metiiod Ii7
-, another ])n)of of
cxjiansion by synthetic di-
PAGKH
123 12.5
12f»
129
vision 131 \(',n
Problems on these rules 131 |3I
Greatest common me;isure 13"J
Least common multiple 1;$/;
Alyehraic Fractions 1,37
Reduction I37
Addition 141
Subtraction \\2
Multiplication 143
Division I44
Continued fractions 145
Involiitiwi a?id evolution 147
Powers and roots of mono-
mials 147
Square root of compound
expressions 148
Cube root of ditto 149
Reduction of monomial surds... 1.50
binomial surds ... 157
Progressions
Arithmetical progression 159
Geometrical progression 162
Circulating decimals 164
Geometrical proportion 165
Harmonical progression 167
Equations 169
Simple equations 170
. , simultaneous 174
■ , problems ... 181
Quadratic equations, problems 185
, simulta-
neous 189
Quadratic equations, problems ....j^l93
Cubic and f/iquadratic equations 1 98
Cardan's solution of the cubic 198
Simpson's, of the biquadratic... 200
Solution of equations by trial
i<- and error 202
7'Ae numerical Solution of E'fuations 206
Definitions and notation 206
Calculation of an expression.... 208
To increase ordiminish the roots 209
To change the signs of the roots 212* 227
To multiply or divide the roots 213 227
To form an equation from given
roots 214
Expression for the transformed
equation 21,5
Equations with equal roots 216
Imaginary roots are in pairs 217
Irration.al roots are in pairs 218
Harriot's rule of signs 219
Effect of successivesubslitutions 221
Limits of the roots 222
132
1.32
137
138
i:w
140
141
141
146
147
153
155
162
164
163
165
167
177
183
191
195^
199 £%.j,^^ft
206 Ka.r~*^>
IS '•'!"
212
214
219
219
223
224 f .
It CiMf^t^
228
J30^U.?>
228
228
221
•>>)
2.30
Theorem on approximation 223 233
vm
CONTENTS.
PAGES
De Gua's criterion 224 230
Sudan's criteriou 226 231
Sturm's criterion 228
Homer's method of approxima-
tion 232 233
Recapitulation of processes 234
Indeterminaie Co-efficients 236 238
Piling of balls 240 158
. triangular pile 240 158
square pile 240 159
rectangular pile 241 161
. incomplete piles 241
Binomial theorem 242 240
Method of working.. 245 172
Exercises for expan-
sion by 245 173
Approximation to the
roots of numbers 247 174
Exponential theorem 247 242
Theory of logarithms
Definitions and properties... 248 244
Logarithmic series 250 246
Computations of logarithms 251 247
Logarithmic tables 252 248
Definitions 252
Tabular theorems 253
Description of tables 254 248
Use of the tables 256 248
Logarithmic operations 257 251
Exponential equations 261 255
Simple interest 264 258
Compound interest 265 258
Annuities 267 261
Series by subtraction 270 265
Reversion of series 272 267
Method of finite differences 273 267
Definitions, notation, and .
principles 273 268
General term of, with order
of differences 274 270
General term of a series 277 277
To find the several orders... 278 270
To continue a given series .. 279 279
To transform powers into
factorials 279
Integration of general term 281 271
To find the sum of a series.. 283 273
285 279
Interpolation of series 287 281
Scholium on piles of balls... 289 284
THEORETICAL GEOMETRY.
Definitions in plane geometry 290 286
Axioms 296 292
Theorem independent of ratio 296 292
Ratios and proportions 3*22 317
Definitions 322 317
Principle employed (note) 323
Theorems depending on ratio... 324 320
Theorems depending on ratio 328 322
Exercises in plane geometry 343 331
Definitions of solid geometry 347 334
Theorems on lines and angles 349 336
Exercises 359 342
Theorems on solid angles 360 349
on volumes of solids 364 343
for exercise 369 352
PRACTICAL GEOMETRY.
PAGES
Introductory explanations 370 353
Construction of plane problems 372 354
A line nearly equal to the circle .. ^ 400
To measure an angle by the com- 402
passes only 402
To find the diameter of a sphere ...
Field problems 403
Application of algebra to geometry.. 413 377
PLANE TRIGONOMETRY.
Introduction 421
Definitions and notation 422 383
Relations of the functions of an arc 423 388
Functions of two arcs 425 401
Particular relations of arcs 427
Values of functions of arcs 429
Trigonometrical tables, their con-
struction and usage 431
Expansion of sin J, cos J 435 386
Euler's and Demoivre's theorems... 437. 36*
Multiple arcs 437 37*
Subsidiary angles 438
Change of the radius 441
Inverse functions 442
Exercises on arcs 443 •{ o-«
Right-angled triangles 445 399
, calculation of 446 400
Oblique-angled triangles 447 390
case 1 (calculation) 450 390
case 2 453 394
case 3 455 397
App. to heights and distances 457 | <«,
Miscellaneous exercises 468 34*
MENSURATION.
Areas of rectilineal figures 471 415
Length and area of the circle 475 420
Problems on plane surfaces 478 414
Surfaces and volumes of solid figures
with plane boundaries 486 426
The cone, cylinder, and sphere 487 427
Problems on surfaces and volumes.. 489
Land surveying 492 432
Bricklayers' work 510 4.55
Masons' 512 456
Carpenters' 512 457
Slaters' and tilers' 513 458
Plasterers' 514 458
Painters' 514 459
Glaziers' 515 460
Pavers' 515 460
Plumbers' 516 460
Timber measuring 516 461
Practical exercises 518 463
Notes 524
Tables of Squares, Cubes, and
Roots 529 72
A
COURSE
OF
MATHEMATICS,
DEFINITIONS.
1. Quantity, or Magnitude, is that which admits of increase or decrease.
Those kinds of magnitude only which are capable of estimation in comparison
with some unit of the same kind, are the projjcr subjects of mathematical study.
2. Arithmetic is conversant with numbers only in their abstract state. Algebra
contemplates the subjects of arithmetic in a more general form ; and generally
(among other objects) furnishes the rules for the more complex arithmetical
operations. Geometry treats of space, as of the forms, magnitudes, and positions
of figures. The differential and integral calculus, the calculus of functions, &c.
are also branches of Algebra, but of which no definite idea could be conveyed
till the student's progress is considerably extended.
3. The sciences of Arithmetic and Geometry are styled the Pure Mathematics :
whilst all applications of thetn to physical, civil, or social inquiries, (as Me-
chanics, Astronomy, Optics, Life Insurance, Population, &c.) constitute what is
termed the Mixed Mathematics.
4. In mathematics are several general terms or principles; such as. Defi-
nitions, Axioms, Propositions, Theorems, Problems, Lemmas, Corollaries,
Scholia, &c.
5. A Definition is the explication of any term or word in a science ; showing
the sense and meaning in which the term is employed. — Every Definition ought
to be clear, and expressed in words that are common, and perfectly well under-
stood.
6. A Mathematical Proposition refers either to something proposed to be
demonstrated, or to something required to be done ; and is accordingly either a
Theorem or a Problem.
7. A Theorem is a demonstrative Proposition; in which some property is
asserted, and the truth of it required to be proved. Thus, when it ia said that.
2 GENERAL PRINCIPLES.
The sum of the three angles of a plane triangle is equal to two right angles, that
is a Theorem, the truth of which is demonstrated by Geometry. A set or col-
lection of such Theorems constitutes a Theory.
8. A Problem is a proposition or a question requiring something to be done ;
either to investigate some truth or property, or to perform some operation. As,
to find out the quantity or sum of ail the three angles of any triangle, or to draw
one line perpendicular to another. A Limited Problem is that which has but
one answer or solution. An Unlimited Problem is that which has innumerable
answers. And a Determinate Problem is that which has a certain number of
answers.
9 Solution of a Problem, is the resolution or answer given to it. A Numerical
or Numeral Solution is the answer given in numbers. A Geometrical Solution is
the answer given by the principles of Geometry. And a Mechanical Solution
is one which is gained by trials.
10. A Lemma is a preparatory proposition, laid down in order to shorten the
demonstration of the main proposition which follows it.
11. ^ Corollary, or Consectary, is a consequence drawn immediately from
some proposition or other premises.
12. J Scholium is a remark or observation made upon some foregoing pro-
position or premises.
13. An Axiom, or Maxim, is a self-e\'ident proposition ; requiring no formal
demonstration to prove its truth ; but received and assented to as soon as men-
tioned. Such as. The whole of any thing is greater than a part of it ; or, The
whole is equal to all its parts taken together ; or, Two quantities that are each
of them equal to a third quantity, are equal to each other.
14. A Postulate, or Petition, is something required to be done, which is so
easy and evident that no person will hesitate to allow it.
15. An Hypothesis is a supposition assumed to be true, in order to argue from,
or to found upon it the reasoning and demonstration of some proposition.
16. Demonstration is the collecting the several arguments and proofs, and
laying them together in proper order, to establish the truth of the proposition
under consideration.
17. A Direct, Positive, or Affirmative Demonstration, is that which concludes
with the direct and certain proof of the proposition in hand.
18. An Indirect, or Negative Demonstration, is that which shows a proposition
to be true, by pro\nng that some absurdity would necessarily follow if the
proposition advanced were false. This is also sometimes called Reductio ad
absurdum; because it shows the absurdity and falsehood of all suppositions
contrary to that contained in the proposition.
19. Method is the art of disposing a train of arguments in a proper order, to
investigate either the truth or falsity of a proposition, or to demonstrate it to
others when it has been found out. This is either Analytical or Synthetical.
20. Analysis, or the Analytic Method, is the art or mode of finding out the
truth of a proposition, by first supposing the thing to be done, and then reason-
ing back, step by step, till we arrive at some known truth. This is also called
the Method of Invention, or Resolution j and is that which is commonly used in
Algebra.
21. Synthesis, or the Synthetic Method, is the searching out truth, by first
laying down some simple and easy principles, and then pursuing the conse-
quences flowing from them till we arrive at the conclusion. This is also called
the Method of Composition : and is the reverse of the Analytic method, as this
proceeds from known principles to an unknown conclusion ; while the other
NOTATION AND NUMEIL\TION. 3
goes in a retrograde order, from the thing sought, considered as if it were true
to some known principle or fact. Therefore, when any truth has l)ccn diJ
covered by the Analytic method, it may be demonstrated by reversing the proceni
or by Synthesis: and in the solution of geometrical propositions, it is very
instructive to carry through both the analysis and the synthesis.
ARITHMETIC.
Arithmetic may be viewed as a subject of speculation, in which light it ia a
science ; or as a method of practice, in which light it is an art.
As a science, its objects are the properties and relations of numliers under
any assigned hypothesis respecting their mutual relations or methods of com-
parison and combination.
As an art, it proposes to discover and put into a convenient form, compen-
dious methods of obtaining those results which flow from any given methods of
combining given numbers ; but which results could, in the absence of these
compendious methods, only be ascertained by counting the numbers themselves
into one single and continuous series.
When it treats of whole numbers, it is called Vulgar, or Common Arithmetic ;
but when of broken numbers, or parts of numbers, it is called Fractions.
Unity, or an Unit, is that by which every thing is regarded as one ; being the
beginning of number; as, one man, one ball, one gun.
Number is either simply one, or a compound of several units; as, one man,
three men, ten men.
An Integer, or Whole Number, is some certain precise quantity of units ; as,
one, three, ten. These are so called as distinguished from Fractions, which
are broken numbers, or parts of numbers ; as, one-half, two-thirds, or three-
fourths.
A Prime Number is one which has no other divisor than unity; as, 2, 3, 5, 7,
17, 19, &c. A Composite Number is one which is the product of two or more
numbers; as, 4, 6, 8, 9, 28, 112, &c.
A Factor of a composite number, or simply a Factor, is any one of the num-
bers which enters into the composition of that composite number.
NOTATION AND NUMERATION.
These rules teach how to denote or express any proposed number, either by
words or characters : or to read and write down any sum or number.
The Numbers in Arithmetic are expressed by the following ten digits, or
Arabic numeral figures, which were introduced into Europe by the Moors, alniut
eight or nine hundred years since ; viz. 1 one, 2 two, 3 three, 4 four, 5 five,
6 six, 7 seven, 8 eight, 9 nine, 0 cipher, or nothing. These characters or figures
were formerly all called by the general name of Ciphers; whence it came to pass
that the art of Arithmetic was then often called Ciphering. The first nine are
called Significant Figures, as distinguished from the cipher, which is of itself
quite insignificant as a number.
Besides this value of those figures, they have also another, or local value,
B 2
4 ARITHMETIC.
which depends on the place they stand in when joined together; as in the fol-
lowing table :
i
CD
c
o
s
CO
<<-<
(M
3
o
o
O
O)
i
00
•73
c
eo
•^3
5
«^
aj
u^
3)
ui
o
ui
O
t.
'a
.2
•73
3
O
•n
tn
22
%
C
3
CO
C
c
3
C
D
Eh
t:)
9
8
7
6
5
4
3
2
1
9
8
7
6
5
4
3
2
9
8
7
6
5
4
3
9
8
7
6
5
4
9
8
9
7
8
9
6
7
8
5
6
7
9 8
Here, any figure in the first place, reckoning from right to left, denotes only
its own simple value ; but that in the second place, denotes ten times its simple
value ; and that in the third place, a hundred times its simple value ; and so
on : the value of any figure, in each successive place, being always ten times its
former value.
Thus, in the number 1796, the 6 in the first place denotes only six units, or
simply six ; 9 in the second place signifies nine tens, or ninety ; 7 in the third
place, seven hundred ; and the 1 in the fourth place, one thousand : so that the
whole number is read thus, one thousand seven hundred and ninety-six.
As to the cipher, 0, though it signify nothing of itself, yet being joined on
the right-hand side to other figures, it increases their value in the same ten-fold
proportion : thus, 3 signifies only five ; but 50 denotes 5 tens, or fifty ; and 500
is five hundred ; and so on.
For the more easily reading of large numbers, they are divided into periods
and half-periods, each half-period consisting of three figures ; the name of the
first period being units ; of the second, millions ; of the third, millions of mil-
lions, or bi-millions, contracted to billions ; of the fourth, millions of millions
of millions, or tri-millions, contracted to trillions, and so on. Also the first part
of any period is so many units of it, and the latter part so many thousands.
The following Table contains a summary of the whole doctrine.
Periods. Quadril. ; Trillions; Billions; Millions; Units.
Half-per.
Figures
th. un. th. un. th. un. th. un. th. un.
123,456; 789,098; 765,432; 101,234; 567,890;
And the whole may be thus read : — One hundred and twenty-three thousand,
four hundred and fifty-six quadrillions ; seven hundred and eighty-nine thou-
sand, and ninety-eight trillions ; seven hundred and sixty-five thousand, four
hundred and thirty-two billions ; one hundred and one thousand, two hundred
and thirty-four millions ; five hundred and sixty-seven thousand, eight hundred
and ninety.
NOTATION AND NUMERATION. 5
Numeration is the reading of any number in words lliat is propowtl or let
down in figures; which will be easily done by help of the following rule,
deduced from the foregoing tables and observations; viz.
Divide tbe figures in the proposed number, as in the summary above, into
periods and half-periods ; then begin at the left-hand side, and read the figure*
with the names set to them in the two foregoing tables.
EXAMPLES.
Express in words the following numbers ; viz.
34 15080 ] 3405670
96 72033 47050023
380 109026 309025600
704 483500 4723507CS9
6134 2500639 274856390000
9028 7523000 6578600307024
Notation is the setting down in figures any number projwscd in words;
which is done by setting down the figures instead of the words or names belong-
ing to them in the summary above ; supplying the vacant places with cipher*
where any words do not occur.
EXAMPLES.
Set down in figures the following numbers :
Fifty-seven.
Two hundred and eighty-six.
Nine thousand two hundred and ten.
Twenty-seven thousand five hundred and ninety-four.
Six hundred and forty thousand, four hundred and eighty-one.
Three millions, two hundred and sixty thousand, and one hundred and six.
Four hundred and eight millions, two hundred and fifty-five thousand, one
hundred and ninety-two.
Twenty-seven thousand and eight milhons, ninety-six thousand, two hundred
and four.
Two hundred thousand and five hundred and fifty millions, one hundred and ten
thousand, and sixteen.
Twenty-one billions, eight hundred and ten millions, sixty-four thousand, one
hundred and fifty.
OF THE ROMAN NOTATION.
The Romans, like several other nations, expressed their numbers by certain
letters of the alphabet. The Romans used only seven numeral letters, being the
seven following capitals; viz. i for one; v ior five : x for ten; l for fifiy :
c for a hundred: d ior five hundred ; m for a thousand. The other numbers
they expressed by various repetitions and combinations of these, after the fol-
lowing manner :
As often as any character is repeated, to
many times is its value repeated.
A less character before a greater dimi-
nishes its value.
A less character after a greater iocre«8e«
its value.
10 = x
1
=
I
2
=
II
3
=
III
4
=
mi or IV
5
=
V
6
=
VI
7
=
VII
8
=
VIII
9
=
IX
6 ARITHMETIC.
50 = L
100= c
500 = D or 13 For every o annesed, this becomes 10
times as many.
1000 = M or CI3 For every c and o placed one at each
2000 ^ MM end, it becomes 10 times as much.
5000 =\- OT iDD A bar over any number increases it
6000 = VI 1000 fold.
10000 = x^or cciOD
50000 = L or 1033
60000 = Lx
100000 = C or CCCI333
1000000 := M or CCCCI3333
2000000 = MM
&c. &c.*
EXPLANATION OP CERTAIN CHARACTER.^.
There are various characters or marks used in Arithmetic and Algebra, to
denote several of their operations and propositions f ; the chief of which are as
follow :
+ signifies plus, or addition.
— - - minus, or subtraction.
X or . - multiplication.
-r - - division.
• 1 1 • - proportion.
= - - equality.
a/ - - square root.
V' - - cube root, &c.
«5 - - difference between two numbers when it is either not known, or
not necessary to state, which is the greater.
Thus, 5 + 3, denotes that 3 is to be added to 5.
6 — 2, denotes that 2 is to be taken from 6.
7 X 3, or 7 . 3, denotes that 7 is to be multiplied by 3.
8 -T- 4, denotes that 8 is to be divided by 4.
2 : 3 I I 4 : 6, expresses that 2 is to 3 as 4 is to 6.
G + 4 ^ 10, shows that the sum of 6 and 4 is equal to 10.
\/ 3, or 3i, denotes the square root of the number 3.
V 5, or ok denotes the cube root of the number 5.
7^ denotes that the number 7 is to be squared.
8', denotes that the number 8 is to be cubed.
2 c/5 6 signifies the difference between 2 and 6.
&c. &c. &c.
See, farther, the definitions in Algebra.
* To those students whose taste leads them to inquire into the History of Arithmetic, refe-
rence is especially made to Professor Leslie's Philosophy of Arithmetic, to the Rev, Georpe
PeacocJis Treatise on A riihmetic, in the Encyclopedia Meiropolitana, to a paper by the cele-
brated Humboldt, read before the Royal Academy of Berlin, of which a translation is jirinted
in the Journal of tfie Royal Institution, vol. xxix. ; and to a paper in the Bath and Bristol
Moffozine for Oct. 1833 (No. viii.) by Mr. Davies. Other references will be found in those
works which, for want of room, must be omitted here.
+ All such symbols as designate operations to be performed, are called symbols of operation,
and those which designate quantities of any kind are called symbols of quantity.
ADDITION.
OF ADDITION.
AoniTioN is the collecting or putting of several numbers together, in order to
find their sum, or the total amount of the whole. This is done m follows :
Set or place the numbers under each other, so that each figure may stand
exactly under the figures of the same value, that is, units under units, tens under
tens, hundreds under hundreds, &c. and draw a line under the lowest number,
to separate the given numbers from their sum, when it is found. — Then add up
the figures in the column or row of units, and find how many tens are contained
in that sum. Set down exactly below, what remains more than those ten.-*, or
if nothing remains, a cipher, and carry as many ones to the next row as tlicre
are tens. — Next, add up the second row, together with the number carried,
in the same manner as the first : and thus proceed till the whole is finished,
setting down the total amount of the last row.
TO PROVE ADDITION.
First Method. — Begin at the toj), and add together all the rows of numbers
downwards, in the same manner as they were before added upwards ; then if
the two sums agree, it may be presumed the work is right. — This method of
proof is only doing the same work twice over, a little varied.
Second Method. — Draw a line below the uppermost number, and suppose it
cut off. — Then add all the rest of the numbers together in the usual way, and
set their sum under the number to be proved. — Lastly, add this last found num-
ber and the uppermost line together; then if their sum be the same as that
found by the first addition, it may be presumed the work is right. — This method
of proof is founded on the plain axiom, that '• The whole is equal to all its parts
taken together."
Third Method. — Add the figures in the ujjpermost line example i.
together, and find how many nines are contained in their 3497 « 5
sum. — Reject those nines, and set down the remainder Co 12 .5 5
towards the right hand directly even with the figures in ^295 ^ o
the line, as in the annexed example. Do the same . „„ „ _
with each of the proposed lines of numbers, setting all g
these excesses of nines in a column on the right hand, as r^
here, 5, 5, 6. Then, if the excess of gs in this sum, found
as before, be equal to the excess of 9-s ia the total sum 18304, the work is pro-
bably right.— Thus, the sum of the right-hand column, 5, 5, 6, is 16, the excess
of which above 9 is 7. Also thu sum of the figures in the sura total 18304, is
16, the excess of which above 9 is also 7, the same as the former.*
♦ This method of proof depends on a property of the number 9, which, except the number 3,
belongs to no otlier digit whatever; namely, that "any number divided by 9, will leave the
same remainder as the sum of its figures or digits divided by 9 :" which may be demonslntcd
in this manner.
Demonstration.— Let there be any number proposed, as 4fi.58. This, separated into its severU
parts, becomes 4000 + 600 + 50 + 8. But 4000 = 4 X 1<W« = 4 X (''iW + 1 ) = (4 X 999)
+ 4. In like manner 600 = (6 x 99) + 6; and 50 = (5 X 9) + 5. Therefore the given
number 4658 = (4 X 999) -}- 4 + (6 X 99) -|- 6 -h (5 X 9) + 5 + 8 = (4 x 999) -|- (« X
99) -f- (5 X 9) + 4 + 6 -f- 5 -f 8 ; and 4G58 -i- 9 = (4 X 999 -}- 6 X 99 + 5 X 9 -|- 4
^ 6 -f 5 + 8) -^. 9. But (4 X 999) + (6 X 99) 4- (5 X 9) is evidently divi»iblc by 9,
without a remainder ; therefore if the given number 4658 be divided by 9, it will leave the
same remainder as44-6 + 5-|-8 divided by 9. And the same, it is evident, will hold for
any other number whatever.
ARITHMETIC.
OTHER EXAMPLES.
2. 3. 4.
12345 12345 12345
67890 C7890 876
98765 9876 9087
43210 543 56
12345 21 234
67890 9 1012
302445 90684 23610
290100 78339 11265
302445 90684 23610
Ex. 5. Add 3426; 9024; 5106; 8890; 1204, together. Ans. 27630.
6. Add 509267; 235809; 72920; 8392; 420; 21; and 9, together.
Ans. 826833.
7. Add 2; 19; 817; 4293; 50916; 730205; 9180634, together. Ans. 9966891.
8. How many days are in the twelve calendar months ? Ans. 365.
9. How many days are there from the 15th day of April to the 24th day of
November, both days included ? Ans. 224.
10. An army consists of 52714 infantry*, or foot, 5110 horse, 6250 dragoons,
3927 hght-horse, 928 artillery, or gunners, 1410 pioneers, 250 sappers, and 406
miners : what is the whole number of men ? Ans. 70995.
OF SUBTRACTION.
Subtraction teaches to find how much one number exceeds another, called
their difference, or the remainder, by taking the less from the greater. The
method of doing which is as follows :
Place the less number under the greater, in the same manner as in Addition,
that is, units under units, tens under tens, and so on ; and draw a line below
them. — Begin at the right hand, and take each figure in the lower line, or num-
ber, from the figure above it, setting down the remainder below it. — But if the
figure in the lower line be greater than that above it, first borrow, or add, 10 to
In like manner, the same property may be shown to belong to the number 3 ; but the
preference is usually given to the number 9, on account of its being more convenient in
pi'actice. A similar property belongs to the number 11.
Now, from the demonstration above given, the reason of the rule itself is evident : for the
excess of 9s in two or more niimbers being taken separately, and the excess of 9s taken also
out of the sum of the former excesses, it is plain that this last excess must be equal to the
excess of 9s contained in the total sura of all these numbers ; all the parts taken together being
equal to the whole. — This rule was first given by Dr. WaUis in his Arithmetic, published in
the year 1657.
* The whole body of foot soldiers is denoted by the word Infantry ; and all those that charge
on horseback by the word Caxxdry. — Some authors conjecture that the term infantry is derived
from a certain Infanta of Spain, who, finding that the army commanded by the king her father
had been defeated by the Moors, assembled a body of the people together on foot, with wliirh
she engaged and totally routed the enemy. In honour of this event, and to distinguish the
foot soldiers, who were not before held in much estimation, they received the name of Infantry,
from her own title of Infanta.
MULTIPLICATION. g
the upper one, and then take the lower fif^ure from that sum, nettin!? down the
J remainder, and carrying ] , for what was borrowed, to the next lower fijfurc. with
1^ which proceed as before ; and so on till the whole is finished.
*)
TO PROVE SUBTRACTION.
Add the remainder to the less number, or that which is just above it; and if
the sum be equal to the greater or uppermost number, the work is right •.
EXAMPLES.
1- 2. 3.
From 5386427 From 5386427 From 12345G7
Take 2164315 Take 4258792 Take 702973
Rem. 3222112 Rem. 1127635 Rem. 531594
Proof 5386427 Proof 5386427 Proof 1234567
4. From 5331806 take 5073918. Ans. 257888.
5 From 7020974 take 2766809. Ans. 4254165.'
6. From 8503402 take 574271. Ans. 7929131.
7. Sir Isaac Newton was born in the year 1642, and he died in 1727 : how
old was he at the time of his decease ? Ans. 85 years.
8. Homer was born 2568 years ago, and Christ 1835 years ago: then how
long before Christ was the birth of Homer ? Ans. 733 years.
9. Noah's flood happened about the year of the world 1656, and the birth of
Christ about the year 4000 : then how long was the flood before Christ?
Ans. 2344 years.
10. The Arabian or Indian method of notation was first known in England
about the year 1150 : then how long is it since to this present year 1840?
Ans. 690 years.
11. Gunpowder was invented in the year 1330 : how long was that before the
invention of printing, which was in 1441 ? Ans. Ill years.
12. The mariner's compass was invented in Europe in the year 1302 : how
long was that before the discovery of America by Columbus, which happened in
1492? Ans. 190 years.
OF MULTIPLICATION.
Multiplication is a compendious method of Addition, teaching how to find
the amount of any given number when repeated a certain number of times ; as,
4 times 6, which is 24.
The number to be multiplied, or repeated, is called the MuUiplicand.—Tlie
number you multiply by, or the number of repetitions, is the Multiplier.—
And the number found, being the total amount, is called the Product. Also,
both the multiplier and multiplicand are, in general, named the Tervu or
Factors.
* The reason of this method of proof is evident; for if the difference of two numbcrt be
added to the less, it must manifestly make up a sum equal to the grc*ter.
10
ARITHMETIC.
Before proceeding to any operations in this rule, it is necessary to commit
thoroughly to memory the following Table, of all the products of the first 12
numbers, commonly called the Multiphcation Table, or sometimes the Table of
Pythagoras, from its alleged inventor.
MULTIPLICATION TABLE.
1
2
3
2
4
6
3
6
4
5
6
7
8
9
10
11 12
8
10
12
14
16
18
20
22
24
9
12
15
18
21
24
27
30
33
36
4
8
12
16
20
24
28
32
36
40
44
43
5
6
10
12
15
20
25
30
35
40
45
50
55
60
18
24
28
30
36
42
48
54
60
66
72
7
14
21
35
42
49
56
63
70
77
84
8
16
24
32
40
48
56
64
72
80
88
95
9
18
27
36
45
54
63
72
81
90
99
108
10
20
30
40
50
60
70
80
90
100
110
120
11
22
33
44
55
66
77
88
99
110
121
132
12
24
36
48
60
72
84
96
103
120
132
144
To multiply any Given Number by a Single Figure, or by any Number not
exceeding 12.
* Set the multiplier under the imits figure or right hand place of the multi-
plicand, and draw a line below it. Then, beginning at the right hand, multiply
every figure in this by the multiplier. Count how many tens there are in the
product of every single figure, and set down the remainder directly under the
figure that is multiplied ; and if nothing remains, set down a cipher. Carry as
many units or ones as there are tens counted, to the product of the next figures ;
and proceed in the same manner till the whole is finished.
EXAMPLE.
Multiply 9876543210, the Multiplicand.
By - - - - 2, the Multiplier.
19753086420, the Product.
5678
4
* The reason of this rule is the sarae as for the
process in .\d(lition, in which 1 is carried for
every 10, to the next place, gradually as the
several products are produced one after another.
32 = 8x4
280 = 70 X 4
2400 = 600 X 4
instead of setting them all down below each other, 20000= 5000 x 4
as in the annexed example.
•271-2 = 5678 x 4
MULTIPLICATION. H
To multiply by a Number consisting of Several Figures.
* Set the multiplier below the multiplicand, placing them as in Addition,
namely, units under units, tens under tens, &c. drawing a line below it.
Multiply the whole of the multiplicand l)y each figure of the multiplier, as in the
last article ; setting down a line of j)roduct8 for each figure in the multiplier, so
as that the first figure of each line may stand straiglit under the figure multiply-
ing by. Add all the lines of products together, in the order in which they »Und,
and their sum will be the answer or whole product required. It will, of course
be always best to take that number as the multiplier which has the fewest effec-
tive figures.
TO PROVE MULTIPLICATION.
There are three different ways of proving multiplication, which are as below :
First Method. Make the multiplicand and multiplier change places, and
multiply the latter by the former in the same manner as before. Then if the
product found in this way be the same as the former, the number is right.
Second Method, f Cast all the 9s out of the sum of the figures in each of the
two factors, as in Addition, and set down the remainders. Multiply these two
remainders together, and cast the 9s out of the product, as also out of the whole
product or answer of the question, reserving the remainders of these last two,
which remainders must be equal when the work is right.
Note. It is common to set the four remainders within the four angular spaces
of a cross, as in the example below.
Third Method. Multiplication is also very naturally proved by Division ; for
the product divided by either of the factors, will evidently give the other. But
this cannot be practised till the rule of division is learned.
Or thus : —
Having placed the multiplier under the multiplicand as in the previous rule,
multiply by the left-hand figure, setting down the product as if that figure were
* After having found the product of the multiplicand by the first figure of the multiplier, la
in the former case, the multiplier is supposed to be divided into parts, and the product is found
for the second figure in the same manner : but as this figure stands in the place of tens, the
product must be ten times its simple value ; and therefore the first figure of this product mutt
be set in the place of tens; or, which is the same 1234567 the multiplicand,
thing, directly under the figure multiplying by. 4567
And proceeding in this manner separately with all
the figures of the multiplier, it is evident that we 8641969 = 7 time* the mult,
shall multiply all the parts of the multiplicand 74074020 = 60 times ditto.
by all the parts of the multiplier, or the whole of 617283500 = 500 times ditto,
the multiplicand by the whole of the multiplier : 4938268000 = 4000 times ditto.
therefore these several products being added toge- ■
ther, will be equal to the whole required product; 5638267489 = 4567 times ditto,
as in the example annexed.
t This method of proof is derived from the peculiar property of the number 9, mentionrd
in the proof of Addition, and the reason for the one includes that of the other. Another
more ample demonstration of this rule may, however, be as follows:— I^ct P and q denote the
number of 9s in the factors to be multiplied, and a and b what remain ; then 9p -f- « and 9<i
+ b will be the numbers themselves, and their product is (9p X 9q) + {^P X 6) + (9*i X a)
+ {.a X f>); but the first three of these products are each a precise number of 9», because their
factors, either one or both, are so: these therefore being cast away, there remains only a X *;
and if the 9s also be cast out of this, the excess is the excess of 98 in the total product : but <i
and 6 are the excesses in the factors themselves, and a X l> h their product ; therefore the rule
is true. This mode of proof, however, is not an entire check against the errors that might aritc
from a transpos-ition of figures, or other compensation of errors.
12
ARITHMETIC.
the only multiplier. Proceed to the next figure of the multiplier, putting the
first figure one place to the right of the right-hand figure of the last product, in
the line below : proceed to the next, carrying out the first figure one place more
to the right ,• and so on till all the partial products are made. Add up as in the
last rule *.
EXAMPLE.
Mult. 3542
or Mult
. 3542
by 6196
Proof
by
6196
\ 2 /
5 >(^ 4
21252
21252
31878
/2 \
3542
3542
31878
21252
21252
21946232 = Product.
21946232 = Product.
OTHER EXAMPLES.
Multiply 123456789 by 3.
Multiply 123456789 by 4.
Multiply 123456789 by 5.
Multiply 123456789 by 6.
Multiply 823456789 by 7.
Multiply 123456789 by 8.
Multiply 123456789 by 9.
Multiply 123456789 by 11.
Multiply 123456789 by 12.
Multiply 302914603 by 16.
Multiply 273580961 by 23.
Multiply 402097316 by 195.
Multiply 82164973 by 3027.
Multiply 7564900 by 579-
Multiply 8496427 by 874359.
Ans. 370370367.
Ans. 493827156.
Ans. 617283945.
Ans. 740740734.
Ans. 5764197523.
Ans. 987654312.
Ans. 1111111101.
Ans. 1358024679.
Ans. 1481481468.
Ans. 4846633648.
Ans. 6292362103.
Ans. 78408976620.
Ans. 248713373271.
Ans. 4380077100.
Ans. 7428927415293.
Ans. 102330768400.
Multiply 2760325 by 37072.
CONTRACTIONS IN MULTIPLICATION.
I. When there are Ciphers in the Factors.
If the ciphers be at the right-hand of the numbers ; multiply the other figures
only, and annex as many ciphers to the right-hand of the whole product, as are
in both the factors. When the ciphers are in the middle parts of the multiplier;
neglect them as before, only taking care to place the first figure of every line of
products exactly under the figure by which you are multiplying.
EXAMPLES.
1.
Mult. 9001635
by - 70100
9001635
63011445
631014613500 Product.
Mult. 390720400
by - 406000
23443224
15628816
158632482400000 Product.
• This is the eastern mode of putting down the work ; and is evidently productive of the
same final result, only that the progressive partial multiplications are taken in an inverse order.
DIVISION. 13
3. Multiply 81503COO by 7030. Ang. 572970308000
4. Multiply 9030100 by 2100. Ans. 1896J21(X)00 '
5. Multiply 8057069 by 70050. Adh. 564397683450.
II. When the Multiplier is the product of tuo or more Numbers in the Table ;
then
* Multiply by each of those parts successively, instead of the whole number
at once.
EXAMPLES.
1. Multiply 51307298 by 5G, or 7 times 8.
51307298
7
35915108(5
8
287320SG88.
2. Multiply 31704592 by 36. Ans. 1141365312.
3. Multiply 29753804 by 72. Ans. 2142273888.
4. Multiply 7128368 by 96. Ans. 684323328.
5. Multiply 160430800 by 108. Ans. 17326526400.
6. Multiply 61835720 by 1320. Ans. 81623150400.
7. There was an army composed of 104 f battalions, each consisting of 500
men ; what was the number of men contained in the whole .' Aus. 52000.
8. A convoy of ammunition + bread, consisting of 250 waggons, and each
waggon containing 320 loaves, having been intercepted and taken by the enemy,
what is the number of loaves lost? Ans. 80000.
OF DIVISION.
Division is a kind of compendious method of Subtraction, teaching to find
how often one number is contained in another, or may be taken from it, which
is the same thing.
The number to be divided, is called the Dividend. The number to divide by,
is the Divisor : and the number of times the dividend contains the divisor is
called the Qxiotient. Sometimes there is a Remainder left, after the division is
finished.
The usual manner of placing the terms, is, the dividend in the middle, having
the divisor on the left hand, and the quotient on the right, each separated by a
curve line; as, to divide 12 by 4, the quotient is 3,
The chief advantage of this process is, that it assimilates with the method employed in con-
tracted decimals, in the extractions of roots in duodecimals, and in Algebra.
* The reason of this rule is obvious; for any number multiplied by the component p«rt» of
another, must give the same product as if it were multiplied by that number at once. Thu», in
the 1st example, 7 times the product of 8 by the given number, make 5(i times the same num-
ber, as plainly as 7 times 8 make 56.
f A battalion is a body of foot, consisting of 500, or 600, or 700 men, more or 1cm.
+ The ammunition bread is that which is provided for, and distributed to, the toldicrt ; the
usual allowance being a loaf of 6 pounds to every soldier, once in 4 days.
14 ARITHMETIC.
Dividend,
Divisor 4) 12 (3 Quotient; showing that the number 4 is 3 12
times contained in 1 2, or may be 3 times subtracted out of 4 subtr.
it, as in the margin. —
* Rule. Having placed the divisor before the dividend, as 8
above directed, find how often the divisor is contained in as 4 subtr.
many figures of the dividend as are just necessary, and place —
the number on the right in the quotient. Multiply the divisor 4
by this number, and set the product under the figures of the 4 subtr.
dividend before mentioned. — Subtract this product from that —
part of the dividend under which it stands, and bring down 0
the next figure of the dividend, or more if necessary, to join —
on the right of the remainder. — Divide this number, so increased, in the
same manner as before ; and so on, till all the figures are brought down and
used.
Note. If it be necessary to bring down more figures than one to any re-
mainder, in order to make it as large as the divisor, or larger, a cipher must
be set in the quotient for every figure so brought down more than one.
TO PROVE DIVISION.
•f Multiply the quotient by the divisor ; to this product add the remainder,
if there be any ; then the sum will be equal to the dividend, when the work is
right.
* In this way the dividend is resolved into parts, and by trial is found how often the divisor
is contained in each of those parts, one after another, arranging the several figures of the
quotient one after another, into one number.
When there is no remainder to a division, the quotient is the whole and perfect answer to
the question. But when there is a remainder, it goes so much towards another time, as it
approaches to the divisor : so, if the remainder be half the divisor, it will go the half of a
time more; if the fourth part of the divisor, it will go one-fourth of a time more; and so on.
Therefore to complete the quotient, set the remainder at the end of it, above a small line, and
the divisor below it, thus forming a fractional part of the whole quotient.
+ This method of proof is plain enough : for since the quotient is the number of times the
dividend contains the divisor, the quotient multiplied by the divisor must evidently be equal to
the dividend.
There are several other methods sometimes used for proving Division, some of the most
useful of which are as follow :
Second Method. Subtract the remainder from the dividend, and divide what is left by the
quotient ; so shall the new quotient from this last division be equal to the former divbor, when
the work is right.
Third Method. Add together the remainder and all the products of the several quotient
figures by the divisor, according to the order in which they stand in the work ; and the sum
will be equal to the dividend, when the work is right.
Fourth Methrxi, by casting out the nines. Make a cross as in multiplication, and cast out
the nines from the divisor and quotient, and place the respective
remainders, instead of D and Q respectively. Cast the nines also
out of the remainder, and annex it to Q by the sign plus, at R. Mul-
tiply D by Q, and add in the number R ; and from this also cast out
the nines. Place the result at M : and if this last number be the
same as that left after casting out the nines from the dividend, the
work is probably correct.
DIVISION.
15
EXAMPLES.
1.
3) 1234567
12
Quot.
(411522
mult. 3
3
3
1234566
add 1
4
3
1234567 Proof.
15
15
6
6
7
6
1.
Quot.
37) 12345G7a
( 3336C6
111
37
124
2335C62
111
1000998
rem. 36
115
111
12345678 Proof.
246
2-22
247
222
258
222
Rem. 1
Rem 36
3. Divide 73146085 by 4.
4. Divide 5317986027 by 7.
5. Divide 570196382 by 12.
6. Divide 74638105 by 37.
7. Divide 137896254 by 97-
8. Divide 35821649 by 764.
9. Divide 72091365 by 5201.
10. Divide 4637064283 by 57606.
Ans. 1828652IJ.
Ans. 759712289,*.
Ans. 475l636j,V
Ans. 2017246jV
Ans 1421610J,\
Ans. 46886^J,\
Ans. 1386lf;(jV
Ans. 80496iiJJJ.
11. Suppose 471 men are formed into ranks of 3 deep, what is the number in
each rank. Ans. 157.
12. A party, at the distance of 378 miles from the head-quarters, receive orders
to join their corps in 18 days : what number of miles must they march each day
to obey their orders. Ans. 21.
13. The annual revenue of a nobleman being 37960/.; how much per day is
that equivalent to, there being 365 days in the year? Ans. 104/.
CONTRACTIONS IN DIVISION.
Tliere are certain contractions in Division, by which the operation in par-
ticular cases may be performed more concisely : as follows :
I. Division by any Small Number, not greater than 12, may be expeditiously
performed, by multiplying and Eubtracting mentnlly, omitting to set down the
work except only the quotient immediately below the dividend.
IG ARITHMETIC.
EXAMPLES.
3) 56103961 4) 52619675 5) ]?79I92
Quot. 18701320J
6) 38672940 7) 81396627 8) 23718920
9) 43981962 11) 57614230 12) 279S0373
II. * JMien Ciphers are annexed to the Divisors j cut off those ciphers from it,
and cut off the same number of figures from the right-hand of the dividend ;
then divide with the remaining figures, as usual. And if there be any thing
remaining after this division, place the figures cut off from the dividend to the
right of it, and the whole will be the true remainder ; otherwise, the figures cut
off only will be the remainder.
EXAMPLES.
i. Divide 3704196 by 20. 2. Divide 31086901 by 7100.
2,0)3704196 7100)310869,01 ( 4378flg5.
284
Quot. 185209]§
268
213
556
497
599
568
31
3. Divide 7330964 by 23000, Ans. 320*S§51.
4. Divide 2304109 by 5800. Ans. 3971m.
III. When the Divisor is the exact Product of two or more of the Numbers not
greater than 12 ; divide by one of the factors of the divisor, putting down the
remainder to the right of the quotient, but separated by the mark ( ; then this
quotient by the next of the factors, setting down the remainder to the right of
the quotient, as in the former case ; then this quotient by the next factor, and so
on till all the factors have been used. The final quotient is the integer part of
the quotient required.
* This method serves to avoid a needless repetition of ciphers, which would occur in the
common way. And the truth of the principle on which it is founded, is evident; for, cutting off|
the same numher of ciphers, or figures, from each, is the same as dividing each of them hy 10,|
or 100, or 1000, &c. according to the numher of ciphers cut off; and it is evident, that as often
as the whole divisor is contained in the whole dividend, so often must any part of the fonuer bel
contained in a like part of the latter. j
DIVISION.
17
To find the fractional part, proceed thus :
Write the several remainders in a horizontal line from right to left, beginninir
at the left hand with the last ; then write the several factors in the same manner
to the right of these, but separated by a curve, ( . Multiply the first remainder
by the first divisor, and to the product add the second remainder, (this can be
done mentally in all cases to which this method of division applies), ihe sum of
which is to be placed under the second remainder : multiply this sum by the
next divisor, and add the product to the third remainder, putting the sum under
the third remainder : multiply this sum by the next divisor, and so on till the
last sum falls under the last remainder. This will be the entire remainder which
would result from dividing the dividend by the entire divisor •.
EXAMPLES.
Ex. 1. Divide 3672965 by 2x3x4x5x6.
2 3672965
0
10
1836482 1^1
612160 [2
153040 10
30608 1 0
The work of finding the remainder
is placed below, and below the seve-
ral successive numbers is given their
composition. The resulting remainder
is 245, and this combined, as before,
with the quotient and divisor gives
for the entire quotient 5101 |g.
5101 12
Remainders.
0
40
2
122
1
245
Divisors.
(5, 4, 3, 2.
5x2+0 4x10 + 0 3x40 + 2 2x 122+1
• The proof of the truth of this rule may be given as follows; and the example worked will
show the nature of the notation employed.
Let the several remainders (reckoned backwards) be r„ r^, r,, . . . . and the divisors which
gave them be </,, rfj, d^, . . . . Then the preceding fractions being all to be divided by the inc-
cessive divisors (they forming parts of the numbers successively divided) we have
ft. + rf // +
+
+
+ ...
rf, rfj dj d^ d.
. + r^ rf^ . . . + r.
+
</| ' d^d^'^ d^ d^ d^ ' d^ dj rfj d^
Reducing these to a common denominator, we have
r, rfa rfa ^4 rfj . . . + ra </3 rf 4 rfs . . . + tj d^d^
rf, d^djd^d^.. . .
■where the continuing dots express that the multiplication and addition are to be carried to the
extent of embracing all the terms. AVe may suppose them to be five, as in the work written
down, since the process is the same however many there may be, and the steps are continuous.
Then this reduction may be gradually effected thus :
ri + rj + r3 + r4 + r.
r, rfj + rj
^3
'•3
+
r, c/j d, + rj
''s-i-
r, rfj dj d^ +
ds
'•j ^3 '^t
+
rid*
>■*
VOL. I.
r, rfj rf, d^ d^ + i-j //j d^ rfj + r, d^ rf, + r^d^ + r^
C
&.C. &c. Ac.
18 ARITHMETIC.
2. Divide 7014596 by 72. Ans. 97424f?.
3. Divide 5130652 by 132. Ans. 3886873^.
4. Divide 83016572 by 240. Ans. 345502j?g.
IV. Common Division may be performed more concisely, by omitting the several
products, and setting down only the remainders ; namely, multiply the divisor
by the quotient figures as before, and, without setting down the product, subtract
eacli figure of it from the dividend, as it is produced ; always remembering to
carry as many to the next figure as were borrowed before. This is not, however,
to be recommended till considerable practice has conferred on the pupil the
power of carrj'ing on two processes at once ; namely, multiplication and sub-
traction.
EXAMPLES.
]. Divide 3104679 by 833.
833) 3104679 (3727^V
6056
2257
5919
88
2. Divide 79165238 by 238. Ans. 3326275'3V
3. Di^ade 29137062 by 5317- Ans. 54795!i7-
4. Divide 62015735 by 7803. Ans. 7947||^.
OF REDUCTION.
Reduction is the changing of numbers from one name or denomination to
another, without altering their value. This is chiefly concerned in reducing
money, weights, and measures.
When the numbers are to be reduced from a higher name to a lower, it is called
Reduction descending ; but when contrariwise, from a lower name to a higher, it
is Reduction ascending.
Before we proceed to the rules and questions of Reduction, it will be proper to
set down the usual tables of money, weights, and measures, which are as follow.
OF MONEY, WEIGHTS, AND MEASURES.
TABLES OF MONEY*.
2 Farthings ^ 1 Halfpenny J
4 Farthings =: 1 Penny d
12 Pence = 1 Shilling s
20 Shillings = 1 Pound £
qrs d
4=1 s
48 = 12 = 1 £
960 = 240 = 20 = 1
£ denotes pounds, s shillings, and d denotes ])ence.
\ denotes 1 farthing, or one quarter of a j)enny.
i denotes a halfpenny, or the half of a penny.
j denotes 3 farthings, or three quarters of a penny. [The
REDUCTION.
19
PENCE TABLE.
d.
S.
d.
20
are 1
8
30
— 2
6
40
— 3
4
50
— 4
2
60
— 5
0
70
— 5
10
80
— 6
8
90
— 7
6
100
— 8
4
no
— 9
2
120
— 10
0
SHILLIXOS TABLB.
8.
d.
I
are
13
2
—
24
3
—
36
4
—
48
5
—
60
6
—
72
7
—
84
8
—
96
9
—
108
10
—
120
11
"^
132
The full weight and value of the English gold and silver coin, both old and new, are tub-
joined.
GOLD.
Guinea
Half do
Third do. ...
Double Sov.
Sovereign ...
Half do. ...
V.\LLK.
£ S. d.
1 1 0
0 10 6
0 7 0
2 0 0
1 0 0
0 10 0
Weight.
dtct. qr,
5 '9k
2 16|
1 19|
10 6^
5 aft
2 13^
SILVER.
A Crown ..
Half-crown
Shilling
Sixpence ..
Valle.
s. d.
5 0
2 6
1 0
0 6
Old Wt. New Wt.
dtvt.
19
9
3
ffr-
21
1 ooi
I — i
dtrt. qr.
1» i^
9 2,^
1 l!.f,
The usual value of gold is nearly Al. an ounce, or Id. a grain : and that of silver is nearly 5«.
an ounce. Also the value of any quantity of gold, was to the value of the same weight of
standard silver, as 15y^j to 1, in the old coin ; but in the new coin they are 14;f, to 1.
Pure gold, free from mixture with other metals, usually called fine gold, is of so pure k
nature, that it will endure the fire without wasting, though it be kept continually melted. But
silver, not having the purity of gold, will not endure the fire like it : yet fine silver will waste
but a very little by being in the fire any moderate time ; whereas copper, tin, lead, &c. will
not only waste, but may be calcined, or burnt to a powder.
Both gold and silver, in their purity, are so soft and flexible, (like new lead, &c.) that they
are not so useful, either in coin or otherwise (except to beat into leaf gold or silver), as when
they are alloyed, or mi.xed and hardened with copper or brass. And though most nations differ,
more or less, in the quantity of such alloy, as well as in the same place at different times, yet in
England the standard for gold and silver coin has been for a long time as follows : viz. That 22
puts of fine gold, and 2 parts of copper, being melted together, shall be esteemed the true
standai'd for gold coin : And that 11 ounces and 2 pennyweights of fine silver, and 18 penny-
weights of copper, being melted together, be esteemed the true standard for silver coin, called
Sterling silver.
In the old coin the pound of sterling gold was coined into 42^ guineas, of 21 shillings each, of
which the pound of sterling silver was divided into 0"2. The new coin is also of the same
quality or degree of fineness with that of the old sterling gold and silver above described, but
divided into pieces of other names or values ; viz. the pound of the silver into 6tJ shillings, of
course each shilling is the 66th part of a pound ; and 20 pounds of the gold into 9M\ pieces
called sovereigns, or the pound weight into 4CJ3 sovereigns, each equal to 20 of the new shil-
lings. So that the weight of the sovereign is 46Jgths of a pound, which is equal to 5;^ penny-
weighU, or equal to 5 dwt. 3ft gr. very nearly, as sUted in the preceding Ubles. And multiples
and parts of the sovereign and shilling in their several proportions.
c 2
20 ARITHMETIC.
WEIGHTS AND MEASURES,
Agreeably to the Act of Uniformity, which took effect 1st January, 1826.
The term Measure is the most comprehensive of the two, and it is distinguished into six
kinds, viz. : —
1. Length.
Measure of .
12
Inches =:
3
Feet =
5J
Yards :=
40
Poles =
8
Furlongs =
69;.
Miles =
2. Surface.
3. Soliditj-, or Capacity.
4. Force of Gravity, or what is commonly called Weight.
5. Angles.
.6. Time.
The several denominations of these Measures have reference to certain standards, which are
entirely arbitrary, and consequently vary among different nations. In this kingdom
'Length is a Yard.
Surface is a Square Yard, the j^ of an Acre.
The standard of...<] /Solidity is a Cubic Yard.
\ Capacity is a Gallon.
^Weight is a Pound.
The Standards of Angular Measure, and of Time, are the same in all European and most
other countries.
1. MEASURE OF LENGTH.
1 Foot.
1 Yard.
1 Rod, or Pole.
1 Furlong.
] Mile.
1 Degree of a Great Circle of the Earth.
An Inch is the smallest lineal measure to which a name is given ; but subdivisions are used
for many purposes. Among mechanics, the Inch is commonly divided into eiflhtlis. By the
officers of the revenue, and by scientific persons, it is divided into tenths, htotdredtJis, &c. For-
merly it was made to consist of 12 parts, called lines, but these have properly fallen into disuse.
Particular Measures of Length.
^...„..,....„.
Used for the height of horses.
Used in measuring depths.
^ Used in Land Measure to facilitate computa-
/> tion of the content, 10 square chains being
J equal to an acre.
1 Square Foot.
1 Square Yard.
1 Perch, or Rod.
1 Rood.
1 Acre.
1 Square Mile.
3. MEASURES OF SOLIDITY AND CAPACITY.
Division I. Solidity.
1728 Cubic Inches = 1 Cubic Foot.
27 Cubic Feet = 1 Cubic Yard.
Nail
= 2\ Inches,
Quarter
:= 4 Nails,
Yard
= 4 Quarters,
Ell
=: 5 Quarters,
Hand
z=. 4 Indies,
Fathom
= 6 Feet,
Link
= 7-92 Inc.
Chain
= 100 Links,
2. MEAS
144 Square Inches
9 Square Feet
30J Square Yards
40 Perches
4 Roods
640 Acres
TABLES OF WEIGHTS AND MEASURES. £1
Division II.
Imperial Measure of Capacity for all liquids, and for all dry good., except such u arc com-
prised in the third Division.
4 Gills =
1 Pint
zz
34fi5f)25^
2 Pints ==
1 Quart
=
Gfl-318o
4 Quarts =z
2 Gall. =
1 Gallon
1 Peck
i^
277-274
5.54 -.548
> Cubic Inciiet.
8 Gall. =
1 Bushel
z^
22181.92
8 Bushels =
5 Qre. =
1 Quarter
1 Load
=r
10-2()936'
5I-34«81.
Cubic Feet.
r,n, ', ( Cubic Inches, nearlv.
zz 2oio^)
The four last denominations are used for dr>- goods only. For liqui.U, several denominationt
have been heretofore adopted, viz. -.—For Beer, the Firkin of 9 gallons, the Kilderkin of 18,
the Barrel of 36, the Hogshead of 54, and the Butt of 108 gallons. These wll probably con-
tinue to be used in practice. For Wine and Spirits, there arc the Anker, Runlet, fierce
Hogshead, Puncheon, Pipe, Butt, and Tun; but these may be considered rather as the names'
of the casks in which such commodities are imported, tiian as expressing any definite number
of gallons. It is the practice to gauge all such vessels, and to charge them according to their
actual content.
Flour is sold, nominally, by measure, but actually by weight, reckoned at 7 lb. Avoirdupois
to a gallon.
Division III.
Imperial measure of Cap.\CITV for coals, culm, lime, fish, potatoes, fruit, and other goods,
commonly sold by heaped measure :
2 Gallons = 1 Peck =
8 Gallons = 1 Bushel
3 Bushels := 1 Sack = 4S 1 ^ , . „ ^ ,
12 Sacks =lChald. = sgl j Cubic Feet, nearly.
The goods are to be heaped up in the form of a cone, to a height above the rim of the mea-
sure of at least | of its depth. The outside diameter of Measures used for heaped goods are to
be at least double the depth ; consequently, not less than the following dimensions : —
Bushel 19J inches. I Peck 12{ inches.
Half-bushel 15.J I Gallons 9^
Half-gallon, 7J inches.
The Imperial Measures described in the second and third Divisions were established by Act
5 Geo. IV. c. 74. Before that time there were four different measures of capacity used in
England. 1. For wine, spirits, cider, oils, milk, &c. ; this was one-sixth less than the Imperial
Measure. 2. For malt liquor, this was ^ part greater than the Imperial Measure. 3. For
corn, and all other dry goods not heaped, this was ^^ part less than the Imperial Measure. 4.
For coals, which did not differ sensibly from the Imperial measure.
The Imperial Gallon contains exactly 10 lbs. Avoirdupois of pure water ; consequently the
pint will hold I^ lb., and the bushel 80 lbs.
4. MEASURE OF WEIGHT.
Division I. Avoirdupois Wright.
27iJ Grains = 1 Dram = 27iU
16 Drams = 1 Ounce = 437J v Grains.
16 Ounces = 1 Pound (lb.) = 7000 j
28 Pounds = 1 Quarter (qr.)
4 Quarters = 1 Hundredweight (cwt.)
20 Cwt. = 1 Ton.
This weight is used in almost all commercial transactions, and in the common dealing! of
life.
22 TABLES OF WEIGHTS AND MEASURES.
Particular weights belonging to this division : —
cwt. qr.
Used for Meat.
Used in the Wool Trade.
Division II. Troy Weight.
24 Grains =: I Penn}-weight = 24 -j
20 Pennyweights := 1 Ounce =: 480 >• Grains.
12 Ounces =: 1 Pound =■ 5760)
These are the denominations of Troy "Weight when used for weighing gold, silver, and precious
stones (except diamonds). But Troy Weight is also used by Apothecaries in compounding
medicines, and by them the ounce is divided into 8 drams, and the dram into 3 scruples, so that
the latter is equal to 20 grains.
For scientific purposes the grain only is used ; and sets of weights are constnicted in decimal
progression from 10,000 grains downwards to -j^ of a grain.
By comparing the number of grains in the Avoirdupois and Troy pound and ounce respect-
ively, it appears that the Troy pound is less than the Avoirdupois, in the proportion of 144 to
175; but the Troy ounce is greater than the Avoirdupois, in the proportion of 192 to 175.
oz dwts grs
1 lb Avoirdupois =. 14 11 15^ Troy.
1 oz = 0 18 5i
1 dr = 0 1 3i
The carat, used for weighing diamonds, is 3g grains. The term, however, when used to
express the fineness of gold, has a relative meaning only. Every mass of alloyed gold is sup-
posed to be divided into 24 equal parts : thus the standard for coin is 22 carats fine ; that is, it
consists of 22 parts of pure gold and 2 parts of alloy. What is called the new standard, used for
watch cases, &c. is 18 carats fine.
5. ANGULAR MEASURE ; or, DIVISIONS OF THE CIRCLE.
60 Seconds := I Minute.
60 Minutes = 1 Degree.
30 Degrees = 1 Sign.
90 Degrees = 1 Quadrant.
360 Degrees = 1 Circumference.
Formerly, the subdivisions were carried on by sixties ; thus, the second was divided into 60
thirds, the tliird into 60 fourths, &c. At present, the second is more generally di^ded deci-
mally into lOths, lOOths, &c. The degree is frequently so divided.
6. MEASURE OF TIME.
60 Seconds z= 1 Minute.
60 Minutes zr 1 Hour.
24 Hours = 1 Day.
7 Days = 1 Week.
28 Days n 1 Lunar Month.
28, 29, 30, or 31 Days = 1 Calendar Month.
12 Calendar Months = 1 Year.
365 Days = 1 Common Year.
366 Days =: 1 Leap Year.
365| Days =z 1 Julian Year.
365 Days, 5 Hours, 48 M. 45.J Seconds = 1 Solar Year.
In 400 years, 97 are leap years, and 303 common.
The same remark, as in the case of angular measure, applies to the mode of subdividing the
second of time.
TABLES OF WEIGHTS AND ME.\SURES.
in
COMPARISON OF MEASURES.
The old die gallon contained 282 cubic inches.
The old tcine gallon contained 231 cubic inches.
The old Winchester busJiel contained 21.50J cubic inchc*.
The imperial gallon contains 277'274 cubic inches.
The com bushel, eight times the above.
Hence, with respect to Ale, Wine, and Cora, it will be expedient to
posaesa a
TABLE OF F.\CTORS,
For converting old measures into new, and the contrary.
By decimals.
By vulpar Tractiont
nearly.
1
Corn
Measure.
Wine
Measure.
Ale
Measure.
Corn
Mea-
sure.
Wine
Mea-
sure.
Ale
Mea-
sure.
To convert old \ .g^^^^
measures to' new. J
•83311
1-01704 I §J
ii
i
tt
To convert new 1 , rvoi -q
measures to old. / ^ "•^'^'^
1-20032
•983-24 11
]
i
tt
N.B. For reducing the prices, these numbers must all be reversed.
SIZES OF DRAWING-PAPER.
Wove antique 4ft 4 X 2ft 7
Double elephant 3ft 4 X 2 ft 2
Atlas 2ft 9 X 2 ft 2
Columbier 2 ft 9^ X 1 ft H
Elephant 2 f t SJ X 1ft lOJ
Imperial 2 ft 5 x 1ft 9,1
Super royal 2ft 3 X 1ft 7
Royal 2 ft 0 X 1 ft 7
Medium 1 ft 10 X 1ft 6
MISCELLANEOUS INFORMATION.
1 Aum of hock contains 36 gallons.
I Barrel, imperial measure 9981-864 cubic inches.
1 Barrel, anchovies 30 pounds.
soap 256 pounds.
herrings 32 gallons.
salmon or eels ^^ gJ'l'o"*-
1 Bushel of coal 88 poundi.
flour ^ P*"""***-
1 Butt of sherry '30 ^lloni.
I Chaldron of coals, with ingrain 104809-572 cubic inchr..
without mgrain 9981864 cubic inchct.
at Newcastle, is ^3 cwt.
8 Chaldrons of coals at Newcastle are equal to ... 15J London chaldront.
I Clove of wool 7 pounds.
1 Firkin of butter ^ P*'"'"^*-
soap ^* P",'*'"^*-
soap 8 8*llon..
24
TABLES OF WEIGHTS AND MEASURES.
1 Fodder of lead, at Stockton
Newcastle...
London
1 Gross
1 Great gross
1 Hand
1 Hogshead of claret
tent
1 Hundred of salt
1 Keg of sturgeon
1 Last of salt
gunpowder
beer
potash
cod-fish
herrings
meal
soap
pitch and tar
flax
feathers
■wool
1 Pack of wool
1 Palm
1 Pipe of Madeira
Cape Madeira
Teneriffe
Bucellas
Barcelona
Vidonia
Mountain
Port
Lisbon
1 Pole, Woodland
Plantation
Cheshire
1 Sack of wool
I Seam of glass
1 Span
1 Stone of meat
fish
(horseman's weight) ...
glass
wool
I Tun of vegetable oil
animal oil
1 Tod of wool
1 Wey of Cheese, in Suffolk
in Essex
1 Wey of wool
1 Ton or load of rough timber ...
hewn
40 Chaldrons of coal at Newcastle
at London ...
22
cwt.
21
cwt.
19.1
cwt.
12
dozen.
12
gross.
4
inches.
58
gallons.
63
gallons.
7
lasts.
4
or 8 gallons.
18
barrels.
24
barrels.
12
barrels.
12
barrels.
12
barrels.
12
barrels.
12
barrels.
12
barrels.
12
barrels.
17
cwt.
17
cwt.
4368
pounds.
240
pounds.
3
inches.
110
gallons.
110
gallons.
120
gallons.
140
gallons.
120
gallons.
120
gallons.
120
gallons.
138
gallons.
140
gallons.
18
feet.
21
feet.
24
feet.
364
pounds.
124
pounds.
9
inches.
8
pounds.
8
pounds.
14
pounds.
5
pounds.
14
pounds.
236
gallons.
252
gallons.
28
pounds.
256
pounds.
336
pounds.
182
pounds.
40
cubic feet.
50
cubic feet.
106
tons.
55
tons.
DIGGING.
24 Cubic feet of sand, or 18 cubic feet of earth, or 17 cubic feet of clay, make 1 ton.
1 Yard cube of solid gravel or earth contains 18 heaped bushels before digging, and 27 heaped
busbels when dug.
27 Heaped bushels make 1 load.
FRENCH WEIGHTS AND MEASURES.
SJO
FRENCH AND ENGLISH WEIGHTS AND MEASURHS COMPARED.
The following is a comparative Table of the Weights and Mea*um of England and Prwu-r,
•which was published by the Royal and Central Society of Agriculture of l'ari», in tlie Annuaire
for 1829, and founded on a Report made by Mr. Mathieu, to the Royal Aca.lcniy of Science, of
France, on the bill passed the 17th of May, 1824, relative to the Weights and Mraiures termed
" Imperial," which are now used in Great Britain.
MEASURES OF LENGTH.
ENGLISH. FRENCH.
1 Inch {l-36th of a yard) 2o.3y.W4 centimetre*.
1 Foot (l-3d of a yard) 3 047.'»449 dccimctrr*.
Yard imperial 0 9i4;Mm« metre.
Fathom (2yard8) 1-8287G6.% metre.
Pole, or perch, (5^ yards) .5 02911 metres.
Furlong (220 yards) 20116437 metres.
Mile (1760 yards) 16093149 metres.
FRENCH. ENliLISH.
1 Millimetre 0039.37 inch.
1 Centimetre 0-393708 inch.
1 Decimetre 3937079 inches.
r39 37079 inches.
1 Metre J 3-2808.992 feet.
( 1093633 yard.
Myriametre 621382 niiles.
SQUARE MEASURES.
ENGLISH. FRENCH.
1 Yard square 0-836097 metre square.
1 Rod (square perch) 2.5-2919.39 metres square.
I Rood (1210 yards square) 10-11677.5 ares.
1 Acre (4840 yards square) 0404671 hectares.
FRENCH. ENGLISH.
1 Metre square 1- 1960.33 yard square.
1 Are 0 09884.5 rood.
] Hectare 2473614 acres.
SOLID ME.\SURES.
ENGLISH. FRENCH.
1 Pint (l-8th of a gallon) 0 567932 litre.
1 Quart (l-4lh of a gallon) 1 130864 litre.
1 Gallon imperial 4'8434o794 litre*.
1 Peck (2 gallons) 9 0869159 litres.
1 Bushel (8 gallons) 36 347664 litres.
I Sack (3 bushels) 10!»043 hectolitre.
1 Quarter (8 bushels) 2-.9078I3 hectolitre*
1 Chaldron (12 sacks) 13 08516 hectolitre.
FRENCH. ENGLI.SH.
. f 1 "760773 pint.
^ ^'^""^ ( 0-2200967 gallon.
I Decalitre 221)09667 gallons
1 Hectolitre 22 009667 gallons.
WEIGHTS.
ENGLISH TROY. FRENCH.
1 Grain (I-24th of a penny-weight) 006477 gramme,
1 Pennyweight (l-20th of an ounce) 1 55456 gramme.
1 Ounce (l-12th of a pound troy) 310913 grammes.
1 Pound troy imperial 0 3730956 kilogrumme.
26 RULES FOR REDUCTION.
ENGLISH AVOIRDUPOIS. FRENCH.
1 Drachm (l-16th of an ounce) 1"7712 gramme.
1 Ounce (l-16th of a pound) 28-3384 grammes.
1 Pound avoirdupois imperial 0-4534148 kilogramme.
I Hundredweight (112 pounds) 50-78-246 kilogrammes.
1 Ton (20 hundredweight) 1015-649 kilogrammes.
FRENCH. ENGLISH.
r 15-438 grains troy.
1 Gramme -| 0'643 pennyweight.
V 0*03216 ounce troy.
2'680-27 pounds troy.
1 Kilogramme j
2'20548 pounds avoirdupois.
RULES FOR REDUCTION.
I. When the Numbers are to be reduced from a Higher Denomination to a Lower :
Multiply the number in the highest denomination by as many of the next
lower as make an integer, or I, in that higher ; to this product add the number,
if any, which was in this lower denomination before, and set down the amount.
Reduce this amount in like manner, by multiplying it by as many of the next
lower as make an integer of this, taking in the odd parts of this lower, as before.
And so proceed through all the denominations to the lowest ; so shall the
number last found be the value of all the numbers which are in the higher
denominations, taken together*.
EXAMPLE.
1. In 1234? 15* 7d, how many farthings ?
£ s d
1234 15 7
20
24695 Shillings,
12
296347 Pence.
4
Answer 1185388 Farthings.
II. When the Numbers are to be reduced from a Lower Denomination to a Higher:
Divide the given number by as many of that denomination as make 1 of the
next higher, and set down what remains, as well as the quotient.
Divide the quotient by as many of this denomination as make one of the next
higher; setting down the new quotient, and remainder, as before.
Proceed in the same manner through all the denominations to the highest ;
and the quotient last found, together with the several remainders, if any, will be
of the same value as the first number proposed.
• The reason of this rule is very evident ; for pounds are brought into shillings by multi-
plying them by 20 ; shillings into pence, by multiplying them by 12 ; and pence into farthings,
by multiplying by 4 ; and the reverse of this rule by division. And the like, it is evident, will
be true in the reduction of numbers consisting of any denominations whatever.
COMPOUND ADDITION.
27
EXAMPLES.
2. Reduce 1185388 farthings into pounds, shillings, and pence.
4 1185388
12 296347^
20; 2469 5s 7d
An«. 23040.
Ans. 351/ 13*0jrf.
Ans. 36288.
Ans. 36.
Ans. 340157.
Answer 1234/ 15* 7d
3. Reduce 24/ to farthings.
4. Reduce 337587 farthings to pounds, &c.
5. How many farthings are in 36 guineas ?
6. In 36288 farthings how many guineas ?
7. In 59lb 13 dwts 5 gr. how many grains ?
8. In 8012131 grains how many pounds, &c.
Ans. 1390 1b 11 oz 18 dwt 19gr.
9. In 35 tons 17 cwt 1 qr 23 lb 7 oz 13 dr how many drams ? Ans. 20571005.
10. How many barley-corns will reach round the earth, supposing it to be
25000 miles? Ans. 4752000000.
11. How many seconds are in a solar year, or 365 days 5 hrs 48 min
45isec? Ans. 3 1556925 J.
12. In a lunar month, or 29 days 12 hrs 44 min 3 sec, how many seconds i
Ans. 2551443.
COMPOUND ADDITION.
Compound Addition serves to add or collect several numbers of different
denominations into one sum.
Rule. Place the numbers so, that those of the same denomination may stand
directly under each other, and draw a line below them. Add up the figures in
the lowest denomination, and find, by Reduction, how many units, or ones, of
the next higher denomination are contained in their sura. Set down the
remainder below its proper column, and carry those units or ones to the next
denomination, which add up in the same manner as before. Proceed thus
through all the denominations, to the highest, whose sum, together with the
several remainders, will give the answer sought.
The method of proof is the same as in Simple Addition.
EXAMPLES OF MONEY.
1. 2. 3. 4.
£ s d £ s d £ s d £ s d
7 13 3 14 7 5 15 17 10 53 14 8
3 5 10§
6 18 7
0 2 5J
4 0 3
17 15 4|
8 19 2i
7 8 li
21 2 9
7 16 8i
0 4 3
3 14 6
23 6 2J
14 9 4i
15 6 4
6 12 91
5 10 2}
93 11 6
7 5 0
13 2 5
0 18 7
39 15
9J
32 2
6J
39 15
9f
28
ARITHMETIC.
5.
6.
7.
8.
£
s
d
£ s
d
£ s d
£ s
d
14
0
n
37 15
8
61 3 2i
472 15
3
8
15
3
14 12
n
7 16 8
9 2
2^
62
4
7
17 14
9
29 13 10 J
27 12
6i
4
17
8
23 10
9i
12 16 2
370 16
2i
23
0
4|
8 6
0
0 7 5i
13 7
4
6
6
7
14 0
5i
24 13 0
6 10
5i
91
0
lOj
54 2
7i
5 0 10|
30 0
HI
_
_
Exam. 9- A nobleman, going out of town, is informed by his steward, that
his butcher's bill comes to 197^ 13s 7i«?; his baker's bill to 59/5«2|d; his
brewer's to 85Z; his wine-merchant's to 103/ 13s; to his corn-chandler is due
75? 3d; to his tallow-chandler and cheesemonger, 27? 15s ll?c/; and to his
tailor bhlZsb\d\ also for rent, servants' wages, and other charges, 127? 3s:
Now supposing he would take 100? with him, to defray his charges on the road,
for what sum must he send to his banker ? Ans. 830? 14s ^\d.
10. The strength of a regiment of foot, of 10 companies, and the amount of
their subsistence *, for a month of 30 days, according to the annexed Table, are
required :
Numb.
Rank.
Subsistence foraMonlb.
11
30
30
20
2
390
Colonel
Lieutenant-Colonel . .
Major
Captains
£ 27 0 0
19 10 0
17 5 0
78 15 0
57 15 0
40 10 0
7 10 0
4 10 0
5 5 0
4 10 0
4 10 0
45 0 0
30 0 0
20 0 0
2 0 0
292 10 0
Lieutenants
Ensigns
Chaplain ...
Adjutant
Quarter-Master
Surgeon
Surgeon's Mate
Serjeants
Corporals
Drummers
Fifes
Private Men
507
Total . .
£656 10 0
* Subsistence Money is the Money paid to the soldiers weekly; -which is short of their full
pay, because their clothes, accoutrements, &c. are to be accounted for. It is likewise the money
advanced to officers till their accounts are made up, which is commonly once a year, when they
are paid their arrears. The following Table shows the full pay and subsistence of each rank on
the English establishment.
COMPOUND ADDITION.
20
rr
2{
30
iz;
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1
30 ARITHMETIC.
EXAMPLES OF WEIGHTS, MEASURES, &C.
TROY WEIGHT. APOTHECARIES* WEIGHT.
1.
2.
3.
4.
lb
oz
dwt
OZ
dwt
gr
lb OZ dr
■ sc
OZ
dr
sc
RT
17
3
15
37
9
3
3 5 7
2
3
5
1
17
7
9
4
9
5
3
13 7 3
0
7
3
2
5
0
10
7
8
12
12
19 10 6
2
16
7
0
12
9
5
0
17
7
8
0 9 1
2
7
3
2
9
176
2
17
5
9
0
36 3 5
0
4
1
2
18
23
11
12
3
0
19
5 8 6
1
36
4
1
14
AVOIRDUPOIS WEIGHT.
—
LONG MEASURE.
5.
6.
7.
8.
lb
OZ
dr
csvt
qr
lb
mis fur
pis
yds
ft
in
17
10
13
15
2
15
29 3
14
127
1
5
5
14
8
6
3
24
19 6
29
12
2
9
12
9
18
9
1
14
7 0
24
10
0
10
27
]
6
9
1
17
9 1
37
54
1
11
0
4
0
10
2
6
7 0
3
5
2
7
6
14
10
3
0
3
4 5
9
MEASl
23
0
5
MEASURE.
CLOTH
LAND
JRE.
9.
10.
11.
12.
yds
qr
nls
el en
qrs
nls
ac ro
P
ac
ro
P
26
3
1
270
1
0
225 3
37
19
0
16
13
1
2
57
4
3
16 1
25
270
3
29
9
1
2
18
1
2
7 2
18
6
3
13
217
0
3
0
3
2
4 2
9
23
0
34
9
1
0
10
1
0
42 1
19
7
2
16
55
3
1
4
4
1
7 0
6
75
0
23
COMPOUND SUBTRACTION.
Compound Subtraction shows how to find the difference between any two
numbers of different denominations. To perform which, observe the followinji
Rule.
♦ Place the less number below the greater, so that the parts of the same
denomination may stand directly under each other ; and draw a line below
them. Begin at the right hand, and subtract each number or part in the lower
• The reason of tliig rule will easily appear from what has been said in Simple Subtraction •
for the borrowing de|)end8 on the same principle, and is only different as the numbers to be
subtracted are of different denominations.
COMPOUND SUBTRACTION. ,'jl
line, from the one just above it, and set the remainder straight below it : but if
any number in the lower line be greater than that above it, add aa many to the
upper number as make 1 of the next higher denomination ; then take the lower
number from the upper one thus increased, and set down the remainder. Carry
the unit borrowed to the next number in the lower line ; after which subtract
this number from the one above it, as before ; and so proceed till the whole is
finished. Then the several remainders, taken together, will be the whole differ-
ence sought.
The method of proof is the same as in Simple Subtraction.
EXAMPLES OF MONEY.
1.
£ s
From 79 17
Take 35 12
d
8i
4i
2.
£ s
103 3
71 12
31 10
d
n
5|
8J
3.
£ s d
81 10 11
29 13 3i
4.
£ s d
254 12 10
37 9 4i
Rem. 44 5
4^
Proof 79 17
81
103 3
H
5. What is the difference between 73/ 5id and 19/ 13* lOrf?
Ans. 53/ 6* 7i</.
6. A lends to B 100/ ; how much is B in debt after A has taken goods of him
to the amount of 73/ 12s 4|c/? Ans. 26/ 7* 7{d.
7. Suppose that my rent for half a year is 20/ 12s, and that I have laid out
for the land-tax 14s 6c/, and for several repairs 1/ 3s S^t/; what have I to pay of
my half-year's rent? Ans. 18/ 14s 2jd.
8. A trader failing, owes to A 35/ 7s 6c/, to B 91/ 13s Ojc/, to C .53/ 7i</, to D
87/ 5s, and to E 111/ 3s 5^d. When this happened, he had by him in cash,
23/ 7s 5c/, in wares 531 lis lO^d, in household furniture 63/ 17s 7J</, and in
recoverable book-debts 25/ 7s 5d. What will his creditors lose by him, sup
posing these things delivered to them? Ans. 212/ 5s 3^d.
EXAMPLES OF WEIGHTS, MEASURES, &C.
TROY WEIGHT. APOTHECARIES* WEIGHT.
1. 2. 3.
lb OZ dwt gr lb oz dwt gr lb oz dr scr gr
From 9 2 12 10 7 10 4 17 73 4 7 0 14
Take 5 4 6 17 3 7 16 12 29 5 3 1 19
Rem.
P of
LONG
17
11
From
Take
AVOIRDUPOIS
4.
c qrs lb
5 0 17
2 3 10
WEIGHT.
5.
lb OZ dr
71 5 9
17 9 13
6.
m fu
14 3
7 6
MEASURE.
7.
yd ft in
95 0 4
71 2 9
Rem.
Proof
32
ARITHMETIC.
CLOTH MEASURE.
LAND MEASURE.
8.
9.
10.
11.
yd qr nl
yd
qr
nl
ac ro p
ac
ro p
From
17 2 1
9
0
2
71 1 14
57
1 16
Take
9 0 2
7
2
1
17 2 8
22
3 29
Rem.
Proof
DRY MEASURE
TIME.
12.
13.
14.
15.
la qr bu
bu
gal
pt
mo we
da ds
hrs min
From
9 4 7
13
7
1
71 2
5 114
17 26
Take
6 3 5
9
2
7
17 1
6 72
10 37
Rem.
Proof
20. The line of defence in a certain polygon being 236 yards, and that part of
it which is terminated by the curtain and shoulder being 146 yards 1 foot 4
inches ; what then was the length of the face of the bastion ?
Ans. 89 yds ift 8 inches.
COMPOUND MULTIPLICATION.
Compound Multiplication shows how to find the amount of any given
number of different denominations repeated a certain proposed number of times ;
which is performed by the following rule.
Set the multiplier under the lowest denomination of the multiplicand, and
draw a line below it. Multiply the number in the lowest denomination by the
multiplier, and find how many units of the next higher denomination are con-
tained in the product, setting down what remains. In like manner, multiply the
number in the next denomination, and to the product carry or add the units,
before found, and find how many units of the next higher denomination are in
this amount, which carry in like manner to the next product, setting down the
overplus. Proceed thus to the highest denomination proposed : so shall the
last product, with the several remainders, taken as one compound number, be
the whole amount required.
examples of money.
1. To find the amount of Sib of tea, at 5s 8^d per lb.
s d
5 8i
£2 5 8 Answer.
2. 4 lb of tea at 7« 8rf per lb.
3. 6 lb of butter, at 9^rf per lb.
£
s
Ans.
1
10
Ans.
0
4
COMPOUND MULTIPLICATION.
S3
£ i d
4. 7 lb of tobacco, at Is 8 JJ per lb. Xns. 0 U lu
5. 8 stone of beef, at 2s 7^^ per stone. Ans. 1 1 o
6. 10 cwt of cheese, at 2l 17s lOrf per cwt. An«, 28 18 4
7. 12 cwt of sugar, at 3/ 7s 4d per cwt. Ana. 40 8 o
CONTRACTIONS.
I. If the multiplier exceed 12, multiply successively by its component parta,
instead of the whole number at once.
EXAMPLES.
1. 15 cwt of cheese, at I7s 6d per cwt.
£ s d
0 17 6
3
2 12 6
5
£
s
d
Ans.
87
3
4
Ans.
81
U
0
Ans.
3
7
6
Ans.
7
1
9
Ans.
84
0
0
Ans.
137
4
0
Ans.
112
0
0
Ans.
34
10
0
Ans.
96
0
0
£13 2 6 Answer.
2. 20 cwt of hops, at 4l 7s 2d per cwt.
3. 24 tons of hay, at 3/ 7* 6rf per ton.
4. 45 ells of cloth, at Is 6d per ell.
5. 63 gallons of oil, at 2s 3d per gallon.
6. 70 barrels of ale, at 1/ 4s per barrel.
7. 84 quarters of oats, at IZ 12s 8cf per qr.
8. 96 quarters of barley, at IZ 3s id per qr.
9. 120 days' wages, at 5s 9^ per day.
10. 144 reams of paper, at 13s 4d per ream.
II. If the multiplier cannot be exactly produced by the multiplication of sim-
ple numbers, take the nearest number to it, either greater or less, which can be
80 produced, and multiply by its parts, as before. Then multiply the given
multiplicand by the difference between this assumed number and the multiplier,
and add the product to that before found, when the assumed number is less than
the multiplier, but subtract the same when it is greater.
EXAMPLES.
1. 26 yards of cloth, at 3s Of rf per yard.
£ s d
0 3 OJ
5
0
15
3J
5
3
16
6i
3
Of add.
£3 19 7i Answer.
2. 29 quarters of corn, at 2l 5s 3id per qr.
3. 53 loads of hay, at 3/ 15s 2d per load.
4. 79 bushels of wheat, at 1 Is 5f d per bushel.
5. 97 casks of beer, at 1 2s 2d per cask.
6. 1 14 stone of meat, at 15s 3f d per stone.
VOL. I. D
£
5 d
Ans.
65
12 lOi
Ans.
199
3 10
Ans.
45
6 lOi
Ans.
59
0 2
Ans.
87
5 74
34f ARITHMETIC.
EXAMPLES OF WEIGHTS AND MEASURES.
7.
8
9.
lb oz dw't gr
lb
oz dr
sc
gr
cwt qr lb
oz
28 7 14 10
2
6 3
2
10
29 2 16
14
5
8
—
12
10.
11.
"~~
12.
mis fu pis yds
yds
qrs
na
ac
ro po
22 5 29 3^
126
3
1
28
3 27
4
7
9
13.
—
14.
we qr bu pe gal
mo
we
da
ho min
24 2 5 3 1
17-2
3
5
16 49
6
10
COMPOUND DIVISION.
Compound Division teaches how to divide a number of several denomina-
tions by any given number, or into any number of equal parts. It is performed
as follows : —
Place the divisor on the left of the dividend, as in simple division. Begia
at the left hand, and divide the number of the highest denomination by the
divisor, setting down the quotient in its proper place. If there be any remainder
after this division, reduce it to the next lower denomination, which add to the
number, if any, belonging to that denomination, and divide the sum by the
divisor. Set down again this quotient, reduce its remainder to the next lower
denomination again, and so on through all the denominations to the last.
EXAMPLES OF MONEY.
1.
Divide 237/ 8s 6d by 2.
£
s
d
2) 237
S
6
the Quotient.
£118
14
3
£ s d
£ «
d
2.
Divide 432 J2 If by 3.
Ans. 144 4
Oh
3.
Divide 507 3 5 by 4.
Ans. 126 15
lOi
4.
Divide 632 7 6§ by 5.
Ans. 126 9
6
5.
Divide 690 14 3 J by 6.
Ans. 115 2
44
6.
Divide 703 10 2 by 7.
Ans. 100 15
8J
COMPOUND DIVISION.
£ s d
85
£ $ d
7. Divide 760 5 6 by 8. Ana. 95 0 8J
S.Divide 761 5 7i by 9. Ans. 84 11 8{
9. Divide 829 17 10 by 10. Ans. 82 19 9j
10. Divide 937 8 Sf by 11. . Ans. 85 4 5
11. Divide 1145 11 4i by 12. Ans. 95 9 3j
CONTRACTIONS.
I. If the divisor e.xceed 12, find what simple numbers, multiplied together,
will produce it, and divide by them separately, as in simple division, as below.
EXAMPLES.
1. What is cheese per cwt if 16 cwt cost 25/ 14j 8di
£ s d
4 25 14 8
4 6 8 8
£1 12 2 the Answer.
£
2. If 20 cwt of tobacco come to 150/ 6s 8d, what is that per cwt ?
Ans. 7 10
3. Divide 98/ 8s by 36. Ans. 2
4. Divide 71/13* lOd by 56. Ans. 1
5. Divide 44/ 4s by 96. Ans. 0
6. At 31/ 10s per cwt, how much per lb ? Ans. 0
II. If the divisor cannot be produced by the multiplication of small numbers,
divide by the whole divisor at once, after the manner of long division, as
follows : —
EXAMPLES.
1. Divide 59/ 6s 3^rf by 19-
£ s d £ 8 d
19) 59 6 3| (3 2 5i Ans.
57
10
4
14
8
5
7i
9
n
5
ri
2
20
46 (2
3S
8
12
99 (5
93
4
4
19 (1
£
s
d.
2.
Divide 39
14
5iby
57.
3.
Divide 125
4
9
by
43
4.
Divide 542
7
10
by 97.
5.
Divide 123
11
2iby
li.
£ s d
Ans. 0 13 lU
Ans. 2 18 3
Ans 5 11 lU
Ans. 0 19 bk
D 2
3 ARITHMETIC.
EXAMPLES OF WEIGHTS AND MEASURES.
1. Divide 17 lb 9 oz 0 dwts 2 gr by 7- Ans. 2 lb 6 oz 8 dwts 14 gr.
2. Divide 17 lb 5 oz 2 dr 1 scr 4 gr by 12. Ans 1 lb 5 oz 3 dr 1 sc 12 gr.
3. Divide 178 cwt 3 qrs 14 lb by 53. Ans. 3 cwt 1 qr 14 lb.
4. Divide 144 mi 4 fur 20 po 1 yd 2 ft by 39- Ans. 3 mi 5 fur 26 po 2 ft 8 in,
5. Divide 534 yds 2 qrs 2 na by 47. Ans. 11 yds 1 qr 2 na.
6. Divide 77 ac 1 ro 33 po by 51. Ans. 1 ac 2 ro 3 po.
7. Divide 206 mo 4 da by 26. Ans. 7 mo 3 we 5 ds.
THE GOLDEN RULE, or RULE OF THREE.
The Rule of Three enables us to find a fourth proportional to three numbers
given : for which reason it is sometimes called the Rule of Proportion. It is
called the Rule of Three, because three terms or numbers are given, to find a
fourth. And because of its great and extensive usefulness, it was often called,
by early writers on Arithmetic, the Golden Rule. This Rule is usually by prac-
tical men considered as of two kinds, namely. Direct and Inverse. The dis-
tinction, however, as well as the manner of stating, though retained here for
practical purposes, does not well accord with the principles of proportion ; as
will be shown farther on.
The Rule of ITiree Direct is that in which more requires more, or less requires
less. As in this: if 3 men dig 21 yards of trench in a certain time, how much
will 6 men dig in the same time ? Here more requires more, that is, 6 men,
which are more than 3 men, will also perform more work in the same time.
Or when it is thus : if 6 men dig 42 yards, how much will 3 men dig in the
same time ? Here, then, less requires less, or 3 men will perform proportionably
less work than 6 men in the same time. In both these cases, then, the Rule, or
the Proportion, is Direct ; and the stating must be
thus, as 3 : 21 ; : 6 : 42, or as 3 : 6 : : 21 : 42.
And, as 6 : 42 ; : 3 : 21, or as 6 : 3 : ; 42 : 21.
But the Rule of Three Inverse, is when more requires less, or less requires
more. As in this : if 3 men dig a certain quantity of trench in 14 hours, in how
many hours will 6 men dig the like quantity ? Here it is evident that 6 men,
being more than 3, will perform an equal quantity of work in less time, or fewer
hours. Or thus : if 6 men perform a certain quantity of work in 7 hours, in
how many hours will 3 men perform the same ? Here less requires more, for 3
men will take more hours than 6 to perform the same work. In both these
cases, then, the Rule, or the Proportion, is Inverse ; and the stating must be
thus, as 6 : 14 : : 3 : 7, or as 6 : 3 ; ; 14 : 7.
And, as 3 : 7 ! I 6 : 14, or as 3 : 6 ; ; 7 : 14.
And in all these statings, the fourth term is found, by multiplying the 2d and
3d terms together, and dividing the product by the 1st term.
Of the three given numbers : two of them contain the supposition, and the
third a demand. And for stating and working questions of these kinds, observe
the following general Rule :
State the question, by setting down in a straight line the three given numbers,
in the following manner, viz. so that the 2d term be that number of supposition
which is of the same kind that the answer or 4th terra is to be ; making the other
number of supposition the 1st term, and the demanding number the 3d term.
RULE OF THREE.
tn
when the question is in direct proportion ; but contrariwise, the other number
of supposition the 3d term, and the demanding number the Ist term, when the
question has inverse proportion *.
Then, in both cases, multiply the 2d and 3d tenns tojjether, and divide the
c product by the first, which will give the answer, or 4th term sought, viz. of the
same denomination as the second term.
Note I. If the first and third terms consist of different denominations, reduce
them both to the same ; and if the second term be a compound number, it i«
mostly convenient to reduce it to the lowest denomination mentioned. If, after
division, there be any remainder, reduce it to the next lower denomination, and
divide by the same divisor as before, and the quotient will be of this lant deno-
mination. Proceed in the same manner with all the remainders, till they be
reduced to the lowest denomination which the second admits of, and the several
quotients taken together will be the answer required.
Note II. The reason for the foregoing rules will appear when we come to
treat of the nature of proportions. Sometimes two or more statings are neces-
sary, which may always be known from the nature of the question : but in this
case it falls under compound proportion, and may be more easily worked by the
rule for that case.
Note III. When the first term is divisible by any number which also divides
the second or third, we may so divide them, using the quotients instead of the
original terms. This will often diminish the labour of the calculation con-
siderably,
EXAMPLES.
1. If 8 yards of cloth cost 1/4* what will 96 yards cost ?
Or, in accordance with note iii : —
yd £ * yds £ $
yds £
s yds
£
As8 : 1
4 : : 96 :
14
20
24
96
]14
216
8; 2304
2o' 2S|8s
£ 14 8
Answer.
As 1 : 1
12 : 14 8
£ 14 8
Or again,
yd s yds £ s
As 1 : 3 : : 96 : 14 8
3
20)2818
£14 8
• If we adhere to the rigid geometrical principles of ratio, as Mr. Bonnycastle ht« done, we
should put the term which is of the same kind with the answer in the third place instead of tb«
second. It is not, however, with concrete but with abstract numbers that we work ; and, hence,
though the relations of things show us the relations of the numbers by which tbejr are repre-
sented, still we may conceive a ratio between the numbers whilst the things thetnselve. are
dissimilar. Such a restriction was necessary in the geometry of the Greek*, but U not at all
implied, and is therefore not necessary, in the arithmetic of symbols.
It should be added, that this rule is of very great European antiqt.ity, and it hat been
universally given in this form : though the application of Jones's Rule (sec Compound Pro-
portion) is certainly more simple, and upon the whole more easily applied. As, however, the
very form of stating the Rule of Three has been almost universally adopted in writing p^oport1on^
and it has acquired so strong a hold upon the language, habiu, and pncUce of nwakind, it b«
not been considered desirable to alter it here.
38 ARITHMETIC.
Ex. 2. An engineer having raised 100 yards of a certain work in 24 days
with 5 men ; how many men must he employ to finish a like quantity of work
in 15 days ?
ds men ds men
As 15 : 5 : : 2t : 8 Ans.
5
15) 120 (8 Answer.
120
3. What will 72 yards of cloth cost, at the rate of 9 yards for 5l 12s ?
Ans. 44Z l6s.
4. A person's annual income being 1461 ; how much is that per day?
Ans. 8s.
5. If 3 paces or common steps of a certain person be equal to 2 yards, how
many yards will 160 of his paces make ? Ans. 106 yds 2 ft.
6. What length must be cut off a board, that is 9 inches broad, to make a
square foot, or as much as 12 inches in length and 12 in breadth contains ?
Ans. 16 inches.
7. If 750 men require 22500 rations of bread for a month, how many rations
will a garrison of 1200 men require ? Ans. 36000.
8. If 7 cwt 1 qr of sugar cost 26/ 105 id; what will be the price of 43 cwt
2 qrs ? Ans. 159/ 25,
9. The clothing of a regiment of foot of 750 men amounting to 2831/ 5s;
what will the clothing of a body of 3500 men amount to ? Ans. 13212/ 105.
10. How many yards of matting, that is 3 ft. broad, will cover a floor that is
27 feet long and 20 feet broad ? Ans. 60 yards.
11. What is the value of six bushels of coals, at the rate of 1/ 14s 6d the
chaldron ? Ans. 5s Qd.
12. If 6352 stones of 3 feet long complete a certain quantity of walling ; how
many stones of 2 feet long will raise a like quantity ? Ans. 9328.
13. What must be given for a piece of silver weighing 73 lb 5 oz 15 dwts, at
the rate of Ss Qd per ounce ? Ans. 253/ 10s Ofti.
14. A garrison of 536 men having provision for 12 months; how long will
those provisions last, if the garrison be increased to 1124 men?
Ans. 174 i*;jx days.
15. What will be the tax upon 763/ 15s, at the rate of 3s 6c/ per pound
sterhng? Ans. 133/ 13s l^d.
16. A certain work being raised in 12 days, by working 4 hours each day;
how long would it have been in raising by working 6 hours per day ?
Ans. 8 days.
1 7. What quantity of corn can I buy for 90 guineas, at the rate of 6s the
bushel ? Ans. 39 qrs 3 bushels.
18. A person, failing in trade, owes in all 977/; at which time he has, in
money, goods, and recoverable debts, 420/ 6$ 3|</; now supposing these things
delivered to his creditors, how much will they get per pound ? Ans. 8s 7^d.
19. A plain of a certain extent having supplied a body of 3000 horse with
forage for 18 days ; then how many days would the same plain have supplied a
body of 2000 horse ? Ans. 27 days.
20. Suppose a gentleman's income is 600 guineas a year, and that he spends
25s 6d per day, one day with another ; how much will he have saved at the
year's end ? Ans. 164/ 12s 6c/.
RULE OF THREE. 39
21. What cost 30 pieces of lead, each weighing 1 cwt 12 Ih, at the rate of
16* 4rf the cwt ? Ans. 27/ 2* 6rf,
22. The governor of a besieged place having provision for 54 days, at the rate
of 141b of bread ; but being desirous to prolong the siege to 80 days, in expec-
tation of succour, in that case what must the ration of bread be ? Ans 1 ' lb
23. At half-a-guinea per week, how long can I be boarded for 20 jwunds ?
Ans. 38,7, wk«-
24. How much will 75 chaldrons 7 bushels of coals come to, at the rate of
1/ 135 &d per chaldron ? Ans. 1 25/ 1 9* o!J.
25. If the penny loaf weigh 8 ounces when the bushel of wheat cost 7» 3rf,
what ought the penny loaf to weigh when the wheat is at 8» 4e/?
Ans. 60Z 15,^ dr.
26. What rent will 173 acres 2 roods 14 poles of land yield, at the rale of
Il7s8d per acre ? Ans. 240/ 2f Jjgd.
27. To how much amount 73 pieces of lead each weighing 1 cwt 3 qr* 7 lb,
at 10/ 45 per fother of 19^ cwt ? Ans. 69/ 4* '2d l^Jq.
28. How many yards of stuff, of 3 qrs wide, will line a cloak that is IJ yards
in length and 3i yards wide ? Ans. 8 yds. 0 qrs. 2j nl.
29- If 5 yards of cloth cost 14s 2d, what must be given for 9 pieces, contain-
ing each 21 yards 1 quarter? Ans. 27/ la lOjfi
30. If a gentleman's estate be worth 2107/ 12s a year; what may he spend
per day, to save 500/ in the year ? Ans 4/ 8» Iji^d.
31. Wanting just an acre of land cut off from a piece which is IJi poles in
breadth, what length must the piece be ? Ans. 11 po 4 yds 2 ft 0^^ in.
32. At 7s 9id per yard, what is the value of a piece of cloth containing 53
ells English 1 qr. Ans. 25/ 18* Ijrf.
33. If the carriage of 5 cwt 14 lb for 96 miles be 1/ 12s 6d; how far may I
have 3 cwt 1 qr carried for the same money ? Ans. 151 m 3 fur 3,'j [wl.
34. Bought a silver tankard, weighing 1 lb 7 oz 14 dwts ; what did it cost
me at 6s 4c/ the ounce ? Ans. 6/ 4s 9\d.
35. What is the half year's rent of 547 acres of land, at 15s Gd the acre ?
Ans. 211/ 19» 3d.
36. A wall that is to be built to the height of 36 feet, was raised 9 feet high
by 16 men in 6 days ; then how many men must be employed to finish the
wall in 4 days, at the same rate of working ? Ans. 72 men.
37. What will be the charge of keeping 20 horses for a year, at the rate of
H^rf per day for each horse ? Ans. 441/ Os lOrf.
38. If 18 ells of stuff that is J yard wide, cost 39* 6d; what will 50 ells, of
the same quality, cost, being yard wide ? Ans. 7/ 6* 3l1d.
39. How many yards of paper that is 30 inches wide, will hang a room that is
20 yards in circuit and 9 feet high. Ans. 72 yards.
40. If a gentleman's estate be worth 384/ l6s a year, and the land tax be
assessed at 2s d^d per pound, what is his net annual income ?
Ans. 331/ 1* 9\d.
41. The circumference of the earth is about 25000 miles; at what rate per
hour is a person at the middle of its surface carried round, one whole rotation
being made in 23 hours 56 minutes? Ans. 1044iVj*j mile*.
42. If a person drink 20 bottles of wine per month, when it cost 8* a gallon,
how many bottles per month may he drink, without increasing the expense,
when wine costs 10s the gallon ? Ans. 16 bottles.
43. What cost 43 qrs 5 bushels of corn, at 1/ 8* 6rf the quarter ?
Ans. 62/ 3* 3|d:
40 COMPOUND PROPORTION.
44. How many yards of canvass that is ell wide will line 50 yards of say that
is 3 quarters wide ? Ans. 30 yards,
45. If an ounce of gold cost 4 guineas, what is the value of a grain ?
Ans. 2-}^d.
46. If 3 cwt of tea cost 40/ 12s ; at how much a pound must it be retailed, to
gain 10/ by the whole ? Ans. Sj^V-
COMPOUND PROPORTION.
Compound Proportion is a rule by means of which the student may resolve
such questions as require two or more statings in simple proportion.
The general rule for questions of this kind may be exhibited in the following
precepts, viz.
1. Set down the terras that express the conditions of the question in one line.
2. Under each conditional term, set its corresponding one, in another line,
putting the letter q in the (otherwise) blank place of the term required.
3. Multiply the effective terms of one line, and the objective terms of the
other line, continually, and take the result for a dividend.
4. Multiply the remaining terms continually, and let the product be a divisor,
5. The quotient of this division will be q, the term required *.
Note. By effective terms are here meant whatever necessarily and jointly
produce any effect ; as the cause and the time ; length, breadth, and depth ;
buyer and his money; things carried, and their distance, &c. all necessarily
inseparable in producing their several effects. In short, the causes of the effect.
By objective terms, those which express the effect itself.
Thus, if the number of men, the time of the siege, and the daily i-ations, be
the effective terms in producing the consumption of the quantity of food in the
garrison ; then, in reference to the same problem, the quantity of food consti-
tutes the objective term.
In a question where a term is only understood, and not expressed, that term
may always be expressed by unity.
A quotient is represented by the dividend put above a line, and the divisor
put below it.
EXAMPLES.
1. How many men can complete a trench of 135 yards long in 8 days, when
6 men can dig 54 yards of the same trench in 6 days ?
men days yds
16 6 54
Q 8 135
Here 16 men and 6 days are ths effective terms of the first line, and 135 yards
e objective term of the other. Therefore, by the rule,
16x6x135 2x135 „„
^= 8X54 =-9- =^"'
the number of men required.
• This rule, which is as applicable to Simpie as to Compound Proportion, was given, in 1706,
by W. Jones, Etq. F,R.S., the father of the late Sir W. Jones.
VULGAR FRACTIONS. 4|
ANOTHER QUESTION.
If a garrison of 3600 men have bread for 35 days, at 24 oz each day ; how
much a day may be allowed to 4800 men, each for 45 days, that the same
quantity of bread may serve ?
men oz days bread
3600 24 35 1
4800 Q 45 I
3600X24X35 ,^
•»=-r8oo^a5-='^°'=^^ '''"»•
AN EXAMPLE IN SIMPLE PHOPORTION.
If 14 yards of cloth cost 21/ ; how many yards may be bought for 73/ 10»?
yd £ yds
1 21 J4
1 7H Q
73ix 14
Q= — |t — =3 of 73J=49 yards. Answer.
2. If 100/ in one year gain 5/ interest ; what will be the interest of 750/ for
7 years ? Ans. 262/ lOi.
3. If a family of 8 persons expend 200/ in 9 months ; how much will serve a
family of 18 people 12 months ? Ans. 600/.
4. If 27s be the wages of 4 men for 7 days ; what will be the wages of 14 men
for 10 days? Ans. 6/ I5».
5. If a footman travel 130 miles in 3 days, when the days are 12 hours long ;
in how many days, of 10 hours each, may he travel 36() miles ? Ans. gjj days.
6. If 120 bushels of corn can serve 14 horses 56 days ; how many days will
94 bushels serve 6 horses ? Ans. 10,?,Jdays.
7. If 3000 lbs of beef serve 340 men 15 days; how many lbs will serve 120
men for 25 days ? Ans. I7fi4 lb 1 li',oz.
8. If a barrel of beer be sufficient to last a family of 8 persons 12 days ; how
many barrels will be drunk by 16 persons in the space of a year ?
Ans. 60j barrels.
9. If 180 men, in 6 days, of 10 hours each, can dig a trench 200 yards long,
3 wide, and 2 deep ; in how many days of 8 hours long, will 100 men dig a trench
of 360 yards long, 4 wide, and 3 deep? Ans. 48| days.
OF VULGAR FRACTIONS.
A Fraction, or broken number, is an expression of a part, or some parts, of
something considered as a whole.
It is denoted by two numbers, placed one below the other, with a line between
them :
Thus, - , y, which is named S-fourths.
4 denominator j
The Denominator, or number placed below the line, shows how many equal
parts the whole quantity is divided into; and it represents the Divisor in
Division. And the Numerator, or number set above the line, shows how many
of these parts are expressed by the fraction : being the remainder after division.
Also, both these numbers are in general named the Terms of the Fraction.
42 ARITHMETIC.
Fractions are either Proper, Improper, Simple, Compound, Mixed, or
Complex.
A Proper Fraction, is when the numerator is less than the denominator ; as,
3. or I or I
An Improper Fraction, is when the numerator is equal to, or exceeds, the
denominator; as §, or f, or J. In these cases the fraction is called improper,
because it is equal to or exceeds unity.
A Simple Fraction, is a single expression, denoting any number of parts of the
integer ; as, §, or |.
A Compound Fraction, is the fraction of a fraction, or two or more fractions
connected with the word o/" between them ; as ^ of §, or ^ of f of 3.
A Mixed Number, is composed of a whole number and a fraction together ;
as, 3j, or 12i.
A Complex Fraction, is one that has a fraction or a mixed number for its
numerator, or its denominator, or both ;
i 2 I 3^
as,|^,or— ,or|-,or-^-.
A whole or integer number may be expressed like a fraction, by writing I
below it, as a denominator ; so 3 is ^. or 4 is ^.
A fraction denotes division ; and its value is equal to the quotient obtained by
dividing the numerator by the denominator : so 'f is equal to 3, and V is equal
to 41.
Hence then, if the numerator be less than the denominator, the value of the
fraction is less than ] . But if the numerator be the same as the denominator
the fraction is just equal to 1. And if the numerator be greater than the deno-
minator, the fraction is greater than 1.
REDUCTION OF VULGAR FRACTIONS.
Reduction of Vulgar Fractions, is the bringing them out of one form or
denomination into another; commonly to prepare them for the operations of
Addition, Subtraction, &c. ; of which there are several cases.
PROBLEM.
To find the greatest common measure of two or more numbers.
The common measure of two or more numbers, is that number which will
divide them all without remainder: so, 3 is a common measure of IS and 24 ;
the quotient of the former being 6, and of the latter 8. And the greatest number
that will do this is the greatest common measure : so 6 is the greatest common
measure of 18 and 24 ; the quotient of the former being 3, and of the latter 4,
which will not both divide further.
RULES.
If there be two numbers only, divide the greater by the less ; then divide the
divisor by the remainder ; and so on, dividing always the last divisor by the last
remainder, till nothing remains ; so shall the last divisor of all be the greatest
common measure sought.
When there are more than two numbers, find the greatest common measure
of two of them, as before; then do the same for that common measure and
REDUCTION OF VULGAR FRACTIONS.
43
another of the numbers; and so on, through all the numbers; so will the
greatest common measure last found be the answer.
If it happen that the common measure thus found is 1 ; then the numhem are
said to be incommensurable, or not to have any common measure, or lliey are
said to be prime to each other *.
Ex. 1. Find the greatest common measure of 3852 and 762996.
Hut the mo<]e of put-
ting down the work may
be more compactly done
as below f.
19,3852'762896il98,
,200 3852 I
38
32) 762896 (198,
3852
37769
34668
3852
200
(19,
31016
30816
200)
1852
1800
52) 200 (33
156
44)
52
44
8)
(I4
44
40
(5,
1852
(7769
1800 34668
<
I4
52
31016
44
30816
—
26
8
200
3,
8
156
0
44
40
5,
41"
And
4 the
last divisor is
the
gre
atest common
mt'
asure
4) 8 (2,
* It is not absolutely necessary that our products should be less tlian the dividend. All that
the principle requires is, that we should take the difference hettcecn the dividend and the nearrst
multiple of the divisor. The method given in the rule is that most usually employed : though
when the next higher multiple would be nearer to the dividend than the next lower, the actual
Work is considerably lessened by the adoption of the higher multiple. Thus in the example in
the text, had we taken the quotient 4 in the third division, it will be obvious that one division
would have bees saved.
For a proof of this rule see tlie corresponding subject in algebra.
+ The several quotients in both processes are numbered by subscribed figure», a» l^orCj
•bowing that 1 is the 4th quotient, and 2 is the 6th. In tlie new method the remainder it con-
sidered and treated as the divisor of the previous quotient, without being placed (after the fint
step) in the usual place of tlie divisor in common operations. This can occasion no difficulty in
any case, as tlie divisor is not more removed from the place of the succcsiivc products than in
the old method.
One advantage is, that it saves the repetition of the winting of the dividend figures. A more
important one (and which is of great practical convenience in the corresponding algebraical
operation) is the compactness of the work, and the small space it occupies. Both con*idei»-
tions concur in recommending it to general adoption.
The 2nd, 4th, &c. quotients are placed on the left of the work ; the 1st, Sd, &c. on the
right. It will conduce to ready re-exaniinalion of the work to draw the horizontal lines abure
each final remainder throu(jh the tide lines, as in the example.
44 ARITHMETIC.
Ex. 2. To find the greatest common measure of 1908, 936, and 630.
262936 1908 2, 2j'36:630
72 18721 36'36
12161 36
216
0 , 270
252
17
18l
And 36 is the g. c. m of I9O8 and 936 ; and IS the g. c. m. of 36 and 630 is
the g. c. m. of all the three numbers 1908, 936, and 630.
3. What is the greatest common measure of 246 and 372 ? Ans. 6.
4. What is the greatest common measure of 324, 612, and 1032 ? Ans. 12.
CASE I.
To abbreviate or reduce fractions to their lowest terms.
* Divide the terms of the given fraction by any number that will divide
them without a remainder ; then divide these quotients aj^ain in the same man-
ner ; and so on, till it appears that there is no number greater than 1 which will
divide them ; then the fraction will be in its lowest terms.
• That dividing both the terms of the fraction by the same number, whatever it be, -mil give
another fraction equal to the former, is evident. And when these divisions are performed as
often as can be done, or when the common divisor is the greatest possible, the terms of the
resulting fraction must be the least possible.
Note. 1. Any number ending with an even number, or a cipher, is divisible, or can be
divided, by 2.
2. Any number ending with 5, or 0, is divisible by 5.
3. If the right-hand place of any n\imber be 0, the whole is divisible by 10 ; if there be two
ciphers, it is divisible by 100; if three ciphers, by 1000 : and so on; which is only cutting off
those ciphers.
4. If the two right-hand figures of any number be divisible by 4, the whole is divisible
by 4. And if the three right-hand figures be divisible by 8, the whole is divisible by 8.
And so on.
5. If the sum of the digits in any number be divisible by 3, or by 9, the whole is divisible by
3, or by 9.
6. If the right-hand digit be even, and the sum of all the digits be divisible by 6, then the
whole is divisible by 6.
7. A number is divisible by 11, when the sum of the 1st, 3d, 5th, &c. or of all the odd places,
is equal to the sum of the 2d, 4th, 6th, &c. or of all the even places of digits.
8. If a number cannot be divided by some quantity less than the square root of the same,
that number is a prime, or cannot be divided by any niimber whatever.
9. All prime numbers, except 2 and 5, have either 1, 3, 7, or 9, in the place of units ; and all
other numbers are composite, or can be divided. It is not, however, to be inferred that all
numbers which end in 1,3, 7, 9, are prime numbers. No method, indeed, is yet known by
which prime numbers can be either immediately calculated, or assigned, or detected. The best
practical method for numbers not very high, is the siere of Eratosthenes [kokkivov), an account
of which may be seen in the Phil. Trans, by Dr. HursUy, and in most works on the theory of
numbers.
10. When numbers, with the sign of addition or subtraction between them, are to be divided
by any number, then each of those numbers must be divided by it. Tiius, — X =^4"
4—2=7.
11. But if the numbers have the sign of multiplication between them, only one of them must
be divided. 1^.n.:.^^J^^='l^'^='^^S^J^^,.
6x2 6x1 2x1 IXl 1
REDUCTION OF VULGAR FRACTIONS. ^
Or, Divide both the terras of the fraction by their greatest common measure
at once, and the quotients will be the terms of the fraction required, of the same
value as at first.
1 EXAMPLES.
1. Reduce |gf to its least terms.
liS = ii = ?§ = 11 = i = i. the Answer.
Or thus :
216) 288 (1 Therefore 72 is the greatest common measure •
216 a"d 72) l^t = } the Answer, the same ai
before.
72) 21G (3
216
2. Reduce ||g to its lowest terms.
3. Reduce ll'\ to its lowest terms.
4. Reduce ^^j to its lowest terms.
Ans. J.
Ans. }.
Ans. j.
CASE II.
To reduce a mixed number to its equivalent improper fraction.
* Multiply the integer or whole number by the denominator of the fraction,
and to the product add the numerator ; then set that sum above the denominator
for the fraction required.
1. Reduce 23? to a fraction.
23
5
115
2
ll7=numerator,
and the fraction is 'i^-
2. Reduce 12j to a fraction.
3. Reduce 1475 to a fraction.
4. Reduce 183 ^^ to a fraction.
EXAMPLES.
Or the work may be written thus,
when the denominator is capable of
being used at once : —
(23 X 5) 4- 2 117
= - - the Answer.
Ans. 'i».
Ans. VJ.
Ans. »Vi*-
CASE III.
To reduce an improper fraction to its equivalent whole or mixed number.
t Divide the numerator by the denominator, and the quotient will be the
whole or mi.xcd number sought.
EXAMPLES.
1 . Reduce " to its equivalent number.
Here '3' or 12 -f- 3 = 4, the Answer.
• Tins is no more than first miiltiplying a quantity by some number, and tiien dividing the
result back again by the same : wliich it is evident does not alter the ralue ; for any fraction
represents a division of the numerator by the denominator.
f This rule is evidently the reverse of the former ; and the reason of it is manifest from the
nature of common division.
■iMMtJM.mi-^it^'^i^'^/'riiXC^^^ii&f&ii!!^^
46 ARITHMETIC.
2. Reduce ', to its equivalent number.
Here '| or 1 5 -i- 7 = 2 j, the Answer.
3. Reduce '^^^ to its equivalent number.
Thus 17) 749 (44tV
68
69
68
4. Reduce ^^ to its equivalent number. Ans. 8.
5. Reduce '^3' to its equivalent number. Ans. 54.^|.
6. Reduce '|^* to its equivalent number. Ans. I7ln.
CASE IV.
To reduce a whole number to an equivalent fraction, having a given
denominator.
* Multiply the whole number by the given denominator; then set the pro-
duct over the said denominator, and it will form the fraction required.
EXAMPLES.
1. Reduce 9 to a fraction whose denominator shall be 7.
Here 9 x 7 = 63 : then ''^ is the Answer ;
For «7' = 63 -r 7 = 9, the Proof.
2. Reduce 12 to a fraction whose denominator shall be 13. Ans. '/j*.
3. Reduce 27 to a fraction whose denominator shall be ] 1. Ans. %V.
CASE V.
To reduce a compound fraction to an equivalent simple one.
f Multiply all the numerators together for a numerator, and all the deno-
minators together for a denominator, and they will form the simple fraction
sought.
When part of the compound fraction is a whole or mixed number, it must first
be reduced to a fraction by one of the former cases.
And, when it can be done, any two terms of the fraction may be divided by
the same number, and the quotients used instead of them. Or, when there are
terms that are common, they may be omitted, or cancelled.
• Multiplication and Division being here equally used, the result must be the same as the
quantity first proposed.
+ The truth of this rule may be showTi as follows : Let the compound fraction be § of |.
Now J of ^ is ^ -;- 3, which is ^; consequently § of if will be i^ X 2 or .J^ ; that is, the nume-
rators are multiplied together, and also the denominators, as in the Rule. AVhen the compound
fraction consists of more than two single ones ; having first reduced two of them as above, then
the resulting fraction and a third \nll be the same as a compound fraction of two parts ; and so
on to the last of all.
REDUCTIOxN OF VULGAR FRACTIONS. 47
EXAMPLES.
1. Reduce ^ of 3 of J to a simple fraction.
1X2x3 _ ^ _ 1
2 X 3 X 4 ~ 24 ~ 4 ' *^^ Answer.
Here
1 X X X ^ I
^'^' / X ^ X 4 ~ 4 ^^' *^^"<^^^'^°g 2 and 3.
2. Reduce | of J of i^ to a simple fraction.
Here ^ X 3 x 10 _ ^ _ 12 _ 4
3 X 5 X 11 ~ 165 ~ 33 - 11 ^^^ Answer.
2
/^. 2 X ^ X/jz^ 4
/ X /^n I ~ iT ^™® *^ before, by cancelling the 3 and
dividing both terms by 5.
3. Reduce ^ of 3 to a simple fraction. Ans iJ.
4. Reduce § of ? of | to a simple fraction. Ans. |.
5. Reduce 3 of | of 3^ to a simple fraction. Ans. J.
6. Reduce ? of | of ^ of 4 to a simple fraction. Ans. *.
7. Reduce 2j of | to a fraction. Ans. \ or 2.
CASE VI.
To reduce fractions of different denominators to equivalent ft actions having
a common denominator.
* Multiply each numerator by all the denominators except its own, for the
new numerators : and multiply all the denominators together for a common
denominator.
Note. It is evident, that in this and several other operations, when any of the
proposed quantities are integers, or mixed numbers, or compound fractions, thejr
must first be reduced, by their proper rules, to the form of simple fractions.
• This is evidently no more tlian multiplying each numerator and its denominator by the
Hune quantity, and consequently the value of the fraction is not altered.
It is in many cases not only useful, but easy, to reduce fractions to their letut common
denominator.
The rule is this :
Find the least common multiple of all the denominators, and it will be the common deno-
minator required.
Then divide the common denominator by the denominator of each fraction, and multiply the
quotient by the numerator — the several products will be the numerators ; which are to be placed
respectively over the common denominator for the answer.
To find the least common multiple proceed thus :
Divide by any numbers that will divide two or more of the given number* without a i*-
mainder, and set the quotients, together with the undivided numbers, in a line beneath?
Divide the second line, as before, and so on, until tlicre arc no two numbers, beginning with
the lowest numbers, and only prirnes need be used, that can be divided ; then the continuetl pro-
duct of the divisors, quotients, and undivided numbers, will give the multiple required.
EfOMtpU
48 ARITHMETIC.
EXAMPLES.
1. Reduce ^, §, and J, to a common denominator.
1 x3 x4 = 12 the new numerator for ^.
2 X 2 X 4 = 16 ditto §.
3 X 2 X 3 = 18 ditto |.
2 X 3 X 4 = 24 the common denominator.
Therefore the equivalent fractions are ^f, |f , and -Jf.
Or the whole operation of multiplying may often be performed mentally, only
setting down the results and given fractions thus, J, §, f , = 'J, i^% if, = ^, -^^ f^,
by abbreviation.
2. Reduce f and g to fractions of a common denominator. Ans. Jf, g*.
3. Reduce f, ?, and J to a common denominator. Ans. j2, g^, JJ.
4. Reduce |, 2*, and 4 to a common denominator. Ans. ^, ^, ^°.
Note. 1. When the denominators of two given fractions have a common
measure, let them be divided by it ; then multiply the terms of each given frac-
tion by the quotient arising from the other's denominator.
Ex. 2^ and 33 = ^,3 and f^*., by multiplying the former by 7 and the latter by 5.
5 7
2. When the less denominator of two fractions exactly divides the greater,
multiply the terms of that which has the less denominator by the quotient.
Ex. ^ and ^j = i* and ^. by multiplying the terms of the former by 2.
2
3. W'hen more than two fractions are proposed, it is sometimes convenient,
first to reduce two of them to a common denominator ; then these and a third ;
and so on till they be all reduced to their least common denominator.
Ex. I and | and J = § and 3 and J = l\ and l\ and ^\.
CASE VII.
To redttce complex fractions to single ones.
Reduce the two parts both to simple fractions ; then multiply the numerator
of each by the denominator of the other ; which is in fact only increasing each
part by equal multiplications, which makes no difference in the value of the
whole.
So, 1 = 1-. And?i=L Alsol'="-=lixl-=?^.
3 6 4 12 4^ I 5 9 45
Example, Reduce ^, |, |, ^., fsi ''"*^ ii ^° fractions having the least common denominator.
3' 3 4 5 12 15 20
1 4 5 4 5 20
115 15 5
111111
Tlierefore we have 3 X 4 X 5 = 60 ^ least common denominator.
Then 60 -r- 3 = 20, and 20 X 2 = 40
60 -f- 4 = 15, and 15 X 3 = 45
60 -j- 5=12, and 12 X 4 = 48 [ new numerators.
60 -^ 12 = 5, and 5 X 7 = 35
60 -1- 15 = 4, and 4 x 8 = 32
60-1-20= 3, and 3x11=33
Hence |3, JJ, |J, gj, gj. §^ are the fractions required. It is of great importance that the student
should be made familiar with this rule, both on account of the facility which it gives in actual
reductions, and especially in the reductions that occur in algebraic fractions and equations.
REDUCTION OF VULGAR FRACTIONS. 40
CASE VIII.
To find the value of a fraction in parts of the integer.
Multiply the integer by the numerator, and divide the product by the
denominator, by compound multiplication and division, if the integer be a
compound quantity.
Or, if it be a single integer, multiply the numerator by the part« in the next
inferior denomination, and divide the product by the denominator. Then, if any
thing remains, multiply it by the parts in the next inferior denomination, and
divide by the denominator, as before ; and so on as far as necessary ; so shall
the quotients, placed in order, be the value of the fraction required •.
EXAMPLES.
1. What is the Jof 2/6s?
By the former part of the rule,
21 6s
4
5 9 4
£ 1 16 9§ f
What is the value of J of W?
By the 2d part of the rule,
3 I 2/
I
£ 0 13 4
3. Find the value of § of a pound sterling. Ans. 7s 6d.
4. What is the value of | of a guinea ? Ans. 4* Sd.
5. What is the value of J of a half-crown ? Ans. Is lOjrf.
6. What is the value of § of 45 lOd ? Ans. la 1 l^d.
7. What is the value of J lb troy ? Ans. 9 oz 12 dwts.
8. What is the value of /j of a cwt ? Ans. 1 qr 7 lb.
9. What is the value of I of an acre ? Ans. 3 ro 20 po.
10. What is the value of ^^j of a day ? Ans. 7 hrs 12 min.
CASE IX.
To reduce a fraction from one denomination to another.
f Consider how many of the less denomination make one of the greater;
then multiply the numerator by that number, if the reduction be to a less name,
but multiply the denominator, if to a greater.
EXAMPLES.
1. Reduce | of a pound to the fraction of a penny.
I X S" X 'f = T = 'f> the answer.
2. Reduce f of a penny to the fraction of a pound.
f X ^ X j'g = sbs, the answer.
* The numerator of a fraction being considered as a remainder, in division, and the deno-
minator as a divisor, this rule is of tlie same nature as compound division, or the Taluation of
remainders in the rule of three, before explained.
t This is the same as the rule of reduction in whole numbers from one denominatioo io
another.
VOL. I. B
50 ARITHMETIC.
3. Reduce ^V to the fraction of a penny.
' 4. Reduce Iq to the fraction of a pound.
5. Reduce | cwt to the fraction of a lb.
6. Reduce i dwt to the fraction of a lb troy.
7. Reduce I crown to the fraction of a guinea.
8. Reduce § half-crown to the fraction of a shilling.
9. Reduce 2s 6d to the fraction of a ;£.
10. Reduce l7s 7d olq to the fraction of a ^6.
Ans.
?rf.
Ans.
1
5TO5"
Ans.
f.
Ans.
Ams-
Ans.
s
ss-
Ans.
H-
Ans.
A-
Ans.
2119
ADDITION OF VULGAR FRACTIONS.
If the fractions have a common denominator ; add all the numerators
together, then place the sum over the common denominator, and that will be
the sum of the fractions required.
* If the proposed fractions have not a common denominator, they must be
reduced to one. Also compound fractions must be reduced to simple ones, and
fractions of different denominations to those of the same denomination. Then
add the numerators, as before. As to the mixed numbers, they may either be
reduced to improper fractions, and so added with the others ; or else the frac-
tional parts only added, and the integers united afterwards.
EXAMPLES.
1. To add I and | together.
Here i + 3 = 5 = Ifj the answer.
• Before fractions are reduced to a common denominator, they are quite dissimilar, as much
as shillings and pence are, and therefore cannot be incorporated with one another any more than
these can. But when they are reduced to a common denominator, and made parts of the same
thing, their sum, or difference, may then be as properly expressed by the sum or difference of
the numerators, as the sum or difference of any two quantities whatever, by the sum or differ-
ence of their individuals. Whence the reason of the rule is manifest, both for addition and
subtraction.
Note 1. When several fractions are to be collected, it is commonly best first to add two of
them together that most easily reduce to a common denominator ; then add their sum and a
third, and so on.
Note 2. Taking any two fractions whatever, -j^ and §4, for example, after reducing them to
a common denominator, we judge whether they are equal or unequal, by observing whether the
products 35 X H, and 7 X 55, which constitute the new numerators, are equal or unequal. If,
therefore, we have two equal products 35 X 1 1 = 7 X 55, we may compose from them two
equal fractions, as §4 = ^, or ^ = ff .
If, then, we take two equal fractions, such as ^ and ?i, we shall have 35 X H ^ 7 X 55 ;
taking from each of these 7x11, there will remain (35 — 7) X 11 = (55 — 11) X 7, whence
wehaveg^ = -?l,or3| = fr
In like manner, if the terms of -j^ were respectively added to those of ^, we should have
35 4- 7_,,_,
55^11 ~^''~'"'
Or, generally, if - =i _ , it may in a similar way be shown, that ~ = - ^ ^ -- .
b a 0 ±: d b a
Hence, when hco fraction^ are of equal value, the fraction formed by taking the sum or the
difference of their numerators respectively, and of their denominators respectively, is a fraction
equal in value to each of the original fractions. This proposition will be found useful in the
doctrine of proportions.
MULTIPLICATION OF VULGAR FR.\CTIOx\S. 51
2. To add I and ^ together.
s + B = Jo + ?o = 30 = Ij?. the answer.
3. To add | and 7 J and i of | together.
1 + 7i + J of f = 1 + » + i = * + «0 ^ I ^ Y ^ g,
4. To add f and f together. ^^^^ j,
5. To add J and g together. . .?!
6. Add if and /, together. . ' y*'
7. What is the sum of § and ? and f ? ^„,* JJm
8. What is the sum of | and | and 2J ? ^„g' 3^'
9. What is the sum of I and \ of 4, and 9,'i, ? Ana! loL.
10. What is the sum of § of a pound and g of a shiUing ?
„,, . , Ans. ';*»or 13* 10(i2jo.
11. What IS the sum of i of a shilling and ^\ of a penny ? Ans. V^rf or Td \\\q.
12. What is the sura of ^ of a pound, and J of a shilling, and ^^ of a penny >
Ans. ^ or 3* \d \\\q.
SUBTRACTION OF VULGAR FRACTIONS.
Prepare the fractions the same as for addition, when necessary ; then sub-
tract the one numerator from the other, and set the remainder over the common
denominator, for the difference of the fractions sought.
EXAMPLES.
1 . To find the difference between I and J. Here * — J = ^ = », the answer.
2. To find the difference between | and |. J — g = §J = ^ = j',, the answer.
3. What is the difference between I'j and f^j ? Ans. J.
4. What is the difference between -p, and 3^ ? Ans. ^.
5. What is the difference between -^^ and ^3 ? Ans. -f^.
6. What is the difference between 5| and } of 4J ? Ans. A^.
7. What is the difference between 9 of a pound, and 3 of f of a shilling ?
Ans Vs's or 10» 7d IJ9.
8. What is the difference between Ij of 5^ of a pound, and J of a shilling ?
Ans. JY^J/or 1/ 8» 11, y.
MULTIPLICATION OF VULGAR FRACTIONS.
• Reduce mixed numbers, if there be any, to equivalent fractions ; then
multiply all the numerators together for a numerator, and all the denominators
together for a denominator, which will give the product required.
• Multiplication of any thing by a fraction, implies the taking some part or part* of the thing;
it may therefore be truly expressed by a compound fraction ; which is resolved by multipljing
together the numerators and the denominators.
Note. A fraction is best multiplied by an integer, by dividing the denominator by it ; but if
it will not exactly divide, then multiply the numerator by it.
E 2
52 ARITHMETIC.
EXAMPLES.
1. Required the product of f and |.
Here i x 5 = /s ^= b. tbe answer.
Or, i xl = i X i = l
2. Required the continued product of §, 3j, 5, and | of ^.
Here ^x-X^y^X- = llAi = ?9 = 47 Answer.
if 4 14/ 4X2 8
3. Required the product of , and f . Ans. ^.
4. Required the product of ^ and /,. * Ans. j^.
5. Required the product of ^, ^, and i,\. Ans. ,3.
6. Required the product of i, 3, and 3. Ans. 1.
7. Required the product of I, 3, and 4/,. Ans. 23'5.
8. Required the product of 5, and § of ,. Ans. i".
9. Required the product of 6, and § of 5. Ans. 20.
10. Required the product of | of I, and | of 3*. Ans. 1^.
11. Required the product of 3| and 4^5. Ans. 14.}^^
12. Required the product of 5, *, | of I, and 4g. Ans. 2.jV
DIVISION OF VULGAR FRACTIONS.
* Prepare the fractions as before in multiplication : then di\'ide the nume-
rator by the numerator, and the denominator by the denominator, if they will
exactly divide : but if not, invert the terms of the divisor, and multiply the
dividend by it, as in multiplication.
EXAMPLES.
1. Divide ^ by f. Here */ -=- f = 3 = 1?, by the first method.
2. Divide I by ^K. Here I ^ ^% = I x 'i = I X 'i = 'i = 4^.
3. It is required to divide ?J by |. " Ans. i.
4. It is required to divide ^s hy |. Ans. /j-
5. It is required to divide 5* by |. Ans. I3,
6. It is required to divide g by ". Ans. iV
7. It is required to divide "5 by f. Ans. ^.
8. It is required to divide Ij by |. Ans. \^.
9. It is required to divide /^ by 3. Ans. /j.
10. It is required to divide | by 2. Ans. 15.
11. It is required to divide 73 by 9|. Ans. ^3.
12. It is required to divide 3 of 3 by 7 of 7|. Ans. 7^.
RULE OF THREE IN VULGAR FRACTIONS.
Make the necessary preparations as before directed (p. 36, 37) ; then multiply
continually together the second and third terms, and the first with its parts
inverted as in division, for the answer t.
* Division being ihe reverse of multiplication, the reason of the rule is evident.
Note. A fraction is best divided by an integer, by dividing the numerator by it ; but if it
will not exactly di\-idc, then multiply the denominator by it.
+ This is only multiplying the 2d and 3d terms together, »nd dividing the product by the first,
as in the rule of three in whole numbers.
RULE OF THREE IN VULGAR FRACTIONS. 53
EXAMPLES.
1. If g of a yard of velvet cost | of a pound sterling; what will ,', of a yard
cost?
3 2 . . 5 ^ X /
g ' 5 * I 16 * 3 X "^ X y^ = Ji = 6* 8d, Answer.
2. What will 3g oz of silver cost, at 6s 4d an ounce ? An«. 1/ U 4 Jrf.
3. If TO of a ship be worth 273/ 2s Qd; what are ^j of her worth ?
Ans. 227/ 12« Id.
4. What is the purchase of 1230/ bank-stock, at lOSg per cent ?
Ans. 133C/ \s Od.
5. What is the interest of 273/ 15s for a year, at 3^ per cent ?
Ans. 8/ 17 1 Hid.
6. If ^ of a ship be worth 73/ Is 3d; what part of her is worth 250/ 10* ?
Ans. ^.
7. What length must be cut off a board that is 7 J inches broad, to contain a
square foot, or as much as another piece of 12 inches long and 12 broad ?
Ans. 18jf inches.
8. What quantity of shalloon that is J of a yard wide, will line 9J yards of
cloth, that is 2.^ yards wide? Ans. 31J yd«.
9. If the penny loaf weigh 6^*5 oz. when the price of wheat is 5* the bushel ;
what ought it to weigh when the wheat is «« 6d the bushel ? Ans. 4,1, oz.
10. How much in length, of a piece of land that is lli.J poles broad, will
make an acre of land ? Ans. IS/Vj poles.
11. If a courier perform a certain journey in 35^ days, travelling 13j hours a
day; how long would he be in performing the same, travelling only llfj hours
a day ? Ans; 40gVi days.
12. A regiment of soldiers, consisting of 976 men, are to be new clothed ;
each coat to contain 2i yards of cloth that is If yard wide, and lined with
shalloon I yard wide ; how many yards of shalloon will line them ?
Ans. 4351 yds 1 qr 2^ nails.
Scholium. A rule for operations of this nature, where the first term is unity,
was long in great use under the name of Practice, and was broken down into a
variety of separate cases adapted to the peculiar circumstances of each question.
It was doubtless owing to the apparent complexity produced by the number of
cases that it was generally considered very difficult of acquisition, and has now
fallen into very general disuse. It is, however, an exceedingly useful process for
daily purposes, and is, in fact, of very easy acquirement; but though the present
Editor intended to give a page or two on the subject here, he is compelled to
omit it for want of sufficient disposable space. A very systematic view of it is
given by Mr. Rutherford in his edition of Gray's Arithmetic, and to that the
student is referred *.
• The Editor takes tins opportunity also to remark, that the verv best work with which he it
acquainted on Arithmetic for commercial purposes, and apart from ulterior views, is one bearing
the title of " Tlic Quadrantal System of Arithmetic, by Daniel Harrison." The scientific reader,
too, will find some articles worthy of his attention in the work; though from iU object it doc»
not take a strictly scientific form.
54 ARITHMETIC.
DECIMAL FRACTIONS.
A Decimal Fraction is that which has for its denominator an unit (1), with
as many ciphers annexed as the numerator has places ; and it is usually ex-
pressed by setting down the numerator only, with a point before it, on the left
hand. Tlius -^ is '4, and {^^ is -24, and -,5^g is -074, and ■nfe'ij^jg is -00124 ; where
ciphers are prefixed to make up as many places as there are ciphers in the
denominator, when there is a deficiency in the figures. Thus, the understood
denominator of a decimal is always either ten, or some power of ten j whence
its name.
A mixed number is made up of a whole number with some decimal fraction,
the one being separated from the other by a point. Thus, 325 is the same as
•*Tiro» ^'- TOT-
Ciphers on the right-hand of decimals make no alteration in their value ; for
•4, or "40, or '400 are decimals having all the same value, each being = ^^g, or |.
But when they are placed on the left-hand, they decrease the value in a tenfold
proportion : Thus, '4 is -j^, or 4 tenths ; but "04 is only f^g, or 4 hundredths, and
'004 is only fOOT. or four thousandths.
In decimals as well as in whole numbers, the values of the places increase
towards the left-hand, and decrease towards the right, both in the same tenfold
proportion ; as in the following scale or table of notation.
•13 rt 'O
.•5ia> £ ^ -B ^ S '^
B ^ ^ s M 2 ^ B ^ -B s ^ e
3333333-33 3333
ADDITION OF DECIMALS.
Set the numbers under each other according to the value of their places, as
in whole numbers ; in which state the decimal separating points will stand all
exactly under each other. Then, beginning at the right-hand, add up all the
columns of numbers as in integers ; and point off as many places for decimals,
as are in the greatest number of decimal places as any of the lines that are added;
or, place the point directly below all the other points.
EXAMPLES.
1. Add together 290146, and 3146-5, and 2109, and -62417, and 1416.
290146
3146-5
2109-
•62417
14-16
5299-29877 = the sum.
MULTIPLICATION OF DECIMALS.
2. What is the sum of 276, 39-213, 720149, 417, and 5032?
55
Ans. 777791 13.
3. What IS the sum of 7530, 16-201, 30142, 95713, 6 72119, and 03014 ?
Ans. 851309653.
4. What is the sum of 31209, 35711, 71956, 71498, 9739-215, 179, and
'0027? Ans. 17500 9768.
SUBTRACTION OF DECIMALS.
Place the numbers under each other according to the value of their places,
as in the last rule. Then, beginning at the right-hand, subtract as in whole
numbers, and point off the decimals as in addition.
EXAMPLES.
1. Find the difference between 91-73 and 2-138.
91-73
2-138
89592 = the difference,
2. Find the difference between 1-9185 and 273.
3. Subtract 4 90142 from 214-81.
4. Find the difference between 2714 and -916.
Ans. 0-8115.
Ans. 209-90858.
Ans. 2713-084.
MULTIPLICATION OF DECIMALS.
* Place the factors, and multiply them together the same as if they were
whole numbers. Then point off in the product just as many places of decimals
as there are decimals in both the factors. But if there be not so many figures
in the product, then supply the defect by prefixing ciphers.
EXAMPLES.
1. Multiply -321096
by -2465
1605480
1926576
1284384
642192
•0791501640 =
Or, thus, see p. 12.
•321096
•2465
642192
1284384
1926576
1605480
the product = ^0791501640
• The rule will be evident from this example :— Let it be required to multiply -12 by 361 ;
these numbers are equivalent to -^ and -^^ ; the product of which is 1^^ = •0433-2, by the
nature of Notation, which consists of as many places as there are ciphers, that is, of as many
places as there are in both numbers. And in like manner we reason for any other numbers.
As a general investigation, however, let the one factor have m decimal places and the other n ;
and let allthe figures of the first number, taken as integers, be expressed by M, and all tho»«
of the other by N. Then the actual numbers are j^ and jq;^ . Whence, their product U
MN , . , • .
fO^T+T": that. is, there are m + n decimals in the quotient.
5G ARITHMETIC.
2. Multiply 79-347 by 23-15. Ans. 1836-88305.
3. Multiply -63478 by -82C4. Ans. -520773512.
4. Multiply -385746 by -00464. Ans. -00178986144.
CONTRACTION I.
To multiply decimals by 1 with any number of ciphers, as by 10, or 100,
or 1000.
This is done by only removing the decimal point so many places farther to
the right-hand, as there are ciphers in the multiplier; and subjoining ciphers if
need be.
1. The product of 51-3 and 1000 is 51300.
2. The product of 2714 and 100 is
3. The product of 916 and 1000 is
4. The product of 2131 and 10000 is
CONTRACTION II.
To contract the operation so as to retain only as many decimals in the product as
may be thought necessary, when the product would naturally contain several
more places.
Remove the decimal point of the multiplier (if necessary) until the left-hand
figure is an integer in the unit's place ; and so many places as you have moved
the decimal point in the multiplier to the left or to the right, remove, on the
contrary, the decimal point in the multiplicand to the right or to the left. Then,
place the multiplier under the multiplicand in the usual way; and begin to mul-
tiply by the left-hand figure of the multiplier, retaining in the product only so
many decimals as you wish to have at last. Then, multiply by the remaining
figures in the multiplier one by one, from the left towards the right ; as you
proceed, set each product one figure more to the left-hand; and, of course, leave
out one figure more to the right-hand in each successive multiplication. The
sum of these successive lines of products will give the general product required.
It will always be better to calculate one place of decimals more than are required
by the question. See the subsequent example and remarks.
In multiplying be very careful to increase the first right-hand figure retained
in each line by what would be carried on from the figures omitted, in this man-
ner : viz. add 1 if the preceding number fall between 5 and 14, 2 from 15 to 24,
3 from 25 to 34, 4 from 35 to 44, and so on. This process will usually make
the general product true to the last place of decimals.
EXAMPLES.
1. Multiply 2*714986 by 9241035, so as to retain only 4 places of decimals in
the product.
This is evidently the same as to multiply 271 4986 by 9 241035 ; where the
decimal point in the multiplicand is moved 2 places to the rt^A/-hand, and that
in the multipUer 2 to the left.
DIVISION OF DECIMALS.
57
Common method.
2-714986
9241035
13 574930
8144958
2714 986
108599 44
542997 2
24434874
Eastern method.
2-714986
9241035
24434874
5429972
10859944
2714986
8144958
13574930
Contracted mrthod.
271-4986
9*241035
24434874
5429972
108599,4
27I5'0
815
136
1
2508-9280 650510 2508-9280650510 2508-9-28o'7
By a comparison of the common with the eastern method, it will appear upon
inspection that the diflference is only in the arrangement of the work .- and by a
comparison of the eastern with the contracted method, it will be seen that the
only difference is to leave out that portion of the multiplication which does not
contribute to the figures within the limits prescribed for the contraction.
In this contraction a correction column is kept to the rij^fht of the vertical
line, which, in fact, is computing one decimal place more than was required, in
order to insure accuracy in the required number of places. It is always desirable
to do this, as otherwise the last figure cannot be depended on ; and the more so
as such a correction column must always be kept in almost the only place where
the method is of constant occurrence ; viz. in the solution of equations, and its
subordinate class of operations, the extraction of roots.
Here, in the contracted way, we have multiplied first by the left-hand figure,
9 ,- then by the 2, omitting the product of 2 x 6, but regarding the 1 carried on;
then by the 4, omitting the product of 4 x 86, but regarding the 3 carried on.
The rest of the process is, in like manner, conformable to the rule ; and it is
much easier than the usual method of contracted multiplication by inverting the
multiplier.
2. Multiply 480-14936 by 2*72416, retaining only four decimals in the
product.
3. Multiply 2490-3048 by -573286, retaining only five decimals in the pro-
duct.
4. Multiply 325-701428 by -7218393, retaining only three decimals in the
product.
DIVISION OF DECIMALS.
Divide as in whole numbers ; and point oflT in the quotient as many places
for decimals as the decimal places in the dividend exceed those in the divisor *.
• The reason of this rule is evident ; for, since the divisor multiplied by the quotient givct
the dividend, therefore the number of decimal places in the dividend is equal to those in the
divisor and quotient, taken together, by the nature of multiplication ; and consequently tho
quotient itself must contain as many as the dividend exceeds the divisor.
The investigation may, as in the last case, be here given in a gener.il form.
M N
Let — and — be the divisor and dividend respectively, which designates
10- 10- »~ /> o
N__^ M_ _ 10-N _ 10— "N_ N
10- - 10-~~1U"M"" M ""10"--M
When m is greater than n, there will be a removal of the decimal point m — n places to the
right, or, in other words, if the division be complete, m — n ciphers must be added ; but when
« is greater than ;n, there will be w — m decimal places.
58
ARITHMETIC.
Another way to know the place for the decimal point is this : The first figure
of the quotient must be made to occupy the same place of integers or decimals, as
that figure of the dividend which stands over the unit's figure of the first product.
When the places of the quotient are not so many as the rule requires, the
defect is to be supplied by prefixing ciphers.
When there happens to be a remainder after the division ; or when the
decimal places in the divisor are more than those in the dividend ; then ciphers
may be annexed to the dividend, and the quotient carried on as far as required.
EXAMPLES.
1.
178) -48520998 (-00272589
1292
460
1049
1599
1758
156
3. Divide 12370536 by 5425
4. Divide 12 by 7854.
5. Divide 4195 68 by 100.
6. Divide 8297592 by -153.
•2639) 27-00000 (102-3114
6100
8220
3030
3910
12710
2154
Ans. 2-2802.
Ans. 15-278.
Ans. 41-9568.
Ans. 5-4232.
CONTR.VCTION I.
When the divisor is an integer, with any number of ciphers annexed ; cut
off those ciphers, and remove the decimal point in the dividend as many places
farther to the left as there are ciphers cut off, prefixing ciphers, if need be ; then
proceed as before.
Or thus : —
455
•151666 ...
•021666 ...
EXAMPLES.
1. Divide 45 5 by 2100.
21-00) -455 (-0216
35
140
14
2. Divide 41020 by 32000.
3. Divide 953 by 21600.
4. Divide 61 by 79000.
CONTRACTION II.
Hence, if the divisor be one with ciphers, as 10, 100, or 1000 ; then the
quotient will be found by merely moving the decimal point in the dividend so
many places farther to the left, as the divisor has ciphers j prefixing ciphers, if
need be.
EXAMPLES.
So, 217-3 -1. 100 = 2-173
And 5-16 -J- 100 =
And 419 -r 10
And -21 -T- 1000
CONTRACTION III.
When there are many figures in the divisor ; or when only a certain number
of decimals are necessary to be retained in the quotient ; then take only as many
REDUCTION OF DECIMALS.
59
figures of the divisor as will be equal to the number of figures, both integers and
decimals, to be in the quotient, and find how many times they may be contained
in the first figures of the dividend, as usual.
Let each remainder be a new dividend ; and for every such dividend, leave
out one figure more on the right-hand side of the divisor : remembering to
carry for the increase of the figures cut off, as in the second contraction in
multiplication.
Note. When there are not so many figures in the divisor as are required to
be in the quotient, begin the operation with all the figures, and continue it as
usual till the number of figures in the divisor be equal to those remaining to be
found in the quotient ; after which begin the contraction.
EXAMPLES.
1. Divide 250892806 by 92*41035, so as to have only four decimals in the
quotient, in which case the quotient will contain six figures.
1. Common method.
92-41035) 2508-928 06 (27'1498
1848207 0
660721 06
646872 45
13348 610
9241 035
4607 5750
3696 4140
911 16100
83169315
79467850
73 928280
2. Contracted method.
92-4103 5) 2508-928 0,6 (271498
1848 -207 0
660721 1
646872 5
13848 6
92410
4607 6
3696 4
9112
8317
i_
79 5
73 9
5 539570 1 5 6
In this operation, as in multiplication, a correction column to the right of the
vertical line should be kept. The method itself is obviously only a rejection of
the figures which do not contribute to the result within the prescribed limits.
2. Divide 4109 2351 by 230409, so that the quotient may contain only four
decimals. ^"s. 17-8345.
3. Divide 37-10438 by 5713-96, that the quotient may contain only five
decimals. Ans. -00649.
4. Divide 91308 by 21372, that the quotient may contain only three
decimals.
REDUCTION OF DECIMALS.
CASE I.
To reduce a vulgar fraction to Us equivalent decimal.
Divide the numerator by the denominator, as in division of decimals,
annexing ciphers to the numerator, as far as necessary ; so shaU the quotient
be the decimal required *.
* It will frequently happen (indeed always when the fraction in iu lowest term, hw in iw
denominator any factors besides 2 and 5, or powers and producU of these,) that the dirision wiU
60 ARITHMETIC.
EXAMPLES.
I. Reduce ^^ to a decimal.
24 = 4 X 6. Then 47-
6,1750000
•291666
never terminate. In this case, it is commonly suflicient to carry it to some specific number of
quotient places, and neglect the remaining ones as of comparatively no value. If, however, the
question be one in which the accurate result instead of the approximation is required, it will be
necessary to work by vulgar fractions instead of decimals. In this case any other decimal quan-
tities that have presented themselves amongst the data of the question can be readily thrown
into the usual fractional form for greater facility of combination with the fraction above men-
tioned.
A rapid and elegant method of throvdng a vulgar fraction, whose denominator is a prime
number, into a decimal consisting of a great number of figures, is given by Mr. Colson, in page
162 of Sir Isaac Newtans Fluxions. It will be readily understood from the following
example: —
Let ^^ be the fraction which is to be converted into an equivalent decimal.
Then, by dividing in the common way till the remainder becomes a single figure, we shall
have ^ = •03448Jj for the complete quotient, and this equation being multiplied by the nume-
rator 8, will give i^ ^ •275845|, or rather ^| ::= "27586^^ : and if this be substituted instead of
the fraction in the first equation, it will make ^ = ■0344827586._^. Again, let this equation be
multiplied by six, and it will give .% =: "2068965517^ ; and then by substituting as before
% = •034482758620689655175^;
and so on, as far as may be thought proper; each fresh multiplication doubling the number of
figures in the decimal value of the fraction.
In the present instance the decimal circulates in a complete period of 23 figures, i. e. one less
than the denominator of the fraction. This, again, may be divided into equal periods, each of
14 figures, as below :
•03448275862068
■96551 7^24137931
in which it will be found that each figure with the figure vertically below it makes 9; 0 -|-9 = 9;
3 -|- 6 = 9 ; and so on. This circulate also comprehends all the separate values of ;^, .^, ;^
in corresponding circulates of ^28 figures, only each beginning in a distinct place, easily ascer-
tainable. Thus, rjj = "06896 beginning at the 1^2th place of the primitive circulate.
^ = "103448 beginning at the •28tb place. So that, in fact, this circle includes 28 complete
circles.
The property of circulation of some number of the figures in periods is not one peculiar to
this or any other of the interminable decimals, but belongs to them all. Sometimes the period
is comjmsed of a single figure, as "333 wliich is the decimal expansion of J ; sometimes of a
considerable number, as in the example above given from Colson. It is, however, never com-
posed of a number of places so great as is expressed by the denominator : as the cxp.*»nsion of
\ is composed of one term, and one is less tlian the denominator, three ; and the period of the
expansion of 5^ is composed of 28 places, and less therefore than the denominator •29. Some-
times the circulating periods are composed of a half, or a fourth of the number of places that
would be expressed by subtracting one from the denominator.
Whenever in the division a remainder occurs that has occurred before, (the number brought
down from the dividend being necessarily the same in both cases, viz. 0,) then the same quo-
tient figure will again occur, and the same remainders will occur after both ; and hence the
same quotient figures, and the same remainders, will continually succeed, till the first men-
tioned remainder again occurs. A continual circulation of the same process will thence take
place without end.
The same circumstances would obviously take place if, instead of bringing down ciphers, as
above mentioned, we brought dowD any other figure from the dividend, or even the figures of
any circulating dividend.
That there must be a less number of figures in the circulating period, than is expressed by
the denominator, appears from this. Tiie only remainders that can exist are all less than the
denominator, and being integers, their number is less than is expressed by it : and hence the
number of steps in the division that takes place without falling upon the same remainder must
REDUCTION OF DECIMALS. 6|
2. Reduce J, and i, and f , to decimals. Ana. -25, and -5. and 75.
3. Reduce | to a decimal. Ans. -625
4. Reduce ./j to a decimal. Ans. '12
5. Reduce tIj to a decimal. Ans. -03125.
6. Reduce ^^f^ to a decimal. Ans. -143 1 54 ... .
CASE 11.
Tojind the value of a decimal in terms of the inferior denominations.
Multiply the decimal by the number of parts in the next lower denomina-
tion ; and cut off as many places for a remainder, to the right-hand, as there are
places in the given decimal.
Multiply that remainder by the parts in the ne.\t lower denomination again,
cutting off for another remainder as before.
Proceed in the same manner through all the parts of the integer ; then the
several denominations separated on the left-hand will make up the answer.
Note. This operation is the same as reduction descending in whole numbers.
EXAMPLES.
Required to find the value of 775 pounds sterling.
•775
20
s 15-500
12
d 6 000 Ans. 15s 6d.
2. What is the value of -625 shil ? Ans. 7^d.
3. "What is the value of -8635/ ? Ans. 17* 3 24rf.
4. What is the value of -0125 lb troy ? Ans. 3 dwts.
5. What is the value of "4694 lb troy? Ans. 5 oz 12 dwts 15-744 gr.
6. What is the value of 625 cwt ? Ans. 2 qr 14lb.
7. What is the value of -009943 miles? Ans. 17 yd 1 ft 598848 inc.
8. What is the value of 6875 yd ? Ans. 2 qr 3 nls.
9. What is the value of -3375 acre ? Ans. 1 rd 14 poles.
10. What is the value of 2083 hhd of wine ? Ans. 13-1229 gal.
at the most be one less than the number expressed by the denominator ; and hence again the
number of quotient figures must be at most one less than the same number. That i(, the
quotient is composed of such repeating circles as we have stated.
Any further examination of this subject (which, nevertheless, is a very curiou» and a very
important one) would be incompatible with the limits of this work. The inetb<Ml of finding
the value of such a decimal will, however, be found in the Chapter on Geometrinil /'roj/rrxtum :
and we may refer the inquiring student also to Mr. Goodwin's TitbU-s of Ikcimut Circlet, and
to the Ladies' Diary for 1824.
A convenient notation has been used to designate the circulating period which is tliat of
putting a dot over the first and last figures of the period. Thus 7-2-%'8«)'Jo signifies that «^25
is the circulating period. When the period is at only one place, there is but one dot required,
as in the value of J, which is '3.
The reverse problem, of finding the finite fractional value of an interminable decimal, will be
found in Geometrical Progression, in the Algebra.
62 ARITHMETIC.
CASE III.
To reduce integers or decimals to equivalent decimals of higher
denominations.
Divide by the number of parts in the next higher denomination ; continuing
the operation to as many higher denominations as may be necessary, the same as
in reduction ascending of whole numbers.
EXAMPLES.
1 . Reduce 1 dwt to the decimal of a pound troy.
20 j 1 dwt
12 0*05 oz
0004166 lb. Ans.
2. Reduce Qd. to the decimal of a pound. Ans. 0375/.
3. Reduce 7 drams to the decimal of a pound avoird. Ans. -02734375 lb.
4. Reduce -26^ to the decimal of a pound. Ans. •0010S336/.
5. Reduce 2-15 lb to the decimal of a cwt. Ans. "OlQige. , . . cwt.
6. Reduce 24 yards to the decimal of a mile. Ans. "013636 mile
7. Reduce 056 pole to the decimal of an acre. Ans. 00035 acre.
8. Reduce 1*2 pint of wine to the decimal of a hhd. Ans. '00238 hhd.
9. Reduce 14 minutes to the decimal of a day. Ans. "009722 day.
10. Reduce .21 pint to the decimal of a peck. Ans. -013125 peck.
11. Reduce 28^ 12^^ to the decimal of a minute.
Note. IVhen there are several numbers, to be reduced all to the decimal of thi
highest.
Set the given numbers directly under each other, for dividends, proceeding
orderly from the lowest denomination to the highest.
Opposite to each dividend, on the left hand, set such a number for a divisoi
as will bring it to the next higher name ; drawing a perpendicular line betweei
all the divisors and dividends.
Begin at the uppermost, and perform all the divisions : only observing to se
the quotient of each division, as decimal parts, on the right hand of the dividenc
next below it ; so shall the last quotient be the decimal required.
EXAMPLES.
1. Reduce 17 s. Old to the decimal of a pound.
4
12
20
3-
9-75
17-8125
£0-890625 Ans.
2. Reduce 19^ 17* 3\d to pounds. Ans. 19-8635416;.
3. Reduce I5s 6d to the decimal of a pound. Ans. -775/.
4. Reduce 7^d to the decimal of a shiUing. Ans. •625*.
5. Reduce 5 oz 12 dwts 16 gr. to lb. Ans. -4694 lb.
RULE OF THREE IN DECIMALS.
Prepare the terms, by reducing the vulgar fractions to decimals, and an;
compound numbers either to decimals of the higher denominations, or t«
integers of the lower, also the first and third terms to the same name : thei
multiply and dinde as in whole numbers.
DUODECIMALS. 63
Note. Any of the convenient examples in the single or double rule of thr«e
in integers, or vulgar fractions, may be taken as proper examples to the same
rules in decimals. — The following example, which is the first in vulgar fractions,
is wrought out here, to show the method.
If I of a yard of velvet cost II, what will ,*j yd cost ?
I = -375
,*8=3125
yd
•375 :
£
: -4 : :
•375)
yd £
•3125 : -333 ....
•4
« d
. or 6 8
•12500 ( 333333..
1250 20
1*^5
••
*6C6CC6..
12
••
Ans. 6s 8d.
d 7-99999..
. . = 8t/.
DUODECIMALS.
Duodecimals, or Cross Multiplication, is a rule used by workmen and
artificers, in computing the contents of their works.
Dimensions are usually taken in feet, inches, and quarters ; any parts smaller
than these being neglected as of no consequence. And the same in multiplying
them together, or computing the contents. The method is as follows :
Set down the two dimensions to be multiplied together, one under the other,
80 that feet may stand under feet, inches under inches, and so on.
Multiply each term in the multiplicand, beginning at the lowest, by the feet in
the multiplier, and set the result of each directly under its corresponding term,
observing to carry 1 for every 12, from the inches to the feet.
In like manner, multiply all the multiplicand by the inches and parts of the
multiplier, and set the result of each term one place removed to the right-hand
of those in the multiplicand ; omitting, however, what is below parts of inches,
only carrying to these the proper numbers of units from the lowest denomi-
nation.
Or, instead of multiplying by the inches, take such parts of the multiplicand
as these are of a foot.
Then add the two lines together, after the manner of compound addition,
carrying 1 to the feet for every 12 inches, when these come to so many.
EXAMPLES.
1. Multiply 4f 7 inc 2. Multiply 14 f 9 inc
by 6 4 by 4 6
27 6 59 0
1 64 7 4J
Ans. 29 0^ Ans. 66 4 J
3. Multiply 4 feet 7 inches by 9 feet 6 inches. Ans. 43 f 6J inc
4. Multiply 12 f 5 inc by 6 f 8 inc. Ans. 82 Qj
5. Multiply 34 f 4i inc by 12 f 3 inc. Ans. 421 l\
6. Multiply 64 f 6 inc by 8 f 9i inc. Ans. 565 bj
Note. The denomination which occupies the place of inches in these products,
means not square inches, but rectangles of an inch broad and afoot long. Thus,
64 ARITHMETIC.
the answer to the first example is 29 sq. feet, 4 sq. inches ; to the second 66 sq
feet, 54 sq. inches.
If the resulting product be one of three dimensions, length, breadth, an(
thickness, then the first denomination to the right of the feet must be multipliei
by 144, the second by 12, and these products added to the figure in the thin
place, will give cubic inches *.
* Though it is the practice to neglect all the smaller dimensions than inches or half inche
both in actual measuring among artificers, and of course in the computations which are mad
from such sui-vejs ; yet in theory, all the subordinate dimensions are reckoned in a descendin
scale of twelves, as in our common numbers, we employ a descending scale of tens ; and in al
cases where the theoretical result is required, the process must be continued in the same wa}
Instead of the descending denominations below the units, tenths, hundredths, thousandths, &(
the terms, parts or primes, seconds, thirds, &c. descending successively below the inch, ar
employed. We are obliged, however, to keep the denominations in separate columns, c
separated by a blank sjjace, or by dots (,,) in our calculations in Duodecimals, instead of placin
the numbers in each successive denomination in juxta-position as in our common notation. Bi;
if there were distinct and concentrated symbols employed to designate the numbers 10 and 1]
(as 4» or <^ and 11 or ir, as is done by some writers on the Theory of Numbers,) then we migh
dispense with the extra spaces, columns, or dots, and write the results continuously. All th
advantages of the decimal notation in point of simplicity of writing would thus be gained for th
duodecimal : and it is quite obvious that the same method is applicable to anj- other scale (
numbers and its corresponding notation.
The great barrier, however, to any change, except in the particular instance of feet an
inches, is the terminology, or names of numbers, which could not possibly co-exist with a change <
scale. The names are of an origin so decidedly and obviously decimal, that it requires som
degree of fixed attention to ascertain how many dozens there are in any number specific
decimally. All our language and all our ideas of number flow in terms of the decimal scale
and hence, however desirable it might appear in the eyes of some of the most enlightene
mathematicians to adopt the duodecimal or dozen scale, the inveterate adherence which ever
people feels to its old language, and the consequent (in this case) daily practice, forbids eve
the most distant hope of ever realizing such a project.
The intelligent teacher, however, may avail himself of the principle employed in duodecimal
to explain to his more intelligent pupils the nature of numerical scales in general. Such pupil
are now arrived at a stage in their arithmetical studies which renders such knowledge essential.
A single specimen of the process of complete duodecimal multiplication expressed in th
common and contracted notations is here subjoined, which it is hoped will give a clear idea <
the views expressed above.
ft in pr "
Multiply 3 10 11 8
by 2 3 11 10
7 9 11 4
11 8 11 0
3 7 0 8 4
3 3 18 8
Product 9 1 6 6 10 0 8
ft in pr " '" iv v
By the contracted notation.
3</>'7r8 3<^Tr8
2 3 IT <p 23-7r<^
33188 7 9ir4
37084 ttSttO
•jtSttO 37084
7 9Tr 4 3 3 18
9166.^08 9166(^08
The want oi higher classes of twelves prevents our proceeding to the left without cncounterin
tbe decimal notation for 10, 100, feet.
INVOLUTION.
Involution is the raising of Powers from any given number, as a root
A Power is a quantity produced by multiplying any given nurabjr, called tbe
Root, a certain number of times continually by itself. Thus,
2 = 2 is the root, or 1st power of 2 ;
2x2= 4 is the 2d power, or square of 2 ;
2x2x2= 8 is the 3d power, or cube of 2 ;
2 X 2 X 2 X 2 = 10 is the 4th power of 2 ; and so on ;
and in this manner may be calculated the following table of the first nine
powers of the first 9 numbers.
TABLE OF THE FIRST NINE POWERS OF THE FIRST NINE NUMBERS.
]st
2.1
3d
4th
5th
6th
7th
8th
9th
1
1 1
1
1
1
1
1
1
2
4
8
16
3-2
64
128
25«)
512
3
9
27
81
243
729
2187
6561
19683
4
l(i
64
256
1024
4096
16J84
65536
262144
5
25
l-2o 625
3125
15625
78125
390625
19.531-2.5
6
36
216 1296
7776
46656
279936
1679616
100776.%
7
49
343
•-'401
16807
117649
823543
5764801
40353607
8
64
512
4096
32768
262144
2097152
16777216
13421772J!
9
«I
729
6561
59049
531441
4782969
43046721
3;;74-20489
The Index or E.xponent of a power, is the number denoting the height or
degree of that power ; and it is 1 more than the number of multiplications used
in producing the same. Thus 1 is the index or exponent of the Ist power or root,
2 of the 2nd power or square, 3 of the 3rd power or cube, 4 of the 4 th power, or
biquadrate, and so on.
Powers, that are to be raised, are usually denoted by placing the index above
the root or first power.
So 2* ^ 4 is the 2d power of 2.
2' = 8 is the 3d power of 2.
2* ^ 16 is the 4th power of 2.
540'' = 85030560000 is the 4th power of 540.
When two or more powers are multiplied together, their product is that povrtr
■whose index is the sum of the exponents of the factors or powers multiplied.
Or the multiplication of the powers answers to the addition of the indices.
Tims, in the following powers of 2,
1st 2d 3d 4th 5th 6th 7th 8th 9th lOlh
64 128 256 512 1024
2« 2' 2'' 2» 2"
16, and 2 + 2 = 4 its index;
128, and 3 + 4 = 7 its index ;
also 16 X 64 = 1024, and 4 + 6 = 10 its index.
VOL. I. F
2
4 8 16 32
2'
22 23 2* 2'
Here, 4x4 =
and 8 X 16 =
66 . ARITHMETIC.
OTHER EXAMPLES.
1. What is the 2d power of 45 ? Ans. 2025.
2. "What is the square of 416 ? Ans. l7-305fi.
3. What is the Sd power of 35 ? Ans, 42-875.
4. What is the 5th power of -029? Ans. -000000020511149
5. What is the square of 3? Ans. J.
6. What is the 3d power of | ? Ans. |g.
7. What is the 4th power of | ? Ans. ./zj.
EVOLUTION.
Evolution, or the reverse of Involution, is the extracting or finding the
roots of any given powers.
The root of any number, or power, is such a numher, as being multiplied into
itself a certain number of times, will produce that power. Thus, 2 is tlie square
root, or 2d root of 4, because 2' = 2 x 2 = 4; and 3 is the cube root or 3d
root of 27, because 3' = 3 x 3 x 3 = 27.
Any power of a given number or root may be found exactly, namely, by mul-
tiplying the number continually into itself. But there are many numbers of
which a proposed root can never be exactly found ; those numbers beinji them-
selves incapable of being produced by the involution (to the corresponding
power) of any rOot composed of a finite number of integer or decimal places.
Yet, by means of decimals, we may approximate or approach towards the root,
to any assigned degree of exactness.
Those roots which only approximate are called Surd Roots ; but those which
can be found quite exactly, are called Rational Roots. Thus, the square root of
3 is a surd root ; but the square root of 4 is a rational root, being equal to 2 :
also the cube root of 8 is rational, being equal to 2 ; but the cube root of 9 is
surd or irrational.
Roots are sometimes denoted by writing the character \^ before the power,
with the index of the root against it. Tiiu*, the 3d root of 20 is ex|)ressed by
•\/ 20 ; and the square root or 2d root of it is «J 20, the index 2 being always
omitted, when only the square root is designed.
When the power is expressed by several numbers, with the sign i- or — be-
tween them, a line is drawn fiom the top of the sign over all the parts of it :
thus the third root of 45 — 12 is \/ 45 — 12, or thus, V ("t^ — 12), enclosing
the numbers in parentheses, which is, usually, the best way to express it.
But all roots are now often designed like powers, with fractional indices;
thus the square root of 8 is 8i, the cube root of 25 is 25^, and the 4th root of
45 — 18 is (45 — 18)1--
TO i;XTRACT THE SQUARE ROOT.
* Divide the given number into periods of two fiijures each, by setting a
point over the place of units, another over the place of hundreds, and so on,
* The re:iM>ii for sejiaratiiig the fig'ires of tlie diviileiul in:o peiinds or portii)iis of two places
each, is, that the square of any single figure never consists of iDore tlian two places; the square
of a number of two figures, of not more tiian four places, and so on. So tliat there will be as
manr figures in the root as the given nunilur contains periods so divided or parted oflF.
And the reason of the several steps in tlie operation appears frnui the algebraic form of the
square of any nnnilicr of terms, whether two or (liree or inoi-c. Thus
(a -|-/>)- = «2 -f2uA -|-6^=:a*-|- ('Ja -f-'')^, the square of two terms ; where it appears that a
is the first terra of the root, and b the second term ; aUo a the fii-»t divisor, and tiie new divisor
SQUARE ROOT.
67
over every second figure, both to the left hand in integers, and to the right in
decimals.
Always beffitt to point at the place of units ; or, if the number to be extracted
be entirely decimal, put a cipher in the unit's place, and over it put the first point.
Find the greatest square in the first period on the left hand, and set its root on
the right hand of the given number, after the manner of a quotient figure in
Division.
Subtract the square thus found from the same period, and to the remainder
annex the two figures of the next following period, for a dividend.
Double the root above mentioned for a divisor ; and find how often it is con-
tained in the said dividend, exclusive of its right-hand figure ; and act that
quotient figure both in the quotient and divisor.
Multiply the whole augmented divisor by this last quotient figure, and sub-
tract the product from the said dividend, bringing down to it the next period of
the given number, for a new dividend.
Repeat the same process over again, viz. find another new divisor, by doubling
all the figures now found in the root ; from which, and the last dividend, find
the next figure of the root as before ; and so on through all' the periods, to the
last.
Note. The best way of doubling the root, to form the new divisors, is by
adding the last figure always to the last divisor, as appears in the following ex-
amples. Also, after the figures belonging to the given number are all exhausted,
the operation may be continued into decimals at pleasure, by adding any number
of periods of ciphers, two in each period.
EXAMPLES.
1, To find the square root of 2950G624.
29506624 ( 5432 the root.
25
104 450
4 416
1083
3
3466
3249
10862
21724
21724
is 2a r b, or double the first term increased by tlie second. And hence the manner of extrac-
tion is thus :
1st divisor a) a* -f 2ab -\- h' (a -f 6 the root.
2d divisor 2a + /> I 2ah -\- '>*
b\2ab-\- 62
Again, for a root of three pnrts, «, />, c, thus :
(a +i -f c)2 = rt* -f 2<ib + i» -f 2«r -f 2/< -f r' =
a' -|- (2a • b) b -\- (2a -f 2/> -f- c)c, the «qn«ro of three tcrtni,
*liere a is the first term of the root, 6 tlie sctond, and c the third tt-nn ; alw « tlie fir«l divisor,
2a -f- 6 tlie second, and 2a -\- '21) -r c tlie tiiird, each consisting of the double of the root
increased bv the next term of the same. In like manner (« /, -^c-f </)' = "* -|-(2/i -\-b)b-\-
(2a -j- 2/^ -f- f) c -j- (2a -f 2/> -|- 2c -\-d)d\ and so on, to whatever number of terms wc pro-
ceed. And the mode of extraction agrees witii the rule. Sec fartiier, C'a»c 2, of Evolution in
Uic Algebni.
4'»»-|-S;i , . „ .
For an approximation observe that s/'*^ -\-n=.a. yv/ , ^ nearly, la ail caan where a n
«mull in respect of a.
f2
68 ARITHMETIC.
Note. When the root is to be extracted to many places of figures, the work mai
be considerably shortened, thus :
Having proceeded in the extraction after the common method, till there b(
found half the required number of figures in the root, or one figure more ; then
for the rest, divide the last remainder by its corresponding divisor, after th(
manner of the third contraction in division of decimals j thus ;
2. To find the root of 2 to nine places of figures.
2 (1-414211356 the root,
1
24
4
100
96
281
1
400
281
2824
4
11900
11296
28255
2 1 60400
2 1 56564
28284 I 3836:
I 1007 6
I60jl
18i7
1:8
The figures 1356, to the right of the vertical line, being obtained simply by
division.
3. What is the square root of 2025 ? Ans. 45.
4. What is the square root of 17'3056 ? Ans. 4*16.
5. What is the square root of -000729 ? Ans. -027.
6. What is the square root of 3 ? Ans. 1-732050.
7. What is the square root of 5 ? Ans. 2236068.
8. What is the square root of 6 ? Ans. 2-449489.
9. What is the square root of 7 ? Ans. 2-645751.
10. What is the square root of 10 ? Ans. 3* 162277-
11. What is the square root of 1 1 ? Ans. 3316624.
12. What is the square root of 12 ? Ans. 3 464101.
EULES FOR THE SQUARE ROOTS OF VULGAR FRACTIONS AND
MIXED NUMBERS.
First prepare all vulgar fractions, by reducing them to their least terms, both
for this and all other roots. Then
1. Take the root of the numerator and of the denominator for the respective
terms of the root required ; which is the best way if the denominator be a com-
plete power : but if it be not, then
2. Multiply the numerator and denominator together; take the root of the
product : this root being made the numerator to the denominator of the given
fraction, or made the denominator to the numerator of it, will form the frac-
tional root required.
TO,,* ;» .a ^a ^ab a
Thatis,^^=^=-^=^^
This rule will serve, whether the root be finite or infinite : and sometimes one of
these expressions will simplify the operation, sometimes another ; as will be
learnt from a little experience.
CUBE ROOT.
G9
3. Or reduce the vulgar fraction to a decimal, aqd extract itx rt)ot Thit it
generally the best method in practice when the terms of the fraction are not very
low primes or very simple composite numbers.
4. Mixed numbers may be either reduced to improper fractions, and extracted
by the first or second rule, or the vulgar fraction may be reduced to a decimal,
then joined to the integer, and the root of the whole extracted.
«XAMPLCS,
1 . What is the root of l\ ?
2. What is the root of t?, ?
3. What is the root of ^^ ?
4. What is the root of /^ ?
5. What is the root of 17^ ?
Ans. J.
Ans. ^,
Ans. 0-866025.
Ans. 0'645497.
Ans. 4 168333.
By means of the square root also may readily be found the 4th root,* or the 8th
root, or the 16th root, that is, the root of any power whose index is some
power of the number 2 ; namely, by extracting so often the square root as is
denoted by that power of 2 ; that is, two extractions for the 4th root, three for
the 8th root, and so on.
So, to find the 4th root of the number 21035-8, extract the square root two
times in succession, as follows :
21035-8000
1
24T1T6
4 I 96
( 145-037237 ( 12-0431407 the 4th root.
1
285
5
2900
1
1
3
3
435
425
108000
87009
2900
6
20991
687
107
22
2
45
44
2404
4
10372
9616
24083
3
75637
72249
24086
3388
9S0
17
Ex. 2. What is the 8th root of 9741 to four places of decimals ?
Ex. 3. Find the l6th root of 15 to three decimals.
TO EXTRACT THE CUBE ROOT *.
1. Divide the page into three columns, and call them l, m, n, in order from
left to right, and so that m shall be double the breadth of l, and n double the
breadth of m. At the head of n place the number whose root is to be extracted ;
• This method was invented by the late Mr. W. G. Homer, and forms only one single and
simple application of a universal method of extracting all roots whatever; and even his nirihod,
M applied to roots, is only a particular application of bis general method of solving numerical
equations of all orders.
In the form here given it is rather more concise in the operation than as generally applied to
equations of a higher order, and is put in this form for the use of those students who arc not
likely to proceed to the higher equations. It is recommended to those who intend to pureue
the study of the subject to any extent, to adopt the form of work given in the equation* in this
volume, in preference to the present one, on account of the uniformity amongst the procwaet
for all roots whatever.
70
ARITHMETIC.
and mark off the place for the root as the quotient is marked in division, or the
root in the extraction of the square root. Thus : —
3 X root already
found.
3 X square of root
already found.
Number whose root is
to be found.
Root.
2. Divide the number into periods of three figures * each, by setting a point
over the place of units, and also over every third figure from thence to tlie left-
hand in whole numbers, and to the right in decimals. Find the nearest less
cube number to the first (or left-hand) period, and having subtracted it there-
from, annex the next period to the remainder. Call this the resolvend. Also
set the root of the said cube in the place appropriated to the root.
3. In the column l put three times the root already found, and in m put thre«
times its square. With this last number as a trial divisor of the resolvend
omitting the two figures to the right-hand, find the next figure of the root, anc
annex it to the former one, and also to the number in the column l. Multiph
the number now in l by the new figure of the root, and place the product undei
that in m, but having its figures two places more to the right than the number;
already there. Add the two numbers together, and they will form the correctec
divisor.
4. Multiply the corrected divisor by the root figure, and place it under th(
number in n, and subtract it therefrom f .
Note. If the number last found should be greater than the number previously
in N, the subtraction cannot be performed : the woi k dependent upon the roo
figure which produced it must, therefore, be cancelled, and a smaller root figure
tried instead of it. If necessary, the same process must be again repeated, til
a number is found which will admit of being subtracted according to th<
rule.
5. Write twice the last root figure under the number l, and the square of th(
last root figure under m. Add the two last written numbers in l together, ant
* The reason for pointing tlie given number into periods of three figures each, is, because
the cube of one figure never amounts to more than tliree places; the cube of two figures t(
more than six, but always more than three; the cube of three figures never to more than nine
but always more than six; and so on to any extent.
For a similar reason, a given number is pointed into periods of four figures, of five figures
of n figures, when the fourth, fifth, «'•' roots are to be extracted.
+ A little attention to the composition of the algebraic expression for the cube of a binomia
will render the truth of this rule very obvious. For {a -f- 6)3 = a3 -f- 3a% -\- 3ai' -\-b3z=a
+ JSa' -f- (3a -f- h) X l>\ f>, the last form of which is exactly that whose composition is directet
in rules 3 and 4. Thus : —
L
3a
3a -f 6
3<j»
M
f 3a +b)xb
3u'^ -H (3a -I- 6) 6
N
„3 ^ 3asi -I- 3;ii2 + !>» + B + C.. .{a +b
{3a' -f (:v, -|-/,)/,| h
-I- B + C . . . .
when B, c, &c. stand for other quantities to be afterwards brought down for continuing the pr»
cess.
CUBE ROOT.
I
the three last written numbers in m ♦. These numbers will be respectively triple
the new root and triple its square ; and we may proceed with them to find a new
root figure, and com|)lete the operation as in rules 3 and 4.
Note. After about one-third of the number of fiifures in the root have lieen
obtained (or one-third of the number intended to be taken into account if the
root be interminable) we may contract the work very considerably by the foU
lowing method : —
Cut oflf one figure from m, and two from l, by vertical lines. Then with the
new root figure work as in contracted nmltiplication, p. 57, keeping as the cor-
rection columns those immediately to the right of the vertical lines. We hhall
thus avoid all the work which does not contriljute to the derivation of llie figures
of the root to tl.e extent assigned.
In the column si, the correction of the square of the last root figure falling to
the right of the correction column is not i)ut down, a-; in Ex. 2.
Proceeding thus, we shall, in as many steps as we had taken before the con-
traction began, have cut off" all the addends that would arise from l ; and the
process for obtaining the remaining third of the figures will be identical with
that for Contracted Division, p. 59.
Ex. 1. Extract the cube root of 4822854.
9/
96
12
1084
27.
48228544
27
(3/64
2122S
= first resolvend.
19656
1572544
1572544
=: second
resolvend
57(3 -j
327t) >
36 J
3888
43L'6
393136
Statement of the process. Here the number .v being pointed as prescribed in
the rule, the first period is 48, and the greatest integral cul)e contained in it is
27. Its root 3 is put in the root place, and the first resolvend is 21228. The
trial divisor or number in m is 27, or three times the square of the root figure,
and the number l is three times the root figure, or 9- The trial quotient, or
integer part of *^'rj, is 7. Annex this to l, which gives 97, and add 7 x 97 to m,
the figures being carried two i)laces to the right, which gives 3379, the product
of which by 7 gives 23653. This cannot be subtracted from the resolvend : and
hence the whole work dependent on 7 is to be obliterated, and a lower number,
6, tried
Performing the same operations with 6 as has been just described with 7, we
get 96, 3276, and 19656 for l, m, and n, the last of which subtracted gives for
the next resolvend 1572544.
Adding 2x6 and 6^ to the columns l and m respectively, Uking in the 576
* Tliat tliis method produces tlie triple of the new root in t!ie column L, and the triple of
its squ.ire in the column M, may be thus immediately shown : —
3a -r 6
•2l>
3« -r 3/>
= 3(a -f 6)
M
(3a +6)6
3a2 -f (3a -f 4) h
3<j» -r Gu'j + 3fP
= 3 (a + 6) »
(a +6
And we thus start fron^ the last root o -|- 6 as we did at first fiom a.
ARITHMETIC.
of the latter, we have the triple (= 108) and triple the square (= 3888) of 36 in
L and M respectively.
The trial divisor 3888 is found to be contained 4 times in 15725, which is
annexed to the former figures of the root making 364, and to the column i
making 1084. This multiplied by 4 and added to m gives 393136; and this lasl
multiplied by 4 gives 1572544 ; which being equal to the resolvend, can be
subtracted. No remainder occurring, the work is hence terminated at this stepi
and the root is 364.
2. Find the cube root of 9 to fifteen places of decimals.
L M N
6 08 12 ... . 9(208008i3823;051904
16 4864^^ 8
6-2 40 08
16
6-2140 24
12-48 6 4
64,
1-.
•998912
12-97 92
4 99 2 06 4'
12 9796992064
64
12 9801984192
1 8 7 2 07")
12-9802 1 7 1 399J
06
1088
103S375936512
49624063488
38940651419
106834120683
10384192682 4
12-98 02 3 586
499
1298 0 2408 5
12-98 02 4 5 84
12
2992193S5 9
2596049I9!4
39614466J3
38940738 3
673728 2
649012 3
12-9802 4 59
;}
24715 9
12980 2
I.
12-19,8,0 2 4l6!l|0 117357
11682J2
53'5
61J9
The decimal points are marked in this solution, but they are unnecessary, as
the arrangement of the work itself will lead to a correct disposition of the
figures.
The first vertical bar in the root marks where the contraction begins to give
its figures to the root, and the second where the pure division commences. The
root is true in the last figure.
This example contains specimens of the case where ciphers occur in the root ;
and is also a pattern for the method of contraction.
Ex. 3. Extract the cube root of 57148219.
Ex. 4. Extract the cube root of 16281582.
Ex. 5. Extract the cube root of 1332.
Ex. 6. Extract the cube root of 3 to six places; and also the cube root of 6 :
and show how the cube root of 2, of 12, and of 18, may be obtained from these.
I
CUBE ROOT.
73
Ex. 7. Find how many more figures are required to be written in extracting
the cube root of 3 to six places of decimals, than in finding it to three places.
Ex. 8. Find the cube root of ^hsJIJXmt:. and subtract it from the cube root
of 5.
Ex. 9. Extract the cube root of '009009009 ; and thence obtain its sixth and
its ninth roots to six and to four places of decimals respectively.
Ex. 10. Find the square of the cube root of 1000, and likewise of -001 : and
the cube of the square root of 100 and -01 ; and determine the product of these
four results.
TO EXTRACT ANY ROOT WHATEVER*.
Let p be the given number, n the index of the root to be extracted, r the
true root, a the nearest approximation (either greater or less than r) that has
been made by trial or otherwise ; and let a" = a, and r = a + x. Then,
a Jp — .\?
X = K — a =^— ■ . , . , , , V — nearly.
i (n— 1) p -f ^ (n + 1) A '
Add or subtract x, according as p is greater or less than a, which gives a new
and nearer approximation to r. With this new value of a, find a new value of
X, and hence a new approximation to r. Continue this series of a])proximation8
till the required degree of accuracy is obtained.
The number of figures which may be depended on in each successive value of
Xf is generally equal to the number of accurate places in a ; so that each opera-
tion doubles the number of figures already obtained.
* This rule is a modification and extension of Dr. Halley's, and was first given in its present
form by Dr. Hutton. See his Tracts, vol. i. p. 213.
The following is essentially Dr. Hutton's investigation :
p V -\- q \
Let a + J? =: — r -— . a ; or since P = (a -f j) ",
n (n — 1)
(P + 7)« + ;? K a—'j- + ;) . J— 2" «"""•!*+••■
' « (« — I) '^
{p -^ q) a' -\- q n a'-^x + q . j ^ 2" """' ** + •■
p—q p-q U-\ qn i^
= a + — ; — . nx + — ; — . n 1 — -r ; — f + ...
Now since a is an approximation to y/p, x is small in comparison with it, and hence its powers
above the second may be rejected, as of values too small to materially affect the calculation
within the prescribed limits. Then equating the co-efficients of the homologous powers which
remain, we have
p-q n-l qn
a = a, — j — . n = 1, and — n „ . „ = o.
P + q ' ^ P +9
From the second and third of these equations we get the same relation between p and q: viz.
p n — 1
p
Inserting this value of - in the assumed fraction for a -{- j we obtain
(n_l)p + («+!) A 2 a ^p-a}
« + ^=(„+l) P + („_1) A«; °'' = (»+l) P + (»-l) A
which is the working formula in the text, and perhaps the best form which the correction
admits of.
In another place I have given an improvement upon this approximation from a memorandum
of the late Mr. W. G. Horner, but to which I cannot make more special reference here.
71.
ARITHMETIC.
Note. It will always be well to reduce the index into factors as far as possi-
ble, and extract successively the several roots in the manner directed at the head
of the rule. Thus the thirtieth root is the fifth root of the third root of the
square root, since 5 x 3 X 2 = 30. It will also be always least laborious to
commence with the lowest roots, as with the second before the third, and the
third before the fifth, in the case just cited.
EXAMPLE.
Extract the tenth root of 4425048S1-64.
Here extracting the square root of the given number, we shall have to extract
the fifth root of 21035 8.
A few trials will show that the root hes between 73 and 7 4. Takinfj a = 7 3,
we have & = a^ = 20730-71593. Also p = 210358, and n = 5. Hence,
_ 7 3. [21035-8 — 20-3071593]
~~ 2. 21U35 8 + 3. 20730-71593
= 0021360, and hence ■V21035S = 7-321360 nearly.
This result is true in the last figure; but it is seldom that the approximation
proceeds so far correctly ; and even here it would not have been safe to assume
its correctness without verification.
OTHER EXAMPLES.
1.
What
IS the 3d root of 2 ?
Ans.
1-259921.
2.
What
s the 3d root of 3214?
Ans.
1475758.
3.
What
s the 4th root of 2 ?
Ans.
ri89-207.
4
What
s the 4th root of 9741 ?
Ans.
3-1415999
5.
What
s the 5th root of 2 ?
Ans.
1-148699.
6.
What i
s the Gth root of 21035-8?
Ans.
5-254037.
7.
What
s the 6th root of 2 ?
Ans.
1-122462.
8.
What
s the 7th root of 21035-8 ?
Ans.
4145392.
9.
What
s the 7th root of 2 ?
Ans.
1-104089.
10.
What
s the 8th root of 21035 8 ?
Ans.
3 470323.
11.
What
s the 8 th root of 2 ?
Ans.
1090508.
12.
What
s the 9th root of 21035 8 ?
Ans.
3022239.
13.
What
s the 9th root of 2 ?
Ans.
1 080059.
For a simple and ingenious method of constructing tab'es of square and cube
roots, and the reciprocals of numbers, see Dr. Hutton's Tracts on Mat/iematical
and Philosophical Subjects, vol. i. Tract 24, p. 459. By means of the method
there laid down, the tables at the end of the volume were computed.
A method adapted to the square root, in which the root is exhiljited as a simple
vulgar fraction, is also given in the same volume, which is extremely convenient
for the extraction of the roots of isolated numbers ; but where the roots of a
series of consecutive numbers are required, the one above referred to is the most
convenient and rapid one ever discovered. The following is the method for the
square root.
OF RATIOS, PROPORTIONS. &c. 75
Let N denote the given number, and ~ the fraction to which its square root
is approximately equal; then if n be a near integer or fractional value of the root
jl2 J. jj(/2 •
the fraction — - — will denote one still nearer.
Or. for continuing the process, the following is still more convenient. Let
^denote the last approximation: then the next is^"*~' ; or if inlendtd for
final conversion into the decimal scale, it is —
Example.— F.xlract the square root of 920. Here .\ = 920, and n = 30, rf = 1 .
Hence by the first formula V 920 ='1^"^'= -g^" = ^^ ■ Then from the
second formula, putting "=^\ we have ^'-^ ~ ^ =l-i>I' - 1 _ I656j _
° d 3 2(In 2 . 91 . 3 ~ 546 ~
30-33150183, which differs from the truth only by 6 in the tenth place of figures,
the true value being 30-33150177.— Htt//o»'« Tracts, vol. i. p. 457—549.
OF RATIOS, PROPORTIONS, AND PROGRESSIONS
OF NUMBERS.
Numbers are compared to each other in two difFerent ways : the one com-
parison considers the difference of the two numbers, and is named Arithmetical
Relation ; and the difference sometimes the Arithmetical Ratio : the other con-
siders their quotient, which is called Geometrical Relation ; and the quotient
expresses the Geometrical Ratio. So, of these two numbers, 6 and 3, the dif-
ference, or arithmetical ratio is 6 — 3 or 3, but the geometrical ratio is J or 2.
There must be two numbers to form a comparison : the number which is
compared, being placed first, is called the Antecedent ; and that to wliich it is
compared, the Consequent. So, in the two numbers above, 6 is the antecedent,
and 3 the consequent.
If two or more couplets of numbers have equal ratios, or equal diff*erences, the
equality is named Proportion, and the terms of the ratios Proportionals. So, the
two couplets, 4, 2 and 8, 6, are arithmetical proportionals, because 4 — 2 = 8
— 6 =: 2 ; and the two couplets, 4, 2 and 6, 3, are geometrical proportions,
because ;j = 3 = 2, the same ratio.
To denote numbers as being geometrically proportional, a colon is set between
the terms of each couplet, to denote their ratio; and a double colon, or else a
mark of equality, between the couplets or ratios. So, the four proportionals,
4, 2, 6, 3, are set thus, 4 : 2 ; ) 6 : 3, which means, that 4 is to 2, as 6 is to 3 ;
or thus, 4 : 2 = 6 : 3, or thus, .J = §, both which mean, that the ratio of 4 to 2,
is equal to the ratio of 6 to 3 * .
* The test of equal ratios in arillimetic is that the quotients of the antecr<1cnts and conse-
quents in the two alleged ratios is a fraction of the same value. In reference to geomctrr,
however, the test does not involve the idea of quotients or fractions, nor indeed of numbers mt
all. See Dcf. v. hook v. Elements of Euclid. The difficulty felt in treating proportion by
means of arithmetical ideas, and of tlie aritlimcliral definition of ratio, arises from the incom-
mensurability of numbers ; but this difficulty is not encountered in Euclid's method of treating
76 ARITHMETIC.
Proportion is distinguished into Continued and Discontinued. When the
difference or ratio of the consequent of one couplet, and the antecedent of the
next couplet, is not the same as the common difference or ratio of the couplets,
the proportion is discontinued. So, 4, 2, 8, 6, are in discontinued arithmetical
proi)ortion, because 4 — 2 = 8 — C=2, whereas 2 — 8 = — 6 : and 4, 2, 6, 3,
are in discontinued geometrical proportion, because ^ = f = 2, but | = 3, which
is not the same.
But when the difference or ratio of every two succeeding terms is the same
quantity, the proportion is said to be Continued, and the numbers themselves
make a series of Continued Proportionals, or a progression. So, 2, 4, 6, 8, form
an arithmetical progression, because 4 — 2 = 6 — 4=8 — 6 = 2, all the same
common difference; and 2, 4, 8, 16, a geometrical progression, because .^ =: | z=
'/ = 2, all the same ratio.
When the successive terms of progression increase, or exceed each other, it
is called an Ascending Progression, or Series ; but when the terms decrease, it is
a Descending one.
So, 0, 1, 2, 3, 4, . . . is an ascending arithmetical progression, but 9, 7, 5, 3, 1,
.... is a descending arithmetical progression. Also 1, 2, 4, 8, 16 is an
ascending geometrical progression, and 16, 8, 4, 2, 1, .... is a descending
geometrical progression.
ARITHMETICAL PROPORTION AND PROGRESSION.
In Arithmetical Progression, the numbers or terms have all the same common
difference. The first and last terms of a Progression are called the Extremes ;
and the other terms, lying between them, the Means. The most useful part of
arithmetical proportion is contained in the following theorems :
Theorem 1. When four quantities are in arithmetical proportion, the sum of
the two extremes is equal to the sum of the two means. Thus, with regard to
the four, 2, 4, 6, 8, we have 2 + 8=4 + 6 = 10.
Theorem 2. In any continued arithmetical progression, the sum of the two
extremes is equal to the sum of any two means that are equally distant from
them, or equal to double the middle term when there is an odd number of
terms.
Thus, in the terms 1, 3, 5, it is 1 + 5 = 3 + 3 = 6.
And in the series 2, 4, 6, 8, 10, 12, 14, it is 2 + 14 = 4 + 12 = 6 + 10 =
8 + 8 = 16.
the subject. It is, however, necessary to remark, that quantities (for numbers belong to tlic
predicate quantity) which fulfil the arithmetical test (or definition, as most writers term it) of
proportionality, can be readily subjected to Euclid's test : but the proposition is not convertible.
Hence, tlierefore, though we cannot build a system of proportion adapted to geometry upon the
aritlmietical basis, we can establish upon grounds equally valid and convincing with tliose of
geometry, all the properties of proportional quantities as expressed by numbers, to whatever
branch of pure or applied mathematics they may refer. In reading the fifth book of Euclid,
the intelligent teacher will avail himself of the opportunity of doing tliis. From the very early
stage of the student's progress, the common definition and test is necessarily employed in this
place : but by no means with a view to supersede the more logical and satisfactory investigations
just referred to.
f
ARITHMETICAL PROPORTION. 77
Theorem 3. The difference between the extreme terras of an arithmetical
progression, is equal to the common difference of the series muUiplied by une
less than the number of the terms. So, of the ten terms, 2, 4, 6, 8, 10, 12, U,
16, 18, 20, the common difference is 2, and one less than the numl>er of terms 9;
then the difference of the extremes is 20 — 2 = 18, and 2 x 9=18 also.
Consequently the greatest term is equal to the least term added to the product
of the common difference multiplied by one less than the number of terms.
Theorem 4. The sum of all the terms of any arithmetical progression, is
equal to the sum of the two extremes multiplied by the number of terms, and
divided by 2 ; or the sum of the two extremes multi|)lied by the number of the
terms, gives twice the sum of all the terms in the series.
This is made evident by setting the terms of the series in an inverted order,
under the same series in a direct order, and adding the corres^)onding terms
together in that order. Thus,
in the series 1, 3, 5, 7, 9, H, 13, 15;
ditto inverted 15, 13, 1^1, 9. 7, 5, 3, 1;
the sums are 16 + 16 + 16 + 16 +~16 +~\6 +~16 +~16,
which must be double the sum of the single series, and is equal to the sura of
the extremes repeated as often as are the number of the terms.
From these theorems may readily be found any one of these five particulars ;
the two extremes, the number of terms, the common difference, and the sum of
all the terms, when any three of them are given, as in the following problems.
PROBLEM I.
Gtren the extremes, and the number of terms, to find the sitm of all the terms.
Add the extremes together, multiply the sum by the number of terms, and
divide by 2.
examples.
1. The extremes being 3 and 19, and the number of terms 9; required the
sum of the terras.
19 4-3 22
Here X 9 = — X 9 = H X 9 = 99, the answer.
2. It is required to find the number of all the strokes a common clock strikes
in one whole revolution of the index, or in 12 hours. Ans. 78.
3. How many strokes do the clocks of Venice strike in the compass of a day,
which go continually on from 1 to 24 o'clock ? Ans. 300.
4. What debt can be discharged in a year, by weekly payments, in arith-
metical progression, the first payment being Is, and the last or 52J payment
5/ a$> Ans. 133/4*.
PROBLEM II.
Given the extremes, and the number of terms, to find the common differemct.
Subtract the less extreme from the greater, and divide the remainder by I
less than the number of terras, for the coraraon difference.
EXAMPLES.
1. The extremes being 3 and 19, and the number of terms 9 ; required the
common difference.
19—3 16 „ .
Here — — -= -^ = 2, Answer.
78 ARITHMETIC.
2. If the extremes be 10 and 70, and the number of terms 21 ; what is the
common diflference, and the sum of the series ?
Ans., the common difference is 3, and the sum is 840.
3. A certain debt can be discharged in one year, by weekly payments in arith-
metical progression, the first payment being Is, and the last al 3s; what is the
common difference of the terras ? Ans. 2.
PROBLEM III.
Given one of the extremes, the common difference, and the number of terTns ; to find
the other extreme, and the sum of the series.
Multiply the common difference by 1 less than the number of terms, and
the product will be the difference of the extremes : then add the product to
the less extreme to give the greater extreme, or subtract it from the greater, to
give the less.
EX.^MPLES.
1. Given the least term 3, the common difference 2, of an arithmetical series
of 9 terms ; to find the greatest term, and the sum of the series
Here 2 x (9—1) + 3 = 19 = the greatest term : hence (19 + 3) I = '.f =
99 = the sum of the series.
2. If the greatest term be 70, the common difference 3, and the number of
terms 21, what is the least term, and the sum of the series?
Ans., the least term is 10, and the sum is 840.
3. A debt can be discharged in a year, by paying 1 shilling the first week,
3 shillings the second, and so on, always 2 shillings more every week ; what is
the debt, and what will the last payment be ?
Ans., the last payment will be 5/ 3s, and the debt is 135/ 4s.
PROBLEM IV.
To find an arithmdical mean proportional between two given terms.
Add the two given extremes or terms together, and take half their sura for
the arithmetical mean required.
EXAMPLE.
To find an arithmetical mean between the two numbers 4 and 14.
Here — -~ — = 9 = the mean required.
Pi'vOBLRM V.
To find two arithmetical means between two given extremes.
Subtract the less extreme from the greater, and divide the difference by 3, so
will the quotient be the common difference ; which being continually added to
the less extreme, or taken from the greater, will give the means.
EX.\MPLE.
To find two arithmetical means between 2 and 8.
^ 2
Here - — - — = 2 = common difference.
Then 2 + 2 = 4 = the one mean, ami 4 + 2 = 6 = the other mean.
r
GEOMETRICAL PROGRESSION. 79
PROBLEM VI.
To find any number of arithmetical means betveen two given terms or extremes.
Subtract the less extreme from the greater, and divide the difference hy 1
more than the number of means required to lie found, which will give the com-
mon difference; then this bein<? added continually to the least term, or sub-
tracted from the greatest, will give the mean terms required.
EXAMPLE.
To find five arithmetical means between 2 and 14.
14 2
Here ^. — = 2 = common difference.
b
Then by adding this common difference continually, ths means are found to
be 4, 6, 8, 10, 12.
See more of arithmetical progression in the Algebra.
GEOMETRICAL PROPORTION AND PROGRESSION.
• If there he taken two ratios, as those of 6 to 3, and 14 to 7. which, by
what has been already said, (p. 50, 75,) may be expressed fractionally, J and ',' ;
to judge whether they are equal or unequal, we must reduce them to a common
denominator, and we shall have 6x7, and 14 x 3 for the two numerators. If
these are equal, the fractions or ratios are equal. Therefore,
Theorem i. If four quantities be in geometrical proportion, the product of
the two extremes will be equal to the product of the two means.
And hence, if the product of the two means be divided by one of the ex-
tremes, the quotient will give the other extreme. Thus, of the above numbers, if
the product of the means 42 be divided by 6, the quotient 7 is the other ex-
treme ; and if 42 be divided by 7, the quotient 6 is the first extreme. This is
the foundation of the practice in the Rule of Three,
We see, also, that if we have four numbers, (5, 3, 14, 7, such, that the pro-
ducts of the means and of tlie extremes are equal, we may hence infer the
equality of the ratios % = y, or the existence of the proportion 6:3:: 14:7.
Hence
Theokem II. We may always form a proportion of the factors of two equal
products.
If the two means are equal, as in the terms 3, 6, 6, 12, thiir product becomes
a square. Hence
Theore.m iir. The mean proportional between two numbers is the square
root of their product.
We may without destroying the accuracy of a proportion, give to its various
terms all the changes which do not affect the equality of the products of the
means and extremes.
* See ibe note at p. I'u
80 ARITHMETIC.
Thus, with respect to the proportion 6 : 3 ; : 14 : 7, which gives 6x7=3x14,
■we may displace the extremes, or the means, an operation which is denoted bj
the word Alternando.
This will give 6:14:: 3:7
or 7 : 3 : : 14 : 6
or 7 : 14 : : 3:6.
Or, 2dly, we may put the extremes in the places of the means, called Invertendo.
Thus 3 : 6 : : 7 : 14.
Or, 3dly, we may multiply or divide the two antecedents, or the two conse-
quents, by the same number, when proportionality will subsist.
As 6x4:3:: 14x4:7; viz. 24 : 3 : : 36 : 7
and 6 -j- 2 : 3 : : 14 -i. 2 : 7 ; viz. 3:3:: 7:7.
Also, applying the proposition in note 2, Addition of Vulgar Fractions, to the
terms of a proportion, such as 30 : 6 : : 15 : 3, or |" = 3', we shall have
30+15 15 „ , 30 + 15 30—15 „
— -^ = — and —~^ — = Hence
6+3 3 6+3 6—3
Theorem iv. The Eum or the diflFerence of the antecedents, is to that of the
consequents, as any one of the antecedents is to its consequent.
Theorem v. The sum of the antecedents is to their difference, as the sum oi
the consequents is to their diflFerence.
In like manner, if there be a series of equal ratios, 1=5"= y = f^ ; we have
Theorem vi. In any series of equal ratios, the sum of the antecedents is to
that of the consequents, as any one antecedent is to its consequent.
6 + 10^J4 + 30_6_10__14_30 'pjjgrgforg
'3 + 5^^7^15 3 5 y ~ 15" "^ '''■^'
Theorem vii. If two proportions are multiplied, term by term, the products
will constitute a proportion.
Thus, if 30 : 15 :: 6 : 3
and 2 : 3 : : 4 : 6.
Then 30x2:15x3::6x4:3x6
or 60 : 45 : : 24 : 18 ; or f? = ^.
Theorem viii. If four quantities are in proportion, their squares, cubes, and
all other powers will be in proportion.
For this will evidently be nothing else than assuming the proportionality ol
the products, term by term, of two, three, or more identical proportions.
The same properties hold with regard to surd or irrational expressions.
Thus, ^ 720 : v^ 80 : : ^/ 567 : V 63
and V 12 : n/ 3 : : a/ 4 : a/ 1.
For "^"^'^^ _.\A9X80) _ 3 ^jjj ^/.567 _ x^ C9 X 63) _ 3
a/ 80 ^80 1' v/63 a/ 63 1
^. -v/ 12 / 12 _ s/ 4 2
and =r ^ _ =: :^ — =; _.
-v/ 3 3 11
Theorem ix. The quotient of the extreme terms of a geometrical progression
is equal to the common ratio of the series raised to the power denoted by 1 less
than the number of the terms.
So, of the ten terms 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, the common
ratio is 2, one less than the number of terms 9 ; then the quotient of the ex-
tremes is "2*- = 512, and 2»= 512 also.
Consequently the greatest term is equal to the less term multiplied by the said
•power of the ratio, whose index is 1 less than the number of terms.
Theorem x. The sum of all the terms, of any geometrical progression, is
GEOMETRICAL PROGRESSION. gf
found by adding the greatest term to the difference of the extremes divided by
1 less than the ratio.
So, the sum of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, (whose ratio is 2.) is
1024 + 1024j- 2 _ JQ24 _}. io22 = 2046.
This subject will be resumed in the algebraic part of this work. A few
examples may here be added.
EXAMPLES.
1. The least of ten terms in geometrical progression being 1, and the ratio 2 •
what is the greatest term, and the sum of all the terms ?
Ans., the greatest term is 512, and the sum 1023.
2. What debt may be discharged in a year, or 12 months, by paying 1/ the
first month, 2/ the second, 4/ the third, and eo on, each succeeding payment
being double the last ; and what will the last payment be ?
Ans., the debt 4095/ and the last payment 2048/.
PROBLEM I.
To find one geometrical mean proportional between any two numbers.
Multiply the two numbers together, and extract the square root of the pro-
duct, which will give the mean proportional sought.
EXAMPLE.
To find a geometrical mean between the two numbers 3 and 12.
^ (12 X 3) = a/ 36 = 6 = the mean.
PROBLEM II.
To find two geometrical mean proportionals between any two numbers.
Divide the greater number by the less, and extract the cube root of the quo-
tient, which will give the common ratio of the terms. Then multiply the least
given term by the ratio for the first mean, and this mean again by the ratio for
the second mean ; or, divide the greater of the two given terms by the ratio for
the greater mean, and divide this again by the ratio for the less mean.
example.
To find two geometrical means between 3 and 24.
Here 3) 24 (8 ; its cube root 2 is the ratio.
Then 3x2 = 6, and 6x2=12, the two means.
Or 24 -r 2 = 12, and 12 -r 2 = 6, the same.
That is, the two means between 3 and 24, are 6 and 12.
PROBLEM III.
To find any number of geometrical means between two numbers.
Divide the greater number by the less, and extract such root of the quotient
whose index is 1 more than the number of means required ; that is, the 2d root
for one mean, the 3rd root for two means, the fourth root for three means, and so
on ; and that root will be the common ratio of all the terms : then, with the
ratio, multiply continually from the first term, or divide continually from the
last or greatest term.
82 ARITHMETIC.
EXAMPLE.
To find four geometrical means between 3 and 96.
Here 3) 96 (32 ; the 5th root of which is 2, the ratio.
Then 3x2 = 6, and 6 x 2 = 12, and 12x2 = 24, and 24 x 2 = 48.
Or 96 -7- 2 = 48, and 48 -^ 2 = 24, and 24 -i- 2 = 12, and 12 -j- 2 = 6.
That is, 6, 12, 24, 48, are the four means between 3 and 96.
OF HARMONICAL PROPORTION.
There is also a third kind of proportion, called Harmonical or Musical, which
being hut of rare occurrence in questions purely arithmetical, a very short
account of it may here suffice. It will however be again noticed both in algebra
and in geometiy, but especially in the latter.
Musical Proportion is when, of three numbers, the first has the same propor-
tion to the third, as the difference between the first and second has to the dif-
ference between the second and third.
As in these three, 6, 8, 12 ;
where 6 : 12 ; I 8 — 6 : 12 — 8,
that is 6 : 12 : : 2 : 4.
When four numbers are in musical proportion, then the first has the same
ratio to the fourth, as the difference between the first and second has to the dif-
ference between the third and fourth.
As in these, 6, 8, 12, 18;
where 6 : 18 : : 8 — 6 : 18 — 12,
that is 6 : 18 : : 2 : 6.
"When numbers are in musical progression, their reciprocals are in arithmetical
progression ; and the converse, that is, when numbers are in arithmetical pro-
gression, their reciprocals are in musical progression.
So in these musicals 6, 8, 12, their reciprocals 5, ^, ij, are in arithmetical pro-
gression ; for i + Vj = fj = ^ ; and |^ -f- ^ = | = :J ; that is, the sum of the
extremes is equal to double the mean, which is the property of arithmeticals.
The method of finding a series of numbers in musical proportion and progres-
sion is best expressed by algebraic methods and symbols.
FELLOWSHIP OR PARTNERSHIP.
Fellowship is the rule by which any sum or quantity may be divided into
any number of parts which shall be in any given proportion to one another.
By this rule are adjusted the gains or losses or charges of partners in com-
pany ; or the effects of bankrupts, or legacies in case of a deficiency of assets or
effects ; or the shares of prizes ; or the numbers of men to form certain detach-
ments ; or the division of waste lands among a number of proprietors.
Fellowship is either Sinple or Double. It is single, when the shares or por-
tions are to be proportioned each to one given number only ; as when the stocks
of partners are all employed for the same time : and double, when each portion
is to be proportional to two or more numbers ; as when the stocks of partners
are employed for different times.
SIxNGLE FELLOWSHIP.
GENERAL RULE.
Add together the numbers that denote the proportion of the shares : then
say.
As the sura of the said proportional numbers,
is to the whole sura to be parted or divided,
so is each of the several proportional numbers,
to the corresponding share or part.
Or, As the whole stock, is to the whole gain or loss,
so is each raan's particular stock,
to his particular share of gain or loss.
To PROVE THE WORK. Add all the shares or parts together, and the su.n
will be equal to the whole number to be shared, when the work is right.
EXAMPLES.
1. To divide the number 240 into three such parts, as shall be in proportion
to each other as the three numbers, 1, 2, and 3.
Here 1 + 2 4- 3 = 6 = the sura of the numbers.
Then, as 6 : 240 ! ; 1 : 40 = the 1st part,
and as 6 : 240 \\ 2 : 80 = the 2d part,
also as 6 : 210 : ; 3 : 120 = the 3d part.
Sura of all 240, the proof.
2. Three persons, A, B, C. freighted a ship with 340 tuns of wine ; of which
A loaded 110 tuns, B 97, and C the rest : in a storm the seamen were obliged to
throw overboard 85 tuns ; how much must each person sustain of the loss ?
Here 110+ 97 = 207 tuns, loaded by A and B ;
therefore 340 — 207 = 133 tuns, loaded by C.
Hence, as 340 : 85 : : 110
or as 4 : 1 : : 1 10 : 27^ tuns ^ A's loss ;
and as 4 : 1 : : 97 : 24 J tuns = B's loss;
also as 4 : 1 : : 133 : 33| tuns = C's loss;
Sum 85 tuns, the proof.
3. Two merchants, C and D, made a stock of 120/; of which C contributed
75/, and D the rest : by trading they gained 30/; what must each have of it ?
Ans. C 18/ 15*, and D 11/5*.
4. Three raerchants, E, F, G, make a stock of 700/; of which E contributed
123/, F 358/, and G the rest : by trading they gain 125/ 10*; what must each
have of it ? Ans. E must have 22/ 1* Od 2lq^
F 64 3 8 Oji.
G 39 5 3 IjV
5. A General imposing a contribution • of 700/ on four villages, to be paid in
proportion to the number of inhabitants contained in each ; the first containing
250, the 2d 350, the 3d 400, and the 4th 500 persons ; what part must each vil-
lage pay ? Ans. the first to pay 116/ 13* id.
the 2d 163 6 8.
the 3d 186 13 4.
the 4th 233 6 8.
* Contribution is a tax paid by province*, tow-ns, 0/ villagps, to exciiM them from bejn|
•plundered. It is paid in provisions or in money, and sometime* in both.
G 2
84 ARITHMETIC.
6. A piece of ground, consisting of 37 ac 2 ro 14 ps, is to be divided among
three persons, L, M, and N, in proportion to their estates : now if L's estate be
worth 5001 a year, M's 320Z, and N's 75/; what quantity of land must each one
have ? Ans. L must have 20 ac 3 ro SQl?! pis.
M 13 1 30^1,1.
N 3 0 23^g
7. A person is indebted to O 571 15s. to P 108/ 3s 8rf, to Q 22/ lOd, and to R
73/; but at his decease, his effects are found to be worth no more than 170/ 14s;
how must it be divided among his creditors ?
Ans. O must have 37/ 15s 5d 2f^^fsq.
P 70 15 2 2^'fi,.
Q. 14 8 4 O^l/jfg.
R 47 14 11 2,%%.
8. A ship, worth 900/, being entirely lost, of which ^ belonged to S, J to T,
and the rest to V; what loss will each sustain, supposing 540/ of her were
insured ? Ans. S will lose 45/, T 90/, and V 225/.
9. Four persons, W, X, Y, and Z, spend among them 25s, and agree that W
shall pay ^ of it, X J, Y J, and Z 3 ; that is, their shares are to be in proportion
as §, 2, i, and 5 : what are their shares ? Ans. W must pay 9s 8d 3^],q.
X 6 5 3|2.
Y 4 10 m
Z 3 10 37V
10. A detachment, consisting of 5 companies, being sent into garrison, in
which the duty required 76 men a day ; what number of men must be furnished
by each company, in proportion to their strength ; the 1st consisting of 54 men,
the 2d of 51 men, the 3d of 48 men, the 4th of 39, and the 5th of 36 men ?
Ans., the 1st must furnish 18, the 2d 17, the 3d 16, the 4th 13, and
the 5th 12 men*.
DOUBLE FELLOWSHIP.
Double Fellowship, as has been said, is concerned in cases in which the
stocks of partners are employed or continued for different times.
RuLEf. Multiply each person's stock by the time of its continuance; then
divide the quantity, as in Single Fellowship, into shares, in proportion to these
products, by saying.
As the total sum of all the said products.
Is to the whole gain or loss, or quantity to be parted.
So is each particular product,
To the corresponding share of the gain or loss.
• Questions of this nature frequently occurring in military service. General Haviland, .an
officer of great merit, contrived an ingenious instrument, for more expeditiously resolving them ;
■which is distinguished by the name of the inventor, being called a Haviland.
+ The proof of this rule is as follows : When the times are equal, the shares of the gain or
loss are evidently as the stocks, as in Single Fellowship ; and when the stocks are equal, the
iharei are af the times; therefore, when neither are equal, the shares must be as their products.
SIMPLE INTEREST. 85
EXAMPLES.
1 . A had in company 50Z for 4 naonths, and B had 60/ for 5 months ; at the
end of which time they find 241 gained : how must it be divided between them i
Here 50 60
4 5
200 + 300 = 500.
Then as 500 : 24 : : 200 : 9? = 9l i2s = A's share,
and as 500 : 24 : : 300 : 14| = 14 8 = B's share.
2. C and D hold a piece of ground in common, for which they are to pay 54/.
C put in 23 horses for 27 days, and D 21 horses for 39 days ; how much ought
each man to pay for the rent ? Ans. C must pay 23/ 5* 9d.
D 30 14 3.
3. Three persons, E, F, G, hold a pasture in common, for which they are
to pay 30/ per annum ; into which E put 7 oxen for 3 months, F put 9 oxen
for 5 months, and G put in 4 oxen for 12 months ; how much must each
person pay of the rent? Ans. E must pay 5/ 10s 6d if^q.
F 11 16 10 0^.
G 12 12 7 2,V
4. A ship's company take a prize of 1000/, which they agree to divide
among them according to their pay and the time they have been on board :
now the officers and midshipmen have been on board 6 months, and the
sailors 3 months ; the officers have 40s a month, the midshipmen 30s, and
the sailors 22s a month; moreover, there are 4 officers, 12 midshipmen, and
110 sailors: what will each man's share be ?
Ans. each officer must have 23/ 2s 5rf O^q.
each midshipman 17 6 9 3i*,»j.
each seaman 6 7 2 O^^j.
5. H, with a capital of 1000/, began trade the first of January, and, meeting
with success in business, took in I as a partner, with a capital of 1500/, on the
first of March following. Three months after that they admit K as a third
partner, who brought into stock 2800/. After trading together till the end of
the year, they find there has been gained 1776/ 10s; how must this be divided
among the partners ? Ans. H must have 457/ 9* ^l^ ^Iq.
I ... 571 16 8i i7,.
K . . . 747 3 llj JjJ.
6. X, Y, and Z made a joint-stock for 12 months; X at first put in 20/, and
4 months after 20/ more ; Y put in at first 30/, at the end of 3 months he put
in 20/ more, and 2 months after he put in 40/ more ; Z put in at first 60/, and
5 months after he put in 10/ more, 1 month after which he took out 30/ ; during
the 12 months they gained 50/; how much of it must each have ?
Ans. X must have 10/18* 6d 3l\q.
Y ... 22 8 1 Oif.
Z ... 16 13 4 0.
SIMPLE INTEREST.
Interest is the premium or sum allowed for the loan or forbearance of money.
The money lent or forborn is called the Principal ; and the sum of the prmcipal
86 ARITHMETIC.
and its interest is called the Amount. Interest is allowed at so much per cent,
per annum, or interest of 100/ for a year, is called the rate of interest. Thus :
when interest is at 3 per cent, the rate is 3 ;
4 per cent. . . 4 ;
5 per cent. . . 5 ;
6 per cent. . . 6.
But, hy law, interest ought not to be taken higher than at the rate of 5
per cent.
Interest is of two sorts ; Simple and Compound.
Simple Interest is that which is allowed for the principal lent or forbom only,
for the whole time of forbearance. As the interest of any sum, for any time, is
directly proportional to the principal sum, and also to the time of continuance ;
hence arises the following general rule of calculation.
As 100/ is to the rate of interest, so is any given principal to its interest for
one year. And again,
As 1 year is to any given time, so is the interest for a year, just found, to the
interest of the given sum for that time.
Otherwise. Take the interest of 1 pound for a year, which multiply by the
given principal, and this product again by the time of loan or forbearance, in
years and parts, for the interest of the proposed sum for that time.
Note. When there are certain parts of years in the time, as quarters, or
months, or days ; they may be worked for, either by taking the aliquot or like
parts of the interest of a year, or by the Rule of Three in the usual way. Also,
the division by 100 is done by pointing off two figures for decimals.
EXAMPLES.
1. To find the interest of 230/ 10s, for 1 year, at the rate of 4 per cent,
per annum.
Here, as 100 : 4 ; : 230/ 10* : 9/ 4s ild.
4
llOO) C22 0
20
4140
12
4|80 Ans. 9/ 4s 4|i.
4
320
2. To find the interest of 547/ 15s, for 3 years, at 5 per cent, per annum.
As 100 : 5 '. ; 54775
Or 20 : 1 ; ; 547*75 : 273875 interest for 1 year.
3
/ S2 1625 ditto for 3 years.
20
S3-2500
12
rf300 Ans. 82/ 3s 3d.
SIMPLE INTEREST. gy
3. To find the interest of 200 guineas, for 4 years, 7 months and 25 days
at 4^ per cent, per annum. '
210Z
4i
ds £ ds
As 365 : 9-45 : : 25 :
or 73 : 9-45 ; ; 5 :
£
•6472
840
5
105
945 interest for 1 year.
4
73 ) 47-25 ( -6472
345
530
19
37-80 ditto 4 years.
6 roo = J 4-725 ditto 6 months.
1 mo = i -7875 ditto 1 month.
•6472 ditto 25 days.
/ 43 9597
20
8 19-1940
12 Ans. 43/ 19s 2id.
d 2-3280
4
q 1-3120
4. To find the interest of 450/, for a year, at 5 per cent, per annum.
Ans. 22/ 10*.
5. To find the interest of 715/ 12s 6(7, for a year, at A\ per cent, per annum.
Ans 32/ As 0\d.
6. To find the interest of 720/, for 3 years, at 5 per cent, per annum.
Ans. 108/.
7. To find the interest of 355/ 15s, for 4 years, at 4 per cent, per annum.
Ans. 56/ 18s A\d.
8. To find the interest of 32/ 5s 8c/, for 7 years, at 4^ per cent, per annum.
Ans. 9/ 12s Irf.
9. To find the interest of 107/, for 1| year, at 5 per cent, per annum.
Ans. 12/ 15s.
10. To find the insurance on 205/ 15s, for i of a year, at 4 per cent, per
annum. Ans. 2/ Is Ijrf.
11. To find the interest on 319/ 6i, for 5 J years, at 3 J per cent per annum.
Ans. 68/ 15s 9i</.
12. To find the insurance on 107/, for 117 days, at 4 J per cent per annum.
Ans. 1/ l'2s 7d.
13. To find the interest of 17/ 5s, for 117 days, at 4^ per cent, per annum.
Ans. 5s 2d.
14. To find the insurance on 712/ 6s for 8 months, at 7i per cent, per annnm.
Ans. 35/ 12s3irf.
Note. The rules for Simple Interest serve also to calculate Insurances, or
the Purchase of Stocks, or any thing else that is rated at so much per cent.
See also more on the subject of Interest with the algebraical expression and
investigation of the rules, at the end of the Algebra.
88 ARITHMETIC.
COMPOUND INTEREST.
Compound Interest, called also interest upon interest, is that which arises
from the principal and interest, taken together, as it becomes due, at the end of
each stated time of payment. Though it be not lawful to lend money at com-
pound interest, yet in purchasing annuities, pensions, or leases in reversion, it is
usual to allow compound interest to the purchaser for his ready money.
Rule 1. Find the amount of the given principal, for the time of the first
payment, by simple interest. Then consider this amount as a new principal for
the second payment, whose amount calculate as before. Proceed thus through
all the payments to the last, always accounting the last amount as a new prin-
cipal for the next payment. The reason is evident from the definition of com-
pound interest.
Otherwise,
Rule 2. Find the amount of 1 pound for the time of the first payment, and
raise or involve it to the power of whose index is denoted by the number of pay-
ments. Then that power multiplied by the given principal, will produce the
whole amount : from which the said principal being subtracted, leaves the
compound interest of the same. This is evident from the first rule.
EXAMPLES.
1. To find the amount of 720Z, for 4 years, at 5 per cent, per annum.
Here 5 is the 20th part of 100, and the interest of 1/ for a year is ^ or '05,
and its amount 1*05. Therefore,
1. By the 1st rule. 2. By the 2d rule.
£ s d 1*05 amount of ] 2.
20) 720 0 0 1 St year's principal 105
36 0 0 1st year's interest
1-1025 2d power of it.
20) 756 0 0 2d year's principal 1-1025
37 16 0 2d year's interest
20) 793 16 0 3d year's principal 720
39 1 3 92 3d year's interest
20) 833 9 9i 4th year's principal
41 13 5 J 4th year's interest
£875 3 3i the whole amount, 12
or answer required.
1-21550625 4th power of it.
/ 875-1645
20
s 3-2900
d 3-4800
2. To find the amount of 50/ in 5 years, at 5 per cent, per annum, compound
interest. Ans. 63/ l6s ^\d.
3. To find the amount of 50/ in 5 years, or 10 half-years, at 5 per cent, per
annum, compound interest, the interest payable half-yearly. Ans. 64/ Os \d.
4. To find the amount of 50/ in 5 years, or 20 quarters, at 5 per cent, per
annum, compound interest, the interest payable quarterly. Ans. 64/ 2s Q\d.
5. To find the compound interest of 3/0/ forborn for 6 years, at 4 per cent,
per annum. Ans. 98/ 3« \\d.
ALLIGATION MEDIAL. gg
6. To find the compound interest of 410/ forborn for 2§ years, at 4) per cent,
per annum, the interest payable half-yearly. Ans. 48/ 4« llW,
7. To find the amount, at compound interest, of 217/ forborn for 2^ years, at
5 per cent, per annum, the interest payable quarterly. Ans. 242/ 13* 4 id.
ALLIGATION.
Alligation teaches how to compound or mix together several simples of
different qualities, so that the composition may be of some intermediate quality
or rate. It is commonly distinguished into two cases. Alligation Medial, and
Alligation Alternate.
ALLIGATION MEDIAL.
Alligation Medial is the method of finding the rate or quality of the com-
position, from having the quantities and rates or qualities of the several simples
given. It is thus performed : —
* Multiply the quantity of each ingredient by its rate or quality ; then add all
the products together, and add also all the quantities together into another sum ;
then divide the former sum by the latter, that is, the sum of the products by the
sum of the quantities, and the quotient will be the rate or quality of the com-
position required.
EXAMPLES.
1. If three sorts of gunpowder be mixed together, viz. 50lb at I2d a pound,
44lb at 9d, and 261b at 8d a pound ; how much a pound is the composition
worth ?
Here 50, 44, 26 are the quantities,
and 12, 9, 8 the rates or qualities ;
then 50 X 12 = 600
44 X 9 = 396
26 X 8 = 208
120) 1204 (10^^0=103^.
Ans., the rate or price is lOj'jd the pound.
• Demonstration. The rule is thus proved by Algebra.
Let a, 6, c be the quantities of the ingredients,
and w, n, p their rates, or qualities, or prices ;
then am, Im, cp are their several values,
and am -{- bn -\- cp the sum of their values,
also a -)- 6 -|- c is the sum of the quantities,
and if r denote the rate of the whole composition,
then {a -\- I) -j- c) X r will be the value of the whole,
conseq. {a -\- b -\- c) X r ::z am -\- bn -\- cp,
and r = {am -f- '^« + <7') -7- (« + ^ + c)i which is the rule.
Note. If an ounce or any other quantity of pure gold be reduced into 24 equal part*, these
parts ar6 called cai-ats ; but gold is often mixed with sonic base metal, which is called the
alloy, and the mixture is s.iid to be of so many carats fine, according to the proportion of pure
gold contained in it: thus, if 22 carats of pure gold, and 2 of alloy be mixed together, it i* said
to be 22 carats fine.
If any one of the simples be of little or no value with respect to the rest, it« rate U tuppoaed
to be nothing; as water mixed with wine, and alloy with gold and silver.
90 ARITHMETIC.
2. A composition being made of 5lb of tea at 7s per lb, 9lb at 8* 6c? per lb, and
I4|lb at OS lOd per lb ; what is a lb of it worth ? Ans. 6s lOkd.
3. Mixed 4 gallons of wine at 4s lOd per gall, with 7 gallons at 5s 3d per gall,
and 9| gallons at 5s Sd per gall ; what is a gallon of this composition worth ?
Ans. 5s 45^.
4. Having melted together 7 oz of gold of 22 carats fine, 12§ oz of 21 carats
fine, and 17 oz of 19 carats , fine : I would know the fineness of the composi-
tion ? Ans. 20 J3 carats fine.
ALLIGATION ALTERNATE.
Alligation Alternate is the method of finding what quantity of any num-
ber of simples, whose rates are given, will compose a mixture of a given rate.
It is, therefore, the reverse of Alligation Medial, and may be proved by it.
RULE I*.
1. Set the rates of the simples in a column under each other. 2. Connect,
or link with a continued line, the rate of each simple, which is less than that of
the compound, with one, or any number, of those that are greater than the
compound ; and each greater rate with one or any number of the less. 3. Write
the difference between the mixture rate, and that of each of the simples, opposite
• Demonst. By connecting the less rate with the greater, and placing the difference between
them and the rate alternately, the quantities resulting are such, that there is precisely as much
gained by one quantity as is lost by the other, and therefore the gain and loss upon the whole is
equal, and is exactly the proposed rate : and the same will be true of any other two simples
managed according to the rule.
In like manner, whatever the number of simples may be, and with how many soever every
one is linked, since it is always a less with a greater than the mean price, there will be an equal
balance of loss and gain between every two, and consequently an equal balance on the whole.
It is obvious, from the rule, that questions of this sort admit of a great variety of answers ;
for, having found one answer we may find as many more as we please, by only multiplying or
dividing each of the quantities found, by 2, or 8, or any integer; the reason of which is evident :
for, if two quantities, of two simples, make a balance of loss and gain, with respect to the mean
price, so must also the double or treble, the i or J part, or any other ratio of these quantities, and
so on ad infinitum.
These kinds of questions are called by algebraists indeterminaie or tadimiied problems ; and
by an analytical process, theorems may be deduced that will give all the possible answers.
Thus, taking for example the four simples A, B, C, D, which are to be mixed so as to produce
the mean price m. Denote the prices of a, b, c, d hj tn -\- a, m -j- 6, m — c, and ni — d
respectively. Likewise let the quantities taken be x, y, z, v. Then
Mean
Prices.
m -\-a
Quantities. I Then if each quantity be multiplied
by its price, the sum of the pro-
ducts will evidently be the same
as of all the quantities multiplied
by the mean price, viz. :
(m -|- a) * -f- (m + 6) y + (w —c)z-\-(m—d)v=m(^x-^y + z + t).
That is, ax •\- hy =. cz -\- dt.
But as there are four unknown quantities, and but one equation, we are at liberty to assume
any other three conditions we please, and still the true result will be obtained. The three
simplesi that can be taken are those upon which the rule above given is founded, viz.
J? = (/, y = c, I = 6, c =: a ; or x = c, y ^ d, z = a, v — b. [The
ALLIGATION ALTERNATE. 91
the rate with which they are linked. 4. Then if only one difference stand
against any rate, it will be the quantity belonging to that rate ; but if there be
eeveral, their sum will be the quantity.
The examples may be proved by the rule for Alligation Medial.
EXAMPLES.
1. A merchant would mix wines at I6s, at 18s, anil at 22* per gallon, so that
the mixture maybe worth 20* the gallon ; what quantity of each must be taken ?
16 — s. 2 at 16*
. 18 -^ ^ 2 at 18s
Here 20 <
22 — ^ 4 + 2 = 6 at 22s
2. How much sugar at id, at 6c?, and at Urf per lb, must be mixed together,
so that the composition formed by them may be worth 7d per lb ?
Ans. 1 lb. or 1 stone, or 1 cwt, or any other equal quantity of each sort.
3. How much corn at 2s 6d, 3s 8c?, 4s, and 4s 8c? per bushel must be mixed
together, that the compound may be worth 3s 10c? ])er bushel ?
Ans. 2 at 2s 6d, 3 at 3s 8d, 3 at 4s, and 3 at 4* 8rf.
RULE II.
When the whole composition is limited to a certain quantity: find an answer
as before by linking ; then say, as the sum of the quantities, or differences thus
determined, is to the given quantity ; so is each ingredient, found by Imking,
to the required quantity of each.
example.
1. How much gold of 15, 17, 18, and 22 carats fine, must be mixed together,
to form a composition of 40 oz of 20 carats fine.
15 ^ 2
17 0\ 2
Here 20 .J ^^ ^ ^^ 2
22-^ 5 + 3+2= 10
16
Then as 16 : 40 ; ; 2 : 5
and 16 : 40 ; ; 10 : 25
Ans. 5 oz of 15, of 17, and of 18 carats fine, and 25 oz of 22 carats fine •.
^
The work will therefore stand in either of the following forms :
m -\- a d I m -\- a
m + b — s \ c \ m 4- h
) I , or, TO
TO — C ^ J U TO C
m — d a I TO — d -^ 6
which is the rule, and any other three relations would give a different but a less practicable
rule than this. The next in point of simplicity is
X ■=. pd, y = 7^, z =: qh, v = }xi ; or, x ■=■ JK, y ■=. (/d, z ■=. pa, r = q/>.
See the Key to Keith's Arithmetic, by Maynard, from which valuable little treatise the
latter part of this note is taken.
* A great number of questions might be here given relating to the specific gravities of bodic*,
liut one of the most curious may suffice.
Hiero, king of Syracuse, gave ordere for a crown to be m.idc entirely of pure gold ; but,
92 ARITHMETIC.
RULE III*.
When one of the ingredients is limited to a certain quantity : take the dif-
ference between each price, and the mean rate as before ; then say, as the
difference of that simple, whose quantity is given, is to the rest of the differences
severally ; so is the quantity given, to the several quantities required.
EXAMPLES.
1. How much wine at 5*, at 5s 6d, and 6s the gallon, must be mixed with 3
gallons at 4* per gallon, so that the mixture may be worth 5s 4d per gallon ?
(-48 ^ 8 + 2 = 10
J 60 l\ 2 + 8 = 10
^''''']66U) 4 + 16=20
^72 J^ 16+4 = 20
Then 10 : 10 '. : 3 : 3
10 : 20 : : 3 : 6
1 0 : 20 ; ; 3 : 6 Ans. 3 gallons at 5s, 6 at 5s 6d, and 6 at gs.
2. A grocer would mix teas at 12s, at 10s, and at 6s per lb, with 20lb at 4s
per lb : how much of each sort must he take to make the composition worth 8s
per lb ? Ans. 20lb at 4s, lOlb at 6s, lOlb at 10s, and 20lb at 12s.
POSITION.
Position is a rule for performing certain questions, which cannot be resolved
by the common direct rules. It is sometimes called False Position, or False
Supposition, because it makes a supposition of false numbers, to work with in
the same manner as if they were the true ones, and by their means discovers the
true numbers sought. It is sometimes called Trial-and-Error, because it pro-
ceeds by trials of false numbers, and thence finds out the true ones by a com-
parison of the errors. Position is either Single or Double.
suspecting tbe workmen had debased it by mixing it with silver or copper, he recommended the
discovery of the fraud to the femous Archimedes, and desired to know the exact quantity of
alloy in the crown.
Archimedes, in order to detect the imposition, procured two other masses, the one of pure
gold, the other of silver or copper, and each of the same weight with the former ; and by putting
each separately into a vessel full of water, the quantity of water expelled by them determined
their specific gravities : from which, and their given weights, the exact quantities of gold and
alloy in the crown may be determined.
Suppose the weight of each crown to be 101b, and that the water expelled by the copper or
silver was 921b, by the gold 521b, and by the compound crown 641b : what will be the quan-
tities of gold and alloy in the crown ?
64 I ^l^^ 12 of Conner
I 52 / 28 of gold.
And the sum of these is 12 + 28 = 40, which should have been 10 ; therefore by the Rule,
40; 10:: 12: 31b of copper ■»
40 : 10 : : 28 : 71b of gold J ^^^ answer.
* In the very same manner questions may be resolved when several of the ingredients are
limited to certain quantities, by finding first for one limit, and then for another. The last two
rules can need no demonstration, as they evidently result from the first, the reason of which
has been already explained.
r
93
SINGLE POSITION.
Single Position is that by which a question is resolved by means of one
supposition only. Questions which have their result proportional to their sup-
position, belong to single position : such as those which require the mullipli-
cation or division of the number sought by any proposed number; or when it ia
to be increased or diminished by itself, or any parts of itself, a certain proposed
number of times. The rule is as follows :
Take or assume any number for that which is required, and perform the
same operations with it, as are described or performed in the question : then
say, as the result of the said operation is to the position or number assumed ;
so is the result in the question to a 4th term, which will be the number sought*.
EXAMPLES.
1. A person after spending J and i of his money, has yet remaining 60/ ;
what had he at first ?
Suppose he had at first 1 20/. Proof.
Now i of 120 is 40 i of 144 is 48
\ of it is 30 \ of 144 is 36
^, their sum is 70 their sum 84
which taken from 120 taken from 144
leaves 50 leaves 60 as per question.
Then, 50 : 120 ; ; 60 : 144 the answer.
2. What number is that, which being increased by .|, J, and i of itself, the
sum shall be 75 ? Aus. 36.
3. A general, after sending out foraging § and ^ of his men, had yet remaining
1000 : what number had he in command ? Ans. 6000.
4. A gentleman distributed 52 pence among a number of poor people, con-
sisting of men, women, and children ; to each man he gave 6d, to each woman
4d, and to each child 2d : moreover there were twice as many women as men,
and thrice as many children as women : hov^ many were there of each ?
Ans. 2 men, 4 women, and 12 children.
5. One being asked his age, said, if 3 of the years I have lived be multiplied
by 7, and § of them be added to the product, the sum will be 219: what was
his age ? Ans. 45 years.
DOUBLE POSITION.
Double Position is the method of resolving certain questions by means of
two suppositions of false numbers.
• The reason of this rule is evident, because it is supjwsed that the rwulu »re proportional
to the suppositions.
■ '^ . . ' .
I Thus, nala;\nz:z,OT - . a . . ^ . *.
t
94 ARITHMETIC.
To the double rule of position belong such questions as have their results not
proportional to their positions : such are those, in which the number sought, oi
their parts, or their multiples, are increased or diminished by some given abso-
lute number, which is no known part of the number sought.
RULE *.
Take or assume any two convenient numbers, and proceed with each of then
separately, according to the conditions of the question, as in single position ;
and find how much each result is different from the result mentioned in the
question, calling these differences the errors, noting also whether the results an
too great or too little.
Then multiply each of the said errors by the contrary supposition, namely,
the first position by the second error, and the second position by the first error
Then, if the errors are like, divide the difference of the products by the dif-
ference of the errors, and the quotient will be the answer.
But if the er.'-ors are unlike, divide the sum of the products by the sum of th(
errors, for the answer.
Note. The errors are said to be like, when they are either both too great oi
both too little ; and unlike, when one is too great and the other too little.
I. What number is that, which being multiplied by 6, the product increased
by 18, and the sum divided by 9, the quotient should be 20 ?
• Demondr. The rule is founded on this supposition, namelv, that the first error is to th<
second, as the difference between the true and first supposed number, is to the difference be-
tween the true and second supposed n\imber ; when that is not the case, the exact answer to th(
question cannot be found by this rule. The algebraist wll know at once to what class o
questions this property belongs, when it is stated that it is only to such as rise no higher thar
a simple equation. When it is by means of oiie equation, and one unknown, the solution car
be obtained by single position : and when it involves two propositions, one of which cannot b<
mentally eliminated, we must have recourse to double position. When the conditions lead tc
a quadratic or higher equation, the condition above named, upon the hypothesis of which tlu
rule is formed, does not take place, and the solution obtained by it, cannot therefore be mon
than an approximation. The degree of approximation, and some other particulars, may be seer
discussed in a note under this head in the Algebra. That the rule is true, under this limita-
tion, may be thus proved.
Let a and h be the two suppositions, and A and B their results, produced by similar operations
also r and s their errors, or the differences between the results a and B from the true result x ;
and let x denote the number sought, answering to the true result N of the question.
Then is x — A := r, and X — B z: s, or B — A ;= r — s. And, according to the suppositior
on which the rule is founded, r^s^^x — a ', x — h\ hence, by multiplying extremes and
means, rx — rb ■= sx — sa\ then by transposition, rx — sx =. rb — sa; and, by division.
X — ^ =. the number sought, which is the rule when the results are both too little.
r — s °
If the results be both too great, so that a and B are both greater than x; then X — A = — >
and N — B 1= — $, or r and s are both negative ; hence — rj — si;' — a '. x — b, bul
— r I — * ! I -|- '■ I + Sj therefore r ; «;'. x — a ', x — b • and the rest will be exactly a<
in the former c.nse.
But if one result a only be too little, and the other B too great, or one eiror r positive, and
the other s negative, then the theorem becomes x := , ' i which is the rule in this case, oi
r -\- s '
when the errors arc unlike.
DOUBLE POSITION. 95
Suppose the two numbers 18 and 30. ITien,
First Position.
Second Position.
Pro(
18
30
27
6 mult.
G
6
108
180
162
18 add
18
18
9) 126 div.
9) 198
9) 180
14 results
22
20
20 true res.
20
+ 6 errors unlike
— 2
2d pos. 30 mult.
18 1st
pos.
Errors 52 ^^^
( 6 36
36
sum 8) 216 sura of products
27 = number sought.
Or, by single position : — the increase given to the product of the number bcinjf
18, and this as well as the said product divided by 9, will give the question imder
the following form.
Required a number, to six-ninths of which if 2 be added, the sum shall be 20;
or again, more simply, six-ninths of which is 18.
Then suppose 18 the number. Then,
g X 18 = 12, which is too little by 6.
Then 12 : 18 : : 18 : 1' = 27 Answer,
llie advantage in practice of double position is, that instead of requiring any
mental preparations similar to those above mentioned, it renders the whole pro-
cess mechanical ; and which are indeed tantamount to as many algebraical ones,
bearing in fact a great resemblance to the unsymbolic algebra of the Arabians
and Persians and Indians.
RULE II.
Find, by trial, two numbers, as near the true number as convenient, and
work with them as in the question ; marking the errors which arise from each
of them.
Multiply the difference of the two numbers assumed, or found by trial, by one
of the' errors, and divide the product by the difference of the errors, when they
are like, but by their sum when they are unlike. Or thus, by pro|)ortion : as
the difl^erence of the errors, or of the results (which is the same thing), is to the
difference of the assumed numbers, so is either of the errors, to the correction
of the assumed number belonging to that error.
Add the quotient, or correction last found, to the number belonging to the
said error, when that number is too little, but subtract it when too great, and
the result will give the true quantity sought *.
* For since, by the supposition, r ', s;',x — a', x — 6, therefore by di\ bion, r — »;*I!ft — a '.
* — 6, or as B — A ; 6 — o I '. •' I •*■ — '', for B — a is =: r — s ; which is the 2ii(i rule. Of
course the remarks upon the approximation of the first rule apply likewise to the prest-nt one.
y6 ARITHMETIC.
EXAMPLES.
1. Thus, the foregoing example, worked by this 2nd rule, will be as follows
30 positions 1 8 ; their d iff. 12
-2 errors +6; least error 2
sum of errors 8) 24 (3 subtr.
from the position 30
leaves the answer 27
Or, as 22 — 14 : 30 — 18, or as 8 : 12 :: 2 : 3 the correction, as abore.
2. A son asking his father how old he was, received this answer : your age is
now one-third of mine ; but 5 years ago, your age was only one-fourth of mine.
What then are their two ages ? Ans. 15 and 45.
3. A workman was hired for 20 days, at 3* per day, for every day he worked ;
but with this condition, that for every day he did not work, he should forfeit Is.
Now it so happened, that upon the whole he had 2/ 4* to receive : how many
of the days did he work ? Ans. 16.
4. A and B began to play together with equal sums of money : A first won
20 guineas, but afterwards lost back § of what he then had ; after which B had
four times as much as A : what sum did each begin with ?
Ans. 100 guineas.
5. Two persons, A and B, have both the same income. A saves 3 of his ; but
B, by spending 50/ per annum more than A, at the end of 4 years finds himself
lOOl in debt : what does each receive and spend per annum ?
Ans., they receive 125/ per annum; also A spends 100/, and B spends
1 50/ per annum.
PRACTICAL QUESTIONS IN ARITHMETIC.
Quest. 1. The swiftest velocity of a cannon-ball is about 2000 feet in a
second of time. Then in what time, at that rate, would such a ball move from
the earth to the sun, admitting the distance to be 100 millions of miles, and the
year to contain 365 days 6 hours ? Ans. S^^fg years.
Quest. 2. What is the ratio of the velocity of light to that of a cannon-ball,
which issues from the gun with a velocity of 1500 feet per second ; light passing
from the sun to the earth in 83 minutes ? Ans. the ratio of 704000 to 1.
Quest. 3. The slow or parade-step being 70 paces per minute, at 28 inches
each pace, it is required to determine at what rate per hour that movement is ?
Ans. 1^^ miles.
Quest. 4. The quick-time or step, in marching, being 2 paces per second,
or 120 per minute, at 28 inches each ; at what rate per hour does a troop march
on a route, and how long will they be in arriving at a garrison 20 miles distant,
allowing a halt of one hour by the way to refresh ?
. i the rate is 3-ft miles an hour.
' \ and the time 7| hr, or 7h 17^ min.
Quest. 5. A wall was to be built 700 yards long in 29 days. Now, after
12 men had been employed on it for 11 days, it was found that they had com-
PRACTICAL QUESIIONS. 97
jjleted only 220 yards of the wall. It is required to determine how many men
must he added to the former, that the whole number of them may just finiBh the
vail in the time proposed, at the same rate of working.
Ads. 4 men to be added.
Quest. 6. Determine how far 500 millions of guineas will reach, when laid
down in a straight line touching one another ; supposing each guinea to be an
inch in diameter, as it is very nearly. Ans. 7891 miles, 728 yds, 2 ft, 8 in.
Quest. 7. Two persons, A and B, being on opposite sides of a wood, which
is 536 yards about, begin to go round it, both the same way, at the same in-
stant of time ; A goes at the rate of 1 1 yards per minute, and B 34 yards in
3 minutes ; the question is, how many times will the wood be gone round before
the quicker overtake the slower ? Ans. 17 time*.
Quest. 8. A can do a piece of work alone in 12 days, and B alone in 14 ; in
what time will they both together perform a like quantity of work ?
Ans. 6'] days.
Quest. 9. A person who was possessed of a | share of a copper mine, sold \
of his interest in it for 1800/; what was the reputed value of the whole at the
same rate ? Ans. 4000/.
Quest. 10. A person after spending 20/ more than J of his yearly income,
had then remaining 30/ more than the half of it ; what was his income ?
Ans. 200/.
Quest. 1 1 . The hour and minute hands of a clock are exactly together at 12
o'clock; when are they next together ? Ans. l-pj hr or 1 hr 5^ niin.
Quest. 12. If a gentleman whose annual income is 1500/, spend 20 guineas
a week ; whether will he save or run in debt, and how much in the year ?
Ans. he saves 408/.
Quest. 13. A person bought 180 oranges at 2 a penny, and 180 more at 3 a
penny ; after which he sold them out again at 5 for 2 pence : did he gain or lose
by the bargain ? Ans. he lost 6 pence.
Quest. 14. If a quantity of provisions serves 1500 men 12 weeks, at the
rate of 20 ounces a day for each man ; how many men will the same provisions
maintain for 20 weeks, at the rate of 8 ounces a day for each man ?
Ans. 2250 men.
Quest. 15. In the latitude of London, the distance round the earth, mea-
sured on the parallel of latitude, is about 15550 miles ; now as the earth turns
round in 23 hours 56 minutes, at what rate per hour is the city of London carried
by this motion from west to east ? Ans. 6491iiJ noiles an hour.
Quest. 16. A father left his son a fortune, i of which he ran through in
8 months: i} of the remainder lasted him 12 months longer; after which he
had 820/ left : what sum did the father bequeath his son? Ans. 1913/ 6j 8d.
Quest. 17. If 1000 men, besieged in a town, with provisions for 5 weeks,
allowing each man 16 ounces a day, be reinforced with 500 men more; and
supposing that they cannot be relieved till the end of 8 weeks, how many ounces
a day must each man have, that the provision may last that time ?
Ans. 6J ounces.
Quest. 18. A younger brother received 8400/, which was just I of his elder
brother's fortune : what was the father worth at his death ? Ans. 1 9200/.
Quest. I9. A person, looking on his watch, was asked what was the time of
the day, who answered, " It is between 5 and 6 ;" but a more particular answer
being required, he said that " the hour and minute hands are exactly toge-
ther :" what was the time ? Ans. 27 A min. past 5.
VOL. I. H
98 ARITHMETIC.
Quest. 20. If 20 men can jierform a piece of work in 12 days, how many
men will accomplish another thrice as large in one-fifth of the time ? Ans. 300.
Quest. 21. A father devised ^ of his estate to one of his sons, and ^ of the
residue to another, and the surplus to his relict for life. The children's legacies
were found to be 514/ 6s 8d different: what money did he leave the widow the
use of? Ans. 1270/ Is Ql'^d.
Quest. 22. A person, making his will, gave to one child ^g of his estate, and
the rest to another. When these legacies came to be paid, the one turned out
1200/ more than the other : what did the testator die worth ? Ans. 4000/.
Quest. 23. Two persons, A and B, travel between London and Lincoln,
distant 100 miles, A from London, and B from Lincoln at the same instant.
After 7 hours they met on the road, when it appeared that A had rode 1^ miles
an hour more than B. At what rate per hour then did each of the travellers
ride ? Ans. A 755. and B 6^J miles.
Quest. 24. Two psrsons, A and B, travel between London and Exeter. A
leaves Exeter at 8 o'clock in the morning, and walks at the rate of 3 miles an
hour, without intermission ; and B sets out from London at 4 o'clock the same
evening, and walks for Exeter at the rate of 4 miles an hour constantly. Now,
supposing the distance between the two cities to be 130 miles, where will they
meet ? Ans. 69f miles from Exeter.
Quest. 25. One hundred eggs being placed on the ground, in a straight
line, at the distance of a yard from each other : how far will a person travel
who shall bring them one by one to a basket, which is placed at one yard from
the first egg ? Ans. 10100 yards, or 5 miles and 1300 yards.
Quest. 26. The clocks of Italy go on to 24 hours: how many strokes do
they strike in one complete revolution of the index ? Ans 300.
Quest. 27. One Sessa, an Indian, having invented the game of chess, showed
it to his prince, who was so delighted with it, that he promised him any reward
he should ask ; on which Sessa requested that he might be allowed one grain of
wheat for the first square on the chess-board, 2 for the second, 4 for the third,
and so on, doubling continually to 64, the whole number of squares. Now,
supposing a pint to contain 7680 of these grains, and one quarter or 8 bushels to
be worth 27s 6d, it is required to compute the value of all the corn.
Ans 6450468216285/ 17* 3d d^^Hq.
Quest. 28. A person increased his estate annually by 100/ more than the
^ part of its value at the beginning of that year ; and at the end of 4 years
found that his estate amounted to 10342/ 3s 9d : what had he at first ?
Ans. 4000/.
Quest. 29- Paid 1012/ 10s for a principal of 750/, taken in 7 years before : at
what rate per cent, per annum did I pay interest ? Ans. 5 per cent.
Quest. 30. Divide 1000/ among A, B, C ; so as to give A 120 more, and B
95 less than C. Ans A 445, B 230, C 325.
Quest. 31. A person being a.sked the hour of the day, said, the time past
noon is equal to |ths of the time till midnight : what was the time ?
Ans. 20 min. past 5.
Quest. 32. Suppose that I have {g of a ship whose whole worth is 1200/;
what part of her have I left after selling f of j of my share, and what is it worth?
Ans. tf^'j, worth 185/.
Quest. 33. Part 1200 acres of land among A, B, C ; so that B may have 100
more than A, and C 64 more than B. Ans. A 312, B 412, C 476.
Quest. 34. What number is that, from which if there be taken ^ of g. and to
the remainder be added /j of ^^5, the sum will be 10? Ans. Qj?.
PRACTICAL QUESTIONS. 99
Quest. 35. There is a number which, if multiplied by j of | of 1 j, will pro-
duce 1 : what is the square of that number? Ans. 1/..
Quest. 36. What length must be cut off a board, SJ inches broad, to contain
a square foot, or as much as 12 inches in length and 12 in breadth ?
Ans. 16^1} inche*.
Quest. 37- What sura of money will amount to 138/ 2s 6d, in 15 months, at
5 per cent, per annum simple interest ? Ans. 130/.
Quest. 38. A father divided his fortune among his three sons. A, B, C,
giving A 4 as often as B 3, and C 5 as often as B 6 ; what was the whole legacy,
supposing A's share was 4000/ ? Ans. 9500/.
Quest. 39. A young hare starts 40 yards before a grey-hound, and is not
perceived by him till she has been up 40 seconds ; she scuds away at the rate of
10 miles an hour, and the dog, on view, makes after her at the rate of 18 : how
long will the course hold, and what ground will be run over, counting from the
outsetting of the dog ? Ans. 60;^ sec. and 530 yards run.
Quest. 40. Two young gentlemen, without private fortune, obtain commis-
sions at the same time, and at the age of 18. One thoughtlessly spends 10/ a
year more than his pay ; but, shocked at the idea of not paying his debts, gives
his creditor a bond for the money, at the end of every year, and also insures
his life for the amount ; each bond costs him 30 shillings, besides the lawful
interest of 5 per cent, and to insure his life costs him 6 per cent.
The other, having a proper pride, is determined never to run in debt ; and,
that he may assist a friend in need, perseveres in saving 10/ every year, for
which he obtains an interest of 5 per cent, which interest is every year added to
his savings, and laid out, so as to answer the effect of compound interest.
Suppose these two officers to meet at the age of 50, when each receives from
Government 400/ per annum ; that the one, seeing his past errors, is resolved
in future to spend no more than he actually has, after paying the interest for
.■what he owes, and the insurance on his life.
The other, having now something beforehand, means in future to spend his
full income, without increasing his stock.
It is desirable to know how much each has to spend per annum, and what
money the latter has by him to assist the distressed, or leave to those who
deserve it ?
Ans. The reformed officer has to spend 66/ IQs id 2-65839 per annum.
The prudent officer has to spend 437/ 12s lie/ 3-44515' per annum, and
The latter has saved, to dispose of, 752/ 19s 9-2256fl/.
ALGEBRA.
I. Introductory explanation of the character and objects of this branch of
Mathematics.
1. If a series of given numbers be directed to be combined in any specified
manner, tbat specification may eitber be expressed in words at length, or by
means of the usual s)Tnbols of arithmetical operations, and such other con-
trivances as have been already explained in the treatise on arithmetic. When
the symbolic or abbreviated mode of expression is employed, the collection of
numbers and symbols constitute what is called an arithmetical expression. Thus,
if from the sum of six and seven we were directed to take three, and multiply
the remainder by one-half the defect of six from ten, then the arithmetical
expression for this would be
(6+7-3)x'-^^
all the sjnnbols and modes of writing employed in this expression having been
already defined and rendered familiar (page 6).
2. If we actually perform the several operations here indicated or directed, we
shall obtain what is called the value of the expression, which in the present case
is 20, and in each case is dependent upon the arithmetical conditions of the
given expression.
When we express that 20 is the equivalent or value of such an expression, we
form an arithmetical equation, viz.
(6 + 7-3) X ^-^^=20.
3. In the solution of any arithmetical question, we are enabled, for the most
part, with a little consideration, to refer to a class for solving which rules have
been already invented. These rules consist in the substitution of a series of
arithmetical operations of a simple kind, to be performed in a specified order
upon the several numbers given in the conditions of the question. Thus, in
questions which are reducible to " the Rule of Three terms," or simply the
" Rule of Three," the answer is found by arranging the given terms in a
particular and specified order, and then dividing the product of the second and
third by the first of the terms so arranged. Now as this rule is the same what-
ever the first, second, and third terms may happen to be, it could not be ex-
pressed without some symbols to stand for those terms, which, whilst expressing
the fact of their being numbers so arranged, would yet not confine them to any
particular values as numbers, nor to any particular class as objects.
The letters of the Alphabet have been used for this purpose throughout
Europe, and those regions which have received their science from Europeans,
without a single exception. Sometimes they have been so chosen as to be the
initial letter of the kind of quantity whose numbers they stood in place of, as
t, s, V, for time, space, and velocity : but they have been generally so selected,
that the earlier letters of the alphabet, a, b,c, ... should stand in the place of the
numbers which in every actual question are given, or express the given conditions j
INTRODUCTION.
101
whilst the final letters, z, y, x,w,v, ... have been employed to occupy the
place in the immediate expression of the question, of those numbers which till
the rule has been put into execution are unknovm, and which there/ore it is the object
of the problem to discover. Thus, in the Rule of Three, if a, b, c taken in order
be selected to designate generally the first, second, and third of the given terms
taken in order, and which are all given in each actual case, and if the answer, as
yet unknown, be denoted by x, we shall have the condition, when simplified from
all extraneous considerations, expressed thus :
a : b ', [ c : X,
and the formula or rule for solution would take this form :
bye
X = .
a
4. So likewise, if in the case of the numbers 10, 6, 7, 3, which occur in the
arithmetical formula in the beginning of this chapter, these numbers had been
the particular values of four quantities which in the solution of some specific
class of questions were always given amongst its conditions, and that by some
means or other, the rule for solution had been discovered to be, that the third
number (taking them in the order of their occurrence in the question, or of
some arrangement to which the rule always subjected them) must be taken from
the sum of the second and fourth, and the remainder multiplied by half the
defect of the second from the first : then, writing as the symbols of the first,
second, third, and fourth terms of the equation in art 2 . the letters a, b, c, d,
and for the yet unknown number, that is, the answer sought, writing the symbol
X, the expression of this rule in an algebraic formula will be
{b -\- d — c) X ^—^ — X.
5. To take another example, suppose it were proposed to find that number to
the square of which we add the number b, then the sum shall be equal to a times
the number itself; then the formula of solution would be
a. J_a_x^ _
2 ^ V 4 '
and X ^ - — / — b*.
2 V *
The problem itself would take the following form :
xxx + b=:axx:
and it is very obvious, at first sight, that the method of solution falls under no
rule that has been given in the treatise on arithmetic ; and therefore such rule
must be sought for by some new method of investigation, either analogous to
those by which the arithmetical rules were discovered, or possibly by some
process altogether diflferent from them in principle as well as in plan.
• Tlie studeut may be surprised to see that two different methods of calculating the rc«ult
have been giren ; but he will see the reason of this hereafter, and also that often three, four,
or indeed any number whatever of answers to different questions are possible. To take an
instance, the values 4 and 3, as those of a and b, would find x = 2 -\- .^ l7and t^'2 — ^ 1,
or 3 and 1. He can verify these by actual trial.
In trying other numbers he may hit upon some conditions that will not admit of any •olution
whatever; as, if a = 2 and b = 2, the solutions would be J- = 1 + /s/ —17 and jr = 1 — ^/ — 1 ,
results which he is not in possession of methods capable of interpreting. Such c*»et are said to
be impossible, and their explanation will be found in a note on "Quadratic Equations" in the
present volume.
102 ALGEBRA.
6. The investigation of such rules for calculation is one of the two objects of
Algebra.
The other object, which is subservient to the former, is the discovery of
the different operations which may be performed with the same given numbers,
and shall produce the same ultimate numerical results as any given operations
different from these shall produce, without regard to what those numbers may
chance to be. Thus, if the square of the sum of a and b were sought in another
form, it may be exhibited thus axa + 2xax6 + 6x6. And the state-
ment of this fact is thus written :
(a + ft) X (« + 6) = (a X a) + (2 X a X 6) + (A X b.)
Shorter modes of writing it will be exhibited presently; but here the simple
symbols used in the arithmetic have been alone employed, for the purpose of
showing the nature of algebraic notation in its earliest forms, and to illustrate
the objects for which it was devised.
7. The discovery of formulae for the solution of questions constitutes the
algebraical problem ; and the discovery of formulae of transformation, or of those
which give equivalent results independently of the particular value of the quan-
tities which enter into their composition, constitute the algebraical theorem.
8. The motives which gave rise to the use of alphabetic letters as symbols of
number in preference to any other system of symbols, arbitrarily selected for the
same purpose, are principally the following. First, As they have no numerical
signification in themselves, they are subject to no ambiguity, having in reference
to numbers no other signification than they are defined to have in the outset of
each problem, or either defined, or understood from general practice, to have in
each theorem. Secondly, Being familiar to the eye, the tongue, the hand, and the
mind, that is, having a well-known form and name, they are easily read, written,
spoken, remembered, and discriminated from one another, which could not be
the case were they mere arbitrary marks, formed according to the caprice of each
individual who used them, and always different, as in such case they must
almost of necessity be, at each different time that the same person required to
use them. Thirdly, The order in which the letters are arranged in the alphabet,
facilitates the classification of them into groups much more easy to survey and
comprehend in the expressions which arise from the performance of any assigned
operations, and thereby renders the investigator much less likely to omit any
of them by an imperfect enumeration, than if they were composed of marks that
were used for that purpose only, and selected for each individual occasion from
the various combinations that could be formed of such simple linear elements
as the hand could readily trace, and the eye readily distinguish from all other
combinations.
II. Definitions, Notation, and Fundamental Principles.
The principal symbols which are employed to designate the operations of
algebra and arithmetic, and the relations which subsist between quantities, are
the following. Their object is to abbreviate.
1. 1 . + signifies addition, and is read plus. Thus 2+3 or a + b + c
respectively signify that 3 is to be added to 2, and that b is to be added to a,
and that then c is to be added to the sum of a and b.
A quantity to which the symbol + is prefixed, is called a positive or affirmative
quantity.
2. — signifies subtraction, and is read minus. Thus, 3 — 1, or 6 — a, signify
DEFINITIONS, NOTATION, AND PRINCIPLES. 103
respectively that 1 is to be subtracted from 3, and a from b. The number to b«
subtracted is always placed after the symbol.
A quantity to which the sign — is prefi.xed is called a negative quantity •.
3. cr- signifies the difference of the quantities between which it is placed ; and
is used either when it is not known or is not necessary to specify which is the
greater of them. In this case a cr> b, or b ^ a, si;/;nify the same thing.
4. X is the symbol of multiplication, and is placed between the factors which
are to be multiplied together. Sometimes a point . (placed at the lower part of
the line, to distinguish it from the decimal jjoint, which is placed at the upper
part of the line,) is employed for the same purpose, and especially between the
numerical factors, as 3 . 5 . 7, or 1 . 2 . 3 . 4 : and in the case of simple
literal factors, the practice is now almost universal to drop all marks between
the simple factors, and write them in consecutive juxta-position. Thus a x b
X c X X, or a . If . c . X, or abcx designate the same thing, viz. the continued
product of the numbers which a, b, c, i.nd x are put to represent f.
When one of the factors is a number, it is called a coefficient : thus in 2 x a
X & or 2ab, the 2 is called the coefficient of ab, and in 53^:^;, 53 is called the
coefficient of xyz. When no coefficient is written, 1 is understood to be meant,
the quantity being taken once.
Also, in some cases where letters are put for numbers, the letters which re-
present given or known numbers are likewise called coefficients ; as in 3 axz, 3a
is called the coefficient of xz. In the case of a number being actually given, the
coefficient is said to be a numeral coefficient ; but when it is given in literal sym-
bols, it is called a literal coefficient.
Moreover it may be remarked that cases of algebraical investigation sometimes
present themselves in which even the symbols of the unknown quantities are
conveniently considered as coefficients : but these will be pointed out when they
arise.
Though, as is proved in the note, the order of the factors in multiplication, so
• Quantities affected with the signs -f and -f or — and — , are said to have like signs ; rad
those affected with — and -j-, or -j- and — , are said to have unlilce siffns.
It is manifest from the nature of addition and subtraction, that the disposition of the quantitiea
as to order is immateiial ; for a -f 6, or 6 -j- «. '» ^^^ **™c thing, and a -\- b — c,a — e + It,
or ~~ c -\- b -\- a, express the same quantity, only under a different arrangement, as to relative
position.
+ It may be readily shown that it is immaterial in what order the factors are Uken for the
pnrpose of multiplication : that is, which is made the multiplicand and which the multiplier.
For if a number of dots (or units) be placed horizontally equal in number to the units in the
fector selected as the multiplicand, and tliis be repeated under this horizontal bend till there are
as many bands as there are units in the other factor : then the same number of dots considered
as forming vertical columns will be constituted of as many timet the number there is in the
multiplier as there are uniu in the multiplicand, and representing therefore the result of a luul-
tiplicand with the order of the fi»ctor« inverted. Thus if we take four time* three, the dott will
stand
• ••
and if we turn the column which is vertical into a horizontal position, it bcconea
• •••
• ••■
• •••
And in the same way it is shown of m and n as factors.
104 ALGEBRA.
far as the value of the product is concerned, is immaterial ; yet in the disposition
of them as algebraic symbols, it is found convenient generally to arrange them
in alphabetic order. Thus the quantity abcxy is the same in point of value as
bcxya, or any other arrangement that can be made of them, still the form abcxy
is preferable, for many reasons, to any other that can be given to the same
quantity.
5. -T- is the symbol of division. Sometimes the dividend is put before and
the divisor after the mark, and sometimes they are placed respectively above and
below the line in the place of the two dots, after the manner of an arithmetical
fraction. Thus a -j- J or - alike signify that the number a is to be divided by
b
the number b.
6. = is the symbol for the words " is equal to," and is generally read
" equals." It is used to signify that the value of the aggregate of the terms
which precede it is equal to the value of the aggregate of those which follow it.
The whole expression is called an equation, and the quantities which stand to
the left of the symbol are said to constitute the first side of the equation, and
in like manner those which stand on the other constitute the second side of the
equation. Thus ax + b = cxx -\- dx — e is an equation, the first side of which
is ax -\- b, and the second side cxx -\- dx — e.
7. The symbols >, <, and sometimes :^, are used to express the inequality
of the quantities between which they are placed. The opening of the symbol
is always turned towards the greater quantity. Thus a > b signifies, that a is
greater than b ; and/ < g signifies, that/ is less than g.
When it either is not known, or is not necessary to state which is the greater
of the two quantities, which are nevertheless unequal, the symbol zf: is used.
8. : signifies that it is the ratio of the two quantities between which it stands
which is the subject of consideration. Thus x : y designates the ratio of x to y.
It is read x to y, or the ratio of x to y.
When two ratios are equal ; that is, when some two numbers have the same
ratio that two others have, it is expressed in one of these two ways : —
X '.y '.: w. V ox X '. y =^u \ V
The former is most usual, and adhered to in this Course. The phrases which
they represent are
a? is to y as « is to t? :
or, X has the same ratio to y that u has to r ;
or, the ratio of x to y, is the same as that of « to r.
or, again, the ratio of a; to y is equal to that of « to v.
9. It is often found necessary to class the quantities of which an expression
is composed, into sets combined in some particular way. This classification is
effected by enclosing them under a horizontal bar (called a vinculum), or between
parentheses, or braces, or brackets. This is always done when the actual opera-
tions indicated are to be clianged for others which shall produce equivalent
results, and which are more easy to perform than those which were originally
indicated. Nearly the whole of algebra consists in discovering these equivalent
iterations.
1 hus, referring to the example given in the Introductory Chapter, we might
have put it in the more complicated but equivalent form,
6x10+7x10 — 3x 10 — 6x6 — 6x7 + 6x3:
2
but by taking the form there given, the actual labour of computation is not
above a fourth of that which would attend upon all these. Or, again, in general
DEFINITIONS. NOTATIOxN, AND PRINCIPLES. 105
symbols, the expression {a + 2b — 3c) (4a — 2b + 3c) indicates that the sum
of a and twice b being taken, and three times c being subtracted from the sum,
and that to the difference of four times a and twice 6, three times c is added
then the product of these two results is to be taken, 'lliis effect might be pro-
duced by other means much more complicated, but which are avoided by the
contrivance indicated above.
When the terras which are collected together are also employed to form the
numerator and denominator of an expression in a fractional form, no ambiguity
can arise, except the said numerator and denominator are composed of vincu-
lated factors, in dropping the vinculum. Thus ^^t*^ may be written simply
(a — 6) *^ '
g + ^
a — b
10. When several of the factors which compose a quantity are equal to one
another, a considerable abbreviation of the trouble of writing, and of the space
occupied, has been devised, by simply writing the common factor in its place,
and a small figure above it and to the right hand, expressing how many times
that factor occurs in the product. Thus, instead of axaxaxa or aaaa, it
is usual to write a*, the a signifying the common factor, and the 4 the number
of times of its occurrence. So likewise, instead of the expression Zaaabbxxxz,
is written 3a'6V ; and (a -\- x) {a -\- x) (a + x) is written (a + xf : and so on
for any number or form of the component equal factors of an expression.
The number affixed is called the index or exponent, and the quantity is said to
be raised to the power denoted by that exponent. Thus a* signifies the fourth
power of a, and 4 is called the index or exponent of the power of a : and a" is
called the nth power of a, and n is the exponent of that power.
In conformity with this notation, when no index is annexed to a quantity, 1
is understood : thus, by a, the first power of a is meant, that is, a'.
It will obviously follow, that when two different powers of the same quantity
are to be multiplied together, the symbolical result may be written as a power
whose exponent is the sum of the exponents of the several factors. Thus
a* X a' X a' X a^ = a'*+"'''^+^ = 0+'" ; for it is the same thing as aaaa xaaxax aaa,
that is, aaaaaaaaaa or a'". On the same principle a^b^c multiplied by a'6c* may
be more briefly written a'+'fe^'c "^^j or still more briefly a*6^c*.
Similarly, in general a" x a" x a'... = a"+"'^ •, -where the dots after the
quantities express the continuation of the same class of quantities to which they
are respectively annexed to any assignable extent.
It moreover follows, that the quotient of one power of a quantity by another is
symbolically expressible by a power of that quantity whose exponent is equal to
that denoted by the remainder left after subtracting the exponent of the divisor
from that of the dividend. For since multiplication increases the number of
a«
factors, division will decrease them. Thus, since o* x o'= a*, so also = a*;
and in like manner generally — = a"~"
The two modes of notation just employed to designate the division of one
power of a quantity by another power of the same quantity may be used indif-
ferently. By remarking, however, the corresponding forms in some par-
ticular cases, we shall be led to some simple but important conclusions. And
first.
When m = n, we have - = 1, and o"— = a% which are of course iden-
a
106 ALGEBRA.
tical in signification. Hence we learn that a° always signifies 1, whatever a may
be. Indeed a" signifying 1 a", if m = o we have 1 a° = 1, or 1 x a° := 1, or
1 time a taken of no power * whatever. The result is therefore perfectly con-
sistent with first principles and the adopted notations.
When m < n, then if we put m zzzn — r, we have =: a"~"~'' = a""""
a"
= a~\ But this also signifies, when n = m + r f , the following expression,
— x^ = =: -. Hence, a"' = —-. We learn from this, that when in
any assigned series of operations upon the powers of a quantity, we arrive at
an index of the form — r, then the expression signifies the reciprocal of the rth
1 2
power of the factor a. Thus o "^ signifies —^ , and 2 x 10 ~ ^ signifies — -^ or
2
, or '0002, and so on.
10000
11. As we have occasion to calculate ihci powers of given numbers, considered
as roots, so we have often to find the roots of given numbers considered as
powers. The operations are considered as the inverse of each, and are denoted
by inverse notations.
Thus to cube the quantity aa or a? we have aa x aa x aa or a^ . a- . a^ or a^ ;
so likewise to find the cube root of a* we have to separate it into three equal
factors. This operation, in all such cases, is indicated by dividing the exponent
6
of the power by the exponent of the root : thus the cube root of a^ is a^, where
6 is the exponent of the given quantity, and 3 the exponent of the root. In the
same manner the square root of a ^b* c* d~*, is a* b^ c- d *, or a^ b"^ c d J. If
there be a numerical coefiicient, the root of that is either actually extracted or
i
indicated like the rest ; as in the square root of 25a'' we may either put 2o^a^
or 5o*. Most commonly the numerical root is actually extracted when the given
number admits of an accurate root, but indicated when the value cannot be
assigned in a finite form.
It will be obvious from the signification given above to this notation, that the
numerator of a fractional exponent expresses the power to which a root is raised,
and the denominator the root which is to be taken of that result.
It also follows from the nature and relation of roots and powers, that it is
immaterial whether we first extract the indicated root, and then raise it to the
indicated power, or conversely, we first raise the indicated power, and then
• The term dimension is oficn employed instead of the word potcer. It is derived from the
analogy which the dimensions of line, square, and cuhe in geometry, when they are expressed
numerically, have to the first, second, and third powers. Beyond the third power, geometry
furnishes nothing analogous to the powers of quantities ; and hence the terms fourth, fifth, &c.
dimensions though generally used, are hardly accurate.
+ This is the same as the former, m =: w — r, having r added to both sides ; and hence
m -\- r =. ti — r -\- r'=. n.
X When tlie index of the power is not divisible by the index of the root, the fractional form
3 7
of the index wiiich results is retained ; as in the cube root of a* hi, we write it a' h^ ; or some-
times the given quantity is conceived to be put under the form a* b^ t)\ or simply, a* i* A, and
the root written a* b^ /A The ultimate purpose of the particular inquiry in question is the only
guide to the Algebraist in his choice amongst these notations.
DEFINITIONS, NOTATION. AND PRINCIPLES. 107
extract the indicated root. In actual numerical practice, however, it i8 always
best to e.Ttract the root first, when the number admits of an exact root : but
when it does not admit of an exact one, it is better to raise the indicated power
first, and then extract the indicated root. The reason is, that the number of
figures necessary is always less than would be required in taking the contrary
course.
Respecting the verbal enunciation of these expressions, it should be remarked
that
a* is generally read the third root of a, or the one-third power of a.
V a is always read the third root of a.
a' is read the two-thirds power of a, or the cube root of a', or the sauare of
the cube root of a,
and so on with other expressions.
It ought also to be understood that the fractional indices may be, and often
are, converted into decimals; as for a^ may be written a^, for a~* may be
— 14 \
written either a , or — .
a'*
12. Another method of indicating the extraction of roots is by the symbol v/~
prefixed (being only a modification of the form of the old manuscript r, the initial
letter of the word radix) ; and the order of the root expressed by a small num<
ber written over it, and lying a little to the left: as '{/a, \/b, \/(cr+ x),
\/a^ — b'^ ; which signify respectively the same thing as a\ b\ (a + x)^,
(a^— b^)' , or in words, the square (or second) root of a, the cube (or third)
root of b, the fourth root oi a •\- x, and the nth root of a* — b^.
'ITie distinction of rational and irrational, in respect to quantities whose roots
can or cannot be respectively extracted, is a very convenient one. An irrational
quantity is often called a surd. Rational quantities are sometimes put under
an irrational form, to facilitate their combination with irrational quantities into
one expression,
13. Quantities receive different designations which are found useful in alge-
braical enunciations, according to peculiar circumstances. The following are the
principal ones :
A simple quantity * is that which consists of a single terra, or of several fac-
tors only, each of which is a single term. As 3a, or bab, or Qa^b^c^, or 3a~*. It
is often called a monome or a monomial quantity.
A compound quantity is composed of the aggregation of two or more simple
quantities connected together by addition or subtraction : as a + 6, 2a — 3e, or
a -\- 2b — 3e.
Of compound quantities, that which is composed of two terms, as x -|- y,
xy + ab, or x^ — y"f, or 3x + ^f^by is called a binomial quantity ; when three
terms, or -\- ax -{- b, it is called a trinomial quantity ; when four or more terms,
z polynomial quantity , or simply 2l polynomial.
When the binomial, trinomial, or polynomial expressions are so related that
their terms counterbalance or mutually destroy one another in the a^regate, the
* The term expression is often applied to any combination of algebraical quantities.
+ The teim residual was formerly applied to that form of the binomial exprcMion when they
were connected by the sign — , as a — 6, .r' — t/^. Tlie distinction is now, however, fallen
greatly into disuse.
108 ' ALGEBRA.
total expression is put equal to zero, and the expression in this form is called
a binomial equation, a trinomial ecjuation, or a polynomial equation. Sometimes
the word equation is understood, and the terms binome or binomial, trinome or
trinomial, polynome or polynomial, are used instead of the compound phrase.
14. When any general numerical relation is exhibited in the form of an equa-
tion, the expression is called 2i formula.
15. When the numerator and denominator of any quantity are interchanged,
the resulting expression is called the reciprocal of the former. If the quantity
be in the integer form, it may be put in a fractional form, and its denominator
is understood to be unity or 1. In that case the result or reciprocal is 1 divided
by the given quantity. Thus j- is the reciprocal of — , and — ris the
reciprocal of cd.
16. There is also a distinction to be made between the notation of factors
when writing algebraically and numerically. It was explained in the third
definition, that when algebraical symbols of factors were written, they were
generally put in juxtaposition, without any mark between them. In arithmetic,
the figures placed in juxtaposition have each a relative value derived from the
particular position they occupy, which is generally called the local value, and
are thereby rendered units, tens, hundreds, &c. or tenths, hundredths, &c. To
represent the compound number, then, when the component figures are supplied
in literal notation, each one must have for its coefficient such a power of 10 as
will raise or lower it to its proper locality in the decimal scale. Thus to repre-
sent 5305-2906 without employing the artifice of local value, we must write it
thus: 5 X 103 + 3 X 102 + 0 X 101 -H 5 X 10" + 2 X ^q-' -|- 9 X lO'^ +
0 X 10-3 + 6 X 10-*.
Or, to designate a number whose digits to the left of the decimal point were
X, y, z, and the right of it were t and u, we should have
lO^a; + lO'y + lO^z -{- IQ-H + lO-^u.
17. An algebraical expression is said to be ranged according to the powers (or
dimensions) of some quantity, when the term containing the highest power of
that letter is placed first, the term containing the next power next to it in order,
and so on to the lowest. Sometimes also it is so ranged as to begin with the
lowest and ascend to the higher, in order. In expressions containing a finite
number of terms the former plan is most commonly adopted ; but in expressions
where the terms run out to infinity (as it will appear in the course of the work
is often the case), the latter is necessarily employed. Thus, in the expression
a^ + Zxf^ — 9a? + 10 — 8a?"' + 3a;"^ the arrangement is according to the powers
of X, and so also is 2x~^ — 8a?-' +10 — 9a? + Sa;^ -|- a^, so arranged. Tlie former
in a descending series of powers, the latter in an ascending series of powers. In
a?' -f- 2a?y + y^ they are ranged according to the descending powers of x, and in
y* + 2a?y + x^ in descending powers of y. Of the infinite series of ascending
powers the following is an example —
I — 2x + 2a^ — 2x' + ad inf.
and of descending powers, the following —
c c^ c^
1 — — + ^+—3 +.-, or 1 — car-' + c^x-^ — C'X'^ + ...
EXERCISES IN NOTATION AND ITS INTERPRETATION. 109
III. Exercises in Notalion and its interpretation.
I. Write in words the signification of the expressions x' — 10* = Ug •
y^ — X- = 40 i X Vy — ^ab = c^ i {a + X — y)'' X I a — X + ^y
s/x' + y- ± V:^ — y^ i a" ± a~'; and x» — y* = 0.
II. Write in the common numerical forms the quantities 10* x 3 + H)" x 7
+ 10-^ X 2; and form an expression whose digits are decimals, beginning
at the third place from the unit's place, are z, o, y, o, x, o o t.
III. How are the following expressions to be interpreted verbally ?
(a^ ) ; (a r ; V (a — *) = (a — bx)^ ; 3a — 66 = — 4a + 1 56 ;
— a—b = — i,a->rh);6{ax — by) = &ax—5by; y'a'+b ; aF;
(42)" ; 4" z" ; {abcf, a^b"ccc, {abfab(r. And point out any which are
identical in value though different in their forms.
IV. Put into algebraical symbols the following statements :
1. There is a certain number at present unknown, to the square root of which
if we add its square we shall obtain 18 : and another whose half exceeds its
third part by three-tenths of an unit.
2. Of three unknown numbers, the sum of the first two is equal to 5, the
difference of the second and third is 1, and the sum of all three is 9.
3. Add three given numbers together, and indicate the cube of the square
root of their sum, multiplied by the product of the three numbers themselves.
4. Express that three times the square of a certain unknown quantity is equal
to half the product last mentioned.
5. The sum of two unknown numbers is equal to the cube of the square root
of their product; and their difference is equal to 10.
6. The number 6 is subtracted from 5, and the square root of the remainder
is taken from the cube root of their sum ; how is this to be expressed algebraically,
and how in the condensed form of arithmetical notation ?
7. Express the cube of the cube root of a known quantity, and the cube root
of the cube of the same quantity.
8. Express that the sum of a geometrical series whose first term, ratio, and
number of terms at present unknown is to be computed : — by both methods of
notation.
9. The three leading terms of a proportion are always supposed given :
express that the fourth (yet unknown) is equal to the product of the second and
third divided by the first. And express that the square roots of four given
quantities are reciprocally proportional : and also that the reciprocals of the
squares of the fifth roots of four other quantities are directly projwrtional.
10. Of three given quantities express the sum of each two diminished by the
remaining one : and the product of the three retsulting quantities.
11. Denote that the difference of the cubes of two unknown quantities divided
by the difference of the quantities themselves, is equal to the sum of the squares
added to the product of those two quantities.
12. Express the theorem that if the sum of any number of quantities (first
supposed given and then supposed unknown) be multiplied by another given
quantity, the product is equal to the sum of the products made by multiplying
each of the first-named (whether given or unknown) quantities by the one last-
named.
no ALGEBRA.
13. The following formulae are to be explained in words
(g + ft-c)'; (2-4 + 6)^ = 2; Vx" + 2xy + y- =z ±(x+y)
In what no these last expressions differ ?
14. How is the following theorem to be expressed : — the product of two
known powers of an unknown quantity is equal to that power of the same quan-
tity which is equal to the sum of the two known powers ? Express the same
when all the quantities are known, when all are unknown, and when two
unknown powers of a known quantity are substituted in the theorem.
15. Suppose that in the conditions of some particular example that was pro-
posed, the known indices were found to be 3 and 2, and from some additional
conditions it was otherwise found that the base a was '001, what would be the
value of the expression ? And what if the given powers were 3 and "2 ?
16. Indicate the extraction of the 10th root of the 3d root of , and of
10^ X ffl
.^ , , both by radicals and indices.
17. Express the product of the sum of the square roots of three given quan-
tities into the negative square of an unknown quantity being equal to the
product of all the quantities mentioned.
5 s
18. Interpret the expression a : x',',P : c^; and admitting the rules given
for the " rule of three" to be true, how is the solution to be expressed in letters?
and, also, if a = 10, i = '008, and c ^ 27000, assign the value of x.
19- A certain number is unknown : but it is known that if three times its
defect from 10 be divided by 2, the quotient will be equal to one-third of double
the square of the number multiplied by its square root. Express this statement
in appropriate symbols.
20. What kind of symbol is V ? ^Vhat kind is ( )' ? Point out and dis-
tinguish the symbols of operation from those of quantity in the expression
{\/{a + by — (a — b)']l X 4 ccZr -r (a + 6)i (a — b)h = x^.
21. If the ^^ures which compose a number were s, r, q ...d, c, b, a, reckoning
a that to the right hand, and suppose the decimal point fall between e and d :
How is that number to be represented algebraically ?
V. Range — x^ — 10a? + 15iP — 8x* — 3xf + 4x-' — 6x~^, according to powers
of X, both ascending and descending; and 4x^y — 3xy^ -|- 9x^y^ -j- 4x* — 4y^,
according to ascending and descending powers first of x and then of y.
Arrange x^yz -f xy-z^ -f x*z + x*z — 3x-y^z + Ax^yz -f 9^y°z^ — ]Oa^y'' +
3z*x + z*y, according to powers of z, and the several multipliers in terms of x
and y collected in vincula, and each of these arranged respectively in ascending
terms of y, and in ascending terms of x.
The intelligent teacher can select a few, (such as may suit his purpose, and the
defects he observes in his pupils) from the questions in the application of simple
and quadratic equations : by which means the nature of algebraical notation will
be completely illustrated.
EXAMPLES FOR PBACTICB.
In finding the numeral values of various expressions, or combinations, of quantities.
Supposing a = 6, 6 = 5, c = 4, d = 1, and e = 0. Then will
1. a- -h 3a6 — c2 = 36 + 90 — 16 = 1 10.
DEFINITIONS, NOTATION. &c. 1 H
2. And 2a' — Za'^b + c^ = 432 — 540 + 64 = — 44.
3. And a- X (a + b) — 2abc = 36 x 1 1 — 240 = 1 56.
^•^"^ ^ + '^'=1^-+^^ = 12 + 16 = 28.
5. And x/2ac -j- c^ or (2ae + c^)^ = ^Z 64 = 8.
6. And Vc H ; — s,= 2 + =7.
. , a" — -v/6'' ~ «<? _ 36 — 1 _ 35 _
2a- a/62 ^-^c — 12 -"7 ~ 5 ~ ■
8. And Vb'^ — ac + \/2ac + c- = I +8 = 9.
9. And v^i^ — ac + v'2ac + e- = ^/25 — 24 + 8 = 3.
10. And a^'fi + c — d = 183 ; and 9ab — lOb^ + c = 24.
,, ... . a^b J _ J a -}- b b , , a +b n — b
11. Likewise x ti = 45 ; and — x , = 131; and - =
c c d * c d
1| ; H e = 45 ; — x e = 0.
* c c
2a; + 1 7x
12. Show which of the two quantities - — and — is the greater, when x
o 2
is used to signify the numbers — •001, — '1, — "lO, — 100; + -001, + -1,
+ 10, and + 100, successively.
13. Also keeping the old values of a, b, c, d, e, show that {b — c) x (.d—e)
= 1 ; (a + 6) — (c — d) = 8 ; and (a + 6) — c — rf = 6.
14. Also that a'^c x d^ = 144 ; acd — d = 23 ; a-e + b^e + d = I ; and
b — e a + b
d—e c — d •*
15. And that ^/a^ + b^ — sja^ — V^ = 4*4936249 ; 3oc=' + \/cc' — 6» =
292 497942 ; and 40^ — 3a Va^ — §a6 = 72.
16. Suppose a = 6xlO^;5=5x 10'; c = 4 X 10*; d = 8 X 10' ; and
e =; 1 : what will be the values of the expressions in e.xaniples 11, 12, 13, and
14 ?
17. If we have in any algebraical problem reason to know that a = 0, 6 = 6,
c = -1, rf = 001, and e = ^c, show what the values of the expressions 1, 2, 3,
4, 5, and 6, would then be.
18. If a = -6 ; 6 = -5 ; c = 4 ; rf = 1 ; e = 0 ; show what the results of
the first eleven expressions would be. Likewise when a, b, c, d, e, were half
these last-named values, and also four-fifths of them ; three times as much, and
ten thousand times as much ; and write down these results adapted to each of
these cases.
19. Find the values of the expressions in 11, 12, 13, when a = J ; A = J ;
c = 4 ; rf = f ; and e = i.
20. Let a = 10, what are the values of 3 x a*, 2 X a', 6 x a', 3 x a"', and
9 X a~'? Also assign their sum.
112 ALGEBRA.
ADDITION.
Addition, in Algebra, is the connecting the quantities together by their
proper signs, and incorporating or uniting into one term or sum, such as are
similar, and can be united. As 3fl + 26 — 2a -^ a -\- 2b, the sum.
The rule of addition, in algebra, may be divided into three cases :
(1.) When the quantities are like, and have like signs ;
(2.) When the quantities are like, and have unlike signs :
(3.) When the quantities are unlike*.
CASE I.
When the quantities are like, and have like signs.
Add the co-efficients together, and set down the sum ; after which set the
common letter or letters of the like quantities, and prefix the common sign -|-
or — .
Thus 3a added to 5 a, makes 8a.
And — 2ab added to — 7ab^ makes — Qab.
And 5a + 7b added to 7a + 3b, makes 12a + 106.
* The rcisons on which these operations are founded will readily appear, by a little reflection
on the nature of the quantities to be added or collected together; for, with regard to the first
example, where the quantities are 3a and 5a, whatever a represents in the one term, it will
represent the same thing in the other ; so that 3 times any thing and 5 times the same thine,
collected together, will make 8 times that thing. Thus, if a denote a shilling ; then 3a is
3 shillings, and 5a is 5 shillings, and their sum 8 shillings. In like manner, — 2ah and lab,
or — 2 times any thing, and — 7 times the same thing, make — 9 times that thing.
As to the second case, in which the quantities are like, but the signs unlike ; the reason of
its operation will easily appear, by reflecting, that addition means only the uniting of quantities
together by means of the arithmetical operations denoted by their signs -\- and — , or of addition
and sul)traction ; which, being of contrary or opposite natures, the one co-efficient must be
subtracted from the other, to obtain the incorporated or united mass.
As to the third case, where the quantities are unlike, it is pLiin that such quantities cannot
be united into one, or otherwise added, than by means of their signs : thus, for example, if o be
supposed to represent a crown, and b a shilling ; then the sum of a and b can be neither 2a nor
2/i, that is, neither 2 crowns nor 2 shillings, but only 1 crown plus 1 shilling, that is, a -\- h.
In this rule, the word addition is not very properly used ; being much too limited to express
the operation here performed. The business of this operation is to incor()orate into one mass
or .algebraic expression, different algebraic quantities, as far as an actual incorporation or union
is possible ; and to retain the algebraic marks for doing it, in cases where the former is not
possible. When we have several quantities, some aflBrmative and some negative ; and the
relation of these quantities can in the whole or in p.irt be discovered ; such incorporation of two
or more quantities into one, is plainly effected by the foregoing rules.
It may seem a paradox, that what is called addition in algebra should sometimes mean
addition, sometimes subtr.iction, and sometimes both. But the paradox wholly arises from the
scantiness of the name given to the algebraic process; from employing an old term in a new and
more enlarged sense. Instead of addition, call it incorporation, or union, or strikinff a balance,
or give it any name to which a more extensive idea may be annexed, than that which is usually
implied by the word addition, and the paradox will vanish.
1.
3a
5a
12a
a
2a
ADDITION
•
OTHER EXAMPLES
FOR
PRACTICE.
2.
3.
— 36x
bxy
~bbx
2bxy
— 46a;
bbxy
— 26x
bxy
—7bx
Zbxy
— bx
dbxy
US
4.
3x
9« — 5bx 26a?y iz
2x
3x
z
4x
32a —226a;
5. 6.
2aa:— 4y Sar^^g^^
4aar— y x*+ xy
ax—3y 2x^+4xy
5ax—5y 5x' + 2xy
7ax—2y 4a:f^+3xy
19ax—]5y
9. 10.
— 12yy 30 — 13a;* — Sary
— -Ty^ 23 — lOa:^ — 4xy
— 2y« li — U^x — Txy
— •4yy 10 — 16x2 _ 5^,^
— y^ 16 — 20^3: — xy
— 3yy
7.
8.
5xy
4a — 46
14xy
5a — 56
22xy
6a— 6
17xy
3a — 26
Hxy
2a — 76
i^
8a— 6
11.
5xy-
• 3x + 4a6
8xy —
4x + 3a6
3xy-
■ 5« + 5a6
xy —
2ar + a6
Axy —
X + 7a6 '
CASE II.
When the quantities are like, but have unlike signs.
Add all the affirmative coefficients into one sum, and all the neji^ative ones
into another, when there are several of a kind : then subtract the less sum, or
the less coefficient, from the greater, and to the remainder prefix the sign of the
greater, and subjoin the common quantity or letters.
Thus, + 5a and — 3a, united, make + 2a.
And — 5a and + 3fl, united, make — 2a.
OTHER EXAMPLES FOR PRACTICE.
1. 2. 3. 4.
— 5a + 8x^ + 3y + 3aar' — 3a»
+ 4a — 5x3 + 4y -|- 4flx» — 5a'
+ 6a — 16x3 ^ 5y _ 80x2 _ joa'
— 3a 4- 3x3 — 7y — 6ox» + lOo*
+ a + 2x3 — 2y ^ Sox' + 14a«
+ 3a — 8x3 + 3y
VOL. I. I
114
ALGEBRA.
5.
6.
7.
8.
+ 4ab + 4
— 3aa;^
+ 10^/a X Vm
+ 3y + 4aa;2
— 4ab + 12
+ a\/x
— 3v^aa!
— y — 5ax'^
+ Tab — 14
+ 5ax^
+ 4^/aar
+ 4y + 2aa?-
+ flJ + 3
+ 6ax^
+ 12aM
— 2y + 6aV-
— 5ab — 10
5x%/a+y—2x*^/y+ a/2
3a; {a-\-y)^+6xy* + 2^
-8a? (,a+y)^--4xy^ +3^/2
7x^s/a+y+ 3aVy + 2 \/2
2a7(a+y)*+5a?Vy+iX2*
-gajVo+y— 8a?y* —8^/2
10.
— 3 (ax+by+cz) i — \/x^-\-y'^ -j-a — b
2\/ax+by+cz +i.x^+y^)^—3 (fl—b)
7 ax+by+cz* —^x^+y^+2 {a—b)
3V {ax+by+cz) +(ar'+y2)i +a— 6
-5V (,ax+by-\-cz) + x'^-^y-'^ —2{a—b)
{ax+by-\-cz)'^—^/x^-Jt-y^ — 3{a—b)
CASE III.
When the quantities are unlike.
Having collected together all the like quantities, as in the two foregoing
cases, set down those that are unlike, one after another, with their proper signs.
EXAMPLES.
1.
2.
3.
3xt/
2ax
6yx — 1 2a;-
— 4x^ + 3a;y
4fla;— 130 + Sa'^
5a?2 + 3ax + 9x^
Syx
6ax
+ 43^^ — 2ya;
— 3a;y + 4a;*
7xy — 4x^ -f 90
/v/a; + 40 — Gar"
- 2xy + 8na?
4xy — 8x^
7ax + Sx^ + 7xy
4.
9x^y^
— 7x'y
+ 3aa:y
— 4a^y
5.
14aa; — 2a?^
5ax + 3xy
8y^ — 4fla:
3jf2 + 26
6.
9 + 10 s/ax — 5y
2a; + 7 Vxy + by
by + 3a; ^^a — 4y
10 — 4a*a; + 4y
4yx^
4 ^x —
3y
— 6xy'
2^xy +
14x
+ 3y»ar
3x +
2y
— rx'y
— 9 + 3yVx
SUBTRACTION. Hft
8. 9.
3a« + 9 + X* — 4
2a — 8 + 2«* — 3*
3*2 — 2a»+ 18—7
— 12 + a — 3 x' — 2y
Add a + 6 and 3a — 5b together ; and also 5a — 8x and 3a — 4x together
Add 6x — 56 + a + 8 to — 5a — 4x + 4b — 3,
Add a 4- 26 — 3c — 10 to 36 — 4a + 5c + 10 and 56 — c.
Add 0 + 6 and a — 6 together ; and ^x + iy to — ^x — ^.
Add 3a -|- 6 — 10 to c — d — a and — 4c + 2a — 36 — 7.
Add 3a2 + 6^ — c to 2a6 — 3a= -f 6c — 6.
Add a» + 62c — 6^ to 06^ — 06c + 6^.
Add 9a — 86 + lOx — 6<i — 7c + 50 to 2x — 3a — 5c + 46 + M — 10.
Add a-f-6 + c, — a + 6 + c, a — b + c, and a + 6 — c together ; and
likewise a + 6 + c — rf, cf + a + 6 — c, c + d + a — 6, and b + c + d — a
together.
Note. It often happens that some one letter is considered the principal
object in the calculation, and that the others are viewed as coefficients of this
one. In this case their sums will assume a compound form : as in the following
examples.
2 ax 4- 36y- ax + Zdy^
cdx + Sarfy^ 26y — 3d!x
— 66x — cy^ — 6y2 + 4>ny
{2a+cd— 66) x+(36+8arf— c) y^ (a— 3cO x+(3(f— 6) yH(26+4m) y
In these cases, 2a, erf, — 66, &c. instead of being considered to form part of
the components of the respective terms in which they appear, are collected under
vincula, and the collection under each vinculum treated as a single quantity.
Two other examples are added for the student's exercise.
— a^JT — y^ + b x/x^ + y^ (a + 6) a/x — (2 + m) Vy
A A. fn J. r\ x^
— 5c^x^-t-y- — 3d y/jp- + y* 4y^ + (a + c) '^
— / («^ + y^)^ — 2c (x2 — y2) * 2x sjy — C2rf — «) «*
2 v'x* + y« + 4a v'x^ — y* (m + n) y* + (6 + 2c) V*
^x^ — y* — (x^ + y')^ — 2x V'x + 12a ^y
Sometimes these literal coefficients, instead of being collected horizontally,
are placed vertically under each other. Specimens may be seen in muUiplica-
tioD and division.
SUBTRACTION.
Set down in one line the first quantities from which the subtraction is to be
made ; and underneath them place all the other quantities composing the sub-
trahend; ranging the like quantities under each other, as in addition.
I 2
116 ALGEBRA.
Then change all the signs ( + and — ) of the lower line, or conceive them
to be changed ; after which, collect all the terms together as in the cases of
addition *.
Note. When the sign — is prefixed to a compound expression, it indicates
that if the parenthesis is removed, the signs of all the terms must be changed.
Thus, — {ax — bx ■{- 2cx'^ — Sefe') = — ax -^ bx — 2cx' -\- 3cb^. For
otherwise the sign — would not affect all the terms within the parenthesis
equally.
EXAMPLES.
1.
2.
3.
From 7a^ — 36
Take 2a^ — 8b
9aj2 — 4y-\- 8
6x2 + 5y _ 4
8xy — 3 + 6a? — y
Axy — 7 — 6a; — Ay
Rem. 5a^ + 5b
3x^ — 9y + 12
4xy + 4 + 12x + 3y
4.
5.
6.
From 5xy — 6
Take — 2xy + 6
4t,2 — 3y — 4
2f 4- 2y + 4
— 20 — 6a; — 5a;y
3xy — 9a; — 8 — 2ay
Rem.
7.
From 8a?-y + 6
8.
5 a/ xy -\- 2x >y xy
9.
7x^ + 2 -x/x— 18 + 36
Take — 2x^y + 2
7x3
2x2
7 //xy 4- 3 — 2xy
9x=» — (12 — 56) + x^
Rem.
10.
bxy — 30
7xy — 50
11.
- 2 (a + 6)
- 4 (a -f 6)
12.
4xy» -1- 20a -/ (xy + 10)
3z^y2+ 12a V (^ — 10)
Note. If literal coefficients occur, they must be collected (the subtractive
ones with their signs changed) as directed in note upon case iii. of addition.
From a + 6, lake a — 6.
From 4a + 46, take 6 -|- a.
From 4a — 46, take 3a + 56.
From 8a — I2x, take 4a + 3a?.
From 2x — (4a + 26 — 5), take 8 — (56 — a — 6x).
• This rule is founded on the consideration, that addition and subtraction arc opposite to each
other in tlieir nature and operation, as are the signs -|- and — , by which they are expressed and
represented : hence, since to unite a negative quantity with a positive one of the same kind,
has the effect of diminishing it, or subducting an equal positive one from it, therefore to sub-
tract a positive (which is the opposite of uniting or adding) is to add the equal negative quantity.
In like manner, to subtract a negative quantity, is the same in effect as to add or unite an equal
positive one. So that, changing the sign of a quantity from -|- to — , or from — to -}-, changes
its nature from a subductive quantity to an additive one; and any quantity is in effect sub-
tracted by barely changing its sign.
MULTIPLICATION. II7
From 3a + b + c — (d + 10), take c + 2a — d.
From 3a + b + c — {d — 10), take i — (10 _ 3a).
From 2ab + b'^ — 4c + be — b, take 3a^ — c + b-.
From a3 + 3h^c + ab^ — abc, take 6^ ^ ab'^ — abc.
From 12ar + 6a — (46 — 40), take 46 — (3a — 4x — 6rf) — 10.
From 2x — (3a — 46) + 6c — 50, take Qa + x + (C6 — 6c — 40).
From 6a — (46 + 12c — 12a'). take 2x — (8a — 46) — 5c.
From i {a + b + c), take a, 6, c, separately and successively.
Subtract — 3 Va + x + 4 (x* — y-)^ — 1, from ^ or' + y'—2 (a+x)* + 3;
and — I7flxy + lla6c — x'^x + y, from 2x (x + y)^ — 3axy.
Subtract ss^ — pxy + qx — c, from ax^ + mxy + nx + 6 ; and
mxy — pqxz — n (2^ + a), from pxy -f grx — r {z^ + a).
From a (n — y)* + 6xy + c (a + xf, take {n—y)^ — bxy -if{a + c)(a + x)«; and
take (a — 6) (x + y) -f- (c — d) (x — y) — n, from (a + 6) (x + y) — (c + 1/)
(a' + y) + »».
From (a— 6) xy—hx^ subtract {2p—3q) (x+y)^ ; and from — {p Jf q) ^/x~+y.
subtract — oxy — Sx^ — Ax* : and add the two remainders together.
MULTIPLICATION.
This consists of several cases, according as the factors are simple or compound
quantities.
CASE I.
When both the factors are simple quantities.
1. Multiply the co-efficients of the two terms together; then, to the pro-
duct annex all the letters in those terms, which will give the whole product
required.
2. When the same letter is repeated in the product, the result may be simpli-
fied by writing the sum of the indices instead of the separate factors, agreeably
to def. p. 105. Thus, for a^ x a^ write a».
3. The same rule holds good if there be fractional indices, or their equivalent
radicals, in the product: as for a^a^, write a' "^ % or a* ; and for a' x a%
write a' or a.
This is true whatever a may represent, as for instance, if a = — 1, then
(— 1)* X (— l)*or V^Tf X a/— r= — 1.
Note *. Like signs, in the factors, produce -|-, and unlike signs — , in the
products.
• That this rule for the si^s is true, may be thus shown.
1. When -f o is to be multiplied by -f c; the meaning is, that -f- a is to be Uken a* many
times as there are units in c; and since the sum of any number of positive term* is positive, it
follows that -j- a X + c makes -f- ac.
2. When two quantities are to be multiplied together, the result will be exactly the same, ia
whatever order they are placed ; for a times c is the same as c times a, and therefore, when — ti
118 ALGEBRA.
EXAMPLES
1.
2.
3.
4.
10a
— 3a
7a
~6x
2b
+ 2b
— 4c
— 4a
QOab
— 6ab
— 28flC
+ 24ax
5.
6.
7.
8.
4ac
9a^x
— 2x'^y
— 4ay
— Sab
3
4x^
3x^y^
— xy^
9.
10.
11.
12.
— ax-'
( + 3a.y)^
— 3ax s/ —
1
— 5xyzs/ — I
4a? J^^
1
— 6c
(-
-4y
— 4aa? ;^ -^
Though only two factors have been proposed, there may be any number. The
process is however still the same, repeating the operation with every successive
term upon the result of all the preceding.
When, however, the factors are all equal, the literal parts may be more readily
assigned ; viz. multiply the index of each letter in the common factor by the
index of the number of factors. The products of these ai-e the indices of the
literal parts. Thus,
a" b" (f X a" b" c' X a" b" (f =z a^" b^' c^', or {a" b" c')^.
"When there are numeral coefficients, the powers of these must be found as at
p 65. Thus, suppose we sought the products of
i 3 i 3 i 3
3a-b X 3d^b, and of — 2a^ c* x — 2a^ c^ x — 2a^ c*, we should have
3 9
3 X 3 a* b^, and — 8a^ c^
The signs being regulated by the number of factors when those factors are — .
This is a case of Involution.
is to be multiplied by -|- c, or -(- c by — a: this is the same thing as taking — a as many
times as there are units in -|- c; and as the sum of any number of negative terms is negative, it
follows that — a X +c, or-|-«X — c make or produce — ac.
3. Wlien — a is to be multiplied by — c : here — a is to be subtracted as often as tliere are
units in c : but subtracting negatives is the same thing as adding affirmatives, by the demon-
stration of the rule for subtraction ; consequently, the product is c times a, or -|- ac.
Otherwise. Since a — a = 0, therefore (a — a) X — c '8 also ^ 0, because 0 multiplied
by any quantity, is still but 0; and since the first term of the product, or a X — c is rr — ac,
by the second case; therefore the last term of the product, or — a X — c, must be ac, to
make the sum =: 0, or — ac -\- dc =z 0 ; that is, — a X — c = -\- ac.
Other demonstrations upon the principles of proportion, or by means of geometrical diagrams,
have also been given ; but the above is more natural, simple, and satisfactory.
MULTIPLICATION. j jg
KXAMPLE8 FOB PRACTICE.
1. Required the cube or 3d power of 3a' ( — 1).
2. Required the 4th power of 2a'66-'.
3. Required the 3d power of — 4a"b-^.
4. To find the biquadrate of ^ ( + 1).
5. To find the 6th power of + 2a^^''±^i.
6. To find the 7th power of (+ l)» x (+ a') (— a-').
CASE ir.
When one of the factors is a compound quantity.
Multiply every term of the multiplicand, or compound quantity, separately,
by the multiplier, as in the former case ; placing the products one after another,
with the proper signs ; and the result will be the whole product required.
EXAMPLES.
1.
5a — 3c
2a
2.
3ac — 4b
— 3a
3.
2aP _ 3c + 5
be
10a2 — 6ac
—9a^c + I2ab
2a^bc — 3bc^ + bbc
4.
12a? — 2ac
4a
5.
25c — 7b
— 2a
6.
Ax — b -\- 2ab
2ab
7.
3c^ + x^-l
ixy V—l
8,
10a;2 — 2y^
-4ar»
9.
30^ — 23?=^ — 66
2ax*
CASE III.
When both the factors are compound quantities.
Multiply every term of the multiplier by every term of the multiplicand
separately; setting down the products one after or under another, witii their
proper signs ; and add the several partial products together for the whole pro-
duct required.
1.
a + b
a + b
2.
3x + 2y
4x — by
- uy
- \0y''
3.
2x* + «y - 2y^
3a? — 3y
a2 + ab
+ ab + b^
a^ + 2ab + J2
12a?' + 8a;y
— \bxy
12x2 — jxy -
6*3 + 3x»y — &cy«
— 63?y — 3afy' + 6y^
6x' — 3x»y — 9xy^ + 6y'
120
4.
a + b
a-b
P
ALGE
5.
x^ + y
x' + y
.BRA.
6.
a^ + ab + ¥
a-b
a^ + ab
-ab-
X* + yaP
— a'b — ab^ -
-6*
a2 * _
■i2
X* + 2yx^
+ y'
flS • * -
-ir«
Nofc I. When the factors are all equal, and composed of two terms, as a + x,
or 3a — 5x, the operation is more readily performed by the Binomial
Theorem, the rule of which may be referred to at once. For any purposes
likely to occur in the earlier stajjes of Algebra, the result may be obtained by
actual multiplication, as above. When there are more than two terms, there is
a general theorem for finding the coefficients, called the Multinomial Theouem :
but as occasion for its use occurs comparatively seldom, it will not be given in
this work. Reference may therefore be made to Young's Algebra, p. 262.
When the factors are all equal, as we have here supposed, the operation
is called Involution.
Ex. Raise a — x to the third power, and 2a — Zxy to the fourth power.
'Note II. The following is a specimen of the method of disposing of the
literal coefficients in vertical columns. It has not only the advantage of keeping
an operation of considerable extent within the limits of the breadth of the page,
but it dispenses with the collecting those coefficients together, after the multipli-
cations are developed, on account of its readily disposing them in their places
as we proceed.
Multiply together the binomials x — a, x — b, x — c.
x — a
X — b
X' — a
— b
X — c
X + ab.
sfi — a
— b
X — abc.
3^+ ab
+ be
+ ca
•which in the horizontal disposition of the terms of the coefficients would stand
thus : x^ — {a -j- b -\- c) x^ + {ab + be -\- ca) X — abe.
Had the number of factors been greater, the advantage would have been still
more apparent.
In the case just considered, where the given quantities are numerically given,
this disposition of them is the most favourable to their actual addition into one
numerical sum.
MULTIPLICATION BY DETACHED COEFFICIENTS.
It is not necessary to write down the powers of the quantity according to
which the work is arranged, till we have performed the whole of the arithme-
tical determination of the coefficients : since the same powers of that letter, if
generated by the multiplication of factors in which none of the terms are wanting,
will always, in the above arrangement, fall in the same vertical column. Also,
since the indices of the powers of that letter in going from term to term, either
MULTIPLICATION. ]oi
decrease by units or increase by units, according: as we begin at the highest or
lowest powers, we may write them in juxtaposition with the coefficients found
as above indicated, without the slightest degree of attention beyond the most
ordinary kind.
In finite expressions, it is most usual, though not essential, to begin with the
highest, and in series, it is, from their interminable character, necessary to begin
with the smallest index, whether positive or negative.
The following example will render the practice obvious : and the student i«
earnestly recommended to adopt it, not only on account of economy of time and
room ; but to familiarize his hand, his eye, and his thoughts, to the processes
by which the roots of equations are (in the most improved state of that difficult
problem) found or approximated to.
Ex. 1. Multiply 2w> — 4a + 15 by 3a- + 4.
Here the coefficients of a^ in the first, and of a in the second factor, are each
equal to zero. Hence the work will stand thus :
2 + 0 — 4 + 15
3 + 0 + 4
6 + 0 — 12 + 45
+ 0* + 8 + 0 — 16 + 60
6 + 0 — 4 + 45 — 16 + 60
And the highest power of a being a^ x a^ = a*, we may write a*, a*, a', a^, a',
in juxtaposition with the above coefficients, which gives for the product
6a' + Oa* — 40^ + 450^ — l6a + 60.
The same result would have been obtained, but in an inverted order, by
writing (15 — 4a + 2a^) (4 + 3a^).
When the first term of the multiplier is also unity, the work may be shortened
still more, by allowing the line of coefficients which constitute the multiplicand,
as the first partial product of the work. Thus, were x' — 6x^ + \0x — 9
given to be multiplied by ic^ — 3a; + 2 ; the general method would require it to
be executed as follows.
(a) 1—6 + 10—9
1 — 3+2
(b) 1—6 + 10—9.
— 3 + 18 — 30 + 27
2 — 12 + 20—18
1 _ 9 + 30 — 51 + 47 — 18
And x^.x^ = a^ is the highest power or first terra. Whence, attaching the
literal parts, we get as the product
xi —gx* + 30a:' — 51x* + 47x — 18.
But as the line (b) is the same with the line (a) we may put it thus :
1—6 + 10-9 r-3 + 2
— 3+18—30 + 27
2 — 12 + 20 — 18
1—9 + 30 — 51+47-18
• Here, instead of a horizontal column of ciphers, the first term of the multiplication by 4 it
made to commeuce under the 4, as in common arithmetical multiplication.
122 ALGEBRA.
The coefScients of the remaininf; terms of the multiplier being placed either in
curve to the right, as " (— 3 + 2," or in any other way that may be thought
convenient.
It may be further remarked, for the purpose of connecting the identity of
arithmetic, in its usual form, with the practice of algebra, in the student's mind,
that if all the signs were + (for this is always so considered in arithmetic)
and we make a = 10, the above process would be precisely the same as is
practised in ordinary multiplication, except that the order is inverted.
Note I. In the multiplication of compound quantities, it is usually best to set
them down in order, according to the powers and the letters of the alphabet.
And in the actual operation, begin at the left-hand side, and multiply from the
left-hand towards the right, in the manner that we write, which is contrary to
the usual way, though analogous to the Indian, of multiplying numbers. But
in setting down the several products, as they arise, in the second and following
lines, range them under the like terms in the lines above, when there are such
like quantities ; which is the easiest way for adding them together.
In many cases, the multiplication of compound quantities need only be indi-
cated by setting them down one after another, each within or under a vinculum ;
and either with a sign of multiplication between them, as (a + b) x {a — b) x
3ab, or in juxtaposition, (a + b) {a — b) 3ab.
Note II. The operations in multiplication will often be greatly facilitated, by
fixing the following rules and formulae well in the recollection.
The square of any polynomial is equal to the sum of the squares of its terms
+ twice the product of every two of its terms taken in all their different com-
binations.
Thus, ia + b + c + d)(ai-b + c+d)
= a2 -f 6^ + c=» + rf2
+ 2ab + 2ac + 2ad
+ 26c -I- 2bd
-f 2cd
and (a + & + c -f rf + e +/) {a + b + c -\- d + e + f)
= a' + b' + (P + d'' + e^ +/2
+ 2ab + 2ac + 2ad + 2ae + 2a/
+ 2bc + 2bd -\- 2be -\- 26/
-f- 2cd -\- 2ce + 2cf
+ 2de + 2df
+ 2e/
In all such cases the arrangement of the products is very simple, and the con-
tinuation of the process very obvious.
Note III. From the principle that the rectangle of the sum and difl'erence of
two quantities is equal to the difference of their squares, some useful theorems
obviously ^ov/ : viz.
1. a-^ — b- = {a 4- 6) (a — b).
2. a* — 6* = (a^ -^ Ir) (a» - 6^) = (a' + b"^ {a + b) (a — b).
3. a? — b^— {a* + ¥) (a* — ¥) = (o* -|- b*) {a' -f b^) (o + b) (a — b).
4. a'6- 6'«= (o» + b^ {a* + 6<) (a^ ^ b^ (,a + b) (a — b).
5. a^ — 63 = (a2 -|. a6 -|- 6=^) (a — 6).
6. a' + b^ = (a* — ab + b^ (a -f 6).
7. (I -H a) (X + 6) = ar' -h (a -f 6) a? -f ab.
8. (X — a) (ar — 6) = x» — (a -I- 6) a? -I- a6.
9. (x + a) (x-\- b) {x + c) = sfi + {a + b + c)x^ + iab-j' ac+bc) X + abc.
10. (jc — fl) (a; — 6) (a; — e) ^= a^ — (a -f- 6 -f c) *^ -f- {ab -j- ac + bc) x — abc.
DIVISION. 123
EXAMPLES FOR PRACTICE IN ALL THE CASKS.
1. Multiply lOac by 2a. Ans. 20a'c.
2. Multiply 3a' — 2b by 36. Ans 9a^b - CA».
3. Multiply 3a + 2b by 3a — 2b. Ans. 9a» — 4^^.
4. Multiply afi — xy -\- y^hy X + y. Ans. jfi + y\
5. Multiply a» + d% + a6* + b^ by a — b. Ans. a* — A«.
6. Multiply a'^ + a6 + 6^ by a- ~ ab -\- b^.
7 Multiply 3x^ — 2xy + a by x^ + 2xy — 6.
8. Multiply 3a' — 2ffa; + 5x" by 3a^ — 4ax — 7x^.
9. Multiply 3a^' + 2xY + 3y^ by 2*3 _ Sx^y^ + 3y».
10. Multiply (a2 + ai +b^) y by (a — 26) x.
11. Multiply a- + a— • a? + a"-'^ x^ + .... aV"' + ax—' + x' by a — x.
12. Multiply ax + bx' + cx^ hy I + X + aP + x^, and consider a, b, c, as
coefficients of the powers of x .- as in p. 120.
13. Develope (a; + a) (x + b) (x + c) (x + d) and also (x — a){x — b) (i — c)
(x — d); and attach the combinations of a, b, c, d, to the powers of x as co-
efficients.
14. The student may take in another factor {x + e), and it would be worth
his while to attempt to discover inductively some law or rule by which he could
form the terms seriatim, without the trouble of writinsf down the previous steps.
15. Multiply x^ -\- {a — b)x — ab by sfi + {c — d)x — cd.
16. (1 + r + r^ + r^ r") x (1 - r) = ?
DIVISION.
Division in al}^ebra, like that in numbers, is the converse of multiplication ;
and it is performed like that of numbers also, by beginning at the left-hand side,
and dividing all the parts of the dividend by the divisor, when they can be so
divided ; or else by setting them down like a fraction, tl»e dividend over the
divisor, and then abbreviating the fraction as much as can be done, by cancelling
any quantities which are common both to numerator and denominator. This
may naturally be distinguished into the following particular cases.
CASE I.
When the divisor and dividend are both simple quantUies.
Set the terms both down as in division of numbers, either the divisor before
the dividend, or below it, like the denominator of a fraction. Then abbreviate
these terms as much as can be done, by cancelling or striking out all the letters
that are common to them both, and also dividing the one coefficient by the other,
or abbreviating them after the manner of a vulgar fraction in arithmetic, by
dividing them by their common measure.
It will, of course, be necessary to subtract the index of the less power (whether
it be in the numerator or denominator of the fraction thus formed) from the
124 ALGEBRA.
index of the greater, leaving the difference where the greater index previously
was. If, however, on the contrary, any ulterior purposes render it advantageous
(and this often happens in algebraic investigations) to keep the latter in that
term of the fraction, from which it would thus be expunged, we may subtract
the greater index from the less, and put the difference with the sign — . Thus
a- b^ b^ b^ ar^ 1 a~^
—r-r may be written either -5 , or — — - (that is ¥ar^), or - , , or 7—, . And
so with any other quantities which appear in the result of the indicated division.
Nole. Like signs in the two factors make + in the quotient ; and unlike
signs make — ; the same as in multiplication ♦.
EXAMPLES.
1. To divide 6ab by 3a.
This may be written either 6ab -7- 3a, or — , and the result is in
either case ^ 26.
3a
2.
3.
4.
5.
Also c -r- c = - = 1 ; and abx -
c
Divide l6a^ by 8a?.
Di\'ide 12aV by — 3a^a?.
Divide — ISay^ by 3ay.
- bxy =
abx
bxy
a
~y'
Ans. 2a?.
Ans. — 4a?
Ans. — by
6.
Divide — laax^y by — 8oa?z.
Ans.?^.
CASE II.
When the dividend is a compound quantity, and the divisor a simple one.
Divide every term of the dividend by the divisor, as in the former case.
EXAMPLES.
1 / t . i9n «! ab + b^ a -{- b , , , ,
1. (oi + &«) -J. 26, or ^ = ~~~ = ia + |6.
2. (lOafe + 15aa?) ^ 5a, or ^0°^ + ^^aa? _ gj + 33,.
5a
3. (30flz - 48z) -r z, or ^""^ ~ ^^^ = 30a - 48.
z
4. Divide 6a6 — 8aa; -f a by 2a.
5. Divide 3a?2 — 15 + 6a? + 6a by 3a?.
6. Divide 6a6c + 12a6a? — 9a^6 by 3a6.
7. Divide 10a=^x — 15a^ — 15x by 5a?.
* BeciuM the divisor multiplied by the quotient must produce the dividend. Therefore,
1. When both the terms are +, the quotient must be +; because + in the divisor multi-
plied by -|- in the quotient, produces -j- in the dividend.
2. When the terms are both — , the quotient is also -f ; because — in the divisor multiplied
by — in the quotient, produces -+- in the dividend.
3. When one term is -f- and the other — , the quotient must be — ; because + in the divisor
multiplied by — in the quotient produces — in the dividend, or — in the divisor multiplied
by -f in the quotient gives — in the dividend.
So that the rule is general ; viz. that like signs give +, and unlike signs give — , in the
quotient.
DIVISION. 1^>5
8. Divide 15a'6c — \5acx^ + 5a(Pby — 5ac.
9. Divide 15o + 3ay — I8y^ by 21a.
10. Divide — 20(Pb^ + 60ab^ by — 6ab.
11. Divide - — a?2 _ aj by -. ' ,^ .
12. Find the quotient of 2x^-'^" + y»"-» by 6a?'-*" y-'.
13. Divide --«-« + —a"* — a by 5a^b-^.
14. Divide •001x=' + lOOOx' + -Olx-"' by •6x~^ ; and lO^x' + 3.1oy +
5 X lO^ity by '06 x lOV.
15. Divide 6 (a + ft)"" + 3 x 10^ (a+i)"" by - 3 (a + 6/; and — 6 x
{a + b + c)^ — 4 (a + b + cy by — {a + b + c)».
16. Divide -00015 x 10* + '03 x lO'ar by •0005-'x« ; and 00015-* x lO'V
— -6-^ X 10-3 X -001-2 by 10-*.
17. Divide a' — a?" by a — x, and by a + a?, and also a" + a:" by a + a?.
CASE III.
When the divisor and dividend are both compound quantities.
1 . Set them down as in division of numbers, the divisor before the dividend,
with a small curved line between them, and range the terms according to the
powers of some one of the letters in both, the higher powers before the lower.
2. Divide the first term of the dividend by that of the divisor, as in the first
case, and set the result in the quotient.
3. Multiply the whole divisor by the term thus found, and subtract the result
from the dividend.
4. To this remainder bring down as many terms of the dividend as are re-
quisite for the next operation, dividing as before ; and so on to the end, as in
common arithmetic.
Note I. If the divisor be not exactly contained in the dividend, the quantity
which remains after the operation is finished may be placed over the divisor, like
a vulgar fraction, and set it down at the end of the quotient, as in arithmetic.
EXAMPLES.
a — i) flS — 2a6 + b^ (,a — b
a^ — ab
— ab + b^
~ab + b^
a — c)a^ — 4a^c + 4ac» — c^ (a» — Sere + c»
— 3a^c + 4a(^
— 3a^c + 3ac^
ac^-
1^6 ALGEBRA.
2x*
a + x) a* — 3x* {a? — tt^x -{■ cuiP — 3^ —
a ■\- X
fl* + a^x
— a}x -
— a^x -
- 3x*
a-a^ — 3a;*
aV 4- ax^
— ax^ — Sa;"*
— ax^ — X*
— 2x^
EXAMPLES FOR EXERCISE.
1. Divide a- + Aax -\- Ax^ by a + 2a;. Ans. a + 2x.
2. Divide a^ — 3a^z + 3az^ — z^ hy a — z. Ans. a^ — 2az + z^.
3. Divide 1 by I + a. Ans. ] — a + a^ — a^ + ....
4. Divide 12a;* — 192 by 3a; — 6. Ans. 4a?' + 8x^ + \6x + 32.
5. Divide a^ — 5a*b + lOaW- — ^Oarb^ + Sab* — ¥ by «« — 2a6 + b"^.
Ans. a? — 3a^b + 3ab^ — b^.
6. Divide 48^3 — 960^2 _ 6402^: + nocfi by 2z — 3a.
7. Divide 6« — 3&V + 36V — a;« by ¥ — 3b^x + 3bx^ — a^.
8. Divide aJ — x' by a — x.
9. Divide a' + Sa^a; -f 5aa;^ + a?' by a + ar.
10. Divide a* + 40^62 _ 326^ by a + 2b.
11. Divide 24a* — 6* by 3a — 2b.
3 5 7
12. The quotient found from dividing ~x^ — -oc^ — 8a; + 9 by 3a;* + - x— 9
... ., 9a^^ 25Pm^ , 70dfm _ j, , 3ab , 5/m
13. Divide -^^--|,— +^ -49d* by— + -^g - 7d.
-^. ., a^ 2a , ae , be c* , a c
14. Divide - - - +^^ +- _- by ^-^.
Note II. By observing that the number of terms in any remainder that takes
place after all the terms are brought down from the dividend is always less than
the number of terms in the divisor, it is clear that, however far the operation is
carried, the work can never terminate. The remainder always occurring, the
terms of the quotient may always be increased ; and that without any assignable
limit. The series of terms thus formed is, from its capability of unlimited
extension, called an Infinite Series. By attending to the manner in which
the successive terms are related to the preceding one or preceding ones, the law
of the progression (in Infinite Series resulting from Division) may be always and
very readily discovered : so that when a few of the first terms have been actually
obtained by the prescribed process, the remaining ones may be written out to
any extent we may choose or require, by merely attending to this law of observed
dependence. Examples of these will be found under the head of Infinite Series
in a future part of this volume.
f
DHISION. 127
Note III. Operations in division may often be facilitated by the formulK
given in the last note to multiplication, as well as by the following.
Remember that 2n denotes an even exponent, 2» + I an odd exponent ; then
flS" _ ft!" is divisible by a — b, by a + fc, or by a* — It*.
gin ^ j?» is divisible neither by a — 6, nor by a + 6.
a' — b' is always divisible by a — b.
a'*+' — i'"+' is divisible by a — b, but not by a + 6.
a2«+i ^ ft2-+i is divisible by a + 6, but not by a — b.
Thus,
(a* — b*) u. (a — 6) = a' + a-h + a6» + 6\
(a* — b*) -r (a + 6) = a3 — a-6 + ab^ — b\
(a* _ fts) -J- (a — 6) = a< + a»6 + a*6- + ab^ + b*.
icfi + b^) -i- (a + b) = a* — a^ + a-b^- — al^ + b\
(a* — i*) ^ (a + J) = a< — a'6 + a^ft* — ab^ ^ b* ~r,
a -\- 0
where the latter is evidently not a complete quotient.
These theorems will enable the student to effect important simplifications in
the reduction of fractions, and of equations, and must therefore obtain sufficient
attention before he proceeds further.
DIVISION BY DETACHED COEFFICIENTS.
As it was shown, in the note on multiplication, that the multiplication may be
very conveniently carried on by means of the detached coefficients only, so it
may be readily shown that the same can be done in division ; and its practice
is earnestly inculcated on the student for precisely the same reason as it was
there done, — its economy of time and space, and especially as an introduction to
the recent improvements made in the solution of numerical equations. Thus,
for example, to divide 3a:^ — fix by x^ — bx — c.
1 + 0 — 6 — c) 3 + 0 — 6 + 0 (3 + 0 + 26 + 3c + 26* + 56c . . .
3 + 0 — 36 — 3c
0 4- 26 + 3c
0 + 26 + 0 — 26* — 26c
3c + 26* + 26c
3c + 0 — 36c — 30*
26' + 56c + 3c'
26* + 0 — 26» -
-26*e
+ 56c + 3c» I + 26»c
+ 26»|
And since the highest power of * is ^ = «", we have for the result 3jc» + Oar"
+ 26ar* + 3 car' + 26*a?-' + 56cx-* + ad inf.
This example has been adopted on account of its conUining both literal and
zero, as well as numeral, coeflScienta.
SYNTHETIC DIVISION.
When one algebraic function of a quantity is to be divided by another, the
]2S ALGEBRA.
coefficients of each being given in numbers, the following process, invented by
the late Mr. Horner to subserve the solution of numerical equations, is of the
utmost value.
1. Write down the coefficients of the dividend in a horizontal line with their
proper signs, and where a term is wanting write 0 in the place of its coefficient.
2. Draw a vertical line before the first term, and to the left of this line put
down the coefficients of the divisor, with the same precaution respecting absent
terms, but the signs of these coefficients changed ; and having them so disposed
that the first coefficient * is in a line with the horizontal column spoken of in (1).
3. Bring down the first coefficient of the dividend : this will be the first term
of the quotient.
4. To obtain the others in succession, multiply the immediately preceding
term of the quotient by the remaining terms of the divisor, having their signs
changed ; and place them successively under the corresponding terras of the
dividend in a diagonal column, beginning at the upper line. Add the results in
the second column, which will give the second term of the quotient ; and multi-
ply the terms of x in the divisor by this result, placing the products in a diago-
nal series, as before. Add the next series of results, which will give the next
coefficient of the dividend ; and multiply x by this again, placing the products
as before. This process, persevered in till the results become 0, or till the quo-
tient is determined as far as necessary, will give the same series of terms as the
common mode of division, or as the division by detached coefficients, in the last
article, when carried to an equivalent extent.
Let us take as an example the division of a:* — 6a!* -f 20a;^ — AOx^ + ^^^ —
40x -|- 100 by a?^ — 2a;^ -1- 5a; — 9. Following the prescribed directions with
respect to arrangement, we have the horizontal and vertical columns at once.
1
+ 2
— 5
+ 9
1 _ 6 + 20 — 40 + 50 — 40 -1- 100
+ 2 — 8 -h 14 + 6 — 30 — 44 + 316 + 582
— 5 + 20 — 35 — 15 + 75 + 110 — 790 — 1455
+ 9 — 36 + 63+ 27 — 135 — 198+1422+2619
1 — 4 + 7 + 3 — 15 — 22 + 158 + 291 — 406 — ... .
Multiply each term of the quotient in succession by all the terms of the
divisor, (the first or 1 excepted, the upper line standing for the result of that
step,) carrying the results to the places denoted by the corresponding powers of
the quantity x. This will always be done when the deficient terms are supplied
by zero, to preserve the places as in arithmetic, by carrying them out diagonally
to the right, or moving one step to the right in making the commencement of
each successive row. Thus we obtain the diagonal series 1 + 2 — 5 + 9. Add
the vertical column — 6 + 2, and with the result — 4, multiply all the terms
of the divisor as before, giving the next diagonal series — 8 + 20 — 36. Add
the third column, and obtain the result + 7 ; and by this obtain another diago-
nal column + 14 — 35 + 63, and then another sum + 3. Proceed in the
same manner till the results either terminate in zeros, or have been carried far
enough to answer the purpose in view. In the above work nine terms are
obtained : to which the powers of x (the highest being x^^ = x^) may be at-
tached as they stand, and the quotient is a:^ — 4x' + 7a? + 3 — 15x~' — 22ar*
+ 158ar-3 +291a;-~' — 406ar-* — ad infinitum.
• It is to be understood that the coeflBcient of the leading term of tlie divisor is 1 ; and in
cases where this does not occur, it can be made so, b)' dividing every coefficient of the divisor
and dividend by that coefiScient.
DIVISION.
129
With the view of illustrating the operation, it will be advisable to work the
same question in the usual way, employing, however, only the detocbcd co-
efficients.
1-2 + 5-9) 1-6 + 20-40+50-40+100 (1-4+7 + 3-15-22 + 158 + 291-406
1—2+ 5— 9
—4 + 15—31 + 50
—4+ 8—20 + 36
7—11 + 14—40
7—14 + 35—63
3— 21+23+JOO
3— 6+15— 27
-15+ 8 + 127
■15 + 30— 75 + 135
—22 + 202 — 135
—22+ 44-110+198
158— 25 — 198
158—316 + 790—1422
291—988 + 1422
291—582 + 1455—2619
—406— 33 + 2619
—406+ 812—2030 + 3654
— 845+4649—3654
The connexion between this and the synthetic division will best appear by
taking a form intermediate between the two : viz. by placing the subtrahends in
order, having their signs changed, but still in the horizontal position which they
occupy in the old method.
Divisor. Dividend. Quotient.
1—2 + 5—9) 1 —6 +20 —40 +50 —40 +100 (1—4 + 7 + 3— 15— 22+158 +291— 40C
Remainder
—
l* + 2 —
5 +
3 : : :
. +4*— f
3 +20 —36 : :
— 7* + 14 —35 +63 :
— 3*+ 6 —15 + 27
+ 15*— 30 + 75 —135
; : +22*— 44 +100 —198
: : — 158* + 316 —790 +1422
: : — 291* + 582 —1455 + 2619
: : : : +406*— 812+2030-36.14
0 0
0 0
0
0 — 845 + 4649—3654
The relation of this to the common method is obvious.
Had we, however, left out the numbers marked with the asterisk in this work,
the sums would severally have been the terms of the divisor ; and hence, if we
omit multiplying by — 1 (the first coefficient of the divisor with its sign changed)
the line now marked as " remainder" might have been employed for the terms
VOL. 1. K
130 ALGEBRA.
of the quotient, which are the sums of the several columns. This is in accord-
ance with the rule, which requires the first coeflScient 1 to be omitted ; and the
change in the signs of all the other terms is effected by changing the remaining
signs of the divisor before we begin to operate.
Further, to avoid bringing the work so far down the page, leaving so much
space unoccupied on each side of the diagonal columns, the several products of
the coefficients of the modified divisor by the successive quotient figures, may be
themselves set down in diagonal columns : thus, instead of
1 — 6 -h 20 — 40 + 50 — 40 .. ..-^ rl — 6 -J- 20 — 40 + 50 — 40 . ...
2-5+9 [^^.^^ J 2-8+14 .. ..
— 8 + 20 — 36 V wnie _ 5 _|_ 20 — 35
14 — 35 + 63 J L + 9 — 36 + 63
1-4+7+3
In comparing this mode of working with the preceding, we remark that :
1st. The coeflficients which appear as subtrahends in the old method, appear
as addends having their signs changed, in the new. The change of all the signs
of the divisor except the first, in the new method secures this.
2nd. No coefficient is used till we arrive at the vertical column in which it
appears, and which occurs immediately after that column is completed. This
arises from only completing at each step the first term of what constitutes the
remainder in the old method.
3rd. The work is contracted into a series of horizontal columns, in number
equal to the terms of the divisor, without descending the page continually, as in
the old method. This is effected by carrying the Jirst term of each product to
the upper line, and gradually descending in a diagonal line with the others.
4th. The work besides not descending on the page, does not extend across it
so far as in the old method. This arises from the less breadth occupied by the
divisor in its vertical than in its horizontal arrangement ; and from the quotient
falling beneath the work instead of being placed to the right, as in the ordinary
method.
This last process completes the Algorithm of the method, and brings us to the
rule as above laid down in every particular.
On the ground of economy of time alone, this method does not require half
so much writing as the ordinary one ; and the chances of mistake in the ope-
ration are lessened in a still greater degree *.
• In the example just given, if we compare anj' two corresponding columns, as that belong-
ing to 3r~* for instance in the two methods, they will stand thus :
In the synthetic method. In the old method.
+ 316 + 135a-<
+ 110
— 135 _ VSoa-*
_110.i-<
+ 291
— 2ncr-*
— 3iar-«
291..-<
291,i-«
If again we estimate the t0t.1l saving, beside the compactness pf its disposition and the fewer
chances of error, we sh.ill discover that
X occurs 67 times more in the old than in the new process;
[Indices
DIVISION.
131
EXAMPLES.
1. Divide x^ — y^\)y x — y; or, which is the same thing,
Here 1+0 + 0-fO + O + O— I
+ 1 +1 + 1 + 1 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1 + i
which give the coefficients, and we have a;«-' =3^ for the exterior term.
"Whence the quotient is
ar' + x^y^ + xy" + x^y^ ^ xy* -^ y^.
2. Prove that r = 2i — t — ^ h ..
a+b a ' d^ a^ ^
3. Also that ^ — 7-^ = \ + 2x + 3x^ + 4ofi + 5x* +
4. rind the series for .
1 — 3x + 3a?2 _ 5x^
5. And those for ^f^i^^', and -If^lA^lvL^^
x^ y^ + xy 1 — xyz
6. Expand - -, ( or - ) into an infinite series ; and likewise
1 + 1 \ ^/ 1 — 1
^or -^ into a series, and show that - is the symbol of an infinitely great
quantity,
PROBLEMS AND THEOUEMS ON THE FIRST FOUR RULES OF
ALGEBRA.
1. Half the difFerence of two quantities added to half the sum gives the
greater of them, and subtracted leaves the less.
[Let the student select his own symbols, and illustrate it with bis own
numbers.]
2. If 2s = a + 6 + c, what are the values of * — a, s — b, and » — c ? and
what is half their sum equal to ? Find also their product, and arrange its terms
. systematically.
3. The difference between the square of the sum of two numbers and the
square of their difference is equal to four times their product ; and the sum of
the squares of their sum and difference is double the sum of their squares.
Prove this.
4. The sum of two numbers multiplied by their difference is equal to
Indices without sign, (or + understood,) 32 times more;
Indices with sign — , 2.5 times more ;
to which the system of dctaclicd coefficients has the advantage in common with the »)rnthelic
division.
In tlie detached system there occur 172 figures beddcs the answer ; in the new only 72, or 100
less ; and of the signs + and — , the number in the detached operation is 66, and in the other
32, or rather less than half the number.
K 2
132 ALGEBRA.
5. l{ 2s = a -\- b + c -^ d, what is the sum, and what the product of s — a,
s — b, s — c, s — d? and what is the sum of their squares ?
6. Let several binomial factors, of which x is the first term, and where a, b, c,
d, .. . are the second terms of the several factors, be multiplied together : then
describe the manner in which the coefficients of the several powers of x in the
product are formed of the quantities a, b, c, d, .. .
7. Prove that a^ — {b — cy ^ (a — b + c) {a + b — c), and — a^ + {b + cf
— {^a+b+c) (— a-\-b+c)i and hence that A.tt'b^ — {a^+b^ — c^f = {a->rb-\-c)
(,—a+b-\-c) {a—b+c) {a-'s-b—c).
8. Divide s^ + y^ hy x -\- y , and sfi — y^ by a; — y ; and show that if any
other odd whole number be substituted for 5 in these expressions, the division
will terminate without a remainder.
9. Convert {u -^ x -\- y -\- zY into the form (« + a?)^ + (m + y)' + (w + •^)^
+ {x-\- yf + Car + z)2 + (y + z)2 - 2 (1*2 + 0^ + 3,^ + 2^) ;
and likewise into the form
iJ(«+x+3/)2 + {u+x+zf + {u+y+zf + {x+y+zf - {u'+aP-^y^+z'^] ;
and again into
«2 + (2M+a7) x-\- {2 (a+a?) + y} y + J 2 {u->rX+y) + z}z :
and show that in this last, u may also change its place with either of the other
quantities, x, y, or z.
10. Multiply a + 6 ^/ — 1 by a— b ^ — I, and also by e + d V — 1 ; and
multiply together four factors, each equal to a-\-b <y — 1, and then four others,
each of which is a — b\/ — 1; and lastly, multiply the factors a + 6v^ — \,c-\-d\/ — 1,
e-\-f s/ — 1> i^+A \/ — 1, and i-\-k ^ — 1 together.
11. If the term rectangle of two lines, in the first ten propositions of the
second book of Euclid, be exchanged for the term product of two numbers, and
square on a line for the square of a number ; show that the propositions thus
transformed are also true, whatever those numbers may be.
12. Divide a by — 1 + v^ — 3, and by (— 1 + a/"— ^^ : and show that the
quotient in the latter case is the same as would be obtained if we divide i a by
— 1 — a/ — 3. Show also that a ( — 1 — a/ — 3/ = a ( — 1 + s/ — 3)».
13. Show that the sum oi x {x + y -\- z), y (j/ -{- z -]- x), and z {z -{- x + y)
is {x + y + zf.
14. Simplify to the utmost degree the expressions,
x^ + 2xy + 3/2 _ ^3p^xy—y-—{2xy—x^—y')-^ ;
a—\a-\-b—[a+b + c — {aA-b-irCr\.d)\\ ;
and a2 + ( — 6)2 4- (— c)2 + 2a ( — i) + 2a ( — c) + 2 ( — 6) ( - c).
TUE GREATEST COMMON MEASURE AND LEAST COMMON MULTIPLE OF TWO OR
MORE POLYNOMIALS.
A common measure of two or more quantities, whether expressed algebraically
or arithmetically, is any quantity which will divide them both without a re-
mainder.
The greatest common measure is the greatest quantity which will divide them
without a remainder.
A common multiple of any number of quantities is any quantity which is
divisible by them without remainders.
The least common multiple is the least quantity that is divisible by them all
without remainders.
GREATEST COMMON MEASURE.
133
Quantities are said to be prime to each other which have no common mea-
sure, except unity.
I. To find the greatest common measure of two quantities.
1. If there be any visible factor of one of the terms, whether it be the nume-
rator or denominator, which is not a factor of the other, it may be rejected aa
forming no part of the common measure.
This apphes more especially to monomial factors, as it is not often easy, except
in very simple cases, to detect binomial or higher factors.
On the same principle, if it will facilitate the future operation, any factor may
be brought into either of the terms.
2. Range both expressions in ascending or descending (no matter which, but
descending is most usual) powers of some one quantity concerned in tlie ex-
pressions. Divide the greater by the less and the less by the remainder ; then
the remainder by the previous one, and so on till the work terminates by giving
no remainder. The last divisor is the greatest common measure of the fraction.
3. If more than two expressions be given, of which to find the greatest com-
mon measure, proceed as directed in the corresponding subject in arithmetic,
pp. 42—4.
PROOFS AND REMARKS.
1. Denote them by X and X„ and let the following series of operations be
performed, where Q, Q,, Q^ .... Q„ are the successive quotients, and X„ Xj,
.... X„+2 the corresponding successive remainders. The first column indicates
the operation according to the arithmetical type, and the second expresses con-
tinually that divisor x quotient + remainder := dividend.
X = QX, + X,
X, = Q, Xj ■\- Xj
X, = Q, X, + X,
X
.)X(Q
QX.
X,) X, (Q,
Q.X,
X3) X, (Q,
Q,X3
XJ X3 (Q3
X„+,) X„ (C
Q-. X„+,
X._, = Q._,X_, -l-X.
X._. = Q._, X. + X^.
X« = Q. X^. -h X^
Now suppose X„+2 to be the remainder which becomes 0 : then X, = Q, X^„
and X„+, is the last remainder, which, therefore, divides X, exactly.
Substitute this value of X. in the preceding equation :
then X_i = Q_. Q, X.+. -f- X.+. = X,+, ^Q_, Q- -H Ij
134 ALGEBRA.
From which, as CL_i and Q„ are integral, we see that X„_, is divisible by X„4.,
exactly.
Next substitute the values of X„_, and X„ in the third equation from the end,
in terms of X„ + „ and we have
X„_, = Q„_. JQ„_, Q„ + 1} X„ + , + Q„ X„ + ,
= {Q„-aGi".-. Q-. + 1} + Q™}x„ + ,
Whence as Q„, Q„._,, and Q„_2 are integral, X„_2 is divisible by X„ + j.
In the same way, we find that in succession all the preceding X„_3, X „_^,
X4, Xj, Xj, X,, X, are divisible by X„ + ,. Whence X„ ^ , is a common
measure of the given expressions X and X, ; and it is likewise a common mea-
sure of all the subordinate quantities, or remainders, X^, X^ . . . . X„.
2. In the next place, X^ + i is the greatest common measure. For, let Y
be any other common measure, and put X = mY and X, := nY. Then from the
foregoing equations, and substituting these values of X and X,,
X, = X - QX, = [m -nQi]Y = P,Y,
X3 = X, - Q, X, = [n - mQ, + nQQ,] Y = P,Y,
X, = X3-Q,X3= { } Y = P3Y,
xlV. = xl-i - q11, X„" =' i .".'..'."."..'..] Y = P„Y,
where it is evident from the composition of the coefficients P,, Pj, . . . . P„, that
they are all integers, since they are composed of the products, sums and differ-
ences of the integers m, n, Q, Q,, .... Q.„, by hypothesis. Hence we have
P„Y = X„+,, and P„ integral ; and therefore Y is less than X^+i : that is, X„+,
is the greatest common measure.
3. It also follows from this, that any common measure Y of two terms is also
a measure of their greatest common measure. For since P„Y =: X„+, and P„ is
an integer, X„4., is divisible by Y without remainder.
X X
4. The quotients -^ and r, ' - may be thus formed by a succession of
operations ; and each is a closer approximation than the preceding to the true
value of the fraction.
Write down Q, Q,, Q.^, Q„, in a line as follows : —
Q Q, Q, Q3 Q4
forX;l,Q, QQ.-hl, (QQ,-|-1)Q,+ Q, J(QQ,-f-l) Q.+Q] Q3+CIGI. + I
forX.;0, 1, Q,+o. QQ, + 1 , (QQ.-hl) Q3 + Qi
that is, multiplying each successive value by the quantity under which it stands,
and adding the second preceding one to the quotient.
These expressions have several curious and interesting properties, which there
is not room in this work to touch upon. One or two, however, will be stated
under Continued Fractions, a little further on.
5. As a form of work we may adopt with advantage that given on the corre-
sponding occasion in the arithmetic, p. 43; and, generally speaking, we may avail
ourselves of the method of detached coefficients. Also, to avoid the introduc-
tion of fractional coefficients in the successive divisions, we may, by cross-mul-
tiplication of the coefficients of the highest terms of divisor and dividend, reduce
the leading coefficients to identity.
6. When the given expressions can visibly be resolved into factors, it is
always better to do so to the utmost possible extent. The factors which are
common to both are common measures ; and if the resolution has been com-
plete, all the common measures will thus have been obtained ; and their con-
tinued product will be the greatest common measurg.
GREATEST COMMON MEASURE.
185
EXAMPLES.
Ex. 1. Fiud the greatest common measure of the expressions «* — i^y' +
ar.2 y3 — yS ^jj(j x^ -\- x^ y^ — x^ y^ — y*.
Apply art. (6) : then we have
X=x'-X^y' + X^y^-y^ = (ix^ + y^) (jJ _ yi)
= ix' ~xy + y^-) (x + y){x + y) {x — y) and
X, = a?* + a^y-^ — a^y^ — y= =i (jfi — y^ (x^ + y^)
= (X — y) (x^i + xy + y2) ;
and we see that the only factor common to both terms is x — y ; which is
therefore the greatest common measure.
Ex. 2. To find the greatest common measure of a^ — ah- and a' -|- 2ab -\- b'.
a^ + 2ab + b^) a^ — ab^ (a
a^ + 2a^b + ab-
— 20^6 — 2ai0 a2 + 2ab + 6^ (
or dividing by — 2ab, which is not a divisor of the other quantity,
a + b)a^ -^ 2ab + br (a + 6
a^ -\- ab
ab + 62
ab + 62
Therefore a + ^ is the greatest common divisor.
But by detached coefiicients, and under the indicated arrangement, (art. 5,) it
would stand thus :
1 + 1
1
1
+
+
2
1
+ 1
+ 1
+ 1
1
1
.
1+0—1+011—2
1 + 2+1
—2—2+0
—2—4—2
2 + 2, or dividing by 2,
1 + I
Hence la + 1 J, or a + 6 is the greatest common measure.
Ex. 3. Find the greatest common measure of x" + 3** — 6i^ — Gj:' + 9x^ +
3a: — 4 and Qxr- + 15x< — 24*3 — ISa^ + 18a; + 3.
6 + 15—24 — 18 + 18 + 3
2+ 5— 8— 6+ 6+1
13 multiplier.
26 + 65—104—78 + 78 + 13
26+ 8— 60— 8 + 34
57—44—70 + 44 + 13
26 multiplier
1482—1144—1820 + 1144 + 338
1482+ 456—3420— 456 + 1938
1+ 3— 6— 6+ 9+ 3— 4
4^-12-24— 24 + 36+12 — 16 i2 + l
4+10—16—12+12+ 2
2— 8 — 12 + 24+10—16
2+ 5— 8— 6+ 6+ 1
_13_ 4 + 30+ 4 — 17
— 2 multiplier
26+ 8—60— 8+34
57 multiplier
—1600+ 1600+ 1600—1600 1482+456— 3420— 456— 1938| 1
or dividing by — 1600 or dividing by 114
1—1—1+1
13 + 4 — 30 — 4 + 17
13 — 13 — 13 + 13
17 - 17 — 17 + 17
17 - 17 — 17 + 17
13+17
136 ALGEBRA.
And as there is no remainder, we have by restoring the letters x^ — x^ — x -\- \
for the common measure of X and X,.
Ex. 4. Find the greatest common measure of a* — 4 and ab + 26.
Ex. 5. And of w" — a^b and a* — ¥.
Ex. 6. And of a^x + 2aV + 2ax^ + x^ and 5a> + lOa^x + SaV.
Ex. 7. And of 6ar* + 20x* — \2x'^ — 48x2 ^ 22a; + 12 and x^ + Axr" — Zx*
— I6a?3 + lla^ + 12a; — 9.
II. The least common multiple.
1. If the quantities be prime to each other, their least common multiple is
their product.
2. If one of them be a multiple of all or any of the others, whatever is a mul-
tiple of this is a multiple of those others; and the least common multiple which
takes in this greater quantity will be the least common multiple of all those
others.
3. Let X, X„ X„ .... be the quantities, no one of which is a multiple of any
of the others ; and let the greatest common measure of X and Xi be V, such
that X := mV and X, = nV. Then m and n are prime to each other ; and the
least common multiple M of X and X, is mnV, for it is the least quantity di-
visible by mV and nV, or by X and X,. But
^ „ mV . nV XXi product of the quantities.
V V their greatest com. meas.
Again, the least common multiple of X, X„ and X,„ is the least common
multiple of M and Xj,. Suppose V, to be the greatest common measure of M
and Xn, then the least common multiple of X, X„ and X„, is
^ MXi, XX, Xj[ product of the quantities.
' V, VVj product of their g. c. measures.
Proceeding in the same manner, we find for p quantities
j^ XX, X„ . , . . p factors product of the quantities
~' V Vj .... {p — 1) factors product of their g. c. measures.
EXAMPLES.
Ex. 1. Required the least common multiple of a^ -\- a"b and a- — J-.
Here a -f- 6 = V, is their greatest common measure. Hence
(g^ + o?b) (g^ - b^ _ a^ (a + bf (a - b) _ _
^-+"6 = a~+"6 ^^ ^ ^ ^ ^ ~ ^ •'■
Ex. 2. Required the least common multiple of x^ + x^ -\- x + 1 and x^ — x^
+ X — 1.
Here the greatest common measure is a;- -f 1 = V, and hence
M _ i^-'^ + x-l)ix^ + x' + x+l) _ . ,
^"+1 - ^ - 1.
Ex. 3. Required the least common multiple of a^+ab, a*+a^^ and a^—b".
Here V ^ a, and hence
M = ^^' = '°' + °^)f + °1>') = „.(» + « (a» + J=).
Hence again V„ the greatest common measure of M and X,,, or of a* (a -|- b)
(a* -h 6») and C — b^,is a -\- b ; and therefore
M. = ^'^ a^^a^b)^a^ + b^)ia^-b^) ^ ^, ^^, _ ^,^^
FRACTIONS. 137
Ex. 4. The least common multiple of x* + ax' — 9a'x» + lla^x — 4a* and
X* — ax^ — 3a^3^ + 5a^x — 2a* is a;* + 3ax* — 7a'x' — 7(^2r' + I8a*x — 8a*.
Ex. 5. The least common multiple of x" — a?x — ax» + a', x* — a*, and ax*
-f a'x — a^x^ — a* is ax^ — a^x* — a^x + o*.
The arithmetical process at p. 48 is only a convenient method of putt'ng down
the work, where the divisors or common measures that will suit are found by
inspection or by trials.
ALGEBRAIC FRACTIONS.
Algebraic Fractions have the same names and rules of operation, as
numerical fractions in common arithmetic ; as appears in the following Rules
and Cases.
CASE I.
To reduce a mixed quantity to an improper fraction.
Multiply the integer by the denominator of the fraction, and to the product
add the numerator, or connect it with its proper sign, + or — ; then the deno-
minator being set under this sum, will give the improper fraction required.
EXAMPLES.
1. Reduce 3i, and a to improper fractions.
T,- . o4 (3x51 + 4 15+4 19,,
First, 31 = — = — =^ — = — the answer.
5 5 5
. , b {a X x) — b ax — b ,
And, a = = the answer.
XX X
2. Reduce a + -j^ and a — ~~~ — to improper fractions.
^. , a2 (a xb) + a" ab + a^ .
Fust, a + T = ' r— ' — = — L — the answer.
z^ — a"^ a-' — z' + a^ 2a^ — z- .
And. a -= — = the answer.
a a a
3. Reduce 5^ to an improper fraction.
4. Reduce 1 — — to an improper fraction.
5. Reduce 2a to an improper fraction.
4ji J 8
6. Reduce 12 H to an improper fraction.
5x
7. Reduce x -\ ^^ '- to an improper fraction.
2^ 3a
4 Ix 7 5 . r •
8 Reduce + to an improper fraction.
9 3 5a
8
„ . 30x — 100 X 10-' ^ . - .
9. Reduce 4 — 2 x 10--a? + .T^s^TTo'xTo^ "* improper fraction.
138 ALGEBRA.
6^3? 0^ — 4*r
10. Reduce '001 — — — — . — —-^ to an improper fraction.
100 lOOiT— 5x10"^ ^ ^
x" + *
11. Reduce l-fx + ar^ + ai' + x* ar"H to an im-
\ —X
proper fraction ; and likewise o' — a-x + aar — x^ ; and again the
b P 5"+' a
quantities 1 -\ 1- „ + . . . — xr . y, and I — 2x + 2x" — 23:' + 2x* —
^ a or 0'+' a — b
. , . . H — — to improper fractions *.
CASE ir.
To reduce an improper fraction to a whole or mixed quantity.
Divide the numerator by the denominator, for tlie integral part ; and set the
remainder, if any, over the denominator, for the fractional part ; the two joined
together will be the mixed quantity required.
EXAMPLES.
1. To reduce — - and r — to mixed quantities.
3 0
First, 'I = 16 -7- 3 =: 5], the answer required.
. . ab -{- a^ „ , a^ .
And, T = (,ao + a-) -r- o = a + r- . Answer.
0 0
^ n_ , 2ac — 3a^ , Sax + 4tx^ ^ . ,
2. To reduce and ; to mixed quantities.
c a -\- X ^
_ 2ac—3a^ ,^ .^ 3a-
Jirst, = (2ac — 3a-; -i- c = 2a . Answer.
c c
And, ^""^ "^ ^^' = {3ax + 4x^ ^ {a + x) = 3x + -^. Ans.
a + X a + X
3. Reduce - and to mixed quantities. Ans. G|, and 2x— — .
5 o a
4a^x 2a* + 26
4. Reduce and j— to whole or mixed quantities.
2a a — 6 ^
3j;2 ^y2 2x^ 2v'
5. Reduce ~ and — to whole or mixed quantities.
x + y x — y
- „ , lOa^ _ 4a + 6 ^ . ,
6. Reduce to a mixed quantity.
7. Reduce 3^ _|_ 2a^ _ 2a — 4 *° * ™"^^*^ quantity.
CASE III.
To reduce fractions to a common denominator.
Multiply each numerator by all the denominators except its own for a new
* In such examples as these the multiplication may be advantageously performed by detached
coefScients.
r
FRACTIONS. 139
numerator of the equivalent fraction ; and all the denominators tofjether for a
common denominator to all the fractions equivalent to the given ones •.
It will be convenient in putting down the work to write all the numerators in
succession in a vertical column, and commence the factors which follow by the
denominators of those fractions which succeed them.
EXAMPLES.
n ■, a y , b
1. Keduce -, ,, and — to a common denominator
X 0 c
a X b X c = abc'^
y X c X X = cay I = new numerators.
— bxxxb=z—b^x}
a? X 6 X c = bcx =r common denominator.
And the fractions are , , -r^-, and — r— .
OCX OCX OCX
a ^"~ X Q, "^ X
2. Reduce — ; — and to a common denominator.
a -\- X a — X
{a — x) (a — x) = a^ — lax + ir)
(a + x){a^-x) = a2+ 2ax + x^] = "'^^ numerators.
(a -\- x) {a — x)=. a^ — 3^ =. new denominator.
A J tu f *• a^ — 2ax -\- x" , a= + 2ax + x^
And the fractions are 3 : — and -^ — - .
o- — X' or — x^
3. Reduce and ;^ to a common denominator. Ans. ^^ and .
X 2c 2cx 2cx
4. Reduce -7- and — to a common denominator.
0 2c
. 4ac , 3ab + 26»
Ans. -7- and \ .
26c 26c
5. Reduce -— and — , and — 4c?, to a common denominator.
3a? 2c
. lOac , 9bx . 24cdx
Ans. -z^ — and - — and „ — .
bcx bcx OCX
6. Reduce — - and — and 2b .- to fractions having a common
0 4 o °
, . . . 20b , I8ah , 48b' — 72a
denominator. Ans. — — r and and — , — .
246 -Mb 24b
1 2a- 2a^ 4- b^
7. Reduce - and and , y- to a common denominator.
3 4 a -f- 0
8. Reduce — -, and and to their least common denominator.
4a^ 3a 2a
X It z 2/1 4o oc
9. Reduce ^ -f — = i| and ^ + ;;- = H/g, each to a common
2a 4b 6c * 3x 6y 9z ^'
denominator.
JO. Reduce -Sa — "036, and -0030 — 10~V to a common denominator.
11. Reduce , + ,, and -|- r — , — , to common denominators.
a — b a + b a b 1 1
* For this is only multiplying the numerator and denominator of each fraction by equal
quantities, which does not aher its value.
140 ALGEBRA.
CASE IV.
To reduce fractions to their least common denominator.
Find the least common multiple M of their denominators for a new denomi-
nator. Divide M by each of the denominators D, D,, D, ...., and multiply
the corresponding numerators N, N,, Nj, by these quotients, for the correspond-
ing nt-'w numerators.
EXAMPLES.
Ex. 1. Reduce -r— , - — , ~r. — , -t-> ^^^ — ^ to a common denomi-
00 ac ao c 0 a
nator.
Here abc is the least common multiple of the denominators.
Therefore -,— ^ a, and %- X - ^ ~r-
be be a abc
abc , :, i b b^
— = b, and — x 7 = ,
ac ac b abc r mi
Ihe proposed
abc ^ c c c" \ . ., J
= c, and — X - = ^ I fractions reduced
to the least
com. denom.
abc.
ab ' ab c abc
abc , , ab ab a^b-
— = ab, and — x -r = -r-
c c ab abc
abc , ac ac a"c^
-^- = ac, and -r x ~= -7-
0 0 ac abc
abc , .be be b"c^
— = be, and — x 7- = ,
a a be abc y
Exx. 2, 3, 4. Reduce Ex. 6, 8, 9, of the last case to their least common deno-
minators.
CASE V.
To reduce a fraction to its lowest terms.
Find the greatest common measure of its numerator and denominator. Then
divide both the terms of the fraction by the common measure thus found, and
it will reduce it to its lowest terms at once, as was required. Or, divide the
terms by any quantity which it may appear will divide them both, as explained
in art. (6), p. 134.
EXAMPLES.
1. Reduce -.r1^,-o to its lowest terms.
ac^ + bc^
Here a + 6 is the greatest common measure, by which, dividing the nume-
rator and denominator, we have -„ for the fraction in its lowest terms.
(j3 __ ^2^ ^ ^^
2. To reduce ., , ^, , ,„ to its least terms. Ans. j-.
c^ + 2bc + b- c + b
3. Reduce --. ., „ to its lowest terms. Ans. -— , , , „ .
c* — b^c^ c» + bc^
a^ ~¥ 1
4. Reduce -7 r^ to its lowest terms. Ans.
04 _ H "" *-"-°« »• ""-• fj^i ^ 1,2'
r
FRACTIONS. 141
6. Reduce ^3_^^^,^^_-^^__. to ,ts lowest terms.
;. Simplify 20a^« - 12^ + 16^ - 1 5x3-_14x^-i5x+_4
CASE VI.
To addfy actional quantities together.
If the fractions have a common denominator, add all the numerators together;
then under their sum set the common denominator, and it is done.
If they have not a common denominator, reduce them to one (the least), and
then add them as before.
EXAMPLES.
1. Let „ and be given, to find their sum.
TT a , a 4a , 3a 7f^ ■ , . ,
Here -+-=:,- + --=— IS the sum required.
34121 2 12 *
2. Given r, -, and -„ to find their sum.
be a
,, a b c acd , b'^d , bc^ acd -f b-d -\- b(^ , . ,
Here r H h j = r— , + t— > + t— , = j— t the sum required.
bed bed bed bod bed *
*3. Let a , and b + be added together.
TT 3a?^ , , . 2ax Zcx^ , , , 2abx , 2fl6x— Scar.
Here«--^ + 6+ ^ =«- ^^ + J + -^ =a + 6 + — ^^—
the sum required.
^ . , , 4a! , 2ar ^ . . 20bx + Gax
4. Add -- and -r together. Ans.
3a bb * \bab
5. Add - , — , and , together. Ans. iZa.
3 4 5° "
,, 2a — 3 5a^ . . 9a — 6
6. Add and — together. Ans.
4 8 * 8
* In the addition of mixed quantities, it is best to bring the fractional parts only to a rommon
denominator, and to annex their sum to the sum of the integers, with tlic proper oign. And the
same rule may be observed for mixed quantities in subtraction also. If, however, in the sum
of the fractions thus obtained there should happen to appear integer quantities, these will be
better brought out and united with the integers of the given quantities.
Though, it must be further remarked, it is often the more convenient roelho<l to thus reduce
the quantities to a mixed state, especially in the arithmetical part of a process; yet it al«o
frequently happens (and this more particularly where i\\e formula itself is the object of consi-
deration) that the more elegant result is obtained by reducing the whole to the form of one
fraction.
See also, the note to addition effractions in the arithmetic, p. 50.
142 ALGEBRA.
^ *jj« «+3^ , 2a — 5 AC 140—13
7. Add 2a '— to 4a — . Ans. 6a — •
5 4 20
8. Add — 6a, and r- and — j— together.
40 3o
9. Add — , and — , and — togetber.
4 5 7
10. Add the several results furnished by each of the questions in Case in.,
taking each of the examples in that case as an example in this.
11. Add — 2a, and — , and — ]^ ~ « ( together.
12. Add 8a + "— and — |2a — — [ together.
13. Fmd the sum of and ; and likewise of — \x — —}■
a + X a — X ( X -\- y)
and- \x + ^^].
{ x — y)
14. Ascertain the sum of —-5-7 — ; — :, — v^ ;. and ^ „ ^^ , - — .^
4a^ (a + x) 4a' (x — a) 2a^ x — {a- + x-);
and that of -^ 7., — r— , and .
X- — y X -\- y X — y
15. Express ^ + — — 22. in single fractions adapted to each of
the cases where + and — are taken.
CASE VII.
To subtract one fractional quantity from another.
Reduce the fractions to a common denominator, (the least is the more
elegant) if they have not a common denominator.
Subtract the numerators from each other, and under their difference set the
common denominator.
EXAMPLES.
1. To find the diflerence of — and - .
4 7
_. 3a a 21a I6a 5a . ^, ,.». . ,
Here = = — is the difference required.
4 7 28 28 28 ^
2. To find the difference of and —-7 — .
4c 3o
„ 2a— 6 3a— 46 _ 6a&— 36- _ 12nc— I66c_ 6ab~3b^— 1 2ac-f- 1 6bc
"^ ~4c W~ ~ 126c 126c ^ — 126c
is the difference required.
„ c , 4a , 3o , 3a - 4a
3. Subtract — from — , and — from — .
7447
4. Required the difference of -— and -—'
^ ^ , 2a — b . 3a — 4b , ^ , 2a - „ a — 3b
5. Take - — - — from — -,— , and 4a H from 2a — .
4c 36 c 2c
r
FRACTIONS. 143
6. Subtract — from — , and 6a from
4 3 4
7. Subtract — from — j — and from — -/"— .
c b b
_,,2o+6- 4a + 8 ,3a + c. 26
8. Take — -— from — _■— , and — ~ from —
9 5 6c
9. Take 2a - °-~- from 4a + -.
c c
,„ rp , 9a + b. 6a — A .6. 1
10. Take — from — — - — , and - from — 8 + .
4 J 7 7
11. Subtract - ^^^ from ^^f-, and likewise from - ^^"-±1.
9 5' 5
12. Which is the greater of the quantities (-6 and -" + 6, and what is
their difference ? Likewise when taken with sign of a changed, viz. — " + 6 and
4
-*^ + 6>
13. Subtract — -— - from 5 , and hkewise from - — — ,.
X + I x^ — I X — 1
14. Subtract — — from — ; — , and —^ :; from -„-, — ,.
X — y X -\- y X' — y^ x- + y^
CASE VIII.
To multiply fractional quantities together.
Multiply the numerators together for a new numerator, and the denomina-
tors for a new denominator *.
EXAMPLES.
1. Required to find the product of „ and - .
8 5
„ a X 2a 2a2 a^. , ^ . ,
Here =r — - = -— the product requu-ed.
8 X 5 40 20 ^ ^
2. Required the product of -, - , and —.
a X 3a X 6a ISa-^ 3a^ . , ^ . ,
— — = = — the product required.
3X4X7 84 14 ^ ^
3. Required the product of ,- and
2a + c
• 1. When the numerator of one frartion, and the denominator of the other, can be divided
by some quantity, which is common to both, the quotients may be used instead of them : or,
in other words, the fractions may be reduced to their lowest terms be/ore tbey are multiplied
together.
2. When a fraction is to be multiplied by an integer, the product is found either by multiply-
ing the numerator, or dividing the denominator by it ; and if the integer be the same ■with the
denominator, the numerator may be taken for the product. This may be readily deduced as a
case of the general rule.
144 ALGEBRA.
4. Required the product of — and — .
T, • T , J . r 3a J 462 J . 3a , 4*2
5. Required the product of - and - - ; and of and — — .
^ rr, , ■ n 3a , 8oc , 4a6 ,
6. To multiply r> and -^-, and --— together.
' 0 0 3c
7. Required the product of 2a + -^ and -r- .
o D • A.x, J . f 2a2 — 262 4a2 + 1V^
8. Required the product of — -^- — and , .
9. Required the product of 3a, and and g tTa"
10. Multiply a + 2^-^.by^-2^ + ^,.
,, , . , - 3a— 46 2a— i6 8 , /4ir 15a;\ 12a;
11. Multiply together g^^I^^,- ,^' 3 and - (y^ - ^; " jy
^. , , ■, ,. 9 , , 1 , 6a 3a~'
12. Find the product of - -a, - IQ-^a-s, ~^a\ and — y^ x - ^— 2*
„ , . , ^ 62^c2_a2 a2_|_J2_c2 a2_i2^c2
13. Multiply together 1 ^^^ , 1 ^^^, and 1 - ^^^ 5
1-1 ,. . a+6 + c
and in the result put — - — =^ s.
CASE IX.
To divide one fractional quantity by another.
Divide the numerators by each other, and the denominators by each other, if
they will exactly divide : but, if not, then invert the terms of the divisor, and
multiply by it as directed in multiplication, p. 143 *.
EXAMPLES.
1 . Required to divide - by — .
TT « 3a a 8 8a 2 . . ^
Here --; = x— = =- the quotient.
4 • 8 4 3a 12a 3 ^
• If the fractions to be divided have a common denominator, take the numerator of the
dividend for a new numerator, and the numerator of the divisor for the new denominator. See
3. below.
2. When a fraction is to be divided by any quantity, the value is the same whether the nume-
rator be divided by it, or the denominator multiplied by it.
3. When the two numerators, or the two denominators, can be divided by some common
quantity, let that be done, and the quotients used instead of the fractions first proposed. This
is obvious from the circumstance that if these quantities be suffered to remain, they constitute a
common factor, or a common measure of the quantity which results from the division.
FRACTIONS. 145
3o 5c
2. Required to divide —. by
„ 3a 5c 3a 4d ,^uu u»u ,
3. To divide „ . by
3a— 26 '
„ 2a+6 4a+0 8a=i-6a6+6* , . . ,
""' 3"^=r6 ^ 3a+26 = 9a^-46^~ *^' 1"°^*°* "l"""^'^-
„ 3a2 a+b 3a»x (a+6) 3a • ,. .. ^ . ,
"^■^^ ~3 I ta X = , . . .^- = —: r-r-r5- w the quotient required.
a^+b^ a (a3+63) xa o*_a6+62 ^ ^
, ™ ,. ., 3a;, 11 •,6a;2, ^ j a+* . 3a'+36»
5. To dmde —-by — , and — - by 3x and — !— by —, — .
4 ' 12 5 ' a ^ 3a*
6. To divide — -^by— -, and ■ "" . by ^.
9 '3 2ir— 1 -^ 3
w m J- -J 4«, 3a J 2a — 6, 5ac
7. lo divide — by-y, and , , by . ,.
5 ' 5o 4co ' oa
„ „. ., 5a^— 56* , 6a24-5a6 . 10-*, l-^
8. Divide — ^ , , ,,,by , , and — „ by --.
2a2 — 4a6+262 ' 4a— 46 ' 10" ' 100
2ax-l^ i-a _j ,„u..(«+^)'
flaj-f ]
10. Suppose we divide jt^ + y^ by ar* — y\ and then divide the result by the
quotient of a? + y by a; — y ; what final quantity shall we get, provided i*
11. If we add three-fourths of a number to one-half its square, and divide the
sum by three-eighths of its cube j and moreover, if we subtract three-fourths
of that number from half its square, and divide the result by minus one-eighth of
the square ; what is the quotient of the former by the latter result ?
CASE X.
Continued fractions.
If we recur to the process for finding the greatest common measure, p. 133 ;
and denote for simplicity the functions X, X„ X.j . . . . by a, 6, c . . ., and the
quotients Q, Q,, Qj . . by a, /3, y ...; and if we also denote the divisions on
the left by fractions ; then we shall have those two columns converted into the
following : —
1 . a = a6 -|- c,
9. Divide a H 7 — by -, and 12 by-^^-'-
b=pc + d
c =z yd + e
d = ?e +f
or, as it is more conveniently written,
"■^^ + 7 + 5+.. .
The expression (either form) on the left is called a continued fraction. It has
many curious properties *, and in several inquiries is of great value. The only
* The idea of continued fractions was first started by Lord Brounckcr, the first President of
the Royal Society ; but the method owes its present elegant form to Lagrange, who made ex-
VOL. I. L
146
ALGEBRA.
one to which, however, the student of this course will have occasion to apply it,
is to find a series of fractions converging towards the true value of the given
fraction r, but having its terms expressed in smaller numbers than a and b.
The formation of the values of the terms of the converging fractions has been
stated at p. 134, art. 4, where the upper line expresses the numerator, and the
lower one the denominator of the converging fraction at the first, second, third,
and successive steps.
EXAMPLES.
365
Ex. 1. Represent — - in the form of a continued fraction, and find the con-
u 365 , , 1 ,
Hence- =1 + ^1 ^
3 + „
vergmg
i8 = l
fractions.
b \ a\
224 365 1 = a
141 224
S = 1
83 1411 = y
58; 83
K = 3
25 58 2 = £
24! 50
1
1
88 = JJ
8
0
Or, in Herschel's notation,
365 _ 1 1 1^ 1_ 1 I
m"'^"^! + 1 + 1+2 + 3 + 8
And forming the converging fractions according to the rule, they are
-, ~, ~, -, — , — , — , the last of which is the original fraction itself.
1 1 2 3 8 27 224
The properties of these fractions are : —
1 . That they are alternately less and greater than the true value.
2. That each of them is in its lowest terms.
3. That each of them is a nearer value than any other that can be formed
without taking higher values of the numerator and denominator.
4. That if -^ and -~ be two consecutive converging fractions, then
Pm Qm+i — Pm+i ?« = ± 1- Sbc Hlud's Algebra, pp. 284—306.
314159
Ex. 2, Find the fractions converging to
100000
tensive use of it, both in the solution of algebraical equations with numeral coefficients, and the
solution of indeterminate equations.
The latest improvements in the use of the method as applied to the solution of algebraical
equations, were made by the late Mr. Horner, and published in the Annals of Philosophy and
Quarterly Journal. The second notation above given was proposed by Sir John Herscliel, and
for economy of space, both in writing and printing, is tlie most convenient. For a good view of
the subject in an elementary form, tlie reader who wishes to go further into the subject may
consult Hind's Algebra or Young's Equations : — both of thenoi works deserving of cordial recom-
mendation.
INVOLUTION AND EVOLUTION. I47
By the common measure we have the quotients 3,7, 15, 1 ; and by the
rule for the formation of the terms we have - , — , — - , —
INVOLUTION AND EVOLUTION.
The term Involution has already been explained at pp. 118, 119, as the
multiplication of several equal factors together ; and it has been intimated that,
for the case of the factors being binomial, a much more concise process will be
given under the head of the Binomial Theorem. When there are more than two
terms in the expression to be involved, it will, at least during elementary study,
be better to have recourse to actual multiplication in the few cases that can arise,
than to employ the Multinomial Theorem.
Evolution is the extraction of roots; that is, the inverse operation of raising
powers, or of Involution. It constitutes but a very limited application of a
general process, viz. that of the general solution of algebraic equations *. The
rules usually given for Evolution are identical in substance with that here given
for the square and cube roots ; but in a form, though more briefly expressed,
implying much more actual work in performing them. This rule is in fact only
an expression of Horner's method of solving equations.
CASE I.
To extract any root of an expression composed of one single term.
Divide the indices of all its factors by the index of the root to be extracted,
and extract by any arithmetical process the root of the numerical coefficient.
The continued product of all these roots is the root sought.
Thus the n* root of jja'&'c' is p'a'b'c'; and the cube root of 27ti?1fi<? is
27'a^6'c' or Safi^cS.
Of the signs of the results, it is only necessary to recollect that all odd roots
of a negative quantity are real and negative ; of all positive roots, real and
positive ; that all even roots of a positive quantity are real, and either positive or
negative ; whilst all even roots of a negative quantity are impossible or imagi-
nary. This, however, is to be understood as applying to one individual root in
each single case : since, besides these, there may be several other roots, wholly
imaginary ; and in all instances above the square root there actually are such
roots. This view of the subject, however, belongs to a more advanced stage of
the study of algebra.
EXAMPLES.
1. The square root of 4a', is + 2a.
3
2. The cube root of 8a', is 2a' or 2a.
5fl26' . 'ab ,,
• An extension of the signification of the terms Inrolution and Evolntiem has been recently
proposed by a distinguished mathematician, Professor De Morgan, viz. to the composition and
resolution of algebraic equations. Involution and Evolution being particular cases of those
general problems, the extension is perfectly justifiable ; and in a treatise founded on this idea
there would at least be the advantage pf keeping those subjects together which were naturally
connected with each other. It would, however, interfere too much with the existing arrange-
ment of the work to adopt it here.
l2
148 ALGEBRA.
4. The cube root of W^' ^^ §~ V:2o.
5. The square root of 20^6^ is + ab^ V2.
6. The cube root of — 64a^6^ is — 406^.
nn, r^efib^ ■ , ^ah , 2
7. The square root of — 3- , is + ^ X''
8. The 4th root of 8la*b\ is + Sab s/~±b.
9. The 5th root of — 32a»6^ is — 2ab\yb.
10. The 6th root of 729a^b^\ is + Sail
CASE II.
To extract any root of a compound quantity.
The breadth of our page will not allow us to exhibit an example of a higher
root than the third ; but we shall enunciate the rule for all cases, and give the
form of work for the square and cube roots.
1. Write the given quantity, arranged according to the powers of some one
letter * in the place of the dividend, and the curve to the right for the root, as
in the arithmetical square and cube roots, pp. 66 — 72.
2. Make n columns to the left of the given expression, numbering them back-
wards from that expression as columns (0, (11), (in), .... (n).
3. Extract the N" root of the first term of the given expression, and put that
root in the column to the right of the curve. Denote, for the purpose of con-
tinuing the directions for working, this root by r, whatever that root may be.
4. Put 1 in column n ; 1 x r in col. (n— 1); (I x r) r in column (n— 2);
and continue the process till the last result falls under the given expression :
then evidently this result will be equal to the first term of the given expression.
Subtract this, and bring down N terms for a di^ndend.
5. Form a new horizontal line as follows. Multiply 1 by r, place it in column
(N — 1), and add it to the previous result in that column ; multiply this sum by
r, and add it to the next column ; multiply this sum by r, and add it to the next ;
and so on till the result falls in column (i).
Then form a new horizontal line in the same manner, adding each product
to the result above which it is written ; but stop in each horizontal line one
column sooner than in the preceding. We shall thus obtain a series of results,
one in each column, preparatory to evolving the next term of the root.
6. With the expression in col. (i) as a divisor and the first term of the new
dividend, find a new term of the quotient. This will be the second term of the
root, which is to be used in forming the several columns, as before described.
Proceeding thus, we shall obtain the successive terms of the root, if it be an
exact root, or as many of them as we desire, if it be not exact.
EXAMPLES.
1. Extract the square root a* — 4c'6 -f- 6a-b^ — 4aP -\- b*.
Here the index of the root is 2, and the highest power of a is the fourth.
4
Hence a^ = a", and the condition is fulfilled. Also the powers follow regularly,
and we can work with detached coefficients ; but to show the identity of the
methods, the work is put down here both ways. Also as n = 2, there will be
two columns, and we have
• The rule cannot be applied except the first term is of a power exactly divisible by the index
of the root ; that is, for the «"■ root we must have the first term x^ where m is an integer.
INVOLUTION AND EVOLUTION.
140
(")
(I)
1
a^
0?
(o)
a* — 4o»6 + Ga^fc' — Wh + 6« (a* — lab -\- b'.
a*
2a- — 2ab
— 2ab
— 4a^b + 6a-b^
— 4a»6 + 4a''b'^
2a3 — 4ab + 6»
2a262 — 4a-''6 + b*
2a-b^ — 4a^b + b\
which is worked in strict accordance with the rule, and it is evidently identical
with that given for the square root of numbers, p. 67 .
The same example, by detached coefficients.
(It) (I) (o)
2 — 2
— 2
1-4-1-6 — 4 + 1 (1— 2 + 1
1
— 4+6
— 4 — 4
I
2—4+1
2—4+1
4 + 1
Ex. 2. Extract the square root of a^ — a^.
Here c^ = a, and the condition of the applicability of the rule is fulfilled.
aP
Write it under the form a^ (1 5) : then we have to multiply the square root
of 1 2 by tb^* of a^, or by + a. Considering ^ ^ ^^^ second term, the
detached coefficients are 1 — 1 ; and the process will be as follows : —
(XI)
I
(I)
1
1
2-4
-4
2 — 1 -
-T^
(0)
1-1(1-4-^-1^-....
1
-1 + 0
-i+i
-i
2 — 1 -
-i-
-i + -h + n + ^>
» 1 I .
— tl 81 Hi »
in which the method of work is very simple and easy, and attaching the letters
( x^ X* x^ \
to the coefficients, we have + a ] 1 — ^ — ^ — Tg^ — • • • • j ♦ *"^
( ac' X* x^ \
multiplymg out it becomes + |°~^~^3~ fgjs ~ J •
Ex. 3. Extract the cube root of e« — 6a* + 21a^ — 44a' + 63a= — 54a + 27.
Here there are three columns, and the condition is fulfilled for the appUcation
of the rule. Whence
150
ALGEBRA.
(III) (11)
1 1
1
2
1
-2
-2
(I)
1
2
3
6 +
4
(o)
1—6 + 21—44 + 63 — 54 + 27(1—2 + 3
1
— 6 + 21-44
— 6+12— 8
3-
3 —
6 +
6 +
4
8
9 — 36 + 63 — 54 + 27
9 — 36 + 63 — 54 + 27
3-
-4
-2
3 —
12 +
12
9-
-18 + 9
3 — 6 + 3 3—12 + 21 — 18 + 9,
and attaching the powers, we have the root a^ — 2a + 3.
The several terminated courses of operations are marked by dark lines imme-
diately above their results.
4. The square root oi a* + 'ia'^b + lOa-b'^ + I2ab^ + 9b* is a^ + 2ab + 3h-.
5. To find the square root of a* + 4a^ + 6a^ + 4a + 1, and of 1 + 4a + 6a-
+ 4a3 + a*. Ans. a^ + 2a + 1, and 1 + 2a + a^.
6. Extract the square root oif a* — 2a^ -\- 2a^ — a + i. Ans. a^ — a + i-
b h^ b^
7. The square root of o^ — a6 is a r> .> — &c.
^ 2 8a I6a^
8. Find the square root of (a^ + ar) and of (x^ + a^).
9. The square root of a^ — 2a6 + '2ax + 6^ — 2bx + a?^, is a — b + x,
10. The cube root of a^ — 3a^ + 9a* — 13a^ + 18a2 — 12a + 8, is a= — a + 2.
11. The 4th root of 810" — 2l6a36 + 2l6a-b- — 95a¥ + l6b\ is 3a — 26,
12. The 5th root of a^ — lOa* + 400=* — 80a- + 80a — 32, is a— 2.
13. Find the square root of 1 — y^ and the cube root of 1 — x^.
14. Expand ^/(a^ — 2bx — xr^) and .^a^ — x- into infinite series.
15. Expand ^(1 + 1) = v'2 into a series. Ans. 1 + 5 — ^ + t« — its + ••••
16. Expand \/(l — 1) into an infinite series. Ans. 1 — 3 — ^ — is — m — ••••
17. Expand \/{aP + x) into an infinite series.
In all these cases, however, the binomial theorem will be more convenient and
effective.
SURDS.
[The student is advised to defer this subject till he has read some portion of
the simple and quadratic equations.]
SuHDS are such quantities as have no exact root, and are usually expressed by
fractional indices, or by means of the radical sign ^. Thus, 3% or ^3, denotes
the square root of 3 ; and 2', or V2^, or \y4, the cube root of the square of
2 ; where the numerator shows the power to which the quantity is to be raised,
and the denominator its root. The index may be put also in the form of a
1 -6 i L . .
decimal, and is often so used. As for a^ writing a , or for a~^ 0^, writmg
a-' b^^^ ••• See Definitions, p. 107.
SURDS. 151
PROBLEM I.
To reduce a rational quantity to the form of a surd.
Raise the given quantity to the power denoted by the index of the surd ;
then over or before this new quantity set the radical sign, and it will be of the
form required.
Note. When any radical quantity has a rational coefficient, this coefficient
may be put under the irrational form, and the whole of the factors thereby
brought under the symbol of radicality. Thus, instead of SOyi'A, we may put
//3ax3ax6, or s/9a^b ; and so of others.
EXAMPLES.
1 . To reduce 4 and — 4 to the form of the square root.
First, 4' = 4 X 4 =: 16 ; then + v^l6 is the answer.
Second, (— 4)" = — 4 x — 4 = 16, then — ^16 is the answer.
2. To reduce 3a^ to the form of the cube root.
First, 3a2 X M" X 3a2 = (3a^)^ = 21 a^ ; then we have V27a* or (27a«)^.
3. Reduce 6 to the form of the cube root. Ans. (216)* or V216.
4. Reduce — \ab to the form of the square root. Ans. — s/^a-b-.
5. Reduce + 2 to the form of the 4th root. Ans. + (16)*.
6. Reduce a^ to the form of the 5th root.
7. Reduce o + « and a: + a to the form of the square root.
8. Reduce a — a? to the form of the cube root.
9. Reduce (— 40^ + b) V— a^^ to the form of the sixth root ; and likewise
to the form of the square root.
10. Transform (a — 6) (a + 6) \/a^ — b^ into its simplest form, and likewise
represent them as radicals of the 3d, 4th, 5th, and 6th degrees.
11. Reduce (4 — 1) x 3 to the form of a cube root; and then reduce
+ be ^/^+^l — be V+ I to the simplest form it admits of.
PROBLEM II.
To reduce quantities to a common index.
1. Reduce the indices of the given quantities to a common denominator, and
involve each of them to the power denoted by its numerator ; then 1 set over
the common denominator will form the common index. Or,
2. If the common index be given, divide the indices of the quantities by the
given index, and the quotients will be the new indices for those quantities.—
Then over the said quantities, with their new indices, set the given index, and
they will make the equivalent quantities sought.
examples.
1. Reduce 3^ and 5*^ to a common index.
Here, 5 and J = fj and ^.
Therefore 3^« and S^'* = {,3')"^ and (5^)^'' = 'V3* and '%/5' = iV243 and >V25.
152 ALGEBRA.
I
2. Reduce a^ and 6' to the same common index ^.
Here, ? h- i = i X f = i the 1st index,
and J^i = Jxf = 5 the 2d index.
12 1 5
Therefore (a^)^ and (J^)^ or \/a^ and .^6' are the quantities.
1 i
3. Reduce 4^ and 5^ to the common index ^, and 3^ and 5~* to the common
index "2, and then to the common index —-2.
Ans. (256*)* and 25^ ; (S^'y^and (5-2)2; (3-*^)-'' and (S^)-*.
4. Reduce a^ and x^ to the common index 5. Ans. (a^y and (x^)^.
5. Reduce a* and a;^ to the same radical sign. Ans ^/a* and ^ x^-
6. Reduce (a + xY and (a — a?)^ to a common index,
7. Reduce (a -f- ^) ^ and (a — b)~^^ to a common index.
8. Transform a-^i" Vd^^s to another quantity whose index is — 3*25.
PROBLEM III.
To reduce surds to simpler forms.
Divide the surd, if possible, into two factors, one of which is a power of the
kind that accords with the root sought ; as a complete square, if it be a square
root ; a complete cube, if it be a cube root ; and so on. Set the root of this
complete power before the surd expression which indicates the root of the other
factor ; and the quantity is reduced as required.
If the surd be a fraction, the reduction is effected by multiplying both its
numerator and denominator by some number that will transform the denomi-
nator into a complete square, cube, or other requisite power : its root will be
the denominator to a fraction that will stand before the remaining part, or surd.
See Ex. 3, below.
Recollect that // - = ' «/«: thus ^/^ = J >/2, \^\ = I ^/Z, and so on.
EXAMPLES.
1 . To reduce ^^32 to simpler terms.
Here V'32 = ^16 X 2 = ^16 x a/2 = 4 x -v/2 = 4 a/2.
2. To reduce \/{Z20) to simpler terms.
3^320 = V (64 X 5) = V64 X V5 = 4 X VS = ^V^-
3. Reduce V^ to simpler terms.
,44 .44 y 44 5\ ,4.11.5 ,,22.55 2 ,^.
^75-^T575 = AT5:5- J = ^15715"= '^-15^= Is ^^"^
4. Reduce a/75 to its simplest terms. Ans. 5v^3.
5. Reduce V — 189 to its simplest terms. Ans. — 3V7-
6. Reduce V + W to its simplest terms. Ans. +JVIO.
7. Reduce a/ —750^6 to its simplest terms. Ans. ba^Zb^—l.
8. Express the square root of +a''b^c* in the simplest form.
Note I. There are other cases of reducing algebraic surds to simpler forms,
that are practised on several occasions ; one of which, on account of its simpli-
city and usefulness, may be here noticed, viz. in fractional forms, having com-
pound surds in the denominator, multiply both numerator and denominator
SURDS. 153
by the same terms of the denominator, but havinj? one sign changed, from + to
— or from — to +, which will reduce the fraction to a rational denominator.
Ex. 1. To reduce \^^ — \/3' "'"''•'P^y i*^ '»>' J^. \ /!» *"<^ '^ becomes
*1 * ^ 3^/15 — 4^/5 ,. , . , -v/15— -v/5 ,. ,
Also, to reduce — ..■. . .., , multiply it by -r— -,,, and it becomes
^15+^5 " ' ' V15 — v'a
65—7 n/75 _65 — 35v/3 _ 13 — 7\/3
15 — 5 ~" 10 ~ 2 ■
And the same method may easily be applied to examples with three or more
surds.
Ex. 2. Reduce the fractions ^^ ~ ^! x ^^^, ~ ^-^ and ~ ^ "^ ^^^
a/3 4- ^/5 v/2— 4 15 — v^a:»
to others having rational denominators : and then to such as have rational
numerators.
A'o^e II. In the same manner may any binomial surd be rendered rational in
the denominator, whatever the degree of the radicals may be. If, for instance,
Q
the surd had been ^-^ /W *^^" *^^ multipher would be V c? + V«* + V ^S
and the surd itself become -^ ^7— r — — — •
a ±b _ _
And generally since 7~ — „ , = V""' + V'"""^ ^ + X/a''^ i^ -f
by actual division, the rule may be extended as we have stated above.
2 45
Ex. Rationalize the denominator of -j-— ^T- ' ^^^ °^ , j . ; and of
V < — v 3 2 i V 3
3 , .V3 + V4
V5-V2-^"^"^V3+V4-
PROBLEM IV.
To add surd quantities together.
1. Bring all fractions to a common denominator, and reduce the quantities to
their simplest terms, as in the last problem. 2. Reduce also such quantities as
have unlike indices to other equivalent ones, having a common index. 3. 'llien
if the surd part be the same in them all, annex it to the sum of the rational parts,
with the sign of multiplication, and it will give the total sum required.
But if the surd part be not the same in all the quantities, the addition can
only be indicated by the signs + and — .
EXAMPLES.
1. Required to add \/18 and ^/32 together.
First, v'lS = v'g X 2 = 3 s/2; and a/32 = ^16 X 2 = 4 v^2.
Then, 3 ^2 + 4 a/2 = (3 -f 4) a/2 = 7 -v/2 = sum required.
2. It is required to add V375 and VI 92 together.
First, V375 = Vl25 x 3 = 5V3 ; and V«92 = '^6i X 3 = 4V3.
Then, 5V3 + 4V3 = (5 + 4) V3 = 9V3 = sum required.
3. Required the sum of a/27 and v'48. Ans. 7 >/3.
4. Required the sum of — a/50 and v^72. Ans. ^2.
5. Required the sum of — Vi and a/t^. Ans. — I'j ^15.
154- ALGEBRA.
G. Required the sum of V56 and V — 189. Ans, — V7.
7. Required the sum of Vj and V^- Ans. J V2.
8. Required the sum of 3 ^cP-b and 5 //l6a*6. Ans. (3a-)-20a2) v'i.
9. Find the sum oi^^J —TJa? and + \/64a^. Ans. 5a or —11a.
10. Find the sum of — "v^ , ,^ and - — ~ ^ _,: alsoof a^/8, 6s/18,
^/8 + ^/2 ^5 + ^/15 ^
0^/27, — &\/45, 6v^l25, and a^/147 ; and again of — 2^/32, 9.243% 5.68"5,
17 X 54^ 3V432, Vl28,v'1452, Vl458, 363% and lU/1331.
11. What is the sura of — 48-5a — 25-6, and Ir'^a — 6-*6 ?
PROBLEM V.
To find the difference of surd quantities.
Prepabe the quantities the same way as in the last rule; then suhtract the
rational parts, and to the remainder annex the common surd, for the difference
of the surds required.
But if the quantities have no common surd, the subtraction can only be indi-
cated by means of the sign — .
EXAMPLES.
1. To find the difference between ;y/320 and vi'80.
First, ^320 = a/64 x 5 =8^/5 ; and -v/80 = a/16 X 5 = 4^/5.
Then, 8\/5 — 4^/5 = 4^5 the difference sought.
2. To find the difference between V'l28 and \/54.
First, V128 = Vo4 X 2 = 4V2 ; and V54 = V27 X 2 = 3V2.
Then, 4\/2 — 3V2 = ^\/2, the difference required.
3. Required the difference of a/75 and ^ 48. Ans. ^3.
4. Required the difference of V256 and V32. Ans. 2V4-
5. Required the difference of \/| and V^. Ans. J\/3.
6. Find the difference of */§ and a/|. Ans. -^a/G.
7. Required the difference of Vi and Vf- Ans. fj V75.
8. Find the difference of \/24a'b- and ^^516^. Ans. (36- — 2ab) ^6.
9. Subtract — 9 * from 64~' and add the sum to the difference of a/J and
3-^.
10. From the half of — a/I take a third of ^J.
11. Find the difference of ^S — a/2 and a/252 — a/175. Ans. 1-2315377.
01.. . a/8 — a/2 ^ a/252 + a/175 , a/5 — v/3 .
12. Subtract — .„ , ,^ from /„^^ /.I., and T^^ , ^,,^ from
a/8 -f V^2 a/252 — a/175 a/20 + a/12
n/15 •-V'5 34 , I .„ r - ,.,
Vlil^S ^°^- y. and - 2 n/3 { 2 - Vo].
,0 m 1 a/o + -v/fi , Va — s/b J a/5 — v/1 r '^^^ + ^
13. Take , , from -^^^ — --^,-. , and ., , ^,, from -j-
a/o — v'fi Va + V6 a/3 + a/I a^5 — 1
PROBLEM Vr.
To multiply surd quantities together.
Reduce the surds to the same index, if necessary ; next multiply the rational
quantities together, and the surds together ; then annex the one product to the
SURDS. 155
other for the whole product required ; which may be reduced to more simple
terms if necessary.
EXAMPLES.
1. Required to find the product of 4\/l2 and 3 \/2.
Here 4 x 3 x s/12 x a/2 = 12 ^12 X 2 = 12 V2i = 12 ^^4 X G = 12 x 2
X \/6 = 24 >y6, the product required.
2. Required to multiply i VJ ^>y :! Vi-
Here i x I Vi x Vi = t j X Vli = Vj X Vi5 = li X i x Vl8 = Vs V'S.
the product required.
3. Required the product of 3^/2 and 2^/8. Ans. 24.
4. Required the product of 3V4 and | •V^2. Ans. ^ V^-
5. To find the product of — * a/'I and -i*g V'i. Ans. — ,j}, \/15.
6. Required the product of 2\/— 14 and 3 V*. Ans. — 12 V~'
2 4
7. Required the product of 2a? and a'. Ans. 2u^.
1 a
8. Required the product of (a + b)' and (a -f b)*.
9. Required the product of 2x + <yi and 2x — \/b.
10. Required the product of — (a -f 2 ^/6)^, and (a — 2 V*) •
11. Required the product of — 2a?« and — 3a?»
1 1
12. Required the product of 4x« and 2y«.
13. Multiply-^2+ ^^3by^^^^^^andV3.(s/5-l)by^27(v/5+l);
and add the resulting products together.
14. Required the continued product of — ^, , ' ,- , 7 — 3 \/5
15. Reduce C+a)^ X (+a)~^ X (+a)^ X i+a)'* to a single term.
PROBLEM VII.
To divide one surd quantity by another.
Reduce the surds to the same index, if necessary ; then take the quotient of
the rational quantities, and annex it to the quotient of the surds, and it will ^ive
the whole quotient required ; when the result can be reduced to more simple
terms, it should then be done.
EXAMPLES.
1. Required to divide 6^/96 by 3^8.
Here 6-r3.-v/(96-T-8) = 2A/12 = 2v'(4 X 3) = 2x2v'3=:4^3,
the quotient required.
2. Required to divide 12 V280 by 3 \y5.
Here 12 -7- 3 = 4. and V^G = V^ X V7 = 2 V7 ;
therefore 4 x 2 x V7 = 8 \/7, is the quotient required.
3. Let 4 a/50 be divided by 2-v/5. Ans. 2^10.
4. Let 6 VlOO be divided by 3 V-5- Ana. — 2V20.
156 ALGEBRA.
6. Let I VJl, be divided by |^|. Ans. 1^/5.
6. Let f Vt6 be divided by | V§. Ans. ^ V30.
7. Let i^/a, or |a^, be divided by la\ Ans. |a«.
4 3 1
8. Let — a' be divided by 4a3, and 5a~^ by — a^.
11 11
9. To divide 3a' by 4a", and So— by4o~".
PROBLEM VIII.
To involve or raise surd quantities to any power.
Raise both the rational part and the surd part. Or multiply the index of the
quantity by the index of the power to which it is to be raised, and to the result
annex the power of the rational parts, which will give the power required.
EXAMPLES.
1
1. Required to find the square of |o^.
Fu^t, (J)" = J X I = ■^, and (a*)^ = a^ ^ ' = a^ = a ;
therefore (Ja*)^ = j^go, is the square required.
2. Required to find the square of ^a^.
First, i X i = i, and (aV = a^ = (^Va;
therefore (^a^)'- ^ ^a^Va, is the square required.
3. Required to find the cube of § >/ 6 or | x 6- .
First, (§)3 = § X § X § = 5^ and (6^)^ = 6^ = 6 V 6 ;
therefore (§ ^/6)^ := -»^ x 6 \/6 = '5^ ^^6, the cube required.
4. Required to find the square of 2 V2. Ans. 4 V**-
5. Required the cube of 3", or ^/3. Ans. 3 .^3.
6. Required the 3d power of — ^ //3. Ans. — ^ ^^3.
7. Required to find the 4th power of § \/2. Ans. J.
8. Required to find the — mth power of — a '.
9. Required to find the square of 2 + >/3.
PROBLEM IX.
7b evolve or extract the roots of surd quantities.
I. When the given expression contains but one term, extract both the rational
part and the surd part. Or divide the index of the given quantity by the index
of the root to be extracted ; then to the result annex the root of the rational part,
which will give the root required.
EXAMPLES.
1. Required to find the square root of \6 ^&.
First, Vl6 = 4, and (6^)^ = 6^ "^ ^ = 6* ;
therefore (16 V&r = 4.6* = + 4V6. is the square root required.
2. Required to find the cube root of ^ ^3.
First, VjV = i. and (^/3)^ = 3^ "^ ^ = 3*j
therefore (^ v'^)' = J. 3' = 3 V3> is the cube root required.
SURDS.
157
3. Required the square root of 6*. Ans. + 6 ^6.
4. Required the cube root of Ja^6. Ans. la \/b.
5. Required the 4th root of l6a-. Ans. + 2 V + a.
6. Required to find the — mth root of x".
7. Required the square root of a^ — 6a \/b + Ob.
II. To extract the square root of a binomial quadratic surd as of a + ^b.
T. .. • A + \/a' —b, /a + Va- —b
Its root IS ^ — ^^ + ^ — ! —
(1.) The product of two quadratic surds will be a surd, except when one of
them is some rational multiple (integer or fractional) of the other.
Let \/x . /y/y =: m : then
Or thus: —
Let y =. mx then Vy = s/mx,
and ^/xy ^ ^/mx^ = x^m,
where, except ^.^m be rational, the result is irrational.
Hence, if m be rational, ^y is a rational multiple of Vx ; if not, not. Whence
the proposition is true.
(2). One quadratic surd cannot be the sum or difference either of a rational
quantity and a quadratic surd, or two quadratic surds.
For, first, assume ^Jz = a? + //y; then squaring, r = a^' + 2 xVy + y,
Whence, if we suppose s/z = a; + \/y, we shall have a surd equal to a rational
quantity, which is contrary to the definition of a surd.
Again, secondly, assume ^z = \^x + \/y : then squaring, we have
z = a; + 2Vx -\- y, or -v/a-y ^ + 2 (z — x — y), or again,
a surd equivalent to a rational quantity, which is contrary to definition.
This demonstration may be objected to as incomplete, inasmuch as x^x and ^y
may be the one a rational multiple of the other, in which case ^Jxy will be
rational. Then, however, the expression would take the form \/z=. (1 + m) s/x;
and multiplying both by \/z, we have z = (1 + m) i^xz, and the equation
cannot hold good, except i^/z be a rational multiple of \/x, or the converse. In
this case we have then simply ^z ^ p ^/x ■=■ q ^y, and the equation would
take the form ^Jz = (—-\ ) Vz, or the surd factor s/z is simply a multi-
plier of every term of the equation, and should be rejected, leaving the expres-
sion entirely rational.
It has now been proved that the equations
^z =. X + \/y and \/r = v'x + \/y
cannot have a real existence ; and, therefore, that whenever they occur they are
the result of incompatible conditions amongst the data of the inquiry. They con-
stitute in fact one of the many forms of the imaginary symbol.
It is to be understood that x, y, z, are to be perfectly general in their nature,
and not restricted to special numbers.
It readily follows from this, that in the equation
a + \/6 = X + \/y,
we must have x ■= a, and Vy = \/^-
158 . ALGEBRA.
Assume a = x + a: then
X + a + \/b = X + x/y, and hence,
+ s/y = a + s/b, which has been proved impossible, except
a = 0 : and then x z= a, and + Vy = + Vb.
(3.) To extract the square root of a binomial surd of the form a + \/b.
Assume ^ a + \/b = « + a/v, or squaring we have
a + ,yb = u + p + 2 ^uv ; and hence,
u + V =: a, and + 2^uv = + \/b.
In resolving these equations, we shall have successively,
' 12 r or subtractmg
4uv := br ) o
u- — 2uv -|- p- = a^ — b
and « - p = ± ^/a-^ - 6 U
u -\- V ■=■ a S
Whence, Va -f ^b
It is quite obvious that, except x/a' — 6 be a rational quantity, the formula,
on the tight side of the equation here obtained, is far less convenient than that on
the left. If this criterion be fulfilled, the solution will be more simple, and the
method will be of advantage ; but if not, it is better to calculate a + ^b, and
then extract its square root. As the criterion is always easy to apply, it is de-
sirable to do so in the outset, and then be guided by its result in the choice of
the method to be employed in the actual numerical valuation.
EXAMPLES.
1. WTiat is the square root of 8 + a739; of 10 — ,/96; of 6 + >J20 ; of
6 — 2 ^/5 ; of 7 — 2 ^10 ; and of 42 + 3 v'174| ?
2. What is the sum of ^27, v'48, — 4 ^/\^7, and 3 v'/a ?
a '5
3. Rationalize the denominators of the fractions ^— — , . , . ,^ . . , •
V« + Vy V3 + V4
4. Divide a + 6 — c + 2 >Jab by ^a -\- sjb -\- \/c, and by \^a + ^Jb — ^/c.
I 3
5. Multiply X -\- y — ^/xy by y^/x + ,^y ; and a** -|- a^6' + a'^i-* -\- ab -\-
a^b^ 4- P by a^ — b^.
6. Write the following expressions with fractional and (where possible) de-
lax , ^a? — 3?
cimal indices : ^/a, V (a + ^)% V («' — *^)°' sj by' \/2b — ' ' ^^^
express, by means of radicals, the following quantities: — a'^j {{a — a-)"' j - ;
± (aHa^)"=» ; (a')~' ; (^.)", and (^«)'"-
7. Ascertain whether any of the following expressions are rational :
1 (a + 6^3^)'^ ^ '' - x/«' + *' + ^ + 2a6 qp 2ax'+ 2bx; ^ {a + bx)* xy ;
ARITHMETICAL PROGRESSION. 159
" /(ca? — x^r (a + a?) , (3 (a» — r*) {a^— 4ax + ar\] ^.
V (a + a;)262; V b + x_ ' ^"° I 56=' (a + x) {a - x)»"c^" )
8. Reduce ^/a2 — x2 and V"^ + x* to others having a common index J ; and
(o~' — a:-')~*5 and (a — x) ^5 to others having the common index — 5.
9. Divide a' — b^ by a "^ — 6 ", and by a* — b*.
10. Find the square root of a — 2 -^abx + bx.
11. Express the mth root of the nth root of a' in as many ways as possible;
and show that v v V *" = a-'p ■ •
12. Extract the square root of the expressions x — 2 '/x — 1, of 2 2
(1 — x) Vl -\- 2x + x^ ; of 28 + 5 ^^12; and of -v/32 — ^/24.
Scholium.
Amongst the examples already given in this work, the symbol + ^ — 1 as a
co-efficient has appeared. This is called the imaginary symbol, and expressions
into which it enters are called imaginary quantities.
This symbol is one which indicates an operation that cannot be performed in
real numbers, positive or negative. For since (-|- 1)' and ( — 1)' are alike unity,
— 1 cannot be the result of squaring either of them ; whilst the direction to ex-
tract the square root of — 1 implies that — 1 has been produced by squaring some
quantity, and which quantity it is required to assign. All, then, that can be
done is to prefix the radical symbol, giving the general form + ^ — 1.
In all problems where this symbol appears in the result, there has been some
incongruity in the conditions of the question ; or in other words, the alleged
conditions would not co-exist.
The calculus by means of these is precisely similar to those laid down for
quadratic surds, and no remark seems necessary, except to caution the student to
use due care in the signs of his reduced quantities. The laws, however, are
precisely the same as already laid down for products and quotients.
ARITHMETICAL PROGRESSION, OR PROGRESSION
OF DIFFERENCE.
[Though in accordance with the original arrangement of the Course, the
subject of progressions is retained in this place, it is very desirable that the
study of it should be deferred till the simple and quadratic equations are
thoroughly understood. The investigations at least, as they depend on the
solution of simple equations, will be unintelligible to the student who has read
no farther than the preceding pages.]
An Arithmetical Progression is a series of quantities which either increase or
decrease by the same common difference.
Thus, 1, 3, 5, 7, 9, 11, . . . and a, a + 6, a + 2i, a + 3b, o + 46, ... are
series in arithmetical progression, whose common differences are 2 and + b
respectively.
"When the quantity b is affected by sign +, the progression is said to be
increasing ; and when by the sign — , it is said to be a decreasing progression.
These signs are therefore the indications of the kind of progression.
160 ALGEBRA.
The most useful part of arithmetical progression has been given in the arith-
metic. The same may be exhibited algebraically, thus :
Let a denote the least terra,
z the greatest term,
d the common difference,
n the number of the termp,
and s or s, the sum of n terms ;
then the principal relations are expressed by these equations, viz.
1. 2 = a + d(7i — 1);
2. 0= 2 — d(n — 1);
« = (a + z) ^ i
^. s=\z—ld{n- \)\n;
5. «= [a + 4rf(n — \)\n*.
rf="-«
7. d
8. d =
n — V
2 {s—an) _
«(«— 1)'
1{nz — *)
n(n — !)■
EXAMPLES.
1. The first term of a series of quantities in arithmetical progression is 1,
their difference 2, and the number of terms 21 : what is the sum ?
Here a = 1 ; d ■=. ■\- 1\ and n = 21 : and hence, by formula (5), we have
1 + 41
s,, = -^ X 21 = 441.
2. A decreasing arithmetical series has its first term 199, its difference — 3,
and its number of terms 67 : what is its sum ?
Here z =: \9g, d = — 3, and n = 67 ; hence a ■= z — (n — 1) d = 199 —
199 + 1
66 X 3 = 1, and — - — X 67 = 6700, the answer required.
3. To find the sum of 100 terms of the natural numbers 1, 2, 3, 4, 5, 6, &c. ;
and likewise of "1, *2 . . . . Ans. 5050, and 505.
4. Required the sum of 99 terms of the negative odd numbers — 1, — 3,
— 5, — 7, — 9 Ans. — 9801.
5. The first term of an arithmetical series is 10, the common difference — I,
and the number of terms 21 ; required the sum of the series. Ans. 140.
6. One hundred stones being placed on the ground, in a straight line, at the
distance of two yards from each other : how far will a person travel, who shall
bring them one by one to a basket, which is placed 2 yards from the first stone?
Ans. 11 miles and 840 yards.
7. The first term is a" — 2ax + x*, the last is a* + lax + x', and the number
• For, writing the given series first in a direct line and then in an inverted order, we have
{a 1 + J«+ rfj + fa+ a/J + + Ja+(«_rf)d}
ja+(«-l)rfJ + Ja + (n-2)rf} + Ja + {«-3)rfJ + + [a \
and adding up the columns vertically, we have douhle the sum of the series equal to
J-2a + (n-l)cf] + J2a + (n-l)4 + {2« + (« _ 1)^ + . . + {2a + (n-\)d\
which is equal to « j2a + (fi — I)''?, there being n such terms ;
and hence s_, = Ja + } (n — \)d\n. This is the formula marked (5) above.
That marked (1) is obvious from the formation of the successive terms : the (2) is obtained
from it by transposition. Also from the method of proceeding in the proof of (5) that marked
(3) is evident ; and substituting the value of a in (5) the result (4) is obtained.
From those expressions which involve values of n and d, formulae for finding these may be
derived by the common operations of algebra : but it is unnecessary to annex the work here.
ARITHMETICAL PROGRESSION. KJl
of terms is o* — «» : what is the sum of the series, and the common difference
of its terms ? Ans. Sum =. a* — a:*, com. dif. =: — ~
a* — z* — I'
8. The first term is — a, and the last term is nine times the first, and the
number of terms one-fifth of the sum of the first and last terms : what is the
sum and common difference? Ans. Sum = lOa*. com. dif. = — '*—
2a + r
9. The sum of the numbers, 1, 2, 3, .... n is ^" — ; the sum of the n
numbers, 1, 3, 5, .... (2n— 1) is n^ ; and the sum of n terms each equal to n is
n'. Investigate these theorems.
10. The first term is 16"5, the last is — 6"5, and the sum is 100. Find the
number of terms and the common difference.
11. The tenth term of an arithmetical series is 17"5, and the fiftieth is 1587J.
"What are the separate sums of the first twenty, and of the last thirty terms ?—
Find also the common difference; and the 11th, 21st, 31st, and 4l8t terms.
12. Sum the series — 7, — 55, — 4, — .... to 10 terms,
16, f, 14?, .... to 49 terms,
17, + '3', + 15§, + to 20 terms,
and find the 7th term from each end of each of these series.
13. The common difference is '001 ; the number of terms is one million, and
the greatest term is 0. What is the least ? Find also the sum of the 100th,
200th, 300th, .... terms of which the series is composed.
14. The first term is 1, the common difference is successively taken 1, 2, 3,
. . . write ten terms of the first six series, and express the sum of each series to n
terms.
15. Find the difference between the sum of n terms of the odd numbers 1, 3,
5, .... and n terms of the series of even numbers 0, 2, 4,
16. If the first term be o, and the common difference be 2a (l -\- a+a' + ...a'~*),
show that the sum of the series of a terms is equal to a". And apply this to the
square and cube as values of m.
17. Let the first term be — a, the common difference d, and the number of
terms — n *, what is the expression for the value of the + nth term ? Also, in
numbers where a ^ + 5, d =: I, and n = 10. Ans. z = — a + '~— <i.
18. A triangular battalion f consists of thirty ranks, in which the first rank is
formed of one man only, the second of 3, the third of 5, and so on : what is the
strength of such a triangular battalion ? Ans. 900 men.
19. A detachment having 12 successive days to march, with orders to advance
* As n denotes the number of terras to be t.iken in the direction indicated by the sign of the
common difference, so also — n denotes the term from which we must have ttarted to obtain
— a (the difference still being d) as the 71th term of tiie scries.
+ By triangiiLir battalion, is to be understood, a body of troops ranged in the form of a
triangle, in which the ranks exceed each other by an equal number of men : if the first rank
consist of one man only, and the difference between the ranks be also 1, then its forni is that of
an equilateral triangle ; and when the difference between the ranks is more than one, its form
may then bo an isosceles or scalene triangle. The practice of forming troops in this order,
which is now laid aside, was formerly held in greater esteem than forming them in a solid
square, as admitting of a greater front, especially when the troops were to make simply a stand
on all sides.
162 ALGEBRA.
the first day only 2 leagues, the second SJ, and so on, increasing 1^ league each
day's march : find the length of the whole march, and the last day's march.
Ans. the last day's march is 18^ leagues, and the whole march, 123 leagues.
20. A brigade of sappers * having carried on 15 yards of sap the first night,
the second only 13 yards, and so on, decreasing 2 yards every night, till at last
they carried on in one night only 3 yards : what is the number of nights they
were employed ; and what is the whole length of the sap ?
Ans. they were employed 7 nights, and the whole sap was 63 yards.
21. A number of gabions f being given to be placed in six ranks, one above
the other, in such a manner as that each rank exceeding one another equally,
the first may consist of 4 gabions, and the last of 9 : what is the number of
gabions in the six ranks ; and what is the difference between each rank ?
Ans. the difierence between the ranks is 1, and the number of gabions is 39.
22. Two detachments, distant from each other 37 leagues, and both designing
to occupy an advantageous post equi-distant from each other's camp, set out at
different times ; the first detachment increasing every day's march 1 league and
a half, and the second detachment increasing each day's march 2 leagues : both
the detachments arrive at the same time ; the first after 5 days' march, and the
second after 4 days' march ? What is the number of leagues marched by each
detachment each day ?
Ans. the first marches -^g, 2^%, 3-^, 5-^, 67o, leagues on the successive days,
and the second If, 3|, 5|, 7f, leagues.
23. A triangular course of shot of n in each side is composed of n, n — 1, n — 2,
.... 3, 2, 1 shot in succession. Find the number of shot, c„, in the whole
nf«-|-l)
course. Ans. c, = — - — .
24. A rectangular course of shot has n shot in its shorter side, and m+n in
its longer one. How many shot are there in the course ? Ans. c„ = (m-\-n)n.
GEOMETRICAL PROGRESSION, OR PROGRESSION
BY RATIO.
Ira series of terms (three at least) be so taken that each is the product of the
* A brigade of sappers consists generally of 8 men, divided equally into two parties. "While
one of these parties is advancing the sap, the other is furnishing the gabions, fascines, and other
necessary implements : and when the fii'st party is tired, the second takes its place, and so on,
till each man in turn has been at the head of the sap. A sap is a small ditch, between 3 and
4 feet in breadth and depth ; and is distinguished from the trench by its breadth only, the
trench having between 10 and 15 feet breddth. As an encouragement to sappers, the pay for all
the work carried on by the whole brigade is given to the survivoi-s.
■f Gabions are baskets, open at both ends, made of ozier twigs, and of a cylindrical form :
those made use of at the trenches arc 2 feet wide, and about 3 feet high ; which, being filled
•with earth, serve as a shelter from the enemy's fire : and those made use of to construct batte-
ries are generally higher and broader. Tliere is another sort of gabion, made use of to raise a
low pai-apet : its height is from 1 to 2 feet, and 1 foot wide at top, but somewhat less at bottom,
to give room for placing the .muzzle of a firelock between them : these gabions serve instead of
Siiud bags. A sand bag is generally made to contain about a cubical foot of earth.
GEOMETRICAL PROGRESSION. 1 63
preceding one of the series by some constant factor •, these terras are said to
form a geometrical series. Thus, a, ar, ar^, ar*, ar^'^ form a series of
terms in geometrical progression ; as likewise do 2, 6, 18, 54, ... . and 2, —6,
18, —54, .... of which the constant factors are respectively r, 3, and —3.
These factors are called the ratios of the series.
The following notation is generally employed : —
a for the first, and z for the last term ;
r for the common multiplier or ratio ;
n for the number of terms ; and
s or s, for the sum of n terras of the series.
FORMULA OF SOLUTION.
From the formation of the terms, z == ar—^ (1)
s' = a{r-' + r"-^ + . . .. r^ + r + 1) . ^J = ~^ ^^^
\ 2 (r l)s
L From (1, 2) we have a = _ ^ , and a = , _\" (3, 4)
From (1, 2), or" = rz and ar" — a = (r — IK ;
whence, by subtraction, a = rz— (r — l)s. (5)
j, From (1, 4), z = ar"-', and a = ^ ^_^ ' ; whence z = 'j^. _^ . . (6)
From (2), (r — l)s. = ar" — a ; whence a + (r — 1)5. = ar" =■ rz ; whence
— V <»
_ a + (r — l)s, _ s. — a
~ r ~ ' r
From (1), r^* = - ; whence, extracting, r = f-J"^' (8)
g — a
From (1, 2), rz = ar" and rs, — s, = ar' — a ; hence r = ^ _ ^. ... (9)
17 .o^ . «(''-l) »""' (r--l)ar-'_ (r'-l)z
z^ , . . z"-'-a"-> ._j _ , _ r— -0—
.— I
From (8), r = — ,- ; hence r" — 1 = , and r— 1 =
fl(r"— 1) z^'—a"-' a.aF' g-'- a"^' . .
whence s, = — _ = — , , . — ^^ = "TI H ^ '
^~^ z^^-cF^ cF' z^'-o"-'
Reducing (2) we have r*- f-"Jr+ " =1 (^2)
And from (6) we have r" -— r""' + - , = 0 ('^^
z , s.-a , ._, f^-—''Y~* _
^ From (1, 9) we get r"-' = and r = ^ _^ ; whence r = ^5,^/ ~
^ (14)
a
• The numerical or algebraical character of this factor is of no consequence; as it may bo
positive, negative, or imaginary, integer, fractional, or irrational. Its constancy throughout the
series is the only condition essential to it.
M 2
164/ ALGEBRA.
The only general method of finding n is by logarithms : but when n is known,
r may be found by the solution of the equations 13, 14. When logarithms are
used, equation (14) is a convenient form for n.
When the series runs out ad infinitum, if the ratio be greater than 1, the sum
is necessarily infinite, as r" will become infinitely great. When r = 1, the sum
of the series takes an indeterminate form s, = - . a : but other obvious con-
0 0
siderations show that it is infinite. When r is less than 1, r" continually de-
creases as the value of n increases, till r" ^ 0. In this case s , = — a =
B r — 1
l-r^ a ,..,
. a = (15)
1 — r 1— r '' ^
The doctrine of geometrical progression finds its application in almost every
department of mathematical inquiry ; but one of the most elementary and most
frequently required cases, is that of
CIRCULATING DECIMALS.
It has already been seen in the arithmetic, p. 60, that when certain vulgar
fractions are converted into decimals, the terms of that decimal form a series of
circulating periods of figures, always recurring in the same order. If we convert
these periods into separate terms with the proper denominators indicated by
their places in the decimal scale, we shall find them to constitute a geometrical
series.
Thus, -333 . . . . ad inf. = - + -^ + — ....=- \ 1 + A "Hii, "T- • -I
10 100 1000 10 ( ^ 10 10* J
J „,r^,^ J- r 215 , 215 . 215 215 < , , 1 , 1 ,
and -215215 ... ad mf, = — - + , — ^-,+ - — -, . . .= — - J 1 -^ ^^ + 7:7. + ....
103 ^(103)2^(100' 103 \ 103 ' io«
which, again, may be conveniently written in the following manner : —
3C1 + -12+-13-)- ) and 215(-P+-l6+-l»+ . . .).
Employing the formula (15) for a decreasing infinite series,
s, = -' " , = „ for the first decimal '333
0 10 — ] 3
215. -P. 10' 215 . ^, , „,.„,^
*i ~ — -3 _ = -— for the second '215215 ....
To take a general view of the subject, let us suppose the decimal is composed
of a series of circulating periods preceded by a finite decimal. Let the circu-
lating period be composed of n figures, which, taken integrally, denote the
number N, and the preceding or finite part of the decimal be composed of m
figures, which, together with the integers, all taken integrally, denote the num-
ber M. Then the entire decimal will be represented by
M N N N
3^- + j^;^^ + y^;q^. + Y^,,;^^ + .... ad inf., which may be written
M_, JLI.] . .i- + -l>+-l3--h l_M _N 10:_
10-^ io»+- ^ ' -r A -r A -^ i + •• •• J— 10-^ io"+" * lO"— 1
= 1"/
99 ... (n — 1) places S
When m = 0, "1" = 1, and the circulating period commences at the decimal
N
point, and is represented by the fraction ; — ; .
*^ *^ ' 999 ... (n — 1) places
GEOMETRICAL PROPORTION. J 65
EXAMPLES IN GEOMETRICAL PROGRESSION.
1. The first term is 1, the ratio 2, and the number of terms 12: find the last
term and sum of the series.
z =z af-^ = 1.2" = 2048, the last term ; and
(r"— 1) o (4096 — 1) . 1
*,a = — , — = 7 = 4095, the sum of the senea.
2. Given the first terra \ and the ratio — ^ to find the eighth term and the sum
of eight terms of the geometrical series.
z = ar* = J C— ^)^ = — si, the eighth term ; and
»•* — 1 53S — 1 1 85 , , . ,
*8 = :;rzr\- " — _ 1 _ ^- 3 = -^^> the sum of eight terms.
3. Required the sum of 12 terms of the series, 1, 3, 9, 27, 81, ...; and of 7
terms of the series 1 — 3 + 9 — 27+.... Ans. 265720, and . . .
4. Required the sum of 12 terms of the series, 1, 4, i, jV» A ; a^J 10 of
the series 1 — 3 + i — ^ -\- . . . Ans. f?^^^, and . . .
5. Given r = 2, n = 6, « = 189. Required a and z. Ans. a = 3, z = 96.
6. Given a = — 4, n = 6, z= — 12500. Required r and *.
Ans. r =: 5, s =: — 15624.
7. Find the tenth terms and the sum of the first ten in each of the following
series :
16 , 8 4
,16 8 4
;; \-^ h • .
. . and 1 — — . .
729 243^81 ^
729 243 ' 81
and find the 11th and 12th terms of each series; and then sum each series to 6
and to 7 terms.
8. Find the sum of ten terms, and the diflference between that and the sum
to infinity of the series f + j + J + . . . and likewise of — 4+^ — i + ...
Also of 8 + 4 — 2 + 1 — . . . to si.\ terms, and to infinity.
9. Find the values of -333 ... ad inf. ; and of 25 + "252 + -25' + ... to 5
terms ; and assign the 9th and 10th terms. Likewise the sum of both to infinity.
10. What is the sum of the infinite series — "1* + "1^ — "1* + '1* — ... .
and oi'Va + -IV + -IV +
11. Required the sum of 1 -\ — -|- — _(- ^ + .... to infinity, and likewise
"f^ + ^l+(^^+--- A"^-:r-l
12. ITie first and last terms are § and I, and the number of terms is 5 : what
are the three intermediate terms ? Also insert two and three geometrical means
respectively, between 4-5 and 2.
13. Suppose a body to move for ever in this manner, viz. 20 miles the first
minute, 19 miles the second, 18 05 the third, and so on in geometrical progres-
sion : required the utmost distance it can reach.
14. Suppose the elastic power of a ball which falls from a height of 100 feet
to be such as to cause it to rise '9375 of the height from which it fell ; and to
continue in this way diminishing the height to which it will rise in geometrical
progression, till it comes to rest : how far will it have moved ?
GEOMETRICAL PROPORTION.
If there be four quantities, a, b, C, d, such that ad = 6c, they are said to be
fireompfrirnl nrnnnrf ionole TVio atatptnpnt ia 1i«iin11v writtpn tl ' b ' ' C '. d, the
166 ALGEBRA.
factors of one side of the equation being taken as the extreme, and those of the
other side as the mean terms. In this a and c are called the antecedents, and b
and d the consequents of the ratios a : b and c : d. Homologous terms are either
both the antecedents or both the consequents. With this understanding we
have at once the four following forms of the statement : —
a : b ;: c : d (i)
b : a ',', d : c (2)
a : e '.'. b : d (3)
c : a\\d : b (4)
Again, by division of the original or defining equation,
a c a b
b = d'''c = d ^''^^
From (5, 6) ^ + 1 = ^ + 1, or - + ] = - + 1 ; whence,
0 — a — c a —
a + b f^ + (^^ or a + b : a — b : : c + d : c - d (7)
a — b e — d
a + c b + d
OT a -\- c : a — c [\ b -{- d : b — d (8)
a — c b — d'
and these two will each admit of four varieties of form analogous to those of (1,
2, 3. 4).
Add bd to both sides of the defining equation : then ad -\- bd ^= be -^ bd, or
{a + b)d={c + d)b; Andb : dy, a + b : c + d.
By this and analogous processes we shall obtain
a : c : a + b : c + d (9)
a : b :', a ± c : b ±d (10)
c : d;; a ±b : c ±d (ii)
c : d ;\ a ± c : b ± d (12)*
Again, if we have several ranks of proportionals, the products of all their cor-
responding terms will be proportional quantities. Let them be,
a : b \] c : d ; then a d = b c
Oi '. bi '. '. c^ : d^; a^dy ^= J, c,
flj : ^2 1 I Cj : c?2 ; Ujd.^ = b.^ Cj
«- : 6- I ! c. : rf„ a„d„ ^ 6,c»
and hence aOytt^ . . . a„ x ddyd.^ . . . rf, = bb^b.^ ... 6, X cCyC.^ . . • c,, and we have
00,0^ . . . fl„ : bb,b.2 ... b, '.] cc^c.^ ... c„ : dd^d., ... d, (13)
If there be n of these factors, and all equal, we have
a" : 6" : ; c" : d" (i4)
If four quantities be proportional, their n"" roots are also proportional. For if
ad ■-= be, then a" d" = b' c", and a" : A* *. I c* : dr (14)
nw «« m m m m
Also O" d- ^b^ dT', and a* : 6" ! I c" : cf" (15)
When the second and third terms are equal, we have
a : b W b : c, and b^ =: ac.
In this case b is called a mean proportional between a and c, and c a third pro-
portional to a and b.
In a geometrical series we have the terms continually proportional : viz.
ra ' ' ra '. r'a ") .. . ,
since these fulfil the defining condition of propor-
ir-a^ .
ra : ra^ , ra' . , ,* ( . ,
, . . , ^ I tionals.
r-a : r^a . . r'a : r*a
Several technical terms are used bv eeomcters to distiuiruisli tlicsc diflferent modes of
1G7
HARMONICAL PROGRESSION AND PROPORTION.
When the reciprocals of a series of numbers form an arithmetical progression,
the numbers themselves constitute an harmonical progression. Thus, i, i, i, .. .
c c c c
'^' ^' ^'^ °'' a' ^TTb' ^TTi' r+lb ' ■•■ '^onst't"^* an harmonical
series, or an harmonical progression .
If, therefore, we are required to form an harmonical proj^ression, or Id find
the law from a sufficient number of terms given in any part of it, we have only
to form the reciprocals, and thence the arithmetical progression, and finally to
take the reciprocals of these. The resulting series is the one sought.
Thus, to find a fourth harmonical to the given terms ^, J, J, we have only to
take 2, 3, 4, and find the fourth arithmetical proportional to 2, 3, 4, viz. 5. Then
the reciprocal of this, viz. \, is the fourth term sought. It is, however, suflRcient
to give two of the terms immediately preceding that sought, as the arithmetical
progression of reciprocals follows from these two, without employing any of the
more remotely precedent ones of the series *.
It will be obvious, that as an arithmetical series does not admit of indefinite
extension in one direction, without employing negative numbers ; so also the
harmonical series does not admit of indefinite extension in the other direction,
without also employing negative numbers. Employing, however, as is always
done in Algebra, both positive and negative numbers, all three series admit of
indefinite extension both ways. It is with this explanation that the remark
made by writers on these classes of series is to be understood, when they say
that the arithmetical series admits of indefinite extension only upwards, the
harmonical only downwards, and the geometrical, in both directions, both
increasing and decreasing.
The following properties furnish a specimen of those which belong to numbers
in harmonical progression.
1. If there be three terms in harmonical progression, then, the first is to the
third as the difference of the first and second is to the difference of the second
and third.
For, by the definition, if a, b, c be three numbers in arithmetical progression,
111.,, ... ., ., b — c a — b . . .,
-, . , - will be m harmonical progression. Also, — ;— = — 7 — . >n virtue ol tne
a b c ^ ° abc abc
.,. .rull" — *
arithmetical progression, and by proportion we obtain from this '. * '. — ^^
b — c^ . L 1 . L i
be ' ' b a ' c ~b'
The converse of this is obviously true : that if three given numbers fulfil this
proportion they are harmonicals.
2. The harmonical mean between two numbers is equal to twice the product
of those numbers divided by their sum. For in the preceding proportion w«
have . ; =z , from which — = , (-+-)»
ac ab be ac ac b \a c/
* An extension of the definition, so as to render the fourth term dependent on the three
immediately precedent ones, has been given by several authors, though that definition has no
reference whatever to musical intervals, nor do the terms of it form the reciprocals of an arith-
metical series.
168 ALGEBRA.
1 2.-M
1 ac a c
a c a c
wliere a, h, e are the arithmetical reciprocals of the three harmonical terms.
3. A third harmonical to two given terms is equal to the product of those
terms divided by the difference between twice the first and the second terms.
For from the same we have
J^
1= «^
c "" 2 _ 1
a " b
4. In an harmonical series, any three terms, the extremes of which are equi-
distant from the extremes, are in harmonical progression. For their reciprocals
are in arithmetical progression *.
5. Let a, b, c, .... h, k, be an harmonical series : then
The product of any two adjacent terms is to the product of any other two
adjacent terms, as the difference of the first pair is to the difference of the second
pair.
For — = rby definition.
a b fi k '
_T b — a k — h
Hence, ^^=-^,
6. When the first two terms a and b are given, the n" may be thus expressed :
ab
^ ~ (n— I) a — {n — 2)b
For d =: = ; — : and
0 a ab
1 1 ^ ,x« — ^ (n - 1) a - (n - 2)6
— = — H (n — 1) — J— = 7
z a ab ab
ab
or z '~~
(re — 1) a — (rt — 2) b'
7. We may insert « harmonic means between a, b.
For - = - + (n — I) d, or d =
(n — 1) ab'
* As the doctrine of geometrical proportion, which essentially involves four terms, has heen
applied to three only, by taking the second and third as identical (thereby constituting con-
tinued proportionals), so in the harmonical progression continued (each step of which essentially
contains three terms), the second term has been supposed to be replaced by a different one,
which stands as the third, whilst that which originally stood as the third, thereby becomes the
fourth. In this case the/bur numliers have been said to be in harmonical proportion, when the
first is to the fourth as tlie difference of the first and second is to the difference of the tiiird and
fourth : thus, if a, b, c, d fulfil the condition a \ d \\ a j' b \ c J^ d, x\ie four quantities a, b, c, d
are harmonicals.
In this case, if c be made =: b, and d ■=. c, the harmonicals previously defined will result ; but
the definition of harmonicals there given does not apply to any of the other numbers which ful-
fil this condition. .Some writers also speak of contra-hurmonicals. Into the study of each of
these kinds of proportion, the reader who desires to enter will find ample information in Mal-
colm's Arithmetic, pp. 297— 313, 1730. Further notice of them here would be incompatible
with the plan and objects of this Course of Mathematics.
EQUATIONS.
169
from which the arithmetical progression, and thence the hannonlcal, is readily
found.
EXAMPLES.
1. Find the fifth term of an harmonical series whose first and second terms
are 3 and 4, and likewise of that whose first and second terms are 4 and 3.
In the first case, J and ^ are the first and second terms of an arithmetical pro-
gression descending, since the harmonical progression ascends. Hence 1 = z,
is the greatest term ; and we have J — i = ^'^ = common difiference = t/ : and
2 = i, and n = 5. Therefore the term a, or least term, is i — i"!, = J — J = 0,
and the fifth term is therefore 5 = infinite.
In the second case the harmonical series descends, and hence the arithmetical
one ascends ; and, therefore, as before, d = -f^, and n = 5, and the least terra is
^: the fifth term is, therefore, i + t^ = ^ = J The fifth term of the harmo-
nical series, 4, 3, &c. is therefore |.
2. Find an harmonical mean between 3 and 4, and six harmonical means be-
tween ] and 2.
3. An harmonical series consists of fifteen terms, and the greatest and least
terms are x and y : what is the middle term ?
4. A line, whose length is 10 inches, is divided harmonically, so that the first
section (from the origin) is 3 inches : how far distant is the second point of sec-
tion from the first ?
5. Four terms are in harmonical proportion; the first and last are 6 and 10 :
what is the relation between the second and third ?
Ans. 10a; + 6y = 120 ; where x is the second, and y the third term.
6. The first and second terms of an harmonical proportion are 4, 5 : and the
first, second, and third terms of an harmonic progression are also 4, 5, 6. Find
the fourth term of the proportion, and the fourth term of the progression.
7. Ten terms are in harmonical progression, and the last two are /j and -fj :
what are the terms, and what is their sum ?
EQUATIONS.
An equation is the algebraic expression of the equality of two assemblages of
quantities to one another; and consists in writing =, the sign of equality,
between them *. Thus 10 — 4 = 6 is an equation expressing the equality of
10 — 4 to 6 ; and 4x + b := c — d is an equation expressing that 4* -|- 6 is
equal to c — d.
Equations are designated by different names, according to the manner of their
composition, and the highest power of the unknown quantity which enters into
them. "When the highest power is the first, the equation is called a simple equa-
tion, or an equation of the first degree : when it is of the second, the equation is
* The mark here used vras introduced into algebra by the first English author on the subject,
Robert Recorde, in his "Whetstone of Witte," (sig. Ff. !*>,) 1557. He gives his reason in his
own quaint manner in the following words : " And to aroide the tediousc repetition of these
■woordes : is equalle to : I will sctte as I doe often in woorke use, a paire of parallclcs, or
Gemowe lines of one lengthe, thus : n, bicause noe 2 thyngcs can be more equalle."
For a long period afterwards, the Continental mathematicians employed the symbol « , which
was, doubtless, a rapid formation of the diphthong <e, the initial of the phrase aquaie eat.
170 ALGEBRA.
called a quadratic equation, or an equation of the second degree .- when it is of the
third, the equation is called a cubic equation, or an equation of the third degree :
and so on.
The known, or given quantities, are represented by the earlier letters of the
alphabet, a, b, c, . . . and the unknovm, or quantities whose values are sought, by
the later ones, z, y, x, w, ....
It often happens that equations arise which are composed of only two terms,
in which the power of the unknown is of a higher degree than the first. These
are called binomial equations, as a^ = — iOO ; and otherwise pure equations, to
distinguish thera from adfected (or affected) equations. These are, for the pur-
poses of solution, considered as simple equations, the operations that are requisite
for completing the solution being purely of the arithmetical kind. Examples in
which these occur are therefore classed amongst those of simple equations.
When there are several equations given, the unknown of which in one is
capable of such multiplications or divisions by that in some of the others, or
when they admit of a ready combination with one another, so as to form results
that are known to be powers of some binomial or trinomial expression, they are
frequently classed also amongst the exercises on simple equations. Such a
method of classification is, evidently, very arbitrary; and hence there are several
questions in the following series which are by some authors distributed under a
different denomination : though in general this classification accords with the
most common practice of algebraical wTiters.
The quantities which precede the mark of equality, are often called together
the first member or the first side of the equation j those which follow it, the second
member or the second side.
The resolution of equations, is the finding the value of the unknown quantity,
or in disengaging that quantity from the known ones ; and this consists in so
transforming the equation, that the unknown letter or quantity may stand alone
on one side of the equation, without a coefficient; and all the rest, or the known
quantities, on the other side.
SIMPLE EQUATIONS, WITH ONE UNKNOWN.
In these, the unknown quantity, when properly transformed, is of the first
degree, as ax ^ b, and its solution in this state is obvious : but as they seldom
so occur, we must lay down the principles of transformation so as to disengage
X from all other quantities on one side of the equation.
In general, the unknown quantity is disengaged from the known ones, by perform-
ing always the reverse operations. That is, if the known quantities are connected
with it by +, or addition, they must be subtracted; if by minus, — , or sub-
traction, they must be added ; if by multiplication, we must divide by them ; if
by division, we must multiply ; when it is in any power, we must extract the
corresponding root ; and when in any radical, we must raise it to the cor-
responding power. The following special rules are founded on this general
principle, viz. : that when equivalent operations are performed on equal quan-
tities, the results must still be equal ; whether by equal additions, subtractions,
multiplications, divisions, extractions, or involutions.
I. When known quantities are connected with the unknown by -j- or — ;
transpose them to the other side or member of the equation, and change their
signs. Which is only adding or subtracting the same quantities on both sides.
SIMPLE EQUATIONS.
171
in order to get all the unknown terms on one side of the equation, and all the
known ones on the other *.
The same rule applies whether the known quantities be given in numbers or
in symbols.
Thus, if a; + 5 = 8 ; then transposing 5, gives a? = 8 — 5 = 3.
If a? — a + ft = cd, then by transposing a and A, it is ar = a — b •\- cd.
II. When the unknown term is multiplied by any quantity; divide all the
terms of the equation by it.
• Here it is earnestly recommended that the pupil be acctistotned, at every line or step in the
reduction of the equations, to name the purticulccr operation to lie per/urmed on tlie preceding
equation, in order to produce the next form or state of the equation, in applying each of these
rules, according as the particular form of the equation may require : applying them according to
the order in which they are here placed, and always allotting a single line for each operation and
its description, and ranging the equations under each other, in the several lines, as they are suc-
cessively produced. The master, indeed, nerer otujht to receit-e a stAutionfrom his pupil in tcritinff
in which this ride is not complied tciih, and as much attention given to the proper concatenation of
the verfMl descriptions as to the mere work set dotvn in the uljehra. Due regard being had to
this point would prevent algebra from becoming a mere piece of ingenious mechanism, as it now
too often does become.
The procedure here enforced differs in no respect from that employed by the earlier alge-
braical writers, as may be seen by reference to Wallis, Ronayne, Kersey, Ward, and others. It
was also a useful custom, and one which has been recently revived, to number the several suc-
cessive steps of the process, and to quote the equation by means of the number attarlicd to it.
The older writers ruled a column down the middle of the page in which to put the ordinal num-
bers, and kept the written description of the process on the left, and the work itself on the right
of this column. However, in the extended equations, to which modem physical science gives
rise, the great inequality in the length of the lines renders it more convenient to write the
ordinal numbers, (1), (2), (3) ... at the margin of the page. Tlie mode of taking the ordinal
column down the middle is better, however, for the learner, as his tcork is thereby kept in one
vertical column to the right of it, and is therefore much more easily inspected by himself as well
as by the master. On this account its adoption is adWsed in the earlier stages of study, even
though it may ultimately be laid aside when good and regular habits are formed. Thus, if the
equations x* — y* = «', and ae •\-y=.h, had been given, we should have had
{
1
a-'— y' = a»
Given
2
X ■\-y =b
Divide (1) by (2) then
3
Add (2), (3) together,
4
^ = 1-'
or reducing to common den.
5
2r = «i+^'
b
and dividing (5) by 2
6
X - «' + *'.
2i
Again, subtracting (3) from
(2)
7
.,=.-»•
and reducing (7) to a common denom.
8
,,_«-«.
and dividing (8) by 2.
9
^=-26-
In this notation a figure enclosed in a parenthesis, as (2) or (3) indicate* the words " the
equation marked two," whilst in the case of no parenthesis, it signifies the number 2.
This subject is illustrated and enforced very elegantly in Butler's Course of Mathematics,
vol. ii. p. 17. The author correctly traces the first proposal of the practice to Dr. Pell, aa
eminent analyst of the early English school.
172 ALGEBRA.
Thus, if ax ■=^ ah — 4a ; then dividing by a, gives ic = A — 4.
In like manner, if ax + 3ai = 4c- ; then by dividing by a, it is a; + 36 =
— J and then transposing 3i, gives x =. Zb.
a o,
III. When the unknown term is divided by any quantity, we must then mul-
tiply all the terms of the equation by that divisor ; which takes it away.
[Note. If there be several terms in which fractions appear, it is often best to
multiply the numerator of every term of the equation by the least common mul-
tiple of all the denominators, and divide by the corresponding denominator. All
the terms are thus cleared of fractions at once, whether known or unknown.]
Thus, if - = 3 + 2 : then mult, by 4, gives a; = 12 + 8 = 20.
4
And if - =: 3& + 2c — d : then mult, by a, it gives x = 3ab + 2ac — ad.
a
IV. When the unknown quantity is included under any root or surd expres-
sion : transpose the rest of the terms, if there be any, by rule 1 ; then raise each
side to such a power as is denoted by the index of the surd ; viz. square each
side when it is the square root ; cube each side when it is the cube root ; and so
on, which removes that radical from the equation.
Thus, ii V X — 3 = 4: then transposing 3, gives ^ x = 7 ; and squaring
both sides gives x = 49-
Also, if V^* + 4 + 3 = 6 : then by transposing 3, it is \/3x 4-4 = 3;
and by cubing, it is 3a? -|- 4 =: 27 ; and by rules I. II. x = 7h
V. When that side or member of the equation which contains the unknown
quantity is a complete power, or can easily be reduced to one, by rule I. II. or
3 ; then extract the root of the said power on both sides of the equation ; that
is, extract the square root when it is a square power, or the cube root when it is
a cube, and so on.
Thus, if a?2 + 8a? -t- 16 = 36, or {x -\- 4)^ = 36; then by extracting the
root, it is a? -|- 4 = 6 ; whence x = 2.
Also, if f a?2 — 6 = 24 : then transposing 6, gives f a?^ = 30 ; and multiply-
ing by 4, gives 3a:- = 120; then dividing by 3, gives a:^ = 40; and, lastly,
extracting the root, gives x =: ^ AO = 6*324555.
VI. When there is any analogy or proportion, it is to be changed into an
equation, by multiplying the two extreme terms together, and the two means
together, and making the one product equal to the other.
Thus, if 2a; : 9 : : 3 : 5, gives 10a; = 27 ; and by rule II. x = 2^.
And if la; : a : : 56 : 2c; then 'icx= Sab : hence by rules II. III. a;= .
^ ■' 3c
VII. WTien the same quantity is found on both sides of an equation, with the
same sign, either plus or minus, it may be cancelled or left out of both ; and
when every term in an equation is either multiplied or divided by the same
quantity, that quantity may be struck out of them all.
Thus, if §a; — 5 =: 3° — i ; then, cancelling J, it becomes §a; = '3" ; and multi-
plying by 3, it is 2a; ^ 10 ; or a; = 5.
The following example furnishes specimens of all the rules just laid down :
5^2
Let 2a; + 2 v^a^-f x^ = -7-»- -„ be given ; to find x.
va-* -f- or
Multiplying by v'a* -^ a;*, gives 2x ^ a^ + x' + 2a^ + 2a!^ = 5a^ ; trans-
posing 2a^ -f- 2a;% gives 2a? v'o'' + x^ = 3a- — 23?*; squaring both sides, 4x^ x
(a^ -f x2) = (3a^ — 2a;")- ; that is, 4a V + 4a;* = 9o* — 12aV + 4a;'' ; can-
SIMPLE EQUATIONS. 173
celling 4x* from both sides, we have 4a-x' = Oa* — 12aV; transposinf^ I2a'i',
gives l6a-aP ^ Qa^; dividing by d^ gives l6ar = Qa^; dividing by 16, givea
X- = fgo,^; and lastly, extracting the square root, gives x = ja.
EXAMPLES FOR PRACTICE.
1. Given 2x — 5 4- 16 = 21 ; to find x. Ans. « = 5.
2. Given 6x — 15 = x + 6 ; to find x. Ans. x = 4j.
3. Given 8 — 3x + 12 = 30 — 5x + 4 ; to find x. Ans. x = 7.
4. Given x -|- gX — Jx = 13; to find x. Ans. x = 12.
5. Given — Zx — \x — 2 = — 5x + 4 ; to find *. Ans. x = + 4.
6. Given 4ax + ia — 2 := ax — ftx; to find x. Ans. x =
7. Given ^x — \x •\- \x ■=■ \; to find x. Ans. x ^ ?^.
8. Given ^4 + x = 4 — sjx : to find x. Ans. x = 225.
9. Given 4a + x = ; — ; to determine x. Ans. x =: — 2a.
4a + X
10. Given ^40^ + x- = V^*^ + a^^ ; to find x.
11. Given ^x + ^/2a + x = —. J to find x.
V 2a -|- X
12. Given, . 1 - =26; to find x^.
I -h 2x 1 — 2x
13. Given a + x = v/a^ + a? '\/46^ + x- ; to find x.
14. Given v/4 + -v/a^ — x* = x — 2.
15. Given (a + x) (6 + x) — a(6 + c) = -^ + x^
16.
In ^/a + X - V a^jT^ = V2a + X, X = -y^^
17. Fmd X in '^±^JlJ^^^^ = ^j. Ans. x = ^-\^*
Va + x— Va — X i + o
18. Find x in -ISx + -2 — •875x + -375 = •0625x — 1. Ans. x = 2.
_.. 2x + 3-5 13x — 22 , X Ix x + 16 a„„ , _ .
10 Given ' ^ - = ^ — Ans. x ^ 4.
ly. uiven ^ I7x - 32 ^ 3 12 36
20. Tmd X from \/a + a/* + V a — V'x = V*-
8a3 + 15q'6 4- 6fli» — 6»
276
Ans. X =
21. In 3-25X — 5 007 — x = "2 — •34x, what is x? Ans. 2 010424 . . .
22. Given (2 + x)* + a?^ = 4 (2 + xH to find x. Ans. * = J.
23. Given - ° ~ ^- + y = a + 2y to find y. Ans. y = 1 — a.
Va — y
24. Given x + v/7^^ = a (a — x)- i to find x. Ans. x = o — 1.
25. Given ^4-!)^^^,:3333^.^^dinf. ^ ^^ ^^ ^^^ ^ ^^^ , ^ .
X
26. Given x + ? x + ^ x = m to find *. Ans. x = ^y:^^-
174 ALGEBRA.
27. Given V V Va + x =b; to find a-; and (z — If = 100 to find «*.
Ans. X = b'* — a, and z^ = 2906923 ....
28. When — — = -r, what is the value of x~"^ ?
29. If the recurring decimal "082082 ... be multiplied by x^, and the square
of the result divided by x^, gives the same value, '082 . . . ; what is the value
of a: ? . 999
Ans. - — .
82
30. Resolve (^ + ') %- ') _ 3« = "^^ - 2x + «^.
a — 0 a + 0 0
Ans. X
_ «■» + 3a^6 + 4a-b^ — 6ab^ + 2b*
~ 2b{a + b) (2a — b)
1 i
31. Given — j^ -\ Y =2^; to find v. Ans. v = 4.
2v — V^2v + v^
SIMULTANEOUS OR COEXISTING EQUATIONS.
When an equation contains two or more unknown quantities, it is obviously
insufficient for the determination of the value of any one of them. The methods
hitherto laid down enable us to obtain the value of any one symbol which is in-
volved in the equation in terms of the remaining ones, whether they be numeral
or literal, known or unknown ; but if there be more than one unknown quan-
tity, the expression of the value of any one of them that we may select will in-
volve the remaining unknowns, and such value will therefore be indeterminable
till such other conditions are added as shall enable us to assign the values of all
these last-named unknowns.
If now a second equation, different in its composition from the former, but
involving the same unknowns, be given ; then also the value of the selected one
can be obtained as in the preceding case ; and if this second equation also express
a second condition to which the relation of the unknowns is subjected, and
which, therefore, must exist simultaneously with the former, the two values of
the unknown must be identical. There may hence be formed a third equation,
which will also be true simultaneously with the two former. This equation will
involve one unknown less than either of the others.
If there remain more than one unknown in this equation, it is still incapable
of furnishing the value of either of them, as before; and there must hence be
still other relations given to render the problem determinate. Suppose then a
third equation to express a third condition, which is simultaneous with the other
two. Then from this also we can obtain a value of the same unknown that we
at first selected, in terms of the remaining ones, and this value equated to either
of the other values, will furnish a second equation, containing only the remain-
ing unknowns. Having thus two equations containing the remaining unknowns,
we can find two values of a second unknown, and equate them ; which will give
us one equation which is freed from both the forementioned unknowns. If this
yet contain more than one unknown, we shall still want other conditions, and
must proceed in the same manner to eliminate them one by one from each pair
of equations that is either given, or which results from the previous elimina-
SIMULTANEOUS EQUATIONS. 175
tions ; till, at last, we arrive at an equation which contains only one unknown,
the value of which must be determined as has been already explained and prac-
tised.
It will be quite apparent from this reasoning, that there must be as many
equations simultaneously given as there are unknown quantities involved in all
of them together ; and that though some of the equations should not contain all
the unknowns, yet they may be conceived to do so by considering 0 as the
coefficient of any one that is absent.
Other processes besides that explained above can sometimes be used more
advantageously ; and as facility in the practice of elimination is best attained by
exercise upon the simple cases, the following rules have been adapted to the case
of two unknowns. The extension of the same kind of processes to three or
more simultaneous equations will then become easy and obvious.
It is, however, to be understood, that any involutions, transpositions, multi-
plications, or divisions by which the equations can be reduced to simpler forms
than the given ones, must be executed previous to the api)lication of any of the
special rules here laid down.
TWO SIMULTANEOUS EQUATIONS.
To exterminate or eliminate one of the two unknown quantities from two simul-
taneous equations: that is, to reduce the two simple equations containing them,
to a single one.
I, Find the value of one of the unknown letters, in terms of the other quan-
tities, in each of the equations, by the methods already explained. Then put
those two values equal to each other for the new equation, involving only one
unknown. The value of this is to be found as before.
It is evidently best to begin by determining the values of that letter which is
easiest to be found from the two proposed equations.
EXAMPLES.
1. Given 2x + 3y = 17, and 5x—2y = 14, to find x and y.
17 3« 14 + 2y
From (1) we have x-= — „— . and from (2), x= — - — .
t:. 1. • 14 + 2y 17 — 3y „
Jbquatug these gives — ;: — - = — - — , or y = 3
., 17 — 3y
Also X = — - — =' = 4.
2
Or, again, by finding two values ofy.
1 7 — *2 J* 5«r ^— 1 4
From (1) we have y = — — , and from (2), y = — ^ — '
^ . , 17 — 2x 5a?— 14
Equating these, — = — — , or x = 4.
3 '*•
., 17 — 2a; „
Also y = = 3.
* 3
2. Given \x-^2y=a, and \x — 1y—h; to find x and y.
Ans. x=.a-\-h, and y=-\a — \h.
Ans. X = — ; f- ; y = -=-4 r-
176 ALGEBRA.
3. In 3x + y = 22, and 3y -f a? = 18, find x and y. Ans. a; = 6, and y = 4.
4. In i ar + ] y = 4, and 3X + iy = 3i; x = 6, y = 3.
^ _. 2x , 3y 22 , 3a; , 2y 67 . , ,
5. Given y + -^ = ~ , and — + -^= — . Ans. a; = 3, and y = 4.
S2 -L. (P s" d^
6. In a? + 2y = «, and x^ — 4y^ = d^; x= , and y = .
' ^ > ' 2s ^ 45
7. In x — 2y = d, and a; : y : : a : 6 ; a; = -r, and y = -7.
8. Given b {x -\-y)^ a Qc — y), and x^ — y^ z=. c^ ; to find x and y.
Ans. :r = ^-^i.^-^ and y = ^-^il^
9. Given a,a; + b,y = c/, and a,,x + i„y = c,,^ ; to find x and y.
i i i i
10. Given -Zx- — •03y2 = 300, and 300y2 + 30a;- = 30000 ; to find x'^
and yl Ans. y^ = 0, x-^ = lO-^^.
11. Find the value of one of tlie unknown letters, in one of the equations, as
in the former rule, and substitute this value instead of that unknown quantity in
the other equation : then there will arise a new equation, with only one un-
known quantity, whose value is to be found as before.
It is evidently best to begin with that letter whose value is most easily found
in either of the given equations.
EXAMPLES.
1. Given 2a; + 3y ^ 17, and 5a; — 2y = 14, to find x and y.
From (1), X = — - — -, which substituted in (2) gives - — — 2y = 14,
2 «
or y = 3 ; and a; = 4.
Or, finding xfrom the second equation.
Here x = — - — -, which, substituted in (1) gives — - + 3y = 1 7, or
5 5
y = 3 ; and hence a; = 4.
In a similar way we may begin to operate by finding y from either of the
equations and substituting its value in the other.
2. In 2a; + 3y = 29, and 3a; — 2y ^ 1 1 , we have a; = 7, and y = 5.
3. In X ->r y = 14, and a; — y= 2, we have x = 8, and y =: 6.
4. In a; : y : : 3 : 2, and a^ — y" = 20, we have x = 6, and y = 4.
5. In ^ + 3y = 21, and ^ + 3a; = 29; x = 9, andy = 6.
6. Given 10 — | = | + 4, and *-^ + ? - 2 = ^-^-^ - 1 ; to find x
and y. Ans. x = 8, y = 6.
7. Given x : y ; : 4 : 3, and x^ — y^ — 37 , ^q fljjd the product of ar and y',
and the difference of x and y. Ans. x- y' = 432, x — y = 1.
8. From x + y ^ a(x — y), and x- + y^ = b' ; find ar and y^.
Ans x»-^° + ^^'^'- (a-l)^i^
^°'-'^- 2(a2+l)' y -2(a2+l)'
SIMULTANEOUS EQUATIONS. 177
9. In bx — cy = 0, and a;^ _ ys _ ^s . ^jj^^ ^^^ ^j^g yalues of x* and y^ ?
Ans ar'= ,,y* = _.
10. Given 7ar + ?j/ = 41 1^, and 39a: — 14y = — gzbf^ ; to find x and y.
Ans. x= I7i, and y= 115J.
1 1 . Given (x + 5) 0 + 7) = (a^ + 1) (y — 9) + 1 1 2, and 2x + 10 = 3y + 1 ;
to find X and y. Ans. 5 and 3 : which is x ?
12. From the equations r- ," = „ — ; — . and ax + 2by = d: to find x and v.
^ o + y3a-|-x ' y > y
Anc „ ^a^-b^^d , 2b^-6a^ + d
Ans. y = -r , and x = '—.
36 3a
III. Let the given equations be so multiplied, or divided, by such numbers
or quantities, as will make the terras which contain one of the unknown quan>
tities the same in both equations.
Then by adding or subtracting the equations, according as the signs may
require, there will result a new equation, with only one unknown quantity, as
before : that is, add the two equations when the signs are unlike, but subtract
them when the signs are alike, to cancel that common term.
The best multipliers generally are those of the selected term in the alternate
equations ; as in the example, ax -\- by =^ c-, and a,x ■+■ b,y = c,^, where the first
equation being multiplied by a^, and the second by a, we get the co-efficients of
X equal.
Again, it will often happen that a and a, have a common measure ; and in
this case, instead of a and O/, take the quotients of them by that common mea-
sure for the cross-multipliers. This will always, when it can be done, lessen the
arithmetical labour.
Let us take as a numerical example 4x + 6y = 2, and lOx — 3y = 59.
Here in eliminating x, the co-efficients 4 and 10 have the common measure 2,
and hence 5 and 2 are the multipliers. Hence we get 20x + 30y = 10, and
20x — 6y = 118.
Subtracting, we have 36y = — 108, or y = — 3 ; and hence x = 5.
Or again, multiply (2) by 2, and (1) by 1, then 4x -f 6y = 2, and — 6y +
20x =118; and adding 24x = 120, or x = 5 ; and hence again y = — 3.
EXAMPLES.
1. In^^ + 6y =: 21, and ?^^ -f 5x = 23; x = 4, andy = 3.
2. la ?^ -H 10 = 13, and ^-^— + 5 = 12 ; x"' = J, and y"" = J.
5 4 o 0
4. In 3x — 4y = 38, and 4x -f- 3y = 9 ; x = 6, and y = — 5.
X 3v
6. Given (x + i) (r, + 7) = (a? + I'j) (y — '9) + lU. and - + 1 = - + 'l ;
to find X and y. Ans. x = 5 ^ and y = 6 ?g.
VOL. I. N
178 ALGEBRA.
7. From - + ^ = 1 , and - -|- - = 1 + - ; find the values of x and y.
a 0 c a o c
ahc {ac + ab — be) ^ abc (ac — ab — be)
Ans. ^ — „2 ^2 ^ a2 c2 - 6^ c^' ^"^ ^ ~ a^'^ +a^~^^P'c-''
8. In X {be — ocy) = y {xy — ae), and xy {ay -\- bx — xy) = abc {x -\-y — c) ;
find a:^ and y^. Ans. x^=±V ± V ± ac, and y' = + V^ ± ^/ ± be.
In the examples given under the different rules for elimination, those have not
always been chosen which are most simply solved by the rule under which they
are given. This was purposely done for the sake of leading the student to
examine the different questions by other of the rules, so as to afiford him the
means of judging in some degree from the appearance of any given equations,
which rule will be most applicable to their solution. Ihe improvement in his
judgment will amply repay him for the trouble of solving each of them by all
the methods.
THREE OR MORE SIMULTANEOUS EQUATIONS.
These are in their nature and mode of solution precisely similar to those
already treated of : they are, however, generally longer, and often the particular
process to be employed is less obvious.
When it happens that the co-efficients of several of the equations are alike with
the same or opposite signs, it will be found advantageous to add the equations
together, and divide the sum by the number of them ; then subtracting each
equation from this quotient will give the value of each of the unknowns in suc-
cession. Sometimes when they are combined in factors, it will be advantageous
to multiply or to divide only the others, and thus get equations in a simpler
form. These, however, are special rules, and can only be acquired by observa-
tion and practice.
Ex. 1. Given x-\-y+z — Q,x-\-2y-\-2,z= 16, and x-\-Zy + Az= 21.
By the first method we have three values of x, viz. :
a?= 9 — y — z, x-= 16 — 2y — 3z, and a?= 21 — 3y — 42.
Equating the second value with each of the others, we have
9 — y — z = l6 — 2y — 3z,ory = 7 — 2z, and
16 — 2y — 3z=2l — 3y — 4^, or y = 5 — z.
Equating these values of y, we get 5 — z = 7 — 22r, or z = 2.
Hence y := 5 — 2^3, and a?=: 9 — y — r^4.
Or, by the second method, we have from the first equation x:=9 — y — z ;
and substituted in the two others gives
9 +y + 2z= 16, ory = 7 — 2z; and 9 + 2y -f 3z = 21.
Substitute the former of these in the latter; then we get 23 — 2 = 21, or
z = 2 ; and X and y will be found as in the last case.
Again, by the third method, the co-efficients of x are already equal ; hence the
equations are prepared for subtraction. Let (1) be taken from (2), and (2) from
(3), then y + 2z = 7, and y -|- z = 5.
The coeflScients of y are here equal, hence subtracting, z = 2 ; and hence x
and y may be found.
Ex. 2. Given x + y = 10, y + z = 23, and z -|- ar = 19.
Add all three together and divide by 2 : then we get
x + y + z = 26.
SIMULTANEOUS EQUATIONS. 179
Subtract each of the given equations from this, and we find
X = 3, y = 7, and z = 16.
The student should also solve this by the other methods.
Ex. 3. Given xy = o', yz = ¥, and zx = c^.
Multiply all three together : then x^y^z^ = w'b-c', or xyz = + abc.
Divide this by each of the given equations : then there will result
, ac ab , he
a; = + -r, y = H , and ■? = H .
— 0 — c — a
This example is an instance of the remark on classification at p. 170; and
would, in strict theory, like some that have gone before, be classed as a
quadratic.
Ex. 4. There are given the three following equations for solution :
ax ■{- by -\- cz =i dp (1)
a,x + b,y -\- c,z = d,^ (2)
a,.x + b,y + c,,z = d,K... (3)
Multiply (1) by a, and (2) by a, and subtract : then we get
(a,b — abi) y + (a,c — aC/) z = fl,e?' — ad,^ (4)
Multiply (2) by a„ and (3) by a,, and subtract : then we have
(a^,6, — a,bi,) y + ia,iC, — a,c,) z = a„di^ — a,d/ (5)
Multiply (4) by a„b — a,6y,, and (5) by a,b — abi, and subtract : then
resolving for z, and substituting in the values of x and y, we have
_ d^ {a„b, — g,6„) + d^ (ab„ — a„b) + d,,^ {a,b — ab,)
c {a,tb, — a,b,i) + c, (^ab„ — a,,b) + c,, (jajb — ab)
d^ (ai,c, — a.c„) + d/' {ac„ — a,,c) + dj^ (a,c — ac,)
^ ~ b {a„c, — a,c,,) + b, {ac„ — a„c) + b„ (a,c — ac,)
_ d? ib.,cj — b,c„) + d,HbCu^— b„c) + rf^^&.c — be)
"' ~ a{bi,c, — b,c„) + a, {bc„ — b„c) + a,i {b,c — be,)
This is the general solution for three unknowns, and by substituting any given
numbers for the co-efficients in these, the corresponding values of x, y, z, would
be obtained.
Ex. 5. Equations of the following forms are of very frequent occurrence in
the subsequent parts of algebra, and hence it may be desirable to indicate the
best mode of resolving them.
+ 1»« + l^a? + I'y + ^ = 0,
.... + 23u + 2'^x + 2^y + z = a,
+ 3% + S^a; + 3'y + z = 0,
+ 4'k + 4-x + 4'y + r = a.
where there are as many such equations as there are unknowns; and the
second side of each equation given.
Subtract the first from the second, the second from the third, the third from
the fourth, and so on to the end. This will give n — I equations clear of *.
Pursue the same course with these n — 1 equations, and we shall obtain « — 2
equations clear of z and y. Pursue again the same course, and n — 3 equations
will be obtained clear of z, y, and x. Proceeding thus, we shall at last obtain a
single equation involving only the letter to the left, whose value is thus found.
Substitute this value in either of the two results obtained by the above process
immediately before the last, and we obtain the value of the second letter. Sub-
stitute these two in either of the three next preceding results, and we shall get
the value next unknown. We may thus obtain the whole very simply and con-
veniently.
N 2
180
ALGEBRA.
The following example, adapted to four unknowns, may serve to illustrate the
process and mode of writing the successive steps of the work.
First differences.
Second diffs. 1 Third diff.
7u+3x+y= 22
19M+5a;+y= 84
37u+7x+y = 212
I2u+2x= 62
lSu+2x= 128
6m = 66
The given equations.
«+ x+ y+2= 2
8u+ 4x+2y+z= 24
27«4- 9x+3i/+z = l08
64u + l6x+4y+z=320
Whence u = 11 ; which substituted in either of the second differences (that
with the least co-efficient will of course be most convenient) gives x = 31 —
6u := — 35 ; and these values of u and x in the first difference gives y ^ 22 —
3x — 7u = 22 + 105 — 77 = 50 ; and lastly, z = 2 — y — x — u = 2 —
50 + 35 — 11 = — 24.
The student may solve the following examples.
«+ x+ y+z= 1
8u+ 4x+2y+z=. 4
27m+ 9x + 3y+z=10
64M + l6a? + 4y+z=20
Ans. a=5, x=:h, y=3, z= 0
u-\- x+ y+z= 1
8u+ 4X'{-2y+z= 5
27u+ 9x+3y+z=l4
6iu + l6x-\-4y+z=30
Ans. «=J, x=^, y=5, z=0.
EXAMPLES FOR PRACTICE.
fX + y + Z= 18 1
. When •lx + 3y + 2z = 38>; then a; = 4, y = 6,
U + Jy + i^ = 10 3
(x + iy + hz = 27)
. l{{x+ ly + iz=9.0};
U + iv + lz=l6)
= 3,5
)= 45)
)= 75 V;
) = 105 J
then X = 1, y = \2, z = 60.
5, z
+ iy + i^ = 27 ■
+ ly + '
+ iy + li
3. If a? — y ■= 2y X — r = 3, and y -j- z = 9 ; then x ^ 7,y
ix{x + y + z)=z
4. When -^ y (jT -f- y + z) = 75 J- ; then a: = + 3, y = + 5, and r = + 7.
U (X + y + z)
5. Given vxyz = o, tury = b, tuxz = c, tuyz = d, txyz = c; to find t, u, x, y, z.
Ans. t =
\/ abode *\/abcde _i/ abode _\/ abode
and z =:
\y abode
a '"- b ' c '" ~ d
6. Given a:+y-|-2=:a, my = nx, and pz =z gx ; to find x, y, z.
Ans. X =
7 ■ Given
amp
y =
anp
mp + np + mq' ^ mp + np+ mq'
and z =
amq
xy
= 1,
yz _
ay -\- bx * cz -\- dy
= m, and
ez -\-fx
mp -\- np + mq
— ^ n ; to find x, y, z.
A _ ^'">» (.bde+aof) __ Imn {bde+aof) _ Imn {bde+acf)
cfjnn—bfln-\-bdlm ajln+demn — adlm' beln — ceinn-\-aclm'
!x + y + z + t + u = a\ a-\-b+c+d+e+f
x + y + zi-u + v}=b\ Ans. put s = ^ ;
x-fy + z-f/-ftP = ci
x + y + u + t+w=d ^1^6° w==s — a, t = s — b,
x + z + u-\-t-l-w^e\ U=:S — 0,Z=lS — d,
y + z + u + t+w=/{ y = s — e,x = s—f.
9. Given x (y + z) = a^, y (x + z) = 6», and z (x + y) zz c^ ; to find the
unknowns.
10. Given x — ay + a^z — a^ = 0, x — by + Pz — b^ = 0, and a; — cy +
c^z — c^ = 0.
SIMULTANEOUS EQUATIONS. ' Igj
1 1. Given xy = a (x + y), xz = b (x 4- z), and yz = c (y + z) ; to find the
recij)rocals of x, y, z.
fu-\-x-\-y-\-z = A '\
12. Given<"^ + «y + "^+^i' + ^^+/^ = H.tofind« x v z
J uxy + uxz + vyz + xyz = 4 I
Luxye = 1 J
A COLLECTION OF QUESTIONS PRODUCING SIMPLE EQUATIONS.
Quest. 1. To find two numbers, such, that their sum shall be 10, and their
difl^erence 6.
Let X denote the greater number, and y the less •.
Then the first and second conditions are x + y =: 10, and x — y = 6, x = 10
— y ; whence x ^ 8, y = 2.
Quest. 2. Divide 100/ among A, B, C, so that A may have 20/ more than B,
and B 10/ more than C.
Let A's share = x, B's := y, and C's = z.
Then x + y + z= 100, x = y + 20, and y = z -\- 10.
From which x= 50, y =■ 30, and z = 20.
Quest. 3. A prize of 500/ is to be divided between two persons, so that their
shares may be in proportion as 7 to 8 ; required the share of each.
Put X and y for the two shares ; then the conditions are
7 : S ',[ X : y, or 7y ~ 8x, and x + y = 500 ; hence x = 2383 and y = 266}.
Quest. 4. What fraction is that, to the numerator of which if 1 be added,
the value will be ^ ; but if 1 be added to the denominator, its value will be \ ?
X X ~^~ 1 X
Denote the fraction by - : then = ^, and — = J.
^y y ^ y+1 '
X 3
'From which a? = 3, u = 8, and - = -.
" y 8
Quest. 5. A labourer engaged to serve for 30 days on these conditions : that
for every day he worked, he was to receive 20d, but for every day he played, or
was absent, he was to forfeit lOd. Now at the end of the time he had to receive
just 20 shillings, or 240 pence. It is required to find how many days he worked,
and how many he was idle ?
Let X be the days worked, and y the days idled.
Then 20a? are the pence earned, and lOy the forfeits ;
Hence, by the question a: + y = 30, and 20a; — lOy := 240 ;
"Whence x = 18, the days worked ; and y = 12, the days idled.
Quest. 6. Out of a cask of wine which had leaked away J, 30 gallons were
drawn, and then, being gauged, it appeared to be half full : how much did it
hold ?
Suppose it held x gallons ; then it leaked \x gallons.
Hence there had been ^x + 30 gallons taken away, and by the question, J* =
^a? + 30; and a:= 120, the gallons it held.
* In these sohitions, as many unknown letters are always used as there arc unknown num-
bers to be found, purposelv for exercise in the modes of rcducina; the equations : avoiding; the
short ways of notation, which, though they may give neater solutions, afford less exercise in
practising several rules in reducing equations. It is also considered unnecessary to carry out the
solutions to their completion, as the steps arc so familiar to the student from the exercise in
reduction which has preceded. The examples, indeed, are given principally with a view to
practice in the translation of the verbal conditions of a question into the symbolical language of
algebra.
183 ALGEBRA.
Quest. 7- To divide 20 into two such parts, that 3 times the one part added
to 5 times the other may make 76.
Let X and y denote the two parts.
Then, by the question, x + y ^ 20, and 3x -\- 5y •=. 76.
From which a? = 12, and y ^= 8.
Quest. 8. A market woman bought in a certain number of eggs at 2 a penny,
and as many more at 3 a penny, and sold them all out again at the rate of 5 for
two-pence, and by so doing, contrary to expectation, found she lost three-pence ;
what number of eggs did she buy ?
Suppose she bought x eggs of each kind : then the cost of the first lot was ix.
and that of the second lot was Ix. Also in selling 2x eggs at 5 for two-pence,
she received f . 2x pence : and by the question, this was three-pence less than she
gave for them. Hence - + ^ — 3 = ^r » ^°^ therefore x ^ 90, the number in
each lot, or 2x = 180, the whole number.
Quest. 9. Two persons, A and B, engage at play. Before they begin, A has
80 guineas, and B has 60 : but after a certain number of games won and lost
between them, A rises with three times as many guineas as B : how many
guineas did A win of B ?
Denote by x the number of guineas won by A. Then they rise with 80 -j- a?
and 60 — x respectively. But by the question SO + a? = 3(60 — x) ; hence
X = 25, the guineas won by A.
Quest. 10. The sum of the three digits composing a certain number is 16 ;
the sura of the left and middle digits is to the sum of the middle and right ones
as 3 to 3|; and if 198 be added to the number, the digits will be inverted in the
expression of this sum.
Let X, y, z denote the digits ; then 100a? -\- lOy -\- z will express" the number
itself, and lOOz -f \0y -\- z will express the number having the same digits in an
inverted order. Whence the three conditions are
a? -I- y + z = 16, a- + y : y -I- z : : 3 : 3?, and
100a? -f lOy -1- 2 + 193 = lOOr-f lOy -\- x, or z — x = 2.
Whence the solution is a; = 5, y = 4, and z = 7 ; and the number itself is
5 . 100 + 4 . 10 + 7, or 547.
Quest 11. If N men of a certain degree of skill can do a piece of work in n
days, N, others of different skill in n, days, N,, others in n,, days, and so on for
m sets of men : in how many days would one of each set be able to do - of the
work, supposing they all worked together, without impeding each other's
operations ?
Each of the first set would do :j^ of the work per day ;
second _J
ith
N.
SupjMJse that one of each of these men working together could execute the
jjth part of the work in x days : then they would execute the whole in px days,
or — in one day. Hence equating the two expressions for their total work
in one day, we have
SIMULTANEOUS EQUATIONS. 183
N^ "^ N.n, + N.,n„ + " " * " N^^ =^ = »'«''" "
Nn N, n, ^„ «„ ^ '•• >j_ „_
QUESTIONS FOR PEACTICE.
J/ is recommended that the student should also solve these questions generally, by
taking literal symbols instead of the given numbers.
1. Determine two numbers such, that their difference may be 4, and the
difference of their squares 64. Ans. 6 and 10.
2. Find two numbers with these conditions, viz. that half the first with a thiid
part of the second may make 9, and that a fourth part of the first with a fifth
part of the second may make 5. Ans. 8 and 15.
3. Divide the number 2 into two such parts, that a third of the one part
added to a fifth of the other may make I. Ans. To and "5.
4. Find three numbers such, that the sum of the 1st and 2d shall be 7, the
sura of the 1st and 3d 8, and the sum of the 2d and 3d 9. Ans. 3, 4, 5.
5. A father, dying, bequeathed his fortune, which was 2800/, to his son and
daughter, in this manner; that for every half-crown the son might have, the
daughter was to have a shilling : what then were their two shares ?
Ans. the son 2000/, and the daughter 800/.
6. Three persons. A, B, C, make a joint contribution, which in the whole
amounts to 400/ : of which sura B contributes twice as much as A and 20/ more ;
and C as much as A and B together : what sum did each contribute ?
Ans. A 60/, B 140/, and C 200/.
7. A person paid a bill of 100/ with half-guineas and crowns, using in all 202
pieces ; how many pieces were there of each sort ?
Ans. 180 half-guineas and 22 crowns.
8. Says A to B, if you give me 10 guineas of your money, I shall then have
twice as much as you will have left; but says B to A, give me 10 of your
guineas, and then I shall have 3 times as many as you : how many had each ?
Ans. A 22, B 26.
9. A person goes to a tavern with a certain quantity of money in his pocket,
where he spends 2 shillings ; he then borrows as much money as he had left,
and going to another tavern, he there spends 2 shillings also ; then borrowing
again as much money as was left, he went to a third tavern, where likewise he
spent 2 shillings ; and thus repeating the same at a fourth tavern, he then had
nothing remaining : what sum had he at first, and what was he in debt ?
Ans. at first 3* Qd, and had borrowed 4s 3d.
10. A man with his wife and child dine together at an inn. The landlord
charged 1 shilling for the child ; for the woman as much as for the child, and i
as much as for the man ; and for the man as much as for the woman and child
together : how much was that for each ?
Ans. the woman 20<f, and the man 32c/.
11. A cask, which held 60 gallons, was filled with a mixture of brandy, wine,
and cyder, in this manner, viz. the cyder was 6 gallons more than the brandy,
and the wine was as much as the cyder and i of the brandy : how much was
there of each ? Ans, brandy 15, cyder 21, wine 24.
184 ALGEBRA.
12. A general, disposing his army into a square form, finds that he has 284
men more than a perfect square ; but increasing the side by 1 man, he then
wants 25 men to be a complete square : how many men had he under his
command? Ans. 24000.
13. What number is that, to which if 3, 5, and 8 be severally added, the
three sums shall be in geometrical progression ? Ans. 1 .
14. The stock of three traders amounted to 760l : the shares of the first
and second exceeded that of the third by 240/; and the sum of the second and
third exceeded the first by 360/ : what was the share of each ?
Ans. the 1st 200/, the 2d 300/, and the 3d 260/.
15. What two numbers are those, which, being in the ratio of 3 to 4, their
product is equal to 12 times their sum ? Ans. 21 and 28.
16. A certain company at a tavern, when they came to settle their reckoning,
found that had there been 4 more in company, they might have paid a shiUing
each less than they did ; but that if there had been 3 fewer in company, they
must have paid a shilling each more than they did : what then was the number
of persons in company, what did each pay, and what was the whole reckoning ?
Ans. '24 persons, each paid 7s, and the whole reckoning was 8 guineas.
17. A jockey has two horses; and also two saddles, the one valued at 18/,
the other at 3/. Now when he sets the better saddle on the 1st horse, and the
worse on the 2d, it makes the 1st horse worth double the 2d; but when he
places the better saddle on the 2d horse, and the worse on the 1st, it makes the
2d horse worth three times the 1st : what were the values of the two horses ?
Ans. the 1st 6/, and the 2nd 9/.
18. What two numbers are as 2 to 3, to each of which if 6 be added, the sums
will be as 4 to 5 ? Ans. 6 and 9.
19. What are those two numbers, of which the greater is to the less as their
sum is to 20, and as their difference is to 10 ? Ans. 15 and 45.
20. What two numbers are those, whose difference, sum, and product, are to
each other, as the three numbers 2, 3, 5 ? Ans. 2 and 10.
21. Find three numbers in arithmetical progression, of which the first is to the
third as 5 to 9, and the sum of all three is 63. Ans. 15, 21, 27.
22. It is required to divide the number 24 into two such parts, that the quo-
tient of the greater part divided by the less, may be to the quotient of the less
part divided by the greater, as 4 to 1. Ans. 16 and 8.
23. A gentleman being asked the age of his two sons, answered, that if to the
sum of their ages 18 be added, the result will be double the age of the elder;
but if 6 be taken from the difference of their ages, the remainder will be equal
to the age of the younger : what then were their ages ? Ans. 30 and 12.
24. Find four numbers such, that the sum of the 1st. 2d, and 3d shall be 13 ;
the sum of the 1st, 2d, and 4th, 15 ; the sum of the 1st, 3d, and 4th, 18 ; and
lastly, the sum of the 2d, 3d, and 4th, 20. Ans. 2, 4, 7, 9.
25. Divide 48 into 4 such parts, that the first increased by 3, the second
diminished by 3, the third multiphed by 3, and the fourth divided by 3, may be
all equal to each other. Ans. 6, 12, 3, 27.
26. A cistern is to be filled with water from three different cocks : from the
first it can be filled in 8 hours, from the second in 10, and from the third in 14 :
how soon would they all together fill it ? Ans. in 3 h 22 min 24^| sec.
27. Show at what periods the hands of a watch will be together during a
complete revolution of the hour-hand.
Ans. at - -, where m is 1, 2, ... 11 successively.
QUADRATIC EQUATIONS. 185
28. A labourer engages to work for 3s 6d a day and his board, but to allow
grf for his board each day that he is unemployed. At the end of 24 days he has
to receive 3/ 2s 9^^ : how many days did he work ? Ans. 19 days.
29. Three workmen are employed to dig a ditch of 191 yards in length. If
A can dig 27 yards in 4 days, B 35 yards in 6 days, and C 40 yards in 12 days,
in what time could they do it if they worked simultaneously ? Ans. 12 days.
QUADRATIC EQUATIONS.
A QUADRATIC cquatiou is that in which the unknown quantity is of the second
degree, and is generally represented by oar + 6x + c = 0, where a, b, c may
be any numbers positive or negative, integer, fractional or irrational.
When 6 = 0, it takes the form aar + c = 0, and it is called a pure quadratic.
It is treated as a simple equation, since in the solution no operation is required
c
but the arithmetical one of extracting the square root of .
When all the terms are present, the equation is called an adfected quadratic.
There are two methods of solving such equations ; one due to the Hindus,
the other to the early Italian algebraists. They are alike in principle, which is
that of so modifying the first side as to render it a complete square ; and by
corresponding additions, subtractions, multiplications, or divisions, applied to
the other side, to still retain the original truth of the equality. The operation is
technically called completing the square.
1 . The Italian or common method.
Transpose e, and divide every term by a : then we have
X' A X = — -
a a
Add the square of half the co-efficient of x, viz. of — , to both sides : tnen we
have
b b^ b^ c ¥ — Aac
a 4a'' 4a'' a 4a-
and the first side is a complete square. E.xtract the roots, and resolve the
resulting simple equation. This gives successively
a?+;r-=± „ , and a? = — — „ .
2o 2a 2a
ITiis operation is often troublesome, on account of the reduction of the second
side of the complete square, in actual numbers. When, however, — is an even
c . . .
number, positive or negative, and also integer, it is the most convenient : but
as this is seldom the case in the quadratics that arise in practice, we shall give a
preferable one, viz. :
2. The Hindu method.
Let the equation he ax^ + bx =■ — c. Multiply by 4a, and add i' to the
product on each side. Then we have
186 ALGEBRA.
4a^ar -f 4abx + b^ = b- — 4ac, and extracting
— J + \^b'^ — 4ec
2ax -f i = + ^b' — 4ac, or a: = — ,
which is precisely the same result as before obtained by the Italian method,
though by a process which is drithmetically simpler *.
It has already been explained, that the square root of any quantity a^ is either
-)- a or — a, and marked by writing the sigu thus, + a, signifying that the root
has both values + o and — a. Hence the answers above given are to be under-
stood as twofold in each case, viz. : x = , and ,
2a 2a
either of which substituted for x in the given equation, will render all the terms
on one side equivalent to those on the other, taken collectively.
Also, since the square root of a negative quantity cannot be actually extracted
in real numbers, positive or negative, when b^ — 4ac is negative, the equation
does not admit of resolution in numbers, or it is only symbolical. The roots, or
values of x, are said iu this case to be imaginary j and wherever such a result
appears, it is the indication of contradictory conditions involved in the condi-
tions of the problem which gave rise to the equation.
All equations, in which there are two terms including the unknown quantity,
and which have the index of the one just double that of the other, are resolved like
quadratics, by completing the square, as above.
Thus, X* + ax- =: b, or x-' -\- ax^ = b, or x -{■ ax- = b, or (x^ + or)* + m
(x^ + ax) = b, are analogous to quadratics, and the value of the unknown quan-
tity may be determined accordingly.
It may also on some occasions be useful to remark, that any quadratic equa-
tion in which four times the product of the co-efficient of the first term by the
third term is equal to the square of the coefficient of the second term, is already
a complete square. For let ax- -f 5a? -|- c = 0 be the equation : then if 4oc = b-,
we shall have c = — , and the equation becomes ax- -{• b -\ = 0, the root
4a 4a
of which is x \/a -\-s—r = 0.
2va
The same of course is true if ax- + bx -\- c=^ k, where k designates any quan-
tity or expression whatever.
EXAMPLES.
1. Given x- -\- 4x = 60, to find the values oi x.
The Italian method apphes to this example, and
a^ -f- 4a? -h 4 = 64, or extracting, a? -|- 2 = +8.
Whence x = 6, and a? = — 10 are the values of x.
2. Resolve the equation of 3x — 5x= 12.
The Hindu method is applicable in this case.
Then 36x^ — 60a? -}- 25 = 144 -h 25 = l69, and
4
6a? — 5 = + 1 3 : hence x = 3, and x= — „.
• When the co-efficient of the second term is an even number, it will be sufficient to multiply
all the terms by the co-efficient of the first, and add the square of half that of the second to both
the products.
For let a^ -\- 2b/c =: c, : then the completed equation is o,V -|- 2a,h^ -\- b/^ = b,^ -j- a/r„
the first side of which is the square of a/c -\- b,.
QUADRATIC EQUATIONS. 187
3. Given the equation ^x^ — Ja? + 30^ = 52§.
Transposing and cancelling the denominators gives
3x- — 2x= 133.
Hence the special rule in the note on the HindG method gives
9x- — 6x + 1 = 400, or 3a; — 1 = + 20. Whence
X := 7, and x = — GJ.
4. Given x* — 2ax" = b, to find x. The Italian method can be used here.
Complete the square, then x* — 2ax- 4- a' := a' + 6 ;
Extract the roots, then x^ — a = + ^a'^ 4- b ;
Resolve simple equation x^ = a + ^/a- + b ;
Resolve pure quadratic x = + v a + >/**' + *•
That is, X has the four values.
+ \/a+ Va^+b,— \/a+ Va^+b, + \/ a— Va^+b, and - \/a— x/a^+b.
5. Given / x 1- / I — =a;, to find x.
V a? V a:
Transpose : then^/ 1 = x — I x — ;
square and cancel : then l=a;" — 2x I x V x;
divide by x : then -=a? — 2 I x 1-1;
"' x S/ X
transpose : then fa?—-) — 2 /x — - -|- 1 = 0 ;
^0.- i_ 1 =0,or J^-l=^-r
extract : then
square : then x — - = 1, or a:* — a; := 1 ;
complete the square by the HindA method, and we get
4a?2 — 4a; 4- 1 = 5, or a; = - | i ± V 5 j .
6. Given ^x — 4 { v'a; + 13} ^ + 7 = a; — 2^/a; - 9, to find x.
Add 10 to both sides : then we get both sides squares, viz. :
^a; + 13 — 4V'^s/x + 13 + 4 = a; — 2 v'a; + 1, and extracting,
VVx+ 13 — 2 = + J\/a; — iL
To continue the solution, take the results + and — separately.
(I.) Take + : then v'a? — 1 — n/V^; + 13 = — 2, or adding 14 to both,
\^x + 13 — \/ sjx + 13 = 12, which is again of the quadratic form.
Complete by the Hindu method, and extract, which gives
\ \^x -\- \Z\ — 4 ^^x+ 13 + 1 = 49, ors/VJ? + 13 = - = 4
or — 3.
188 ALGEBRA.
Squaring v a/^ + 13=4, we get sjx + 13 = 16, or v'-e= 3, and a? = Q.
Squaring V Vx + 13 = — 3, we get ya: + 13 = 9, or ^/x~ — 4, and
a?= 16.
(2.) Take — : then •/« + 13 + v ^/a; + 13 = 16; and completing the
square (Hindu method) as before, extracting and reducing, we obtain ultimately
two other values of x. Collecting the four together, we have x = 9, a? = 16, and
57+7a/65
X = —2 •
This example exhibits a class of contrivances for completing the square of
very frequent use ; but no general rule can be laid down respecting it. The only
general remark that can be made is, to endeavour to render the parts without
the radical, the square of the radical itself; and then, if on completing the square
of the side so transformed, the other side is also a square, the method will be
eflfective.
7. Given V (1 + a?)' — V (I — x)^ = V (1 — ^^) ; to find x.
Dividing both members of this equa. by V (1 — ^)» ^^ have
yi + X "/l - a' _ ^
VI— a? Vl + a?
■ Theref. by .ra„sp. (L^^)'^ - (}±{)^ = l.
This is evidently in the form of a quadratic ; by resolving which, we get,
and finally, a: =[[i^^-f;.
EXAMPLES FOR PRACTICE.
1. Given x- — 6a: — 7 = 33; to find x. Ans. a? = 10 or — 4.
2. Given x^ — 5a? — 10 =r 14 ; to find x. Ans ar ^ 8 or — 3.
3. Given 5x^ + 4x — 90 = 114 ; to find x. Ans. a; = 6 or — 6i.
4. Given \x^ — Ja; + 2 = 9 ; to find x. Ans. a; = 4 or — 3^.
5. Given 3a:* — 2a72 = 40 ; to find x. Ans. a; = 2 or — 2.
" 6. Given yo — ^^/x = 1 J ; to find x. Ans. x = 9 or 25^.
7. Given ^x* + ga- := J; to find x. Ans. x ^ — 3 + 8 \^70.
8. Given ar* + 4a:^= 12 ; to find x. Ans. x = \/2, and a; =: — VS-
9. Given a?- + 4a? = a^ + 2 ; to find x. Ans. x= — 2 + ^a- + 6.
,0. In ^^-±4:^%5^x+;^-57i = 0;. = 4,9.or:^^^^-A9.
11. Solve ^- ^^^ = 1. Ans. x= + /L±J^'
x—^l—ar a?+-v/2— ar* ~V 2
,^T/ ,o.«. s « a , a-\- sJa^ — 4
12. In (x — a)2 + - (a: — a) = 2_p - 1 ; x = -, and x = ""^2 '
63 16 8 a — a/a^ — a?"
13. Solve x^ + X' = 7-56 ; Zx^ — 2-5x' -f 592 = 0 ; and — .i__- = b
SIMULTANEOUS EQUATIONS. 189
Vi
14. Resolve 2x Wx — 3a? ^ ^ = 20, and 2a?' + 3x— 5^/2x» + 3ar + 9 = —3.
15. Solve 1(23? + 1)« + a;V — a? = (2ar + 1)* + 90, and x = '^~"\
L J a
16. Resolve (x + 1) (x^ + I) (x^ + 1) = 30a?^ and a?« — 2x« + ar = o.
<:t>' _ .
SIMULTANEOUS EQUATIONS OF THE SECOND AND
HIGHER DEGREES.
The same conditions that had place in the case of simple equations also hold
with respect to those of the higher degrees. The elimination of any number of
them, and the actual determination of the values of any one of them, becomes,
however, much more difficult. Even when the elimination of all but one of the
quantities is effected, the equation in which it is involved is generally of a high
degree, and the trouble of actually determining its numerical value becomes con-
siderable : but except in very particular cases, when more than one of the simul-
taneous equations is above the first degree, the equation which results from the
elimination takes a form which does not admit of any solution in terms of the
literal quantities involved in the given ones, or at most by means of expressions
of extreme complexity. In the case of the simultaneous equations involving two
of the second degree, and the remaining ones simple, the solution is theoretically
possible in the terms of the given symbols ; but still the transformations to be
made in the forms of the equations to effect it are too numerous to render it
capable of practical application.
Into the general reasonings concerning the principles of elimination, the limits
of this Course forbid our entering. Nevertheless, as cases of frequent occur-
rence, in almost every branch of mathematics, involve this problem under more
or less confined conditions, it has been considered necessary to devote a page or
two to the simpler uses of it, and to give a few exercises, as a praxis for the
student, at the close of the chapter.
I. All the operations described in the section on Simultaneous Equations of
the first degree are to be applied to others of a higher order, in any case that
admits of it.
II. If all the terms of an equation contain the same number of unknown
factors (in which case it is called a homogeneovs equation), we may put one of the
factors equal to the other, multiplied by a new unknown, assumed for the
purpose. As for instance, in Zxy + 2y^ = 4xS we may put y = vx, which
gives —
(3» -I- 2p2)a;2 _ 4^^^ or
317 + 2p2 = 4 ;
from which p may be obtained. The same substitution being made in another
equation simultaneously given, but with the value of v, instead of v itself, gives
an equation containing y only, and which may be resolved by the usual methods
and known quantities. If there be three or more unknowns, so many inde-
pendent but simultaneous substitutions must be assumed for them as there are
of quantities besides tlie one selected above.
III. Sometimes we can effect the reduction by substituting for one of the
unknowns the sum, and for the other the difference of two other quantities, of
course unknown too. This method applies to the case of two unknowns, and
190 ALGEBRA.
obviously is confined to it. Methods in some degree analogous to this have,
however, been devised for the case of three or more unknowns.
IV. It often happens that by raising one equation to some power, several of
its terms will be identical with those of some other power of another equation,
and in this case the equations are simplified by subtracting one from the other.
The same is true of multiples of one, and powers of another equation. Some-
times too it happens that adding some quantities to one side of an equation, to
render it a complete square, cube, or higher power of a binomial, the other side,
so increased, becomes also a square, cube, and so on. The roots then being
taken, the equation is reduced to lower dimensions, and the ultimate elimination
more easily eflfected.
Other rules and remarks will occur to the intelligent student as he proceeds ;
and the teacher will often, in the actual solution of individual problems, be able
to enforce a rule, and to point out the circumstances under which it can be
applied, which could scarcely be rendered intelligible in print without extreme
prolixity.
Ex. 1 . Given x -\- y = a, and xy = b-, to find x and y.
By squaring ( 1 ) we have a^ + 2xy -\- y^ = a-,
and multiplying (2) by 4, 4xy = 4^P ;
Subtracting, x''' — 2xy + y" = a^ — 4.b-,
and extracting, x — y := + ^a^ — 46^.
From this and (1) we get
a + ^/a2 — 46^ , a + Va^ — 46«
X = ~ „ , and y = .
2 ^ 2
Ex. 2. (liven x -\- y ^= a, and x^ -\- y^ =: b^, to find x and y.
Cube eq. (1) : then x'^ + 3x^y + 3xy -j- y' = a^.
Subtract (2) from this : then 3xy{x ■{• y) = a^ — J"*,
or by substitutmg (1) m this, xy = ^ = ^^ .
Then by means of this and (1) proceed to find x and y as in the last example.
Or thus. Put 07 =: « + f and y = u — t : then we have a? -f- y = 2m = a, or
u = '-. Insert the assumed values of u and r in ar^ _j_ y3 __ js . i^q^ ^.g get
2u^ -\- 6uv^ = b^, or putting in this the value found for u, we have
fl 4^3 — a' . , , I A hi „3
vr = , or extractmg, v=-rh /— -_.
12a ^' -^V 3a
Putting in the equations x = u -\- v and y = u — v these values, we have the
result required.
Or, thus again. Put y = vx : then the equations become x(l -\- v) = a, and
X3( 1 -f- «3) = b\
Divide the cube of the first by the second of these : then
-r zu -t u _ a ^^ ^^3 _ ^3^^2 _ 3 _^ 2i3)„ 4. a3 _ ^3 ^ 0.
1 — « -j- «•* 0*
Hence u becomes known by the solution of the quadratic equation.
Also, inserting the values of u thus found in a?(l -|- «) = a, we get x = ;
and from this, again, y = ux will be obtained.
115
Ex. 3. Given x^y -\- y^x = 30, and + = -, to find x and y.
a: y 0
Break (1) into factors, and cancel denominators in (2), then xy{x -f y) = 30,
and 6(x + y) = 5xy.
SIMULTANEOUS EQUATIONS. 191
Multiply the second of these by xy and the first by 6, and subtract : then
x-y"^ = 36, or a:y := + 6.
As there is subsequent work to perform, it will be advisable to work with
a?y = 6 and xy ■=■ — 6 separately *.
First, take ary = 6 : then inserting this in (1) we get a? + y ^ 5 ; and re-
solving this as in Ex. 1, we find x — y = + 1 ; and this again combined with
(1) by addition and subtraction gives a? =: 3 or 2, and y = 2 or 3.
Secondly. Take xy = — 6 : then, proceeding as before, we have x = 6 or — 1,
and y ^ — 1 or 6.
The latter pair of results not fulfilling the equation, do not come properly
under the denomination of answers. They come into the work from the ambi-
guous root of x^y^ = 36 ; but it might have been inferred from this being
{-|- xy) . (-f xy) = 36, that only xy = -\- (S was admissible.
Or thus. Put x =1 u •\- V and y = u — v. Then substituting in the given
equations and reducing, we find u{u^ — v^) = 15, and 12w = 5(h' — p*).
5
Divide the first of these equations by the second : then 4m' = 25, or u ^ -
(see remark on last solution).
Substitute this in either of the last equations : then we get p = + -. Whence
5 + 1
we obtain the same results as before, x = u •\- v = —~ = 3 or 2, and y = u
5 + 1
— V = [ — = 2 or 3.
2
Or, thus again. Put y = ux. Then the equations reduce to uy' (1 + u) = 30,
and 6(1 -\- u) = buy. Equating the values of y' derived from these we obtain
36 (1 + m)^ = 625a2, or 6(1 + m)' = + 25m : that is, Qu' — 13« + 6 = 0,
and 6k2 + 37« + 6 = 0.
The former gives « = „ and - , and the latter m = — , and — ^.
Take m = + - : then y = ^' = 3, and a: = «y = 2.
.... u = + : then y =z =2, and a? =: «y ^ 3
.... M = — 6 : then y = . . . . = — 30, and x = uy = 180.
1
....«=-g:
The two last results are, as in the other case, the consequence of the ambi-
guous sign in 6(1 + m)- = + 25. and their inapplicability might have been
inferred at the outset, as in the former solutions.
Ex. 4. Given ^ -4- - -I- ~ = 9, ^ + ^ = 13, and Sx + 3y = 5.
X y z X y
Q q 3 1 3x 2
From (3), - = — - — . and from (2), = ; equating these values
^ " y b — d,x' y X * **
3
of , and reducing, we have
y
104x2 — 72x + 10 = 0 ; whence x = , and x = ^^.
2 2o
• In questibns of this kind, these separations should, for the »ake of secure working, be
always employed.
then y = = — 6, and x = uy = 1 .
\92 ALGEBRA.
., 5 — 8a? 1 15
Also y = — ;; — =: „ or -- .
^ 3 3 13
Insert these values in (1) and reduce; then we obtain
111 1 ,
- = 9 = 9 — 2 — 3 = 4, or 2 = -, and
z X y 4
26 13 44 15
= 9--5--r5=r5'"''" = 4T
77 rr.- 4,4+y 8+ 4m, 12y" , , ,
Ex. 5. Given -^ -\ !— ^ = — - — ^ -\ f-, and 4y- — xy = x.
y y XX
The first equation is convertible into one having both sides squares, viz,
x^ (2 -\- yY — 4 xy"^ (2 + y) + 4y* = l6y^, and extracting a" (2 + y) 2y^ = + 4y" ;
or taking them separately, and reducing, 6y^ — xy = 2x, and 2y^ -j- ay = — 2a;.
Combining each of these with the second given equation, we shall readily
50 5
obtain x = 2 and y from the first combination ; and x = —, and y = — —
from the second.
Or thus. From the second given equation - = ^.
Substitute this in the first, and reduce; then Zy^ + 2y = 5.
wi, — 1 + 4 ^ 5
Whence y = = — = 1 or .
^3 3
From the second equation x = —^ — = 2 or — „ , the same as before.
1 + y 3
EXAMPLES FOR PRACTICE.
1. Given xf -\- y' := a", and ary = b^, to find x and y.
Ans. x=z [h d" :h i s/^'" — ib^"]', and y = [^ a" + ^/c?' — 4i*"]".
2. Given a; + y = a, and x'' -\- y^ ■=■ d^, to find x and y.
Ansa;-"- / 3aV /rfM^ , a, / 3a^ - /d^l^*
Ans.a._2^^__+ ^— ^— ,andy=2± V "T + V^^'
3. In the following equations find the values of x and y, viz.
X + y + s/x^ —y^ 9 , , , ,
X -\- y — ■y/x^ — y^ °y
(x^ + yf + X — y = 2x {x- + y) + 506.
/x hi 6 1
4. In / -+ ^ - = —, f- 1. and ^sj3?y + V^ry' = 78, find x and y.
5. Given - 4- - = - , and -5 + — = — , to find x and y.
a: y m a?'' y- «^ ^
6. Given x"- — y"^ ■=. a?, and {x + y + bf + {x — y -\- bf = c^, to find x, y.
7- Find the values of x, y, z in the three equations,
X {y ■{■ z) =1 a\ y (z + x) = b", and z {x + y) = c^ ;
And likewise also from the three,
X f y + z = 10, a;2 + y2 + ^2 = 38, and y^ ^ 1 = xz.
SIMULTANEOUS EQUATIONS. 193
8. Find x,y,z n the three following equations,
x^ + xt/ + y^ = c^, x^ + xz + z^- = b-, and y^ + yz + z^ = a'.
9. Given = — , and 3xy + 2x + y — 485=z o, to find x and y.
y X '
_. aa? iy ,
10. Given — = -, and cxy + dx + ey = h, to find x and v.
y X J ' y
11- ^'v^'^ :r?^, = «' TTir-, = *. and -^ = e, to find «, y, « ;
X + y y + z z + X >y»»
Also, iry = 10, z (4 — y) = 12, and (5 — a-) (6 — 2) = 21.
12. Resolve the following pairs of simultaneous equations.
(1). x^ + y^ -^ X -\- y = a, and x- — y- + x^ — y = b.
(2). X -\- y =: xy, and x -\- y + x^ + y^ ■= a.
(3). 19 + lli = lii(, and 7x - 4y - -18- '-^ = 0.
y X y
(4.) 5y + 1 Jx2 — I5y — 14]2 = 1 a?2 _ 36,
8 3 ( 3 4J 2
(5). » + M^ — V "JJ = 2f^ + i (21 — m), and u^ + r^ = 6.
13. Given x + y -\- z ■= a, a;^ + y^ + 2^ = i^, and y =: ^xz.
14. Find tt, a?, y, 2, from the equations u -\- x -\- y -\- z =■ \5, u^ -j-a^ -|- y«
+ 2' = 85, a^ ^ My, and y^ =1 a;^.
15. M» + a^2 = 444, ux -^ vz ■= 180, M2 + rx = 156, uvxz = 5184.
16. Given x -\- y =i a, and aH* + y' = i' 1
A 1 It ^ 2. .; 1 to find a; and y in both pairs.
Also X — y =■ a^, and a* — y* = 6,^ j :' v •
QUESTIONS PRODUCING QUADRATIC EQUATIONS.
As none of these questions are difficult of solution after the equations are ob-
tained, it has been considered unnecessary in general to do more than form the
equations and put down the answers.
1. To find two numbers whose diflTerence is 2, and product 80.
Let X and y denote the numbers : then the conditions are x — y ^ 2, and
a;y = 80.
Resolving, we have a; = 10 and y = 8, or x = — 8 and y = — 10.
2. To divide the number 14 into two such parts that their product may be 48.
Let X and y be the numbers : then the conditions are a; -j- y = 14 and
xy = 48.
The answers are 6 and 8.
3. A\Tiat two numbers are those whose sum, product, and difference of their
squares are all equal.
Let X and y be the numbers : then x -{■ y ■= xy, and x 4- y = a^ — y^,
which are easy to resolve, and give a? = - ~ — , and y = — — -^^-
194 ALGEBRA.
4. There are four numbers in arithmetical proo;ression the product of whose
extremes is 22, and that of whose means is 40. What are those numbers ?
Let X be the less extreme, and y the common difference : then the numbers
are x,x-\-y,x-\-2y,x-\- Zy. Whence the conditions are x^ -\- "^ xy :=
22, and x^ -\- Z xy + 2y^ = 40,
The answers are 2, 5, 8, 11, or — 11, — 8, — 5, — 2.
5. To find four numbers in geometrical progression whose sum shall be 80,
and the sum of whose squares shall be 3280.
Here, taking u, x, y, z for the numbers, we shall have, by geometrical pro-
gression, uy =■ 3^ , xz =^ y^ \ and by the given conditions u -\- x + y -\- z
= 80, and u^ -\- x^ -\- y'^ + z' = 3280,
These equations are precisely cases of Ex, 14, p, 192, and may be resolved as
those were.
But the problem admits of being expressed by only two equations. For let x
be the first term, and u the ratio of the proportion ; then x, ux, vrx, and u^x are
the four numbers, and the conditions are expressed by
X {\ + u -{- u" + u^) — 80, and x^ (1 + u^ + ^4 _|_ „6) _ 3230.
By geometrical progression these are convertible into
x{l — u*) , x^ (1 — M«) ^^„„
— ^ = 80 and — ^ r-^ = 3280,
1 — u 1 — It-
Divide the second by the square of the first ; then
1 — mS (1 — m)2 _3280
1 — w- ' (1 — u*)" ~6400'
(1 + m") (!—«)_ 1 + u* 41
^^' (1 - u*) (1 +u) - (1 + M) (1 + M + -w2 _,_ „3) — ^' or again
By multiplication and transposition this becomes 39 — 82m — 82u' — 8214^
4- 39w* = 0, and dividing by u^ it becomes 39 (u^ + -2) — 82 (m + -) = 82,
or 39 (m + -f — 82 (m + -) = 160,
Resolving by the HindA method, considering m + - as one quantity, we get
u
, 1 10 16
« + - = ^r or — — .
M 3 13
From these we have four values of u, viz. 3, |, and — , the lat-
ter pair of which are imaginary, and therefore imply that all the four numbers
are imaginary also. The two former values of u give the four numbers 2, 6,
18, 54, and 54, 18, 6, 2.
In very nearly the same way may the problem be solved when there are five
numbers instead of four,
6, There is a number composed of four digits which are in arithmetical pro-
gression. The sum of the digits is 20. If 6174 be subtracted from the number,
the remainder will have the same digits in an inverse order ; and the product of
the extreme digits is two-thirds of the product of the intermediate ones. What
is the number ?
Let u, X, y, z be the digits in order from right to left : then lOOOu -f lOOa? +
lOy -I- z denotes the number, and lOOOz -f- lOOy -H 10a? -j- u denotes the num-
ber when the digits are inverted. But by the question we have 1000a + 100a?
SIMULTANEOUS EQUATIONS. 195
+ ICy + 2 — 6174 = lOOOz + lOOy + 10a? + « ; or, 999« + 90* — 90y -
999z = 6174 ; whence
Ul{u — 2) + 10 (a? — y) = 686 (1)
The other conditions are readily formed, and are
u + 2 = X + y, u + X + y + z = 20, and U2 = - srtf,
■which admits of solution as before.
7. To find a number such that if 7 be subtracted from its square, and this root,
be added to twice the number, the sum shall be 5.
Let X be the number; then the conditions are expressed by the equation
2x + ^x^— 7 = 5.
By transp, ^aP — 7 = b — 2x, and squaring Zx^ — 20* + 32 = 0, which,
resolved by the Indian method, gives a; = 4 and x == 2§.
By substituting these values in the equation, they are found not to fulfil the
condition, though they do fulfil the condition 2a; — a/'x- — 7 = 5. No numbers,
indeed, can be found to fulfil the given condition ; and it must be carefully borne
in mind that except the expressed condition be given free from radicals, a solution,
even a symbolical one, cannot be depended on without subsequent verification.
This circumstance has been a source of much perplexity to mathematicians,
and w%» never cleared up till Mr. Horner addressed a letter to the present Editor
of the Course, and which was published in the Philosophical Magazine for Jan.
1836. To this letter the reader is referred, as any intelligible account of the prin-
ciples of his exposition would require more space than can be allotted to it here.
After the student has solved the following questions for the particular data,
he should be required to solve them when, instead of the given numbers, literal
symbols are substituted. He should also, as he proceeds through the numerical
solutions, put down all the results, even though they be imaginary ; he should be
accustomed to seek the interpretation of those results, and especially to point
out those which are true solutions in the form proposed, and which are solutions
of some collateral problems that are involved in the same algebraical expression,
as well as to assign those which involve contradictory data, and which give
therefore merely symbolical results.
Sometimes questions are proposed, in which, though the equation has real
answers, yet the question has not, from the answers, though real, being incon-
sistent with some condition either implied or expressed, which is not taken into
account in the equation ; as when a fractional number of men, or of terras of a
progression, &c. results from the question ; or when the number of things to be
added turns out to be subtractive, or — ; and so on.
In the literal solution, the student must point out the conditions that render
a problem impossible ; that is, which give rise to imaginary roots, or are incon-
gruous with the ideas implied in the subject of the problem.
QUESTIONS FOR PRACTICE.
1 . What number being added to its square will make 42 ? Ans. 6, or — 7.
2. Find two numbers such, that the less may be to the greater as the greater
is to 12, and that the sum of their squares may be 45. Ans. 3 and 6.
3. What two numbers are those, whose difference is 2, and the difference of
their cubes 98 ? Ans. 3 and 5.
4. What two numbers are those, whose sum is 6, and the sura of their cubes
72 ? Ans. 2 and 4.
5. What two numbers are those, whose product is 20, and the difference of
their cubes 61 ? Ans. 4 and 5.
o2
196 ALGEBRA.
6. Divide the number 1 1 into two such parts, that the product of their squares
maj' he 784. Ans. 4 and 7*.
7. Divide the number 5 into two such parts, that the sum of their alternate
quotients may be 4^, that is, of the two quotients of each part divided by the
other. Ans. 1 and 4.
8. Divide 12 into two such parts, that their product may be equal to 8 times
their difference. Ans. 4 and 8.
9. Divide the number 10 into two such parts, that the square of 4 times the
less part may be 112 more than the square of twice the greater. Ans. 4 and 6.
10. Find two numbers such, that the sum of their squares may be 89, and
their sum multiplied by the greater may produce 104. Ans. 5 and 8.
11. What number is that, which being divided by the product of its two digits,
the quotient is 5J ; but when 9 is subtracted from it, there remains a number
having the same digits inverted ? Ans. 32, or — 23.
12. Dinde 20 into three parts, such that the continual product of all three
may be 270, and that the difference of the first and second may be 2 less than
the difference of the second and third. Ans. 5, 6, 9.
13. Find three numbers in arithmetical progression, such that the sum of
their squares may be 56, and -the sum arising by adding together 3 times the
first, and twice the second, and 3 times the third, may be 32. Ans. 2, 4, 6.
14. Divide the number 13 into three such parts, that their squares shall have
equal differences, and that the sum of those squares shall be 75. Ans. 1, 5, 7.
15. Find three numbers having equal differences, so that their sum shall be
12, and the sum of their fourth powers shall be 962. Ans. 3, 4, 5.
16. Find three numbers having equal differences, and such that the square of
the least added to the product of the two greater shall make 28, but the square
of the greatest added to the product of the two less shall make 44. Ans. 2, 4, 6.
17. Three merchants. A, B, C, on comparing their gains, find that among
them all they have gained 1444/; and that B's gain added to the square root of
A's made 920/ ; but if added to the square root of C's it made 912/. What were
their several gains? Ans. A 400, B 900, C 144.
18. Find three numbers in arithmetical progression, so that the sum of their
squares shall be 93 ; also if the first be multiplied by 3, the second by 4, and the
third by 5, the sum of the products may be 66.* Ans. 2, 5, 8.
19. Find two numbers, such that their product added to their sum may make
47, and their sum taken from the sum of thgr squares may leave 62.
Ans. 5 and 7-
20. (1) The sum of two numbers is 2, and their product is also 2; what are
they ? (2) Also find two numbers whose sum is a and whose product is b'-.
Ans. (1) impossible; (2) x = — = ; y = — '
2 A
21. The greatest term of an arithmetical series, the common difference, and
• In a great number of the folio-wing questions the answers will apjjear in two forms or in
four forms, and it will often happen in even more numerous ones still. The present question
11 ±3 11+3 11 ± v'iSS II + v^-233
lag for its solutions .r = — s — i^^ — o — ; and j" :^ .-, , y :^ g '
Though for the most part only the positive answers are set down to enable the student to verify
his actual work, he should be required to give all the solutions that the equation admits of, in
eymbols at least, and reduced where they admit of reduction.
(n
(2)
(3)
(4)
Greatest term ....
•25
8
•25
400
Common diflference
•02
•02
•02
4
Sum
10-25
10-25
1-25
20200
SIMULTANEOUS EQUATIONS. I97
the sum of the series, in five several cases, as below, are given, to find the number
of terms : —
(5)
400
4
201 Go
22. (1) There are two numbers whose sum, sum of their squares, and their
product, are all equal; and (2) two others whose sura, product, and difierence of
their squares, are all equal. What are these pairs of numbers ?
Ans. (1) impossibles (2) § (3 ± s/5) and ^ (.1 ± ^/5).
23. A gentleman bought a horse for a certain sum, and having re-sold it for
119/. found that he had gained as much per cent, by the transaction as the horse
cost him ; what was the prime cost of the horse ?
Ans. 701. or — 170/. The latter is incongruous.
24. The arithmetical mean of two numbers e.xceeds the geometrical mean by
13, and the geometrical exceeds the harmonical mean by 12. What are those
numbers ?
25. A traveller sets out for a certain place, and travels one mile the first day,
two miles the second, three the third, and so on : and five days afterwards
another sets out and travels 12 miles a day. Show how far he must travel before
be overtakes the first, and explain the other answer.
26. A wine merchant sold 7 dozen of sherry and 12 dozen of claret for 50/.
and finds that he sold 3 dozen more of sherry for lOl. than of claret for 6/. What
was the price of each ? Explain the double answer.
27. A parcel contained 24 coins, valued 18s, part of them silver and the
other copper. Each silver coin is worth as many pence as there are copper
coins, and each copper coin is worth as many pence as there are silver coins.
How many more were there of copper than of silver ?
28. Find four numbers which exceed one another by unity, such that their
continued product may be 120.
29. There is a number consisting of two digits, which, when divided by the
sum of its digits, gives a quotient greater by 2 than the first digit ; but if the
digits be inverted,' and the resulting number be divided by a number greater by
unity than the sura of the digits, the quotient is greater by 2 than the preceding
quotient. Find the congruous answer.
30. '• Some bees were sitting on a tree ; at one time the square root of half
their number flew away. Again, eight-ninths of the whole flew away the second
time ; two bees remained. How many were there • ?"
31. D sets out from F towards G, and travels 8 miles a day; after he had
gone 27 miles, E sets out from G towards F, and goes every day ^ of the whole
journey ; and, after he had travelled as many days as he goes miles in one day,
he met D. What is the distance of the two places ?
32. There is a number con-sisting of three digits, of which the first is to the
second as the second to the third ; the number itself is to the sum of its digits
as 124 to 7; and if 594 be added to it, the digits will be inverted. Required the
number.
* From Strachey's translation of the Bija Gatiita, the work from which the nilc for the
solution of Quadratic Equations, given at p. 197, was taken. The scientific world is indebted
for the publication of this very ciirit)us and interesting work to my lamented friend, the late
Professor Leyboum, of the Royal Military College, Sandhurst.
198 ALGEBRA.
33. Bacchus having caught Silenus asleep by the side of a full cask, seized
the opportunity of drinking, which he continued for two-thirds of the time that
Silenus would have taken to empty the whole cask. After that, Silenus awakes,
and drinks what Bacchus had left. Had they both drunk together it would
have been emptied two hours sooner, and Bacchus would have drunk only half
of what he left for Silenus. Required the time in which each would have emptied
the cask separately. Ans. Bacchus in 6, and Silenus in 3 hours.
34. A and B travelled on the same road, and at the same rate, from H to L.
At the 50th milestone L, A overtook a drove of geese, which were proceeding
at the rate of 3 miles in 2 hours, and 2 hours afterwards met a stage-waggon
which was moving at the rate of 9 miles in 4 hours. B overtook the same drove
of geese at the 45th mile-stone, and met the same stage- waggon exactly 40 minutes
before he came to the 31st milestone. Where was B when A reached L ?
Ans. at the 25th milestone.
THE SOLUTION OF CUBIC AND BIQUADRATIC
EQUATIONS.
Although Horner's general method of approximating to ihe roots of numerical
equations of all degrees supersedes the special methods adapted to particular
classes, yet there are many occasions in the higher departments of mathematical
inquiry in which it would be of great advantage to possess the symbolical values
of the roots in terms of the literal coefficients of any given equation. Beyond
those of the third and fourth degrees the labours of mathematicians have been
altogether unsuccessful in assigning these symbolical values : and, indeed, therd
are strong reasons for believing that this want of success arises from circum-
stances that render the solution of the problem altogether impossible. Were it,
however, otherwise, it is evident from a comparison of the rapid increase in the
complexity of the expressions of the roots of equations of the first, second,
third, and fourth degrees, that the roots of an equation of the fifth degree in
terms of the coefficients would be so unwieldy as to preclude the possibility of
substituting them in any other expression, and effecting a sufficient degree of
reduction in the result to be of the slightest mathematical utility.
I. cardan's solution of the cubic equation.
1. To prepare the equation, it is necessary to transform it into another whose
second term has the coefficient 0.
Let ox' -f bx^ -1- ex -|- ff = 0 be the given equation. Assume two unknown
quantities u and z, such that w 4- 2 = x. Substitute this value of x in the given
equation ; then it reduces to
az^ + (3au + b) z^ + {Zau^ + 2bu + c) z + ai^ + bu^ + cu + d = 0.
Now that the coefficient of z^ may be 0, we must have 3au -f 6 = 0, or
u := — —. Put this value for «, and reduce the expression; then we have
finally for the transformed equation
27d^2r' — (9ab^ — 27a^c) z + 2b^ — 9abc + 27a^d = 0.
When a = 1, and - an integer, the expression takes a much simpler form.
SOLUTION OF CUBIC AND BIQUADRATIC EQUATIONS. 199
When the equation is of any higher degree, as the nth, the same process will
give for the value of u which will remove the second term u^= •.
na
As an example of the cubic, let the equation x^ — 6x^ -\- lOx — 8 ^ 0 be
transformed into one wanting the second term.
Here a = 1, 6 = — G, and hence u= ^ = 2, and x = z+ 2; which
uj)on substitution in the given equation gives z^ — 2r — 4^0.
Again, for the biquadratic x* — 6x^ — 21a;- -f 146x — 120 = 0, where
5 3 3
a=l, n = 4, b =: — 6; hence u = = -, and x = z + -. Substitute
na 2 2
this, and there will result z* — 34-52^ + 56z + 36'5625 = 0.
EXAMPLES.
Remove the second terms from each of the following equations :
1. X* — 4x^ — 8x^ + 32 = 0, and from y* + 4y' + 8y^ — 32 = 0
2. z^ + 3z^ + 9r — 13 = 0, and from ar» + J ar* + 1-5 = 0
3. y* — 2-5 y* + 16-25 y^ — 18 375^^ _ 18575 = 0
4.x^-^,x + 3 = 0; a^ + ^l,x + 3 (-1)3 = 0.
5. Transform y-'+12y-'-64y-> + 15y'=0, and z^—loJ+loZ'—lSz^^Of,
into equations deficient of the second term.
2. To resolve the transformed cubic equation.
Let the transformed equation be z^ + 3ez — 2/ = 0, where e and / denote
the coefficients of the transformed equation, divided by the coefficient of z^.
Assume the two unknowns x and y to fulfil the equations x + y ^ z, and
xy = — e. Substitute these in the given equation, we have
a^ •{■ y^ = 2f; or squaring,
x^ + 2z^f + y* = 4/^ Also ^r'f = — 4e3.
Whence by subtraction, x^ — 2a^y^ -j- y* = 4(/^ + e^), and by extraction,
a:3 _ y3 _ 4. 2 ^/2 ^T^.
From these values of a?' -f- y^> and a^ — y', we have
a^ =/+ ^7^+13, and y=» =/+ vi/M^.
Hence extracting the cube roots, and putting the values of x and y in x + y
=■ z, we have
z = yf± vr~+^ + >//+ ^//■+^,
which are both contained in the single expression
z = Vf+ v7m^ + v^/— v7*T^-
And 2 + w gives the value of the root of the original or untransformed equation,
« being determined according to the process already explained.
• As a process for numerical work, it may be stated, that a much shorter one may be given
than that of actual substitution ; but as it will be detailed further on, (in Homer's method of
solution,) it will be unnecessary to do more than refer to it here.
+ In the first of these two equations, it will be necessary to reduce the equations to the form
of integer indices before the application of the method ; and in the second, to consider ji» as the
quantity according to which the arrangement is made. This last, however, naay be put in a still
different form, though not a more advantageous form.
0X) ALGEBRA.
When e is negative, and c* is greater than f^, the radical v^^ + e^ becomes
imaginary, and hence the numerical calculation of the root becomes impossible
■without some further contrivance. Many have been proposed ; as the expan-
sion of {/+ ^]' "^ \.f " ^i ^°'° series, by which the odd powers of k or
n//^ + e^ mutually cancel, and leave only terms which are real ; also by means
of trigonometrical tables, and by special tables devoted to the purpose. All
these methods are now so completely superseded by Horner's process as to need
no remark here.
It will only be requisite to state, that the case now supposed is called TAe
Irreducible Case of Cardan's Cubic ; and that it is known from other considera-
tions, that in this case all the three roots of the equation are real, whilst in that
to which Cardan's formula applies, two of the roots are imaginary, and the real
one is given by that formula.
As an example, let us take the equation x^ — Gx^ + \0x — 8 = 0, whose
second term we had eliminated by our previous transformation, giving z^ —
2
2z — 4 =: 0, Here e = and /"= 2. Hence
3 "^
z = 4/2 + s/T^. + v/2 — vT^ = V2TTv^ + V2 - '5 V3
= (1 + ^ ^/3) + (1 — ^ ^3) = 2.
Hence x == z + 2 ^ 4, which is the real value of x in the given equation.
Other Examples for Practice.
2. Find the value of x in the equations xr^ — 6xr^ -f 18x = 22, and x^ — 7x^
+ I4x = 20. Ans. X ^ 2 "32748, and x = 5 respectively.
3. Find Cardan's roots of the equations x^ -\- 6x = 20, and x^ — 12ir^ -|- 36x
= 7. Ans. 2 and 7 respectively.
4. Given x^ — ISa;^ -|- 71a; = 297; and a^ — 12j?2 -j- 57a; — 94 = 0, to find
Cardan's roots. Ans. 11 ; and 4 + \/3 — ■ V9.
5. Given y^ + 18y^ + 2l6y^ = 3392, to find Cardan's value of y. Ans. 2 *.
3 JL
6. Find Cardan's root of y'^ + •2iy'^ = '245. Ans. y = "25.
7. Resolve x^' — 36a;" = 91, and ic — 36ar« = 91.
Ans. X = 7", and x = 7' respectively.
8. Solve those of the equations given at p. 199 for transformation, that come
under the form adapted to Cardan's solution.
II. Simpson's solution of the biquadratic equation f .
Let the given equation be a;^ + 2aa^ + bx"^ + ex + d = 0 . . . . {\), and assume
• In all cases where there arc higher powers of the unknown, if they be in the ratio of 1, 2, 3,
the problem is treated as a cubic. That here referred to is one in which j/ is not directly found
by the formula, but y*, which is := 8, and then from this again 5/ = 2. The same is true of any
inferior powers, where the indices are related in the same way, as in the next question.
t This method of solution was invented by Mr. Thomas Simpson, F.R.S. Professor of
Mathematics in the Royal Military Academy from 1743 to 1761. It has often been erroneously
ascribed to Dr. Waring, Lucasian Professor of Mathematics in the Uuiversity of Cambridge,
and even called by his name.
SIMULTANEOUS EQUATIONS. 201
it equal to (aP + ax + A)^ — (Bx + Cy = 0 (2), where A, B, C are un-
known. Expand (2) and equate the coefficients of the powers of x in the result
with those of (1). This gives,
2A + a2 — B2 = 6, or B2 = 2A + 0= - 6 . . . . (3)
2aA — 2BC = c, or 2BC = 2aA — c .... (4)
A2 - C2 = rf, or C2 = A2 — d.... (5)
Also multiplying four times (3) and (5) together, and equating the product to the
square of (4) we have
8A3 — 46A2 + 4 (ac — 2rf) A = 4a''d — 4bd + c^ .... (6)
Suppose A to be found from this cubic equation ; then
from (3) we have B = + -v/2Ar^f~a2"^^ (7)
/^^ ^ 2aA — c . 2flA — c
•••• ^'^ ^ = -^B- = ± 2V2A+J^-b •••• («)
But from (2) we have x^ + ax + A = + {Bx + C), or
a^ + {a+ B)x + {A + C) = 0 (9)
Inserting (7), (8), in (9) it becomes
_ 2aA — c
«' + (« + V2A + a'-b)x + A + 2 ^2 A + a'-^ =0 • • • • C»0)
in which the doubtful sign is to be taken the same in the second and third
terms.
Note I. Whenever, by taking away the second terra of a biquadratic, after
the manner described at page 199, the fourth term also vanishes, the roots may
immediately be obtained by the solution of a quadratic only.
Note II. A biquadratic may also be solved independently of cubics, in the
following cases : —
1. When the difference between the coefficient of the third term, and the
square of half that of the second term, is equal to the coefficient of the fourth
term, divided by half that of the second. Then if p be the coefficient of the
second term, the equation will be reduced to a quadratic by dividing it by a*
± hP^-
2. When the last term is negative, and equal to the square of the coefficient
of the fourth term divided by 4 times that of the third term, minus the square
of that of the second : then to complete the square, subtract the terms of the
proposed biquadratic from (x^ + ipx)', and add the remainder to both its sides.
3. When the coefficient of the fourth term divided by that of the second
term gives for a quotient the square root of the last terra ; then, to complete
the square, add the square of half the coefficient of the second term to twice the
square root of the last term, multiply the sum by x", from the product take the
third term, and add the remainder to both sides of the biquadratic.
4. The fourth term will be made to go out by the usual operation for taking
away the second term, when the difference between the cube of half the co-
efficient of the second term and half the product of the coefficients of the second
and third terms, is equal to the coefficient of the fourth term.
EXAMPLES.
Ex. 1. Given a?'» + 12a? ^ 17 to find x by Simpson's method.
Here the reducing cubic is A' + 17A = 18, or A = I. Hence
202 ALGEBRA.
2aA — c . — 12
B = ±^-2A+a'-i= + ^2. and C = ±^^^-^—= ± ^^^
= -J- 3v'2. Consequently the two quadratics become
x^ — a?^/2 + 1 + 3^2 = 0, and a^ + a?^/2 + 1 — 3-v/2 = 0.
The roots of these are x = f \/2 + v — i — 3\/2, and
a; = — *v'2 + V— ^ + 3^/2, respectively.
EiT. 2. Let a;^ — 6a?3 — 58a?- — 114a: — 11 = 0 be given to find x.
Here the reducing cubic is A^ + 29A" + 182A — 1256 = 0, and from this
A = 4. Whence B=i 5^/3, and C = + 3-v/3.
Inserting these and reducing the quadratics, we get
ar = I + i V3 + \/l7 + |' ^3.
Ex. 3. Given z* + 2az^ — 3"aV — 38a^z + a* = 0, to find z.
Dividing all by a*, and putting - = a;, we have x* + 2x^ — 37a^ — 38x + 1
= 0.
The reducing cubic is 2A^ + 27 A.^ — 40A — 399 = 0, and the roots (ob-
tained by a method not yet explained, but they may be verified by substi-
tution) are 3-5, —3, and — 19. Either of these may be used in the quadratic
equations, and the final result will in all cases be the same ; and a; can be ob-
tained as before, and thence z =: ax.
Ex. 4. Given x* — Aax^ + Sa^a;^ — 4a^a; -f a** = 0, to find x.
Ex. 5. Solve the three following equations by the methods explained in the
notes, as well as by the general method.
(1). ar* — 25a^-|-60a;=36. Ans. 1, 2, 3, — 6.
(2). X* + 2qx^ + 35V 4. 2q^x — r*=0.
Ans. — hq± v—i q- ± Vq* + r\
(3). a^ — 9a^ + 15«^ — 2ra; H- 9 = 0.
Ans. 9 +3 V5 + \/78± 54 -v/5
4
Ex. 6. The following equations are proposed for solution : —
(1). x^ — 8x3 _ 12x2 _|_ 84aj = 63. Ans. 2 ± V7 ± \/ll + ^7.
(2). x^ 4- 36x3 _ 4ooa?2 — 3168a' + 7744 = 0.
Ans. — 9 ± v/137 ± 3a/2V17 + ^/137.
(3). a^ + 24x3 _ 1 i4a^ _ 24x + 1 = 0. Ans. + x/197 — 14, and 2 + v'S.
(4). ar* — 12x = 5. Ans. 1 + ^/2, and — 1 + 2 ^ — \..
(5). X* = 12x3 ^ 4^3^ _ 72x + 36 = 0. Ans. 1, 2, 3, 6.
THE SOLUTION OF EQUATIONS BY DOUBLE
POSITION.
The method of double position has been explained in the arithmetic, under
the form best adapted to its use there ; which is, as the student will perceive by
SOLUTION OF EQUATIONS BY DOUBLE POSITION. 203
solving the same questions algebraically, in those cases where the equation of
the question is of the first degree. In all other cases it furnishes but gradual
approximations to the true answer ; and in these, according to the circumstances
of the given equation, there will be very different degrees of rapidity in the
approaches to the three values of the unknown. In algebraical equations, the
rate of appro.\imation is generally considered to be such as to give at each opera-
tion about as many figures more as we had already obtained. In other classes
of equations*, however, the approximation is much more slow; and were it not
the only method which we have the means to employ in such cases, its great
slowness, and its being superseded in all respects in its application to algebraical
equations, would call for its total omission from this work : but as the modus
operandi is so easily perceived in the application of it to algebraic equations, and
some practice in the use of it so essential in future inquiries, it will be retained
in nearly the same form as heretofore f .
Let af + a«"-i + buT-^ + hx = k (1).
Suppose two numbers a?, and x^ are found by trial, or otherwise nearly equal
to X, and substitute them in (1) ; giving
X" + axi"-^ + bx'-^ 4- ... hx^=z k^ .... (2)
x^ + ax^~^ + bx^~^ 4- . . . Axj = A:, .... (3)
Subtract (3) from (1) and from (2), and divide the results. Then
(a?" — x^) + a{xr-^ — x^^) + b{x"-^ — x^'-'^) + .. . hQe — x^ _ k — k.^ , .
{xi" — ajj") + a{x^'-'^ — a?j— 0 + 6(a?,"-2 — a:./-^) + . . . h{x^— x^) ~ k^ — k^'
But each compound term of the numerator is divisible by x — x^, and each
one of the denominator by a?, — a:^. Whence
(*,_— /rOJ^^-j?,) _ g,"-' + a^,""" (a?a + a) + x,—^ {x^^ + gj?, + &) + . . .
'{k — k.^ (a^i — x^) x'-' + a?"-2 {x^ + a) + x"-^ (a?,^ + ax, + 6) + ^ '
Now if the second side of this equation were unity, the solution of (5) for x
would be accurate ; but as it is only an approximation, x and a;, being only
nearly equal, the value of x determined on the hypothesis oi x ■=. x^ will give a
result differing from the truth by a small quantity dependent on this inequality.
If a;, Xi, and a?j, agree io p places of figures, the numerator and denominator
will generally agree iB np — 2 places, and hence the quotient will not differ from
unity till about np — 2 places. This, with equations of a low degree, and at
the outset of the work, gives nearly the rate of approximation already spoken
of: but it is evidently much more rapid in higher equations at the outset, and
in all after a few steps in the approximation have been made. When, however,
the given equations involve radicals, it is difficult to investigate without consi-
derable detail the extent of the approximation.
Adopting then as an approximation that the right side of (5) is unity, we
have
k — k
x — x^ = j-zrr (^1 — ^2) ^^)
* Such as, for instince, lO* = 50, in which x = 169897; or r* zz 100, in which * =
3'597285. This application of the method requires, however, a knowledge of the principles of
logarithms and the use of tables, the foundation and structure of which have not yet been
explained. On this account, the solution of such equations will be reserved for a Supplement
to the use of the tables referred to.
+ This method is due to John Bernoulli, see Butler's Mathematics, vol. ii. p. 155.
204 ALGEBRA.
From this we have a closer approximation to x than either x^ or x^ were, viz.
jt k
^- ki -k, (.Xi-x,) + x, (7)
Hence, putting this formula into words, we have the following
RULE.
1. Find, by trial, two numbers, as near to the true root as you can, and sub-
stitute them separately in the equation instead of the given quantity ; and find
how much the terms collected together, according to their signs + or — , differ
from the absolute known term of the equation, marking each error + if in
excess, and — if in defect.
2. Multiply the difference of the two numbers, found or taken by trial, by
either of the errors, and divide the product by the difference of the errors,
having regard to the algebraical laws of the signs.
3. Add the quotient last found to the number belonging to that error, when
Its supposed number is too little, but subtract it when too great, and the result
will give the true root nearly.
4. Take this root and the nearest of the two former, or any other that may be
found nesu-er ; and, by proceeding in like manner as above, a root will be
obtained still nearer than before. And so on, to any degree of exactness
required -f.
Notes.
1 . It is best to employ always two assumed numbers that shall differ from
each other only by unity in the last figure on the right hand ; because then the
difference, or multiplier, is only 1. It is also best to use always the least error
in the above operation,
2. It will be convenient also to begin with a single figure at first, trying
several single figures till there be found the two nearest the truth, the one too
little, and the other too great ; and in working with them, find only one more
figure. Then substitute this corrected result in the equation, for the unknown
letter, and if the result prove too little, substitute also the number next greater
for the second supposition ; but contrariwise, if the former prove too great, then
take the next less number for the second supposition ; and in working with the
second pair of errors, continue the quotient only so far as to have the corrected
number to four places of figures. Then repeat the same process again with this
last corrected number, and the next greater or less, as the case may require,
carrying the third corrected number to eight figures ; because each new opera-
tion commonly doubles the number of true figures. And thus proceed to any
extent that may be wanted.
3. The actual labour of finding the errors which result from the suppositions
will be greatly abridged by the application of Problem I. of the Chapter on the
Solution of Equations, in those cases where all, or nearly all, the terms are pre-
sent, and the equation ordered according to descending powers of the unknown
* The rule in this form was first given by Mr. Bonnycastle, late Professor of Mathematics in
the Royal Military Ac.ideuiy, in the 8vo. edition of his Arithmetic (1810). It is a better form
in cases of approaimaiuin than the one given in the article on Arithmetic, in the present vol.
p. 93.
SOLUTION OF EQUATIONS BY DOUBLE POSITION. 2()5
quantity. To that rule, for the sake of avoiding repetition, the student is at
once referred. See p. 208.
EXAMPLES.
Ex. 1. Find the value of x in the equation ifi + x^ -\- x — 100 = 0. Here
* = 100, and it is easily discovered that the root lies between 4 and 5. Sub-
stitute these numbers for x^ and x„ and adopt the algorithm referred to in
note 3. Then,
1 + 1 + 1 — 100 (4 = X,
4 20 + 84 = Jtj
5 21 —l6 = k, — k
1 + 1 + 1 — 100 (5 = ar,
5 30 155 = kt
31
55 = *. — t
Hence x^ — Xt= I, ki — k^ = 71, k — k^= 16, and we have x =z x, +
^ — K . V . 16-1
^ _ ^-^ (a?, — iTj) = 4 + — - = 4-2 nearly.
Again, suppose a?, = 4*2 and Xi = 4-3. Then, proceeding as before, we have
1 -h 1 + 1 — 100 (4-2
4-2 20 8 91-36
104 4-568
1 + 1 + 1—100 (4-3 = X,
4-3 21-2 9516
5-3 1-59 7137
5-2 22-84— 4-072 = *2 — i 2379+ 2 297 = k^ — k
Hence a?i — a?2 = -1, *, — k.^ = 6 369, and k — k.^ = 4072. Hence, proceed-
ing as before, we have x = 4*264 nearly.
Working, thirdly, with 4-264 and4'265 in the same manner, we have x =
4-2644299 more nearly.
So long as the first three figures only appear in the approximations, and the
equation is only cubic, the table of squares and cubes at the end of this volume
will facilitate the work of finding the values of the substitutions.
Ex. 2. Find the value of a? in a;^ — 1 5a?- + 63a; = 50.
Ans. 1-0280392317 ...
Ex. 3. Let it be required to find the value of a; in the equation
A/I44a7- — (x^ + 20)2 _^ Vigex^' — (ar-' + 24)^ =114.
By a few trials it is soon found that the value of x is but Uttle above 7. Sup-
pose, therefore, first that x is = 7, and then a; = 8.
First, when ar, = 7- Second, when x, := 8.
47906 . . Vll4x* — (ar + 20)=' . . 46 476
65-384 . . V 196x2 — (ar»~+ 24)^ . . 69-283
*, = 113-290 115-759 = *,
/t= 114-000 114-000=*
k., — k = — 0-710 + 1-759 = *,
*, — *,= + 1-759
ki — *j = 2-469- Hence by the formula we find x = 7 2, nearly.
206 ALGEBRA.
Suppose again x = 7'2, and then, since this turns out too great, take 7'1 and
7"2 for the trial numbers, as follows :
First, when x^ = 7 2, Second, when a?i = 7'1
47-990 = ^/144a~' — {x' + 20)2 _ 47.973
66-402 = A/196a?2 — {x^ + 24)» = 65-904
A:,= 114-392 113-877 = *i
Ar= 114-000 114-000 = it
A:, — it = -392 — -123 z=ki — k
jt Jc "l . "123
Hence x = x^ + t r (^1 — ^2) = ''^'^ H — .,,> = 7-124 nearly.
i^x. 4. Resolve the equation .r^+l Ox- H-5x= 2 60. Ans. ar=4-ll798574108.
£ar. 5. To find the value of x in the equation x^ — 2a:=50. Ans. 3 864854.
Ex. 6. To find x in the equation afi + 23r—23x=-70. Ans. x=5- 13457.
Ex. 7. To find X in the equation x^ — l7ar+54a;=350. Ans. a:=: 14-95407.
Ex. 8. To find X in the equation a?*— 3a:2_75x=10000. Ans. ar=10 2609.
Ex. 9. Find x in 2a;*— l6x^ + 40x-— 30x= —1 . Ans. a?= 1-284724 •
Ex.10. Resolve a^+2x*+3x^+4x^-\-5x= 5432 1. Ans. a?=S 414455.
Ex.U. Given 2a?*— 7ar^ + lla;''— 3x=ll, to find x. Ans. x = 1-8375506.
Ex. 12. To find the value of x in the equation
(3^2 _ 2 -v/a; + 1)^ —(ar—ix ^/x + 3 ^xf = 56. Ans. x = 18-360877..
THE NUMERICAL SOLUTION OF ALGEBRAIC
EQUATIONS. •
The solution of algebraic equations having general or literal coefficients has
never been eflfected beyond the fourth degree ; and methods to this extent have
already been explained in this work, at pages 170, 185, 198, and 200. A con-
siderable number of the properties of the roots of algebraic equation without
limitation as to degree, when taken in connexion with the coefficients, have
been investigated ; but as these have been viewed more with a prospect of literal
than mere numerical solution, they become in reference to our present object
rather matters of curiosity than of utility. It is not proposed, then, to enter
upon the general theory of equations in this work further than it conduces to
the solution of those whose coefficients are numerically given.
As, whatever may be the number of equations simultaneously given and con-
taining the same number of unknowns, these equations can be reduced to a
single one containing only one unknown, the inquiry will in this place be
restricted to this single equation, whether originally so given or obtained from
the system of coexisting equations by elimination.
DEFINITIONS AND NOTATION.
1. An algebraic equation is any one which contains positive integer powers of
the unknown quantity. The following is the general type, n being a positive
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIOxXS. 207
integer, and A, B, C, . . . . L, M, N, any numbers whatever, either positive or
negative, or zero, but all free from the imaginary symbol.
Ax" + Bar-' + Cic"-2 + . . . . + Lx» + Ma; + N = 0.
Some classes of reasonings, however, are facilitated by having unity for the
coefficient of x", and the equation is at once reduced to this form by division of
all the terms by A.
2. For brevity of writing, this is often put in the contracted form
/ (x) = 0, or X = 0.
The symbols / (x) and X are in this case called functions of x ; meaning an
expression into the composition of which x enters, or which depends upon the
value of X.
3. A root of an equation is any number or expression which, on being substi-
tuted in the given equation, and all reductions being performed, fulfils the ex-
pressed condition of making both sides equal by the mutual cancel of all the
terms on the first side.
4. If r,, r^, r^, be roots of an algebraic equation / (j) = 0, then the
successive quotients of / (a?) by x — r^, x — r^, .... are called the depressed equa-
tions. It is seldom that these depressed equations require any special notation ;
but when they do, 0, (a?) = 0, (^^ (a;) = 0, (p-j {x) =z 0 .... are found con-
venient.
. 5. If / (x) be divided by x — x, these quotients are called the frst, second,
third, .... del ivative functions of x, or simply derivatives. They are respectively
denoted by/, (x), f^ {x), f^ {x), .... fn'\x) according to the number of successive
divisions performed.
6. An equation is said to be transformed when it is changed into another whose
roots have any assigned relation to those of the given one : as, for instance, when
an equation is given, and another is formed from it whose roots shall be triple,
one half, or any multiple or part of those of the given one ; or again when the
roots shall be all greater or less than those of the given equation by some given
quantity ; and so on. In the last-mentioned case, the new equation is called the
reduced equation.
7. By a permanence of signs, or simply a permanence, is meant that two con-
secutive terms of a complete equation have the same sign prefixed ; as -f -f or
: and by a variation, that two consecutive terms have unlike signs pre-
fixed, as -| or \-.
Theorem 1. If r, r,, r^, .... be roots of an equation, / (x) = 0, then f{x) is
exactly divisible by x — r, x — r,, x—r.^, .... without remainder.
Let /(a;) = Aa:" -|- Baf"' -|- CaT^^ 4- -|- Ma- -(- N = 0 be the given equa-
tion, and let the first side be divided by x—r by the synthetic method, p. 128.
|A-fB +C +D + +L +U +N
■!■ r I -t- Ar -I- B,r + C,r -I- + K,r + h,r + M,r
i A + B, + C, -h D, + -h L, + M, -h N,
where A, B,, C,, .... L,, M,, N, are the coefficients of the quotient, as far as
the term ar' ; and we have to show that N, = 0, and hence that the quotient
terminates at M,.
By attending to the formation of the coefficients of the quotient, we see that
B, = Ar + B,
C, = B,r + C = Ar2 + Br + C,
D, = C,r -h D = Ar' + Br2 + Cr + D,
208 ALGEBRA.
M, = L,r 4- M = Ar-^i + Br"-2 + Cf-^ + . • + Lr + M,
Ni = M.r + N = Ar" + Br"-i + Cr"-^ + . . + Lr^ + Mr + N.
But by hypothesis r is a root of the equation f(x) = 0 ; hence substituting
it for X we have
Ar" + Br"-i + Cr"-" + .... + Lr^ + Mr + N = 0,
and as this is exactly the value of N, found above we have Nj =■ 0, and the
division terminates.
In the same manner it may be shown that f (a?) is divisible by x — r^, x — r^,. .
Cor. Since /(«) is divisible separately by x — r, a? — r„ x — r^, . . it is also divisi-
ble by their product.
Theorem II. The derivatives may be formed by inspection in the following
manner :
Multiply each term by the index of the power of x in it, and diminish that
index by unity, which will give the first derivative : operate upon the first de-
rivative in the same manner to produce the second ; upon the second to produce
the third ; and so on to the end.
For the division by x—x gives the same coefficients in terms of x that the divi-
sion by X — r gives in terms of r ; hence restoring x for r in the values of B,, C„. .
Lj, M,, we have
Aa;°~^ + Aaf"* -f Aa;"~^ + (to ra terms) = raAa?"-^
+ Ba?"-2 + Ba^2 ^ (jq „_i terms) = («— 1) Baf-^
+ Caf-3 + (to w— 2 terms) = (n— 2) Ba;"-^
Hence, l^= raAa;"-' + («— 1) Ba;'-^ + + 3Ka?2 + 21^ + M =/, {x)
The second derivative f^ (a?) will obviously be found from this in the same
manner, since the same reasoning applies to all the derivatives in succession.
Hence the several derivatives are
/, (a;) = wAa^i + (n— 1) ^^"-^ + in— 2) Cx'-^ + . . . .
/j {X) = n (w— 1) Aa;"-2 + (n— 1) (re— 2) Ba?"-^ + (re— 2) (re— 3) Ca?"-<+ ....
/3 {x) — n (n— 1) (n— 2) Aa^-^ + (re— 1) (n— 2) (re— 3) Ba;""* + . . . .
/„_2 {x) = n (w— 1) 3Aa;2 + (re- 1) (w— 2) . . . . 2Ba; + (n— 2) (n— 3) 2.1
/„_, {x) = re («— 1) 2Aa; + (re— 1) (re— 2) 1 .B
/.(a;) = w(ra— 1) 2.1, A
PROBLEM I.
To calculate the value of an algebraical function when the value of x is given ;
that is, to find the value of kx' + Ba:"-' +.... + Ma; + N, when a; is a given
number, the coefficients being also given numericEdly.
Range the coefficients in a horizontal line, as for synthetic division, with their
proper signs prefixed, taking care when any coefficients are absent to fill their
places with ciphers. Multiply A by the given value of x, and add the result to
B, making the sum B, ; multiply B, by a and add the result to C giving C„ and
so on till we arrive at N,. Then N, is the value of the expression.
For this is precisely the operation performed in theorem 1 ; and by the reason-
ing of that theorem
N, = Aa" + Ba"-> +.... + La^ -|- Ma -|- N.
Thus if the expression or function were 4a;^ — 5a;'' + 6a;^ -\- 4a;^ — 10a; +15,
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. oqq
and we were required to find its value when ar = 5 and when j; = — 2, then the
work would stand thus :
4+0— 5+ 6+ 4— 10+ 15
20 + 100 + 475 + 2405 + 12045 + 60175
L^
20 + 95 + 481 + 2409 + 12035 + 60190 = value when x = 5.
4 + 0— 5+ 6+ 4 — 10+ 15 I— 2
— 8 + 16 — 22 + 32 — 72 + 164
— 8 + U — 16 + 36 — 82 + 179 = value when x = — 2.
EXAMPLES FOR PRACTICE.
6a; + 10, when x has severally
- 10a:3 + 30x2 + 53^ _ 120,
1 . Find the values of the function x^ -\- 4x^ —
the values 1, 10, 19, and — 100.
2. Find the values of the function x^ — Ix*
when X has the values 1,2, — 1, — 2, and —3.
3. Find the values of x* — 25^2 + 60a; — 36 when x is 3, 2, 1, and — 6.
When x = ± {i V2 ± \/ + 3 v'2 — ^, what is x* + 12a; — 17 equal to?
What is the value of x^ — 7x^ + 14a; — 20, when a; is 5, or 1 + v/-^ ?
PROBLEM II.
To transform an algebraic equation, having its roots less by a given quantity a
than the roots of the given equation.
Divide the given function continually, employing the 8)mthetic method, by
a? — a, always stopping at the term where (x — a)~' occurs : the several terminal
quotients will be the coefficients of the reduced equation. The following is its
general type. *
Let Aa;" + Bx"^ + Ca?"-^ + . . . I^ + Ma; + N = 0 be the given equation :
then, dividing synthetically, we have the several operations as follow :
A + B +C +D +.. + K +L +M +N
+ Aa + B,a + C,a + . . 4- H,a + K,a + L,a + M,a
A + B. + C. + D. +
+ Aa + B^a + C,a +
+ K, + L, + M, + N,
+ Um + K,a + Lm
A + B, + C, + D, +
+ Aa + B,a +C>-|-
+ K, + L, + M,
+ H,a + K^a
A + B3 + C3 + D3 + . . + K, + L3
A + B,_, + C._, + D,_,
Aa + B,_,a
A + B,_, + C_.
Aa
A + B„
and the transformed equation is
A (a?— ar+B. (a;— a)"-i + C^x (x— 0)-'+. ,
. . Lj (x-a)'+M, (x-o)+N,=0.
* It will be more convenient to place the transforming number to the right of the coefficients,
separated by «he curve employed in division or extraction of roots, as in the numerical illustni'
tions which follow.
VOL. I. P
210
ALGEBRA.
For the results of the successive divisions performed as above, are
kx'-^ + Bjc^-^ + Cyx'-^ + . . . . + Lia; + M, + -^ = 0.
Ai"-= + B^ar^^ ^ c^-^ + . . . . + L, +
M,
+
X — a
X — a (x — a)-
= 0.
Ax-^ + B^'-* + C^x'-^ + .... + i + ^^
+
N,
X — a {x — a)- {x — a)
= 0.
A +
B.
+
+
+
^ {x-a)'-' ^ (.x-aT
L,Cx—aT-+M, (j:-fl) + N,=0,
-a ' {x — a)^ ' {x — a)"'
and multiplying at once by (x — a)', it becomes
A ix—ay+B. (a?— a)"-i + C._, (x—a)'--+.. .
which establishes the truth of the rule.*
Cor. 1. If we diminish the roots of an equation by a quantity a, which is greater
than p of the positive roots a, a^, a^, ... flp_i, then in the reduced equation, p of
the positive roots corresponding to these will be negative, viz. a — Op, a, — a^,
.... Op-i — ffp, all of which are negative, since a, is the greatest of all the quan-
tities. In a similar manner, if the roots of an equation are increased by a quan-
tity a, which is greater than a, a^, ... ap_„ ( — a, — a„
a^i, being p
negative roots,) then the p roots of the reduced equation corresponding to these
will become positive. Of course the transformed roots corresponding to those
which were negative before the diminution, or positive before the increase of the
roots, retain still the same character as their original corresponding roots.
Cor. 2. Had we been required to form an equation whose roots are greater than
those of a given equation, the process would obviously have been the same, only
dividing continually by x -|- a instead of i — a ; that is, by using the factor
— a instead of -f a used above in forming the several successive courses of co-
efficients.
As examples, let 2x* — lOa^ + 20x- — 15a? -}- 10 =: 0 be transformed into an
equation whose roots are less by 4, and the equation 2y^ -f- 22 j/^ -\- Q2y- -j- 177y
+ 142 = 0 into one whose roots shall be greater by 4. The work will stand
thus : —
2 — 10 + 20 —
8—8-1-
15 -f- 10
48 -(- 132
l^
- 2 -1- 12 4-
8 + 24 -H
33
144
+
142
6 -f 36 + 177
8-1-56
14 -f 92
8
22
Hence the transformed equation is
2 (a?— 4)^ -I- 22 {x—Af + 92 (a'— 4)^ +
177 (ar— 4) + 142 = 0.
2 -f- 22 -(- 92 -f- 177 + 142 \ — 4
— 8 — 56 — 144 — 132 ^—
14 -1- 36 +
— 8 — 24 —
33 + 10
48
6 -f- 12 —
— 8 -f 8
]5
2 -I- 20
— 10
Hence the transformed equation is
2 (y + 4)^ - 10 {y + 4f -\- 20 (y-|-4)«
15 (y-l-4) + 10 = 0.
• Otlicr methods requiring rather less woik but involving principles rather less comprehensi-
ble by tlie student, (and given, too, without investigation,) may be seen in Leyboum's Reposi-
tory, vol. V. pp. 42 — 44. The demonstrations of them will appear in my forthcoming publication
of Mr. Homer's works on Equations. — Editoe.
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 21 1
It will at once appear that the second given equation is hut a transformation
of the first, and rice versa. The restoration of the original result proves the
truth of the transformation and re-transformation.
When the transformation is to be made by a number comprising more than
one figure, it may first be transformed by the first of them (regard being had to
its place in the decimal scale), then this transformed equation again transformed
by the next figure, then again this by the next figure, to any assigned extent, the
same precaution respecting decimal place being observed in all. Thus, to reduce
the roots of the equation a?' + Sa;^ -j- 3x — 140 = 0 by 4-23, we shall have
1 +
3 + 3 — 140
4 28 124
v^_
7 + 31 — 16
4 + 44
11 + 75
4
15
1 + 15+75 — 16 \-2
•2 304 15-608
15-2 78-04 —
-2 3-08
81-12
•392
1 + 15-6 + 81-12
•03 -4689
15-4
•2
15-6
•392 \-03
2-447667
15-63
03
81-5889
4698
2055667
15-66
03
82-0587
15-69
And the transformed equation is
(a? — 4-23)' + 15-69 {x — 4-23)- + 820587 (a? — 4^23) + 2-055667 = 0 *.
But in many cases, especially where the process does involve much intermix-
ture of the signs + and — , the whole work may be more advantageously per-
formed at once according to the following method :
1+3 +3 — 140 (4-23
4-23 23-92 134-3316
1-446 6-71658
2169 1-007487
1 + 7-23
4-23
33-5829
45-84
2-292
•3438
1 + 11-46 82-0587
4-23
2-055667
15-69
* Though the decimal points are marked in this process, they will, after a little practice, be
easily dispensed with by the pupil, as the regular arrangement and increase of the places to the
right will always secure the figures falling rightly.
p2
212 ALGEBRA.
EXERCISES ON THE REDUCTION OF EQUATIONS,
1. Transform the equation x' + 4ar + 2a? — 2328 = 0, into one whose roots
shall be less by 10; and this into another whose roots shall be still less by 2.
Then transform the result into one whose roots shall be greater by 12.
2. Reduce the roots oi x^ + 8x — 346485S4 = 0* successively by 300, 20,
and 6.
3. Reduce the roots of the equation a^ — 18609625 = 0, successively by the
numbers 200, 60, and 5 ; and then verify the process by increasing those of the
result by 60, 200, and 5.
PROBLEM III.
To transform a given equation into another the roots of which shall be the same
as those of the given one, but having all their signs reversed.
Change the signs of the alternate terms, beginning with either the first or
second ; then this will be the equation required.
For first, let the degree of the equation be even, as
Ax2. _f. B^:^i _,_ 02-2 + . . . . Kj2 + Lr + M = 0,
then writing — x instead o( x, we shall have
A(— x)2'+B (— a;)2»-i + C(— a?)2'-2+.... +K (— a^)HL (— a;)+M=0; or,
Ax^' - Ba;2"-> + Ca^-^ — . . . . + Ka^ - Lx + M = 0.
And, secondly, let it be of an odd degree, as
Ax2"+i + Bx=" + Cx2"-i + Dx-"-" + ,... Kx^ + Lx^ + yix + 2^ =0,
in which, writing — x for x we get
A(-a;)2"+i + B(-x)-"+C(-x)='-i + . . . . +Ki-xf+L(i—xy-+U(i-x)+y=zO;
or,
— Aa:2'+i + Ba^" — O^-i + . . . . -^ Kx^ _j_ l^s _ ;Mj, ^ ]Sr = o ; or,
Aa:2"+i _ Ba^. _j. Cx2-» — + Ka^ — La^ + Ma: — N = 0.
Thus if the roots of the equation x* + Ox^ — 25a:" + 60x — 36 := 0 be 3, 2,
1, — 6 ; then those of x* + Oa;^ — 2:)X^ — COa: — 36 = 0 are — 3, — 2, — 1,
and + 6.
EXERCISES.
Change signs of the roots of the equations given in problems I. and II. the
functions in problem I. being equated to 0.
PROBLEM IV.
To transform an eqyation into another whose roots shall be the reciprocals of
the roots of the given one.
Reverse the order of the co-efficients : these will be the co-efficients of the
new equation sought.
For, if in Ax" -f Bx"^ + + hiP -f Mx -f- N = 0, Ve write y =
, or X = -, then substituting we get
• When any terms are deficient, their places must be kept and filled with 0; that being in
euch case the value of the coeflBcient of that term, as in the SyntAetic Division.
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 213
A . B , C , , L , M . ^,
-, + -i^l + -^^=2 + . .. + -,+— + N = 0, or
A + By + Cy' + + Ly-2 + My-> + Ny" = 0 • ;
EXERCISES.
Find the equations whose roots are the reciprocals of those of the equations
given in problems I. and II. equating the functions in the former to 0.
Scholium.
Any other proposed relations between the roots of a given equation and those
of another to be determined, may be effected in an analogous manner, viz. : by
first expressing the assigned relation between x and y, resolving for x in terms
of y, and substituting for x its value in the given equation. After the simpli-
fication of the expression to the utmost degree, we shall obtain the equation
sought.
For instance, to form an equation whose roots shall be m times those of the
given equation, put y = mx, or if = - : then the equation
* When the coefficients of the direct and retiprocil equations are alike, that is, N = A,
M = B, L = C, ... it IS evident, that upon knowing the value of half the roots, the other half
will he known from their being tlie reciprocals of these ; or, iriptlicr words, if Tj, r , ... be roots,
then also will — — ... be roots also. This circumsttince, as it lessens the work of solving
an equation of such "form, is important to be remarked.
1. Let the equation be of an even degree, having the jibove relation; then it may be written
Aa^' + Bo-z--! -f Ci2'-» -4- Cj* + Bj + A = 0, or again,
A {x^- + 1 } + B {x2»-' + 3] + C Ja*-2 -I- j^2^ + ... = 0 ; and dividing by f ,
a{.-+~}+b|.-+J^,}+c|^-«+^,}+...=0 (I)
1 r 1 ) C 1 > («-l)(n-2) f _. , IT
and so on.
By these successive reductions we can convert (1) into the foi-m
Au'+ BV-' + C'tt"-2+ =0, (where « = a: + -j (2)
Then resolving (2) we find n values of «, and each of these substituted in * + - = u gives two
recipr0c.1l values ofx, nml thereby furnish the 2n roots of tlie equation.
The solution of Ex. 5, p. 194, is an exemplification of tiiis circumstance.
2. Let the equation be of an odd degree, as the (2« + l)th: then, since it is the same
thing as
A^^.. + . + 1] + B{.v^-i + Ifr + C{a^-3+ ]]x^ + = 0,
in which, as ail the powers of r within the brackets are odd, every bracketted term is divisible
by X + 1 Hence j- -f 1 = 0 is one of the component factors of the equation ; and the given
equation being divided bv x + 1, gives a depressed equation of the 2«th degree ; and hence also
this is soluble bv means of an equation of tl>e «tli degree, an.l a quadratic as in the last case. It
would be just the same if the latter half of coefficients were written - instead of +, since it
would onlv change the sign between e.ich two bracketted terms, and every term would then be
divisible bv a; _ 1 = 0, and j' = 1 would be the corresponding single root.
Tlie problem suggested at p. 194 may be taken as an illustration. Equations m which this
relation exists are called reciprocal recurrents.
214 ALGEBRA.
ksf + Bar*-i ^ + Lar + Ma; + N = 0, becomes
A,- B|^ Cjqa 3^. My _ ^^
Ay' + mBy'~^ + m.-Cy'"^ + . . . . n»"~- . hy" + »n"~^ My + »»" • N = 0.
Or again, to form an equation M'hose roots shall be — th of those of the given
equation, we have y = ~, or x ■= my, and the transformation leads to
Am"y" + Bm"~' y'-^ + hm^y' + Mmy + N = 0.
These transformations are often useful in the solution of equations, and the
arrangement of the work for effecting them is sufficiently obvious, without any
examples.
PROBLEM V.
TO FORM THE EQUATION WHOSE ROOTS ARE ANY GIVEN' NUMBERS.
1 . Change the signs of all the given roots.
2. Put down 1, having annexed to it by its proper sign one of these changed
roots ; and multiply this binomial by another of the changed roots, beginning
the products one place to the right, and add up the columns into one single
horizontal row. These will be the co-efficients with their proper signs of the
quadratic equation whose roots are those two roots already used.
3. Multiply this horizontal row by another of the changed roots, placing as
before the first product in the second column. The several sums will be the
co-efficients with their proper signs of a cubic equation, whose roots are the
three roots already used.
4. Proceed thus through all the roots : then the equation will be completed.
For this is evidently only working by detached co-efficients, and availing our-
selves of the contrivance of allowing the multiplicand to stand as the product of
the multiplicand by 1, and also of omitting the actual exhibition of the multiplier
beneath the multipHcand. The second term of the multiplier may be put in the
margin, as in the example annexed to the rule. This will appear quite clearly
upon working out one example at length.
For this is the only application of the method of detached coefficients to the
multiplication of x — a = 0, x — b =.0, together : and as (x—a) {x—b)
= 0, is fulfilled hy X = a, X = b, ... ., therefore a, b, .. . . are roots of the
equation, by the definition of a root.
Thus, suppose it were required to form the equation whose roots are — 1,1,
3, 4, and — 5, the process would be as follows : —
1 + 1 (^ 1
— 1 — 1
1
— 0 —
-3 +
1
0 +
3
3
4
3
1
— 3 —
-4 +
1 +
12 +
— 12
1
-7 + 11 +
5 — 35 +
7
55
- 12 [4
+ 35 — 6C
1—2 — 24 + 62 + 23—60
Hence the equation is x* — '2x* — 24^^ + 62** + 23x — 60 = 0.
NUMERICAL SOLUTIOxN OF ALGEBRAIC EQUATIONS. 215
EXAMPLES.
1. Form the equation whose roots are 1, 5, — 4, — 3, — 1, and compare it
with the example worked above by means of Problem III.
2. Form the equation whose roots are the reciprocals of each of these, viz. of
T, J, — i' — 3» aofl — T ; and also of — j, — |, \, I, and \, by means of Pro-
blem IV,
3. Form the equation whose roots are severally the first nine digits, taken
alternately + and — , viz. + 1, — 2, + 3, , . . . ; and then form that whose
roots are the reciprocals of these.
4. The three roots of a cubic equation are in arithmetical progression, they are
all integers, and their sum is equal to their continued product. What is that
cubic equation ?
5. If any number of pairs of roots be of the form a -f bs/ — 1 and a — b^ — 1,
Oi + b, a/ — T, and a, — b, v^— 1, show that the co-efficients of the equation will
be real ; or, in other words, that the imaginary symbols will disappear from the
result.
Theorem III. If an equation given in terms of a; be transformed into one in
terms of a* — a, then the several coefficients will be
/.(«) /.-I («) /.-, (a) /»(«) /,(«) f(.
1.2...n' 1.2..(n— 1/ 1 . 2 . . (n— 2)' 1.2' 1 ''^^''
wherein a takes the place of x in the given equation and its successive n deriva-
tives.
This is established at once by completing the transformation in terms of a,
and resolving the numerical coefficients into factors in terms of n. The only
difficulty in giving the successive steps of the work in this place, is the extent of
page which it would require ; but as this does not hold in the student's practice,
he should exercise himself in the actual reduction of four or five of these coeffi-
cients *.
• This theorem admits of an elegant demonstration, by means of the binomial theorem, a*
follows :
Put X = a -\- z: then expanding the function, A.r" + Rr"-' -|- ..., so that like power* of z
stand in the same vertical columns, we have
A « , «Aa-i . »(» — ]) Aa-» , , n{n -\){n-2) Aa-' , ,
a + z)' = Aa« H J— . z H ^-^ . z» + iT^Ts • «" "t" •
a + .)- .:Ca- + (^2) C.^^ . ,^(»-^) (^^W^. ,,^ (n^2)in^(n^ 4)Ca^s _ ^, ^
and so on to the end.
In this we see at once that by adding vertically, and attending to the forms taken by the
derivatives, the several coefficients of r arc as stated in tlic text; and that we have
/(a-t.)-/(a)-\- J .--t-j.2---h + 1.2..(«— 1) ^ 1.2... » •
If we restore the value of ~, this becomes
or, again, reversing the order of the terms, it it
/w=/(«)+^^— «)+4^(— a)' + + rTr7. (—)"•'
216 ALGEBRA.
Theorem IV. If the equation/(a?) = 0 have^ equal roots, then/, (x) = 0
will have p — 1 of them, fiQe) = 0 will have p — 2 of them, and so on ; till
fp_i{x) = 0 will have p — {p — 1) = 1 of them, and y), (a?) = 0 will have
p — p = 0 of them.
For, since /(a?) = 0 contains p roots each equal to r, the function f{x) is
divisible p times successively by a? — r without remainder (theor. 1). Hence
the/j last coefficients, viz. N„ Ma, L3 in the operation of Problem II. will
be zero. But these coefficients are respectively
MrJ f^ 400 /.-I (r) .
1 ' 1.2 ' 1.2.3' 1.2 ...{p— I)'
and as the denominators are all finite, it follows that the sfeveral numerators are
equal to 0, That is,/, (r) = 0, /^ (r) = 0, / (r) = 0 ; and therefore the
value of r, which fulfils the equation f(x) = 0, fulfils also the p — 1 equations
f,ix) = 0,f,ix) = 0, ....f^,ix) = 0*.
•^("^ = 1^ (^-^^'' + 1.{":(1-1-) (-^-^)"~' + -• I?! ("-")' +^ ("-^^ +^(")'
which is the same result as above stated.
One important use of this problem in tlie older methods of solution of equations, vras to
enable us to remove .any specified term from the equation by rendering its coefficient equ.il to
zero. It only required us to solve an equation of an inferior degree, viz. of the first degree to
remove the second term, of the second degree to remove the third term, and so on.
For evidently, if we find such a value of a in these several coefficients as would render them
respectively zero, our object would be accomplished. That is, to solve the equation /p(a) =: 0,
which is of the {n — •jo)"' degree, for the unknown quantity a.
Thus,/„_i{a)=(KAa+B)(w— 1)(»-2) 2.1=0, ornAa+B=0; ora=:-^.
/,_j(a)=[n(«-l)Aa2+(«-l)Ba + C}(«-2) 2,1:^0; or,
w(«-l)Aa2+(n-l)Ba4-C=0.
Hence there are two v.ilues of a (either real or imaginary), which will remove the third term.
In the same way by solving/_3(a)=:0,/,_4(a)=0, and so on, we may remove any term wliat-
ever. It may be remarked, that to remove the last term we must solve /(a)zzO; or the given
equation.
This use is, however, superseded by improved methods of solution ; but it is still applicable
to many others of great importance, some of which will be made apparent in this work.
* The following method may perhaps be found more intelligible to some students than that
in the text.
Since f{cc) = 0 contains p roots equal to r, the function /(.r) is divisible by a? — r suc-
cessively/» times ; and as r is a root of the depressed equations till the last, w-e may write a;
for it in each of them successively. But this transforms the several depressed equations into the
several derivatives ; and thus we have
Aa- + Ba-»-' -(- C*"-^ + + 1^2 + Ma; -|- N = 0 = /(x)
TzAa--' -f- (n _ 1 ) Ba--2 + (n — 2) Ca— 3 + ....+ 2La- + M = 0 = /,(x)
«(n — 1) Aa'-2-|-(n — 1)(» — 2) Ba;"-3 + + 2L z=0 = /^{jr)
n{n—\) {n—p + 1) Ax'-p -\- =0 = /,{r)
Now each depression by a; — r removes one of these roots, it will follow that
/(a?) =: 0 hasp roots equal to r
/(a-) =z 0 has {p — 1) roots equal to r
/J^x) ■=. 0 has (p — 2) roots equal to r;
yj^i (a) := 0 has Jp — ( p — 1 )] or one root equal to r.
/p{,v) = 0 has (p — p) or 0 root equal to r.
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 217
Cor. Should there be other equal roots, as p^ roots each equal to r„ p, roots
each equal to r,, and so on : then the derivatives will contain p, — 1, /), — 2, . .
and Pj — 1, j3j — 2, . . . roots equal to r, and r, respectively.
PROBLEM vr.
To ascertain whether a given equation has equal roots, and if so to find
them.
Form the equation /, (x) = 0, and perform the operation of finding the
greatest common measure of this and the given function. If this process leaves
no final common measure in terms of x, there are no equal roots. If, on the
contrary, there should be such a final divisor, it will always be of the fo#m
(a: — ry. (x — r,)'', . . . . ; and there will be /) + 1 roots equal to r, /», -f- 1 equal
to r^, and so on.
For, if the given equation/(x) = 0 contain p roots equal to r, /?, roots equal to
r,, and so on : then /, {x) = 0 will contain p — 1 roots equal to r, pi — 1 roots
equal to r,, and so on (Theor. 4). Whence {x — r) '^^ . (a: — r,) 'i~' .... will be
a common measure of f{x) := 0 and /, {x) ■=■ 0. Whence is deduced the above
rule.
As an example, let f{x) ^= ifi -{■ Zx^ — Qx* — 6a;^ + 9x" 4- 3x — 4 ^ 0 be
the given equation. Find whether it has equal roots.
Here/(a-) = a;« + 3a^ — &x* — 6^^ + 9a?^ + 3x — 4
/,(x) = 6j;^ + Ibx* — 24x3 _ igjfi ^ isa, _|. 3,
Of these two functions we find, as at p. 135, that the greatest common mea-
sure is x' — x^ — x + 1 ; which resolved into factors is {x — 1)* {x -j- 1), and
hence /(x) contains the factors (x — 1)^ and (a? + 1)*, or three roots equal to
-H 1, and two roots equal to — 1.
EXAMPLES FOR PRACTICE.
Find the equal roots of the following equations, if such exist :
1. x^ — 4x^ + 5x — 2 = 0.
2. x^ — 3a-x — 2a' = 0.
3. x' + 5x^ + 6a?* — 6x* — -ISa:' — 3x» + 8* + 4 = 0.
4. x' — x*^ + 4x^ + 4x* + 5x^ — St' — 2x + 2 = 0.
Theorem V. If an equation whose coefficients are not imaginary have one
root of the form a + 6 \/ — 1, it will have another of the form a — b \/ — 1.
For we have seen, (Theor. 3,) that if we put a + z for x, we shall get
And this may be written, putting 6 >/ — 1 for z, as follows :
^^ 1-2 ^ 1-2-3-4 j \ 1 1-2-3 ^1-2. .5 )
Now, as the first member can never be equal to the second, except the brack-
etted coefficients themselves fulfil the condition, and as by hypothesis a+b^—l
is a root of the equation, this equality must be fulfilled as a consequence of that
hypothesis, we have simultaneously
218 ALGEBRA.
-'^ ■' 1-2 ^ 1-2-3-4
1 1- 2-3 ^ 1-2. .5
which is altogether independent of the si'jn of b ^ — 1. Hence if one sign
-\- b \/ — 1 fulfil the condition of the equation, the other sign — b \/ ^^^Iwill
also fulfil it : that is, if a + ^ a^ — 1 be a root of the equation, a — b^J — ] is
also a root.
Cor. 1. Roots of such forms, (generally termed cov jugate roots,) if they enter
into an equation at all, enter it in pairs, and their number is always even.
Cor. 2. If an equation be of an odd degree, there is at least one root free from
the imaginary sign.
Scholium.
The same kind of reasoning will prove, that if one root be of the form
a 4- V'i, there will be another of the form a — ^yb. For the same reason will
exist for the separate bracketted coefficients being zero in this case as in the
other.
Theorem VI. Change the signs of all the roots r„ r,, r^, .... r^ of an
equatioa of the form
af + Asf-^ + Bx"-» + -f- Mx + X = 0,
and combine the roots so changed by way of multiplication, in twos, threes,
fours, and so on : then the sum of these changed roots will be equal to A ; the
sum of their products in twos will be equal to B ; the sum of their products in
threes will be equal to C ; and so on, till the coefficient of the nth term is the
sum of all the products of the roots (n — 1) at a time, and their continued pro-
duct will be equal to N *.
For, take two roots, r, and r, : then x — r, = 0, and x — r^ = 0, which mul-
tiplied together, give
x^ — (r, + r.,)x + r, r^ = 0,
in which the theorem holds good.
Next take three, and we get
"^ — (J■^ + r^ + r^) x^ + (r, r, + r,.r, + r,, rj x — r, r, fj = 0,
in which, again, the statement is true.
Proceeding thus to any extent, and observing the formation of the coefficients,
we see that the theorem is generally true.
Cor. 1. If any pair of roots had been conjugates, whether irrational or
imaginary, we see that these several coefficients would have become rational or
real, provided A, B. C, were so : for since {{a -^ b \^ — 1) -|- (a — 6 ^ — 1)}
and [a + b ^ — 1^ {a — b ^ — l} are both real, the imaginary parts which
would have come into the values of A, B, C . . . disappear from the result.
• Some ■writers have considered it necessary to complete tins proof, to show that, generally, if
it be true for the />th coefficient, it ^rill also be true for the {p-{-^ )th. This is easily done by
assuming the first, and proceeding by actual multiplication (literal) to the next; and the only
reason for omitting it here is the space which the printing would occupy.
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS, ojg
Cor. 2. It may hence be inferred, that all the roots of an algebraic equation
may be represented either by real quantities, r, r^ , or imaginary ones,
a + b \/ — I, a, + b^ x/ — ], . . .
Theorem VII. Every algebraic equation contains as many roots, either real
or imaginary, as it has dimensions, and no more.
For by Theorem VI. Cor. 2, every root may be represented by r , r , .. .,
and a ±b ^ — I, a,±b, ^ — 1; hence so many factors of the first degree
may be formed of it as there are units in the degree : and such equation admits
of no other binomial factors but these, as then one of the binomial factors
would be divisible by some other binomial factor, which is obviously absurd.
Theorem VIII. No equation can have a greater number of positive roots
than there are changes of sign from + to — , and from — to +, in the terms of
its first member ; nor can it have a greater number of negative roots than of
permanences, or successive repetitions of the same sign.
To demonstrate this proposition, it will be necessary merely to show, that, if
any polynomial, whatever be the signs of its terms, be multiplied by a factor
X — a, corresponding to a positive root, the resulting polynomial will present at
least one more change of sign than the original ; and that if it be multiplied by
X + a corresponding to a negative root, the result will exhibit at least one more
permanence of sign than the original.
Suppose the signs of the proposed polynomial to succeed each other in any
given order, as, for instance,
+ + — + + + +;
then the multiplication of the polynomial, hy x — a, will give rise to two rows
of terms, which, added vertically, furnish the product. The first row will,
obviously, present the same lines of signs as the original ; and the second,
arising from the multiplication by the negative term — a, will present the same
lines of signs as we should get by changing every one of the signs of the first
row. In fact, the two rows of signs would be
+ + - + + + +
— ++ — + + + -
and signs of prod. -\ +H ^■±± — +H
We have written the ambiguous sign + in the product when the addition of
unlike signs in the partial products occurs, and it is very plain that these ambi-
guities will, in this and in every other arrangement, be just as numerous as
permanences in the proposed ; thus, in the present arrangement, the proposed
furnishes four permanences, viz. , + +, -f -|-, ; and there are,
accordingly, in the product four ambiguities, the other signs remaining the same
as in the proposed, with the exception of the final sign, which is superadded,
and which is always contrary to the final sign in the proposed. It is an easy
matter, therefore, when the signs of the terms of any polynomial are given to
write down immediately the signs in the product of that polynomial, by x — o,
as far, at least, as these signs are determinable, without knowing the values of
the quantities employed ; for we shall merely have to change every permanency
in the proposed into a sign of ambiguity, and to superadd the final sign changed.
For instance, if the proposed arrangement were
+ -++ + -+ + +,
the signs of the product would be
+ - + + -. + ± + - + ±±-.
ooQ ALGEBRA.
Again, if the signs of the proposed equation were in order
+ ++- + - + ,
the signs of the product would be in the order
+ ±±- + - + -±±+-
As, therefore, in passing from the multiplicand to the product, it is the per-
manences only of the former can suffer any change, it is impossible that the
variations can ever be diminished, however they may be increased ; conse-
quently, the most unfavourable supposition for our purpose is, that the perma-
nences (omitting the superadded sign) remain the same in number ; and, in this
case, if the proposed terminate with a variation, the superadded sign in the pro-
duct will introduce another variation ; but if it terminate with a permanency,
then the corresponding ambiguity in the result will, obviously, substitute for it
what sign we will, form a variation, either with the preceding, or with the super-
added sign. It follows, therefore, that no equation can have a greater number
of positive roots than variations of signs.
To demonstrate the second part of the proposition, it will sufEce to remark,
that, if we change all the signs in an equation, we change the roots from posi-
tive to negative, and rice versa (Theor. IV.) The equation thus changed would
have its permanences replaced by variations, and its variations by permanences ;
and since by the foregoing the changed equation cannot have a greater number
of positive roots than variations, the proposed cannot have a greater number of
negative roots than permanences.
This proposition constitutes the rule of Harriot*, and serves to point out
limits which the number of the positive and negative roots of an equation can
never exceed. It does not, however, furnish us with the means of ascertaining
how many real roots, of either kind, any proposed equation may involve ; nor,
indeed, does it enable us to affirm that even one positive or negative root actually
exists in any equation ; it merely shows, that j/real roots exist, those which are
positive, or those which are negative, cannot exceed a certain number ; they
may, however, fall greatly short of its number, and, indeed, all be imaginary.
But the rule is not without its use, even in detecting imaginary roots, as it
sometimes discovers discrepancies incompatible with the existence of real roots,
in those equations which are incomplete, or have terms wanting. This will be
* Bj' the foreign writers this rule is always attributed to Descartes, and most English writers
follow their example. There is, however, undeniable evidence that tlie rule was obtained tnJi-
rectly bj- Descartes from Harriot ; and it may be mentioned in support of this view, that Har-
riot gives a reason for the rule, wliile Descartes gives none.
On the otiier hand, it has been alleged that the failure of its generality in consequence of the
existence of imaginary roots was not perceived by H.irriot, and that there is no evidence that he
■was even acquainted with the existence of imaginary roots. It must, however, be replied, that
the ^rs/>ram ^l«a/v/ic« was a posthumous work, edited by Wanier, who does not appear to
liave fully understood H.nrrioi's views, and who, therefore, thought he exercised a sound and
kind discretion towards his friend in suppressing certain parts of tlie work ; a suppression which
we know did take place. We cannot, therefore, say more as to the views which Harriot enter-
tained on this subject, till some of his papers, still in existence, are more completely examined
than they have been. With respect, however, to his knoxcledge of imaginary roots, we have
sufficient proof that he understood their forms .ind their meaning too. In tiie Supplement to
the Works of Bradley, published by my estimable friend, the late Professor Rigaud, plate 5, will
be seen a solution of the equation 1 — 00^::=. — 2<i -j- 34, and tiie solutions are separately put
down ; viz. a = 1 -f- ^ — 3'2, and a ■=.\ — ^ — 32. Even this, were this all, would remove
the imputatioa of his ii/norance of the existence of imaffiiiary roots. — Editor.
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 221
made apparent in the proof, of Do Gua's Criterion of imaginary roots, a little
further on (see p. 224).
Theouem IX. If r,, Tj, Tj r._jp he the real roots of the equation
f{x) = 0 of the nth degree, in the descending order of their magnitudes, and
quantities p, p„ pj, p„_ip taken so that p, r„ p,, r„ p.^,. . ..r^„, p._„ be also
in descending order of magnitude : then we shall have
/(p) = + k,f{p,) = - *„/(p,) = + k„
and so on, the results being alternately + and — .
Let thej9 pairs of imaginary roots of/(x)=0 he a, + b, a/ — 1, flj+^jv' — 1, ...
flp + ftp V— 1 : then the portion of/(x) depending upon these is
Fix) =[ia,- xY + b,^ [{a, - x)^ + b,-] .... [(a,- x^ + b,'] ;
in which, since every factor is essentially positive, their product will be +,
whatever be the value given to x.
Cx — r,) {x — r.j) {x — Tj,.,) . F(ar) = 0 is the general form of /(j) = 0 ;
and of this ¥{x) being always +, it will not affect the signs depending on the
values of x in the other factors.
Substituting then p, p„ .... p„_2p in the other part of the function, we have
successively
(p — r,) (p — fj) (p — fj) ... (p — r„_jp) z= + k, since all the factors are +.
(Pi— J*!) (pi — O (pi — '■3) • • • • (pi — r„_„p) = — ki. since only one factor is — .
(Pa — ^i) (Pa — ''2^ (P2 — '■3) • • • • (Ps — '"n-zp) = + *:» since only two factors are — .
and so on through all the substitutions.
Cor. If in the results of any two substitutions p' and p" in /(ar) we find dif-
ferent signs, there are 1, 3, 5, or some odd number of roots in the interval p' and
p"; and if the signs of those results be alike, then 0, 2, 4, or some even number
of roots are situated in that interval.
Theorem X. In any function / (x) proceeding by decreasing powers of x, a
value may be found for x which shall render the sign of the result the same with
that of the first term.
Let/(x) ^ a?° + Ax°~' + .... + Lr + M ; and suppose the most unfavour-
able case, where all the coefficients after the first, are different from the first,
and K that which is numerically greatest. Then we shall have
K^x"-' + x—2 + X + 1] greater than [Ax"' + Baf-* + . . . Lx -f M].
J." I ^" J
But x°~* + ir°~' -\- . . . X + 1 = ; and hence K . is greater than
all the terms of the function after the first. It will, therefore, be sufficient to
show that such a value can be found for x as shall render x' greater than
a^ I
K . - , or (x — l)j'' greater than K(x" — 1). Now the value K -|- 1 given
to X will render (x — I) x' = K(K + 1)°, and K(x" — 1) = K^ K + 1)" — l]:
but K(K + 1)" is greater than K J(K + 1)° — 1^, and hence such a value of x
as was asserted possible has been found.
Theorem XI. In any function, as N + I\Ix -f- Lx- + • • • • +Bx"-' + Ax",
values of x may be found which will render the result of the entire function of
the same sign as the first term N.
Take, as before, the most unfavourable case, where all the terms after N have
their signs different from that of N. Then if K be greater than either of these
coefficients, we shall have
Kxjl + X + x^ + ... x°-'] greater than x JM + Lx + . . . Bx— * + Ax— '|.
Kx(l x') Kx
Now the greater of these is — -; and if x be less than unity, ——^ —
222 • ALGEBRA.
is crreater than — ^^ ; and hence still more is :; greater than
° 1 — X 1 — X
x{M + Lx+ .... Bx'-' + Ax'^-'] .
If, therefore, we can find a value of x such that N is equal to, or greater than,
Kx • N
, we shall have eflfected the proof of the proposition. Now if x = ^-^^ — ^,
1 — x jN -|- K.
Kx
we shall have = N; and as this is always greater than all the terms fol-
1 — X
lowing N in the given equation, the condition is entirely fulfilled;
Theorem XII. The consecutive roots of/, (x) = 0 lie each in succession
between the consecutive roots of/(j) = 0.
Let r,, Tj, . . . .r, be the roots affix) = Ax* -\- Bx'~* + . . . = 0,
and p„ Pa. p.- 1. be those of/, (x) = tiAx'-^ + (n— 1) Bx'-^ + . . . = 0.
Reduce the roots of/ (a?) = 0 by the indeterminate quantity r, (prob.II.) which
will give the roots of the transformed equation respectively equal to r, — r,
fj — r, .... r, — r. But this transformation gives M, ■=. f^{r), and N, :=/(r).
Now the value of M, in this reduced equation is (theor. VI.)
M, = (r—r,) (r—r^) (r—r^) {r—r,_0
+ (r— r.) (r-r3) {r—r,) . . ..(r—r.)
+ (r—r,) (r—r^) {r—r^) (r— O
+ (.r—r^) (r—r;) (.r—r,). . . . (r—r,)
Now in this expression there is but one group of factors from which any one
of the given factors is absent, as, for instance, r — r,. If then in M, or/(r) we
give the indeterminate quantity r the successive values r,, r,,. . . . r,, the several
results will comprehend only one set of factors, viz. that in which ri, r^,. ...r*
are thus rendered successively absent ; and we will suppose them so ranged that
r,, fj r, are in the order of descending values. This will give
(r, — r,) (r, — r^) . . . .(r, — r.) = -\- k, since all the factors are +.
(rj— r,) (r^—r^) .... (r^ — r.) = — k, since one factor only is — .
(rg — r,) (r^ — r,). . . .{r^ — r,) = -f /c, since two factors only are — .
And so on through the entire series of results.
But when a series of quantities r,, Tj, .... r. are substituted in an equation
/, (x) = 0, which give results alternately + and — , there is in all cases one root
of the equation/, (x) = 0 comprehended between those numbers (theor. IX.) But
pi, pi, ... . p,_, are the roots of /, (x) := 0 ; and hence these values lie between
r„ Tj r„ the roots of/(x) = 0. That is, the roots of both equations being
ranged in the order of descending magnitude follow each other thus :
ri, pi, r^, Pj, Tj, r,_„ p,_„ r,.
Cor. In the same way it may be shown, that the roots of / (a) =: 0 lie
between those of /, (x) = 0 ; those of /, (x) = 0 between those of /j (x) = 0,
and so on.
Scholium.
These properties admit of various applications in the higher theory of alge-
braic equations, and are popularly known as the limiting equations of Netcton.
In actual numerical solution, however, they are now of little use ; and they are
only given here in justification of one or two processes employed. The equa-
tions are evidently the same which have been before treated under the name of
the derivative equations.
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 223
There is one remarkable property of the hmiting equation : viz. that when
/, (p) = 0, then /(p) has a greater or less value than it has when /, (p) is either
a positive or negative quantity a little different from 0 ; or, in more technical
language, (though belonging to a more advanced subject of study,) /(p) is a
maximum or a minimum. *
Theorem XIII. If one of the roots r, of an equation Ax" -f- Bx"~' + ....
lar^ + Mx -f N ^ 0, be very small in comparison with all the others, we shall
have, nearly, r, = — ri-
For, let Tj, r.j r, be the other roots ; then (theor. VI.) we have
M = ± Jr,(rjr3....r,_, + r3r,....r„ + r,r^....r.r,+ ....) + r,r,r,. ... r.J
N := + r, r, Tj . . . . r,.
Now since r, is very small in comparison with the other roots, the vinculated
term which contains it as a factor m M is small in comparison with the term
r^Tj... .r„ which does not contain it. Neglecting, therefore, this term, we have
N +r,r,r^....r, N
*r= T = — ^i> or r, = — :^ .
Scholium.
This theorem enables us, after we have obtained a first distinct approximation,
to obtain a closer one by mere division; and thence to still further reduce the
roots of the equation, and especially that to which in any case we may be
approximating.
PROBLEM VII.
To find the limits between which are situated the roots of any given equation,
Aaf + Bx"-' + . . . + Lr^ + Ma; + N = 0.
1. Find a reducing number k which will render the signs of all the coefficients
positive : for then the roots of the transformed equation will be negative,
(Theor. VIII.) and hence k is greater than the greatest root of the equation,
(Cor. 1. Prob. III.) or it is the superior limit of the positive roots.
2. Find similarly a number which will render all the coefficients of the trans-
formed reciprocal equation positive ; the reciprocal of this number will be less
than the least positive root, or will be an inferior limit of the positive roots.
3. Change the alternate signs, and find the superior and inferior limits of the
positive roots of this transformed equation : these limits, taken with negative
signs, will be the limits between which all the uegative roots lie.
4. To find how many roots lie within any given limits, a and b, of which a is
the greater, reduce the roots of the given equation by the less of those numbers
b; then, again, reduce the roots of this transformed equation by a — b. The
difl^erence of the number of variations of sign in these two transformed equa-
tions indicates the number of positive roots in the interval.
In a similar manner, after changing the signs of the alternate terms, and of
the two negative limits, we may find the numbers of variations in each reduced
equation ; the diflFerence of which will be the number of negative roots in that
interval.
The first part of this rule becomes evident from Theor. VIII. and Cor. I.
Prob. III. : and the latter, from combining it with Prob. III. itself.
Scholium 1.
As a practical course of procedure, it will be advisable to reduce by 1, 10, 100,
224< ALGEBRA.
. . . rather than by intermediate numbers to these, till the utmost limit is ob-
tained; and then to work with such intermediate numbers, as may be thought,
from the state of the coefficients, most likely to make a small number of changes
in the state of the coefficients as to order and number of signs.
Proceeding with these till we^ave obtained two limiting consecutive num-
bers for one or more roots, the object of this problem will be attained.
Scholium 2.
When by narrowing the intervals of the substituted numbers, we find more
than one variation continually disappearing in each of the substitutions made ;
these roots may be equal, or they may have minute differences, or any even
number of them may be imaginary.
If there be equal roots, the process of Problem VI. will find them.
The only question, then, is to find whether an equation known to have only
unequal roots, has any number of them imaginary, and how many ; the remain-
ing ones, of course, being real, and having differences less than that of the limit-
ing substituted numbers between which they are indicated. Several methods of
solving this problem have been proposed ; but we shall here give only three of
these criterions, those of De Gua, Budan, and Sturm ; though those of Lagrange
and Fourier well deserve to be studied by every one whose time and incUnation
lead him to pursue the subject further. See Lagrange, Resolution des Equations
Numeriques, p. 6, and Fourier, Analyse des Equations Determinees, p. 87.
Theorem XIV. De Gua's Criterion of Imaginary Roots *.
This criterion is generally stated incorrectly. It should also be expressed
more in detail than is usually done. Before stating it, however, it will be de-
sirable to enter upon the examination of the principles from which it flows.
1. It is very clear from the reasoning in theor. VIII. that the rule of Harriot
is true for all cases in which the roots of an equation are real, or constituted
merely hy the signs + or — prefixed to a real number, either integer, fractional,
or irrational.
2. It follows, then, that all cases in which this assumption being made leads to
contradictory results, indicate, according to the number of those contradictions,
so many of the roots not being constituted as above expressed j that is, so many
of the roots are imaginary.
3. When we have a cipher-coefficient, such as Ox"", it is either -f Ox' or — Ox" }
the values of the expression in which it occurs being precisely the same in both
cases.
4. The greatest number of negative roots in an equation which contains cipher-
coefficients between any two actual coefficients, will be when all the ciphers are
written with the same signs as one of the terms which form the extremes between
which the ciphers are situated.
5. The least number of negative roots under the same circumstances will be
* Tliis property of cipher coefficients was first given by the Abbe de Gua, in the Memoirs of
the French Academy, for 1743. All writers, after tlie original author, have, however, com-
mitted an oversight in estimating the number of conditions inaplied in this theorem, which thej
uniformly assert to be ^ — - for n cipher coefficients. The conditions may, indeed, appear
under different simultaneous forms : but their number cannot be more than as stated in the
text above.
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 225
when the ciphers are taken with alternate signs, the first cipher being taken with
the contrary sign to its adjacent actual coefficient.
6. ITie difference between the greatest and least number of negative roots indi-
cated by taking the ciphers as in (4), (5), is the number of imaginary roots indi-
cated by the sequences of the cipher-coefficients in the equation.
7. If cipher-coefficients occur in any of the transformed equations, the same
rules apply, since no kind of transformation by real numbers can eliminate the
imaginary part of the root.
It is, however, to be carefully kept in view, that no proof is offered of the
imaginary roots, indicated by the transformed equation, being different from, or
the same with, those indicated by the given equation.
The statement, then, of De Gua's rule will be as follows : —
I. If between terms having like signs, 2w or 2ra — 1 cipher-coefficients inter-
vene, there will be 2« imaginary roots indicated thereby.
II. If between terms having different signs 2n -|- 1 or 2n cipher-coefficients
intervene, there will be 2m imaginary roots indicated thereby.
(1) Let there be 2n ciphers between lilie signs; then writing them as expressed
in (7) we have
+ h + 0 + 0 + 0+ + 0 + k, giving 2n + I negative roots.
+ h — 0-|-0 — 0-|-.... -^ 0 + k, giving 1 negative root.
Whence since only one root is negative in the latter case, and 2n -f- 1 roots in the
former, there is a contradiction in the sign of 2n roots; which are, therefore,
imaginary.
Had the signs of the e.\treme terms, h, k, been both — instead of +, the series
would have taken the form
— A — 0 — 0 — 0— .... — 0 — it, giving 2n + I negative roots.
— h + 0 — 0 + 0— .... — 0 — k, giving 1 negative root.
And the same contradiction with respect to 2b of these roots would have resulted.
(2) Ne.xt, let there be 2ra — 1 ciphers between like signs : then
-j-ft_j_0-fO + 0-f.... + 0 + 0 + k, giving 2n negative roots.
^ fi — o-f-0 — 0-f .... -j-0 — O-j-it, giving 0 negative root.
Hence there are, as before, 2n imaginary roots.
(3) Let there be 2n ciphers between unlike signs : then
^^-j-O-l-O-fO-}- -l-O-l-O — A:, giving 2n negative roots.
^ ji — o-f-0 — 0-j-.... — 0-1-0 — ^, giving 0 negative root.
Hence, as before, there are 2n imaginary roots.
(4) Let there be 2n + 1 ciphers between unlike signs : then
_). }i^o+0 + 0+ -fO-l-O-l-0 — A:, giving 2n -f 1 negative roots.
^^_0-|-0 — 0-f- — -0-1-0 — 0 — k, giving 1 negative root.
Hence, again, there are 2n imaginary.
The theorem of De Gua is, therefore, established universally, and the state-
ment given in its most general form *.
* It has been well remarked by Mr. Homer, (Math. Repos. vol. v. p. 27,) that though the
direct application of this " criterion can only occur incidentally, yet its application is capable of
an extension beyond what is at once apparent. To cite an example ; when the mth coefficient
changes its sign in passing from one set to another, while those which immediately precede and
follow it are and continue to be identical, the existence of zero between like signs somewhere in
vol.. I. Q
226 ALGEBRA.
For example, ar^ -f ar + 1 = 0 having the coefficients 1 + 0+1 +1, has
two imaginary roots ; and the equation ar^ + 5 z= 0 having the coefficients
1 + 0 + 0 + 5, has also two imaginary roots.
EXAMPLES FOR PRACTICE.
1. How many imaginary roots are there in the equations x* — 6x- + a? = 0,
ars + 1 = 0, and x^ — 6x^ + 2x +12 = 0.
2. The equation x^ — 5x* + 20a:^ -\- x = 100 has one pair of imaginary roots
indicated by its present state, and reducing the roots by a certain number to be
found, will show another pair. What is that reducing number, and which are
the places at which the imaginary roots are indicated ?
3. Has the equation x'^ — 4x^ + 8x^ — 16a; + 20 = 0 any imaginary roots?
Theorem XV. Sudan's Criterion *, as arranged by Homer.
The theorem may be stated thus : —
If in transforming an equation by any number r, there be n variations hst, and
if in transforming the reciprocal equation by - there be m variations left ; then
there will be at least n — m imaginary roots in the inten'al 0, r.
(1) There can be no root of an equation infinitely great except the absolute
term be also infinite. The roots of equations, such as generally occur, are, there-
fore, finite.
(2) The reduction of the roots of an equation by - or infinity, must therefore
render all the roots negative; and hence give only permanencies of sign in the
transformed equation.
(3) Imaginary roots enter an equation in pairs, as has been shown in
Theor. V.
the interval may be suspected. Should all the previous signs in these sets be alike, the proba-
bility is increased." He might, indeed, have spoken even more decidedly ; as by other means we
can show that it amounts to all but absolute certainty.
In successive transformatious this remark will often be of considerable utility to be borne in
mind.
* This criterion was first given by Budan in his Nouvelle Meihode pour la Resolution des
Equalions Numeriques, p. 36; but the form in which it now appears is due to Mr. Homer, and
in any other it is next to useless. It would almost appear from a note to Lagrange's Traite de
Resolution des Eq. Num. (p. 169,) that he did not seize its import: at all events, he formed an
inadequate notion of it, and raised an objection to it which is altogether invalid.
The following observations upon using it will be useful to the student : —
1. It will always in practice be most convenient to take the transforming interval r
equal to 1 ; as then the reciprocal is also equal to 1. When the interval is a prime number
different from 2 or 5, it often becomes troublesome to reduce the fractional remainders; and it
Is not often safe to neglect them, as the wliole force of the criterion may be destroyed by a very
small quantity. Besides, though imaginary roots can never be indicated by this criterion e.xcept
they exist, they may exists without being indicated by a specified interval. The smaller the in-
terval, therefore, the?greater the probability of their detection. Hence, except in rare cases, it
is better to take the interval 1.
2. When no indication_is supplied by the interval 1, it will be most convenient to tise
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 227
(4) There are as many positive roots in the interval 0, r, of the direct eqiiation,
as there are between - and - of the reciprocal equation. For the roots of the
reciprocal equation are the reciprocals of the roots of the direct equation ; and
hence must lie between the reciprocals of those limits of the direct equation.
(5) If then the number of variations n, lost in the direct equation by passing
over the interval r, be greater than the number left m in the reciprocal equation,
after transformation by -, there will be a contradiction with respect to the cha-
racter of a number of them, equal to the difference n — m. These roots, there-
fore, are imaginary.
To take one example, find whether the equation x^ — 8x* + liar* — Sgr' -f-
31a? = 101 has imaginary roots.
Reduce by 1 ; then the work will stand thus :
1 — 8 + 11 — 39 + 31 — 101 (1
— 7 + 4 — 35 — 4 — 105
— 6 — 2 . . . . where four variations are already lost.
•5, which narrows the limit still more, as we have the intervals 0, '5 and '5, 1 ; and the recipro-
cal of 5 being 2, we have an easy working number for the reciprocal equation, if both roots be
still left doubtful. If there be one variation, and one only, lost, the roots are separated, one
being between 0 and '5, and the other between "5 and 1. Should it still be uncertain, it will b«
convenient to reduce either the given equation or that in '5 by "2, since the reciprocal 5 is an
easy number for the reciprocal transformation.
3. It will sometimes conduce to clearness of conception to multiply the roots by 10, 100,
1000, &c., in accordance wiih what is said at p. 213 of this work. This, however, is only the
case in early practice of the method, since the process itself is not virtually altered by it.
This step simply adds one cipher to the second coefficient, two to the third, three to the
fourth, &c.
4. It will also be desirable in these successive transformations to keep De Gna's Criterion in
view, both in its strict form, and with the modification suggested at the foot of page 225.
5. It will generally happen when we have taken too wide an interval that the ambiguity may
be very simply removed as follows: —
Reduce the reduced reciprocal equation by unity at a time till the variations are lost in
pairs, or the roots separated belonging to the doubtful interval. Apply the criterion to this
reciprocal equation ; and if the roots be indicated as imaginary, they are so ; and then the roots
of the original equation, which are functions of these, are also imaginary.
To take an example, let x* -\- x^ -\- 4^^ — 4r + 1 = 0 be given.
Direct trans/ormalion. Reciprocal tranfformation.
1 + 1+4-4 + 1(1 l_4 + 4 + l + l|l
+ o_|. (j_j. 2 + 3 ^ —3+1+2+3^
3+9 + 11 _2— 1 +1
4+13 —1—2
5 ±0
In the direct transformation two variations are lost, and in the second two variations are left;
hence no conclusion whether they are or are not imaginary can be drawn from this result. Take
the reduced reciprocal equation, therefore, and we shall have
14-0-2 + 1+ 3(1 3+1-2+0 + 11
l_l+0 + 3^ 4+2+2 + 3^^
2 + 1+1 7+9 + 11
3 + 4 10+19
4 13
In these two transformations we have now two imaginary roots by Sudan's criterion ; and
hence the two corresponding positive roots of the given equation are imaginaiy.
Q2
228 ALGEBRA.
Reduce the reciprocal by 1 : then
— 101 + 31 — 39 + 11 — 8 + 1 (1
— 70 — 109 — 98 — 106 — 105,
where there are no variations left. Hence n — m = 4 — 0 = 4 imaginary
roots.
EXAMPLES FOR PRACTICE.
1. a^ — 2x = 5.
2. a^ — 7x + 7 = 0.
See other examples at p. 232.
Theorem XVI. Sturm's Criterion.
Let X = 0 be an equation of the nth degree, and let X, = 0 be its first deriva-
tive ; and let the given equation be free from equal roots. Perform the operation
employed in finding the greatest common measure, always, however, changing
the signs of the several divisor-remainders : and denote the series of functions
which result, together with the given ones, by X, Xi, Xj, . . . X„ . . . . X.. Then
if a and b be any two numbers substituted in these several functions, the differ-
ence of the numbers of variations of sign in the results of these substitutions
expresses the number of real roots lying between a and b.
1. Denote the successive quotients by Q„ Q^ Q,_, : then
X =Q, X, — X,
X, = Q, X, - X3
Xj = 03X3 — X^
X,_2 — vi,_i A,_i — X,.
Now as the degree of the function is diminished a unit at each stage, the final
remainder is clear of x : and as the equation X = 0 contains no equal roots, X,
cannot be zero, or in other words, X. is a number.
2. No two consecutive functions can become zero for the same value of x.
For if it be possible, let them beX„_i and X„. Then since
X„_i = Q„ X„ — X^i,
and X„_, ^ 0, and X„ = 0 at the same time, we also have X.+, =: 0. But X.
:= Q.+1 X,^, — X„45, and hence, for the same reason, X.^^, = 0; and similarly
all which follow X„ become zero for the same value of x, amongst which is X,.
Now it has already been shown in (1) that X, cannot become zero : and lience
also X, cannot become zero at the same time with X,^,. No two consecutive
functions can, therefore, vanish with the same value of x.
3. If any value of x cause one of the functions, as X,, to vanish, it gives to
the two adjacent ones equal values with contrary signs. This is evident from
the connecting equation X,^i = Q« X„ — X^i, which in this case becomes
X„_i =: — X^i.
4. If such a value be given to x as shall make one of the intermediary func-
tions Xp Xj .... X^i, vanish, without making X ^ 0, there will be a change
in the order of the signs produced at this stage of the variable values of x, but
no change in the number of variations.
For let X, be that in which any special value of x makes the result zero.
Then, considering the three consecutive functions, X^,, X. and X^„ we have
seen that in such case X„_, and X^, have contrary signs (3) ; and the series of
signs will therefore be
+ X_„ ± 0, — X^, or - X^:, + 0, + X.+,.
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 229
That is, writing only the signs, we have the following comhinations :
In the first case -\ — | , or H ; (n
In the second ^ _|-^ or [- (2)
Now denoting by a„ a root of X„ = 0, it is established, (Theor. IX.) that if
numbers greater and less than a„ be substituted in X., the results will have con-
trary signs ; and hence, in passing through X. = o, the signs will undergo the
changes indicated in (1) or (2): but so far as the number of variations is con-
cerned, these are all precisely alike, and differ only in the order of their succes-
sion. No variation, therefore, is lost or gained in the passage of X. from a value
greater than a„ to one less, or the converse.
Moreover, that which holds true for X, holds true for any other function
X^, ; and hence it holds true universally. No variation, therefore, can be lost
or gained amongst the intermediate functions, though their order of succession
should be changed in any way whatever.
5. Every time a value of a? coincides with a real root r, of the equation X = 0,
one variation, and one only, is gained in a descending series of values, or lost in
an ascending series.
Since (4) no variation can be lost or gained amongst the intermediate series of
functions X,, Xj, .... X„, it follows that the only case in which a change in the
number of variations arises, is by its occurring between X and X,.
Now, by hypothesis /(r) = 0; and hence, taking the quantities r — h and
r + A to represent the substituted numbers less and greater respectively than r,
in the equations X = 0, and Xi = 0, we get
f(r-h) = -f^r)l+Mr)^^ -f^^'^YJTs + '■"
Ur-h) =f,(r) -/,(r)J +fAr)~-A{r) ^^ + ....
/(r + A)= +/'('-'I+/«('-)/|^+/.Wl.2'3 + ----
Mr+h) =/,(r) +/,(r) * + fAr) ^^^ ^/Jr) ^-^^ +
But in all these functions we may find a value of h so small, that the value of
the whole series shall be the same with that of its first term. If, then, we take
an ascending series of values for the substitutions, the signs before and after the
passage of x through r will be 1- and + + ; that is, a variation will be con-
verted into a permanence, or one variation is lost. It has also been shown, that
no variation can be lost amongst the intermediate functions ; and hence only one
variation can be lost in passing through a root of X = 0, from less to greater
values of x.
Had we taken the descending series of values for x, the reverse would have
taken place; viz. + + converted into 1-> or one variation only be gained in
the passage through r *.
* During the passage in the v.ilncs of x in the several functions from one root of X ^ 0 to
another, we have seen that any difference of order of succession may occur, but no difference in
the total number of variations: and it may conduce to the rleamess of our conception how a
variation may be introduced or lost in passing through the next lower or higher root of X ^ 0.
We liave seen (theor. IX.) that if r,, rj, r,, ranged in the order of m.-tgnitudc, be the
roots of X = 0, and p,, pj, P"— i, be those of X, ^ 0, ranged similarly, then these Beveral
values ranged in the order of magnitude will be
■ri,Pi,r2,P2,r3,P3, r._„p._|,r..
It hence follows, that in passing from Vi to rj, we pass through p^ a root of X, = 0; and
230 ALGEBRA.
6. Let now a and b be any two numbers substituted in the series of functions,
there will be one variation, and one only, gained or lost, according as we use
ascending or descending values of x in passing from one value to the other,
every time we pass through a real root of X = 0. There will hence be as many
variations gained or lost as there are real roots between a and b, and no more
than these can be gained or lost. There will be as many more variations in the
series of signs of the functions arising from one substitution, than there are in
the series of signs arising from the other, as there are real roots, neither more
nor less. The difference in the number of variations in the two series of func-
tions under these two substitutions, expresses, therefore, exactly the number of
real roots in that interval. Sturm's criterion is, hence, fully established.
Cor. 1. Applying this to find at once the entire number of imaginary roots of
an equation, we have only to take a positive limit greater than the greatest posi-
tive root of an equation, and a negative limit numerically greater than the
greatest negative root of the same equation : then the difference in the number
of variations of the two series of results of the substitution of these limits in X,
X„ Xj, . . ., X„ will be the number of real roots of the equation, the remaining
ones being imaginary.
But as -)- - and , though limits too wide for approximation, are suffi-
ciently near for the present purpose : and as in this, like the more restricted
limits above spoken of, the signs of the entire functions will be the same with
those of their first terms, we may at once obtain the number of real roots, which
is the difference of the number of variations in the true series.
For the particular character of the roots in any given interval, it will, however,
be necessary to form the values of the functions for the two numbers which form
the limits.
Scholium.
It is verj' obvious, that except the changed signs, Sturm's functions are pre-
cisely those which occur in seeking for equal roots ; and hence, if such occur,
they will be made apparent, and the given equation depressed by these roots
may be resumed as an original equation, which, from its being of lower dimen-
sions, wUl create less difficulty in finding the functions *.
Let us take as an example the following equation :
Given x^ -|- 4a;'» — 2a^ 4- IOj;^ — 2x — 962 = 0.
hence the signs of X,, when quantities less and greater than p^ are substituted for x, must be
changed, till we arrive at pj. But before arriving at pj we pass through rj, and hence in this
interval X r: 0. From the value, therefore, ever so little above r,, to the value ever so liltle
below p,, X and X, retain their signs unchanged ; but at this point, p,, the sign of X,, changes
to the opposite, and continues to retain it till we arrive at pj, when it again chauges. The series,
therefore, \\i\\ be, supposing r, the greatest root of X :^ 0,
greater
than r,
I +
between
between
between
between
between
r, and p,
p, and rj
r.^ and p^
Pj and r.
Tj and p3
—
—
+
+
—
+
—
—
+
+
And so on, as the two last
are a repetition of the first
two signs, and the same or-
der will continue to the end.
• Sturm investigates a modification of his method, which dispenses with the resumption of the
process with respect to the depressed equation. Nothing in point of labour is, however, saved
by it.
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 231
Here X = x» f 4x* — 2x3 + lOx' — 2* — 962 j
X,= 5x* + iCar" — 6x' + 20x — 2;
Xj= 14*3 — 29a''^ + 20x + 4007 ;
X3= — 325x* + 11358X + 59159 ;
X^= — 10052526* — 47309473 ;
• X A, Xj Xj x^ x^
When X = — ', we have the signs 1 f-4-, or3 variations ;
X = + . , we have the signs + + H 1-» or 2 variations.
Hence, the difference of the number of variations of sign being 3 — 2 =: 1,
the equation has but one real root.
However, it will be necessary to narrow the limits between which the substi-
tutions are made in order to effect the entire solution.
If X = 0 we have the signs h -\ h, or three variations; and hence
there is no negative real root : but by Harriot's law of signs, there are two of
the roots negative, there being two permanencies of signs in the given equation.
These two negative roots are therefore imaginary.
Again, take 3 and 4 for the values of x in the functions : then, when x ^ 3
the signs are 1- -|- -\ f-, and there are no variations lost in the series
of signs ; hence there is no real root between 0 and 3.
When X = 4, the signs are + -t- + H f-. and there is one variation
lost ; hence there is one real root between 3 and 4.
Moreover, the signs undergoing no change for any greater value of x, there is
no real root greater than 4 ; or the only real root lies between 3 and 4. We
shall resume this example presently for the purpose of completion of the entire
process of solution *.
EXAMPLES FOR PRACTICB.
1. Given x^ + llx- — 102x + 181 = 0.
2. Given x^ — 2x = 5, and x^ — 7j? + 7 = 0, to find whether the roots are
real or not in each.
3. In x^ — 2x' — 7x2 + lOx + 10 = 0, all the roots are real.
4. In 2x5 -f- 2x< — 13x3 — 3x2 — 9x = 19, tbere is one real root, two equal
roots, and two imaginary roots.
• It may be useful to the student to compare the determination of the character of the
roots by Budan's criterion with that in the text by Sturm's method.
(1) Reduce the roots by 1 ; then we have + + + + H , or 2 lost.
(2) Reduce tlie reciprocal by 1 ; then we have , or 0 left.
Hence there are two imaginary roots in the interval 0, 1.
(3) Reduce the roots of (1) by 2 : then the signs are + + + + H , or 0 lost.
(4) Reduce the roots of (3) by 1 : then we get + + + + + +. or 1 lost.
Hence there is one real root between 3 and 4.
To find the negative roots, change the alternate signs : then
(5) Reduce by 3; and we have the signs + + + -| +, or 0 lost.
(6) Reduce (5) by 1 ; and we get + + + + + +, or 2 lost.
(7) Reduce the reciprocal of (5) by 1 ; then + + + + + +, or 0 left.
There are hence two negative rooto between 3 and 4, and they are imaginary.
232 ALGEBRA.
5. Given X := x^ + px + q = 0, to find the conditions which will render all
the roots real.
6. For the same purpose, given x^ + psc^ -\- qx + r =■ 0.
MISCELLANEOUS EXAMPLES ON CRITERIA, AND INITIAL
APPROXIMATION.
1. a?8 — 12a?* + 60a;* + 123a;2 _|_ 45573; _ 89012 = 0.
2. a?* — 8a:* + 32a7* — 7ix^ + I04x^ — 80a; + 25 = 0.
3. a?7 _ 9a;« + 40a;* — 106a?* + l78x^ — 184a;2 + 105x = 25.
4. 89012a;« — 4567a;* — 123a;3 — 60a?2 + 12a; = 1.
5. 21* + 2a;* + 3x^ — 2a^ — 3a7 — 1 = 0.
6. X* — 6x^ + \23^ — 10a; + 3 = 0.
7. a;3 — 15a;2 ^ 533, _ 50 =; q.
8. a;6 — 7a;* 4- 22a;* — 53a;3 + ll7x^ — l60x + 150 = 0.
9. a?8 — 1 = 0, a;« + a;3 ^ 1 _ 0^ and x^^ + 3a;* — 2a; = 0.
PROBLEM VIII.
Homer^s method of Continuous Approximation.
Let the given equation be Aa;" + Bx"-^ + Cx"-- + . . . . La;^ + Ma; + N = 0 ;
and let a, be a distinct approximation * to one of its roots ; it is required to
evolve the remaining figures in succession.
1. Reduce the roots of the equation by a,, and denote the reduced equa-
tion by
A,a;," + B,a;,— 1 + C,a;i"-^ + . . . . L^x^" + Mja;, + N, = 0".
2. Find a new root-figure from — ^i- = Oj. that is, from — :~^i = a, : and
reduce the last equation by this, giving
A^a;/ + B^^'-^ + C^^-- + Laa;.^ + M, x., + No = 0.
If, however, upon transforming, the sign of N^ should prove to be different
from that of N,, whilst that of Mj is the same with that of M„ the value a.j is
too great, and the next smaller number of the same decimal denomination should
be tried, tiU one is found which fulfils the requisite condition.
N
3. From — , ,* = a^ find a new root-figure, and transform as before : and
proceed thus till all the figures are found, if the root terminate, or as many as
may be necessary if the root be interminable.
4. When the root is negative, change the signs of all the roots (Prob. IV.)
* By a dktind approximation is meant a value a, which is nearer to r;, a root of /"(jt) = 0,
than the corresponding root p,, of y,(a*) =: 0 is to r,. In this casey^a) and f\{a) have contrary
signs ; and for the most part (almost always after transformation by the first decimal of the
root) the quotient — /"(a) -r-fAO') gi'^es the next figure of the root accurately, and the more
especially if none of the intermediate cocfEcients are comparatively large numbers. This is
obvious from Theor. XIII. as then one of the roots of the reduced equation is relatively very
small ; and as we proceed to diminish the roots still further, by succeeding decimals of the root
to which we are approximating, it becomes accurate to several decimal places.
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 233
and find the positive roots above. These written negatively, will be the negative
roots of the equation,
5. In actually working out the transformations, it will be convenient to mark
the resulting coefficients of the transformed equation by oblique lines, as in the
example on p. 234, instead of recommencing the work by writing them anew in
a horizontal line.
6. After obtaining two or three decimals of the root, the work may be very
much contracted, analogously to that employed in contracted multiplication and
division of decimals, in the following manner :
(a). Let A, a^/ + B^ x;-' + .... + L, a?/ + M, a?, + N, = 0 be the re-
duced equation after which the contractions are to commence : then draw a
vertical line on the right of the figures in N, ; a vertical line cutting off one
figure from M^ ; a vertical line cutting off two figures from L, ; and so on, till
n — 1 figures are cut off from B,, and n figures from A,.
(b). Find with this contraction the ne.xt figure of the root a^^ ; and reduce by
this, taking the figures to the left of the vertical lines, with one of those on their
right as the multiplicand in each case, (taking care, however, to estimate the
effect for the purpose of " carrying" of the rejected ones, as near as possible,)
and put the results down in the corresponding places of the next column ; viz.
beginning with that on the right of the vertical line. The additions to be per-
formed as when there was no contraction of the work. This will give A^, af^i
+ B^x x»-'^, + . . . M^i x^, + Nh-, = 0.
(c). In the next transformation, cut off one, two, three, .... (n — 1) figures
from the contracted coefficients, M^„ L^, + Kp+„ .... B^, ; and proceed as
before.
In these processes, as the greatest number of figures is cut off from those
columns which originally contained the fewest, these will diminish very rapidly;
and after a few transformations, an equation of a high degree is reduced in point
of simplicity to one of a low degree ; and generally the last half of the entire
figures of the root are obtained by contracted division only.
Moreover, if p figures have been found, and n be the degree of the equation,
the number of figures of the root which may be trusted to as quite accurate, will
he np — 1. Thus in an equation of the 5th degree, if three decimals have been
found, the contraction will give .3.5 — 1=14 places true in all.
The theory of this contraction is very simple, but it does not admit of being
concisely laid down in words : but a little consideration will enable the student
to perceive that the effect of the parts cut off falls entirely to the right, in all the
columns of the correction column, or of that which follows the column of figures
to the right of the vertical Unes.
In illustration of the entire process, let the equation j* + 4x'' — 2t^ + lOar' —
2x — 962 = 0 be proposed, in which a distinct approximation to one of the
roots is 3. Then, performing the reduction by 3, we have
V + igx.^ + 136a?,' + 478a:,» + 841*. — 365 = 0.
365
To find Oj, we have a^ = --t- = '4 ; but upon reducing the equation
by -4, we find N, = + 4068064, or the sign of the final term is changed.
Whence try a^ =. "3, and the condition is fulfilled. In all the subsequent stages
— -? = Op is found upon performing the reductions, to fulfil the required
condition. The following is the process.
234
ALGEBRA.
— 962(3-354848699
597
This result is rather too small, owing to the contracted corrections of the
coefficients, but especially of the fourth, being kept uniformly above the truth :
but they have been retained to show the manner of conducting the operation,
instead of throwing out the zeros which would have taken the place of the units
now to the left of the line.
GENERAL RECAPITULATION AND REMARKS.
1. Count the number of variations and the number of permanencies of sign in
the given equation ; there will be as many positive roots as variations of sign,
and as many negative roots as permanencies of sign. (Theor. VIII.)
2. If there be an odd number of positive roots, one at least of them will be
real ; and if an odd number of negative roots, one at least of these will be
real.
3. As an initial experiment, reduce the roots of the equation, both as it is
given, and with its alternate signs changed by the factors of the absolute term ;
since, if there be any integer roots, they are factors of that term, and in such
case, the first horizontal line of operations will render N, =. 0. If any such be
NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS. 2S5
found, then employ in like manner the factors of M, of the reduced equation ;
and so on, as long as the division terminates. The integer roots will all be thus
easily found. See Example 4, p. 236.
4. No equation having all its coefficients integers, and A different from unity,
can have a fractional root : all such equations must, therefore, have their real
roots either integers or interminable decimals.
5. If one variation be lost in passing from a transformation from an integer a
to its consecutive integer a + I, then the number represented by a is the prin-
cipal part, or " first figure" of the root.
6. If two, four, or any even number of variations be lost in the transition,
there are two, four, or some corresponding even number of roots, in the in-
terval, of which one, two, or some corresponding number of pairs may be
imaginary, and the remainder real.
7. In this case consider which criteria are most likely to be applicable to the
determination of the number of real roots.
If De Gua's apply, it will be the most simple ; or if any inference can be
drawn from it under the aspect presented in the note on p. 225, let it be done.
In case of still doubting the character of the roots, apply Sudan's criterion
first of all, as directed in the statement, p. 226, 1 being the reciprocal of 1. If
there be still uncertain roots, (or m less than n,) proceed as in the foot-notes,
using either the reduced reciprocal equation, or a narrower interval in the
original direct one.
Should there still be any doubt, which can never be the case except there be
equal roots, or roots having very minute differences, have recourse to Sturm's
Criterion. This in its progress, by giving some one of the modified remainders
Xp = 0, will furnish the component equation containing the equal roots ; and
if there be not equal roots, the functions so derived will furnish a complete
criterion for every part of the series of values between r, the greatest, and r.,
the least, of the roots *.
8. When there are two real roots in a small interval, it will always be more
convenient to seek one or two figures of the corresponding roots of its reciprocal
equation in the outset, as suppose a and b : then the leading parts of the roots
of the given equation will be found from - and ; , and the approximation conti-
nued as usual from these.
9. When the root of an equation has been accurately determined, use the
depressed equation for finding the other roots : but when a root has been ap-
• This may be in some degree an apparent inversion of the natural order of proceeding to
obtain a complete solution of the equation. The most obvious course would be : (I), to form
Sturm's functions X, X,, Xj, ... X,, which in their progress would detect the equal roots
(2), to apply the limiting integers a and a -|- 1 to these functions in all such cases as presented
a doubt : (3), to employ narrower intervals for finding the distinct approximation : and (4), to
develop the roots whose initial values had been found by Horner's method. However, the very
great labour attendant on finding Sturm's functions, renders it desirable to evade their use if it
can possibly be dispensed with ; and this can almost always be done by taking narrower limits
for the transformation, as we thereby, for the most part, separate \he pairs of roots which occur
at small distances in the numerical scale from each other ; and there is never any difficulty in
determining by Sudan's criterion, as modified in the notes, whether these be real or not. The
order of working, therefore, pointed out in the text, contributes greatly to expedition, whilst it
is much less likely to be productive of numerical error than the complicated operations and
unwieldy numbers that are essential to Sturm's operation.
23G ALGEBRA.
proximately determined, return to the original one (or an accurately depressed
equation, if such has been found, by means of an accurate root) to find the other
roots.
10. Equations whose coefficients are rational, cannot have irrational or
imaginary equal roots, without their conjugates j or, in other words, if there be
equal roots of the form a + b ^/ + 1, there will also be as many of the form
a — b V + r.
EXAMPLES FOR PRACTICE.
1. Solve the equation ofi — 13089034a;2 + 26178063a? — 13OS903O = 0.
Ans. 1, 1, 235, and the roots of a:^ + 237a; + 55698 = 0.
2. Find the roots oi x^ + Ax^ — 8a;'' — 25a;3 _^ 35^2 ^ 2\x — 28 = 0,
Ans. —4,-1, 1, 1-356896, 1-692021, and — 3-048917.
3. Find the roots of a?"- — 1-01 = 0, and of a;^ — 10a? = 100, by the general
method.
Ans. + 1-00498756 in the former, and 16*18034, and — 618034 in the latter.
4. Solve a?8— 6a?7— l-2a?« + 134a;5— 289a;-* + 480a;3_ egoa;^— 608a; + 96O = 0.
Ans. 1, 3, 4, 4, — 1, — 5, and + 2 ^ — I.
5. Solve the equation x^ + Ix*' + 20x^ + \bbx^ — 10000 = 0.
Ans. 454419552, and four imaginary roots.
6. Determine completely the characters of the roots, and the values of the real
ones in the following equations :
(1) x^ — 12a;2 + 12a> — 3 = 0; (all real.)
(2) X* — igofi + I32xr — 302a; + 200 = 0; (two imaginary.)
(3) a;^ — I7a;^ + 54a; — 350 = 0 ; (two imaginary.)
7. Find what roots are imaginary, and find the real ones, in
(1) a;* — 3a;'' — 24a;3 _|_ 95^2 _ ^q^, — jqi^
(2) a;"* — 4a?3 — 3a; + 23 = 0,
(3) a;3 _|_ 2a;2 — 3a; + 2 = 0,
(4) X* — x^ + 4x^ + X = 4,
(5) x! — 23T' — Sa!^ + 4a;2 — 5a; + 6 = 0,
(6) a;3 + 3a,4 _|_ ^x^ — 3a?2 — 2a; — 2 = 0,
(7) a?* — 10a;3 + 6a? + 1 = 0.
8. To find the values of a? and y, there are given the two following :
4a;2 + hxy — lOy- — 4a; — lOy + 250 = 0
— 10a;2 4- 100a?y — 8y^ + 100a? + lOOy + 2356 = 0.
9. Given xy^ + x^y + 1-75 = 0, and x^ + y^ = 4-25, to find a? and y.
10. Given x- + yz = 16, y^ -{- zx ^ 17, and z^ + xy =1 18, to find x, y, z.
INDETERMINATE COEFFICIENTS.
In all the inquiries which have preceded the present, in this course, we have
been required to find the special values of one or more unknown quantities, so
as to satisfy the given equations, or the conditions which they expressed. There
is, however, a distinct class of inquiries, in which we are required to change the
form of any given compound expression into a series of single terms. If the
indicated operation be one which we know how to perform, this change may in
general be effected by an actual performance of those operations : but it will
INDETERMINATE COEFFICIENTS. 237
generally happen that the labour attendant upon it will be intolerably great. It
may happen, too, that we may be in possession of no rule for actually per-
forming the indicated operation, and in this case the problem could not be
solved at all. As, for instance, to extract the fifth root of a + x, or more gene-
rally the nth root of a + a:, to which no rule of extraction given in the earlier
part of this work is applicable. A general method of expanding such e.xpres-
sions in a series of single terms, will be given a little farther on, by means of
indeterminate coefficients, as well as solutions of one or two other problems
which cannot be dispensed with in our future investigations.
In this method, the development is assumed to be of a particular form, so far
as indices are concerned, and equations by which the corresponding coefficients
are obtained, are deduced from considerations presently to be explained. If these
equations give real values of the coefficients, the development is effected by the
solution of them : though, independently of an investigation specially directed
to the decision of the point, we cannot be justified in affirming that the develop-
ment so obtained is the only one that can possibly be made.
The principle upon which the doctrine of indeterminate coefficients turns is
this :
That in all developments that are made according to successive powers of any
quantity (whether integer or fractional), in whatever form the coefficients may
appear, the one which belongs to any specified power of that quantity in one
form must be equal in value to that which belongs to the same power in the
other development. Thus, if the function fx could be written in two different
ways, or so that the coefficients took different forms in the two developments,
the following equation must be fulfilled independently of the specific value of x,
viz. :
A + Bx + Cx2 + = A, + B,x + C,x2 + ...,
For, by transposition, we have
(A — A.) + (B — B,) X + (C — C,) x2 + . . . . = 0.
Now, if we have not simultaneously A — A, = 0, B — B, = 0, C — C, = 0,
.. . ., we should have an equation in x of some degree, n, and x would in this
case have n values only, and not admit of all values indiscriminately.
But since in development we have only to change the forms of the expressions,
the quantities themselves must be capable of all possible values, and hence can-
not be restricted to the special ones which constitute the roots of any given
equations. It follows, therefore, that unrestricted values of x in the preceding
equation are essential to the idea of a development, and hence that the coeffi-
cients of the several powers of x in the simplified equation are separately equal
to zero.
We may also make another remark which will hereafter be found useful. If
the following equation be true for n + 1 values of x, it is true for all values of
X, viz. :
A + Bx + Cx^ + . . . . Nx- = A, -I- B,x + C.x2 -)- ... N,x-.
For if it be true for m + 1 values of x, it is true for a value which is not a
root of the equation of the nth degree ; and hence it can only be true when
A — A, = 0, B — B, = 0, . . . . N — N, = 0. But in this case the equation
is true whatever x may be, or for all values of x.
It will at once appear, that to render this method effective, the development
in symbolic coefficients must be of essentially different forms ; and that in all
cases where this cannot be effected, it will indicate that the powers of x, whose
coefficients they are, do not enter into the true developement ; or, in other
238 ALGEBRA.
words, that the assumed law of the indices of x is not possible. The same is
true if any of the values of the coefficients become imaginary.
A few examples will be necessary to illustrate the nature of the method.
Ex. 1. Extract the square root of I -{• x^.
Assume the root to be in positive powers of a;; or \/l + afi ^= A + Bx +
Cx^ + ...... Then squaring both sides,
1 + a?2 = A2 + ABx + ACa?2 + ADa^ + AEa:< + . . . .
+ ABa? + BV ^ BCa^ + BDa;^ +
+ ACx2 + BCV + C"x* + ....
+ ADa;' + BDa;* + . . .
+ AEa;^ + • • •
The coefficients of the two sides of the equation being of different forms, and
those of the first side given, we may proceed to equate those of the like powers
of X. We have, therefore.
In x", A2 = 1, or A = + 1.
In a?^ 2AB = 0, or B = 0.
In a^, 2AC + B^ = 1, or C = ^ ~,^' = J- = + I.
2A + 2 — 2
Ini«, 2AD + 2BC = 0, orD= — ^ = 0.
In ^. 2AE+2BD+C==0, or E= - ?5^= - g^J = + {.
and so on. WTience the expansion is
vrT^=± {i + l^'-l^'+ }
as may be easily verified by actual extraction, according to the method pointed
out at p. 148.
Moreover, had we foreseen that only even powers entered into the expansion,
we might have obtained twice as many terms of it as above with the same quan-
tity of work, by assuming
a/1 + x^ = A, + B,ar + C,x^ +
fl—x + aP)(l+x — x^).
Ex. 2. Expand ^^ r-^ — z — 2 ^^^ ^ series.
Assume it equal to A + Bx + Cx' + . . . . ; then multiply both sides by
1 + X + ar^, and multiply the two factors of the numerator together. This
gives
l—0x — x^ + 2x^ — x* = A + Bx + Cx^ + Dx^ + Ex* + ...
+ Ax + Bx'^ + C3^ + 'Dx* + ...
+ Aa^ + Bx^ + Cx* + ...
Equating the coefficients of the like powers of x, we have
A=l, A + B = 0, A + B + C = — 1, B + C + D = 2,
C + D + E = — 1, D + E + F = 0, E + F + G = 0,
and so on. Resolving these equations, we have
A=l, B = — 1, C = — 1, D=4, E = — 4, F = 0,
and so on. Whence the development is
1 —x — x^ + ix'' — 4x* + Ox^ + 4x^ — ....
as may be easily verified by Si/nthetic Division.
These questions, and others of similar kinds, admit, however, of easier solu-
tion by other methods, and they are only instanced here for showing the nature
of the operations to be performed, whilst their results admit of verification by
INDETERMINATE COEFFICIENTS. 239
those easier methods. The following is one of a class to which this method is
the only one that can be applied without great labour ; and it is one of per-
petual occurrence in integration.
^ ^ „ «"-' + oa?'-2 + kx + I , , , .
Ex. 3. Given _. , — p- -. ^--r> where the denommator is re-
af + ttiX"^ + .... k,x + li
solvable into factors of the first, second, third, &c. degrees : to transform it into
a series of partial fractions, each of which shall be one of those factors in suc-
cession.
As the method will be quite as well seen by a numerical example^ it will be
suflScient to so transform —. :, — .
a;-* — x' — -Ix
Here the denominator is x(x + 1) (x — 2) we may assume its partial fractions
to be
A ^B C
x^ — x^ — 2x~ X + 1 ^ X ^ X — 2'
or reducing to a common denominator,
_ (A + B + C) ar- — (2A + B — C) a? — 2B
~~ x(x + 1) Cx — 2)
Whence, equating the homologous coefficients of the numerator, we get
A + B + C = 0]
2A + B - C = 0 Ur, A = i B = - i C = '
-2B = lJ ^ 2 0
and hence -r ^ — = — — , — -r — r- +
a^ — a;"'^ — 2a; 3(x +1) 2a? ^ 6{x — 2)'
which may be easily verified by reducing to a common denominator.
It is necessary to remark, that the power of x in the assumed numerators
must in all cases be one degree lower than in the denominator.
EXAMPLES FOH EXERCISE.
4. Expand ,, } ,, or ,— — - — r— » in a series according to positive, and
^ {I -\- xy 1 + 2ar + a?*
then to negative, powers of x.
J 1 — ^ J 1 — ar'
5. Expand ,-— — , and — ;— ;-— rr
*^ 1 + a; 1 + ar
^ „ , 2ab a a' j o*
6. Expand — — r, ■ — , i — -rj. and -.
^ a + o I — a a* — o^ a — x
7. Extract the square root of 4 — 6a; + 5x* — 9x^.
8. Expand V^^^. Vl+2x -H 3ar«, ^2 - 3a; + Sx^ and s/l+x + x' + x'.
9. ResolTO into partial fractions the following expressions :
lgS_gx* + !l3 — 3X+5
a;« + 4a^ — Sx* — 25xr^ + 35a;* + 21x — 28
and
3x* — 9a^ + 6a:2 _f. 4a. _ 20
xs _ Qx'' — 12a;« +l.Mx^ — 289a;* + 480a;* — 660ar^ — 608a; + 960
the denominators being the same as those in Examples 2 and 4, p. 236.
The preceding simple applications of the method will suffice to show its cha-
racter ; and we pass on to some of its applications in the investigation of general
theorems for summation and expansion.
240
ALGEBRA.
PILING OF BALLS.
The usual forms of piles of balls are the triangular, the square, and the
rectangular, and they take their names from the figure of their lowest courses.
In all cases the successive courses have one ball less in each of their sides than
the one upon which they respectively rest; the highest course being in the
triangular and square piles a single ball, and in the rectangular, a line of balls.
A pile is said to be incomplete or broken when it has either not been finished,
or when some of the balls have been removed from a finished pile. The follow-
ing figures represent the three kinds, as named below them respectively. The
number of courses, therefore, is the same as the number of balls in the shortest
side of the lowest course.
Rectangular pile.
Square pile.
Trian. pile.
1. To find the number of balls in a triangular pile ofn courses.
Here we have C„ = — I- -
p _ n2 n
"-' ~ 2 ~ 2
c-.=t
3n
+ 1.
Hence, as there are ra courses expressible in the same manner, we may infer
that the highest power of n that enters the expression for the sum ±* be n^ Let,
then, S, = a + a_, + C_j + . . . C^ + C, = An' + Bn^ + Cb + D.
Then, substituting in this the first four values of n, we shall have
S, = A + B + C + D = 1, for the first course ;
Sj = 8A + 4B + 2C + D = 4, for the first two courses ;
53 = 27A + 9B + 3C + D = 10, for the first three courses.
54 = 64A + 16B + 4C + D = 20, for the first four courses.
Hence, resolving these equations, as at p. 180, we have
A = ^, B = 2. C = -J, and D = 0.
Whence S. = '^ + |' + " = "i^L±iH" + 2)^
which gives the sura of the n courses.
2. To find the number of balls in a square pile ofn courses.
In this we have
C, =«^
C,_, = n* — 2n + 1
C,_j — n^ — 4n + 4
for the ra courses : and for the same reason as before the form of the function
will be S, = An^ + Bn^ 4- C« + D. Substitute for n the successive values
1, 2, 3, 4, as in the last case : then
PILING OF BALLS. 241
S, = A4- B4-C + D= 1, for the first course;
S, = 8A + 43 4- 2C + D = 5, for the first two courses ;
S3 = 27A + 9B + 3C + D = 14, for the first three courses ;
S, = 64A + 16B + 4C + D = 30, for the first four courses.
From these equations we have A = -, B= -, C= -,D = 0; and therefore
5>. — g- + -^ + g = g , the number m a square pile.
3. TTie rectangular pile. Let n be the number of balls in the shorter side, and
n -\- m the number in the longer side of the lowest rectangular course.
Then it will be obvious, by reference to the structure of the pile, that it is
composed of a square pile of n courses, with m oblique triangular courses added
successively to one of the oblique faces. Hence the entire pile is
S, = square pile + m triangular courses.
- "(» + 1) (2« + 1) , w(« + 1)
6 + " • —2 —
_ n(n + 1) (3ot + 2n 4- 1)
_ ^ ,
which gives the balls in the rectangular pile.
It may be well to recollect that the ridge of the rectangular pile has m + 1
balls.
4. The incomplete pile. This will require the whole pile to be computed, and
then the partial pile removed, both by the appropriate formula for the kind of
pile. The difference is the number in the incomplete pile. Formulae, indeed,
may be given for all the cases in terms either of the number of courses taken off
or the courses left, and the original number of courses : but these would be
much more complex, and the implied numerical operations more laborious, than
the unreduced formula and the work which it involves. Such formulae, there-
fore, would be without utility, and from having little mathematical elegance,
would be destitute of sufficient interest to justify their introduction here.
Ex. 1. Find the number of balls in a triangular and in a square pile, each
composed of twelve courses.
Triangular pile. Square pile.
c 12 . 13 . 14 „^ £, 12. 13.25 .^^
S12 = g- = 364. Sj-i = — = 650.
Ex. 2. In a rectanfjular pile are IS courses, and the number in the ridge is
45. How many balls are there in the entire pile ?
Here m + 1 = 45, or m = 44, and n = 18. Hence by the formula,
S _^3- ^9. ^3.44 + 2. 18 + 1|^ 18. 19. I69_gg33
18 g 6 '
Ex. 3. Of the preceding rectangular pile there are to be taken away 1031
balls; how many complete courses must be removed ?
Let X be the number : then
ir(a: +'l) (3.44 + 2x -f- 1) ,^^, , , .
-i — ^ — — — - — ■ — - = 1031, or by reduction,
2x^ + 135x2 + 133x = 6186.
The real root of this is between 6 and 7 ; and hence six complete combes,
together with 96 balls more, must be removed ; for
2 + 135 + 133 — 6186 |6^
12 + 882 6090
147 1015 —96
VOL. I. a
242 ALGEBRA.
EXAMPLES FOR EXERCISE.
1. How many more balls are there in a square pile of 15 courses than in a
triangular one ? Ans. 560.
2. A square pile which has as many courses as a triangular pile, contains half
as many more balls. How many balls were there in both ?
Ans. 4 courses in each, and 50 balls in all.
3. The upper and lower courses of an incomplete square pile have 15 and 25
balls in each side. Find the number in the original pile, and the number left.
Other examples may be easily formed to suit the ability of the pupil.
THE BINOMI.\L THEOREM.
This theorem affirms that every expression of the form (a + ^T can be
developed in a series of positive integer powers of either a or x, and assigns the
coefficients of those powers of a or a" in the development. It does not, however,
affirm that this is the only form of development possible. The expression is
either
ia + xy = a' + -. a"-' x + '^ ^ a'-^ ar^ + -^ — p" 2 3 '^+ •••
or
(x + ay=x'-\- - x--' a + ^ ^ x"-' a- + - "j 2 3 — a^+. • •
according as we consider a or x the leading term of the given expression.
1. To Jind the Jirst term of the development.
Assume (a + it)" = A. + B,x + C.x^ + B^ + . . . . (a)
Then, using similar notation where 2n is written for n, we have
(a + x)^ = A„ + B^ + C,.x" + D^x^ + .... (i)
But (a + ^y = J(« + ^T]', in which, substituting from (a) and (b), we
get
A.2 + 2A,B^ +....= A,. + B^ + ... •
whence, equating the coefficients of x", we get A,- = Aj, : which is fulfilled by
A, = a", and the first term is found to be a" universally.
2. To find the second term of the development.
It will simplify the process to write the expression (a -}- a?)" = a" ■< 1 + - r
■= a'(l + r)" ; and it will evidently be sufficient to expand the compound factor,
and multiply every term of the expansion by a' to obtain the complete develop-
ment.
First, let n be a positive integer: then we may assume as before,
(1 + »)' = 1 + A,p + B,»- + C.p3 -I- . . . . (c).
For since a" p + A.v + ....? gives a' + o'A.f + , the first term of the
bracketted series being 1, fulfils the condition respecting the development de-
duced iu the former part of the investigation.
Divide, synthetically, equation (c) continually by 1 + v. then
BINOMIAL THEOREM. 245
(1 + p)-' = 1 + (A. - 1) r + B._, f^ + C._, v^+...
(1 + v)-^ = 1 + (A. - 2) p + B_, t>2 + C_, r^ ^ . . .
(1 + p)-3 =l + (.A. — 3)v+B^,v^+C^,v' + ...
(1 4- p)"— = 1 + (A. - m) p + B^ r' + C_. tr" + . . . .
and if we take Tn = n — 1 , we have
1 + p = 1 + {A. — (n — 1)] p + B, p- + C, p3 +
Equating the coefficients of the homologous powers of v, we have for those of
the first powers, or p^,
1 ^ A, — (n — 1), or A, = n.
When, therefore, n is a positive integer. A, = n, and the coefficient of the
second term is found.
Secondly, let n be a negative integer; or the expression to be developed be
(1 -f v)~" . This is the same thing as
„ ! ,.= rx ^p\, =1 — np + B_.p2+C_r«+....
(1 -f- p)" 1 -j- np + B, p'' + . . . .
by actual division. Whence in this case we have (1 + p)"" = 1 — nv +
and the coefficient of the second term is found.
Thirdly, let n be a fraction denoted by + -. Then, as before, assume that
(1 + P)-» = 1 + A. p + B, p2 + C, p3 + . . . .
liaise both sides to the 9th power : then we have
1 ±iJr+B, p2 + . . . = I + 5P (A.+B, p+. . .) + B, p2 (A.+ B. p+ ..)»+... .
and equating the homologous coefficients of p, we have for that of the first power,
P
+ /J = q\n, or A, = + ^.
Whence in this case also the coefficient of the second term is found : and in all
cases it is equal to the index of the power.
3. To find the third and subsequent terms of the development.
Put v = y + z: then we have, denoting the coefficient of (y + ^Y by A,
throughout,
(1 +y + 2)-=: 1 +niy+z)-\-A^ (y+2)2+A, (.y+2f+ . . . A^-, (y+«)'+'+ . • •
But we have also
(ii+y + zr = {i+yy{i + p~}
= (i+y) {1 + rr-y + (TtjO^^+ ••• CTT^' "^ -)•
Now by indeterminate coefficients, the coefficients of each separate power of *
in these two developments are equal. Equate them for r' : then
n + 2A,y + 3A3y- + ...(r+l)A, + ,j^ + ... = (l+y)'. ^ "^ ^;
or multiplying out, and changing sides,
n(H-y)'=(l +y) Jn-|-2A,y + 3A3y2 + ...Cr+l) A, + ,y'+ ....|;
or again reducing, the two sides of the equation become respectively
n[l + ny + A.,y^+ A,y' + ...| and
n+(2A,+»0y-l-(3A3+2Aj) y2+(4A4 + 3A,) y'^ . . . . {(r+l) A, + ,+rA,Jy'+. . .
r2
244 ALGEBRA.
Again, equating the homologous coefficients in these two expansions, we get,
generally,
mA, = rAr + (r + 1) A,+ „ or (r + 1) A,+ , = (w — r) A,.
Substituting in this the successive values of r, we have
rA,= {n-(r-l)}A,_,
(r-l)A_, = {n-(r-2)}A._,
(r — 2) A,_, = {n - (r - 3)} A._3
3A3 = (n — 2) A,
2Aj=(n— 1)A,
lA, = nAo,
where, by the preceding investigation of the second term, we have Ag = 1.
Multiply these columns vertically, and there results
1 . 2 . 3 rA, = n(n — 1) (n — 2) {n — (r — 1)} , or
_„(„_!) („_2) .... {n-ir-Vj}
^- 1.2.3.... r '
which is the general form for the coefficient of the r"" power of r ; and giving
to r the successive values, 1, 2, 3, ... . we have the corresponding coefficients,
n n(« — 1) n(n — 1) (n — 2) , . • j • .1. • ^- f
-, -^^ -, — — -, . . . . the forms assigned m the enunciation of
11.2 1.2.3' ^
the theorem. Inserting these, restoring the value of r, ^iz. -, and multiplying
all the terms by a", we have the formula as there given. Thus we obtain
(a + xy = a' + - a-'x + '^ ^ a'-^ x" + ^ ^ 3 0 ^ + • • •
for the expansion sought.
Similarly, since a — x ^ a + ( — x), we have in the expansion all the odd
n
powers of x negative ; that is, Ja + (— a:)]" = (a — a:)" = a" — - a'~^ x +
f^n^)^^, _ nin-l)in-2) „._3 ^ ^ . . .
1.2 1.2.3
4. When n is a positive integer, this series terminates with the (n + 1)""
term, and the coefficients of the terms reckoned from either extremity of the
series are equal. In all other cases the series will never terminate.
For the (n + 2)"" term contains the factor 0, and this factor entering into all
the succeeding terms, they also become 0. Whence the series terminates with
the (» + 1)"" term.
Again, the r"" term from the end is the (^n — r + 2)"" from the beginning ;
and therefore its coefficient is
p _ n(n — 1) (« — 2) r
1 .2 .3 ... (ra — r+ 1)'
and the coefficient of the r"" term from the beginning is
n(n- 1) (n — 2) .... (.n — r + 2)
1.2.3 (r— 1)
It, therefore, only remains to show that P = Q. Now we have very obviously
F__n{n—l) in — 2) r 1.2.3 (r— 1)
Q 1.2.3 ... Cn — r+ 1) ' n(n — 1) (n — 2) . . . (n — r + 2)
__ n(n— 1) (n — 2) ....r(r— l)(r — 2) ....3.2. I
1.2.3 („_r+ 1) (n — r + 2) (n — r + 3) ....(n— l)n
in which the numerator and denominator are manifestly equal. Hence P =: Q,
as above affirmed.
BINOMIAL THEOREM.
245
5. It still remains to point out the arithmetical forms which are most con-
venient in the practical application of this theorem.
Suppose, as an instance, we had to develop (a + a:)^'", the calculations will be
as follow :
Write the terms without coefficients a'" a^x aV aV with spaces
between to receive the coefficients and their signs. Then when,
n = 10 n = — 10
n — 1= 9 n— 1 = — 11
290
45 ..
2 110
55
n — 2 = 8
n — 2 =
-- — 12
3 360
3
— 660
120 ..
— 220
: — 13
n — 3 = 7
n — 3 =
_n(n-l)
1.2
_n(n— l)(n— 2)
~ 1.2.3
4 840
210
4 2860
715 ^n(n-l)(n-2)(n-3)
1.2.3.4 '
and so on to the required extent : the literal parts of this throughout being
merely explanatory, and need not be put down in actual working. The process
is simply multiplying by the decreasing series, and dividing by the increasing
one alternately. Each successive quotient is the successive coefficient of the
series, which inserted in its place, gives the expansion sought.
This vertical alineation is not, however, convenient when n is a fraction, a
horizontal one being much preferable. Thus, to expand (1 + r) ^, we may con-
tinually work as follows :
1st coefF.
1
2
1*. —
2nd coeflF.
1* _ 3
2 ' 2
3rd coeff.
3* _ 5
8 ■ 2
4th coeff.
5* _7
16 * 2
5th coeff.
128 '
12 3 4
where the asterisks are placed at the several successive coefficients. The appa-
rent continuity of equality may be, were it necessary, cut off, by drawing vertical
lines after each sign of equality that is to be destroyed by the next operation, as
in the example above. This cutting off is better than crossing out : but neither
of them is absolutely necessary.
EXAMPLES IN POSITIVE INTEGER POWERS.
Ex. 1. Raise a — a; to the 10th power.
Ex. 2. Find the sixth power of a — x.
Ex. 3. Find the fourth power of a — x.
Ex. 4. Involve a — a? to the ninth, and a + 6 — c to the fourth power.
Ex. 5. Find the cube of ^/a — ^b.
Ex. 6. Find the fifth term of (3w — 2a?)'.
Ex. 7. Raise — a — b to the fourth power.
Ex. 8. Find the fifth power of — a — b.
246 ALGEBRA.
EXAMPLES OF NEGATIVE AND FRACTIONAL POWERS.
Ex. 1. Extract the square root of a- + 6^ in a series ; or evolve (a" + 6*)*.
,62 b* be 5^,8
Ans.a-f ^-g^3+Y6^-I^^7+ •'•
Ex. 2. Show that , ^ ,,= i + ?5 + ^ + ....
(a — x)- a} a? a*^
Ex. 3 *. Also that ~ — = a + a? + - + ^+....
a — X a a'
x' . A 1 1- . /I 1 1 1 ar 3x* 5x^ ,
Ex. 4. And that ^ /-^ -„ or ^-s- — ^.i = :r-,+ .z—.— 7^—7 + • • •
TT r: rnu • f a^ . , , 26 , 3*2 463 5J4
Ex. 5. llie expansion of jr, is 1 H 5- H 5- H 7- + . . .
(a — 6)2 a (V- a^ a*
Ex. 6. And that of ^/a- — x^ is a — „—_ — -^ . — ,-zw^7 — •• ■
2a 8a' l6a' 128a'
Ex. 7. Show that (a? - ¥f = « - — „ — -r - ^ - . . . .
^ 3a- 9a' 81a^
XT' 2x1° gjjlo
Ea;. 8. The expansion of ^/a^ + a?' is o + --, — — — r +
5a^ 2ba^ 125a"
^a;. 9. And that of — -— , isl — :r t- +
a + 6 a a^ a^
Ex.10. The cuhe root of -^^, is 1 _ ^, + ^^ _ il^^ + .. .
a' + 63 3a^ 9a« Sla^
Ex. 11. Expand v a + 'v/6 — c + \/a — s/b — c.
. „Jf, d^ lOd* 154d« 935ti« > , ,T ,
Ans. 2a ■< 1 — -— — ^ ^ . — ^ . , . -^ x — ... f , where v6— c=o.
I 9a* 243a* 656la« 590490^ S ' ^
Ex 12. Assign the first eleven coefficients of (1 + zY *".
JBa;. 13. Expand ; — , — 3i> and into series: and find the sum
^ 1 + a; 1 a + a: ^/- 1
and difference of — ; — ; — tt and ~, in their expanded states.
^a + v6 va — ,/6 ^
Ex. 14. It is required to find the square and the cube of the expression
^—x — y sj^^ — V^— a? + y a/^^.
Scholium.
The binomial development may be employed also in the extraction of roots of
numbers, and sometimes with considerable advantage. It is especially the case
when the number whose root is to be extracted does not diflfer greatly from the
same power of some whole number, as in such cases the convergency is so rapid
as to give eight or ten figures of the root true, by means of three or four terms
of the development.
• Examples 3, 5, 9, may be verified by Synthetic Division ; and in all cases where it can be
applied, and the result merely is required, this method is rather simpler than the binomial
theorem.
Hence (^1 - 33^ _ 1 _ ^^-^3- .- .-.^- _ -^-^
EXPONENTIAL THEOREM. 0J.7
Let us, for instance, seek the 11th root of 2044. Here 2" = 2048, and hence
we have
2044 = 2048 — 4 = 2048 ( 1 ~ )= 2" ( 1 — -, )
\ 2048/ \ 8V
the 11 th root of which is 2 x ( 1 — -3 ) .
» n— 1 1 10 21 32 , . c .,
The values of -, -^, ... are -^, ~ ^' — g^* — 74' • • • ad infinitum.
10 21 1 _
;1 22*33'89
By calculating only the terms here written down, we obtain ^1^2044 ^
1 •999644570706, true, probably, within one unit in the last figure.
When, however, the quantity r is nearly = 1, or indeed above '5, the con-
vergency becomes very slow ; as it is obvious that the successive factors of the
several coefficients continually increase and approach towards unity as their
common limit, in all the roots. The convergency depends then on the smallness
of V, which causes its powers to diminish rapidly in value.
This difficulty, however, may be completely evaded by taking two figures
instead of one for the first approximation. It will be well to take 4, 5, 6, or 7»
for the second figure, according as the first taken with it shall form a number
decomposable into factors never greater than 12, and such as shall be supposed
most likely to approximate closely to the true root. Thus, for instance, the
sqiiare root of 6 gives, whilst we take only one figure, 2 (1 + 2)^, or 3 (1 — J)',
which would converge slowly : but if we take two figures, as 2*5, we have
25 (1 — 4)^, which converges very rapidly. And, in all cases, the binomial
theorem enables us to secure this rapid convergency *.
Ex. Let the student extract the roots which follow by this method. \/7 i V9i
»y]7 ; V'246 ; and calculate to six decimals the values of the following binomial
surds : 3 + 'V9 ; J — V"006564 ; and ^/ — 1 x V 16.
THE EXPONENTIAL THEOREM.
The expression a' takes the form of a simple term : but it is of great import-
ance to develop it in a series proceeding according to powers of x, as in the last
case. That is, to find the coefficients of the series in
a- = Ao + A.a: + A^^ ^ ^^3 + .,.. (1)
Assuming this particular form, we have also
ay = A, + A,y + Aj/--\-A.y + .... (2)
But a'a" = a'+» ; and
ar + y = Ao + A,ix + y) + A,{x + y)- + AjCx + y?+ ... (3)
Multiply (1) (2) together, and equate it to (3) : then equating the coeflScients
of y' in the result, we have
A.{Ao + A,x + A^ + A3X3 + . . .]= A, + 2A^ + -SA^ + 4A,«i + . . .
Again equating the homologous coefficients of x in this, we have the results
* This is the method most commonly employed by foreign mathematicians for approximating
to the roots of numbers, when more figures are required thaa can be obtained by the logarithmic
tables. Bourdon, Alyebre, p. 290.
248 ALGEBRA.
A,Ao = A„ or Ao = 1
2A,= A,^orA, = ^
3A3 = AjAg, or A3 = — ^ := - - -
A A A ■*
4A, = A.A3.orA, = ^^=:- ^'
2.3.4
mA„ = A,A„_,, or A„ = ■ ' " ' = , — - — .
' m 1.2 m
Whence, omitting the subscribed accent from A,, the development is
, Aa; , AV , A^a^ , AV
a' = 1 + — • H h • -1 ■ +
It Still remains to determine the value of A in terms of a, which may be thus
effected.
Put a'=|l + (a — 1)|'; then expanding the binomial, we have
1 1 ^ /• ,^ I a?(« — 1)/ ,N2 , a-C*— !)(« — 2)
a'=l + £(« — !) + -j-^ (a — 1)^ + TTTTs (« — 1^ + ••••
Now the coefficients of x^ in the several terms are as follows :
+ (a — 1) in the second terra,
— i (a — 1)2 third ,
+ lia—lf fourth . . .,
— iia — iy fifth .... ,
Whence A = (a — 1) — i (a — 1)^ + J (« — 1)' — i (« — 1)* + • •• •
For the purposes of calculation, this expression is generally useless, on account
of its want of convergency. As an analytical expression, however, it is an
essential element in the deduction of the formula for logarithms ; and the neces-
sity of its calculation here is avoided by taking a different subsequent course.
LOGARITHMS.
I. DEFINITIONS AND ELEMENTARY PROPERTIES.
In the equation a' = N, a is called the base of the system, N the number,
and X the logarithm of N to the base a. This is generally denoted by the
equation,
X = log„N, or a? = 1„N,
where the base of the system is written as a subscribed letter to the contractions
" log" or " 1" of the word logarithm. Logarithms are said to be of different
systems, according to the value of the base a.
As logarithms are, by the definition, only indices of the powers of the base a,
it will be obvious that the fundamental operations will be the same as those of
indices already explained. It will, nevertheless, be advantageous to collect into
one place, and with appropriate phraseology, the simple propositions relative to
these indices which we shall have occasion to employ. They are, in fact, the
rules for the use of logarithms ; and the only difficulty in the inquiry is the
actual calculation of the logarithms themselves.
1. The sum of the logarithms of two numbers is equal to the logarithm of
their product.
LOGARITHMS. 240
For let (f = N, and a" = N, : then a'a? — (f + i = N N,, or * + y =
log. N N..
2. The difference of the logarithms of two numbers ia equal to the logarithm
of their quotient.
Let a* = N, and a' = N,. Then — = a'-' = ^ ; or x — y = log. 4^ •
a' N, ' *• N,
3. The logarithm of the nth power of any number is equal to n times the
logarithm of that number.
Let a' = N ; then o" = N", or nx ^ log, N*.
4. The logarithm of the rath root of any number is the nth part of the
logarithm of that number.
- i a; ^
Let a' = N ; then a' = N" ; or - = log. N".
5. If a series of numbers be taken in geometrical progression, their logarithms
are in arithmetical progression.
For any number may be represented by a". Let a" be the first term of the
geometrical series, and a' the ratio : then the series are
For the numbers a", a""*"", a'" + ^, a" + **,
and for the logs, m, m -\- n, m + 2n, m + 3n . . . .
and it is obvious that these logarithms are in arithmetical progression, whatever
the base of the system may be.
6. The logarithm of the base in every system is I.
For a' = a, or log. a = I.
7. The logarithm of 1 in every system is 0.
For a9 = 1, or log. 1 = 0.
8. If a table of logarithms be calculated to any one system, those for another
given system can be obtained from these by the use of a constant multiplier for
all the logarithms of the first table.
For let a'* = Oi: then a^' = a^' = N.
Whence, taking log of N in both systems, we have log, N = x, and log,N
= Mx, where M depends upon the bases a, and a, and is constant for all values
of X, so long as the systems remain the same.
It will therefore follow, that if we can more easily compute logarithms to one
base a,, than to any other a, we may avail ourselves of it, and convert them to
another system by means of the proper multiplier M.
9. As a general mode of finding M., we have, from the last equation,
• log. N-
I
Whence, if we can compute the logs, of any one number N in the two systems,
we can obtain the requisite multiplier for all the other transformations.
The number M is called the modulus of the system of logarithms ; and refer-
ing to the base a, it is written M., signifying the modulus to the base a.
10. If a, b be the bases of two systems, and N, N, any numbers whatever :
. log. N log* X
then , " „ = -. v^.
log. N, log» Xi
For let e be the base whose modulus is unity : then we have
log.N = M. log. N, log. N, = M. log. X„ log .N = M» logj^,
and logjN, = Mj log, N,.
log, N _ M. log. N _ log. N _ M> log. N _ log, N
Hence also ^^^^ ^-^ - ^^ j^^^ j^- - j^^^ j^^ - ^,^ ,^^^ ^^,^ - j^^^ ^.
250 ALGEBRA.
II. LOGARITHMIC SERIES.
In the equation a* =: N, to find an expression for the value of x io terms of a
and N.
Raise both sides to the zth power ; then we have a" = N'. Develop both
sides by the exponential theorem : then we obtain
in which
A=(a— 1)— i(a— Ij'+Ka-l)'— ••■;andA,=(N-l)-i(N-nHK>'— 1)'— -
Equating the homologous coefficients of the indeterminate quantity z, we have
from any one of the resulting equations, as that of z^, for instance,
A. = A.,or. = A' = (^-^^-*^^-^^^ + ^^^-^^^---
A (a— l)-i(a — ])2 + ^(a— If— ...
which is an expression for x, the logarithm of N to the base a.
It is more usual to write n instead of N — 1, and M„ instead of
; — — ; r:^—; — r^ and this reduccs the expression to
(a _ 1) _ i (a — 1)- + Ha — 1)^ — • • .
X = log. (1 + n) = M, ^n — 4 71= + J n3 — J n* + ]
This series is not in a form well adapted for calculation, except when n is a
small fraction. The following process will transform it into another adapted to
any number whatever.
Substitute — n for n, and write the two equations,
log. (I + n) = M4+ n — * n2 + J n^ — i n^ + . . . . |
log. (1 — n) = M.^— n — I n? — i n' — i- n* — ]
Hence by subtraction,
log.(l+n)-log.(l— n) = log. [±^=2M.Jn + ^n5 + in5+ ....]
X 1 il+n/' + lii
Let now n = - — -— : then , = ■' , and we have
2p + \ 1 — n p
or finally, log.(;.+ 1) =lor.p + 2M.{^j + ^^^+ ^^^^\-^, + ...}
Hence, whenever we can calculate log./), we can, by means of this series,
calculate log.(j»-|-l) ; and the series converges the more rapidly as p becomes
greater.*
* Several improvements of this formula, at least in respect of practical application, have been
proposed by different \vriters; but as the tables have already been computed and verified, they
arc, in this ])oint of view, of little importance. Nevertheless, it may not be out of place to
merely indicate one or two of them, referring for more ample details to the eleg^ant little treatise
of Professor Young, on the "Computation of Logarithms," second edition, 183.5.
). Putn =. -- —'- : then ~~ =- — — ^, and we have log ( n-|-2) = log (;)-2)
wliich is Borda's theorem, and essentially the same as Leslie's. 2. Put
LOGARITHMS. 251
III. ON THE COMPUTATION OF LOGARITHMS.
1. To find the value & of a. which will render M, = 1 ;= . , or A ^ 1.
A.
In the exponential e* where A= 1, we have - := I = M„ and A = 1.
Whence, whatever x may be, we shall have
X oc^ a?
Whencee' = l + ;+^+~^-3 + ....
Computing this series to thirteen terms, we have e =■ 2718281828
Logarithms calculated for the base e, or modulus 1, are called napierean,
from their inventor Lord Napier. They are also often called hyperbolic
logarithms, from an analogy which exists between them and the spaces con-
tained by the rectangular hyperbola and its asymptotes. Under the latter name
they are given in Hutton's Tables.
2. To calculate the hyperbolic logarithms.
The general series for the logarithm of p + 1 is, in this case.
Log. (p + 1) = log. ;, + 2 [^~ + 3-(~|rr)3 + 5(^^Vl? +••••}
Now, (theor. 7, p- 249) log. 1=0, and hence
Log.2= 0 + 2 {^- + ^3- + ^, + .... }= -6931472
Log. 3 = -6931472 + 2 J J + ^53 + 5 ^5 + ..••}= 10986123
Log. 4 = 10986123 + 2 {^ + ~j^ + ^-y. + ....}= 1-3862944
Log. 10 = 21972246 + 2 {1 + ^-1^3+ _!_+....}= 2-3025851
And in the same way, the series may be continued to any extent required.
But since (oO" = «" (O" = a', oT a* = (f + », and — = o-»
the logarithms of numbers which are either products or quotients, powers or
2, Put — for p in the equation of the text ; and then if we make m =: j^ — 250^ and m-\-n
= a^ — 25a^ + 144, we get log (x-\-b) = log (x+3) + log (*-3) + log (*+4) + log (*_4)
_ log (^-5) - 2 log a-- 2 M { a4_2L»+ 72 + 3 {x*--26^'+7-2) + } ''^^^ '" ^''°
formula of Haros.
3. Put^ = -]-t^; then we get n =^^^. Put also /) = a« — 9ar* + 2401a^, and 0 = 46 _
<l 1— « ° p-tq '^ • ' /
9ar^ +2401a2 — 14400; from which is obtained the formula of Lavemide, 2 log x + 2 log
(^•+7) + 2 log (a'— 7) — log (^+8) _ log (a^-8) - log (j+5) — log (f— 5) — log (J-+3) — log
( ia_9mJ 7-300 1/ 7200 y ^
V^— oj_-i>i \ ^.6 _ 9ai4 _^ 240U> - 7200 "^ 3 Vj« _ D&l* + -2401*"— 72007 "•" J
252 ALGEBRA.
roots, of numbers whose logarithms are already computed, may be obtained by
much simpler means. Thus since
g 69314/2 _ 2, and e^°^^^^^ = 3, we have gS^^i^rs x e'<»s6i23 = g = 517917595^
or the log of 6 is obtained by adding together the logarithms of 2 and 3. Hence,
instead of computing the logarithms of 4, 6, 8, 9, and 10, by the series above,
they may be computed by simple addition, or by doubling, tripling, &c. when
the number is the square, cube, &c. of a number whose logarithm has been
already computed ; or by subtraction or division by two, three, four, &c. in the
converse cases.
Thus, log 4 =2 log 2.
log 6 = log 2 + log 3.
log 8 =3 log 2, or log 4 + log 2.
log 9 =2 log 3.
log 10 = log 2 + log 5.
3. To compute the logarithms to any other base, as where a = 10.
HereM.o=^. Put N = 10 : then M.„ = 1^-^ = -—i-—
log, ^ log. 10 2-3025851
•43429448.
Hence Mjo log. 2 = log.o 2 = -43429448 X -6931472 = -3010300
M,o log. 3 = log.o 3 = -43429448 X 1 0986123 = -4771213
M,o log. 4 = logjo 4 = -43429448 X 1-3862914 = 6020600
Miolog, 10 = logio 10 = -43429448 X 2-3025851 = 1-0000000,
That is, 10" =1
2Q-3010300 __ 2
10<"iii3 _ 3
10* = 10
IV. ON TABLES OF LOGARITHMS.
To render logarithms really useful in computation, we must have them
registered in tables. To compress them into the smallest possible space, and at
the same time render them convenient for use, several contrivances have been
adopted. The editors of such tables differ in the minutiae of their arrangements ;
but the general principles of their construction are alike in all. Those in most
general use are Dr. Hutton's, and hence this description will have reference
especially to the last edition of that work. The great accuracy, too, of these
tables, independent of the convenience of their arrangement, is a strong reason
for this choice.
1. Definitions.
(1.) The significant figure of a number N is the figure which stands highest in
the numerical scale. Thus 5 is the significant figure of 54*69, of 5-469j of -5469,
of -05469, of -005469, &c.
LOGARITHMS. 253
(2). The distance of the significant figure is the number of places in the decimal
scale which it is distant from the units' figure : it is considered positive when to
the left, and negative when to the right of the units' place. Thus in 5469, the
significant figure is -f 3, being in the third place to the left of the units' place
9 : in 005469 it is — 2, (or as more conveniently written, 2) being in the
second place to the right of the units' place 0-.
(3). The integer part of a logarithm is called the characteristic or index of that
logarithm.
(4). The decimal part, which is always positive, is called the mantissa of the
logarithm.
(5). The arithmetical complement of a logarithm is its defect from 10.
2. Tabular theorems.
(1). The removal of the unit-place, whilst the effective figures composing N
remain the same, will alter the characteristic but not the mantissa.
For let logio N = m + d, m being any integer number whatever, positive or
negative, and d the decimal part, always positive. Then the removal of the
decimal point in N, p places will be the same as multiplying N by lO', where p is
positive if the removal of the unit-place be to the right, and negative if to the
left. In this case we have
log.o 10"N = log.o lO' 4- log N =p + log N = j9 -f m -f rf;
and since ^ is an integer, p -\- mis an integer. Whence the decimal d, or man-
tissa, remains the same, whilst the characteristic is increased or diminished by
p, according as ^ is + or — .
This, of course, is to be understood in reference to the fourth definition ; that
the mantissa is always to be taken positive.
(2). The characteristic of a logarithm is that number which expresses the dis-
tance of the significant figure from the unit-place.
Let the number N lie between 10' + ' and lO'; then its logarithm lies between
p + 1 and p ; or it is p -\- decimal. But the number N is composed of ^ -|- 1
places, or its significant figure is p places to the left of the unit-place.
If p be negative, then the number lies between 10"'"'"' and lO"'; and hence
the logarithm is — p + decimal. But the number lO"' commences in the^th
decimal place, and hence p places to the right of the unit-place.
(3). When we have to subtract a logarithm from another, we may add its
arithmetical complement and subtract 10 from the sum.
For log a — log * = log a -I- (10 — log b) — 10.
(4). When m is very small with respect to N, we shall have, very nearly,
log, {l -f ^,}=mlog. {l +;^}.
For developing these we have
r . "il »» ("» m^ , m^ )
log. |l I- ^\ = M.{j^- ^ +-3j^3 -••••}
"•log. {l + f}=^-{n-2^^+§M^-----}
Now since -, is a very small fraction, the succeeding terms ^ , -rp-. ...will
be still smaller, and have their significant figures further and further removed
from the unit-place : and if they be so taken that the significant figure fall more
254. ALGEBRA.
remotely from that place than the extent to which we calculate our table, these
terms may be rejected as insensible. In this case we have
log. {l + ^.} = M..^, and m log, {l + ^} = M.. L.
and since the second terms are equal, the first are so, or
log.{l+5}=-log.{l+^}.
3. Description of the Tables.
1. The hyperbolic or napierean logarithms are given for numbers from 101 to
10 to a mantissa of seven places with the proper characteristics, in Table V.
pp. 219 — 223, Hutton's Tables. Then in Table VI. are given those from 1 to
1200 for every unit.
The number is given in the column headed " N," and the corresponding
logarithm in the adjacent column headed " Logar."
These logarithms are only used in calculating integrals : those used for all
other purposes being to the base a = 10.
2. The common, or Briggs's logarithms, are given, characteristics and mantissas,
for the numbers from 1 to 100 in Table 1, p. 2. These occupy the first two
pairs of columns.
The numbers from 100 to 999 occupy to the bottom of p. 5 ; but the charac-
teristics are not inserted, they being always determinable by inspection, theorems
(1), (2).
The mantissas of the logarithms of all numbers composed of four places follow
these, forming the columns headed "0," from p. 6 to 185; and tabulated as
the last, without the characteristics.
The three leading figures of the mantissae are omitted from those of all the
logarithms after the first in which they occur, and the places they would occupy
left blank. These spaces are, therefore, to be understood as occupied by the
three figures which occur above them. Thus, mantissa of log. 1091 (p. 7) is to
be read •03/8248.
The mantissae of those for numbers of^pc places, as far as 10799, are given in
the same manner from p. 186 to 201.
The mantissae of the logarithms of all numbers of five places are given on the
same pages as those of four. This method has been adopted on account of the
three leading figures of these mantissae being the same for several succeeding
logarithms ; and thereby rendering it only requisite to repeat in the table the
last four.
Thus mantissa of log. 10500 is -0211893,
10501 is -0212307,
10502 is -0212720,
and so on. Hence the first four figures 1050 are given on the left margin, and
the mantissa of its logarithm (which is the same with that of 10500, theor. 1) is
given in the column under the fifth figure 0 at the head. The first four figures
of the mantissa of 10501 being the same, and the mantissa itself being the same
in the first three places, these figures -021 are taken from the first column
headed "0," and the remaining four, viz. 2307, from that headed " 1" : thus
giving mantissa of log. 10501 equal to 0212307, as above.
Whenever there is a change of the third figure of the mantissa in any of the
columns not headed " 0", the circumstance is indicated by a small line drawn
LOGARITHMS. 055
over the fourth figure of that mantissa, and in this case the first three figure*
will be taken from column 0 in the horizontal line immediately below. 'ITius
mant. log. 10544 is 0230054, and not 0220054. In like manner, mant. log.
10545 is 0230466, and so on to mant. log. 10549 : and at 10550 the first three
figures take their regular position in the horizontal lines.
The small tables on the right hand of the page, marked " Dif. and pro. pts.,"
enable us to obtain the mantissae of logarithms to numbers of six places of
figures. They are constructed on the principle of theorem 4, p. 253.
It will be seen, that the differences between the logarithms of two consecutive
numbers of five places vary very slowly, or are nearly the same for several num-
bers together. It amounts to this, that of several consecutive (and therefore in
arithmetical progression) numbers of five figures, the logarithms are also nearly
in arithmetical progression ; and hence also must the logarithms of numbers in
arithmetical progression, and lying between any two consecutive numbers, be
also in arithmetical progression But to appeal to theorem 4, we have
logio (N + m) = log.o N + log.o yi + ]JJ) = logio N + M,o . ^,
or logio (N + m) — log,o N = M,o- ^^ = m - ~^°.
The tablets referred to are composed of the values of this expression for dif-
ferent values of N and m, in the manner of the following example.
Let N = 10000, and m = 1. Then M,o = -43429448.
Hence log.,o 10001 — log.i, 10000 = -000043429
or to seven places taking the nearest number, it is -0000434, the eflfective figures
of which are put down at the head of the tablet-column at p. 6 of the tables.
In the same manner are the headings of all the tablets calculable. They were,
however, not found in this manner; but by subtracting the calculated loga-
rithms, that of each number from that of its next higher consecutive number dif-
fering by 1 in the unit-place.
The parts against the numbers 1, 2, 3 .... 9, in the tablets are the values of
]og,„ (N + m')— logio N for the several values -1, -2, -3, . . . -9, or for the num-
bers 1, 2, 3, in the number whose sixth figures are 1, 2, .... 9 in the unit-place.
These are called proportional parts of the logarithm for the sixth figure, and are
inserted for the purpose of being taken out by inspection, instead of having to
compute them in each individual case. These corrections are additive to the loga-
rithm if taken to the first five figures of N, and subtraclive if to the first five figures
of N -f 1. The former is the most convenient, and most generally adopted.
Thus, to recur to the first tablet, and dropping the ciphers, we have
1
434 X "1 = 43, givmg
434 X -2 = 87
434 X '3 = 130
43,
87,
130, and so on.
As for the smaller values of N to sLx places, especially under 107999, the
values of =^ . Mjo vary more rapidly than in other higher numbers of the same
local extent, the mantissa for six places have been given in the tables from
p. 186 to 201, with their tablets of proportional parts for the seventh figure in
the unit-place. These are to be used where great accuracy of approximation is
sought, (which, however, is rarely necessary) and their structure is the same as
already described.
Of the tables to twenty places, it is unnecessary to say anything here, as well
256 ALGEBRA.
as of those that follow : since they are well described in the introductory matter
of the volume.
4. The usage of the Tables.
The direct use, viz. taking out the logarithms of numbers, is mainly implied in
what has been said on the structure of the tables. It will, however, be well to
recapitulate briefly in a didactic form the processes.
1. The characteristic. Count how many places to the right or left of the
unit-place the significant figure stands. This number is the characteristic
(theor. 2, p. 253) ; and is marked tniniis if to the right, and considered ^/jw if to
the left,
II. The mantissa. 1. If the eflJective figures be not more than four, the man-
tissae of their logarithms will be found in juxta-position with them in the tables,
and may be taken out at once *.
2. If the effective figures be five, find the first four in the column marked X,
and the fifth in the horizontal line at the top. The last four figures of the man-
tissa are found at the angle formed by the horizontal and vertical lines in which
the first four and the fifth figures are situated, meet : and the first three figures
of the mantissa adjacent to the first four figures in the horizontal line, or in that
immediately below, according to the explanation already given.
Thus, to find the mantissa of log. 74695, look for 7469 (p. 135, Tables) in the
column X, and for 5 in the horizontal line at the top. We find at the angle of
the lines in which 7469 and 5 stand, the last four figures of the mantissa, viz.
2915, and adjacent (the blank expressing the number above) to it, the first
three, viz. 873. So that the mantissa is •8732915.
Or again, had we sought the logarithm of 74819, the last four figures are
0119 ; and the dash over the 0 signifying that instead of 873 we must take 874
from the line immediately below that which contains 7481, for the first three
figures of the mantissa. Hence the mantissa of 74819 is '8740119.
3. If the number be composed of six effective places of figures, find for the
first five as just directed. In the marginal tablet marked "pro," look for the
sixth figure, and place the adjacent number below the number already found :
add them together : then the sum is the mantissa of the six figures. This is
obvious from theor. 4, and from what is there said on the subject.
4. If the given number be a vulgar fraction or mixed number, the fractional
part may be reduced to a decimal, and the logarithm of the expression then
taken. But if the decimal be of many places, it will be better reduced to a
vulgar fraction, and the process adapted to division by logarithms followed.
Note. In actual practice, it is better to write down the first three figures
before looking out the remaining four; though, for convenience of explanation,
they have been spoken of in a reverse order.
The inverse use, that of finding the number when its logarithm is given, will,
obviously, be equally simple and easy.
5. If the mantissa appear exactly in the table, we have but to write it down,
and assign the unit-place in conformity with the rule already laid down accord-
ing to the characteristic.
• In tlie losrarithms of the numbers from 1 to 99, the characteristics are given also, on the
supposition of the number being entirely integer. When this is not the case, the characteriBtic
must be assigned according to the general rule.
LOGARITHMS. 257
6. If the mantissa be not found exactly in the table, take out that next less
than the given one. Write this under the given one, at least as far as the figures
are not common, and subtract it. With the difference thus found, enter the
second column of the tablet marked " Pro." and the number adjacent to it ia
the sixth figure of N, and the first five those belonging to the next lower man-
tissa. The unit-place is, as before, to be assigned from the characteristic.
Thus, if log N = 287301 58, and we require N ; take out the mantissa, and
write it under the given one, according to the following type.
"28730158
24 = mant. of 74647
34 = pp. 6 nearly, from the tablet ;
and we have 746476 for the number, so far as the mantissa is concerned ; and
from the characteristic we have the first figure in the second place to the right
of the unit-place. Hence the number sought is •076476 very nearly.
V. LOGARITHMIC OPERATIONS.
1. Multiplication by logarithms. Add the logarithms of all the factors
together, and the number whose logarithm is the sum will be the product.
Theor. 1, p. 249.
Note 1. When some or all the characteristics are negative, the rules of alge-
braic addition must be employed.
Note 2. When the given numbers are of six places, it will not be necessary to
do more to each of them in the way of correction separately, than to write the
corrections down beneath the corresponding logarithms of five places, and at
last add all the corrections and logarithms into one sum.
Ex. Find the product of -002356, 47-2985, -32986, 42-7579, and -00004965.
log -002356 = 3-3721753 p. 33, Tables.
log 47-2980 = 1-67484281 .^ - , x „ on
° ' „ r(to five places) p. 80.
prop, part for 5 = _ 47 J ^ '^ ' ^
log -32986 = T-5183297 P- 51.
log 42-7570 = 1-6310072 { ^^^ ^^
pp. for 9= 91^ ^
log -00004965 = 5-6959193 p. 85.
5-8922881 log of product.
39 = mantissa of 78034.
42 = pp. for 8 nearly.
Hence, by the characteristic, the- fifth decimal place is the significant figure,
and the number composed of the digits 780348, the product itself is 0000780348
nearly.
This example contains instances of every possible variety of case that can
occur.
EXAMPLES FOR EXERCISE.
1. Multiply 498-256 by 41-2467- . Ans. 20551-4.
2. Multiply 402674 by 0123456. Ans. 0497125.
3. Multiply 3-12567 by -02868 by -12379. Ans. -01109705.
4. Find the product 2876-9 X -10674 x '098762 x -0031598. Ans. -095830.
VOL. I. a
258 ALGEBRA.
2. Division by logarithms. Subtract the logarithm of the divisor from that of the
dividend : the remainder is the logarithm of the quotient. See Theor. 2, p. 249.
Note. In accordance with what is shown at p. 253, we may use the arith-
metical complement of the subtractive logarithms, which will often much facili-
tate the operation. To effect this, instead of taking the several figures of the
logarithm from the table, write (which can be, with little practice, done by
inspection) the complement of each figure of the logarithm from 9 except the
last, and the complement of this from 1 0.
For example, find the arithmetical complement of log. 37"5 and of •00375.
10-0000000
log 37-5 = 1-5740313
10-0000000
-00375 = 3-5740313
ac. log 37-5 = 8-4259687 ac. log -00375 = 12-4259687
in which, without writing down either of the lines, the arith. comp. may be
written down from the inspection of the logarithms themselves in the table.
Had there been six figures, the correction for the sixth might have been sub-
tracted from the result of the addition, as in example 1. which follows.
This method of work should be early and regularly practised, on account of
its almost constant occurrence in trigonometrical calculations.
When, however, there are several subtractive logarithms, it will be better for
the most part to add them into one sum, and place the arithmetical complement
of the whole under the column of additive ones, as in Ex. 2.
Ex. I. Divide 1728 95 by -110678,
Log 1728-9 = 3-2377699
pp. 5 = 126
ac. log -11067 = 10-9559701
4] 937526, subtracting 10
pp. 8 = — 317
4-1937209
088 = mantissa 15621
121 = pp. 4 nearly,
and characteristic 4 gives five places of integers : hence the quotient is 15621-4
nearly.
7? o 17- J *v. 1 r *i, • 3-1416 X 82 X 5?.
Jix. 2. tma the value of the expression -—.
^ -02912 X 751-3 X SIT
m,- • • vc A . ^ ■ 3-1416 X 82 X 73 X 941
This, in a simphfied state, is ^
^ ' -02912 X 751-3 X 6 X 41
log 3-1416 = 0-4971509
log 82 = 1-9138139
log 73 = 1-8633229
log 941 = 2-9735896
7' 2478773 = log numerator.
log -02912 = 2-4641914
log 751-3 = 2-8758134
log 6 = 0-7781513
log 41 = 1-6127839
37309400 = log denominator.
3-5169373 = differ. = log quotient.
LOGARITHMS.
S59
Or thus, and better, by the arithmetical complements ;
log 31416= 0-4971509
log 82= 1-9138139
log 73 = 1-8633229
log 941 = 2-9735896
ac. log -02912 = 11-5358086*
ac. log 751-3= 7-1241866
ac. log 6 = 9-2218487
ac. log 41 = 8-3872161
3-5169373 = log. quotient, as before.
18 = mant. 32880
55 = pp.
Hence the quotient is 328804 nearly.
4 nearly.
EXAMPLES FOH EXERCISE.
Ex. 3. Divide -06314 by -007241. Ans. 871979.
4. Divide -7438 by 12-9476. Ans. -057447.
5. Divide -102367 by 496523. Ans. •0-206168.
6. Divide -06314 X -7438 x -102367 by -007241 X 129476 X -496523,
and compare the result with the product 8-71979 X '057447 x
-0206168.
Ans. They ought to be identical, or within a unit in the last place.
7. Divide -0067859 by 123459. Ans. -0000000549648.
3. Proportion by logarithms. This is only the apphcation of logarithms to the
operations of multipUcation and division implied in finding the fourth term.
be
Thus, a a : b ', ' c : X, then x = — , and we have
a
log x = log b + log c + ac. log a — 10.
EXAMPLES FOR PRACTICE.
Ex. 1. If 12-678 : 14-065 : ; 100-979 : X, then x = 112027.
2. If 19864 : -4678 I '. 50*4567 : x, then x = 11-8826.
3. If -498621 : 2-9587 I '. 29587 : x, then x = 17'5562.
4. Involution by logarithms. In conformity with what is shown in theor. 3,
p. 249, we have log a" ^ n log a ; which gives the process :
Multiply the logarithm of the base by the index of the power : the product is
the log. of the power.
EXAMPLES.
Raise -09163 to the 4th power,
log -09163 = 2-9620377
index =: 4
5-S481508
460
48
Hence (09163)* = -0000704938.
Or, in a more illustrative form,
log -09163 = — 2 4- -9620377
4
— 8 + 3-848 1508
which shows the process more at length,
though obviously it need not be so put
down.
• Here the 11 comes from 9 — 2 = 9 + 2.
s2
260 ALGEBRA.
Ex. 2. Find the value of (r0045)3«.
log 1-0045 = -00194994 tables, p. 186
365 index
584982
1169964
974970
•71172810
29 = mant. 51490
52 = pp.
Hence the answer is 1-0045^^^ = 5-14906 nearly.
EXAMPLES FOR EXERCISE.
Expressions, 6-05987^ '1765463, •076543^ l-096S4^ — 1-06524^
Values 36-72203, -00550267, -0000343259, 1-90986, — 1-20877.
5. Evolution by logarithms. Divide the log. of the number by the index of
the root : the quotient is the log of the root : theor. 4, p. 249, where it is shown
1 1
that - log a = log a".
The only difficulty that can present itself is where the characteristic is nega-
tive, and not divisible by the index of the root. To remove this, add a negative
number to the characteristic sufficient to render it the next higher multiple of the
index, and add the same number taken positively to the positive part of the
logarithm, that is to the mantissa. Tlie quotient of the characteristic will in
this case be a negative integer, and the quotient of the positive part of the
expression will be decimal, and form the mantissa of the required logarithm.
The following example will illustrate this.
Find the cube root of -000486296.
log -00048629 = 4-6868953
pp. 6 = 53
3 4-6869006
2-8956335
Here 4-6869006 = — 4 + -6869006 = — 6 + 2-6869006;
and each of the terms divided by 3 gives
log root = — 2 + -8956335 = 2 8956335.
When the index is fractional, as a", we have
loga» = -logo;
and the process obvious. In fact, involution and evolution by logarithms are
the same rule, just as in common algebra under the same circumstances.
EXAMPLES FOR PRACTICE.
Expressions, 365-567^ 2-98763^ -967845*, -098674^ (^^^ Y,and (^il^¥.
\373/ \1727/
Values 19-1198, 1-44027, -9918624, -718315, -146895, and -1937115.
EXPONENTIAL EQUATIONS. 261
MISCELLANEOUS EXERCISES ON LOGARITHMS.
1. Find the values of 31416 x 82 x — , and -02912 x 7513 x — .
41 941
13
2 1
3. Find the square root of , and the cube root of
123 3-141
2. Find X in 7241 : 3-58 ; : 2046 : x, and in ^^724 : / -,
6-927
3-14159
^i (i\^
4. Find the values of ® '}}' , and { v'a X -012 V^.
173
5. Assign the values of ^sVtt X •03 V 15^ ^^^ j27* . ^-^/-^'
7^Vl2i X -19 V17J 141^-11
V35^
V28§-
6. Find the fifth root of -00065, the tenth root of — -001, and the third root
of — -00006 : £md show whether they be all real or not.
72'i 1
— f X 7^ and the 5th root of
3172 . 3^ . 5^
251
o r- J .1. 1 r /60\-' V — 3 X — 53
8. Find the value of - (^- j x ^jzr^~~ , and of ^ Vl X -034 ^/,V
9. Of how many figures does the number represented by 2^^ consist ? And
of how many does 9^ consist ?
10. Which is the greatest and which the least of the three numbers, 10'", 9",
or IP? And show how to determine generally which is the greater, a' or b',
supposing a greater than b.
11. Find the logarithm of 22-5, having given the logs of 2 and 3.
12. Having given the logs of 6 and 15 to find those of 8 and 9.
9 2
13. Given the logs of 2 and 3 to find those of . and — -,.
° 16 375
14. Given the logs of 2, 3, 13, to find those of (.^e)' ^"<^ Vl'625.
15. Given log, 15 = 27080502, log. 5 = 16094379, to find log, 25.
16. Given log. 2 = 6931472, log. 5 = 16094379, and log,o 1*9 = '2787536,
to find log. -0019.
THE SOLUTION OF EXPONENTIAL EQUATIONS.
An exponential equation is one in which the unknown appears in the form of an
exponent or index. When in this form, unmixed with other combinations, the
solution is readily obtained by means of logarithms. Thus, if c* = N, then
a loe a = log N, and x = -~ — , which is the general form of solution.
0 o ' log a
Moreover, it makes no analytical difference how complexly the base or the
index be given, the same method of solution applying to these as to the simpler
form given above, provided the combinations be by multiplication, division,
involution, or evolution only.
262 ALGEBRA.
When, however, the unknown appears as an index, and in any other cha-
racter, for instance, as a base, or as an addend ; or if two exponentials be con-
nected by other signs than those of multiplication or division : then the solution
can only be attained, in general, by means of trial and error, or some mode of
approximation tantamount to it, having previously simplified the expressions by
means of the logarithmic formulae.
The following examples will illustrate the processes to be employed.
Ex. 1. Given 2' = 769, to find x. Here we have, as in formula above,
_ log769_2-8859263 _
''- log 2 -0-3010300 -^'^^^^^'
Ex. 2. Find x from the equation cT" b" ■=^ c.
In this mx log a -\- nx log h ^ log c, or ar = — ^ ° — ; r .
° ° ° ' w» log a + n log 6
Ex. 3. Given (-^ = 54§, to find x.
TT 1 5 , 109 log 54-5 ,„„,»_
Here a; log - = log — , or x = ^^^j:^ = 17-9177.
Ex. 4. Given a ^ c to find x *.
Put b" ■= z : then a' =. c, and hence z ^ , .
logo
Again, since If -^ z, x\ = ,— -, = * Vlog a / .
log 6 ,
log 6
Ex. 5. Given a* 6' ^ ^, and x \ z \\t : s,\.o find x and z.
These give x log a •{■ z log 6 = log A:, and 2 = — .
Insert the value of 2 from the second of these results in the first : then
Jlog a + - log b\x ■=■ log k, and resoUnng, we get
r log Jt , , 5x s log ^
a; = — -. J — -. j; ; and hence z = — = — f — - — ^ .
r log a + s log 0 r r log a + s log 0
i^x. 6. Given c* ^ 6' ^ A:, and x '. z\\ t ; s,\.o find x and z.
This gives, in the same way as before,
r log k T 5 log k
-— ^ and 2 = ^
r log a — « log b r log a — s log 6 '
Ex. 7. Given m .n^ = a, and x : r [ | r : s, to find x- and z-.
• In reading expressions of this form, it ■will be very important to keep in mind the significi-
tion of the notation. In the present case, it signifies a raised to the power of b* ; and however
many, n, successive exponents there be, the valuation is supposed by raising the (n — l)th
exponent to the power denoted by the nth ; then raising the (« — 2) exponent to the power
denoted by the last-mentioned : and so on, till we come to the base, which is raised to a power
denoted by the result of all the previous involutions.
+ In numerical solutions of such quantities, it will be convtnient to c.ilculate, in all cases, the
value of z before proceeding to the final equation which gives the value of x : but in certain
cases it will be essential to do so, as it may happen that c and a, being numbers less than unity,
the real values of their logs, being then negative, we cannot express log ^a and log *c without
the introduction of the imaginary symbol. The solution is, however, generally expressible
symbolically by j- = l"g'<^— l"g'« .
log 6
EXPONENllAL EQUATIONS. 263
be second
2r^ lojf a
Here, taking the log of the first equation, and reducing the second, we get,
after substitution.
2r^x^ log m + s^x' log n = 2r" log a, or a^ = ,. ^'' '">< ,
27^ log m + s- log n
and z^ = ?^f = 2*^ bg a
r- 2r2 log m + «2 log n
£a:. 8. Given m n =. a, and a:' : z ; ; r : *, to find x and r.
Here we have x- log m + z- log n = log a, and z =. —.
T
By substitution r^ a;'' log m -j- s^ 3,2 jog „ = r" log a, or
, / r- log a , , , .
ic = + ^ / -vi ; — ^-j ; and from this again,
— sj r^ log m -\- s^ log n -e. »
sx , / s- log a
~ T ~ — V ;
r^ log m -\- s- log n *
J5a;. 9. Find the value of a? in the equation af = 100.
As an initial experiment, take ar, = 3, and x.^ = 4. Then 3' = 27, and 4^ =
256, one of which being considerably too little, and the other considerably too
great, we may take for x^ or x.^ a number midway between 3 and 4, with a
prospect of a near approximation ; and from the result, judge of the other value
whether greater or less than 3*5. Also taking logs, we have x log. x ^ log.
100 = 2.
First let a^, = 3*5 ; then,
3-5 log 3-5 = 19042380
true no. = 2 0000000
too little by 095762
Second, let x.^ = 36 ; then,
3-6 log 3-6 = 2-002689
true no. = 2 000000
too great by -002689
Hence, pp. 203, 204, a? ^ 3-59727 nearly ; the extent of the approximation,
however, being less clear in equations of this class, than in purely algebraical
ones. It has been often the case, that approximations have been trusted to too
far; as in the example above given, for instance, though a rather favourable
one for rapidity of approximation. By forming the value of the expression for
x = 3-5973, we find it too great by -0000149, and for x = 3-5972, too little by
•000084 1 . From these values we may obtain a further correction by a repetition
of the use of the formula.
This method may, for practical purposes, be a little improved. For since
of = a, we have, as before, x log x = log a. Put now log x = y, and log a
=: b : then we have xy =z b; and taking the logs we get log x + log y =
log b, or since log x ■=■ y, this is y -j- log y = log b. This may be solved for
y by Trial and Error, and the value thus found will be log x, and x becomes
known. The following example is given in illustration.
Ex. 10. Given x" = 123456789 to find the value of x.
Here x log x = xy = log 1^3456789 = 8*0915148 = b; and y + log y =
log b = -9080298.
Taking y., = 1, we have 1 + log 1 = 1, too great by -0919702.
Taking y, = -9, we have 9 + log "9 = -8542425, too little by -0537873.
Hence by the rule we have y ^ -93 nearly; and we may proceed to a second
approximation.
Taking y^ = 93, we have y + log y = -8984829, too little by 0095469.
Taking y^ = '94, we have y + log y = -9131279, too great by -0050981.
264 ALGEBRA.
Hence again by the rule we have y ^ '93652 nearly ; and we may again pro-
ceed to approximate still more closely.
The next step gives y = -gsesiSOS; and from this we have a? = 8-640026,
which is true in the last figure.
This method of solution becomes inapplicable when the equation is a:^ = - ,
and a not less than unity ; but it is easy to transform it so as to find the
reciprocal of x, and thence x itself.
Put 37 = - : then we get (^ - V" = -, and thence a' = z. Take the logs put-
ting log 2 = M, and we get z log a = log z = u ; and taking the logs again, we
find log z + log^ a = log u, or log u — u = u log- a ; an equation of the same
nature as that of the last example, and soluble by a similar process *.
EXERCISES IN EXPONENTIAL EQUATIONS.
rj^) = 2^. e""^. Ans. a; = 2-01374.
2. Fmd y from 2^" = 10, and x from (2^)' . 9' = 4^^
Ans. y = 1-1073093, and x = -371606.
3. Resolve the equation 2^ = 2*. Ans. x = 1-26186.
4. Given- = - and 3' = (4^)^ ' Ans. x = -222885, y = -297181.
y 4
5. li V : z ', '. z : 4:, and 4^^ = 9^^, what are the values of v and z ?
^ „ , 16 /log. 2 V 8 log. 2
Ans. v = 0,z = 0; and v= — i , -^ I ,2 = -—r^ — .
9 ^log. 3/ 3 log. 3
6. In ( rt- ) ' = 1-75, show that x = 9-677291.
V673/
7. Find x and y from a'' + »' = b^' + »\ and a,'*-*' = 6/'-*'.
Ans..= * |;-^ + J?^|.andy = 4 f-^-J^?^.); andx = 0,y=0.
- (log. 6 log. a J ' ^ Mlog. 6 log. aj' '"
8. Find X from x' = 5. Ans. 2 129372.
9. Solve the equation z' = 2000. Ans. 4-827822.
10. Solve the equation 0?*= 123456789. Ans. 8-640027.
1
» I 3
11. Given x' = 10*, and a?"' = 5" , to find x and y.
Ans. y = 0 and x indeterminate; and x = -00117937, y = — 292836.
z^ z — *
12. Given z = 1000 ; and r~ = -5, to find z in both cases.
Ans. z = 2384917 in the former ; and z is imaginary in the latter.
SIMPLE INTEREST.
The interest of any sum for any time being proportional to that sum and the
time, the interest of 1/. for 1 year, being multiplied by the principal and time,
will give the interest for that time at the specified rate per cent.
• An elegant method of solving such equations may also he seen in Mr, Charles Bonny-
castle's Appendix to his Father's Algehra, published in 18-23. A very elegant, though practi-
cally more laborious method of solution, may also be seen in foreign elementary works on
algebra, by means of continued /ractions.
COMPOUND INTEREST. 265
For the sake of expressing this algebraically, put p for the principal lent at
interest, t the time of its continuance, r the rate of interest, or periodical interest
upon 1/., and a the amount of the principal and interest at the end of the given
time, and lastly, i the interest itself.
Then, obviously, we have the fundamental theorems, prt = i, and p + prt =
p{\ + rt) = a.
From these equations we may find any one of the values in terms of the other
three : and taking all together, we have
a=.p •{■ prt , . (1)
" = 1^71 (^'
a — n
t = —~ (4)
Ex. Find in what time any principal will double itself at any given rate of
simple interest.
Here by equation (1) 2p =:a =:p-{- prt, or p^=. prt, and < = - ;
2/j — p 1
or, by equation (4), t =. — — ^= -, as before.
COMPOUND INTEREST.
If r be the interest of \l. for a given period, the amount at the end of that
period will be 1 + r ; and this put out for an equal period at the same rate, will
amount at the end of this period to (1 + O + (1 + r)r, or (1 + r)^; and this
again put out for a third period equal to each of the former, becomes (1 + r)'
+ (1 + rYr, or (1 + r)' ; and so on, till after t periods, the amount is
(I + r)'.
As this is the amount of \l. for t terms, the amount of p£ will be p times as
much, viz. p(l + '')' ; since each of the p pounds produces the same final
amount. Hence, generally, adopting the notations of simple interest so far as
they are common to both, and putting R = I + r,
- = Ji «
a=pK' (1)
f = w <-^'
P
_ logo — log j>
logR •••• ^*^
The equations (2, 3, 4) being obtained from (1) by mere common processes.
It is usual to call R the ratio, meaning the ratio of increase at compound
interest.
Example. Suppose it be required to find in how many years any principal
sum will double itself, at any proposed rate of compound interest.
In this case the 4th theorem must be employed, making a = 2p ; and then
it is
log a — log p log 2p — log p log 2
log R log R log R*
Thus, if the rate of interest be 5 percent, per annum; then R=ro5; and hence,
log 2 -301030 , „„,^ ,
t = T — '—- = - — - = 14-2067 nearly;
log 105 -021189 ^
that is, any sum doubles itself in less than 14j years, at the rate of 5 j)er cent.
per annum compound interest.
The following Table will very much facilitate calculations of compound
266
algebra;
interest on any sum, for any number of years, not exceeding 38, at various
rates of interest : it being the value of (1 -|- r)' for various values of r and t.
The Amount of IZ per annum in any number of Years.
Yrs.
■2iperCent.
3 per Cent.
3.1 perCent.
4 per Cent.
4^ perCent.
1-04500
.5 per Cent.
6 perCent.
1
1-02500
1-03000
1-03500
1-04000
1-05000
1-06000
2
05063
106090
1-07123
1-08160
1-09203
1-10250
1-12360
3
07689
1-09273
1-10872
1-12486
1-14117
1-15763
1-19102
4
10381
1-12551
1-14752
1-16986
1-19252
1-21551
1-26-248
5
131411 M5927
1-18769
1-21665
1-24618
1-27628
1-33823
6
15969' 1-19405
1-22926
1-26532
1-30226
1-34010
1-41852
7
18869 j 1-22987
1-27228
1-31593
1-360S6
1-40710
1-50363
8
21840 1-26677
1-31681
1-36857
1-42210
1-47746
1-59385
9
24886 1-30477
1-36290
1-42331
1-48610
1-55133
1-68948
10
28008 1-34392
1-41060
1-48024
1-55297
1-62889
1-79085
11
31209
1-38423
1-45997
1-53945
1-62285
1-71034
1-89830
12
34489
1-42576
1-51107
1-60103
1-69588
1-79586
2-01220
13
37851
1-46853
1-56396
1-66507
1-77220
1-88565
213293
14
41297
1-51259
1-61869
1-73168
1 85194
1 97993
2-26090
15
44830' 1-55797
1-67535
1-80094
1-93528
207893
2 39656
16
48451 i 1-60471
1-73399
1-87298
2.02237
2-18287
2-54035
17
5216-2 1-65285
1-79468
1-94790
2-11338
2-29-202
2-6Q277
18
55966
1-70243
1-85749
2-02582
2-20848
2-40662
2-85434
19
59865
1-75351
1-92250
2-10685
2-30786
2-52695
3 02560
20
63862 1-80611
1-98979
2-19112
2-41171
2-65330
320714
21
67958
1-86029
205943
2-27877
2-5-2024
2-78596
3-39956
22
72157
1-91610
2 13151
2-36992
2-63365
2-925-26
3-60354
23
76461
1-97359
2-20611
2-46472
2-75217
3-07152
3-81975
24
80873
2-03279
2-28333
2-56330
2-87601
3 22510
404893
25
85394
2-09378
2-36324
2-66584
300543
3-38635
4-29187
26
90029
2-15659
2-44596
2-77-247
3-14068
3-55567
4-54938
27
94780
2-22129
2-53157
2-88337
3-28201
3-73346
4-82235
28
99650
2-28793
2-62017
2-99870
3-42970
3-92013
5-11169
29
2
04641
2-35657
2-71188
3-11865
3-58404
4-11614
5-41839
30
2
09757
2-42726
2-80679
3-24340
3-74532
4-32194
5-74349
31
2
15001
2-5000S
2-90503
3-37313
3-91386
4-53804 6-0881 0|
32
2
20376
2-57508
3-00671
3-50806
4-08998
4-76494
6-45339
33
2
25885
2-65234
3-11194
3-64838
4-27403
5-00319
6-84059
34
2
31532
2-73191
3-22086
3-79432
4-46636
5-25335
7-25103
35
2
37321
2-81386
3-33359
3-94609
4 66735
5-51602
7-68609
36
2
43254
2-89828
3-45027
4-10393
4-87738
5 79182
8-14725
37
2
49335
2-98523
3-57103
4-26809
509686
6-08141
8-63609
38
2-55568 3-07478
3 69601
4-43881
5-32622
6-38548 9-15425 1
For example, let it be required to find, to how much 5231. will amount in 15
years, at the rate of 5 per cent, per annum compound interest.
In the table, on the line 15, and in the column 5 per cent, is the amount of
ll, viz. 2-0789 ; and this multiplied by the principal 523, gives the amount
1087-2647, or 1087/ 5s 3Jd, and therefore the interest 564/ 5s 3ld.
Note 1. When the rate of interest is to be determined to any other time than
a year ; as, suppose to ^ a year, or i of a year ; the rules are still the same : but
then t will express that time, and R must be taken the amount for that time
also.
Note 2. When the compound interest, or amount, of any sum, is required for
the parts of a year ; it may be determined in the following manner :
Is^ For any time which is some aliquot part of a year. Find the amount of
ANNUITIES. 267
l/for 1 year, as before; then that root of it which is denoted by the aliquot
part, will be the amount of 1/. This amount being multiplied by the principal
sum, will produce the amount of the given sura as required.
2c?. When the time is not an aliquot part of a year. Reduce the time into
days, and take the 365th root of the amount of 1 / for 1 year, which will give
the amount of the same for 1 day. Then raise this amount to that power whose
index is equal to the number of days, and it will be the amount for that time.
Which amount, being multiplied by the principal sum, will produce the amount
of that sum, as in the former cases.
ANNUITIES.
Annuity is a term used for any periodical income, arising from money lent,
or from houses, lands, salaries, pensions, &c. payable from time to time, but
mostly by annual payments.
Annuities are divided into those that are in possession, and those in reversion :
the former meaning such as have already commenced ; and the latter such as
will not begin till some particular event has happened, or till after some certain
time has elapsed.
When an annuity is forborne for some years, or the payments not made for
that time, the annuity is said to be in arrears, or in reversion.
An annuity may also be for a certain number of years ; or it may be without
any limit, and then it is called a perpetuity.
The amount of an annuity, forborne for any number of years, is the sum
arising from the addition of all the annuities for that number of years, together
with the interest due upon each after it becomes due.
The present worth, or value, of an annuity, is the price or sum which ought to
be given for it, supposing it to be bought off, or paid all at once.
Let a = the annuity, pension, or yearly rent ;
n = the number of years forborne, or lent for ;
R = the amount of 1/ for 1 year ;
m = the amount of the annuity ;
V = its value, or its present worth.
Now by compound interest, (theor. 2,) we have p = -rr^ . Hence giving to
^ a a a a .
t the successive values 1, 2, 3, ... n, we get -p, 5-,, Kj. • • vr,. as the pre-
sent values of a due at the end of 1, 2, 3, . . «, years respectively. Therefore,
the sum of all these will be the present value of the n years' annuities ; and if n
be infinite, it will be the present value of a perpetud annuity of a £ per term.
Now by summing this geometrical series, we have »=-i+ 5i + ui+'-'"l"
^= ^ . -^ , the present value of the annuity which is to terminate in n
R" R" R — I
years; and if n be infinite, v = 5 -, for the value of the annuity in per-
petuity.
Again, because the amount of £1 in n years is R", the increase in that time is
R" — 1 J but its amount in one year, or the annuity answering to that increase
268
ALGEBRA.
is R — 1 : and as these are in the ratio of a to m, we have m ^
R"
. a, and
R— 1
the several cases relating to annuities in reversion are easily found to be as
follow :
R-
— 1
R
-1
R'
— 1
a = pR"
R-1
= (---
IR" R"
a
)
R"
j ' R-1'
(1)
(2)
(3)
R— 1
R
1
R'— 1
vR'
R-— 1
m(R— n+a
...(4)
log m — log V
loor
log. R =
logR
log m
logR
■logr V
(5)
(6)
In theorem (3), r denotes the present value of an annuity in reversion, after p
years, or not commencing till after the first p years ; and it is found by taking
the diflference between the two values -?; — . :fr- and -^Fi . ^=r, for
R — 1 R" R — 1 R'
n years and/) years. The other formulae are derived from those in compound
interest taken in connexion with the fundamental theorem deduced above.
However, for practical purposes the amount and present value of any annuity
for any number of years, up to 2], will be most readily found by the two follow-
ing tables. In works professedly devoted to the subject, these tables are carried
to a much greater extent.
TABLE I.
The Amount of an Annuity of IZ at Compound Interest.
Yrs.
1
At 3 per C.
^ per C.
4perC.
41 per C.
5perC.
6 per C.
10000
1-0000
1-0000
1-0000
1-0000
1-0000
2
20300
2 0350
2-0400
2-0450
2-0500
20600
3
30909
31062
3 1216
3-1370
3 1525
3-1836
4
41836
4-2149
4-2465
4-2782
4-3101
4-3746
5
5-3091
5 3625
54163
5-4707
5-5256
5-6371
6
6-4684
6-5502
6-6330
6-7169
6 8019
6-9753
7
7-6625
7-7794
7-8983
80192
8-14-20
8-3938
8
8-8923
9-0517
9-2142
9-3800
95491
98975
9
10-1591
10-3685
10-58-28
10-8021
11-0266
11-4913
10 11-4639 1 117314
12-0061
12-2882
12-5779
13-1808
11
12-8078 13-1420
13 4864
13-8412
14-2068
14-9716
12
141920 14 6020
15-0258
15-4640
15-9171
168699
13
15-6178 16-1130
16-6268
17-1599
17-7130
18 8821
14
17-0863 17-6770
18-2919
18-9321
195986
210151
15
18-5989 19-2957
20-3236
20-7841
21-5786
23-2760
16
20 1569 20-9710
21-8245
22-7193
23-6575
25-6725
17
21-7616 22-7050
23-6975
24-7417
25-8404
28-2129
18
23-4144
24 4997
25-6454
26-8551
28-1324
30-9057
19
251169
263572
27 6712
290636
30-5390
33-7600
20
26-8704
28-2797
297781
31-3714
33-0660
367856
21
28-6765
30-2695
31 9692
33-7831
35-7193
39-9927
ANNUITIES.
TABLE
II.
The Present Value of
an Annuity of 1/.
Yrs.
At 3 per C.
31 per C.
4 per C.
4.1 per C.
5 per C.
6 |>er C.
1
0-9709
0-9662
0-9615
09569
0-9524
0-9524
2
19135
1-8997
1 8861
1-8727
1-8594
1 8334
3
2-8286
2-8016
27751
2-7490
2-7233
2-6730
4
3-7171
3-6731
36299
3 5875
3-5460
3-4651
5
4-5797
4-5151
4-4518
4-3900
43295
4-2124
6
5-4172
5-3286
5-2421
5-1579
5-0757
49173
7
6-2303
61145
6-0020
5-8927
5-7864
5-5824
8
7-0197
6-8740
7-7327
6-5959
6-4632
6-2098
9
7-7861
7-6077
7-4353
7-2688
7-1078
6-8017
10
8-5302
8-3166
8-1109
7-9127
7-7217
7-3601
11
9-5256
90016
8-7605
8-5889
8-3054
7-8869
12
9-9540
9-6633
9-3851
9-1186
8-8633
S-3838
13
10-6350
10-3027
9-9857
96829
9-3936
8-8527
14
11-2961
10-9-205
10-5631
10-2-2-28
9-8986
9-2950
15
11-9379
11-5174
11-1184
10-7396
10-3797
y-7123
16
12-5611
12-0941
11-6523
11-2340
10-8378
10 1059
17
13-1661
12-6513
121657
11-7072
11-2741
10-4773
18
13-7535
13-1897
12 6593
12-1600
11 6896
10-8276
19
14-3238
13-7098
13-1339
12-5933
12-0853
11-1581
20
14-8775
14-2124
13-5903
13-0079
12-4622
11-4699
21
15-4150
14-6980
14-0292
13-4047
12-8212
11-7641
To find the amount of any annuity forborne a certain number of years.
Take the amount of 1/ from the first table, for the proposed rate and tiaie;
then multiply it by the given annuity ; and the product will be the amount, for
the same number of years, and rate of interest. Also, the converse to find
either the rate or the time.
Ex. To find how much an annuity of 50/ will amount to in 20 years, at 3§
per cent, compound interest.
On the line of 20 years, and in the column of Z\ per cent, stands 28-2797,
which is the amount of an annuity of 1/ for the 20 years. Then 28-2797 X 50,
gives 1413985/ = 1413/ 19s Srf for the answer required.
To find the present value of any annuity for any number of years.
Proceed here by the second table, in the same manner as above for the first
table, and the present worth required will be found.
Ex. 1. To find the present value of an annuity of 50/, which is to continue -20
years, at 3| per cent. By the table, the present value of 1/ for the given rate
and time, is 14-2124; therefore 14-2124 x 50 = 710-62/, or 710/ 12s 4c/, is the
present value required.
Ex. 2. To find the present value of an annuity of 20/, to commence 10 years
hence, and then to continue for 11 years longer, or to terminate 21 years hence,
at 4 per cent, interest. In such cases as this, we have to find the difference
between the present values of two equal annuities, for the two given times ;
-which, therefore, will be done by subtracting the tabular value of the one period
from that of the other, and then multiplying by the given annuity. Thus, the
tabular value for 21 years is 14 0292, and that for 10 years is 8-1109. Then, the
difference 59183 multiplied by 20 gives 118-366/, or 118/ 7s S^rf, the answer.
270
SERIES BY SUBTRACTION.
This method is most readily applicable to the cases where the several terms of
the series are the differences (or the same multiple of the differences) between
two equi-distant corresponding terms of some other series. A few sinaple exam-
ples will sufficiently illustrate the practice, whilst the principle of these processes
is self-evident. The only difficulty is to find the series whose differences are the
terms of the given one ; and for this no general and simple rule exists.
Ex. 1. Let 1 + 5 + ^ + ^ + ad inf. = s
then ^+ 1 + 2+1 + .... ad inf. = s- 1
2 3 4 5
Ex. 2. Let 1 + | + l+l + .,. = s.
then -+-+- + -+ ... =._-.
1, I, 2 , 2 , 2 , 2 , 3
by sub. r -+ ...= -.
' 1.3^2.4^3.5^4.6 2
1:3 + 2:^ + 3:5+0 + --- = i-
then -1-+ J_ + -l,+ ...=s-l.
2.3 3.4 ^ 4.5 ^ 2
V I. 2 , 2 , 2 , 1
by sub. h ■ h \- ... = - .
' 1.2.3^2.3.4^3.4.5 ^ 2
1.1,1. 1
or + u . . . = -.
1.2.3^ 2.3.4 ^3.4.5 ^ 4
Ex. 4. Find the sum of the series x + —^- + . ^ „ + . . . .
2.4.6 4.6.8 6.8.10
Take away the last factor out of each denominator, and assume
J_ , J_ , 1 , _
2.4 ■•■ 4.6 "^ 6.8 "^ • • • • ~*
♦I, 1 j^ 1 1 1 . 1
^^^'^ ?:6+6:8 + 8-:io+--- =*-8-
^y «"^- ^ + 4-X8 + eii^ + • • • = 8-
^''''' 2X6+iii + 6-Xl5 + ---=-8^4=3'2-
Ex. 5. Find the sum of 1 .. . ad inf.
1.3 2.4 ^ 3.5 "^
1 1 1 1 1 , *i, * .11
assvune - — n+o \- .. . = s, then transposmg - — -,
wehavel-l + l-l + ... = 5-l;
2 2 2 1 1
hence by subtraction T-;r — r-T+;rT — ••• =:^. or* = -.
' 1.3 2.4 ^ 3.5 2' 4
Ex. 6. Find — 1 1 ? h — ^ 1- . . . adir^. Ans. -1-.
2.4.6.8 ^ 4.6.8.10 ^ 6.8.10.12 T^ • • • »" »'!'• '~^»' 288
SERIES BY SUBTRACTION. 271
^^- ^- ^^^° °^ 2:5x0 + ixrn-4 + • • • «^ *■»/• Ans. yj- .
Ex. 8. Sum the series -— — „ + -__... Ans. I.
1.5 3.7 5.9 0
JVo<e. The upper line (carried to any extent) contains one term at the end,
under which there is no term in the lower line. But in the above examples this
circumstance creates no difference, since the terms being infinitely distant, and
continually converging towards 0, that last term itself is virtually 0. However,
if the law of the series be such, that the terms converge towards a limit different
in value from 0, then this value being that of the last term of the upper line,
must be added to the sum obtained as above,
Ex.9. Thus, 1^77^+^-; + tt + ••• were purposed to be summed; the
JitO o*4 4*3
process would be, if performed according to the preceding type,
12 3 4
assume - + - + " + , + •••=*> then, transposing,
2 3 4 5 » o
I, 2 . 3 , 4 , 1
we have - + - + - + =s — -;
3 4 5 2
hence by subtraction ^ + ^^+^"5 + = — 2"
But as the values of the several terms of the series, whose value is s, con-
verge towards 1, the uncompensated term itself is 1, which added to the value
already found, gives 1 — J = § for the true sum of the series. This final term,
in such cases, may be called the correction of the sum.
Ex. 10. Find the sum of -—+—- + -— -f -— - + . . . Ans. ^.
3.5 57 7.9 9.11 6
MISCELLANEOUS EXAMPLES.
Ex. 1. Find the sum of ^ - ^ + 3^ - j^ + ^ - • • • Ans. ^
Ex. 2. Show that {i-3 + 3i^+i^ + . . .} {I + -3 + J- + fo + ....}
isequaltothesumof ^-^ + ^^+3^^-^+ ....
Ex. 3. The sum of the series -^^ + ^^^.^ '^^^r) +(^27o'(m+3r)
+ ... ad inf. = — .
' •' mr
Ex. 4. The sum of n terms • of o + 2ar + 3ar^ + 4ar' + .... is
„ a — or" nar* , ^, ^ • c •» • c "
S, = _ >2~l~_:7~' ^^^ "^^ ^"°* *° mfinity is is. = . ^.
12 3 J /■ o a-\-b a+2b
Ex. 5. Sum n terms of - + - + ^3 + .... and of - + — — + -— j- + . ..
r r' r^ m mr mr'
and likewise the sum of each series to infinity.
* This and such questions differ from the preceding in no respect but this: that the ex-
pression for the nth term must be taken as the last of the assumed series instead of the infinitely
distant term towards which in that case the succeeding terms more directly point. The correction
for this must be made as directed ia the note above.
REVERSION OF SERIES.
When the powers of an unknown quantity are contained in the terms of a
series, the finding the value of the unknown quantity in another series, which
involves the powers of the quantity to which the given series is equal, and known
quantities only, is called reverting the series *.
Rule 1. Assume a series for the value of the unknown quantity, of the same
form with the series which is required to be reverted.
2. Substitute this series and its powers, for the unknown quantity and its
powers, in the given series.
3. Make the resulting coefficients equal to the corresponding coefficients of
the given series, whence the values of the assumed coefficients will be obtained.
4. When the series is expressed by means of another, as ax + br' + ca?^
+ . . . = ay + 6y^ + ^' + • • • • the value is to be obtained in the same
manner, by assuming x =A y + By^ + Cy^ + . . .
. EXAMPLES.
Ex. 1. Let 0 = aa; + bx^ •\- ca^ -^^ dx^ + ... be given, to find the value of
X in terms of z and known quantities.
Assume x = Kz -\- Bz^ + Cz^ -\- . . ., and substitute for the powers of x in
the given series, the powers of this assumed series. Then we shall have
z = aAz + aB -) -f aC "l + aD
+ 6A2 U2 + 26AB S-z^ + 2bAC ,
J + cA3 J 4- 6B2 )>z^ + ...
+ ScA'B
+ dA*
By equating the coefficients of the homologous terms of e, we shall have
«A = 1, or A =-;
a
aB + JA2 = 0, or B = — ^^= --, ;
aC + 2bAB + cA' = 0, or C = ^*'~ ""
a"
, . .. , T^ 5abc — 5b^ — a^d
and similarly, D ^ ^ ;
and so on to any extent. Hence,
1 b „ 2b' — ac , 56' — 5abc + a-d , ,
a a^ a^ a'
This conclusion forms a general theorem for every similar series, involving
the like powers of the unknown quantity.
• Other methods of reversion are given by different mathematicians. The above is selected
for its simplicity, most of the others depending for their evidence on principles more recondite
than have yet been laid before the student, or being more difficult of application, or more con-
fined as to generality. This is, evidently, only au application of the Method of Indeterminate
Coefficients.
THE METHOD OF FINITE DIFFERENCES. 273
Ex. 2. Let the series z = x ± x' + x' ± x* + .... be proposed for
reversion.
Here a= l,b= ± I, c = I, d = ± 1, and so on ; these values being sub-
stitutedjn the theorem derived from the j»receding example, we shall obtain
X = z + z' + z^ + z* + . . ., the answer required.
fcT** X^ U*'^
Ex. 3. Let y = ir— --f — — T + - • '^ given for reversion.
Substituting as before, we have a = 1, i = — i, c = J, rf = — i, and so on.
These values being substituted, we shall have x = v 4- ^-i-^ -^^ -l.
^^2^6^ 24 ^•'■'
from which if y be given, and sufficiently small for the series to converge, the
value of X will be known.
Ex. 4. Given the series y = ar — ^ a-s + J a?* — ^ a;? -|- . . . to find the value
of X in terms of y. Ans. « = y + i y^ + -^ y* + a'fi y^ + . . .
Ex. 5. Given the series y = l+a: + |+— -+ -^ -f ... to find x in
terms of y. Ans. x = (y- 1) - ^^ " ^^' + C^ " J)' _ (yj^ll' ^. . . .
2 3 4
Ex. 6. It is required to find x and y from the two following equations :
30y = x + ^+^ + — + - + — + ..1
-^1.3-^2.3-^2.5-^3.5 + 3.7+ Ans. x = 10, y = I
9« 1 10 ( 3
Yo=3 + y + ^y'+ jy' + ^y*-^- ■ • J
Ex.7. Giveny = a?+^ + - + — + ;?^ + • • • . to find x.
^ ^ 6 ^ 24 ^ 5040 ^ 725678 ^ •••'■"""" '•
Ans. x-y g + 24 ^^^^ + ^^5678
THE METHOD OF FINITE DIFFERENCES.
I. Definitions, notation, and principles.
1. If the successive integers, 1, 2, 3, 4, . . . be given to x in any expression, a
series of numbers will be produced, which are called the successive values of that
expression : and, conversely, that expression is called the general term of the
series of numbers thus produced.
2. The general term or function of x from which the series of numbers is
formed, is sometimes written, as in algebraic equations, /(x) ; but more com-
monly u, or v^ the letter u or p being called the characteristic of the function.
Thus, if x^ + 6x' — 5x + 10 were the general term, it would be written u, :
and the values of this function, when 1, 2, 3, 4, .. are wTitten for x, (viz. 12,
32, 76, 150, . . .) are written «,, «.j, Uj, u^, . .
Also, if — 1,-2, — 3, . . . were put for x, the several results, 20, 36, 52, . . .
are written u_„ u
— l> »— 2> " — 3»
• In this equation y expresses the meridional parts corresponding to the latitude x : and it is
remarkable that the numeral coefBcients are the same in the direct and reverted series, and
differ only in the signs of the even-numbered terms. Let/bourn's Rep. II. 44.
VOL. I. T i^ hi' l^cU^t^ M.\iC^fe.j^
274 ' ALGEBRA.
3. If a series of n terms be given, the (n + 1)* is called the increment of the
series, or the increment of the sum of the series. The increment is the same func-
tion of n + 1 that the last term of the series is of n ; or in symbols, if «. be the
last term, «. + , is the increment.
4. If the first term of a series be subtracted from the second, the second from
the third, the third from the fourth, and so on ; the several remainders constitute
a new series, which is called the first order of differences. Taking the differences
of the successive terms of this series, we obtain the second order of differences j
and from this again, the third order of differences ; and so on, as long as re-
mainders result from such operations.
5. The general expression for the difference of the two consecutive terms, the
(a? + 1)"" and the a;"", will, according to the preceding notation, be written
u^i — M, : but it is often convenient to write it simply Au„ where A is called the
sign of differencing. The second difference, or A(u^i — u,), is written A"k„ the
third A^M„ and so on, as long as any differences exist.
6. The symbol S prefixed to an expression w,, signifies the operation of finding
another expression r, such that r,+, — r, = «»• In other words, it expresses an
operation directly the reverse of taking the difference of a function ; and hence
A and 2 are indicative of operations each of which neutralises the effect of the
other. The symbol 2 is called the sign of integration.
7- A factorial is an expression composed of factors in arithmetical progression,
as x(x + a) {x + 2a) .... (x + no) ; where every factor differs from the pre-
ceding by the common quantity + a.
In respect of notation, the following has the advantage of concentrated
writing, viz. a^+Ml* : where the first factor of the series is written down ; then
the index of the number of factors in the factorial, in the manner of the bino-
mial index ; and lastly, separated from the index by a vertical line, the common
difference of the several successive factors *.
II. To find the general term of the successive orders of differences of a given
function u,.
1 . The expression for the a?" term of the n"' order of differences is
n „(„_!) „(„_]) (n_2)
AX = «.+.- J . «^: + —^.u,^, j-^^ . «^,+ ...
the coeflScients being those of the expanded binomial (1 — 1)'.
For Au, = M^i — u„ by def. 5
Aht, = A {u^^ — u,] = {«^, — «,+, j — {«,+, — M, j
= «.+! — 2m^, -I- u,
A^u, = A [u^ — 2u,+, + «,}
Proceeding thus we find the fourth, fifth, and subsequent differences, to any
extent, to retain the assigned form : but to complete the proof, we must establish
the necessary continuity of the law.
• This very elegant notation was invented by M. Kramp, Professor of Mathematics at Slras-
burgh. See his Elimens eTArith. Un. p. 347, a work of great originality and value.
^^1 •■' , m(jn—\) m(m— 1) m— 2)
,+„+» ^ x-H. J 2 '^"-' 12.3 ".+— sT-
THE xMETHOD OF FINITE DIFFERENCES. 275
Now this will be done, if, supposing it true for the m" difference, we prove
that it will be so for the (m + 1)"" difl'erence. In this case we have
m . mim — V) m(m— 1)(ot — 2)
A"«. = «,+. - y . «H— . + j-^— U.-U-, - '- ^^^ «,+_, + . .
And writing x -\- \ ior x, and subtracting A"«, from the result, we have
m(m— 1)(
1,2.
m{m — 1)
— Y^ «-+— » +.
-., ("» + 1) „ m(m + l) (CT-l)(w)(m+l)„
— Wru+i 1 — ««+- -< f^ — «.+— I YYs «-f— j+'
which establishes the general and necessary continuity of the law of the series.
For example, if the form of the function were u, = aa^, then
Au, = «rf 1 — Mr = o{x + ly — ax^
A-u, = u,+i — 2u^+, 4- w, = a(a; + 2)^ — 2a(x + 1)' + ax^.
A^u, = «,+3 — 3«,+2 + 3Krf, — Wx = a(a? + 3)3 — 3a(x + 2)^ + 3a(x + 1)' — acs
and so on, to any required extent.
2. The preceding formula applies to any form of the function u, : but when
that function is one of an algebraic form, the actual process is simpler when the
reductions are made, pari passu, from one step to another of the differencing.
Thus, if the function were u, = ax^ — bxi^ -\- ex + d, we should have
A«, =a{{x + ly — x^\— b{ix + \y — x^]+c{(x + I) — x} + {d — d}
= 3ax- + {3a — 2b)x + a — b + c
aX= 3a[{x+iy—aP} + (3a — 2b){(x+\) — x} + {{a — b+c) — {a — b+c)}
= 6ax + {6a — 26)
A^u,= 6a {{x + \) — x\ + {{6a — 26) — (6a — 26) J
= 6a
Ahi^ = 0, and so on for all higher orders of differences.
It may also be remarked, that though d — d, {a — b -\- c) — (a — b ■{• c),
and so on, are written in the expressions above, they are unnecessary in prac-
tice, as the absolute term is always cancelled in differencing.
Scholium.
It is very clear, since {x + !)• — x" is an expression of the (n — 1)* degree,
that the difference of an algebraic function is one degree lower than the function
itself. In like manner, the second difference is one degree lower than the first
difference, or two degrees lower than the given function. Proceeding thus, we
shall find the n'" difference constant, and the (n + 1)'S (n + 2)'S and all sub-
sequent differences, severally equal to 0.
EXAMPLES.
Ex. 1. Find the general term of the several orders of differences of
(1). a;»o + 6a^ + 3^2 — lOj (3). .IJT^ + 4a;-» + 6a;-' - 3«»;
(2). ax +bx\cx+...; C^)- J^' - '^^^ + "OOl^'- "
Ex. 2. Show that Aa' = {a^ — 1) a', and Aa"* = -, (^^ — IJ-
T 2
276 ALGEBRA.
u 1
Ex. 3. The first differences of «,|' v„ of -^, and of ; where u, and r,
express different, but given, functions of x.
The solution of the first of these examples is subjoined.
A (u, »,) = «rf 1 »,+i — «. tJ,
= (a, + £iU,) (o, + Ap,) — «, c,
= C, . A«, + Wi • ^"x + ^w. • Ap,.
[To facilitate the student's comprehension of the signification of these results,
let him apply them to the following particular examples :
Let (a^ — 4a? + 6) (Sa;^ _ 2x) = u, v,. Then
Am^ =: 2a; — 4, and Ac, =: 9x^ — 2. Hence,
A (u, r,) = (3a;3— 2a;) (2a7— 4) + {3^—\x-\-&) (9x^—2) + (2a;— 4) (9ar— 2).
Reduce this, and then actually multiply the values of «^ and p, together, and
take the difference : they will be found identical.
Let again u^ = — and p, = 4 ; and let •3"'^ be divided by 10"', and the first
difference found.]
Simuarly, we find A — ^ — ; ^ ; and
fx fxJPx + Ap^J
1 w, Ap, + p, A«^ + Au, Ap,
«x Vx Ux fxf Mx + Au,? iv, -f Ap,?
Ex. 4. Let a series of factors in arithmetical progression be given,
«x = x{x + a) {x + 2a) (a? + 3a) .... (a; + na).
For X write its succeeding value, x -{• a, and subtract u^ from it :
«,+, — «, = (a; + a) (a; + 2a) . . . -[a? 4- (» + l)a} — a?(a7 + a) . . . (ar + na),
or Au,= (x -\- a) {x -\- 2a) . . . . (x + na) {a; + (n + 1) a — x}
= (n + 1) (a? + a) (a; + 2a) {x + na) a *
That is, the difference sought is the product of all the factors except the first, by
the common difference and the number of factors.
Ex. 5. Let «x = ; — ; ;^ ; : thcu we have
{x + a) (x + 2a) .... {x -\- na)
^„^= _(„+!)„*
x{x + a) (a; + 2a) .... {^x -\- na)
That is, the difference is found by multiplying the denominator, the preceding value
of the function, and the numerator, by the common difference, into the number of
factors in the denominator so increased.
Ex. 6. It is required to determine the first differences of x(f, xcr', and x'cr*"'
Ex. 7. Find the numerical values of the first ten terms of the first order of
differences in the functions whose coeflScients are numerical in Ex. 1.
Ex. 8. Find the second, third, and fourth differences of the functions given in
Ex. 6.
* Since a z= Ar, this latter expression may be written for it in the answers above. The
farters are also often written aa-, t^ t^ .... Jr„; and hcnco the results obtained in the text may bo
put (n + 1) J, Tj ... x„ A* and — v^Jlii^, respectively.
**, a?a ... Xn
THE METHOD OF FINITE DIFFERENCES. 27'
III. Having given an adequate number of terms of a series, to find the general term
of the series.
Suppose the general form of the term to be
air + a,a;"-i + a^-^ + -|- «_,« + a. ;
then there is required the index m and coefficients a, a^, a^, . . . a., a^+ .
1. In the first place, we have seen (II, Schol.) that in an expression of the
m"" degree, the m*^ differences are all equal, and that all higher orders of dif-
ferences become 0. Hence to find m, we have only to take the successive orders
of differences, till we find one order all whose terms are equal. The number of
these operations which are thus performed, gives the value of m.
2. Let Mj, Uj, M,, be the several given terms of the series : then, as these
are the values of the general term when m is 1, 2, 3, ... we have
l"a + l-» a, + i-« a,+ ... + ia_, -)- a. = «,
2-a + 2-' a, + 2-2 a, + .... + 2a_, + a. = «,
3"o + 3— » a, + 3"-2 a^ + ....+ 3a_, + a. =u,
(m + l)-a + (m + 1)— i a, + (m + I) a^^ + a„ = «.+, ;
in which there are as many equations as there are unknown coefficients a, a,, a,,
a., viz. m + 1. All these equations are, with respect to the unknowns,
of the first degree ; and the method most readily applicable to the process of
solution, is that pointed out at pp. 179, 180, of this work.
Ex. 1. Let the series 7, 33, 79, 145, 231, be given to find its general term.
7|33|79!l45 231
26|46 66 86
20| 20 20
fiven series
rst differences
second differences.
Hence, as the second differences are all equal, we have m = 2, and the general
term is of the form ax^ + a^x + a,. Hence, substituting I, 2, 3, in this for
X, we have
and, as at p. 180, we get
a = 10, a, = — 4, o, = 1.
a + a^ + a^= 7
4a + 2a, + a, = 33
9a + 3a, + a^ = 79
The required expression is, therefore, «, ^ lOx' — 4a: -|- 1.
Ex. 2. Find the general terra of the series 2, 24, 108, 320. Also ascertain
whether any of the terms 1512, 4668, 7290, and 11011, belong to the series;
and if so, assign their places.
Ex. 3. Given 2, 14, 66, to find the general term, and hence the next term •.
* This example was actually formed from the expression tc* — ** -j- ** -|- j ; that is, one of
the fourth degree : but as the tenns, so far as they are actually given, can be formed from one
of the second degree, this latter ought to be considered the determinale solution of the question,
in contradistinction to the indeterminate ones, which the solutions of the third, fourth, ...
degrees do really become. These considerations surest, that when a series of m terms is given,
such that the ;«"■ difference is not constant, then we may fulfil the condition to which these
given terms are subject, by taking A"+* k = 0, and therefore also d^^u^ = 0 : though at the
same time the general term which results is only a particular case of a more general solution
•which would have been obtained by supposing the dimension of the general term to be higher.
The assumptioH of the dimeusion of the general term is, therefore, in fact altogether arbitrary,
278 ALGEBRA.
Ex. 4. Given 2, 14, 66, 212, ... and likewise 2, 14, 66, 212, 530 ...., to
find the general term.
IV. Having given a series of terms, to find the several orders of differences.
1. Subtract the first term from the second, the second from the third, and so
on, to get the entire first order of diflFerences ; employ the same process upon
the first order to obtain the second ; upon the second to obtain the tliird. The
first terms of these several orders are those sought ; as follows from definition 4.
2. Generally, however, it will be more convenient to employ the formula
deduced in II. p. 2/4, making in all the expressions a; = 1, and w =: 1, 2, 3, . . .
in succession. The expression so modified becomes
n n(re — 1) n _
A-U, = U,+i — - «n H — «,_, — + y Wj + M,
Scholium.
It is evident from both the methods, that there must be given one term more
than the number indicated by the order of the difference sought, in order to
render the problem determinate.
EXAMPLES.
Ex. 1. To find the first term of the third order of differences of 1, 4, 8, 13,
19, ...
Here m, = 1, Wg = 4, Mj = 8, m^ = 13, Mb = 19, . . .
Hence A^Wj = u^ — 3U3 + 3Mj — ?i, = 13 — 24 + 12 — 1 = 0.
Ex. 2. The first * term of the seventh order of differences of 1, 4, 8, 16, 32,
64, 128,
Here A^Mj = Ug — 7«7 + 21m6 — 35mj + 35u^ — 2IM3 + 7u.^ — Mi ;
in which, inserting the given values of u,, u^, u^, . . ., we have
A^M, = 256 — 896 + 1344 — 1120 + 560 — I68 -f 28 — 1 = 3.
Ex. 3. Given 1, 2, 4, 8, 16, . . . to find the first term of the seventh order of
differences. Ans. 1.
Ex. 4. Find the several orders of differences of 1, 2, 3, 4, 5, . . .
Ex. 5. Find the several orders of differences of 1, 4, 9, 16, 25, . . .
Ex. 6. Find the several orders of differences of 1, 8, 27, 64, 125, . . .
Ex. 7. Find the first four orders of differences of the logarithms of 101, 102,
103, 104, 105,
Ex. 8. Given 1, 6, 20, 50, 105, ... to find the first four orders of differences.
80 tliat the index be at least equal to the number of terms ; these terms being always understood
to be consecutive in the scale, beginning at unity.
When, however, the calculation is of one single term, however distant, it will be effected more
e.isily by the following process, without determining the general term of tlie series. If several
be required, the general term, (that is, the most simple general term which can be found from
the given numbers,) is indispensable.
• When only the first term of a single order of differences is required, as in the first three
examples, the second method is the preferable, as in the text ; but when all are required in
succession, it will be more convenient to employ the former method. Thus, if the several orders
of this example were sought, the work would stand as below :
THE METHOD OF FINITE DIFFERENCES.
279
Ex. 9. Given 1, 4, 8, 16, 32, 64, 128, 256,
differences.
to find the several orders of
8 16
32 64
128
256
4 8
16 39
64
128
1 4
8 1C
32
64
3
4)8
16
32
1|4
8
16
3
4
8
1
4
3
114 8|16 32 64 1281256 . . . .the f?iven series
34| 81639 64128 ... .first differences
...second differences
, . . third differences
. . fourth differences
. . . fifth differences
. . . sixth differences
, . . seventh differences.
"V. Having given the first terms of the first n orders of differences, to find the
(n + 1)"" term of the series.- or in symbols, given u^, Au^, A'^u,, .... A"u,, to
find Un^ P
The sohition is m^, = u^ + - Au, -j- ~ a'm, + . . . carried to the
term involving V^ A "
For by II. we have, putting a? = 1, in all cases.
Am, = Mj — u^, or ti^ = M, + AMi.
«3 = «2 + AMj = M, + Am, + A[ui + AmiJ
= M, + 2Am, 4- A^Mi
"4 = «3 + A«3 = «i + 2Am, + A\ + A[u^ + 2A«i + A=«,^
= M, + 3A«, + 3A2m, -I- A% ;
and so on to any extent required, the coefficients being those arising from the
expansion of the binomial (1 + 1)", and the necessary continuity of the law
being capable of establishment nearly as in II.
EXAMPLES.
Ex. 1. To find the twentieth term of the series, 2, 6, 12, 20, 30, . . .
, . the functions
, . their given values
, . first order of differences
. second order of differences
. third order of differences.
Hence m, = 2, Am, = 4, A-u, = 2, A^u, = 0, and so on, all the subsequent
orders of differences being zero. Hence
«, Mj
u.
«4
Ms
2 6
12
20
30
4
6
8
10
2
2
2
^
0
19
19.18
A\ = 2 + 76 + 342 = 420,
U^ = U^ + Y Au^ + ^2
which is the twentieth term of the series sought.
Ex. 2. Find the tenth term of the series, 2, 5, 9, 14, 20, . . Ans. 65.
Ex. 3. Required the fifth term of 1, 3, 6, 10 Ans. 15.
Ex. 4. To find the tenth term of the series, 1, 4, 8, 13, 19, . . Ans. 64.
Ex. 5. Required the twentieth term of 1. 8, 27, 64, 125, . . . Ans. 8000.
Ex. 6. Required the sixth term of 101, 108^, 118, 129i, . . . Ans. 158^.
VI. To convert a given function in powers of x to one which shall have every term
composed of factors x + a, x + b, and so on.
Divide synthetically hy x + a; the last coefficient is the coefficient of {x + a)"
of the transformed expression : divide again in the same manner hy x + b,
stopping one step sooner ; the last coefficient is that of (x + a) in the trans-
280 ALGEBRA.
formed expression : proceed similarly with x + c, stopping one step sooner than
in the preceding ; then the last coefficient is that of (x + a) {x + b) in the
transformed expression. Proceed thus with all the factors, then the coefficients
of the transformed expression will be all determined.
For let Aa;" + Ba?"~^ + Ca?"""- + . . . . + hx^ + Ma; + N be the function
given in powers of x : then dividing by the several given factors, we have in
succession
Aa;"-i + B,a;»-2 + C,a?-2 + + L.a; + M, + -^
X ±a
Ax'-^ + B^-3 + C,x'-* + + L, + -^. + , , f' ,^
^ x + b{x+a)(x+b)
Ar»-3 4- B a;— 4 4- Cje"-^ 4- -^-4- '^ -I — '
AX -t ii,x -t^^ -t ^_j.^ + (^+j)(a;+c) + {x±a) (x±b) {.x±cy
and so on, till we arrive at
A 4- ^-^i 4 ^ii? i_
^ X ±m^ ix ±1) {x ±m)^""
Multiplying now hy {x + a) (x + b) {x + c) {^ + 0 (^ i "*)» ^^ ^^^'^
A{x±a){x±b) ....{x±m) + B^^(,x ±a)(,x±b)...ix±l) + ....+^,
for the transformed expression.
If these factors be taken in arithmetical progression, the result is a transforma-
tion into an expression oi factorials. See def. 7, p. 274.
EXAMPLES.
Ex. 1. Transform the function 3a;* — Qa^ + 2a:^ — 5a; — 9 into factorials
involving a; + 1, a; + 2, a; + 3, a; + 4.
3—6+ 2—5—9
— 1
— 2
3 +
11 + 16
3
—
9 +
6 +
11 —
30 —
16 +
82
7
3
—
15 +
9 +
41 —
72
98
3
—
24 +
12
113
3 — 36
and the transformed function becomes, in the notation of Kramp, p. 274.
3(a; + 1/n — 36 (a; + \f\^ + 113 (a; + If ^ — g^ (^n ^ l)i|i _)_ 7.
Of, in the common notation,
3 (a; + 1) (a; + 2) {x + 3) (a? + 4) — 36 {x + 1) (x + 2) (a; + 3) + 113 (a; + 1)
(a; + 2) — 98 (a: + 1) + 7-
When the given expression itself is a combination of binomial factors, and it
is required to transform it into some other combination, as a factorial one, the
given expression may be first reduced to powers, and then transformed by the
general rule ; as in the next example.
Ex. 2. Given (a? + 1) (a? — 3) {x + 3) (a; + 5) to be converted into factors
involving x + 1, a; + 2, a; + 3, a^ + 4.
Since a; + 1 is a factor of the given and the sought expressions, it need not
be attended to, as it multiplies all the terms in both. However, for illustration
THE METHOD OF FINITE DIFFERENCES.
281
we shall work out as though no two factors in the two expressions agreed with
each other.
1 + l(-3
— 3 — 3
I — 2 — 3 (3
3 — 6 — 9
1 + 1 — 9 — 9 (5
5 + 5 — 45—45
1 +6 — 4 — 54
45
— 1
1 -f. 6 — 4 — 54—45
- 1 — 5 + 9 + 45
— 2
1+5— 9 — 45 + 0
— 2 — 6 + 30
— 3
1 4- 3 _ 15 _ 15
— 3—0
— 4
1+0—15
— 4
1 —4
Hence the given expression is x* + 6ofi — 4x^ — 54x — 45, and the factorial
is (a; + l)*ii — A{x + l)^!' — I5(a; + l)2li — I5(x + 1).
Ex. 3. Show that x^ z= {x — I) x {x + 1) + a? by this method.
Ex. 4. Transform x* to factorials involving x — 4, a; — 3, x — 2, x — 1.
VII. To integrate the general term of a series, or to find the expression tchose
first difference constitutes that general term.
1 . When the compression is composed of factors in arithmetical progression.
Multiply the increment (or given general term) by the preceding value of the
first factor, and divide the result by the number of terms thus obtained, and by
the common difference of the factors. This result, when corrected, gives the
sum required.
2. When the expression to be integrated is the reciprocal of such a series of factors
in arithmetical progression.
Expunge the last factor from the denominator ; divide the resulting fraction
by the number of factors remaining, and the common difference of the factors.
Then this result, written minus, will be the integral sought.
These being precisely the reverse processes by which the differences or incre-
ments were found, their truth is evident.
The correction arises from this cause : that A (2 + a) = Az, and hence it
cannot a priori be ascertained whether the integral is z + 0, or simply z.
The correction is found from this consideration. If from any circumstance we
can find what the aggregate value of a certain number of the terms is, and at the
same time ascertain what value the integral gives of the same number ; then the
difference of these two results is the correction. Most frequently, putting x = 0
is the best method : but examples will render this process much plainer than
precept could do.
The student will find little difficulty in reducing all expressions which involve .
only positive integer powers of x, that he commonly meets with, to one or more
of these forms.
EXAMPLES.
Ex. 1. Integrate x^' ^ or x (x+1) (x+2) (x+3).
282 ALGEBRA.
The preceding value is x — 1, the resulting number of factors is 5, and their
difference 1 : hence by the rule,
2a, = 1 {x—l)x(x+l) {x+2)(x+3) +c= , (x— l)5ii -1- c.
5 5
Ex. 2. Integrate (5ir+l) (3a;+6.) Ans. — (5a;— 4) (5x+l) (5x+6) + e.
15
Ex. 3. Given {f-s} {|-^-} {|-2} . Ans. ± (.-7)^ - + c.
(4x ") (4x \) 1
Ex. 4. Given jy— U |y~5J • ■^°^- 300 ^^^~^^ (4a;— 5) (4a?— 1) + c.
Ex. 5. Integrate the expression (x+1) (x+3).
This expression not being composed of successive values, (the increment of x
being generally taken in unity on account of the most frequent application of
this method,) it must be reduced to such a form. We write, for this purpose,
(x+l) (x+3) = (x+2) (a;+3) — (a;+3) each term of which is integrable by the
rule ; and we have
Su, = S (x+2) (a+3) — S (a?+3)
= 1 (x+l) (x+2) {x+3) - i (x+2) ix+3) + c.
= -^(2a;-l)(a;+2)(ar+3) + c.
Ex. 6. Integrate (2a?4-3) (2a:+7) ; that is, (2a:+5) (2a?+7) —2 (2«+7).
Ans. ^u,= -x (2a;+5) (2a:+7).
Ex. 7. Prove (l)S(2x+\) (2af+3)2= ^ i6x+7) (2x— 1) (2a:+l) (2a?+3) + c.
(2) 2a? = ^ a? («-!) + c.
(3) Sar' = ^ ix—l) X (247—1) + c.
(4) 2a?3 = - [xix—\)]' + c.
(5) 2«^ = — (6a:*— 15a?* 4- lOa?*— a;) + c.
30
Ex. 8. Let (!)«, = (^^:y^^) ; then 2«. = ^ + c.
^'^ «' = .(.+ l)Cx+2)^ ^^^° ^«' = 2^) + '^•
^^^ "' ~ (3a?+2) (3a;+5) (3a?+8) ' ^ ®° ^"' ~ 6 (3a?+2) (3a?+5) + *'•
Ex. 9. Integrate (^x+l) (2:r+3) (2x+5)- ""' ^' ^"'''' ^ ^'^°'""'
^_ i(2x+l) — i _ i h
(2x+l) C2ar+3) (2a?+5) (2j?+3) (2a?+5) (2x+l) (2*+3) (2a?+5)
»^"^^^"— -8 (2x^1)^(2^+3) +^-
4x+3A
Ex. 10. Integrate — ; — —rr- — rrnr, where Aa:=A.
* X (x+A) (X+2A)
THE METHOD OF FINITE DIFFERENCES.
Ex. 1 1 , Show that the integral of
1 2*+3
(x+l)(a:+3) 2(a?+l)(ar+2)
and that 2 -^ — - = c +
x^—l ~ ^ 2xix—iy
3. The integration of the exponentials (f and cr'.
By II. Ex. 2, Ao* = a' {a^'—\) ; and hence (f = —^"' ; whence
a-^' — 1
taking the integrals, '2a' = — 1- c.
Similarly Aff-' = — ^7- a"" ; and a-' = ^ Aar' ; and, integrating,
Y.ar' = a-' + c.
l—a'^'
For in both cases Ax is constant. When Ax =: 1, we have simply,
a^ , a
20* = + c, and So-* = a"* + c.
a—1 i—a
VIII. To find the sum of a series, whose general term is given.
Write M + 1 for n in the general term: then the integral is the sum of the
series.
For let the series be «j + u.^ + M3 + . . . . + m. = S, : then u, + u, + «, 4-
.... + «, + «.+, = S^,. Hence by subtraction AS. = S.+, — S, = u^f, ; and
integrating, S, = Sw^+i.
EXAMPLES.
Ex. 1. Find the sum of the series lH-24-3 + -- + «-
Here «, = n, and «,,+, = n+1. Hence aS, = n+1.
By integration S. = S (n+ 1) = 1- c.
To find c, which is the same value whatever n may be, put n ^ 0; then S, = 0
also ; and we have 0 = 0 + c, or c = 0. Whence the sum of the series of n
terms is — ^-- — -, as at p. 16 1.
Ex. 2. Find the sum of the series 1^ + 2- + 3^^ + +n'.
Here m, = n^ and «^, = (n + 1)-. Whence as before
AS. = (n+iy, and integrating, S. = 2 (n+ 1)= = ''(^+^n^^+^\
the correction being found 0 as in the last example.
Ex. 3. Sum the series of cubes 1' + 2' + 3^ + n».
Here AS. = (n + l)^ and S.» = |"i^y.
Ex. 4. Find the sum ofn terms of the series 1.2 + 2.5 + 3.8 + 4.11 +....
Here the general term of the first factor of the several terms is obviously n,
and the second (found by III.) is 3n— 1. Hence the general term of the series
• By comparing the solutions of examples 1 and 3 wc see that the sum of n terms of the
cubes of the natu);al numbers is equal to the square of the sum of those numbers themselves.
284 ALGEBRA.
is u, = n (3b— I) ; and the increment or (n+l)* term is tt^, = (n+1) (3»-|-2)
= 3n(nH-l) + 2(ra+l). "Whence integrating, S. = n^ (n+l).
Ex.5. Show that 12 + 32 _|. 52 _^ 72 ^ ^ (2n— 1)2 = -n(4n-— 1).
Ex. 6. Find the sum of n terms of each of the following series :
(1.) 1 +3 + 5 + 7 +.... I (3.) 1 + 5+ 9 + 13+....
(2.) 1 + 4 + 7 + 10+.... j (4.) 1+6+11+16+....
Ex. 7. Find the sum of the 2'"', 3"*, and 4"" powers of the preceding series of
numbers.
Ex. 8. Sum the series of n terms of — - + — 1 (-••••
1.2.3 2.3.4 3.4.5
Here S. = 2 * ^
(n + l)(n+2)(n+3) 2(n+l) (n+2)
To correct, put n = 0, then S. = c - ^^^^p^j^^^p^ = c - i.
But S„ =: 0 ; whence c — - = 0, orc=-; and the corrected integral is
g _1 1 _ "(« + 3)
4 2(n + l)(ra + 2) 4Cn+l)(n+2)*
Ex.9.Smn^+^ + ^+....to„terms.
Here «. = (2„+i) (2n+3) (2n+5) ' ^""^ ""+' = (2n+3) (2n+5)"(2MT)"
_ i(2n+3)-i _ 1 1
(2n+3) (2»+5) (2«+7) 2(2n+5) (2n+7) 2(2n+3) (2«+5) (2n+7)
Whence integrating u^. and correcting, we get S,= ^— - — , \^ ,J , ,,.
° ° ' 00 6(2»i+3) (2n+5)
Ex. 10. Find the sum of 21 terms of ^-A_ + _^ + _13^_ + . . . .
_, 4n+l J 4n+5
Here u. = — ., „ ,^ — , ., ,^ — — — and u,+, =
(2n— 1) 2n(2n+l) (2n+2) ""^' (2«+l) (2n+2) (2n+3) (2»+4)*
Now the denominator of the function «,+, not being composed of consecutive
integer values of n, it must be so transformed as to answer that condition, or
else into some other form which can be integrated. The latter plan is adopted
here.
Put 2w + 1 = M, then 2n + 3 = «, -i
and 2n + 2 = r, then 2n + 4 = », J " *
where Ui and »i are the next values of u and v.
XT 1 "-^P + ^^^ + Au ap , ., - , , , -
rsow A — ^ : and if we find the values of u, v, u,,v,,
UP uu^ vo^ ' ' '' "
Att and Ap in (1) and insert them in this equation, we get
^ 1^_ 2(4^:5) ^_
UP uu^ pp,
Hence integrating, restoring the values of m, p, h,, r,, and correcting we finally
obtain the sum of the given series, viz. : —
n(2n+3)
S, =
2(2«+l)(2n + 2)'
THE METHOD OF FINITE DIFFERENCES. 286
Ex. 11. Let the several series below be summed to n terms :
1 + 1+1+1+1+ l-f 1+ 1+ ....
1 + 2+ 3+ 4+ 5+ 6+ ;+ 8+....
1+3+ 6 + 10+15+ 21+ 28+ 36 + ....
1+4 + 10 + 20 + 35+ 56+ 84+120+....
1 + 5 + 15 + 35 + 70 + 126 + 210 + 330 + ....
Ans. Generally the m'" series is S. = "(" + !)(" + 2)- ••• ^n+(m-l)|*
1.2.3 .. .. m
OTHER EXAMPLES FOR PRACTICE.
Ex. 1. Sum the series a + ar + ar^ + .... +ar" hy integration; and likewise
3 + 6 + 12 + 24 + to ten terms.
Ex. 2. Fifty terms of the series — + — + ^ + =^
' 2.7 7.12^12.17 42'
Ex. 3. Sum the series j-y^ + ^-^^ + g-gy^ + .... to fifteen terms,
and to infinity. Ans. — and -
992 2
Ex. 4. Sum iTg + 4 o + -^-r^ + to n terms and to infinity
Ex. 5. Find the sum of fifteen t
5 6 7
^ ^ 1.2.3''" 2.3.4 "*" 3X5 "*" "
2.6^4.8 ^ 6.10
Ex. 5. Find the sum of fifteen terms of each of the following series : —
1 1 L _ 1_
^ ^ 5^-1 "*" 6'-l ■•■ 7»-l
^^^32_2»''" 72—22"^ 11-— 2' "*"
IX. The summation of series whose general term is not knoum.
The sum is obtained from the following expression, by the insertion of the
values of n and the first terms of the several orders of differences, found as in
IV.
n(n— 1) „(„_!)(„_ 2)
S. = n«. + -y-^ A«. + j-^^^ A'a.+
For we have seen in V. p. 279, that in all cases
«, = «,
«, = «, + A«,
«3 = M, + 2AM, + A'«,
«^ = «, + 3AM, + 3A''u, + Ahti
tt^ = M, + 4Am, + 6A=^m, + 4A^, + A<«,
ttg = u, + 5 AM, + lOA-'u, + IOAX + 5A*tt, + A*u,
* The several scries in this example have, in connection with each other, very remarkable
properties, and htive been raucli used in mathematical research. They are called the figurate
series, or series of fiyunde numlters ; and are thus derived from each other b_v successive addi-
tions. The first row is a scries of units, and the others are fonncd in succession b_v adding each
term of the m*^ row to the (/h+I)"" term of the (w — 1)" row. The summation shows that
S, in the successive series is expressed by the successive terms of the expanded binomial
(1 + 1)" or (1 — 1)~"; though only that of the m'* series is put down. Sec also p. 289.
286 ALGEBRA.
"Whence, adding the several vertical columns, (which are the sanne with the
series in Ex. 11, VIII. p. 285,) we have
S. = «, + «2 + M3 + + M.
, n(n— 1) n(n— l)(n— 2) ,
= n«. + ^^^ A«, + -^ ^^ A\ + ....
Ex. 1. To find the sum of n terms ofl +2 + 3 + 4
1 1 2 I 3 4 . . . . given series
1 I 1 1 . . . . first differences
0 0 second and all higher differences.
Hence «, ^ 1, Au, = 1, A'tt, = 0, and so on : and we have
„ n(n — 1) ^ , rtfra — 1) nin + 1)
as was obtained by integration at p. 283.
Ex. 2. Find the sum of 1^ + 2' + 3^ + . . . + 10-.
16 25 . . . .the given terms.
Here 1 | 4 I 9
3 I 5
2
9 . . - .first differences,
2 . . . second differences,
0 third differences.
Hence Ui = 1, A«i = 3, A^Uj = 2, A'u, = 0, and so on ; and
u(n - 1) (n-i)(n — 2) _ n(n+ 1) (2n + 1)
^" = " + ~W • ^ + 1:2:3 • ^ - 1:2:3 — •
„ , „ 10.11.21
Put n = 10, then S,o = g = 385.
Ex. 3. Find the sums of 50 terms of the series in IV. Examples 2, 3, 6.
It has been seen (p. 275) that an expression of the n"" degree has its »** dif-
ferences equal, and all the higher orders in succession become 0 : but when an
expression is of any other form, (as (f or log. x for instance,) the successive
differences never vanish ; though in many of them, these differences become
very small, and the assumption of their becoming 0, leads to a very small
amount of error in the final numerical result. When the sum of n terms of such
a series, therefore, whose differences in successive orders become very small, is
required, we are at liberty to assume those differences as actually vanishing.
Ex. 4. Required the sum of all the logarithms on the fifteen pages of Button's
Tables, from p. 186 to p. 201, inclusive, the entire numbers being integer.
Here n=S000, w,=5, AMi= -00000435, A-«,=-00000000005 *. Hence Sgooo =
8000.7999 80007999-7998
8000.5 H . -00000437 H . 00000000005
= 40143-4476668 ; which is, probably, true to five decimal places.
X. The interpolation of seiies.
Let «,, «,, Uj, M„ be the given values of an unknown function «, when
equidistant values of x are substituted, and any number,^, of them be absent :
it is required to supply the absent terms, and to find the value of the function
», for any value of x intermediate between its extreme given values.
• The second difference is taken at ( :^ ) of the interval -00000001, this being the dif-
ference at the 220i>, 210 , 200'», 190<*, and 180^, tenns : or nearly a mean of all.
THE METHOD OF FINITE DIFFERENCES. 287
1. To supply the deficient terms of the regular series.
Suppose m of the terms given, then the m'* difference, as derived from these,
must be taken equal to zero, in conformity with the principle of the last solu-
tion. But in general
A"H-i«, = «x+« — p . «,+._, H — — uh^, — • • • ± "i
and if in this we make x equal to 1, 2, (n — m) we shall have, putting
A"tt, = 0, A'Mj = 0, . . . A"«,_, = 0, the equations requisite for the determina-
tion of the n — m unknown terms, viz. :
m , mCm — 1)
A'-'w, = M„+, — J- M™ H Y2 "'-' "~ ± "1 = °
m , m(m — 1) ,
A"U^ = M,„+j — Y M„+, H —^ M« — ± «, = 0
m , Tn(m — 1)
A"«3 = M„+3 - - .u„+5 H j-^— ' «.+, — . . . . ± «3 = 0,
and so on, till we obtain as many equations as there are unknown or absent
terms of the series.
For illustration, suppose Mi and u^ were given to find k,. Then, as there is
but one deficient term, we have simply
A^u, = ttj — 21*2 -|- tt, = 0, or Mj = - {M3 + tt, } .
Again, if Wi, u.j, u^, u^, were given to find u^ : then
A''«, = % — 4m^ + 6M3 — 4Mj + m, = 0, or t/3 = g {4(Uj + mJ — («, + u^} .
Thirdly, suppose m„ Mj, Mj, v^, were given to find 1*3 and «<.
Here A'*m, =: 0 and A^u^ =: 0 ; or putting their values,
«5 — 4m^ + 6?<3 — 4W2 + M, = 0 ; Wg — 4Uj + 6«< — 4M3 + «, = 0,
which two equations resolved for u^ and u^, we obtain
a^ = ^ {— 3m, + lOUj + SMj — 2wa } ; M4 = ^ {— 2m, + 5m, + lOtt^— 3mj} .
EXAMPLES.
Ex. 1. Given the logarithms of 101, 102, 104, 105, to find that of 103. In
this, M, = 20043214, M, = 2-0086002, M^ = 2 0170333, ttj = 2 0211893. By
the method explained we have
«3 = g{4(«, + «,) - (M. + «5)} = 2-0128372.
Ex. 2. Given the logs, of 510, 511, 513, 514, to find that of 512.
Ex. 3. Given the square roots of 1, 2, 3, 5, 6, 7, 8, to find the square roots of
4 and 9. Likewise determine the cube of 10 from those of 7, 8, 9, 11, 12 ; and
from those of 8, 9, H, 12, 13.
Ex. 4. Given the logs, of 10, 11, 12, 14, 16, and 19, to interpolate the logs,
of 13, 15, 17, 18.
Ex. 5. The expression which gave the following values has its terms at the
asterisks deficient : it is required to supply them, as nearly as can be done, by
interpolation. They are 3-9956352, 3 9956396, *, *, 3-9956527, 39956571, *,
3-9956659, 3-9956703, 3 9956659.
288 '
ALGEBRA.
2. When the terms to be inserted are not those belonging to the equidistant
values of x.
The value will be approximately given by the general formula
X , ^(^— 1)^2 I ^(^ — 1) C-^ — 2) ^, ,
«^i = M, + Y '^ "' "I r2~~ ' "^ TTa ' "^ —
the series being continued till A"!*, = 0 occurs as before.
For this is only assuming that the series preserves the same law in passing
through the intermediate stages between any two terms, that it does in pEissing
from term to term by single steps.
EXAMPLES.
Ex. 1. Given log sines of 3°4', 3°5', 3°6', 3°7', 3°S', to find that of 3'=6'15".
ngles.
log. sines.
3°4'
8-7'283366
3 5
8-7306882
36
8-7330272
37
87353535
3 8
8-7376675
first diflfs.
•0023516
-0023390
•0023263
•0023140
second diffs.
— •0000126
— •0000127
— -0000123
third diffs.
— -0000001
+ 0000004
Whence, «i = 8-7283366, Am, = -0023516, ^H^ = — -0000126 and A^Uj =
g
— -0000001. Alsoa? = 3°6']5" — 3°4' = 2'15" = -,the equidistant interval being
9 45 15
1'. Consequently, by the formula, Ug = «, -j- - A «, + -^ A^u^^ -f -— A^«i=
% 4 32 128
8-733609993 = log sin 3°6'15" nearly.
Ex. 2. Given log sines of 1°, 1°1', 1°2', 1°3', to find that of 1°1'40".
Ex. 3. From the series — , — , — , — , — , find the term which lies
D\J oX Ojm do o4
in the middle between -— and --
52 53
Ans.
3S552503
2024006400"
Ex. 4. Given the log tangents of 68°54', 68°55', 68°56', 68°57', to find that
of 68°56'20".
Ex. 5. Given the natural tangents of the same arcs, to find the natural tangent
of 68°56'20": and compare its logarithm with the answer to the last example.
Ex. 6. Given the sines of 7°34', 7°35', 7°36', 7°37', 7°38', 7°39', to find the
sine of 7°37'30".
Ex. 7- Find also the same sine, supposing, first, that the sine of 7°34' had not
been given in the data; and secondly, that sine 7°40' had been given in addition
to the data.
3. When the first differences of a series of n equidistant terms are very small,
any intermediate term may be interpolated by the following formula :
n , n(n —
For(l-ir=l-«+«-^).
1) „(n-l)(n — 2)
— u, ^^ u. +
^ 1.2.3 ' ^
n(n — 1) (n — 2)
= 0.
1.2.3
+ .... =0;
and as u„ u,, u^,
n
1 •' ■ J. 2
are by hypothesis nearly equal, we have
n
I
n n(n— 1) n n{n — l) „ ,
1.2
THE METHOD OF FINITE DIFFERENCES. 2H9
EXAMPLES.
Ex. I. Given the square roots of 10, 11, 12, 13, 15, to find that of U.
Denote them by «„ u.„ «,„ m„ u^, and Wg, of which u, is the term sought, and
n = 5 the number of terms given. Hence we have
tt, — 5«, + 10«3 — IOm^ + 5«^ — Me = 0,
and hence "5 = 5 {«6 + 10m, — lOu.^ + 5«, — «,} , or
V14 = ^ {^/15 + 10^/13 - 10^/12 + 5^11 - VIO}
1870S3257 ^ ,,
= , =3-74166514 nearly.
Ex. 2. Given the square roots of 37, 38, 39, 41, and 42, to find that of 40.
Ex. 3. Given the cube roots of 45, 46, 47, 48, 49, to find that of 50.
Scholium.
In Ex. 11, p. 285, the general expression for the sum of n terms of the m"
order oijigurate numbers is given. These numbers are so called, from the cir-
cumstance of their capability of being arranged so as to form equilateral tri-
angles, squares, regular pentagons, hexagons, . . . . ; and they are accordingly
called triangular, square, pentagonal, hexagonal, . . . numbers. When the pro-
gression isl + 2 + 3 + 4 + . ..n, a triangle of that number of balls n in each
side may be formed of the sum of the series : when it is 1 + 3 + 5 + 7...
(2n — 1), then a square of the side n may be formed : when 1 +4 + 7 + 10 +
. . . (3n — 2) constitutes the series, it will form a pentagon of n balls in each
side : when 1 + 5 + 9 + 13 + .. . (4n — 3), a hexagon : and generally, when
the general term 'vi \qn — {q — \)\, then a (5 + 2)-gonal figure will result
for every integer value of q.
If, now, in any one case q receive all possible integer values from 1 to any
specified numbers, a series of polygons, of g + 2 sides each will be formed ; and
if they were balls, laid stratum on stratum, they would altogether form a
pyramid on a (g + 2)-gonal base ; and the integration of the general term of
the (7 + 2)-gonal polygonal series would give the number of balls in the
pyramid.
Amongst all these there are only two forms that are capable of being used for
the piling of balls, on account of the balls pressing unequally on the contiguous
ones, and strictly being incapable of a stable or permanent position : these are
the pyramid on the triangular hose, and the pyramid on the square base, p. 240.
First. Here (p. 283) 1 + 2 + 3 + . . . n = ** "^ - = number in the base
r , , M , n(fi + 1) nfn + 1) (n + 2) u • .1, j •
of the triangular pile ; and S ^ ~: '— 1 03 = number in the tri-
angular pile : agreeing with pages I6I and 283.
Second. For the square pile, we have n' for the general term, or number in
the base ; and (p. 283) the sum of all these strata, or courses, from 1 to n, is
w(» + )C2n + l) _ jjyjjj^gj. Qf ^^jg jjj ^jjg square pile ; again, as in pages 162
and 283.
GEOMETRY.
DEFINITIONS AND PRINCIPLES.
1. Geometry treats oi the forms, magnitudes, and positions of bodies produced
according to any specified method of construction. All the other qualities of
body are left out of consideration, and the attention confined exclusively to these,
either singly or in combination. In this case the body is called a figure.
2. The object of geometrical inquiry is twofold : — first, theoretical or specula-
tive ; and secondly, practical or operative. In one case it is a science ; in the
other it is an art founded upon science.
3. As a body of knowledge it is divided into sections, each of which is called
a proposition j and every proposition is either a theorem, a problem, or a porism.
4. A theorem is a proposition in which some statement is made concerning a
specified figure, in addition to the conditions of its construction, but inevitably
flowing from those conditions. In order to its obtaining reception as a truth, it
must be proved or demonstrated.
5. A problem is a proposition in which when certain figures are given, or
already exhibited in construction, some other figure dependent upon these is to
be found which shall fulfil some assigned conditions. It requires for its com-
pletion a discovery of the method of constructing the figure sought, and a
demonstration that the method does effect the proposed object *.
6. The words in which a proposition is stated constitute the enunciatioji ; it is
called the general enunciation when there is no reference made to an exhibited
figure ; and the particular enunciation when such a figure is referred to.
7. The general term solution is often applied to the whole series of construc-
tions and reasonings that are necessary to complete a proposition after the
general enunciation is given.
* The peculiar shade which distinguishes a porism from a problem and a theorem (of the
nature of each of which it in some degree partakes) cannot here be explained to the student
intelligibly. See, however, a note to Piob. 1. of this work; and for a liistory of the methods
devised for investigating them, the article PoriDVts in the Penny Cyclopa;dia, by J. O. Halliwell,
Esq. F.R.S.
To these may be added, thiit a lemma is a proposition which is premised, or solved before-
hand, in order to render what follows more simple and perspicuous.
A corolla?-!/ is a consequent truth, gained immediately from some preceding truth, or demon-
stration.
A scholium is a remark or observation made upon something going before it : generally of a
collateral rather than of a direct application.
291
DEFINITIONS AND PRINCIPLES.
8. Every kind of body that can be supposed actually to
exi§t has three dimensions, length, breadth, and thickness;
and the branch of geometry which takes all these dimen-
sions simultaneously into consideration is hence called the
geometry of three dimensions. It is also very frequently
called solid geometry ; and often again, though not in strict
propriety, the geometry of planes and solids.
9. The boundary of a body is the superficies or surface.
It includes the dimensions, length and breadih, but not that
of thickness. We may speak and think, and reason about
the superficies, without taking note of the thickness of the
body, whose superficies it is. Hence, for all the purposes of
geometry, the figure which is called a superficies has no
thickness.
10. The boundaries of a surface are lines. The line has
therefore no breadth nor thickness, but length only.
11. The extremities of lines, and their mutual intersec-
tions, are called points. A point, therefore, has no dimen-
sions, but position only. •
12. A straight line, or a right line, is that which preserves, at all its points, the
same direction *.
13. A curve line is one which, at its successive points, changes its direction.
14. A plane surface is one in which any two points being taken, the straight
line passing through them shall lie wholly in that plane. If this be not the case,
the superficies is said to be a curved surface.
15. A circle is a figure lying wholly upon a plane super-
ficies, and is composed of one curved line (called the circuM'
ference) such that all straight lines dra\vn from a certain
point in that surface to the circumference are equal to one
another. This point is called the centre of the circle.
The circumference itself is often called a circle, and also
sometimes the periphery.
16. The radius of a circle is a line drawn from the centre
to the circumference.
17. The diameter of a circle is a line drawn throogh the
centre, and terminating at the circumference on both sides.
18. An arc of a circle is any part of the circumference.
\^.y
• When the term line is used in the following treatise, it designates a straight line : and the
cuned line is called siinplv a curve. In more general reasoning, the generic term line is applied
to all kinds of lines, and another mode of classification of them adopted, which wiU be explamed
in a future stage of the work.
u 2
292 GEOMETRY.
19. A chord is a right line joining the extremities of an
arc.
20. A segment is any part of a circle bounded by an arc
and its chord.
21. A semicircle is half the circle, or a segment cut oflF by
a diameter.
The half circumference is sometimes called the Semicircle.
22. A sector is any part of a circle which is bounded by
an arc, and two radii drawn to its extremities.
23. A quadrant, or quarter of a circle, is a sector having
a quarter of the circumference for its arc, and its two radii
are perpendicular to each other. A quarter of the circum-
ference is sometimes called a quadrant.
24. An anffle is the inclination or opening of two lines,
having different directions, and meeting in a point.
According to circumstances, they are distinguished into
right or oblique; and the oblique are again distinguished
into acute or obtuse.
25. When one line standing on another line makes the
adjacent angles equal to one another, each of them is called
a right angle, and the straight lines are said to be perpen-
dicular to one another.
26. An oblique angle is that which is made by two
oblique lines ; and is either less or greater than a right
angle.
Of oblique angles, that which is less than a right angle,
is called acute ; and that which is greater than a right angle,
is called an obtuse angle.
27. Plane figures are bounded either by right lines or curves.
28. Plane figures that are bounded by right lines have names according to
the number of their sides, or of their angles ; for they have as many sides as
angles ; the least number being three.
29. A figure of three sides (and consequently three angles) is called a triangle :
and it receives particular denominations from the relations of its sides and
angles.
When defined according to its sides, it is equilateral, isosceles, or scalene.
30. An equilateral triangle is that whose three sides are
all equal.
31. An isosceles triangle is that which has two equal sides
\J
DEFINITIONS AND PRINCIPLES. 29S
32. A scalene triangle is that whose three sides are all unequal.
When defined according to its angles, it is either right-angled, obtuse-angled,
or acute-angled.
33. A right-angled triangle is that which has one right y\
angle. y^ \
All other triangles are oblique-angled, and are either ^
obtuse or acute. ^^j
34. An obtuse-angled triangle has one obtuse angle.
35. An acute-angled triangle has all its three angles acute. / \
36. A figure of four sides and angles is called a quadrangle, or a quadri-
lateral, or a trapezium.
37. A parallelogram is a quadrilateral which has both its pairs of opposite
sides parallel. And it takes the following particular names, viz. rectangle,
square, rhombus, rhomboid.
38. A rectangle is a parallelogram, having a right angle.
39. A square is an equilateral rectangle ; having its length
and breadth equal, or all its sides equal, and all its angles
equal. | j
40. A rhomboid is an oblique-angled parallelogram. / 7
4 1 . A rhombus is an equilateral rhomboid ; having all its . 7
sides equal, but its angles oblique. / /
42. A trapezium is a quadrilateral which has not its oppo-
site sides parallel.
43. A trapezoid has only one pair of opposite sides
parallel.
44. A diagonal is a line joining any two opposite angles
of a quadrilateral, or of any other right lined figure.
/3
45. Plane figures that have more than four sides are, in general, called poly'
gons; and they receive other particular names, according to the number of their
sides or angles. Thus,
46. A pentagon is a polygon of five sides ; a hexagon, of six sides ; a heptagon,
seven ; an octagon, eight ; a nonagon, nine ; a decagon, ten ; an undecagon,
eleven ; and a dodecagon, or duodecagon, twelve sides.
47- A regular polygon has all its sides and all its angles equal. If they are
not both equal, the polygon is irregular.
48. An equilateral triangle is also a regular figure of three sides, and the
square is one of four : the former being also called a trigon, and the latter a
tetragon.
49. Any figure is equilateral when all its sides are equal : and it is equiangular,
when all its angles are equal. When both these are equal, it is called a regular
figure.
50. By the distance of a point from a line is meant the shortest line that can
be drawn from the point to the line. It is shown that this is the perpendicular.
See th. 21.
294
GEOMETRY.
51. When two or more lines are considered in relation to one another, they
take different names, either parallel, oblique, perpendicular, or tangential.
52. Parallel lines are always at the same distance; and ~
they never meet, though ever so far produced.
53. Oblique lines change their distance, and would meet,
if produced on the side of the least distance.
54. One line is perpendicular to another, when it inclines
not more on the one side than the other, or when the angles
on both sides of it are equal.
55. A line or circle is tangential, or is a tangent to a
circle, or other curve, when it touches it, without cutting,
although both are produced.
56. The height or altitude of a figure is a perppndicular
let fall from an angle, or its vertex, to the opposite side,
called the base.
57- In a right-angled triangle, the side opposite the right angle is called the
hypothenuse; and the other two sides are called the legs, and sometimes the
base and perpendicular.
58. When an angle is denoted by three letters, of which
one stands at the angular point, and the other two on the
two sides, that which stands at the angular point is read in
the middle : thus BAD signifies the angle contained by the
lines BA and AD, and so of the other angles DAE and EAC.
50. For the purpose of calculation the circumference of every circle is
posed to be divided into 360 equal parts called degrees ; and each degree
60 minutes, each minute into 60 seconds, and so on. Hence a semicircle
tains 180 degrees, and a quadrant 90 degrees.
60. The measure of an angle, is an arc of any circle con-
tained between the two lines which form that angle, the
angular point being the centre ; and it is estimated by the
number of degrees contained in that arc, it being shown at
prop, 4, that such a mode of admeasurement is consistent
with the principles and truths of geometry.
61. Lines, or chords, are said to be equi-distant from the
centre of a circle, when perpendiculars drawn to them from
the centre are equal.
62. And the right line on which the greater perpendicular
falls, is said to be farther from the centre.
63. An angle in a segment is that which is contained by
two lines, drawn from any point in the arc of the segment,
to the two extremities of that arc.
sup-
into
con-
DEFINITIONS AND PRINCIPLES.
295
64. An angle on a segment, or an arc, is that which is contained by two
lines, drawn from any point in the opposite or 8upi>lemental part of the circum-
ference, to the extremities of the arc, and containing the arc between them.
65. An angle at the circumference, is that whose angular
point or summit is any where in the circumference : and
an ayigle at the centre, is that whose angular point is at the
centre.
66. A right-lined figure is inscribed in a circle, or the
circle circumscribes it, when all the angular points of the
figure are in the circumference of the circle.
67- A right-lined figure circumscribes a circle, or the circle
is inscribed in it, when all the sides of the figure touch the
circumference of the circle.
68. One right-lined figure is inscribed in another, or the
latter circumscribes the former, when all the angular points
of the former are placed in the sides of the latter.
69. A secant is a line that cuts a circle, lying partly
within, and partly without it.
70. Two triangles, or other right-lined figures, are said to be mutually equi-
lateral, when all the sides of the one are equal to the corresponding sides of the
other, each to each : and they are said to be mutually equiangular, when the
angles of the one are respectively equal to those of tbe other.
71. Identical figures, are such as are both mutually equilateral and equi-
angular ; or that have all the sides and all the angles of the one, respectively
equal to all the sides and all the angles of the other, each to each ; so that if the
one figure were applied to, or laid upon the other, all the sides of the one would
exactly fall upon and cover all the sides of the other j the two becoming as it
were but one and the same figure.
72. Similar figures, are those that have all the angles of the one equal to all
the angles of the other, each to each, and the sides about the equal angles pro-
portional.
73. The perimeter of a figure, is the sum of all its sides taken together.
74. The first part of geometry has respect to figures traced upon a plain
superficies, or in which only two dimensions are concerned. It is called, there-
fore, the geometry of two dimensions; and often also plane geometry, though
general usage has restricted the term, plane geometry, to a particular class of
such figures as may be traced upon a plane. These figures are entirely com-
posed of straight lines and circles, which are therefore called geometrical lines ;
whereas lines constructed any other way are called mechanical curves, or lines of
the higher orders.
296
AXIOMS.
Axioms are those fundamental truths, which, from their simplicity, are evident to
every mind, and are essential in a body of science as the foundation of a system
of reasoning.
1. Things which are equal to the same thing are equal to each other.
2. "When equals are added to equals, the wholes are equal.
3. When equals are taken from equals, the remainders are equal.
4. When equals are added to unequals, the wholes are unequal.
5. "When equals are taken from unequals, the remainders are unequal.
6. Things which are double of the same thing, or equal things, are equal to
each other.
7. Things which are halves of the same thing, are equal.
8. Every whole is equal to all its parts taken together, and greater than any
of them.
9. Things which coincide, or fill the same space, are identical, or mutually
equal in all their parts.
10. All right angles are equal to one another.
11. Angles that have equal measures, or arcs, are equal.
THEOREM I. *
If two triangles have two sides and the included angle in the one, equal to two sides
and the included angle in the other, the triangles will be identical, or equal in all
respects.
In the two triangles ABC, DEF, if the side AC be
equal to the side DF, and the side BC equal to the side
EF, and the angle C equal to the angle F ; then will
the two triangles be identical, or equal in all respects.
For conceive the triangle ABC to be applied to, or
placed on, the triangle DEF, in such a manner that the point C may coincide
with the point F, and the side AC with the side DF, which is equal to it.
Then, since the angle F is equal to the angle C {hyp.), the side BC will fall on
* In the complete process by which a theorem is enunciated and established, the following
parts are to be always found.
I. The general enunciation, already explained, (see def. 6.) and which comprises
1. The sidtject spoken of, or the hypotliesis to which the theorem is affirmed to be true.
2. The predicate or affirmation made respecting tlie hypothetical figure. These arc alike to
be found in the general and particular enunciations.
II. Sometimes a preliminary construction is required to connect the hypothetical with the
predicated parts of tlie figure.
III. One or more syllogisms, by which the necessary dependence of the predicate upon the
hypothesis is established.
The syllogism is composed of three propositions,
1. The major premise : — an axiom or previously demonstrated truth.
2. The minor premise : — a proposition which agrees in sidtject with this axiom or theorem,
and in predicate with the enunciated theorem.
3. The conclusion, which is the theorem in question, and inferred from the premises.
THEOREMS. 297
the side EF. Also, because AC is equal to DF, and BC equal to EF {hyp.), the
point A will coincide with the point D, and the point B with the point E ; con-
sequently the side AB will coincide with the side DE. Therefore the two
triangles are identical, and have all their other corresponding parts equal (ax. 9),
namely, the side AB equal to the side DE, the angle A to the angle D, and the
angle B to the angle E.
THEOREM II.
When two triangles have two angles and the included side in the one, equal to two
angles and the included side in the other, the triangles are identical, or have
their other sides and angle equal.
Let the two triangles ABC, DEE, have the angle A
equal to the angle D, the angle B equal to the angle
E, and the side AB equal to the side DE; then these
two triangles will be identical.
For, conceive the triangle ABC to be placed on the triangle DEF, in such
manner that the side AB may fall exactly on the equal side DE. Then, since
the angle A is equal to the angle D (hyp.), the side AC naust fall on the side
DF ; and, in like manner, because the angle B is equal to the angle E, the side
BC must fall on the side EF. Thus the three sides of the triangle ABC will be
exactly placed on the three sides of the triangle DEF ; consequently the two
triangles are identical {ax 9). having the other two sides AC, BC, equal to the
two DF, EF, and the remaining angle C equal to the remaining angle F.
THEOREM III.
In an isosceles triangle, the angles at the base are equal : or, if a triangle have
two sides equal, their opposite angles will also be eqiuxl.
If the triangle ABC have the side AC equal to the side
BC : then will the angle B be equal to the angle A.
For, conceive the angle C to be bisected, or divided into
two equal parts, by the line CD, making the angle ACD
equal to the angle BCD.
Then, the two triangles, ACD, BCD, have two sides and the contained angle
of the one, equal to two sides and the contained angle of the other, viz. the side
AC equal to BC, the angle ACD equal to BCD, and the side CD common ;
therefore these two triangles are identical, or equal in all respects (/A. 1) ; and
consequently the angle A equal to the angle B.
Cor. 1 . Hence the line which bisects the vertical angle of an isosceles triangle,
bisects the base, and is also perpendicular to it.
Cor. 2. Hence too it appears, that every equilateral triangle, is also equi-
angular, or has all its angles equal.
THEOREM IV.
When one line meets another, the angles which it makes on the same side of the
other, are together equal to two right angles.
Let the line AB meet the line CD : then will the two
angles ABC, ABD, taken together, be equal to two right
angles.
For, first, when the two angles ABC, ABD, are equal to
each other, they are both of them right angles {def. 23).
298 GEOMETRY.
But when the angles are unequal, suppose BE drawn perpendicular to CD.
Then, since the two angles EBC, EBD, are right angles {def. 25), and the angle
EBD is equal to the two angles EBA, ABD, together (cu7. 8), the three angles,
EBC, EBA, and ABD, are equal to two right angles.
But the two angles EBC, EBA, are together equal to the angle ABC {ax. 8).
Consequently the two angles ABC, ABD, are also equal to two right angles.
Cor. 1. Hence also, conversely, if the two angles ABC, ABD, on both sides
of the line AB, make up together two right angles, then CB and BD form one
continued right line CD.
Cor. 2. Hence, all the angles which can be made, at any point B, by any
number of lines, on the same side of the right line CD, are, when taken all
together, equal to two right angles.
Cor. 3. And, as all the angles that can be made on the other side of the line
CD are also equal to two right angles ; therefore all the angles that can be made
quite roimd a point B, by any number of lines, are equal to four right angles.
Cor. 4. Hence also the whole circumference of a circle,
being the sum of the measures of all the angles that can be
made about the centre F {def. 57), is the measure of four
right angles. Consequently, a semicircle, or 180 degrees,
is the measure of two right angles ; and a quadrant, or 90
degrees, the measure of one right angle.
THEOREM V.
When two lines intersect each other, the opposite angles are equal.
Let the two lines AB, CD, intersect in the point E ; then
will the angle AEC be equal to the angle BED, and the
angle AED equal to the angle CEB
For, since the line CE meets the line AB, the two angles /
AEC, BEC, taken together, are equal to the two right ^
angles {th. 4).
In like manner, the line BE, meetiug the line CD, makes the two angles
BEC, BED, equal to two right angles.
Therefore the sura of the two angles AEC, BEC, is equal to the sum of the
two BEC, BED {ax. 1).
And if the angle BEC, which is common, be taken away from both these, the
remaining angle AEC will be equal to the remaining angle BED {ax. 3).
And in like manner it may be shown, that the angle AED is equal to the
opposite angle BEC.
THEOREM VI.
When one side of a triangle is produced, the outward angle is greater than either
of the two inward opposite angles.
Let ABC be a triangle, having the side AB produced to
D ; then will the outward angle CBD be greater than either
of the inward opposite angles A or C.
For, conceive the side BC to be bisected in the point E,
and draw the line AE, producing it till EF be equal to AE ;
and join BF.
THEOREMS.
S99
Then, since the two triangles AEC, BEF, have the side AE = the side EF,
and the side CE = the side BD {suppos.), and the included or opposite angles at
E also equal {th. 5), therefore those two triangles are equal in all respects (/A. 1),
and have the angle C = the corresponding angle EBF. But the angle CBD is
greater than the angle EBF ; consequently, the said outward angle CBD is also
greater than the angle C.
In like manner, if CB be produced to G, and AB be bisected, it may be
shown that the outward angle ABG, or its equal CBD, is greater than the other
angle A.
THEOREM VII.
When a triangle has two of its angles equal, the sides opposite to them are also
equal.
For draw CD perpendicular to the base ; and conceive the
triangle BCD to revolve about CD till the plane of it coin-
cides with that of CDA. Then since CDA and CDB are
right angles (def. 25), the line BD will coincide with DA.
Now if B coincide with A, the line CB will coincide with
CA, and the angles CBD, CAD coinciding, they will be
equal : but if it be denied that B will coincide with A, let it coincide with some
other point E in the line DA, and join CE, and hence (th. 6) the angle CED
(that is CBD) is greater than CAD, which is contrary to the hypothesis, and
hence cannot be true of the figure respecting which the hypothesis is made.
Hence B cannot but coincide with A, and CA cannot but be equal to CB.
Cor. Hence every equiangular triangle is also equilateral.
THEOREM VIII.
When two triangles have all the three sides in the one, equal to all the three sides
in the other, the triangles are identical, or have also their three angles equal, each
to each.
Let the two triangles ABC, ABD, have their three sides
respectively equal, viz. the side AB equal to AB, AC to
AD, and BC to BD ; then shall the two triangles be iden-
tical, or have their angles equal, viz. those angles that are
opposite to the equal sides ; namely, the angle BAC to the
angle BAD, the angle ABC to the angle ABD, and the
angle C to the angle D.
For, conceive the two triangles to be joined together by their longest equal
sides, and draw the line CD.
Then, in the triangle ACD, because the side AC is equal to AD (hyp.), the
angle ACD is equal to the angle ADC (th. 3). In like manner, in the triangle
BCD, the angle BCD is equal to the angle BDC, because the side BC is equal
to BD. Hence then, the angle ACD being equal to the angle ADC, and the
angle BCD to the angle BDC, by equal additions the sum of the two angles
ACD, BCD, is equal to the sum of the two ADC, BDC (ax. 2), that is the whole
angle ACB equal to the whole angle ADB.
Since then, the two sides AC, CB, are equal to the two sides AD, DB, each
to each (hyp.), and their contained angles ACB, ADB, also equal, the two
triangles ABC, ABD, are identical (th. 1), and have the other angles equal, viz.
the angle BAC to the angle BAD, and the angle ABC to the angle ABD
300.
GEOMETRY.
THEOREM IX.
The greater side of every triangle is opposite to the greater angle ; and the greater
angle opposite to the greater side.
Let ABC be a triangle, having the side AB greater than
the side AC ; then will the angle ACB, opposite the greater
side AB, be greater than the angle B, opposite the less side
AC.
For, on the greater side AB, take the part AD equal to
the less side AC, and join CD. Then, since BCD is a triangle, the outward
angle ADC is greater than the inward opposite angle B (th. 6). But the angle
ACD is equal to the said outward angle ADC, because AD is equal to AC
(/A. 3). Consequently, the angle ACD also is greater than the angle B. And
since the angle ACD is only a part of ACB, much more must the whole angle
ACB be greater than the angle B.
Again, conversely, if the angle C be greater than the angle B, then will the
side AB, opposite the former, be greater than the side AC, opposite the latter.
For, if AB be not greater than AC, it must be either equal to it, or less than
it. But it cannot be equal, for then the angle C would be equal to the angle B
(th. 3), which it is not, by the supposition. Neither can it be less, for then the
angle C would be less than the angle B, by the former part of this; which is
also contrary to the supposition. The side AB, then, being neither equal to AC,
nor less than it, must necessarily be greater.
THEOREM X.
TTie sum of any two sides of a triangle is greater than the third side
Let ABC be a triangle ; then will the sum of any two
of its sides be greater than the third side, as for instance,
AC + CB greater than AB.
For, produce AC till CD be equal to CB, or AD equal
to the sum of the two AC + CB; and join BD : — then,
because CD is equal to CB (constr.), the angle D is equal
to the angle CBD {th. 3). But the angle ABD is greater
than the angle CBD, consequently it must also be greater than the angle D.
And, since the greater side of any triangle is opposite to the greater angle (th. 9),
the side AD (of the triangle ABD) is greater than the side AB. But AD is
equal to AC and CD, or AC and CB, taken together (constr.) ; therefore
AC + CB is also greater than AB.
Cor. The shortest distance between two points is a single right line drawn
from the one point to the other.
THEOREM XI.
The difference of any two sides of a triangle is less than the third side.
Let ABC be a triangle ; then will the difi'erence of any
two sides, as AB — AC, be less than the third side BC.
For, produce the less side AC to D, till AD be equal to
the greater side AB, so that CD may be the difference of the
two sides AB — AC ; and join BD. Then, because AD is
equal to AB (constr.) the opposite angles D and ABD are
THEOREMS. SOJ
equal {th. 3). But the angle CBD is less than the angle ADD, and conse-
quently also less than the equal angle D. And since the greater side of any
triangle is opposite to the greater angle (th. 9), the side CD (of the triangle
BCD) is less than the side BC.
Otherwise. Set off upon AB, a distance A I, equal to AC. Then {th. 20)
AC + CB is greater than AB, that is, greater than AI + IC. From these,
take away the equal parts AC, AI, respectively ; and there remains CB greater
than IC. Consequently, IC is less than CB.
THEOREM XII.
When a line intersects two parallel lines, it makes the alternate angles equal to
each other.
Let the line EF cut the two parallel lines AB, CD ; then
will ihe angle AEF be equal to the alternate angle EFD.
For if they are not equal, one of them must be greater
than the other; let it be EFD for instance, which is the
greater, if possible ; and conceive the line FB to be drawn,
cutting off the part or angle EFB equal to the angle AEF,
and meeting the line AB in the point B.
Then, since the outward angle AEF, of the triangle BEF, is greater than the
inward opposite angle EFB (/A. 6) ; and since these two angles also are equal
(constr.), it follows, that those angles are both equal and unequal at the
same time : which is impossible. Therefore, the angle EFD is not unequal to
the alternate angle AEF, that is, they are equal to each other.
Cor. Right lines which are perpendicular to one of two parallel lines, are also
perpendicular to the other.
THEOREM XIII.
When a line, cutting two other lines, makes the alternate angles equal to
each other, those two lines are parallel.
Let the line EF, cutting the two lines AB, CD, make the
alternate angles AEF, DFE, equal to each other ; then will
AB be parallel to CB.
For if they be not parallel, let some other line, as FG, be
parallel to AB. Then, because of these parallels, the angle
AEF is equal to the alternate angle EFG [th. 12). But the
angle AEF is equal to the angle EFD (hyp.) Therefore the angle EFD is equal
to the angle EFG {ax. 1) ; that is, a part is equal to the whole : which ia impos-
sible. Therefore no line but CD can be parallel to AB.
Cor. Those lines which are perpendicular to the same line, are parallel to each
other.
SOg GEOMETRY.
THEOREM XIV.
When a line cuts two parallel lines, the outward angle is equal to the inward oppo-
site one, on the same side; and the two inward angles, on the same side, are
toyether equal to two right angles.
Let the line EF cut the two parallel lines AB, CD ; then
will the outward angle EGB be equal to the inward opposite
angle GHD, on the same side of the line EF; and the two
inward angles BGH, GHD, taken together, will be equal to
two right angles.
For since the two lines AB, CD, are parallel, the angle
AGH is equal to the alternate angle GHD {th. 12). But the
angle AGH is equal to the opposite angle EGB {ih. 3.) Therefore the angle
EGB is also equal to the angle GHD {ax. 1).
Again, because the two adjacent angles EGB, BGH, are t(^ether equal to two
right angles {th. 4), of which the angle EGB has been shown to be equal to the
angle GHD ; therefore the two angles BGH, GHD, taken together, are also
equal to two right angles.
Cor. 1 . And, conversely, if one line meeting two other lines, make the angles
on the same side of it equal, those two lines are parallels.
Cor. 2. If a line, cutting two other lines, make the sum of the two inward
angles on the same side, less than two right angles, those two lines will not be
parallel, but will meet each other when produced.
THEOREM XV.
Those lines which are parallel to the same line are parallel to each other.
Let the lines AB, CD, be each of them parallel to the line ^
EF ; then shall the lines AB, CD, be parallel to each other.
For, let the line GI be perpendicular to EF. Then will
this line be also perpendicular to both the lines AB, CD
{cor. th. 12), and consequently the two lines AB, CD, are ^
parallels {cor. th. 13).
THEOREM XVI.
When one side of a triangle is produced, the outward angle is equal to both the
inward opposite angles taken together.
Let the side AB, of the triangle ABC, be produced to
D ; then will the outward angle CBD be equal to the
sum of the two inward opposite angles A and C.
For, conceive BE to be drawn parallel to the side AC
of the triangle. Then BC, meeting the two parallels
AC, BE, makes the alternate angles C and CBE equal
{Ih. 12). And AD, cutting the same two parallels AC, BE, makes the inward
and outward angles on the same side, A and EBD, equal to each other {th. 14).
Therefore, by equal additions, the sum of the two angles A and C, is equal to
the sum of the two CBE and EBD, that is, to the whole angle CBD {ax. 2).
THEOREMS. 303
THEOREM XVII.
In any triangle, the sum of all the three angles is equal to two right angles.
Let ABC be any plane triangle ; then the sum of the
three angles, A + B + C, is equal to two right angles.
For, let the side AB be produced to D. Then the out-
ward angle CBD is equal to the sum of the two inward
opposite angles A -f C {Ih. IG). To each of these equals
add the inward angle B, then will the sum of the three
inward angles, A + B + C, be equal to the sum of the two adjacent angles
ABC + CBD {ax. 2). But the sum of these two last adjacent angles is equal
to two right angles {th. 4). Therefore also the sum of the three angles of the
triangle, A + B + C, is equal to two right angles {ax. 1).
Cor. 1. If two angles in one triangle be equal to two angles in another
triangle, the third angles will also be equal {ax. 3), and the two triangles equi-
angular.
Cor. 2. If one angle in one triangle, be equal to one angle in another, the
sums of the remaining angles will also be equal {ax. 3).
Cor. 3. If one angle of a triangle be right, the sum of the other two will also
be equal to a right angle, and each of them singly will be acute, or less than a
right angle.
Cor. 4. The two least angles of every triangle are acute, or each less than a
right angle.
THEOREM XVIII.
In any quadrangle, the sum of all the four inward angles is equal to four right
angles.
Let ABCD be a quadrangle ; then the sum of the four
inward angles, A + B + C + D, is equal to four right angles.
Let the diagonal AC be drawn, dividing the quadrangle
into two triangles, ABC, ADC. Then, because the sum of
the three angles of each of these triangles is equal to two
right angles {th. 17); it follows, that the sum of all the
angles of both triangles, which make up the four angles of the quadrangle, must
be equal to four right angles {ax. 2).
Cor. 1. Hence, if three of the angles be right ones, the fourth will also be a
right angle.
Cor. 1. And if the sum of two of the four angles be equal to two right angles,
the sum of the remaining two will also be equal to two right angles.
THEOREM XIX.
In any figure whatever, the sum of all the inward angles, taken together, is equal to
twice as many right angles, wanting four, as the figure has sides.
Let ABCDE be any figure; then the sum of all its inward
angles, A -f B + C + D + E, is equal to twice as many
right angles, wanting four, as the figure has sides.
For, from any point P, within it, draw lines, PA, PB, PC,
&c. to all the angles, dividing the polygon into as many tri-
angles as it has sides. Now the sum of the three angles of
each of these triangles, is equal to two right angles {th. 17) ;
304 GEOMETRY.
therefore the sum of the angles of all the triangles is equal to twice as many
right angles as the figure has sides. But the sura of all the angles about the
point P, which are so many of the angles of the triangles, but no part of the
inward angles of the polygon, is equal to four right angles, (cor. 3, th. 4,) and
must be deducted out of the former sum. Hence it follows that the sum of all
the inward angles of the polygon alone, A + B + C + D + E, is equal to
twice as many right angles as the figure has sides, wanting the said four right
angles.
THEOREM XX.
When every side of any figure is produced, the sum of all the outward angles
thereby made is equal to four right angles.
Let a, B, C, be the outward angles of any poly-
gon, made by producing all the sides ; then will the sum,
A + B + C + D + E, of all those outward angles, be
equal to four right angles.
For every one of these outward angles, together with /^
its adjacent inward angle, make up two right angles, as
A + a equal to two right angles, being the two angles
made by one line meeting another {th. 4). And there \
being as many outward, or inward angles, as the figure
has sides ; therefore the sum of all the inward and outward angles, is equal to
twice as many right angles as the figure has sides. But the sura of all the
inward angles, with four right angles, is equal to twice as raany right angles as
the figure has sides {th. 19). Therefore the sum of all the inward and all the
outward angles is equal to the sum of all the inward angles and four right
angles {ax. 1). From each of these take away all the inward angles, and
there remain all the outward angles equal to four right angles (by ax. 3).
THEOREM XXI.
A perpendicular is the shortest line that can be drawn from a given point to an in-
definite line: and, of any other lines drawn from the same point, those that are
nearest the perpendicular are less than those more remote.
If AB, AC, AD, .... be lines drawn from the given point ^
A, to the indefinite hne DE, of which AB is perpendicular ;
then shall the perpendicular AB be less than AC, and AC
less than AD, &c.
For, the angle B being a right one, the angle C is acute
{cor. 3, th. 17), and therefore less than the angle B. But
the less angle of a triangle is subtended by the less side {th. 9). Therefore the
side AB is less than the side AC.
Again, the angle ACB being acute, as before, the adjacent angle ACD will be
obtuse {th. 4) ; consequently the angle D is acute {cor. 3, th. 17), and there-
fore is less than the angle C. And since the less side is opposite to the less
angle, therefore the side AC is less than the side AD.
Cor. A perpendicular is the least distance of a given point from a line.
THEOREMS. 305
THEOREM XXII.
The opposite sides and angles of any parallelogram are equal to each other ; and
the diagonal divides it into two equal triangles.
Let ABCD be a parallelogram, of which the diagonal is
BD ; then will its opposite sides and angles be equal to each
other, and the diagonal BD will divide it into two equal
parts, or triangles.
For, since the sides AB and DC are parallel, as also the
sides AD and BC {def. 37), and the line BD meets them ;
therefore the alternate angles are equal {th. 12), namely, the angle ABD to the
angle CDB, and the angle ADB to the angle CBD : hence the two triangles,
having two angles in the one equal to two angles in the other, have also their
third angles equal (cor. 1, th. 17), namely, the angle A equal to the angle C,
which are two of the opposite angles of the parallelogram.
Also, if to the equal angles ABD, CDB, be added the equal angles CBD,
ADB, the wholes will be equal {ax. 2), namely, the whole angle ABC to the
whole ADC, which are the other two opposite angles of the parallelogram.
Again, since the two triangles are mutually equiangular and have a side in
each equal, viz. the common side BD ; therefore the two triangles are identical
(Jth. 2), or equal in all respects, namely, the side AB equal to the opposite side
DC, and AD equal to the opposite side BC, and the whole triangle ABD equal
to the whole triangle BCD.
Cor. 1. Hence, if one angle of a parallelogram be a right angle, all the other
three will also be right angles, and the parallelogram a rectangle.
Cor. 2. Hence also, the sum of any two adjacent angles of a parallelogram is
equal to two. right angles.
THEOREM XXIII.
Every quadrilateral, whose opposite sides are equal, is a parallelogram, or has Us
opposite sides parallel.
Let ABCD be a quadrangle, having the opposite sides
equal, namely, the side AB equal to DC, and AD equal to
BC ; then shall these equal sides be also parallel, and the
figure a parallelogram.
For, let the diagonal BD be drawn. Then, the triangles,
ABD, CBD, being mutually equilateral {hyp.), they are also
mutually equiangular {th. 8), or have their corresponding angles equal ; conse-
quently the opposite sides are parallel {th. 13) ; viz. the side AB parallel to DC,
ind AD parallel to BC, and the figure is a parallelogram.
THEOREM XXIV.
Those lines which join the corresponding extremes of two equal and parallel lines,
are themselves equal and parallel.
, Let AB, DC, be two equal and parallel lines ; then will the lines AD, BC,
vhich join their extremes, be also equal and parallel. {See the fig. above.)
For, draw the diagonal BD. Then, because AB and DC are parallel, {hyp.)
VOL. I. 3^.
306 GEOMETRY.
the angle ABD is equal to the alternate angle BDC (th. 12) : hence then, the
two triangles having two sides and the contained angles equal, viz. the side AB
equal to the side DC, and the side BD common, and the contained angle ABD
equal to the contained angle BDC, they have the remaining sides and angles
also respectively equal {th. 1) ; consequently AD is equal to BC, and also paral-
lel to it {th. 12).
THEOREM XXV.
Parallelograms, as also triangles, standing on the same base, and between the same
parallels, are equal to each other.
Let ABCD, ABEF, be two parallelograms, and ABC,
ABF, two triangles, standing on the same base AB, and
between the same parallels AB, DE; then will the paral-
lelogram ABCD be equal to the parallelogram ABEF,
and the triangle ABC equal to the triangle ABF.
For, since the line DE cuts the two parallels AF, BE,
and the two AD, BC, it makes the angle E equal to the •* »
angle AFD, and the angle D equal to the angle BCE
{th. 14) ; the two triangles ADF, BCE, are therefore equiangular (cor. 1, ih.l7)i
and having the two corresponding sides AD, BC, equal {th. 22), being opposite
sides of a parallelogram, these two triangles are identical, or equal in all respects
{th. 2). If each of these equal triangles then be taken from the whole space
ABED, there will remain the parallelogram ABEF in the one case, equal to the
parallelogram ABCD in the other {ax. 3).
Also the triangles ABC, ABF, on the same base AB, and between the same
parallels, are equal, being the halves of the said equal parallelograms {th. 22).
Cor. 1. Parallelograms, or triangles, having the same base and altitude, are
equal. For the altitude is the same as the perpendicular or distance between the
two parallels, which is every where equal, by the definition of parallels.
Cor. 2. Parallelograms, or triangles, having equal bases and altitudes, are
equal. For, if the one figure be applied with its base on the other, the bases will
coincide or be the same, because they are equal : and so the two figures, having
the same base and altitude, are equal.
THEOREM XXVI.
If a parallelogram and a triangle stand on the same base, and between the same
parallels, the parallelogram will be double the triangle, or the triangle half the
parallelogram.
Let ABCD be a parallelogram, and ABE a triangle,
on the same base AB, and between the same parallels
AB, DE ; then will the parallelogram xVBCD be double
the triangle ABE, or the triangle half the parallelogram.
For, draw the diagonal AC of the parallelogram, divid-
ing it into two equal parts {th. 22). Then because the tri-
angles AHC, ABE, on the same base, and between the
same parallels, are equal {th. 25), and because the one
triangle ABC is half the parallelogram ABCD {th. 22), the other equal triangle
ABE is also equal to half the same parallelogram ABCD.
THEOREMS.
307
Cor. 1. A triangle is equal to half a parallelogram of the same base and altitude,
because the altitude is the perpendicular distance between the parallels, which is
every where equal, by the definition of parallels.
Cor. 2. If the base of a parallelogram be half that of a triangle, of the same
altitude, or the base of the triangle be double that of the parallelogram, the two
figures will be equal to each other.
0
THEOREM XXVII.
Rectangles that are contained by equal lines, are equal to each other.
Let BD, FH, be two rectangles, having the sides AB,
BC, equal to the sides EF, FG, each to each ; then will
the rectangle BD be equal to the rectangle FH.
For, draw the two diagonals AC, EG, dividing the
two parallelograms each into two equal parts. Then the
two triangles ABC, EFG, are equal to each other (/A. 1),
because they have the two sides AB, BC, and the con-
tained angle B, equal to the two sides EF, FG, and the contained angle F. {hyp.)
But these equal triangles are the halves of the respective rectangles : and be-
cause the halves, or the triangles, are equal, the wholes, or the rectangles DB,
HF, are also equal {ax. 6).
Cor. The squares on equal lines are also equal ; for every square is a species
of rectangle.
THEOREM XXVIII.
7%e complements of the parallelograms, which are about the diagonal of any
parallelogram, are equal to each other.
Let AC be a parallelogram, BD a diagonal, EIF paral-
lel to AB or DC, and GIH parallel to AD or BC, making
AL IC, complements to the parallelograms EG, HF,
which are about the diagonal DB : then wiU the comple-
ment AI be equal to the complement IC.
For, since the diagonal DB bisects the three parallelo-
grams AC, EG, HG {th. 22) ; therefore, the whole triangle DAB being equal to
the whole triangle DCB, and the parts DEI, IHB, respectively equal to the
parts DGI, IFB, the remaining parts AI, IC, must also be equal (a*. 3).
THEOREM XXIX.
A trapezoid, or trapezium having two sides parallel, is equal to half a parallelogram,
whose base is the sum of those two sides, and its altitude the perpendicular dis-
tance between them.
Let ABCD be the trapezoid, having its two sides AB,
DC, parallel ; and in AB produced take BE, equal to
DC, so that AK may be the sum of the two parallel
sides; produce DC also, and let EF, GC, BH, be all
three parallel to AD. Then is AF a parallelogram of the
same altitude with the trapezoid ABCD, having its base
AE equal to the sum of the parallel sides of the trapezoid ; and it is to be proved
that the trapezoid ABCD is equal to half the parallelogram AF.
x2
30» GEOMETRY.
Now, since triangles, or parallelograms, of equal bases and altitude, are equal
(fior. 2, th. 25), the parallelogram DG is equal to the parallelogram HE, and the
triangle CGB is equal to the triangle CHB ; consequently the line BC bisects, or
equally divides, the parallelogram AF, and ABCD is the half of it.
THEOREM XXX.
77ie sum of all the rectangles contained under one whole line, and the several parts
of another line, any way divided, is equal to the rectangle contained under the two
whole lines.
Let ad be the one line, and AB the other, divided
into the parts AE, EF, FB ; then will the rectangle con-
tained by AD and AB, be equal to the sum of the rect-
angles of AD and AE, and AD and EF, and AD and
FB : thus expressed, AD.AB = AD.AE + AD EF +
AD.FB.
For, make the rectangle AC of the two whole lines AD, AB ; and draw EG,
FH, perpendicular to AB, or parallel to AD, to which they are equal (th. 22).
Then the whole rectangle AC is made up of all the other rectangles AG, EH,
FC : but these rectangles are contained by AD and AE, EG and EF, FH and
FB ; which are equal to the rectangles of AD and AE, AD and EF, AD and
FB, because AD is equal to each of the two EG, FH. Therefore the rectangle
AD.AB is equal to the sum of all the other rectangles AD.AE, AD.EF,
AD.FB.
Cor. If a right line be divided into any two parts, the square on the whole
line is equal to both the rectangles of the whole line and each of the parts.
E H D
I
& C B
THEOREM XXXr.
TTie square of the sum of two lines is greater than the sum of their squares, by
twice the rectangle of the said lines. Or, the square of a whole line is equal to
the squares of its two parts, together with twice the rectangle of those parts.
Let the line AB be the sum of any two lines AC, CB;
then will the square of AB be equal to the squares of
AC, CB, together with twice the rectangle of AC.CB.
That is, AB2 = AC^ + CB= + 2AC CB.
For, let ABDE be the square on the sum or whole line
AB, and ACFG the square on the part AC. Produce CF
and GF to the other sides at H and L
From the lines CH, GI, which are equal, being each equal to the sides of the
square AB or BD {th. 22), take the parts CF, GF, which are also equal, being
the sides of the square AF, and there remains FH equal to FI, which are also
equal to DH, DI, being the opposite sides of the parallelogram. Hence the
fit{ure HI is equilateral : and it has all its angles right ones {cor. 1, th. 22); it is
therefore a square on the line FI, or the square of its equal CB. Also the figures
EF, FB, are equal to two rectangles under AC and CB, because GF is equal to
AC, and FH or FI equal to CB : but the whole square AD is made up of the
four figures, viz. the two squares AF, FD, and the two equal rectangles EF, FB ;
that is, the square of AB is equal to the squares of AC, CB, together with twice
the rectangle of AC, CB. -.
THEOREMS.
309
Cor. Hence, if a line be divided into two equal parts ; the square of the
whole line will be equal to four times the square of half the line.
THEOREM XXXII.
TTie square of the difference of two lines is less than the sum of their squares, by
twice the rectangle of the said lines.
Let AC, BC, be any two lines, and AB their difference :
then will the square of AB be less than the squares of AC,
BC, by twice the rectangle of AC and BC. Or AB- =
AC2 + BC2 - 2AC.BC.
For, let ABDE be the square on the difference AB, and
ACFG the square on the line AC. Produce ED to H ; also
produce DB and HC, and draw KI, making BI the square
of the other line BC.
Now, it is obvious, that the square AD is less than the two squares AF, BI, by
the two rectangles EF, DI : but OF is equal to the one line AC, and GE or
FH is equal to the other line BC ; consequently the rectangle EF, contained
under EG and GF, is equal to the rectangle of AC and BC.
Again, FH being equal to CI or BC or DH, by adding the common part
HC, the whole HI will be equal to the whole FC, or equal to AC ; and conse-
quently the figure DI is equal to the rectangle contained by AC and BC.
Hence the two figures EF, DI, are two rectangles of the two lines AC, BC;
and consequently the square of AB is less than the squares of AC, BC, by twice
the rectangle AC, BC.
t-i
THEOREM XXXIII.
The rectangle under the sum and difference of two lines, is equal to the difference
of the squares of those lines.*
Let AB, AC, be any two unequal lines; then will the
difference of the squares of AB, AC, be equal to a rectangle
under their sum and difference : that is, AB' — AC* =
(AB + AC) (AB - AC).
For, let ABDE be the square of AB, and ACFG the square
of AC. Produce DB till BH be equal to AC ; draw HI
parallel to AB or ED, and produce FC both ways to
I and K.
Then the difference of the two squares AD, AF, is evi-
i—A
dently the two rectangles EF, KB. But the rectangles EF, BI, are equal, being
* This and the two preceding theorems are evinced algebraically, by the three expressions
{a + by = a* -^ 2al> + 6» = o» + i» + 2a4
(a — by -a^ — 2<xb + b^ = a^-\-b'—2ab
(a + b){a — b) = a'> — b\
Of course it is here assumed that an algebraic product corresponds to a geometrical rectangle,
which is 8ho\vn in the Application of Algebra to G^metry.
310 GEOMETRY.
contained under equal lines; for EK and BH are each equal to AC, and GE is
equal to CB, being equal to the difference between AB and AC, or their equals
AE and AG. Therefore the two EF, KB, are equal to the two KB, BI, or to
the whole KH ; and consequently KH is equal to the difference of the squares
AD, AF. But KH is a rectangle contained by DH, or the sum of AB and AC,
and by KD, or the difference of AB, and AC : therefore the difference of the
squares of AB, AC, is equal to the rectangle under their sum and difference.
THEOREM XXXIV.
In any right-angled triangle, the square of the hypolhenuse is equal to the sum of
the squares of the other two sides.
Let ABC be a right-angled triangle, having the right ^
angle C ; then will the square of the hypothenuse AB,
be equal to the sum of the squares of the other two sides
AC, CB. Or AB- = AC^ + BC".
For, on AB describe the square AE, and on AC, CB,
the squares AG, BH ; then draw CK parallel to AD or
BE; and join AI, BF, CD, CE.
Now, because the line AC meets the two CG, CB, so
as to make two right angles, these two form one straight
line GB {cor. 1, th. 6). And because the angle FAC is equal to the angle
DAB, being each a right angle, or the angle of a square ; to each of these angles
add the common angle EAC, so wiU the whole angle or sum FAB, be equal to
the whole angle or sum CAD : but the line FA is equal to the line AC, and the
line AB to the line AD, being sides of the same square ; so that the two sides
FA, AB, and their included angle FAB, are equal to the two sides CA, AD, and
the contained angle CAD, each to each : therefore the whole triangle AFB is
equal to the whole triangle ACD {th. 1).
But the square AG is double the triangle AFB, on the same base FA, and
between the same parallels FA, GB {th. 26) ; in like manner the parallelogram
AK is double the triangle ACD, on the same base AD, and between the same
parallels AD, CK : and since the doubles of equal things are equal {ax. 6) ;
therefore the square AG is equal to the parallelogram AK.
In like manner, the other square BH is proved equal to the other parallelo-
gram BK : consequently the two squares AG and BH together, are equal to
the two parallelograms AK and BK together, or to the whole square AE; that
is, the sum of the two squares on the two less sides is equal to the squares on
the greatest side.
Cor. 1. Hence, the square of either of the two less sides, is equal to the
difference of the squares of the hypothenuse and the other side {ax. 3); or equal
to the rectangle contained by the sum and difference of the said hypothenuse and
other side {th. 33).
Cor. 2. Hence, also, if two right-angled triangles have two sides of the one
equal to two corresponding sides of the other ; their third sides will also be
equal, and the triangles identical.
THEOREMS.
311
THEOREM XXXV.
In any triangle, the difference of the squares of the two sides is equal to the dif-
ference of the squares of the segments of the base, or of the two lines, or distances »
included between the extremes of the base and the perpendicular.
Let ABC be any triangle, haA'ing CD perpendicu.
lar to AB ; then will the difference of the squares of
AC, BC, be equal to the difference of the squares of
AD, BD ; that is, AC^ — BC^ = AD-' - BD^.
For, since (/A. 34) AC^ = AD^* + DC^ and BC^ =
BD^ + DC-, the differences of these are equal ; that
is, AC2 - BC- = AD2 - BD2.
Cor. The rectangle of the sum and difference of the two sides of any triangle,
is equal to the rectangle of the sum and difference of the distances between the
])erpendicular and the two extremes of the base, or equal to the rectangle of the
base and the difference or sum of the segments, according as the perpendicular
falls within or without the triangle.
That is, (AC + BC) (AC - BC) = (AD + BD) (AD - BD) :
or, (AC + BC) (AC — BC) = AB (AD - BD) in the 2nd Jig. ;
and (AC + BC) (AC — BC) = AB (AD -f BD) in the \stfig.
THEOREM XXXVI.
In any obtuse-angled triangle, the square of the side subtending the obtuse angle, is
greater than the sum of the squares of the other two sides, by twice the rectangle
of the base and the distance of the perpendicular from the obtuse angle.
Let ABC be a triangle, obtuse angled at B, and CD perpendicular to AB ;
then will the square of AC be greater than the squares of AB, BC, by twice the
rectangle of AB, BD : that is, AC^ = AB- + BC- + 2AB . BD. See the 1st
fig. above, or below.
For, AD2 = AB2 + BD^ + 2AB . BD (th. 31),
and AD2 + CD* = AB^ + BD^ + CD^ + 2AB . BD {ax. 2):
But AD2 + CD" = AC2, and BD^ + CD" = BC" {th. 34) ;
therefore AC" = AB" + BC" + 2AB . BD.
THEOREM XXXVir.
In any triangle, the square of the side subtending an acute angle, is less than the
squares of the base and the other side, by twice the rectangle of the base and the
distance of the perpendicular from the acute angle.
Let ABC be a triangle, having the angle
A acute, and CD perpendicular to AB ; then
will the square of BC be less than the squares
of AB, AC, by twice the rectangle of AB,
AD. That is, BC" + 2AD , DB = AB" ^
+ AC".
For BD" = AD" + AB" — 2AD . AB {th. 32),
and BD" + DC" = AD" + DC" + AB" — 2AD . AB {ax. 2);
therefore BC" = AC" + AB" - 2AD . AB {th. 34).
312
GEOMETRY.
THEOREM XXXVIII.
In any triangle, the double of the square of a line drawn from the vertex to the
middle of the base, together with double the square of the half base, is equal to
the sum of the squares of the other two sides.
Let ABC be a triangle, and CD the line drawn from
the vertex to the middle of the base AB, bisecting it into
the two equal parts AD, DB ; then will the sum of the
squares of AC, CB, be equal to twice the sum of the
squares of CD, AD ; or AC^ + CB^ = 2CD2 + ^AD-.
For AC2 = CW + AD2 + 2AD . DE {th. 36),
and BC2 = CD^ + BD- — 2AD . DE {th. 37);
whence AC^ + BC^ = 2CD2 + AD^ + BD- = 2CD2 + 2AD= (ax. 2).
THEOREM XXXIX.
In an isosceles triangle, the square of a line drawn from the vertex to any point in
the base, together with the rectangle of the segments of the base, is equal to the
square of one of the equal sides of the triangle.
Let ABC be the isosceles triangle, and CD a line drawn
from the vertex to any point D in the base : then will the
square of AC be equal to the square of CD, together with
the rectangle of AD and DB, That is, AC'' = CD^ +
AD . DB.
For AC- — CD' = AE^ — DE^ (th. 35) = AD . DB {th. 33) ;
Therefore AC^ = CD^ + AD . DB {ax. 2).
THEOREM XL.
In any parallelogram, the two diagonals bisect each other ; and the sum of their
squares is equal to the sum of the squares of all the four sides of the parallelo-
gram.
Let ABCD be a parallelogram, whose diagonals inter-
sect each other in E : then AE will be equal to EC, and
BE to ED ; and the sum of the squares of AC, BD, will
be equal to the sum of the squares of AB, BC, CD, DA.
That is,
AE = EC, and BE = ED,
and AC* + BD- = AB^ + BC- + CD^ + DA».
For, the triangles, AEB, DEC, are equiangular, because they have the oppo-
site angles at E equal {th. 5), and the two lines AC, BD, meeting the parallels
AB, DC, make the angle BAE equal to the angle DCE, and the angle ABE
equal to the angle CDE, and the side AB equal to the side DC {th. 22) ; there-
fore these two triangles are identical, and have their corresponding sides equal
{th. 2), viz. AE = EC» and BE = EI).
THEOREMS. 31S
Again, since AC and BD are bisected in E, we have (th. 38).
AD* + DC* = 2AE* 4- 2ED2 and AB* + BC* = 2AE* + 2EB*;
hence {ax. 2) AB» + BC- + CD* + DA* = 4AE* + 4ED* = AC* + BD=.
Cor. I. If AD = DC, or the parallelogram be a rhombus ; then AD* = AE*
+ ED», CD* = DE* + CE*. &c.
Cor. 2. Hence, and by th. 34, the diagonals of a rhombus intersect at right
angles.
THEOREM XLI.
If a line, dravm through or from the centre of a circle, bisect a chord, it will be
perpendicular to it : or, if it be perpendicular to the chord, it will bisect both the
chord and the arc of the chord.
Let AB be any chord in a circle, and CD a line drawn
from the centre C to the chord. Then, if the chord be bi-
sected in the point D, CD will be perpendicular to AB.
Draw the two radii CA, CB. Then the two triangles
ACD, BCD, having CA equal to CB (def. 44), and CD
common, also AD equal to DB (hyp.) ; they have all the
three sides of the one, equal to all the three sides of the
other, and so have their angles also equal (th. 8) : hence then, the angle ADC
being equal to the angle BDC, these angles are right angles, and the line CD is
perpendicular to AB (def. 53).
Again, if CD be perpendicular to AB, then will the chord AB be bisected at
the point D, or have AD equal to DB ; and the arc AEB bisected in the point
E, or have AE equal to EB.
For, having drawn CA, CB, as before : then, in the triangle ABC, because the
side CA is equal to the side CB, their opposite angles A and B are also equal
{th. 3) : hence then, in the two triangles ACD, BCD, the angle A is equal to
the angle B, and the angles at D are equal (def. 53) ; therefore their third angles
are also equal (cor. 1, th. 17) ; and having the side CD common, they have also
the side AD equal to the side DB (th. 2).
Also, since the angle ACE is equal to the angle BCE, the arc AE, which
measures the former (def. 57), is equal to the arc BE, which measures the latter,
since equal angles must have equal measures.
Cor. Hence a line bisecting any chord at right angles, passes through the
centre of the circle.
THEOREM XLII.
If more than two equal lines can be drawn from any point within a circle to the
circumference, that point will be the centre.
Let ABC be a circle, and D a point within it : then if any
three lines, DA, DB, DC, drawn from the point D to the
circumference, be equal to each other, the point D will be
the centre.
Draw the chords AB, BC, which let be bisected in the
points E, F, and join DE, DF.
Then, the two triangles, DAE, DBE, have the side DA
equal to the side DB by supposition, and the side AE equal to the side EB by
hypothesis, also the side DE common : therefore these two triangles are iden-
314
GEOMETRY.
tical, and have the angles at E equal to each other {th. 8) ; consequently, DE is
perpendicular to the middle of the chord AB {def. 54), and therefore passes
through the centre of the circle {cor. th. 41).
In like manner, it may he shown that DF passes through the centre : and con-
sequently that the point D is the centre of the circlCj and that the three equal
lines DA, DB, DC, are radii.
THEOREM XLIII.
If two circles placed one within another, touch, the centres of the circle and the
point of contact will be all in the same right line.
Lkt the two circles ABC, ADE, touch one another inter-
nally in the point A ; then will the point A and the centres
of those circles be all in the same right line.
Let F be the centre of the circle ABC, through which
draw the diameter AFC. Then, if the centre of the other
circle can be out of this line AC, let it be supposed in some
other point as G; through which draw the line FG, cutting ^
the two circles in B and D.
Now, in the triangle AFG, the sum of the two sides FG, GA, is greater than
the third side AF (th. 10), or greater than its equal radius FB. From each of
these take away the common part FG, and the remainder GA will be greater
than the remainder GB : but the point G being supposed the centre of the
inner circle, its two radii, GA, GD, are equal to each other ; consequently GD
will also be greater than GB. Again, ADE being the inner circle, GD is neces-
sarily less than GB : so that GD is both greater and less than GB; which is
absurd. To remove this absurdity we must abandon the supposition that pro-
duced it, which was that G might be out of the line AFC : and consequently
the centre G cannot be out of the line AFC ; that is, the line joining the
centre FG, passes through the point of contact A.
THEOREM XLIV.
If two circles touch one another externally, the centres of the circles and the point
of contact will be all in the same right line.
Let the two circles ABC, ADE, touch one another exter-
nally at the point A ; then will the point of contact A and
the centres of the two circles be all in the same right line.
Let F be the centre of the circle ABC, through which
draw the diameter AFC, and produce it to the other circle
at E. Then, if the centre of the other circle ADE can be
out of the line FE, let it, if possible, be supposed in some
other point as G ; and draw the lines AG, FBDG, cutting
the two circles in B and D.
Then, in the triangle AFG, the sum of the two sides AF, AG, is greater than
the third side FG {th. 10) : but, if F and G being the centres of the two circles,
the two radii GA, GD, are equal, as are also the two radii AF, FB. Hence the
sum of GA, AF, is equal to the sum of GD, BF; and therefore this latter sura
also, GD, BF, is greater than GF, which is absurd : and consequently, as in the
former proposition, the centre G cannot be out of the line EF.
THEOREMS. 3] 5
THEOREM XLV.
Any chords in a circle, which are equally distant from the centre, are equal to each
other : or if they be equal to each other, they will be equally distant from the
centre.
Let AB, CD, be any two chords at equal distances from
the centre G : then will these two chords AB, CD, be equal
to each other.
Draw the two radii GA, GC, and the two perpendiculars
GE, GF, which are the equal distances from the centre G.
Then the two right-angled triangles, GAE, GCF, have the
side GA equal the side GC, and the side GE equal the side
GF, and the angle at E equal to the angle at F, therefore those two triangles are
identical (cor. 2. tk. 34), and have the line AE equal to the line CF : but AB is
the double of AE, and CD is the double of CF {th. 41), and therefore AB is
equal to CD (ax. 6).
Again, if the chord AB be equal to the chord CD ; then will their distances
from the centre, GE, GF, also be equal to each other.
For, since AB is equal CD by supposition, the half AE is equal the half CF :
and the radii GA, GC, being equal, as well as the right angles E and F, there-
fore the third sides are equal (,cor. 2, th. 34), or the distance GE equal the dis-
tance GF.
THEOREM XLVI.
A line perpendicular to the extremity of a radius, is a tangent to the circle.
Let the line ADB be perpendicular to the radius CD of a
circle ; then shall AB touch the circle in the point D only.
From any other point E in the line AB draw CFE to the
centre, cutting the circle in F.
Then because the angle D, of the triangle CDE, is a right
angle, the angle at E is acute {cor. 3, th. 17), and conse-
quently less than the angle D : but the greater side is
always opposite to the greater angle {th. 9), and therefore the side CE is greater
than the side CD, or greater than its equal CF. Hence the point E is without
the circle; and the same for every other point in the line AB ; and consequently
the whole line is without the circle, and meets it in the point D only.
THEOREM XLVII.
When a line is tangent to a circle, a radius drawn to the point of contact is
perpendicular to the tangent.
Let the line AB touch the circumference of a circle at the point D ; then will
the radius CD be perpendicular to the tangent AB. See the last figure.
For the line AB being wholly without the circumference except at the point
D, every other line, as CE, drawn from the centre C to the line AB, must pass
out of the circle to arrive at this line. The line CD is, therefore, the shortest that
can be drawn from the point C to the line AB, and consequently {th. 21), it is
perpendicular to that line.
316 GEOMETRY.
Cor. Hence, conversely, a line drawn perpendicular to a tangent, at the point
of contact, passes through the centre of the circle.
THEOREM XLVlir.
The angle formed by a tangent and chord is measured hy half the arc of that
chord.
Let AB be a tangent to a circle, and CD a chord draum from the point of
contact C ; then is the angle BCD measured by half the arc CFD, and the angle
ACD measured by half the arc CGD.
Draw the radius EC to the point of contact, and the radius EF perpendicular
to the chord at H.
Then the radius EF, being perpendicular to the chord
CD, bisects the arc CFD (/A. 41). Therefore CF is half the
arc CFD.
In the triangle CEH, the angle H being a right one, the
sum of the two remaining angles E and C is equal to a right
angle (cor. 3, th. 17), which is equal to the angle BCE,
because the radius CE is perpendicular to the tangent. From each of these
equals take away the common part or angle C, and there remains the angle E
equal to the angle BCD : but the angle E is measured by the arc CF (^def. 60),
which is the half of CFD j therefore the equal angle BCD must also have the
same measure, namely, half the arc CFD of the chord CD.
Again, the line GEF, being perpendicular to the chord CD, bisects the arc
CGD {th. 41) ; and therefore CG is half the arc CGD. Now, since the line CE,
meeting FG, makes the sum of the two angles at E equal to two right angles
{th. 6), and the line CD makes with AB the sum of the two angles at C equal to
two right angles ; if from these two equal sums there be taken away the parts
or angles CEH and BCH, which have been proved equal, there remains the
angle CEG equal to the angle ACH. Now the former of these, CFG, being an
angle at the centre, is measured by the arc CG {def. 60) ; consequently the equal
angle ACD must also have the same measure CG, which is half the arc CGD of
the chord CD.
Cor. 1. The sum of two right angles is measured by half the circumference.
For the two angles BCD, ACD, which makes up two right angles, are measured
by the arcs CF, CG, which make up half the circumference, FG being a
diameter.
Cor. 2. Hence also one right angle must have for its measure a quarter of the
circumference, or 93 degrees.
THEOREM .KLIX.
An angle at the circumference of a circle is measured by half the arc that
subtends it.
Let BAC be an angle at the circumference ; it has for its
measure, half the arc BC which subtends it.
For, suppose the tangent DE to pass through the point of
contact A : then, the angle DAC being measured by half
the arc ABC, and the angle DAB by half the arc AB
{th. 48) ; it follows, by equal subtraction, that the differ-
ence, or angle BAC, must be measured by half the arc BC, which it stands
upon.
THEOREMS.
317
THEOREM L.
All angles in the same segment of a circle, or standing on the same arc, are equal to
each other.
Let C and D be two angles in the same segment ACDB.
or, which is the same thing, standing on the supplemental
arc AEB ; then will the angle C be equal to the angle D.
For each of these angles is measured by half the arc AEB ;
and thus, having equal measures, they are equal to each
other {ax. 11).
THEOREM LI.
An angle at the centre of a circle is double the angle at the circumference, when
both stand on the same arc.
Let C be an angle at the centre C, and D an angle at the
circumference, both standing on the same arc or same chord
AB : then will the angle C be double of the angle D, or the
angle D equal to half the angle C.
For, the angle at the centre C is measured by the whole
arc AEB (def. 60), and the angle at the circumference D is
measured by half the same arc AEB (th. 49) ; therefore the
angle D is only half the angle C, or the angle C double the angle D.
THEOREM LII.
An angle in a semicircle, is a right angle.
If ABC or ADC be a semicircle ; then any angle D in
that semicircle, is a right angle.
For, the angle D, at the circumference, is measured by
half the arc ABC {th. 49), that is, by a quadrant of the cir-
cumference : and a quadrant is the measure of a right angle
(cor. 4, th. 6 ; or cor. 2, th. 48). Therefore the angle D is
a right angle.
THEOREM LIII.
The angle formed by a tangent to a circle, and a chord drawn from the point of
contact, is equal to the angle in the alternate segment.
If AB be a tangent, and AC a chord, and D any angle in
the alternate segment ADC ; then will the angle D be equal
to the angle BAC made by the tangent and chord of the arc
AEC.
For the angle D, at the circumference, is measured by
half the arc AEC {th. 49) ; and the angle BAC, made by the
tangent and chord, is also measured by the same half arc
AEC {th. 48) ; therefore, these two angles are equal {ax. 1 1).
318
GEOMETRY.
THEOREM LIV.
The sum of any two opposite angles of a qimdr angle inscribed in a circle, is equal
to two right angles.
Let ABCD be any quadrilateral inscribed in a circle;
then shall the sum of the two opposite angles A and C, or
B and D, be equal to two right angles.
For the angle A is measured by half the arc DCB, which
it stands upon, and the angle C by half the arc DAB,
{th. 49) ; therefore the sura of the two angles A and C is
measured by half the sum of these two arcs, that is, by half the circumference.
But half the circumference is the measure of two right angles {cor. 4, th. 6) ;
therefore the sum of the two opposite angles A and C is equal to two right
angles. In like manner it is shown, that the sum of the other two opposite
angles, D and B, is equal to two right angles.
THEOREM LV.
If any side of a quadrangle, inscribed in a circle, be produced out, the outward
angle will be equal to the inward opposite angle.
If the side AB, of the quadrilateral ABCD, inscribed in
a circle, be produced to E ; the outward angle DAE will be
equal to the inward opposite angle C.
For, the sum of the two adjacent angles DAE and DAB
is equal to two right angles {th. 4) ; and the sum of the two
opposite angles C and DAB is also equal to two right
angles {th. 54) ; therefore the former sum, of the two angles DAE and DAB, is
equal to the latter sum, of the two C and DAB {ax. 1). From each of these
equals taking away the common angle DAB, there remains the angle DAE equal
the angle C.
THEOREM LVI. •
Any two parallel chords intercept equal arcs.
Let the two chords AB, CD, be parallel : then will the
arcs AC, BD, be equal ; or AC =; BD.
Draw the hue BC. Then, because the lines AE, CD, are
parallel, the alternate angles B and C are equal {th. 12).
But the angle at the circumference B, is measured by half
the arc AC {th. 49) ; and the other equal angle at the cir-
cumference C is measured by half the arc BD ; therefore the halves of the arcs
AC, BD, and consequently the arcs themselves, are also equal.
THEOREM LVir.
When a tangent and chord are parallel to each other, they intercept equal arcs.
Let the tangent ABC be parallel to the chord DF ; then
are the arcs BD, BF, equal ; that is, BD = BF.
Draw the chord BD. Then, because the hnes AB, DF,
are parallel, the alternate angles D and B are equal {th. 12),
But the angle B, formed by a tangent and chord, is mea-
sured by half the arc BD {th. 48); and the other angle at
THEOREMS. 319
the circumference D is measured by half the arc BF (/A. 49) ; therefore the arcs
BD, BF, are equal.
THEOREM LVIII.
The angle formed, within a circle, by the intersection of two chords, is measured by
half the sum of the two intercepted arcs.
Let the two chords AB, CD, intersect at the point E :
then the angle AEC, or DEB, is measured by half the sum
of the two arcs AC, DB.
Draw the chord AF parallel to CD. Then, because the
lines AF, CD, are parallel, and AB cuts them, the angles on
the same side A and DEB are equal {th. 14) : but the angle
at the circumference A is measured by half the arc BF, or of the sum of FD and
DB {th. 49) ; therefore the angle E is also measured by half the sura of FD and
DB.
Again, because the chords AF, CD, are parallel, the arcs AC, FD, are equal
{th. 56) ; therefore the sum of the two arcs AC, DB, is equal to the sum of the
two FD, DB ; and consequently the angle E, which is measured by half the
latter sum, is also measured by half the former.
THEOREM LIX.
The angle formed, out of a circle, by two secants, is measured by half the difference
of the intercepted arcs.
Let the angle E be formed by two secants EAB and
ECD ; this angle is measured by half the difference of the
two arcs AC, DB, intercepted by the two secants.
Draw the chord AF parallel to CD. Then, because the
lines AF, CD, are parallel, and AB cuts them, the angles on
the same side A and BED are equal {Ih. 14) : but the angle
A, at the circumference, is measured by half the arc BF
(th. 49), or of the difference of DF and DB : therefore the equal angle E is also
measured by half the difference of DF, DB.
Again, because the chords AF, CD, are parallel, the arcs AC, FD, are equal
{th. 56) ; therefore the difference of the two arcs AC, DB, is equal to the differ-
ence of the two DF, BD ; and consequently the angle E, which is measured by
half the latter difference, is also measured by half the former.
THEOREM LX.
TTie angle formed by two tangents, is measured by half the dfference of the two
intercepted arcs.
Let EB, ED, be two tangents to a circle at the points A,
C ; then the angle E is measured by half the difference of
the two arcs CFA, CGA,
Draw the chord AF parallel to ED. Then, because the
lines AF, ED, are parallel, and EB meets them, the angles
on the same side A and E are equal {th. 14) : but the angle
A, formed by the chord AF and tangent AB, is measured
320
GEOMETRY.
by half the arc AF, {th. 48) ; therefore the equal angle E is also measured by
half the same arc AF, or half the difference of the arcs CFA and CF, or CGA
{th. 57).
Cor. In like manner it is proved, that the angle E,
formed by a tangent ECD, and a secant EAB, is measured
by half the difference of the two intercepted arcs CA and
CFB.
THEOREM LXI.
When two lines, meeting a circle each in two points, cut one another, either within
it or without it ; the rectangle of the parts of the one, is equal to the rectangle of
the parts of the other; the parts of each being measured from the point of meet-
ing to the two intersections with the circumference.
Let the two lines AB, CD, meet each other in E ; then
the rectangle of AE, EB, will be equal to the rectangle of
CE, ED. Or, AE . EB = CE . ED.
For, through the point E draw the diameter FG ; also,
from the centre H draw the radius DH, and draw HI per-
pendicular to CD.
Then, since DEH is a triangle, and the perp. HI bisects
the chord CD (th. 41), the line CE is equal to the difference
of the segments DI, EI, the sum of them being DE : and
because H is the centre of the circle, and the radii DH, FH,
GH, are all equal, the line EG is equal to the sura of the
sides DH, HE ; and EF is equal to their difference.
But the rectangle of the sum and difference of the two sides of a triangle is
equal to the rectangle of the sum and difference of the segments of the base
{th. 35) ; therefore the rectangle of FE, EG, is equal to the rectangle of CE, ED.
In like manner it is proved, that the same rectangle of FE, EG, is equal to the
rectangle of AE, EB: and consequently, the rectangle of AE, EB, is also equal
to the rectangle of CE, ED (ax. 1).
Cor. 1. When one of the lines in the second case, as DE^
by revolving about the point E, comes into the position of
the tangent EC or ED, the two points C and D running
into one; then the rectangle of CE, ED, becomes the
square of CE, because CE and DE are then equal. Conse-
quently, the rectangle of the parts of the secant AE . EB, is
equal to CE^, the square of the tangent.
Cor. 2. Hence both the tangents EC, EF, drawn from the
same point E, are equal ; since the square of each is equal to the same rectangle
or quantity AE . EB.
THEOREMS.
321
THEOREM LXII.
In equiangular triangles the rectangles of the corresponding or like sides, taken
alternately, are equal.
Let ABC, DEF, be two equiangular triangles, having the
angle A equal to the angle D, the angle B to the angle E. and
the angle C to the angle F ; also the like sides AB, DE, and
AC, DF, being those opposite the equal angles : then will
the rectangle of AB, DF, be equal to the rectangle of AC,
DE.
In BA produced take AG equal to DF ; and through the
three points B, C, G, conceive a circle BCGH to be de-
scribed, meeting CA produced at H, and join GH.
Then the angle G is equal to the angle C on the same arc BH, and the angle
H equal to the angle B on the same arc CG (th. 50); also the opposite angles
at A are equal ith. 7) ■ therefore the triangle AGH is equiangular to the triangle
ACB, and consequently to the triangle DFE also. But the two like sides AG,
DF, are also equal by supposition ; consequently the two triangles AGH, DFE,
are identical {th. 2), having the two sides AG, AH, equal to the two DF, DE,
each to each.
But GA . AB = HA . AC (fA. 61) : consequently, DF ; AB = DE . AC.
THEOREM LXIir.
The rectangle of the two sides of any triangle, is equal to the rectangle of the
perpendicular on the third side and the diameter of the circumscribing circle.
Let CD be the perpendicular, and CE the diameter of the
circle about the triangle ABC ; then CA . CB = CD . CE.
For, join BE : then in the two triangles ACD, ECB, the
angles A and E are equal, standing on the same arc BC
(th. 50) ; also the right angle D is equal the angle B, which
is also a right angle, being in a semicircle {th. 52) : therefore
these two triangles have also their third angles equal, and
are equiangular. Hence, AC, CE, and CD, CB, being like
sides, subtending the equal angles, the rectangle AC . CB, of the first and last
of them, is equal to the rectangle CE . CD, of the other two (th. 62).
THEOREM LXIV.
The square of a line bisecting any angle of a triangle, together with the rectangle of
the two segments of the opposite side, is equal to the rectangle of the two other
sides including the bisected angle.
Let CD bisect the angle C of the triangle ABC; then we
shall have CD* + AD . DB = AC . CB.
For, let CD be produced to meet the circumscribing circle
at E, and join AE.
Then the two triangles ACE, BCD, are equiangular : for
the angles at C are equal by supposition, and the angles B
and E are equal, standing on the same arc AC (th. 50) ;
VOL. I. Y
322 GEOMETRY.
consequently the third angles at A and D are equal {cor. 1, /A. 17): also AC,
CD, and CE, CB, are like or corresponding sides, being opposite to equal
angles : therefore AC . CB = CD . CE {th. 62). But CD . CE = CD^ +
CD . DE {th. 30) ; therefore AC . CB = CD^ + CD . DE, CD= + AD , DB,
since CD . DE = AD . DB {th. 61).
THEOREM LXV.
The rectangle of the two diagonals of any qiiadrilaleral inscribed in a circle, is equal
to the sum of the two rectangles of the opposite sides.
Let ABCD be any quadrilateral inscribed in a circle, and
AC, BD, its two diagonals : then AC . BD = AB . DC +
AD . BC.
For, let CE be drawn, making the angle BCE equal to
the angle DCA. Then the two triangles ACD, BCE, are
equiangular] for the angles A and B are equal, standing on
the same arc DC; and the angles DCA, BCE, are equal by
construction ; consequently, the third angles ADC, BEC, are also equal ; also
AC, BC, and AD, BE, are like or corresponding sides, being opposite to the
equal angles : therefore the rectangle AC.BE is equal to the rectangle AD.BC
(^A. 62).
Again, the two triangles ABC, DEC, are equiangular : for the angles BAC,
BDC, are equal, standing ou the same arc BC ; and the angle DCE is equal to
the angle BCA, by adding the common angle ACE to the two equal angles
DCA, BCE ; therefore the third angles E and ABC are also equal : but AC,
DC, and AB, DE, are the like sides : therefore AC.DE = AB.DC [th. 62).
Hence, by equal additions, AC.BE + AC.DE = AD.BC + AB.DC. But
AC.BE + AC.DE = AC.BD {th. 30) : therefore AC.BD = AD.BC + AB.DC
{ax. 1).
Cor. Hence, if ABD be an equilateral triangle, and C any point in the arc
BCD of the circumscribing circle, we have AC = BC -f DC. For AC.BD
= AD.BC + AB.DC ; and dividing by BD = AB = AD, there results
AC = BC + DC.
RATIOS AND PROPORTIONS.
DEFINITIONS.
76. Ratio is the relation subsisting between two magnitudes of the same
kind, in respect of quantity.
Of the two magnitudes compared, that which is taken as the standard of com-
parison is called the antecedent term of the ratio, or simply, the antecedent : and
that which is compared with it, the consequent. The leading idea of ratio is, the
RATIOS AND PROPORTIONS. S23
number of limes that the antecedent is contained in the consequent ; and hence
the doctrine of ratio becomes, essentially, a branch of arithmetic *.
The manner of writing? a ratio is a : b, where a is the antecedent and b the
consequent. The reading it is, a is to 6 ; and the expression of the fundamental
idea is -, written as a fraction, the numerator and denominator of which are
a
the consequent and antecedent respectively.
77. Proportion is the equality of two ratios, expressed as fractions. Thus, if
- =: - , the magnitudes a, b, c, d, are said to be proportionals, or to be in propor-
tion. In geometrical investigations it is, however, more usual to write them
a : b :: c \ d, the verbal interpretation of which is either
a is to 6 as c is to d, or
a has the same ratio to b that c has to d.
78. When there is any number of magnitudes of the same kind, the ratio of
the first to the last of them is said to be compounded of the ratios of the first to
the second, the second to the third, the third to the fourth, and so on to the
last. This is expressed by the term compound ratio.
79. When all these ratios, viz. that of the first term to the second, the second
to the third, and so on, are all equal, the terms are said to form a geometrical
progression, and are said to be continued proportionals.
80 When the ratios are equal, and there are only three terms, (or two ratios,)
the third is called a third proportional to the first and second : and the first is
said to have to the third the duplicate ratio of that which the first has to the
second. The middle term is called a mean proportional between the first and
third.
In like manner, when there are three equal ratios, the first term is said to
have to the fourth, the triplicate ratio of the first to the second, and so on,
however many equal ratios there may be.
There are other technical terms employed to signify certain modifications
• The method of treating the doctrine of ratio by the Greek geometei-s was precisely similar fi/y^ .
in all its essential characters to their method of treating theoretical arithmetic. The modem ~
method of discussing tlie properties of numbers has superseded the Greek one; but in treating
the doctrine of ratio, the original mode is still adhered to by tlie great majority of geometrical
writei-s, on account of its supposed superiority of logical conclusiveness. Tlie great beauty of
that method of investigating the properties of ratios, no one doubts ; but its superior conclu-
'siveness may be very fairly questioned, and its great complexity renders it a serious obstacle to
the progress of geometrical study.
The great logical difficulty that has been felt in treating ratio directly and formally ag a
branch of arithmetic, has arisen from the possible incommensurability of the two terms of the
ratio. Now if it were essential that the specific ratio itself should be assigned between the two
terms, there would be some force in this objection; but as in all our investigations, and in all
the uses we make of the doctrine in theoretical researches, resolve themselves into investiga-
tions respecting the equality or inequality of two or more ratios, as the result of given condi-
tions, it is obviously sufficient that we should be able to determine the essential equality or
inequality of those several ratios, without discussing the actual values of the fractional expres-
sions themselves. Such ratios themselves may be unknown, indeterminable, or irrational; and I
yet their equality or inequality may be determined as completely by arithmetical considerations, |
as by the method of the Greeks. In fact, all the reasonings in which ratio is employed are con-
ducted altogether independently of the actual value of the fraction . , and which may, therefore,
Mitli perfectly logical accuracy, be denoted by any symbol whatever, as m, or n,f{r), or any
other.
y2
334 GEOMETRY.
under which magnitudes originally proportional will still continue so. These
cases being enunciated and proved in the following series, the several terms or
phrases by which they are designated, are annexed to the propositions them-
selves.
81. A line is said to be divided in extreme and mean ratio, when the whole line
is to the greater segment as the greater segment is to the less ; or conversely to
be extended in extreme and mean ratio, when the extended part is to the original
line as the original line is to the whole line composed of the original one and
the extended part.
S2. The altitude of a triangle or a parallelogram is the perpendicular distance
(or simply the distance, def. 50) of the vertex of the triangle, or the opposite
side of the parallelogram from the base.'
83. Two pairs of magnitudes are said to be reciprocally proportional, when
the first of the first pair is to the first of the second pair, as the latter of the
second pair is to the latter of the first pair. Thus, if a, b, c, d, taken in order
were the two pairs, they are reciprocally proportional when a : c '.'. d : b.
84. A line is said to be divided in harmonical ratio, (or simply divided har-
monically,) when it is divided and extended in the same ratio.
85. A (ranstersal is any straight line or circle which is drawn to cut a system
of straight lines.
86. When a straight line is divided harmonically, and lines are drawn from
the points of division to any fifth point, the four lines so drawn are called an
harmonical fa^ceau.
A convenient mode of writing this is as follows :
Let A, B, C, D, be the four points of the harmonical line, and E the point of
Xhefasceau; then EJABCDJ denotes lines drawn as in the definition.
THEOREM LXVr.
Equimultiples of two magnitudes have the same ratios as the magnitudes
themselves.
Let a, b, be the two magnitudes, and ma, mb, their equimultiples. Then
a; b :: ma : mb. For the ratios - and — are equal, whatever be the value of
a ma '
the multiple m, whether integer, fractional, or irrational.
Cor. Hence any equisubmultiples of two magnitudes have the same ratio as
the magnitudes themselves.
THEOREM LXVII.
If four magnitudes of the same kind be proportional, then the antecedents will have
the same ratio as the consequents.
[This is called alternation or permutation of the terms.]
Let a : 6 : : c : rf ; then a : c :: b : d. Then since (def. 77) - — - gives -
— ^, and this fulfils the definition of proportional terms, or a : c :: b : d.
RATIOS AND PROPORTIONS. 325
THEOREM LXVIII.
If four quantities taken in order be proportionals, then will the first consequent be
to the first antecedent as the second consequent is to the second antecedent.
[This is called proportion by inversion^]
Let a : b '.: c I d, then also we shall have b : a '.: d ; c.
For since -= -, we have I = > or again, by the definition, b : a :: d : c.
THEOREM LXIX.
If four magnitudes be proportional, then the sum or difference of the first and
second will be to the first or second as the difference of the third and fourth is to
the third or fourth.
[This is termed proportion by composition or division, according as the sums or
differences are used].
Let a : b :: c : d; then we shall have to prove that a + b : a : '. c + d : c,
and that a + b : b :: c + d : d.
Now, since - = , we have r= ~„ and hence also 1 + - = 1 + -, and , +
a c b d a — c b
1 =: J + 1, Whence also, = ■ — , and — -p— = -^ — ; that is, agam,
o — a c 0 a
after inversion, a + b '. a '.'. c + d '. c, and a + 6:6::c + rf:t?.
Cor. 1. Hence also, a -\- b '. a — b :'. c -^ d '. c — d. For by the preceding
a 4- b c ■\- d , a — b c — d , . a — b c — d
=: , and = : therefore we get — ; — j =: „ or
a c a c ^ a -\- b c -\- d
a -\- b : a — b:\c-\-d\c — d.
Cor. 2. Also a : c : : a -^ b : c + d, and c '. d '.', a + c '. b ^ d.
THEOREM LXX.
Jf, of four proportional magnitudes there be taken any equimultiples whatever of the
two antecedents, and any whatever of the two consequents, these multiples will be
proportionals.
Let a : b :: c : d, then also ma i nb :: mc : nd. For - =-, and hence —
a c ma
= - : or, which is the same thing, ma : nb ;: mc : nd.
mc
THEOREM LXXI.
If there be four proportional magnitudes, and the two consequents be either aug-
mented or diminished by magnitudes which have the same ratio as the two ante-
cedents, the sums or differences form with the two antecedents a set of propor-
tionals.
Let a : b :: c : d, and e :f '.: a : c; then will a : b + e :: c : d +f.
326 GEOMETRY.
For, from the two given proportions we have - = r and = — , hence b , e
:: d : f, Bind b ± e : d ±f :: b : d :: a : c.
Cor. The variation of this theorem is obvious, viz. : a + e : c +/:: b : d.
THEOREM LXXII.
If any number of magnitudes be proportional, then any one of the antecedents is to
its consequent as all the antecedents taken together are to all the consequents
taken together.
Let a : b :: c : d :: e :f :: g : h Then a : b :: a + c + e + g
.... : b+d+f+h ....
„ b d b + d f b + d+f h b + d+f+h ,
For - =' = T~ = - = — r — -r^ = - = -, ^ •'— ' - , and so on to
a c a + c e a + c + e g a + c + e + g
any extent. Hence the conclusion follows.
THEOREM LXXIII.
If a whole magnitude be to a vihole as a part taken from the first is to a part taken
from the other: then the remainder will be to the remainder as the whole to the
whole.
b : then a : b :: a — — a : b b.
n n
, or in another form as stated in the theorem.
THEOREM LXXIV.
If there be several pairs of ratios which are equal each to each, then the ratio com-
pounded of all the first ratios will be equal to the ratio compounded of all the
others.
Let a '. b
For * —
m
: : - a
n
-'%
n
a
a
m
a
n
Then will the ratio compounded of a : i, a, : 5,,
Oi : bf, a„ : b„, be equal to that compounded
oi c : d, Ci : di, c^ '. d^, .. . c„ ; d„.
Let a : b :: c id
fli : 6| : : c, : rf,
flj '. b.^ :: Cj : ct^
Om '. b„ : : c„ : d„
For* = ^.*i=^',*^ =^%.... *-=^. Whence it follows that
a c Oi c, flj Cj am c„
b bfb^ b^ d d, rfj d„
a'a^'aj' " ' a„ ~c.c^'c^' '" c„'
Cor. 1 . If there be magnitudes common to the numerator and denominator of
either multiplied fraction, they may be cancelled, on the principle of the common
measure.
Cor. 2. If the magnitudes be numerically expressed, we shall have, as at
p. 106,
aa.fla a„ I bbj)^ . . . b„ '.'. ccc^ c„ : dd^d, . . . d„.
RATIOS AND PROPORTIONS. 327
Cor. 3. If all these ratios are equal, and the magnitudes expressed numerically,
then we have - := — , and — ^ - - ; and hence, a" : 6" : : c" : rf", and
a" (T - -
a' c*
THEOREM LXXV.
Of four proportional magnitudes, if the first be greater than the second, the third
is greater than the fourth ; if equal, equal ; and if less, less.
Let a : b :: c : d; then if a be greater than b, c is greater than d ; if equal,
equal, and if less, less.
For - ^ - . Then if a be greater than b, the fraction - is less than unity, and
hence- is also less than unity, or c is greater than rf. If a be equal to b, the
fraction - is equal to unity, and hence - is also unity, or c = rf. In like man-
ner, if a be less than b, c is less than d.
Cor. Since {th. 69) we have a : c :: b : d, the same reasoning will lead to
the conclusion, that if four magnitudes of the same kind be proportionals, then
the second will be less, equal to, or greater than the fourth, according as the
first is less, equal to, or greater than the third.
THEOREM LXXVI.
If any equimultiples whatever of the first and third of four^ magnitudes be taken,
and any whatever of the second and fourth ; then, according as the multiple of
the first be greater than, equal to, or less than that of the second, that of the
third will be greater than, equal to, or less than that of the fourth.
Let a : b :: c : d; then it is to be shown that
(1) if ma be greater than nb, mc is also greater than nd ;
(2) if ma be equal to nb, mc is also equal to nd ;
(3) if ma be less than nb, mc is also less than nd.
For, since a : b :: c : d, ma : nc :: mb : nd {th. 70) ; and since these last
are proportionals, according as ma is greater than, equal to, or less than nc, mb
is greater than, equal to, or less than nd (/A. 75).
THEOREM LXXVII.
If there be four magnitudes such that when any equimvltiples whatever are taken of
the first and third, and any whatever of the second and fourth, and if when the
multiple of the first is greater than, equal to, and less than that of the second,
that of the third is greater than, equal to, and less, respectively, than that of the
fourth : these four magnitudes will be proportional.
Let a, b, c, d, be four magnitudes, and m, n, any numbers whatever ; and
when ma is greater than, equal to, and less than nb, let mc be greater than,
equal to, and less than nd : then we have a : b '.: c '. d.
For, if possible, let the fourth magnitude not be a fourth proportional to a, b, c,
and let the fourth proportional to them be d,. Then U d, = d + d', we have
328 GEOMETRY.
a : b : : c : d -\- d'. Now in this case we have ma : nb : • mc ; nd + nd'.
Hence,
if ma be greater than nb, mc is greater than nd + nd'
equal to equal to
less than less than
But by hypothesis,
if ma be greater than nb, mc is greater than nd
equal to .... equal to
less than less than
Now the second of both of these sets of conditions can only be fulfilled by
d' = 0 ; and of the other two of each set, the first is not necessarily fulfilled by
any other value of d'. Whence that the three conditions may be fulfilled, we
must have d a fourth proportional to a, b, c, d ; that is, under the given circum-
stances, a : b :: c ', d.
THEOREM LXXVIII.
If any number of quantities be continued proportionals, then the ratio of the first to
the last is that power of the ratio which expresses the number of ratios com-
pounded.
Let there be n equal ratios a : b, b : c, c : d, . . . . p : q compounded : then
a \a/
„ b c d 9tt bed Q ^ b , ^ .
For -=7=z- = ... =i. Hence - . , . - ....' = -.-.. (n terms)
a ff c p a b c p a a
or, which is the same thing, ^ = f y .
Cor. Hence the duplicate ratio is the square of the simple ratio, the triplicate
ratio is its cube, and so on.
THEOREMS DEPENDING ON RATIOS.
THEOREM LXXIX.
Parallelograms, or triangles, having equal altitudes, are to one another as their
bases : those having equal bases are to one another as their altitudes ; and those
having neither equal bases nor altitudes, are to one in a ratio compounded of the
ratios of their bases and the ratio of their altitudes.
First. (1). Let BADC, GDEF be any two
parallelograms having the same altitude (and, " T ^ "\ ^ V\ VL-^;^ '
therefore, when their bases AD, DE, are in the \ \ W^ U^r\^
same straii^ht line AE, their opposite sides BC, s i 1 jTskkt
Gi', are in the same straight line BF parallel to
AE) : then they are to each other as their bases AD, DE.
In AD produced, take any number of parts AL, LS, each equal to AD; and
any number VAi, HK, KV, each equal to DE ; and draw LM, ST, EN, HP,
KQ, VW, all parallel to AB or DC, and meeting BF, produced, as in the
figure.
Then each of the parallelograms TL, MA, is equal to BD ; and there are as
THEOREMS. 329
many of them as there were taken lines AL, LS, equal to AD. Hence, what-
ever multiple the^base SD is of the base DA, the same multiple is the parallelo-
gram SC of the parallelogram AC. In like manner, whatever multiple the base
DV is of the base DE, the same multiple is the parallelogram DW of the paral-
lelogram DN, or (th. 25) of the parallelogram DF.
Again, if SD, the multiple of AD, be greater than DV, the multiple of DE,
the multiple SC of AC will be greater than the multiple DW of DN or DF ; if
equal, equal; if less, less. Hence, (th. 77,) parallelogram AC : parallelogram
DN : : base AD : base DE.
(2.) Let ARD, DGE, be two triangles of equal altitudes, they will be to each
other as their bases, AD, DE.
For, being of equal altitudes, they are between the same parallels ; and as
each of the triangles ARD, DGE, is the half of the parallelograms AC, DF,
hanng the same base and altitude, they are to one another as those parallelo-
grams. But the parallelograms have been proved to have the same ratio as their
bases ; and hence triangle ARD : triangle DGE :: base AD : base DE.
Secondly. (1). Let the parallelograms AC, DF, have equal
bases AD, DG : they will be to each other as their alti-
tudes.
For, draw AK, DL, GN, perpendicular to A*^, and pro-
duce BC, EF, to meet them, as in the figure. Then the
lines BC, EF, being parallel to AG, are parallel to each
other (th. 15); and the parallelograms AL, DN, are rect-
angles (th. 22, cor.l), and equal to AC, DF, respectively {th. 25) ; and the rectangle
AM is equal to the rectangle DN. But the rectangles AM, AL, are to one
another as AH, AK, by the former part of the proposition ; that is, as the alti-
tudes of the parallelograms AC, DF. Whence also we have parallelogram AC :
parallelogram DF : : altitude AH : altitude AK.
(2). The triangles ABD, DFG, being the halves of the parallelograms AC,
DF, are to one another in the same ratio ; that is,
triangle ABD : triangle DFG : : base AD : base DG.
Thirdly. (1). Let ABCD, DEFG, be any two parallelo-
grams, having neither their bases nor altitudes equal, and
make the same construction as in the last case : they will
be to each other in a ratio compounded of the ratio of their
bases AD, DG, and the ratio of their altitudes AK, AH.
For, by the first and second cases respectively, we have "*" ^ ^
parellelogram AM AD , parallelogram AL AK , .
parallelogram DN DG* parallelogram AM AH '
AD AK parallelogram AM parallelogram A L parallelogram AC
DG ' AH parallelogram DN ' parallelogram AM parallelogram DF*
the parallelograms AC, DF, have the ratio which is compounded of the ratio of
their bases AD, DG, and the ratio of their altitudes AK, AH.
(2). Since the triangles ABD, DFG, have the same bases and altitudes as the
parallelograms AC, DF, they have the same ratios as the parallelograms them-
selves, and hence the proposition is also true respecting triangles.
^-^
i/r^'
330 GEOMETRY.
THEOREM LXXX.
Parallelograms, or triangles, which have one angle of the one equal to one angle of
the other, have to one another a ratio compounded of the ratios of the sides about
the equal angles.
A — f — ¥
(1). Let ABCD, FCEG, be two parallelograms, having ^ ^f-
the angles DCB, ECF equal to one another : then they /'
shall be to one another in a ratio compounded of the ratios
of DC, CE, and BC, CF.
^ , . ,. » parallelogram AC DC , parallelogram CH BC ,
For (th. 79) ^ — vrr^^ ttu = /^rs. and ..-^ ^^ = -jr-, ; hence
^ parallelogram CH CE parallelogram CG CI*
parallelogram AC parallelogram CH parallelogram AC DC CB
parallelogram CH * parallelogram CG parallelogram CG CE ' CF'
(2). Let BCD, FCE, be triangles having the angles at C equal; they shall
be to one another in a ratio compounded of the ratio of the sides DC, CE, and
BC, CF.
For the triangles BCD, FCE, being the halves of the parallelograms AC, CG,
they have the same ratio as the parallelograms : that is, by the last case, the
ratio compounded of the ratios of the sides.
THEOREM LXXXI.
In parallelograms, or triangles, having one angle of the one equal to one angle of
the other, if the sides about the equal angles are reciprocally proportional, the
parallelograms or triangles are equal; and if they be equal, the sides about the
equal angles are reciprocally proportional.
First, let the parallelograms AC, CG, have their angles
at C equal, and the sides about C reciprocally proportional,
(that is, DC : CE : : CF : CB) then they will be equal.
For by the last proposition we have
parallelogram AC DC CB , . , DC CF
~Tr~i TTT^ = Firi • TTT. i and smce also v^^ = vttj,
parallelogiam CG CE CF CE CB'
, DC CB , . , . parallelogram AC , „ ,
we have -7-; . -;— , = 1, and therefore ^^ ,. . ttt- = 1, or parallelogram
CE Cr parallelogrEim CG ^ °
AC = parallelogram CG.
In like manner, under the same circumstances, the triangles BCD, FCE,
being the halves of the equal parallelograms, are also equal.
Secondly. Let the parallelograms AC, CG, be equal, and have the angles at C
equal : then the sides shall be reciprocally proportional, or DC : CE ' ' CF :
CB.
■Tf parallelogram .\C DC CB j 1 „ ,
ror smce '- ,, , 7^7^ = r^^ . -^n, and the parallelograms are ^qual, we
paraJlelognira CG CE CF r b ^ >
DC CB DC CF
have ^^. jTr. = I, or Qw^'nn » °^ »g^i"> finally, DC : CE ; ; CF : CB.
In the same way, it may be proved for the triangles BCD, FCE.
Cor. 1. If four straight lines be proportional, the rectangle of their extremes
is equal to the rectangle of the means : and if the rectangle of the extremes be
equal to the recUngle of the means, the four straight lines are proportional.
THEOREMS. 331
Let the four straight lines A, B, C, D, be ^
proportionals, that is, A : B | * C : D ; then »
shall the rectangle of A and D be equal to the ^
rectangle of B and C. ■" a
For place A, B, C, D, meeting in a point, and I r
forming four right angles at their point of inter-
section, and draw lines parallel to them to complete the figure; where P is the
rectangle of A, and D, Q, that of B and C, and R that of D and B. Theu the
figures P, Q, R, are rectangles, and the theorem is a case of the proposition, in
which the alleged properties have been generally proved.
Cor. 2. If three straight lines be proportional, the rectangle of the extremes is
equal to the square of the mean ; and if the rectangle of the extremes be equal to
the square of the mean, the three straight lines are proportional.
For in this case B = C, and the rectangle of B and C becomes the square of
B or C, Whence by the last cor. the truth follows.
Scholium.
Since it appears, by the rules of proportion in arithmetic and algebra, that
when four quantities ere proportional, the product of the extremes is equal to
the product of the two means ; and by this theorem, that the rectangle of the
extremes is equal to the rectangle of the two means ; it follows, that the area
or space of a rectangle is represented or expressed by the product of its length
and breadth multiplied together : and, in general, the area of a rectangle in
geometry is represented by the product of the measures of its length and breadth,
or base and height ; and a square is similarly represented by the measure of the
side multiplied by itself. Hence, what is shown of such products, is to be
also understood of the squares and rectangles.
THEOREM LXXXII.
If a line be drawn in a triangle parallel to one of its sides, it will cut the other two
sides proportionally.
Let DE be parallel to the side BC of the triangle ABC ; then
will AD : DB : : AE : EC.
For, draw BE and CD. Then the triangles DBE, DCE, are
equal to each other, because they have the same base DE, and are
between the same parallels DE, BC (^th. 25). But the two trian-
gles ADE, BDE, on the bases AD, DB, have the same altitude; and the two
triangles ADE, CDE, on the bases AE, EC, have also the same altitude ; and
because triangles of the same altitude are to each other as their bases, there-
fore
A ADE : BDE ; : AD : DB, A ADE : CDE : I AE : EC.
But BDE = CDE ; and equals must have to equals the same ratio ; therefore
AD : DB ; : AE : EC. In a similar manner, the theorem is proved when the
sides of the triangle are cut in prolongation beyond either the vertex or the
base.
Cor. Hence, also, the whole lines AB, AC, are proportional to their cor-
responding proportional segments {cor. tk. 66),
viz. AB : AC : : AD : AE,
and AB : AC : : BD : CE.
332 GEOMETRY.
THEOREM LXXXIll.
A line which bisects any angle of a triangle, divides the opposite side into two
segments, which are proportional to the two other adjacent sides.
Let the angle ACB, of the triangle ABC, be bisected by
the line CD, making the angle ACD equal to the angle
DCB : then will the segment AD be to the segment BD, as
the side AC is to the side CB. Or, AD : DB : : AC : CB.
For, let BE be parallel to CD, meeting AC produced at
E. Then, because the line BC cuts the two parallels CD,
BE, it makes the angle CBE equal to the alternate angle DCB (th. 12), and
therefore also equal to the angle ACD, which is equal to DCB by the supposi-
tion. Again, because the line AE cuts the two parallels DC, BE, it makes the
angle E equal to the angle ACD on the same side of it (^th. 14). Hence, in the
triangle BCE, the angles B and E, being each equal to the angle ACD, are
equal to each other, and consequently their opposite sides CB, CE, are also
equal {th. 3).
But now, in the triangle ABE, the line CD, being drawn parallel to the side
BE, cuts the two other sides AB, AE, proportionally (/A. 82), making AD to
DB, as is AC to CE or to its equal CB.
THEOREM LXXXIV.
Equiangular triangles are similar, or have their like sides proportional.
Let ABC, DEF, be two equiangular triangles, having the
angle A equal to the angle D, the angle B to the angle E,
and consequently the angle C to the angle F; then will
AB : AC : : DE : DF.
For, make DG = AB, and DH = AC, and join GH.
Then the two triangles ABC, DGH, having the two sides
AB, AC, equal to the two DG, DH, and the contained
angles A and D also equal, are identical, or equal in all
respects (/A. 1), namely, the angles B and C are equal to the
angles G and H. But the angles B and C are equal to the
angles E and F by the hypothesis ; therefore also the angles G and H are equal
to the angles E and F (flur. 1), and consequently the line GH is parallel to the
side EF(cor. I, th. 14).
Hence then, in the triangle DEF, the line GH, being parallel to the side EF,
divides the two other sides proportionally, making DG : DH :: DE : DF
icor. th. 82). But DG and DH are equal to AB and AC ; therefore also
AB : AC : : DE : DF.
THEOREM LXXXV.
Triangles which have their sides proportional, are equiangular.
In the two triangles ABC, DEF, if AB : DE : : AC : DF : : BC : EF; the
two triangles will have their corresponding angles equal.
THEOREMS. 332
For, if the triangle ABC be not equiangular with the
triangle DEF, suppose some other triangle, as DEG, to be
equiangular with ABC. But this is impossible : for if the
two triangles ABC, DEG, were equiangular, their sides
would be proportional {th. 84). So that, AB being to DE
as AC to DG, and AB to DE as BC to EG, it follows that
DG and EG, being fourth proportionals to the same three
quantities, as well as the two DF, EF, the former, DG, EG,
would be equal to the latter, DF, EF. Thus, then, the two
triangles DEF, DEG, having their three sides equal, would be identical {th. 5) ;
which is absurd, since their angles are unequal.
THEOREM LXXXVI.
Triangles, which have an angle in the one equal to an angle in the other, and the
sides about these angles proportional, are equiangular.
Let ABC, DEF, be two triangles, having the angle A equal
to the angle D, and the sides AB, AC, proportional to the
sides DE, DF : then will the triangle ABC be equiangular
with the triangle DEF.
For, make DG equal to AB, and DH to AC, and join GH.
Then, the two triangles ABC, DGH, having two sides
equal, and the contained angles A and D equal, are identical
and equiangular (th. 1), having the angles G and H equal to
the angles B and C. But, since the sides. DG, DH, are pro-
portional to the sides DE, DF, the line GH is parallel to EF "" *^ ■"
{th. 82) ; hence the angles E and F are equal to the angles G and H {th. 14),
and consequently to their equals B and C,
THEOREM LXXXVII.
In a right-angled triangle, a perpendicular from the right angle, is a mean propor-
tional between the segments of the hypothenuse ; and each of the sides, about the
right angle, is a mean proportional between the hypothenuse and the adjacent
segment.
Let ABC be a right-angled triangle, and CD a perpen-
dicular from the right angle G to the hypothenuse AB ;
then will
CD be a mean proportional between AD and DB ;
AC a mean proportional between AB and AD ;
BC a mean proportional between AB and BD.
Or, AD : CD : : CD : DB ; and AB : BC : : BC : BD; and AB : AC : : AC : AD.
For, the two triangles ABC, ADC, having the right angles at C and D equal,
and the angle A common, have their third angles equal, and are equiangular
{cor. 1, th. 17). In like manner, the two triangles ABC, BDC, having the right
angles at C and D equal, and the angle B common, have their third angles equal,
and are equiangular.
Hence then, all the three triangles, ABC, ADC, BDC, being equiangular,
will have their like sides proportional {th. 84); viz. AD : CD :; CD : DB;
AB : AC : : AC : AD; and AB : BC : : BC : BD.
334
GEOMETRY.
Cor. 1. Because the angle in a semicircle is a right angle {th. 52) ; it follows,
that if, from any point C in the periphery of the semicircle, a perpendicular be
drawn to the diameter AB ; and the two chords CA, CB, be drawn to the ex-
tremities of the diameter ; then are AC, BC, CD, the mean proportionals as in
this theorem, or {th. 77), CD- = AD . DB ; AC^ = AB . AD ; and BC^ =
AB . BD.
Cor. 2. Hence AC^ : BC^ : : AD : BD.
Cor. 3. Hence we have another demonstration of th. 34.
For since AC^ = AB . AD, and BC^ = AB . BD.
By addition AC^ + BC^ = AB (AD + BD) = AB^.
THEOREM LXXXVIir.
Equiangular or similar triangles, are to each other as the squares of their like
sides.
Let ABC, DEF, be two equiangular triangles, AB and
DE being two like sides : then will the triangle ABC be to p
the triangle DEF, as the square of AB is to the square of / \
DE, or as AB^ to DE2.
For, the triangles being similar, they have their like sides
proportional (jlh. 84), and are to each other as the rectangles
of the like pairs of their sides {cor. 4, /A. 81) ;
theref. AB : DE : : AC : DF {th. 84),
and AB : DE : : AB : DE of equality :
theref. AB= : : DE^ : : AB . AC : DE . DF {th. 75).
But A ABC : A DEF : : AB . AC : DE . DF {cor. 4, th. 81)
theref. A ABC : A DEF : : AB^ : DE^.
z_.
iiX
THEOREM LXXXIX.
AU similar figures are to each other, as the squares of their like sides.
Let ABCDE, FGHIK, be any two similar
figures, the like sides being AB, FG, and
BC, GH, and so on in the same order: then
will the figure ABCDE be to the figure
FGHIK, as the square of AB to the square of
FG, or as AB^ to FG-.
For, draw BE, BD, GK, GI, dividing the figures into an equal number of
triangles, by lines from two equal angles B and G.
The two figures being similar, {hypoth.) they are equiangular, and have their
like sides proportional {def. 67).
Then, since the angle A is = the angle F, and the sides AB, AE, proportional
to the sides FG, FK, the triangles ABE, FGK, are equiangular {th. 86). In
like manner, the two triangles BCD, GHI, having the angle C = the angle H,
and the sides BC, CD, proportional to the sides GH, HI, are also equiangular.
Also, if from the equal angles AED, FKI, there be taken the equal angles AEB,
FKG, there will remain the equals BED, GKI ; and if from the equal angles
CDE, HIK, be taken away the equals CDB, HIG, there will remain the equals
BDE, GIK ; so that the two triangles BDE, GIK, having two angles equal, are
THEOREMS. 335
also equiangular. Hence each triangle of the one figure is equiangular with
each corresponding triangle of the other.
But equiangular triangles are similar, and are proportional to the squares of
their like sides (/A. 88).
Therefore A ABE : A FGK : : AB^ : FG^ A BCD : A GHI : : BC^ : GH^,
A BDE : A GIK : : DE^ : IK^.
But as the two polygons are similar, their like sides are proportional, and con-
sequently their squares also proportional ; so that all the ratios AB^ to FG*, and
BC* to GH*, and DE"' to IK*, are equal among themselves, and consequently
the corresponding triangles also, ABE to FGH, and BCD to GHI, and BDE
to GIK, have all the same ratio, viz. that of AB^ to FG" : and hence all the ante-
cedents, or the figure ABCDE, have to all the consequents, or the figure FGHIK,
still the same ratio, viz. that of AB^ to FG^ {th. 72).
THEOREM XC.
Similar figures inscribed in circles, have their like sides, and also their whole
perimeters, in the same ratio as the diameters of the circles in which they are
inscribed.
Let ABCDE, FGHIK, be two similar
figures, inscribed in the circles whose diame-
ters are AL and FM ; then will each side
AB, BC, .... of the one figure be to the like
side GF, GH, .... of the other figure, or the
whole perimeter AB -f- BC -|- .... of the one
figure, to the whole perimeter FG + GH +
of the other figure, as the diameter AL to the diameter FM.
For, draw the two corresponding diagonals AC, FH, as also the lines BL, GM.
Then, since the polygons are similar, they are equiangular, and their like sides
have the same ratio (def. 67); therefore the two triangles, ABC, FGH, have the
angle B = the angle G, and the sides AB, BC, proportional to the two sides
FG, GH ; consequently these two triangles are equiangular {th. 86), and have
the angle ACB =: FHG. But the angle ACB =: ALB, standing on the same
arc AB ; and the angle FHG = FMG, standing on the same arc FG ; therefore
the angle ALB = FMG {ax. 1). And since the angle ABL = FGM, being both
right angles, because in a semicircle ; therefore the two triangles ABL, FGM,
having two angles equal, are equiangular ; and consequently their like sides are
proportional {th. 84) ; hence AB : FG : : the diameter AL : the diameter FM.
In like manner, each side BC, CD, .... has to each side GH, HI, .... the
same ratio of AL to FM : and consequently the sums of them are still in the
same ratio, viz. AB -|- BC -(- CD . . . . is to FG + GH -|- HI .... is to diameter
AL as diameter FM {th. 72).
THEOREM XCI.
Similar figures inscribed in circles, are to each other as the squares of the diameters
of those circles.
Let ABCDE, FGHIK, be two similar figures, inscribed in the circles whose
diameters are AL and FM ; then the surface of the polygon ABCDE will be to
the surface of the polygon FGHIK, as AL^ to FM^ iSeefig. th. 90.]
336
GEOMETRY.
For, the figures being similar, are to each other as the squares of their like
sides, AB" to FG- {th. 88). But, by the last theorem, the sides AB, FG, are as
the diameters AL, FM ; and therefore the squares of the sides AB- to FG', as
the squares of the diameters AL^ to FM- {th. 74). Consequently the polygons
ABODE, FGHIK, are also to each other as the squares of the diameters AL- to
FM2 {ax. 1).
THEOREM XCII.
The area of any circle is equal to the rectangle of half its circumference and half
its diameter.
The area of any circle ABD is equal to the rectangle contained by the radius,
and a straight line equal to half the circumference.
If not, let the rectangle be less than the circle ABD,
or equal to the circle FNH ; and imagine ED drawn to
touch the interior circle in F, and meet the circumference
ABD in E and D. Join CD, cutting the arc of the m .
terior circle in K. Let FH be a quadrantal arc of the
inner circle, and from it take its half, from the remainder
its half, and so on, until an arc FI is obtained, less than
FK. Join CI, produce it to cut ED in L, and make
FG = FL ; so shall LG be the side of a regular polygon circumscribing the
circle FNH. It is manifest that this polygon is less than the circle ABD, be-
cause it is contained within it. Because the triangle GCL is half the rectangle of
base GL and altitude CF, the whole polygon of which GCL is a constituent tri-
angle, is equal to half the rectangle whose base is the perimeter of that polygon
and altitude CF. But that perimeter is less than the circumference ABD, be-
cause each portion of it, such as GL, is less than the corresponding arch of cir-
cle having radius CL, and therefore, h fortiori, less than the corresponding arch
of circle with radius CA. Also CE is less than CA. Therefore the poly^^on of
which one side is GL, is less than the rectangle whose base is half the circum-
ference ABD and altitude CA ; that is, {hyp.) less than the circle FNH, which it
contains; which is absurd. Therefore, the rectangle under the radius and half
the circumference is not less than the circle ABD. And by a similar process it
may be shown that it is not greater. Consequently, it is equal to that rectangle.
THEOREM XCIII.
The circumferences of circles are to each other as their radii.
The circumferences of two circles ABD, abd, are as their radii.
If possible, let the radius AC, be to the
radius ac, as the circumference ABD to a
circumference ihk less than abd. Draw the
radius cie, and the straight line fig a chord
to the circle abd, and a tangent to the circle
ihk in t. From eb, a quarter of the circum-
ference of abd, take away its half, and then
the half of the remainder, and so on, until
there be obtained an arc ed less than eg ; and from d draw ad parallel to fg, it
will be the side of a re^^ular polygon inscribed in the circle abd, yet evidently
greater than the circle ihk, because each of its constituent triangles, as acd, con-
THEOREMS. 337
tains the corresponding circular sector cno. Let AD be the side of a similar
polygon inscribed in the circle ADB, and draw the radii AC, CD, ac, cd. The
similar triangles ACD, acd, give AC : ac :: AD : ad, and : ; perim. of polygon
in ABD ; perim. of polygon in abd. But, by the preceding theorem, AC : ac ::
circumf. ABD : circumf. abd. The perimeters of the polygons are, therefore, as
the circumferences of the circles. But this is impossible; because {hyp.) the
perimeter of polygon in ABD is less than the circumference ; while, on the con-
trary, the perimeter of polygon in adb is greater than the circumference ihk.
Consequently, AC is not to ac, as circumference ADB, to a circumference less
than adb. And by a similar process it may be shown, that ac is not to AC, as
the circumference abd, to a circumference less than ABD. Therefore AC '. ac ::
circumference ABD : circumference abd.
Cor. If R, r be the radii, D, d the diameters, and C, c the circumferences, we
have, by this theorem, C : c : : R : r ; or, if C = tt R, c = tt r ; and, by the
former, area (A) : area (o) : : ^RC : ^rc: we have A : a : : ^ttR^ : ^xr^ : : R^
: r- : : D2 : d2 : : C- : c\
THEOREM XCIV.
Angles at the centre of a circle, angles at the circumference of a circle, and sectors
of circles, have all the same ratio as the arcs by which they are subtended.
Let AB, BD be two arcs of a circle subtending the
angles ACB, BCD at the centre, the angles AMB, BFD
at the circumference, and the sectors : then,
(1) angle ACB : angle BCD : : arc AB : arc BD,
(2) angle AMB : angle BFD : : arc AB : arc BD,
(3) sector ACB : sector BCD : : arc AB : arc BD,
First. Take any number of arcs AL, LE, each equal to AB, and any number
DG, GH, HK, each equal to BD, and draw CL, CE, CG, CH, CK.
Then since BA, AL, LE, are all equal, the angles ACB, LCA, ECL, are all
equal {ax. 11), and hence whatever multiple the arc EB is of the arc AB, the
same multiple is the angle ECB of the angle ACB. In like manner, whatever
multiple the arc BK is of the arc BD, the same multiple is the angle KCB of
the angle DCB.
Again, if the arc BE be greater than the arc BK, the angle ECB is greater
than the angle BCK ; if equal, equal ; if less, less : and these are equimultiples
of AB, and ACB the first and third, and of BD, BCD the second and fourth.
Hence {th. 77) it follows that
angle ACB : angle BCD : : arc AB : arc BD.
Secondly. The angles AMB, BFD, at the circumference being the halves
of the angles ACB, BCD, respectively, at the centres, have the same ratios;
that is,
angle AMB : angle BFD ; : angle ACB : angle BCD : ; arc AB : arc BD,
Thirdly. The sectors ECL, LCA, ACB, are equal, and also the sectors BCD,
DCG, GCH, HCK, are equal. Conceive the sector ACB to be placed upon
ECL, so that CB shall coincide with CL ; then the angles ACB, ECL, being
equal, as before, the side AC will coincide with CE. Then the arc AB will
coincide with the arc LE. For if not, let it take some other position, as EPL,
and draw CPQ, cutting the arcs in P and Q ; then since PC and QC are radii
of the same circle they are equal : whence it is impossible that the arcs AB, EL
should not coincide. The sector ECL is therefore equal to the sector ACB. In
VOL. I. z
GEOMETRY.
like manner, the sector LCA is equal to ACB ; and the sectors DCG, GCH,
HCK, are each equal to the sector BCD. Hence it may be proved, as in the
first case, that the sectors are to one another as the arcs on which they stand.
THEOREM XCV.
If the three sides of a triangle be cut by any straight line, any one side will be
divided in a ratio compounded of the ratios of the segments of the other two.
Let ABC be a triangle cut by any line, straight or transversal, in D, E, F :
then
AE : EC : : AF . BD : FB . DC I BF : FA : : CE . BD : AE . CD
CD : DB : : AF . EC : FB . AE I AF . BD . CE = FB . DC . AE.
For, draw BH parallel to AC meeting EDF in H. Then,
by sim. trian. HBF, AFE, AE : BH : : AF : FB, and
by sim. trian. HBD, EDC, BH : EC : : BD : DC ;
heuce, by composition, cancelling BH from the first and second terms,
AE : EC : : AF . BD : FB . DC.
The two next are obtained in a similar manner ; and the last by th. 81, p. 230.
THEOREM XCVI.
If three straight lines be drawn from, the angles of a triangle through any point, to
meet the opposite sides, the segments of any one side will be divided in a ratio
compounded of the ratios of the segments of the other two.
Let P be any point, through which AE
lines AP, BP, CP, from the angles of CD
the triangle ABC are drawn to meet BF
the sides in D, E, F, respectively : then AF
EC
DB
FA
BD .
: AF . BD
: AF . EC
: CE . BD
CE = FB
: FB . DC
: FB . AE
: AE . CD
CD . EA.
For, through B draw HG parallel to AC meeting AP, CP, in H and G.
Then
by parallels AC, HG, cut by HP, BP, GP, AE : EC : : BH : BG,
sim. trian. FAC, FBG, AC : BG : : FA : FB, and
sim. trian. DCA, DBH, HB : AC : : BD : DC;
hence, by composition of ratios, we get
AE : EC : : AF . BD : FB . DC.
Similarly the two next may be obtained ; and the last as the corresponding one
of the last theorem.
f
THEOREMS.
ScTiolium.
Though the sides of the triangles in these two propositions (<A. 95, 96) have
the same relation as to ratio amongst their segments, yet there is an essential dif-
ference as to the numl)er of intersections made in the sides produced and unpro-
duced. In the latter theorem, there are always two or none produced, but never
one singly or all three ; that is, an even number : in the former, always one or
three, but never two or none; that is, an odd number. This distinction will
appear to be of importance in the next theorem, which is the converse of the two
last.
THEOREM XCVII.
If the sides of a triangle be divided so that the segments of one side have the ratio
compounded of the ratios of the segments of the other two sides, then :
1 . If two or more of these sides be so divided in prolongation, lines drawn from the
points of section to the opposite angles, will all pass through the same point.
2. If one or three of the sides be divided in prolongation, the three points of section
will be in one straight line.
I
First, ifigs. th. 96.) Let AE : EC : : AF . ED : FB . DC, two sides or more
being produced, then AD, BE, CF, pass through one point.
For if not, let CF. AD, intersect in P, and draw BP to meet AC in E', Then
{th. 96) AE' : E'C : : AF . BD : FB . DC. Whence AE' : EC : : AE : EC,
and AE' + E'C : AE' : : AE + EC : AE. But AE' + EC = AE + EC ;
hence AE' = AE the less to the greater, which is impossible. The three lines
AD, BE, CF, therefore pass through the same point P.
Second, {figs, th 95.) Let AE : EC : : AF . BD : FB . DC, one side or three
being produced ; then D, E, F, are in one straight line.
For if not, let FD meet AC in E'. Then, {th. 95,) we have AE' : E'C ; :
AF . BD : FB . DC, and hence AE' : E'C ; : AE : EC, and AE' + EC :
AE' : : AE + EC : AE; but AE' + 'EC = AE + EC, and hence AE' = AE,
the less to the greater, which is impossible. Whence D, E, F, are in one line.
Scholium.
This proposition furnishes a ready method of proving a great number of
elegant theorems. For instance the following :
(1). The three perpendiculars from the angles of a triangle to the opposite
sides, meet in one point.
(2). The lines which bisect the angles of a triangle, either all internally, or
one internally and the other two externally, meet in one point.
(3). The lines which bisect the sides of a triangle meet in one point.
(4). Lines drawn from the angles of a triangle to the points of contact of its
inscribed circle, meet in one point.
(5). Lines drawn from the angles of a triangle to the points of contact of
circles each touching one side exteriorly and the other two produced, meet iu one
point.
(6). If the exterior angles of a triangle be bisected by lines which are pro-
duced to cut the opposite sides, the bisecting lines intersect those sides in three
points which lie in one straight line.
z 2
340
GEOMETRY.
THEOREM XCVIII.
Jf through any point lines be drawn from the angles of a triangle to cut the opposite
sides, and the points of section be joined two and two by lines which are produced
to cut the remaining sides of the triangle ; then the sides of the triangle will all
be divided in harmonical proportion, each in the two points in which it is cut by
the lines described, and the last-mentioned three points of section will be in one
straight line.
Let ABC be a triangle, from the
angles A, B, C, of which lines are
drawn through any point P to meet
the opposite sides in D, E, F, re-
spectively ; and let EF be drawn to
meet BC in D', FD to meet AC in
E', and DE to meet AB in F' : then
First, BD : DC : : BD' : D'C,
CE : EA : : CE' : E'A,
and FA : BF : : F'A : BF'.
For, {th. 96,) AE : EC : : AF . BD : FB . CD,
and, (th. 95,) AE' : E'C : : AF . BD : FB . CD.
"Whence AE : EC : : AE' : E'C ; and similarly of the other two proportions.
The lines are therefore divided harmonically (def. 84).
Second. The points, D', E', F', are in one straight line. For, compounding
the three ratios, we have
BD . CE . FA : DC . EA . FB : : BD' . CE' . FA : DC . E'A . FB ;
and since the first term is equal to the second, the third term is equal to the
fourth, and CE' : E'A : : F'B .CD' : F'A . BD'. Hence {th. 97) the points D',
E', F', are in one straight line.
Cor. 1. If from one angle, as B, of a triangle ABC, a line BE be drawn to cut
the opposite side in E, and from the other angles any number of pairs of lines
be drawn to meet in BE and cut the opposite sides, as AD, CF, meeting AE in
P, AH, CG, meeting it in Q, and so on ; then DF, HG, and so on, will all meet
AC in the same point E'. For they all divide, with E, the side AC harmonically
in tlieir points of intersection. These points, must, therefore, coincide.
Cor. 2. If from any point E' in one side of a triangle lines be drawn to cut the
other two, as E'GH, E'FD, and so on ; and lines be drawn from these points to
the angles opposite, they will two and two cut each other in points, all of which
lie in one straight line.
For {th. 97) BQ, BP, and so on, all cut the side AC in points such that with
E' divide it harmonically. Hence these points must coincide, and therefore
BQ, BP, .... must also coincide, or the points P, Q, .... all lie in one straight
line *.
* Several simple and yet very important properties of liarmonical lines are thrown together
in this note, chiefly to avoid the necessity of formal enunciations, which, if given in words, would
occupy considerable space.
1. Let AC ; CB ; ; AD ; DB be the general division of the line.
Bisect AB in M and CD in M'. Then m
2. CA : AD : : CB : BD,or the line CD is harmonically divided ^ ''~c T il
in A and B ; and the same points of division result from supposing
either AB or CD to be the given line.
3. From
THEOREMS. 341
THEOREM XCIX.
If a straight line be divided harmonically, and from the four points of section,
straight lines be drawn through any point in the same plane : then
1. Any straight line drawn parallel to one of those four lines will be bisected by the
other three.
2. Any line oblique to them all will be harmonically divided by them at the points
of intersection.
First. Let AC : CB : : AD : DB, and the lines be
drawn from A, B, C, D, to meet at F : then, if a line
MKL be drawn parallel to AF, one of the extreme lines
of the fasceau, it will be bisected in K.
Through B draw HBE parallel to ML or FA. Then,
by parallels AD : DB : : AF : BH, and AC : CB : : AF
: BE ; but by hyi^othesis, AD : DB : : AC : CB, and therefore AF : BE : :
AF : BH, or BE = BH. Again, by parallels, ML, HE, HB : BE : : MK : KL,
or MK = KL.
Next, let the line L'K'M' be parallel to one of the in-
termediate lines FC, it will be bisected in M'.
Through B draw BI'H' parallel to L'K' or FC, cutting
the extreme lines of the fasceau in I', H'; and draw HI /_Jv^\/~'^''^^^~^^
through B parallel to AF. ' u^\\
Then, because BH' is parallel to FH, and BH to FH', ^.x'
we have FH' = BH : but by the previous case, BH = BI ; and hence FH' is
equal and parallel to BI, and the diagonal H'B is bisected in I' by the diagonal
FI, or extreme sector FD. Also, by parallels, HI' : I'B : : L'M' : M'K', and
HI' = IB ; hence L'M' = MK'.
Cor. Conversely, if from any point two lines be drawn to the extremities of a
given line, a third to bisect that line, and a fourth parallel to it, they will form
an harmonical fasceau.
Second. If any line cut an harmonical fasceau FJABCD] it will be divided
harmonically in the points A', B', C, D', of intersection.
For, through B' draw KI parallel to AF : then by the ^
former part of the proposition, KI is bisected in B'; and
by parallels
AF : BT : : A'D' : B'D', and
A'F . B'K : AC : CB' : whence
AD' : D'B' : : AC : CB', and A'B' is harmonically divided in C', D'.
3. Fiom (1) we have AC — CB : AC -f CB : : AD — DB ; AD + DB ; that is, 2MC ;
2MB ; '. 2MB : ^MD, or MC * MB ; ; MB ,* MD; or MB is a mean proportional between
MC and MD.
4. In like manner, M'B : M'C *. *. M'C '. M'A, or M'C is a mean proportional between M'A
and M'B.
.5. From (3) we have MD : MD ± MD ; *. MB : MB ± MC, or which is the same thing,
MD : DB : : MB ; bc.
6. By th. 33, AD . DB = MD* — MB*, and AC . CB = MB* — MC« ; hence the differ-
ence gives ns at once
AD . DB — AC . CB = MD> — 2MB« + MC», or by (3)
= UD^ — 2MC . MD -f MC«, or th. 32,
= (MD — MC)» = CD».
342 GEOMETRY.
THEOREM C.
If two tangents from one point to a circle, and the chord joining the points of con-
tact, be draicn; then any line drawn from the intersection of the tangents which
culs the circle will be harmonically divided at its intersection with the circle and
its chord.
Let the tangents DE, DF, drawn from D, touch the circle
in E and F, and let EF be joined : then any line DA cutting /
the circle in A and B and the chord in C will be harmo-
nically divided, such that AC : CB :: AD : DB.
For DC2 + EC . CF = DE^ (/A. 39) = AD . DB {th. 6l, cor. 1). Whence
DC='' = AD.DB - EC.CF = AD.DB— AC.CB. Hence (cone, o/ «o/e 6, A
p. 341) the line AB is harmonically divided in C and D.
THEOREM CI.
If each of the angles at the base of an isosceles triangle be double of the vertical
angle, the base is the greater segment of the side divided in extreme and mean
ratio : and if the base of an isosceles triangle be the greater segment of the side
divided in extreme and mean ratio, each of the angles at the base is double of the
vertical angle.
1. Let the angles BAC, BCA, in the isosceles triangle
ABC be each double of the third, ABC; then AC will be
the greater segment of AB or BC divided in extreme and
mean ratio.
For, draw AD bisecting the angle BAC ; and since BAC
is double of ABC, its half, BAD, is equal to ABD or DAC :
but ADC is equal to BAD and ABD; that is, to BAC or
BCA, and likewise AD, DB. are equal, since the angles
BAD, ABD, are e(]ual. Hence the triangle DAC is similar
to ABC, and AC, AD, equal: and therefore AB : AC :: AC : CD ; or since
AD, AC, BD, are all equal, CB : BD :: BD : DC.
2. Let the base AC of the isosceles triangle ABC be the greater segment of
the side BC divided in extreme and mean ratio; then each of the angles BAC,
BCA, will be double of ABC.
For make BD equal to AC, and join AD; and since CB : BD :: BD : DC,
we have AB : AC :: AC : CD, and the angles BAC, ACD, equal, and therefore
the triangles B.\C, ACD. similar. Whence, since the sides AB, BC, are equal,
the sides AC, AD, are also equal, and the angles ACD, ADC, also equal. Now
by construction, BD is equal to AC ; and therefore, also, to AD, or ADB is
isosceles; and the angle ADC being equal to ABD and BAD, is double of one
of them ABD : but ADC is equal to the angle DC.\, and therefore to each of
the angles BAC, BC.\, of the triangle ABC; and hence each of these angles is
double of ihe angle ABC at the vertex.
343
MISCELLANEOUS EXERCISES IN PLANE
GEOMETRY*.
1. From two given points on the same side of a line given in position, to draw
two lines which shall meet in that line, and make equal angles with it.
2. If two circles cut each other, and from either point of intersection diameters
be drawn; the extremities of these diameters and the other point of intersection
shall be in the same straight line.
3. If a line touching two circles cut another line joining their centres, the
segments of the latter will be to each other as the diameters of the circles.
4. If a straight line touch the interior of two concentric circles, and be placed
in the outer, it will be bisected at the point of contact.
5. If from the extremities of the diameter of a semicircle perpendiculars be
let fall on any line cutting the semicircle, the parts intercepted between those
perpendiculars and the circumference are equal.
6. If on each side of any point in a circle any number of equal arcs be taken,
and the extremities of each pair joined ; the sum of the chords so drawn will be
equal to the last chord produced to meet a line drawn from the given point
through the extremity of the first arc.
7. If two circles touch each other externally or internally, any straight line
drawn through the point of contact will cut off similar segments.
8. If two circles touch each other externally or internally, two straight lines
drawn through the point of contact will intercept arcs, the chords of which are
parallel.
9. If two circles touch each other, and also touch a straight line ; the part of
the line between the points of contact is a mean proportional between the diame-
ters of the circles.
10. If a common tangent be drawn to any number of circles which touch
each other internally, and from any point in this tangent as a centre a circle be
described cutting the others, and from this centre lines be drawn through the
intersections of the circles respectively ; the segments of them within each circle
will be equal.
11. If the radius of a circle be a mean proportional to two distances from the
centre in the same straight line, the lines drawn from their extremities to any
point in the circumference will have the same ratio that the distances of these
points from the circumference have.
12. In a circle to place a straight line of a given length, so that perpen-
diculars drawn to it from two given points in the circumference may have a
given ratio.
13. If any two chords be drawn in a circle, to intersect at right angles, then
will the squares upon the four segments of those chords be together equal to the
square upon the diameter of the circle.
14. If the tangents drawn to every two of three unequal circles be produced
till they meet, the points of intersection will be in a straight line.
• It is not expected that the student should go through all these exercises iu bis first study
of geometry; but that the tutor shouM select from them fewer or more according to the capa-
city and tiilent of his pupil ; requiring demonstrations of the theorems, and both construction*
and demonstrations of the problems, thus selected.
344 GEOMETRY.
15. If the points of bisection of the sides of a given triangle be joined, the
triangle so formed will be one-fourth of the given triangle.
16. The three straight lines which bisect the three angles of a triangle meet
in the same point.
17. If from the angles of a triangle, lines, each equal to a given line, be
drawn to the opposite sides (produced if necessary); and from any point within,
lines be drawn parallel to these, and meeting the sides of the triangle; these
lines will together be equal to the given line.
18. The two triangles, formed by drawing straight lines from any point within
a parallelogram to the extremities of two opposite sides, are together half of the
parallelogram.
19. If in the sides of a square, at equal distances from the four angles, four
other points be taken, one in each side; the figure contained by the straight
lines which join them shall also be a square.
20. Determine the figure formed by joining the points of bisection of the sides
of a trapezium, and its ratio to the trapezium.
21. Determine the figure formed by joining the points where the diagonals of
the trapezium cut the parallelogram (in the last problem), and its ratio to the
trapezium.
22. If the sides of any pentagon be produced to meet, the angles formed by
these lines are together equal to two right angles.
23. If the sides of any hexagon be produced to meet, the angles formed by
these lines are together equal to four right angles.
24. If squares be described on the three sides of a right-angled triangle, and
the extremities of the adjacent sides be joined; the triangles so formed are equal
to the given triangle and to each other.
25. If from the angular points of the squares described upon the sides of a
right-angled triangle, perpendiculars be let fall upon the hypothenuse produced,
they will cut off equal segments ; and the perpendiculars will together be equal
to the hypothenuse.
26. If squares be described on the hypothenuse and sides of a right-angled
triangle, and the extremities of the sides of the former and the adjacent sides of
the others be joined; the sum of the squares of the lines joining them will be
equal to five times the square of the hypothenuse.
27. If through any point within a triangle lines be drawn from the angles to
cut the opposite sides, the segments of any one side will be to each other in the
ratio compounded of the ratios of the segments of the other sides.
28. If a line be drawn from the vertex to any point in the base of a triangle,
the sum of the two solids under the squares of the two sides and the alternate
segments of the base will be equal to the solid under the whole base and its two
segments, together with the solid under the same base and the square of the
dividing line.
29. Determine a point in a line given in position, to which lines drawn from
two given points may have the greatest diflFerence possible.
30. Divide a given triangle into any number of parts, having a given ratio to
each other, by lines drawn parallel to one of the sides of the triangle.
31 . Through a given point between two straight lines containing a given angle,
to draw a line which shall cut off a triangle equal to a given figure.
32. Divide a circle into any number of concentric equal annuli.
33. Divide it into annuli which shall have a given ratio.
34. In any quadrilateral figure circumscribing a circle, the opposite sides are
equal to half the perimeter.
EXERCISES IN PLANE GEOMETRY. 345
35. Inscribe a square in a given right-angled isosceles triangle.
36. Inscribe a square in a given quadrant of a circle.
37. Inscribe a square in a given semicircle.
38. Inscribe a square in a given segment of a circle.
39. Having given the distance of the centres of two equal circles which cut
each other, inscribe a square in the space included between the two circum-
ferences.
40. In a given segment of a circle inscribe a rectangular parallelogram whose
sides shall have a given ratio.
41. In a given triangle inscribe a triangle similar to a given triangle.
42. In a given equilateral and equiangular pentagon inscribe a square.
43. In a given triangle inscribe a rhombus, one of whose angles shall be in a
given point in the side of the triangle.
44. Inscribe a circle in a given quadrant.
45. If on the diameter of a semicircle two equal circles be described, and in
the curvilinear space included by the three circumferences a circle be inscribed;
its diameter will be to that of the equal circles in the proportion of two to three.
46. If through the middle point of any chord of a circle two chords be drawn,
the lines joining their extremities will intersect the first chord at equal distances
from the middle point.
47. If in a right-angled triangle a perpendicular be drawn from the right
angle to the hypothenuse, and circles inscribed within the triangles on each side
of it, their diameters will be to each other as the subtending sides of the right-
angled triangle.
48. If in a right-angled triangle a perpendicular be drawn from the right
angle to the hypothenuse, and circles inscribed within the triangles on each side
of it, they will be to each other as the segments of the hypothenuses made by
the perpendicular.
49. In any triangle, if perpendiculars be drawn from the angles to the oppo-
site sides, they will all meet in a point.
50. Three equal circles touch each other; compare the area of the triangle
formed by joining their centres with the area of the triangle formed by joining
the points of contact.
51. If a four-sided rectilinear figure be described about a circle, the angles
subtended at the centre of the circle, by any two opposite sides of the figure,
are together equal to two right angles.
52. If two given straight lines touch a circle, and if any number of other tan-
gents be drawn, all on the same side of the centre, and all terminated by the
two given tangents, the angles which they subtend at the centre of the circle
shall be equal to one another.
53. If two circles cut each other, and from any point in the prolongation of
the straight line which joins their intersections, two tangents be drawn, one to
each circle, they shall be equal to each other.
54. To cut off from a given parallelogram a similar parallelogram which shall
be any given part of it.
55. If there be any right-lined hexagonal figure, and two contiguous sides be
in succession equal and parallel to two other contiguous and opposite sides, each
to each ; then, first, the two remaining sides will be respectively equal and paral-
lel ; secondly, the opposite angles (viz. the first and fourth, second and fifth,
third and sixth,) will be equal to one another; thirdly, any diagonal joining two
of those opposite angles, will divide the figure into two equal parts.
56. In any pentagonal right-Uned figure, thrice the sum of the squares of the
346 GEOMETRY.
sides will be equal to the sum of the squares of the diagonals, together with
four times the sum of the squares of the five right lines joining, in order, the
middle points of those diagonals.
57. If there be any rectilinear figure having an even number of sides in a
circle, the sum of all the angles of those angles of the figure, beginning at any-
one, which succeed one another, according to the odd numbers, will be eqaal to
the sum of all the angles which succeed one another according to the even
numbers.
58. If each side of any rectilinear figure, whose sides are even in number,
touch a circle, the sum of the first, third, fifth, &c., beginning at any one side,
and proceeding in order according to the odd numbers, will be equal to the sum
of the remaining second, fourth, sixth, &c., sides, proceeding according to the
even numbers.
59. If two circles intersect one another, and any right line be drawn
cutting the circles, it will be proportionally divided by the circumferences of
the circles.
60. Given the perpendicular drawn from the vertical angle to the base, and
the difference between each side and the adjacent segment of the base made by
the perpendicular ; to construct the triangle.
61. Given the vertical angle, the perpendicular drawn from it to the base, and
the ratio of the segments of the base made by it; to construct the triangle.
62. Given the vertical angle, the difference of the two sides containing it, and
the difference of the segments of the base made by a perpendicular from the
verte.x ; to construct the triangle.
63. Given the lengths of three lines drawn from the angles to the points of
bisection of the opposite sides ; to construct the triangle.
64. The sum of the descending infinite series a + b -\ 1 — ;+ .... is well
a a'
known to be expressed by 7, or a third proportional to a — b and a. De-
monstrate this upon geometrical principles.
65. Every equilateral polygon circumscribed about a circle, or inscribed in a
circle, is equiangular ; and every equiangular polygon so circumscribed or in-
scribed is equilateral.
66. If from the points of contact of a regular circumscribed polygon, lines be
drawn from each point of contact to its adjacent ones, the polygon so described
will be regular; and if to the circle at the angular points of a regular inscribed
polygon, tangents be drawn, these, by their successive adjacent intersections,
will form a regular circumscribed polygon.
67. Every regular polygon is capable of inscription and circumscription by
circles.
68. If in a regular inscribed polygon of an odd number of sides, parallels to
each side be drawn through the angles opposite to those sides respectively, they
will form by their intersections a regular circumscribed polygon.
69. If in a regular inscribed polygon of an even number of sides, lines be
drawn parallel to those which join every two adjacent sides through the angle,
most distant from these lines, the lines so drawn will be tangents to the circle,
and ihs'ir assemblage will constitute a regular circumscribed polygon.
70. If straight lines be drawn through any point to cut a circle, and the fourth
harmonical points in each of them, (the given point and the intersections with the
circle being the other three,) be found : all these fourth points will be in one
straight line.
OF PLANES AND SOLIDS. 347
71. If two lines be drawn from any point without a circle to intersect it, and
lines be drawn to the alternate points of intersection, these will always intersect
in the chord which joins the points of contact of the tangents drawn to the circle
from the point without it.
72. If the radius of a circle be divided in extreme and mean ratio, the greater
segment is the side of the regular decagon inscril)e(l in that circle ; and the sum
of the squares of the radius and its greater segment is equal to the square of the
side of the inscribed regular pentagon.
73. If a tangent be drawn to a circle equal to its diameter, and from the
extremity of the tangent a line be drawn through the centre, and from the
points of intersection of this line with the circle, lines be drawn to the point of
contact : the greater of these will be the radius of a circle in which the less will
be the side of the inscribed decagon, and in which the tangent will be the side
of the inscribed pentagon.
OF PLANES AND SOLIDS.
The figures which we have hitherto considered are such as lie entirely on one
plane : in those which follow, the intersections of different planes with one an-
other, or with given straight lines, the volumes of space enclosed with certain
combinations of planes, and other topics of the same kind, are the objects of
research. The conception of the figures and of their properties is greatly facili-
tated by the use of models, and no student shouhl proceed without them ;
though, of course, no great regard to extreme precision is requisite in their con-
struction.
DEFINITIONS.*
1. (88.) The common section of two planes is the line in which they meet or
cut each other.
2. (89.) A line is perpendicular to a plane, when it is perpendicular to every
line in that plane which meets it ; and the point in which the perpendicular
meets the plane is called the foot of the perpendicular.
3. (90.) One plane is perpendicular to another plane, when every line of the
one, which is perpendicular to the line of their common section, is perpendicular
to the other.
4. (91.) The inclination of one plane to another, or the angle they form be-
tween them, is the angle contained by two lines, drawn from any point in the
common section and at right angles to it, one of these lines in each plane. This
is often called a dihedral angle.
5. (9 1 a.) If from a point in a line which meets a plane, a perpendicular be drawn
to the plane, and the points of intersection of these two lines with the plane be
joined, the angle formed by the line in the plane and the line which meets the
plane is called the inclination of the line to the plane.
6. (92.) Parallel planes are such as being produced ever so far in every direc-
tion will never meet.
7. (93.) A solid angle is that which is made by three or more plane angles
meeting each other in the same point.
• A modified arningement of the definitions and propositions of this suhject has rendered it
necessary to cotninence both as with a new subject. The nuiubers of the last edition, however,
for obvious reasons, being desirable to be retained, they are here marked in parentheses.
348 GEOMETRY.
8. (94.) Similar solids, contained by plane figures, are such as have all their
solid angles equal, each to each, and are bounded by the same number of similar
planes, and placed in the same consecutive order.
9. (95.) A prism is a solid whose ends are parallel, equal, and like plane
figures ; and its sides, connecting those ends, are parallelograms.
10. (96). A prism takes particular names according to the figure of its base or
ends, whether triangular prism, square prism, rectangular prism, pentagonal
prism, hexagonal prism, and so on.
11. (97) A right prism is that which has the planes of the sides perpendicular
to the planes of the ends or base.
12. (98.) A parallelopiped, or parallelopipedon, is a prism bounded by six
parallelograms, every opposite two of which are equal and parallel.
13. (99.) A rectangular parallelopipedon is that whose bounding planes are all
rectangles.
14. (100.) A cube is a square prism, being bounded by six equal square sides
or faces.
15. (101.) A cylinder is a round prism, having circles for its ends. It is
conceived to be formed by the rotation of a right line about the circumfer-
ences of two equal and parallel circles, always parallel to the axis.
16. (102.) The axis of a cylinder is the right line joining the centres of the two
parallel circles about which the figure is described.
When the axis of the cylinder is at right angles to the planes of the parallel
ends, the cylinder is called a right, and when oblique to them an oblique
cylinder.
17. (103.) A pyramid is a solid, whose base is any right lined plane figure,
and its sides triangles, having all their vertices meeting together in a point
without the plane of the base, called the vertex of the pyramid.
18. (KM.) A pyramid, like the prism, takes its particular name from the
figure of the base ; as a triangular, quadrangular, etc. pyramid.
19. (105.) A cone is a round pyramid, having a circular base. It is conceived
to be generated by the rotation of a right line about the circumference of a
circle, one end of which is fixed at a point without the plane of that circle.
20. (106.) The axis of a cone is the right line joining the vertex, or fixed point,
and the centre of the circle about which the figure is described.
When the axis of the cone is at right angles to its base, the cone is said to be
a right, and when oblique to the base, an oblique cone.
21. (107.) Similar cones and similar cylinders are such as have their alti-
tudes, the diameters of their bases, and their axes, proportional.
22. (108.) A sphere is a solid bounded by one curve surface, which is every
where equally distant from a certain point called the centre. It is sometimes
conceived to be generated by the rotation of a semicircle about its diameter,
which remains fixed.
23. (109.) The axis of a sphere is the right line about which the semicircle
revolves, and the centre is the same as that of the revolving semicircle.
24. (110.) The diameter of a sphere is any right line passing through the
centre, and terminated both ways by the surface.
25. (111.) The altitude of a solid is the perpendicular drawn from the vertex
to the opposite face, considered as its base.
26. (112.) By the distance of a point from a plane is meant the shortest line
that can be drawn from that point to meet the plane. It is subsequently shown
that this is the perpendicular {th. 3, cor. 1).
A^'
, r r
349
THEOREMS.
SECT. I.— OF LINES AND PLANES.
THEOREM I. (96.)
Two straight lines which meet each other; the three sides of a triangle; any three
points in space ; or two parallel lines; — are in the same plane, and being given
determine its position.
First. Let AB, AC, be two straight lines which intersect
each other in A. A plane may be made to pass through AB
in any direction, and hence it may be turned about AB till
it also passes through C. Then the line AC which has
two of its points, A and C, in this plane, lies wholly in the
plane, and the plane itself is fixed in its position.
Second. A triangle ABC, or any three points in space not in the same right
line, determine the position of a plane.
Third. Also two parallels, AB, CD, determine the position of the plane in
■which they are situated. For the plane may be turned about one of them to
touch a point of the other, and the second line being in the same plane as the
first, and passing through a point in it, the plane must be that just determined.
THEOREM II. (97.)
The common section of two planes is a right line.
(Same figure.)
Let ACBDA, AEBFA, be two planes cutting each other, and A, B, two
points in which the two planes meet ; drawing the line AB, this line will be the
common intersection of the two planes.
For, because the right line AB touches each of the planes in the points A and
B, it touches them in all other points (def. 20) : this line is therefore common
to the two planes. That is, the common intersection of the two planes is a right
line.
THEOREM III, (98.)
Jf a straight line be perpendicular to any two other straight lines in their point of
intersection, it shall also be at right angles to the plane which passes through
them, that is, to the plane in which they are.
Let PC, PB, be two lines intersecting in P, and AP be
another line passing through P at right angles to PB and
PC, then AP will be perpendicular to any line PQ in the
plane BPC.
For, in PQ take any point Q, and draw QR parallel to
PC, meeting PB in R ; and take RB equal to RP, and draw
BQ, to meet PC in C.
Join AC, AQ, AB. Then, since the line QR is drawn
350
GEOMETRY.
parallel to the side PC of the triangle PCB, we have (th. 82) PR : RB : : CQ :
QB, and PR = RB ; hence BC is bisected in Q.
Then, {(h. 38.) PC^ + PB^ = 2PQ2 + 2CQ2, and AC^ + AB^ = 2AQ2 +
2CQ- ; and taking the first equation from the second, we shall have {th. 34)
AP- + AP2 = 2AQ2 - 2PQ^ or AQ^ = AP- + PQ^.
Whence (1 cor. th. 34) APQ is right angled at P, or AP is perpendicular to
PQ. And the same may be proved for any other line drawn in the plane MN
through the point P. The line AP is, therefore, perpendicular to every straight
line in the plane MN passing through P ; and hence to the plane in which those
lines are {def. 2).
Cor. 1. The perpendicular AP is the shortest line that can be drawn from A
to the plane. See def. 26.
Cor. 2. Oblique lines which meet the plane at ths same distance from the foot
of the perpendicular and proceed from the same point in the perpendicular^ are
equal to one another ; and that which meets the plane at a less distance from P
is less than that which meets it at a more remote distance.
THEOREM IV. (99.)
T%ere can only be one line perpendicular to a given plane, and passing through a
given point, whether that point be in the plane or without it.
For suppose there can be two.
First. Let P be in the plane HK, and the two
perpendiculars be PQ, PR. Through QPR let a
plane pass, cutting HK in PN. Then the angles
QPN, RPN will both be right angles {def. 89), and
hence equal to one another {ax. 10) : that is, a part
equal to the whole, which is absurd. Hence PR is
perpendicular to the plane HK.
Second. Let P be without the plane HK, and let
PQ, PR be the two perpendiculars admitted for
the moment to be drawn from P to HK. Let the
plane MN passing PQ, and PR, cut the plane HK
in QR. Then the angles PQR, PRQ are both right
angles, which is impossible {th. 17). Hence, there
cannot be two perpendiculars drawn to the same
plane from a point without it.
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1
iiiiii
THEOREM V. (100.)
If a straight line be perpendicular to one of two parallel planes, it will be perpen-
dicular to the other.
Let HK, LM, be two parallel i)lanes, and AB
be perpendicular to HK, it shall also be perpendi-
cular to LM.
For if not, from A draw AC perpendicular to
LM, meeting it in C. and through ABC draw a
plane cntiing the planes HK, LM, in AD and
BCE. Then, since AC is perpendicular to LM,
the angle ACB is a right angle, and hence ABC is
THEOREMS.
351
less than a rijjht angle. Hence, since BAD is a right angle, the two angles
DAB and ABC are less than two right angles, and hence the lines AD, BE, in
the same plane will meet if sufficiently produced. But AD is in the plane HK,
and BC in LM, hence HK also meets LM : which is impossible, since by hypo-
thesis they are parallel.
THEOREM VI. (101.)
\^Seefiywe to theorem 5].
If two planes be perpendicular to the same straight line, they are parallel to one
another.
Let the planes HK and LM be perpendicular to the line AB, they will be
parallel to one another. For if they be not parallel they must meet. Let N be
a point in their common intersection and join NA, NB. Then since AB is per-
pendicular to the plane HK, it is perpendicular to NA drawn through A in that
plane, and NAB is a right angle. In like manner, NBA is a right angle. But
NAB being a triangle, the two angles NAB, NBA, are together less than two
right angles. Hence the perpendiculars from A, B, in the planes HK and LM,
do not meet at N. In the same manner it can be proved that they do not meet
at any other point; and hence that the planes HK, LM, have not any point
common, and are therefore parallel.
THEOREM VII. (102.)
If from the foot of the perpendicular to any plane a line be drawn at right angles to
a line in that plane, any line drawn from the point of intersection to a point in
the line which is perpendicular to the plane will also be perpendicular to the line
which lies in the plane.
Let AP be perpendicular to the plane MN, and
BC be a line situated in that plane : if from P, the
foot of the perpendicular, the line PQ be drawn
perpendicular to BC, then QA drawn to any point
A, in AP, will also be perpendicular to BC.
Take BQ = QC, and join PB, PC, AB, AC.
Then since BQ = QC, and PQB = PQC, and
PQ common, we have also PB := PC. Again,
since PB = PC, APB = APC (def. 89), and AP
common, we have also AB = AC. Then AQ, QB, being equal to AQ, QC,
each to each, and AB equal to AC, the angle AQB = AQC, and hence they
are right angles.
Cor. BC is perpendicular to the plane APQ, for BC is perpendicular to AQ
and PQ, which determine that plane {ih. 1).
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352
GEOMETRY.
THEOREM VIII. (103.)
If a plane be perpendicular to one of two parallel lines, it will be perpendicular to
the other ; and if two straight lines are perpendicular to the same plane, they are
parallel to one another.
First. Let AP be perpendicular to the plane
HK, and DE be parallel to AP; DE will also be
perpendicular to HK.
Let the plane APDE which includes the paral-
lels AP and DE, intersect the plane HK in PD ;
and in the plane HK let the line BC be drawn
perpendicular to PD, and join AD. " '^
By cor. ih. 102, BC is perpendicular to the
plane APDE, and therefore the angle BDE is a right angle. Also, since AP,
DE, are parallels, and APD is a right angle {def. 89), the angle EDP is a right
angle. Hence ED is perpendicular to the two straight lines DP, DB, and hence
is perpendicular to the plane HK in which they lie.
Second. Let AP, DE be perpendicular to the plane HK, they shall be parallel
to one another.
For if this be denied, let some other line DF through D be supposed parallel
to AP. Then DF will (by the former part of the proposition) be perpendicular
to HK. But, by hypothesis, DE, also passing through D, is perpendicular to
HK : and hence through the same point D there can be two perpendiculars
drawn to the plane, HK, which, {th. 99,) is impossible. Hence AP, DE, are
parallel.
THEOREM IX.
(104.)
If each of two lines which intersect one another be parallel to a plane, the plane in
whic^ these lines are situated, will also be parallel tc it.
Let the two straight lines AR, AC, be each of
them parallel to the plane MN ; then the plane
HQ, in which AR, AC, are situated, will be paral-
lel to MX.
For if HQ be not parallel to MN they will meet,
and their intersection will be a straight line {th. 2).
Let NP be their intersection. Then, since NP is
in the same plane with AR and AC, which meet in
A, it cannot be parallel to both of them, and therefore will cut one at least, as
AR, in the point R. Now the point R is situated also in the plane MN, and
hence AR meets MN : but AR is parallel (hijp.) to the plane MN, which is im-
possible. Whence the plane in which AR, AC, are situated, cannot meet the
plane MN to which AR, AC, are parallel; that is, HQ is parallel to MN.
Cor. Hence, through any given line which is parallel to a plane, a second
plane may always be made to pass, that shall be parallel to that plane. For
through any point in the line another line parallel to the given plane may be
drawn ; and the plane of these two Unes will be parallel to the other plane.
THEOREMS.
353
THEOREM X. (105.)
[See figure to theorem 1 .]
If one plane meet another plane, it will make angles with that other plane, which are
together equal to two right angles.
Let the plane ACBD meet the plane AEBF; these planes make with each
other two angles whose sum is equal to two right angles.
For, through any point G, in the common section AB, draw CG, EF, per-
pendicular to AB. Then, the line CG makes with EF two angles together equal
to two right angles. But these two angles are (def. 91) the angles of inclina-
tion of the two planes. Therefore, the two planes make angles with each other,
which are together equal to two right angles.
Scholium.
In like manner it may be demonstrated, that planes which intersect have their
vertical or opposite angles equal ; also, that parallel planes cut by a third make
their alternate angles equal ; and so on, as in parallel lines ; but with obvious
hmitations. See pp. 301, 302.
THEOREM XI. (106.)
If two lines be parallel to a third line, though not in the same plane with it : they
will be parallel to each other.
Let the lines AB, CD, be each of them parallel to the
third line EF, though not in the same plane with it; then
will AB be parallel to CD.
For, from any point G in the line EF, let GH, GI, be
drawn perpendicular to EF, in the planes EB, ED, of the
parallels AB, EF, and EF, CD.
Then, since the line EF is perpendicular to the two lines
GH, GI, it is perpendicular to the plane GHI of those lines
(th. 98). And because EF is perpendicular to the plane GHI, its parallel AB is
also perpendicular to that plane (cor. th. 103). For the same reason, the line
CD is perpendicular to the same plane GHI. Hence, because the two lines
AB, CD, are perpendicular to the same plane, these two lines are parallel
ith. 103).
THEOREM XII. (107.)
If two lines, that meet each other, be parallel to two other lines that meet each other,
though not in the same plane with them ; the angles contained by those lines toiU
be equal.
Let the two lines AB, BC, be parallel to the two lines
DE, EF ; then will the angle ABC be equal to the angle
DEF.
For, make the lines AB, BC, DE, EF, all equal to each
other, and join AC, DF, AD, BE, CF.
Then, the lines AD, BE, joining the equal and parallel
lines AB, DE, are equal and parallel (th. 24). For the
same reason, CF, BE, are equal and parallel. ITierefore
VOL. I. A
351.
GEOMETRY.
AD, CF, are equal and parallel {th. 106) ; and consequently also AC, DF {th. 24).
Hence, the two triangles ABC, DEF, having all their sides equal, each to each,
have their angles also equal, and consequently the angle ABC is equal to the
angle DEF.
THEOREM XIII.
(108.)
'ITie sections made by a plane cutting two parallel planes are parallel to one
another.
Let the parallel planes HK, MN, be cut by the plane EG
in the lines EF, GD ; then EF will be parallel to GD.
For if EF, GD, be not parallel, since they are in the same
plane EG, they would, if produced, meet : but as EF is in
the plane HK, and GD in MN, these planes would in that
case also meet. But these planes cannot meet, since, by
hypothesis, they are parallel : and hence EF, GD, cannot
meet, or they are parallel.
Cor. 1. The parallels ED, FG, comprehended between parallel planes HK,
MN, are equal.
Let the plane in which the parallels ED, FG, lie, cut the parallel planes in EF,
GD. Then these are also parallel : hence EG is a parallelogram, and its sides
FG, ED, therefore equal.
Cor, 2. Hence two parallel planes are every where equi-distant. For in this
case FG, ED, are perpendicular to the two planes {th. 100), and hence are parallel
{th. 101), and hence again {cor. 1, th. 108) are equal to one another.
THEOREM XIV. (109.)
Straight lines being cut by parallel planes are divided proportionally .
Let there be, for instance, two straight lines AB, CD, cut
in A, E, B, C, F, D, by the three paiallel planes HK, MN,
PQ : then we shall have AE : EB ; ; CF : FD.
Draw AD meeting the plane MN in G, and join AC, EG,
GF, BD ; the intersections EG, BD, of the plane ABD with
MN, PQ, are parallel {th. 108) ; and hence AE : EB : ; AG :
GD. In like manner, AC, GF, are parallel, and hence AG:
GD : : CF : FD. Hence AE : EB I ) CF : FD. In the
same way the property is established if there be more lines or more planes, or
both.
Scholium.
If three or more straight lines meeting a plane be divided proportionally, and
in the same order, all the sets of corresponding points of section will lie in
planes parallel to the first : but if there be only two lines, the planes through the
corresponding points may or may not be parallel (p. 360). When the planes are
not parallel, however, their sections with each other will either coincide or be
parallel to each other.
THEOREMS.
355
THEOREM XV. (110.)
1. If through a line which is perpendicular to a plane another plane be made to
pass, this plane will be perpendicular to the former.
2. If two planes be perpendicular to one another, and in one of them a line be
drawn perpendicular to the common section, it will be perpendicular to the other
plane.
3. If a plane be perpendicular to a plane, and if at a point in their intersection a
perpendicular be erected to the former plane, it will lie wholly in the latter,
4. If each of two planes be perpendicular to a third plane, their intersection will be
perpendicular to it also.
First. If the line AP be perpendicular to the
plane MN, any plane through AP will be perpen-
dicular to MN.
Let the planes AB, MN intersect in BC, and in
the plane MN draw DE perpendicular to BP. Then
the line AP, being perpendicular to the plane MN,
will be perpendicular to each of the straight lines
BC, DE : but the angle formed by the two per-
pendiculars PA, PD, at the common intersection
measures the angle of the two planes, (def 91) ; and hence ^def. 90), since the
angle is right, the two planes are perpendicular to each other.
Second. If the plane AB is perpendicular to the plane MN, and in AB the
line AP is drawn perpendicular to the common section PB of the planes MN,
AB, it will be perpendicular to the plane MN.
For, in the plane MN draw PD perpendicular to PB ; then, because the
planes are perpendicular, the angle APD is a right one : therefore the line AP is
perpendicular to the two straight lines PB, PD. Hence it is perpendicular to
their plane MN.
Third. If the plane AB be perpendicular to the plane MN, and if at a point
P of the common intersection a perpendicular be erected to the plane MN, that
perpendicular will be in the plane AB.
For if not, then in the plane AB a perpendicular AP might be drawn to the
common intersection PB, which at the same time would be perpendicular to the
plane MN. Hence two perpendiculars may be drawn from the same point to
the same plane, which is impossible {th. ]09).
Fourth. If each of two planes be perpendicular to a third plane, their common
intersection will be perpendicular to it also.
That is, if the planes AB, AD, are perpendicular to a third plane MN, their
common intersection AP will be also perpendicular to MN.
For at the point P, let a perpendicular be drawn to the plane MN. That per-
pendicular must be at the same time in the plane AB and in the plane AD, and
hence it is their common intersection AP.
Scholium.
The properties in this theorem are the foundation of the method of co-ordinates
in space, and of the principles and practice of Descriptive Geometry.
A a 2
356
GEOMETRY.
THEOREM XVI. (111.)
1/ any prism be cut by a plane parallel to its base, the section will be equal and
similar to the base.
Let AG be any prism, and IL a plane parallel to the base,
AC ; then will the plane IL be equal and similar to the base
AC. or the two planes will have all their sides and all their
angles equal.
For, the two planes AC, IL, being parallel by hypothesis :
and two parallel planes, cut by a third plane, ha\'ing parallel
sections (th. 108) ; therefore IK is parallel to AB, and KL
to BC, and LM to CD, MP to DN, and IP to AX. But AI
and BK are parallels (def. 95) ; consequently AK is a
parallelogram; and the opposite sides BA, IK, are equal {th. 22). In like man-
ner, it is shown that KL = BC, and LM = CD, MP = DX, and IP = AX,
or the two planes AC, IL, are mutually equilateral. But these two planes having
their corresponding sides parallel, have the angles contained by them also equal
(fA. 107), namely, the angle A := the angle I, the angle B = the angle K, the
angle C = the angle L, and the angle D = the angle M. So that the two planes
AC, IL, have all their corresponding sides and angles equal, or they are equal
and similar.
THEOREM XVII. (112.)
If a cylinder be cut by a plane parallel to its base, the section will be a circle, equal
to the base.
Let AF be a cylinder, and GHI any section parallel to
the base ABC; then will GHI be a circle, equal to ABC.
For, let the planes KE, KF, pass through the axis of the
cylinder MK, and meet the section GHI in the three points
H, I, L ; and join the points as in the figure.
Then, since KL, CI, are parallel {def. 102) ; and the
plane KI, meeting the two parallel planes ABC, GHI,
makes the two sections KC, LI, parallel {th. lOS) ; the
figure KLIC is therefore a parallelogram, and consequently
has the opposite sides LI, KC, equal, where KC is a radius of the circular
base.
In like manner it is shown that LH is equal to the radius KB ; and that any
other hnes, drawn from the point L to the circumference of the section GHI,
are all equal to radii of the base; consequently GHI is a circle, and equal to
ABC.
Scholium.
Had the base been any other curve whatever, it may be shown in the same
manner, that the section parallel to the base will be a figure equal and similar to
the base.
THEOREMS.
357
THEOREM XVIII. (117.)
In any pyramid, a section parallel to the base is similar to the base : and these tvoo
planes are to each other as the squares of their distances from the vertex.
Let ABCD be a pyramid, and EFG a section parallel to
the base BCD, also AIH a line perpendicular to the two
planes at H and I ; then will BD, EG, be two similar
planes, and the plane BD will be to the plane EG, as AH^
to AP.
For, join CH, FI. Then, because a plane cutting two
parallel planes, makes parallel sections {th. 108), therefore
the plane ABC, meeting the two parallel planes BD, EG,
makes the sections BC, EF, parallel. In like manner, the plane ACD makes
the sections CD, FG, parallel. Again, because two pairs of parallel lines make
equal angles {th. 107), the two EF, FG, which are parallel to BC, CD, make the
angle EFG equal the angle BCD. And in like manner it is shown, that each
angle in the plane EG is equal to each angle in the plane BD, and consequently
those two planes are equiangular.
Again, the three lines AB, AC, AD, making with the parallels BC, EF, and
CD, FG, equal angles (th. 14), and the angles at A being common, the two
triangles ABC, AEF, are equiangular, as also the two triangles ACD, AFG, and
have therefore their like sides proportional, namely, AC:AF: :BC:EF: :CD:FG.
And in like manner it may be shown, that all the lines in the plane FG, are
proportional to all the corresponding lines in the base BD. Hence these two
})lanes, having their angles equal and their sides proportional, are similar, by
def. 68.
But, similar planes being to each other as the squares of their like sides, the
plane BD : EG :*. BC^ : EF-, or : : AC^ : AF^, by what is shown above. Also,
the two triangles AHC, AIF, having the angles H and I right ones {th. 98), and
the angle A common, are equiangular, and have therefore their like sides propor-
tional, namely, AC : AF : : AH : AI, or AC^ : AF^ : : AH^ : AI-\ Conse-
quently, the two planes BD, EG, which are as the former squares AC, AF*, will
be also as the latter squares AH^, AP, that is, BD : EG : : AH^ : AP.
THEOREM XIX. (118.)
In a cone, any section parallel to the base is a circle ; and this section is to the
base, as the squares of their distances from the vertex.
Let ABCD be a cone, and GHI a section parallel to the
base BCD; then will GHI be a circle, and BCD, GHI, will
be to each other, as the squares of their distances from the
vertex.
For, draw ALF perpendicular to the two parallel planes,
and let the planes ACE, ADE, pass through the axis of the
cone AKE, meeting the section in the three points H, I, K.
Then, since the section GHI is parallel to the base BCD,
and the planes CK, DK, meet them, HK is parallel to CE,
and IK to DE {th. 108). And because the triangles formed
358
GEOMETRY.
by these lines are equiangular, KH : EC : : AK : AE : : KI : ED. But EC
is equal to ED, being radii of the same circle ; therefore KI is also equal to KH.
And the same may be shown of any other lines drawn from the point K to the
perimeter of the section GHI^ which is therefore a circle {def. 44).
Again, by similar triangles, AL : AF :: AK : AE, or : : KI : ED, hence
m/: AF- : : KI- : ED^; but KP : ED^ : : circle GHI : circle BCD {cor. th.
93) ; therefore AL- : AF^ : : circle GHI : circle BCD.
THEOREM XX. (121.)
If a sphere be cut by a plane, the section will be a circle *.
Let the sphere AEBF be cut by the plane ADB ;
then will the section ADB be a circle.
If the section pass through the centre of the sphere,
then will the distance from the centre to every point in
the periphery of that section be equal to the radius of
the sphere, and consequently such section is a circle.
Such, in truth, is the circle EAFB in the figure.
Draw the chord AB, or diameter of the section ADB ;
perpendicular to which, or the said section, draw the axis of the sphere ECGF,
through the centre C, which will bisect the chord AB in the point G (th. 41).
Also, join CA, CB ; and draw CD, GD, to any point D in the perimeter of the
section ADB.
Then, because CG is perpendicular to the plane ADB, it is perpendicular both
to GA and GD {def. 89). So that CGA, CGD are two right-angled triangles,
having the perpendicular CG common, and the two hypothenuses CA, CD,
equal, being both radii of the sjjhere ; therefore the third sides GA, GD, are also
equal (cor. 2, th. 34). In like manner it is shown, that any other line, drawn
from the centre G to the circumference of the section ADB, is equal to GA or
GB ; consequently that section is a circle.
THEOREM XXI.
Jf two spheres intersect one another, the common section is a circle.
Let a, B, be the centres : draw
any two planes through AB cutting
the spheres in the circles KCNG,
MCLG, and KDNH, MHLD.
Join CG meeting AB in E ; and
join AC, CB, AG, GB, AD, DB,
AH, HB. DE, EH.
Then, since CA = AD, and CB =
BD, and AB common, the angles
ACB, ADB, are equal ; as are likewise
the angles CAB, DAB, and the angles CBA, DBA.
* Tlic gcction tlirongh the centre, having the same centre and diameter as the sphere, is
called a great circle of the sphere ; the other pKine sections being called less circles of the
•phcre.
THEOREMS. 359
Again, the angles CAE, DAE, being equal, the sides CA and DA being also
equal, and AE common; hence the angle DEA = CEA.
But the line joining AB bisects CG at right angles in E ; hence also the angle
DEA is a right angle.
In the same manner NEH is a right angle, and EH =; EG. Hence the lines
DE, EH, being in the same plane, and making the angles AED, BEH, right
angles, they are in the same straight line at right angles to AB.
Also, since the two lines CG, DH, are perpendicular to AB, the plane in
M'hich they are is also perpendicular to AB.
Hence all the intersections, C, G, &c. are in a plane perpendicular to AB, and
are, therefore, in a circle.
Again, since CE = EG, DE = CE, and EH = EG, the point E is the centre
of the circle CDGH.
Theorems on the foregoing subjects for the student to demonstrate.
1. If two planes be parallel to the same plane, or to the same straight lines,
thev are parallel to one another.
2. If a plane and a straight line be parallel to the same plane, or to the same
straight line, they are parallel to one another.
3. Two parallel straight lines make equal angles with the same or with parallel
planes : and two parallel planes make equal angles with the same or with parallel
straight lines.
4. When two parallel planes are cut by a third plane, or by a straight line,
the exterior angle is equal to the interior opposite, the alternate angles are equal,
and the two interior angles are together equal to two right angles : and show
whether generally the converse of either of these three conditions takes place, in
the case of the planes being cut by a plane and by a line, the planes would be
parallel.
5. ^VTien two straight lines are not parallel, the planes which are drawn per-
pendicular to them will intersect one another.
6. When a straight line and a plane intersect one another, every straight line
perpendicular to the plane will intersect every plane which is drawn perpendicular
to the straight line.
7. If a plane bisect a dihedral angle, and from any point in it perpendiculars
be drawn to the planes which contain the dihedral angle, these perpendiculars
will be equal, and the plane passing through them will be perpendicular to the
line in which all the planes meet.
8. If the perpendiculars from a point to two planes forming a dihedral angle
be equal, the plane passing through this point and the line in which the planes
containing the dihedral angle intersect, will bisect the dihedral angle.
9. If a plane bisect a line at right angles, lines drav^-n from any point in the
plane to the extremities of the bisected line, will be equal.
10. If from the extremities of a line, two equal lines be drawn to meet in a
point, then a plane drawn through this point perpendicular to the first line will
bisect this line.
11. If through the middle of a given line a plane be drawn at right angles to
that line, it shall pass through the vertex of every isosceles triangle having that
line for a base.
12. If on a given line as base, any number of equal triangles be constituted,
having their equal sides terminated in the same extremity of the line, their
360 GEOMETRY.
vertices will all lie in the circumference, or one cube whose plane is perpen-
dicular to the common base *.
SECTION II.— OF SOLID ANGLES.
THEOREM XXII. (123.)
If a solid angle be contained by three plane angles, any two of them together are
greater than the third; and the difference of any two of them is less than the
third.
Let the solid angle at S be contained by the three
plane angles ASB, BSC, CSA : then any two of them, as
ASC, CSB, shall be greater than the third BSA.
In the plane ASB make the angle BSD equal to ESC,
and draw any line ADB from A in the same plane,
meeting SB in B. Take SC = SD, and join AC, BC,
Because the two sides BS, SD, are equal to the two BS, SC, and the angle
BSD to the angle BSC. the bases BD, BC, are equal. But AC + CB are
greater than BA {th. 10); and hence taking the equals BD, BC, from these, AC
is greater than AD.
But the sides AS, SD, being equal to AS, SC, and the base AC greater than
AD, the angle ASC is greater than ASD. Hence adding to the unequals ASC,
ASD, the equals DSB, BSC, the angles ASC and SCB are together greater than
ASB.
In the same way may the other part of the theorem be proved by means of
th. 11.
• Verbal analogy, and analogy in the forms of enunciation, often le.ids to error. The great
sitniliirity in the general form of the enunciation of several theorems respecting lines in one
plane, and of the lines and planes variously disposed in space, has often caused propositions
respecting the latter to be tacitly assumed by the inexperienced student as truths, merely in
consequence of the analogy in the form of expression, but which a little attention to the circum-
stances involved in the hypothesis would have shown at once to be fallacious. A few such are
annexed : and the student should be required to distinctly demonstrate tchii they are false, and
under what limited circumstances they are true.
1. When two planes intersect one another, straight lines which are perpendicular to them will
also intersect.
2. Wlien two planes intersect one another, any planes which are perpendicular to these will
also intersect.
3. When two straight lines intersect, any straight lines perpendicular to these will also inter-
sect.
4. When a straight line meets a plane, every straight line perpendicular to the given straight
line will intersect every plane perpendicular to the given plane.
5. Two straight lines equally inclined to the same plane are parallel to one another.
6. Two planes equally inclined to the same straight line are parallel to one another.
/. Two planes equally inclined to the same plane are parallel to one another.
8. Two straight lines which make equal angles with the same straight line are parallel to one
another.
9. Two straight lines p.irallel to the same plane are parallel to one another.
10. Two planes whicli arc p.irallel to the s.ime straight line are parallel to one another.
\\. Vs hen two straight lines are cut proportionally by three planes, these three planes are
parallel.
THEOREMS. 361
THEOREM XXIII. (124.)
Every solid is contained by plane angles, which are together less than four right
angles.
Let the solid angle at S be contained by the plane
angles ASB, BSC, CSD, DSE, ESA : then the sum of
these angles is always less than four right angles.
Let the planes which contain the several plane angles
be cut by a plane ABODE, these letters representing its
intersections with the sides of the several plane angles, or
with the edges of the solid angle.
Because the solid angle at A is contained by the three plane angles SAB,
BAE, EAS, any two of which are greater than the third, the angles SAB +
EAS are greater than EAB. For a similar reason, the two plane angles at B, C,
D, E, which are the bases of the triangles having the common vertex S, are
severally greater than the third angle at the same point, which is one of the
triangles of the polygon ABCDE. Hence all the angles at the bases of the
triangles SAB, SBA, &c. are together greater than all the angles EAB, &c. of
the polygon ABCDE. And because all the angles at the bases of the triangles,
viz. SAB, SBA, &c. together with the plane angles at S, are equal to all the
angles of the polygonal base, together with four right angles, (being in each
case equal to twice as many right angles as there are sides AB, BC, . . .), the
angles at S are less than four right angles.
THEOREM XXIV. (125.)
If each of two solid angles be contained by three plane angles equal to one another,
each to each .- then the planes in which the angles are have the same inclination
to one another.
Let the solid angles at A and B be contained by
three plane angles CAD. DAE, EAC, and FBG,
GBH, HBF, respectively; and let CAD = FBG,
DAE = GBH, and EAC = HBF : then shall the
dihedral angle formed by DAC and EAC be equal to
the dihedral angle formed by GBF and HBF.
In the lines CA and FB take AK = BM, and from
the points K and M draw the perpendiculars KD, KL, MG, MN, in the planes
of the angles whose edges are AC and FB. Then the dihedral angles made by
these planes are measured by the angles DKL and GMN respectively. Join
LD, NG.
Then the triangles KAD, MBG, having KAD = MBG, and AKD = BMG,
and the included sides equal, viz. AK ^ BM, therefore KD = MG, and AD
= BG. In like manner, in the triangles KAL, MBN, it may be proved that
KL = MN, and LA = NB.
Again, the triangles LAD, NBG, having LA = NB, AD = BG, and angle
LAD = angle NBG, the bases are equal, viz. LD ^ NG.
Lastly, in the triangles KLD, NMG, since the sides are equal, viz. KL =
MN, KD = MG, and the bases also equal, viz. LD = NG ; therefore the
angles included, viz. LKD and NMG are also equal. But these are the measures
of the inclination of the faces of the solid angle^ or the measure of the dihedral
jjgo GEOMETRY.
angle specified in the enunciation. In like manner may the other dihedral angles
be shown to be equal.
THEOREM XXV. (126.)
If two solid angles be contained, each by three plane angles which are equal to one
another, each to each, and follow each other in the same order, these solid angles
are equal.
Let there be two solid angles at A and B, contained
by the three plane angles CAD, DAE, EAC, taken in
order, and FBH, HBG, GBF, also taken in the same
order, such that CAD = FBH, DAE = HBG, and
EAC = GBF. Then will the solid angle at A be equal
to the solid angle at B.
Let the solid angle at A be applied to the solid angle
at B, so that the plane angle CAD coincide with the plane angle FBH. Then
since CA coincides with BF, and the dihedral angle made by CAD, CAE, is
equal to the dihedral angle made by FBH and FBG {th. 125), the plane CAE
will coincide with the plane FBG. Also, since the angles CAE, FBG, are equal,
the line AE coincides with BG. Wherefore the plane angle EAD coincides
with the plane angle GBH; and the solid angles at A and B thus coinciding,
are equal to one another.
THEOREM XXVI. (127.)
If every two of three plane angles be greater than the third, and if the straight
lines which contain them be all equal, a triangle may be made of the straight
lines which join the extremities of those containing lines.
Let ABC, DEF, GHK, be the
plane angles, any two of which are
greater than the third ; and take AB
= BC = DE = EF = GH = HK :
then the three lines AC, DF, GK,
will form a triangle.
At the point B, in the straight line
AB, make the angle ABL equal to
GHK, and BL equal either of the equal lines AB, &c. Join AL, LC.
Then the triangles GHK, ABL, are equal, and therefore AL = GK. But
the angles DEF and GHK, being greater than ABC, and GHK = ABL, there-
fore DEF is greater than LC, and the base DF is greater than LC. But AL
and LC are together greater than AC ; much more then are AL and DF (that is,
GK and Dl^ together greater than AC. In the same way it may be shown, that
AC + DF are greater than GK, and AC -f GK greater than DF. Hence of the
three lines AC, DF, GK, any two are greater than the third; and therefore a
triangle can be constituted of them.
Cor. If from the three edges of a triangular pyramid a perpendicular be
drawn to the plane of the base, it will meet the base in the centre of the circle
which circumscribes the base. See the next figure.
THEOREMS.
363
THEOREM XXVir.
If about a triangle ABC a circle be described, and from its centre E a perpen-
dicular ED be drawn perpendicular to the plane of the base, any point D in this
perpendicular will be the vertex of a triangular pyramid, whose three edges are
equal.
Let the three edges DA, DB, DC, of the tetrahedron
DABC be all equal ; and from D draw DE perpendicular to
the plane ABC of the base. Then E will be the centre of a
circle described about the triangular base ABC.
For join EA, EB, EC. Then since DE is perpendicular
to ABC, the three angles DEA, DEB, DEC, are right
angles.
Whence AD^ - DE= = EA^ BD^ - DE^ = EB^, and CD^ — DE^ = EC^.
But by hy-pothesis AD- = BD^ = CD", and hence also AE = BE = CE ;
and E is the centre of the circle passing through the points A, B, C.
THEOREM XXVIII.
If two solid angles be contained each by three plane angles which are equal two and
two: then in one solid angle any edge will have the same inclination to the plane
of the two others, that the homologous edge of the other has to the homologous
plane of the other two.
Let the solid angles A and B be contained by
plane angles CAE, EAD, DAC, and FBH, HBG,
GBF, which are two and two equal respectively; then
the edge AC will be inclined to the plane EAD in
the same angle that the homologous edge BF is to
the homologous plane HBG.
Take AC equal to BF, and draw the perpendiculars CK, FL, to the planes
DAE, GBH, meeting them in K, L; and from K, L, draw KP, LR, perpen-
dicular to the homologous lines AD, BG ; and join CP, FR.
Then {th. 102) CP is perpendicular to AD, and FR to BG. Hence CPK,
FRL, are the inclinations of the planes CAD, DAE, and FBG, GBH, re-
spectively ; and these inclinations (/A. 125) are equal to one another, and there-
fore the angles CPK, FRL, are equal.
Again, since in the right-angled triangles CAP, FBR, the homologous sides
AC, BF, are equal, and the angles CAD, FAG, also equal, the sides CP, FR,
are also equal. Whence in the right-angled triangles CKP, FLR, the homo-
logous sides CP, FR, are equal, and the angles CPK, FRL, also equal, the side
CK is equal to the side FL.
Hence, we have in the right-angled triangles CKA, FLB, the two sides AC,
CK, equal to the two BF, FL, each to each, the angle CAK is equal to FBL :
and these are the inclinations of AC to DAE, and of FB to GBH. Hence these
inclinations are equal.
364 GEOMETRY.
SECTION III.— THE VOLUMES OF SOLIDS *.
THEOREM XXIX. (114.)
Jf a prism be cut hy a plane parallel to the ends, it will be divided in the same ratio
as any one of its parallel edges is divided by the same plane.
Let the plane EF cut the prism AD into
two parts AF, ED, and any one of its parallel
edges AC into the parts AE, EC : then
prism AF : prism ED :: AE : EC. ^ „ ^ k c q «. s
For produce the planes AK, AN, BN, BK ; and in the produced edge AC
take any number of lines AO, OP, each equal to AE, and any number CQ,
QR, RS, each equal to EC, and through the several points O, P, Q, R, S, let
planes OU, PW, QX, RY, SZ, be drawn parallel to AB, CD, or EF.
Then since TO is parallel to AG, and OV parallel to AL, the three plane
angles TOY, YOA, AOT, are equal to the three GAL, LAE, EAG, each to
each. Hence if the point O were applied to A, the line OA to the line AE, and
the plane AT to the plane EG, the line OT would coincide with the line AG,
and the line OV with the line AL.
For a similar reason, the lines TU, UV, UB, would coincide with GB, BL,
BF ; the lines BL, BG, with FM, FH. and TG, GA, with GH, HE. Hence, aU
the edges of the prism OB would coincide with all the homologous edges of the
prism AF ; and therefore all the faces of the one coincide with aU the faces of
the other, and the two prisms are equal.
In the same manner, it may be proved that 0^V is equal to AF ; and simi-
larly, that each of the prisms CX, QY, RZ, is equal to ED. Wherefore, what-
ever multiple the line PE is of the line AE, the same multiple is the prism EW
of the prism EB ; and whatever multiple the line ES is of the line CE, the same
multiple is the prism EZ of the prism ED.
Again, if the line PE be greater than ES, the prism EW is greater than the
prism EZ ; if equal, equal ; if less, less : and these are any equimultiples of the
first and third AE, AF, and any equimultiples of the second and fourth EC, ED,
each of each. Hence
prism AF : prism ED :: AE : EC.
Tlio following propositions are mainly proved by means of the method of Cavallerius,
which he calls the arithmetic of infinites. It consists in assuming that all plane figures are made
up of an infinite nuniWr of lines parallel to e.ich other, and connected by a certain law accord-
ing to the particular figure under consideration ; and similarly, solids are assumed to be com-
posed of an infinite number of indefinitely thin lamina;, or mere plane figures, all parallel to
each other, and connected by the properties of parallel sections of the solid. This method is
not rigorously conclusive : but the great length of tlie proofs by the metJuxl of exhaustions, ren-
der* them unsuitable to the space allowed to the subject in this course; and, as they all admit
of a ready and brief investigation by means of the integral calculus, it does not appear to be
csMntial to give a more rigid system of investigation in this place.
THEOREMS.
365
Scholium.
This demonstration being general for all prisms, the particular case of the
rectangiilar parallopiped is included in it. The property as often used in refer-
ence to this particular figure is : — rectangular parallelopipeds of the same altitude
are to one another as their bases.
THEOREM XXX. (113.)
Prisms and cylinders of the same altitude, or between the same parallel planes, are
equal to one another.
Let ABT, DEV, FGXW, be prisms and a cylinder on equal bases ABC, DE,
FG, respectively, and between the same parallel planes MN, PQ (or of the same
altitude H I) ; they shall be equal to each other.
Parallel to the plane MN draw any
plane M'N' cutting the prisms and cy-
linder in the sections A'B'C, D'E', F'G'.
Then these are respectively equal to the
sections ABC, DE, FG ; and as these
latter are equal by hypothesis, the for-
mer are also equal. But the prisms and
cylinder being respectively made up of
these equal laminae, and equally nume-
rous, they must also be equal.
Cor. Every prism and cylinder is equal to a rectangular parallelepiped, of
equal base and altitude with it.
THEOREM XXXI. (115.)
Prisms of equal bases are to one another as their altitudes.
Let AL, ER, be two prisms on equal bases
AC, EG ; they shall be to one another as their
altitudes TU, VW.
For, if the bases be not similar as well as equal,
make the base EG similar to AC, and the prism
ER upon it equiangular to the prism AL. Make
AP' equal to EP, and draw the plane P'R' parallel
to AC. Then the prism AR' is equal to ER ; and
since the prism AL is cut by the plane FR' parallel
to AC into parts proportional to AF, FI, the whole
prism AL : prism AR' :: AI : AF :: UT : UX :: UT : VW :: altitude
AL : altitude of prism ER. That is, the prisms upon the equal bases
are to one another as their altitudes.
of prism
AC, EG,
S66
GEOMETRY.
THEOREM XXXII. (1 15. CoT. 2.)
Prisms, neither whose bases nor altitudes are equal, are to one another in the ratio
compounded of the ratio of their bases and the ratio of their altitudes.
Let AL, EZ, be two prisms, neither of
whose bases AC, EO, nor altitudes TU,
VW, are equal; they will be to one another
in the ratio compounded of the ratio of AC
to EO, and the ratio of TU to VW.
For take AP' equal to EP, and draw the
plane P'R' parallel to AC; and take EF
equal to AB, and draw the plane FR parallel
to EQ. Then the prism AR' will be equal to the prism ER'. Hence
prism ER base EG , prism AL UX , 1 1 r >
^- i?^ = c CvS. and ^ 7-„, = ,p^, {th. i 14, 1 15 ;)
prism EZ base EO prism AR 1 U
whence, compounding these ratios, recollecting the equality of the prisms ER,
AR', and of the altitudes UX, VW, we have
prism AL base EG UX
prism EZ ~ base EO ' TU
; and hence the theorem itself.
THEOREM XXXIII. (119.)
All pyramids and cones upon equal bases, and between parallel planes, or having
equal altitudes, are equal.
Let the pyramids VABC, WDEF, and cone
XGH, have equal bases ABC, DEF, GH, and
have equal altitudes VR, WS, XT, or lie between
the same parallel planes PQ, MN; they will be
equal to each other.
For, draw any plane M'N' parallel to the plane
in which their bases are situated, cutting them
in the sections A'B'C, D'E'F', and G'H', re-
spectively, and the perpendiculars VR, WS, XT,
from their vertices to their bases in R', S', T'.
Then {th. 109) the perpendiculars are divided proportionally in R', S', T' ;
and {th. 117) the sections have to the bases the duplicate ratio of the perpen-
diculars VR', VR; and, therefore, since the bases are equal, the sections are
equal.
Now this is true for all the sections that can be made parallel to the base, and
hence for the sums of all such sections : but these taken together make up the
entire solids ; and hence the solids themselves are equal.
THEOREM XXXIV. (120.)
Every pyramid is a third part of a prism of equal base, and lying between the
same parallel planes.
Let, first, the ])rism and pyramid have triangular bases ; viz. the prism whose
ends are BAG, EOF, and the pyramid whose base is DEF and verte.x B : then
the pyramid will be one-third part of the prism.
THEOREMS.
367
In the planes of the faces of the prism draw CD, DB,
BF; and the planes BDF, BCD, divide the whole prism
into three pyramids BDEF, DABC, and DBCF.
Since the opposite ends of the prism are equal to each
other, the pyramid whose base is ABC and vertex D, is
equal to the pyramid whose base is DEF and vertex B
(/A. 119), being p)Tamids of equal base and altitude. But
the latter pyramid, whose base is DEF and vertex B, is the
same sohd as the pyramid whose base is BEF and vertex D, and this is equal to
the third pyramid whose base is BCF and vertex D, being pyramids of the same
altitude and equal bases BEF, BCF. Consequently all the three pyramids,
which compose the prism, are equal to each other, and each pyramid is the third
part of the prism, or the prism is triple of the pyramid.
In the second place, if the base have four or more sides, it can be divided into
triangles, and the prism and pyramid on that base into corresponding triangular
prisms and pyramids ; it will follow, since each partial pyramid is one third part
of the corresponding partial prism, that the entire partial pyramids will be one
third part of the entire partial prisms. Hence, generally, every pyramid is one
third part of the prism, having the same base, and lying between the same
parallel planes.
Cor. Any cone is the third part of a cylinder, or of a prism, of equal base and
altitude ; since it has been proved that a cylinder is equal to a prism, and a cone
equal to a pjTamid, of equal base and altitude.
THEOREM XXXV. (116.)
Similar solids are to one another in the triplicate ratio of their homologous edges.
First. Let DABC, HEFG, be two similar
prisms, the corresponding letters being at
homologous points ; they shall be to one
another in the triplicate ratio of the homo-
logous sides.
From D, H, homologous points of the
ends opposite to ABC and EFG, draw the
perpendiculars DK, HL, to those bases, and
join KA, EL.
Then, since DABC, HEFG, are similar
prisms, the angles at A are equal to those at
E, and the angle DAK is equal to the angle
HEL, (th. xxviii.) and the angles DKA, HLE, are right angles, since DK, HL,
are perpendicular to the bases. Hence the triangles DKA, HLE, are similar;
and DA : HE :: DK : HL. Again, the two prisms are to each other in the
ratio compounded of the ratio of their bases and the ratio of their altitudes ; and
hence, in the ratio compounded of the ratio of their bases and homologous sides
AD, HE.
Now the bases themselves, being by hypothesis similar figures, are to one
another in the duplicate ratio of their homologous sides AB, EF ; and the sides
AD, HE, are in the ratio of the sides AB, EF. The prisms, therefore, are in
the ratio compounded of the ratio of AB to EF, and the duplicate ratio of AB to
EF ; that is, in the triplicate ratio of AB to EF.
368
GEOMETRY.
Second. Let ABCDK, EFGHL, be two similar
prisms : they shall be to one another in the tri-
plicate ratio of their homologous sides AD, EH.
Draw the perpendiculars DK, HL, from the
vertices to the bases, and join AK, EL. Then
it may be proved, as in the last case, that AD :
EH :: DK : HL, and the remaining part of the
demonstration will be exactly similar.
Lastly. As any two similar polyhedrons are
divisable into the same number of similar trian-
gular pyramids, these partial pyramids will have the same ratios as the entire
polyhedrons : but the partial pyramids have the triplicate ratio of any one of the
homologous sides ; hence the entire polyhedrons are in the same ratio. Like
reasoning appUes also to similar cones and similar cylinders.
Cor. 1. Since cubes are included in the demonstration for the prism, the
cubes described on two homologous edges of the polyhedrons will be to each
other in the triplicate ratio of those edges. Hence any two similar polyhedrons
have the same ratio as the cubes of the homologous sides.
Cor. 2. Similar cones and similar cylinders are also to each other as the cubes
of the diameters of their bases. For if in and about the two similar cyhnders,
similar prisms upon regular polygonal bases be described, of 4, 8, 16,. . . . 2*
sides successively, the circumscribed prisms will be diminished in magnitude
continually as m becomes greater and greater ; whilst under the same circum-
stances, the inscribed prisms will be increased in magnitude. There is, also,
no conceivable limit to the diminution of the magnitude of the circumscribed
prism, nor to the increase of the inscribed prism, besides the cylinders them-
selves. The cylinder, therefore, is the limit towards which the inscribed and cir-
cumscribed prisms continually tend, and ultimately to be equal to each of them.
Whatever, therefore, is proved respecting similar prisms is true, whatever be
the number of their faces, and therefore is ultimately true, when they, by their
number becoming infinite, resolve themselves virtually (so far as magnitude is
concerned) into cylinders. The same reasoning is applicable to cones.
THEOREM XXXVI. (122.)
Every sphere is two thirds of its circumscribing cylinder.
Let CDR be a cylinder circumscribing the sphere
EFGH; the sphere shall be two-thirds of the cyUnder. ^i^
Let FH be the axis of the cylinder, which will also be
a diameter of the sphere, since the sphere is inscribed in
the cylinder ; and let the centre of the sphere I be made
the vertex of a cone whose base is ARB. Through the
axis FH let any plane pass cutting the cyhnder in the
square AC, the sphere in the circle FGH, and the cone
in the triangle AHB. Also draw a plane parallel to the d^
base of the cylinder, cutting the cylinder in the circle
OLS, the sphere in the circle PNT, and the cone in the circle QMU; and
join IN.
Now the triangles IFB, IKM, being similar, and IF equal to FB, the side IK
is equal to KM; and the triangles IKM, IKN, IKL, are right angle, and KL
equal to IG equal to IN, we have
IK- = KM^ and KL^ = IN- = IK^ -f KN-^ = KN^ + KM-.
T I 7 r
THEOREMS. 369
Let now the circle whose radius is r be denoted by r^v {th. 92) : then, mul-
tiplying the terms of the last equation by ir, we have ir.KI> = ir.KN' + tt.KM^
or the circle OTL equal to the two circles FfN and QUM. Hence the cor-
responding sections of the sphere and cone will be equal to the corresponding
section of the cylinder : and as this is the case in all the parallel sections, it is true
of the sums of all the corresponding sections, that is, of the figures themselves;
and the cylinder AG is equal to the hemisphere EFG and cone AIB together.
Again, the cone HABR is double the cone lABR, (/A. 126, 129, Sckol.,) the
sphere EFGHZ is double the hemisphere EFGZ, and the cylinder AC is double
the cylinder AG. Hence, the cylinder AG is equal to the sphere EFGHZ and
the cone AHB together : but the cone AHB is one third of the cylinder AC
(th. 129), and hence the sphere is two thirds of the same cylinder.
Cor. 1. A cone, hemisphere, and cylinder of the same base and altitude, are to
each other as the numbers 1, 2, 3.
Cor. 2. All spheres are to each other as the cubes of their diameters ; all
these being like parts of their circumscribing cylinders.
Cor. 3. From the foregoing demonstration it also appears that the spherical
zone or frustrum EGNP is equal to the difference between the cylinder EGLO
and the cone IMQ, all of the same common height IK. And that the spherical
segment PFN is equal to the difference between the cylinder ABLO and the
conic frustrum AQMB, all of the same common altitude FK.
Theorems for exercise in demonstration.
1. If two great circles of the sphere intersect one another, and tangents be
drawn to them at the point of intersection, these tangents will contain an angle
equal to the dihedral angle of the planes of the great circles.
2. The square of the diagonal of any rectangidar parallelopipedon is equal to
the sum of the squares of the three edges.
3. If a plane be drawn to touch a right cone in one of its edges, and through
the axis and this edge another plane be drawn : this plane will be at right angles
to the plane which touches the cone.
4. Planes bisecting the dihedral angles of a tetrahedron meet in one point;
and that point is the centre of a sphere inscribed in the tetrahedron.
5. If each two of four given spheres be enveloped by tangent-cones, the
vertices of the six cones thus formed will lie in one plane.
6. If a cone have its base coincident with a circular section of a sphere, it
will again cut the sphere in another circular section.
7. If the three edges about any solid angle of a tetrahedron be equal to one
another, and a perpendicular be dra\vn to the plane of the opposite face, it will
meet that plane in the centre of the circle which circumscribes that face.
8. Through the centre of a sphere draw three lines, each at right angles
to the other two : then the six points of intersection will be the angles of a
regular octahedron ; and the lines joining each of them to its adjacent points
will be the edges, and the three diameters of the sphere its diagonals.
9. If a sphere be inscribed in a right cone, its centre is in the axis of the cone,
and the surface of the sphere touches the centre of the base of the cone.
10. Lines drawn on the face of any tetrahedron from the angles to the middles
of the opposite edges, all meet in one point : and if lines be drawn from the
four points thus determined to the opposite solid angles, these four lines inter-
sect in the same point, and divide one another in the same ratio.
11. If three straight lines intersect at any point within a sphere, each at right
VOL. I. , B b
370 GEOMETRY.
angles to the plane of the other two : then the sum of the squares of their six
segments is equal to the square of the diameter of the sphere, together with
twice the rectangle of the segments of the diameter made at the point of inter-
section.
12. If lines be drawn joining the centres of the faces of a cube : these will be
the edges and diagonals of a regular octahedron ; and the square of the diagonal
is double the square of the edge.
13. If the edges of a regular tetrahedron be bisected, and the four solid angles
cut oflf by planes passing through these points, the nucleus left will be a regular
octahedron.
14. Draw lines from the middle of each side of the base of a triangular pyra-
mid to the middles of the opposite edges : the three hnes thus drawn meet in one
point and bisect each other.
15. The sum of the squares of the three faces of a triangular pyramid whose
plane angles at the vertex are all right angles, is equal to the square of the base.
16. If four lines form a quadrilateral figure, but not in the same plane : then
the lines which bisect the opposite sides, and that which bisects the diagonals,
all pass through in one point ; and the sum of the squares of the four sides is
equal to the sura of the squares of the diagonals together with four times the
square of the line joining their middle points.
Also, the sum of the squares of the four sides and of the two diagonals, is
equal to four times the sum of the squares of the lines which join the middles of
the opposite sides, and of the line which joins the middles of the diagonals.
17. Planes drawn perpendicularly from the middles of those six hnes, viz. the
four sides and two diagonals, all intersect in one point.
18. Let a right pyramid be erected on any parallelogram as its base, and be
cut by a plane : then the sum of the reciprocals of the edges, reckoned from the
vertex, which are opposite to one another, is to the sum of the reciprocals of the
other two opposite edges, in a determinable constant ratio.
19. If a circle of the sphere be made the base of a cone, whose vertex is any-
where in the superficies of the sphere, and if a tangent plane to the sphere be
drawn at the vertex of the cone : then any plane parallel to this tangent plane
will cut the cone in a circular section.
20. If any three unequal lines be placed parallel to one another in space, then
lines joining the extremities of these, two and two, (forming three trapezoids, or
one trapezoid and the diagonals of two others,) the three points of intersection
will all be in one straight line : and if four such lines be taken, the six points of
intersection of the lines so drawn will lie in one plane.
21. If three spheres mutually intersect, they will do so in a straight hne
at right angles to the plane passing through their centres.
PRACTICAL GEOMETRY.
The preceding part of this Course contains the most important theorems of
plane and solid geometr)-, demonstrated as briefly as the nature of the subject
would admit of, on valid principles : this section will comprise the constructions
of the most usually required problems that occur in geometrical practice. For
the most part, the demonstrations are either omitted altogether, or only indi-
cated ; 80 that they may serve as additional exercises in demonstration. One or
PROBLEMS. 371
two definitional remarks, however, appear to he necessary in this place, in order
to enahle the student to proceed systematically.
A problem * is a proposition in which certain things (points, lines, circles, or
other curves, or any combination of them) are given or exhibited ; and from these
some other things are required to be found, so as to fulfil certain specified con-
ditions.
The complete statement and solution of a problem comprises the following
parts : —
1. The enunciation, or statement in words, of the things which are given and
required.
When the enunciation is given in words only, it is called the general enuncia-
tion : and when in reference to a particular figure, the particular enunciation.
2. The operations to be performed to obtain the things sought (or queesita)
from those given (the data) ; which is called the construction of the problem.
3. The demonstration of the construction consists in proving, that if all the
operations spoken of be performed, the result will be that which constitutes the
quaesitum of the problem. This assumes, as the hypothesis of a theorem, the
data of the problem and all the subsequent operations j the conclusion of the
theorem being the same with the quaesitum of the problem.
Certain preliminary problems, which are called postulates, are assumed as pos-
sible to be constructed. They merely imply the separate and independent use of
the ruler and compasses : but in the following constructions, other instruments
are used for the sake of facility; though, in all cases, other methods are
• At p. 290 another kind of proposition besides the Theorem and the Problem was spoken
of — viz. a PORisM. It is not proposed, here, to treat of this class of propositions, which are in
some degree intermediate between the two. It constituted a very important and difficult branch
of the Greek Geometry ; but for two thousand years the true view of the subject was totally
lost ; and it was only during the last century that it was re-discovered by Dr. Robert Simson,
the editor of Euclid, and of some other ancient geometrical works. His researches are con-
tained in his Opera Melujua, — a book which is now extremely scarce and valuable. An
excellent dissertation on the same subject was published by the late Professor Playfair, in
the Transactions of the Royal Society of Edinburgh, Vol. III. to which the inquiring student
is especially referred.
Playfair's definition is : — " A porism is a proposition aflSrming the possibility of finding such
conditions [amongst the data] as shall render a certain problem indeterminate or capable of
innumerable solutions."
Simson's is : — " Porisraa est propositio in qua proponitur demonstrare rem aliquam, vel
plurcs datas esse, cui, vel quibus, ut et cuilibet ex rebus innumeris non quidem datis sed qu89
ad ea qua; data sunt eandem habent relationem, convenire ostendendum est aiFectionem quan-
dam commimem in propositione descriptam." — Op. Rel. p. 323.
Dugald Stewart, Professor of Moral Philosophy (and formerly of Mathematics) in the Uni-
versity of Edinburgh, defines it : — " A proposition affirming the possibility of finding one or
more of the conditions of an indeterminate theorem."
Tlie note given at p. 290 of this edition should be cancelled, as the history there referred to,
though written, was not printed. Tliat history was drawn up by the editor of this work, as an
appendix to the very incomplete article which was actually printed ; and included a description
of every theory respecting the nature and mode of investigating porisms, yet made public. The
following is the definition extracted from those unpublished papere ; and it applies equally to the
geometrical and algebraical mode of treating the subject.
A porism is a proposition in whicli is affirmed the possibility of finding such relations amongst
the data of a problem as shall render the general solution indeterminate ; and which also re-
quires the investigation of those relations, and the construction of the problem subject to those
conditions of relation.
B b 2
37^ GEOMETRY.
given dependent only upon the postulates. These postulates are the three
following : —
1. That a straight line may be drawn from any one point to any other
point.
2. That a terminated straight line may be produced to any length in a
straight line.
3. That a circle may be described about any point as a centre, and at any
distance from that centre.
PROBLEM I.
Three straight lines, A, B, C, each two of which are greater than the third, being
given, to construct a triangle whose sides shall be respectively equal to them.
Make DE equal to A, and with centres D and
E and radii equal to B and C respectively, describe
circles intersecting in F, G. Join DF, FE, and
likewise DG, GE. Then either of the triangles
DFE or DGE will be that required *.
Scholium, ^^^^en any two of the three lines are equal, the triangle is isosceles,
and when all three are equal, then it is equilateral. These particular cases are,
therefore, comprised in the general construction.
PROBLEM II.
At a given point D in a given line DE to make an angle equal to the given angle
BAC.
With any radius describe arcs from the centres
A and D ; the first BC meeting AB, AC, in B and \c/ "X?'/'
C, and the second EH meeting DE in E. With
centre E. and radius equal to BC, describe an arc
to meet HE in F. The line DF being drawn, will
make the angle EDF equal to BAC.
This is only an application of the last problem, and the equality of the angles
will be evident from Geom. th. 8 +.
PROBLEM III.
Through a given point D to draw a line parallel to a given line AB.
First method. Draw any line CD through C to
meet AB in C, and produce it backwards, till DF
is equal to DC. With centre F describe the arcs
CKE and DHG, the former meeting AB in E.
Join FE meeting the arc DHG in G. Then the
line DG being drawn and produced if necessary,
will be the parallel required.
In the drawing, only the small portions of the circles in the estimated regions of F and G
need be actually traced.
t The arc BC need not be drawn at all, it being sufBcient that the points B,C,E, be marked,
and the circles meeting at F need only be drawn in the estimated region of their intersection.
PROBLEMS. 373
For the line DG divides the sides of the triangle CFE proportionally, (bisects
them,) and hence it is parallel to the base CE. Geom. th. 82.
Second method. Draw any line DC, meeting
AB in C, and with centres C and D, and radius
CD, describe arcs DB, CH, the former meeting
AB in B. Then with centre C and radius BD
describe an arc to meet CH at E. The line CE
being drawn will be parallel to AB.
For by the construction of the second problem, the alternate angles EDC,
DCB, are equal ; and hence, Geom. th. 13, the line EF is parallel to AB *.
Third method. Draw any line CD through D,
and with centre D describe the circle CHF cut-
ting AB in H and CD in F, and with centres H
and F, and the same radii as before, describe arcs
cutting in E. Then DE will be parallel to AB.
For, since CDH is isosceles, each of the angles
at C and H is half the external angle FDH. But since FD, DE, are equal
to HD, DE, and FE equal to HE, the angles FDE, HDE, are equal, and
hence each of them is equal to half the angle FDH. Whence FDE is equal to
DCH, and therefore the lines CH, DE, are parallel f .
Scholium. The frequent occurrence of this, a paper problem, has given rise
to the construction of instruments for facilitating the operation. They are,
however, reducible, as to general principle of construction, to two,— the parallel
ruler and the parallel scales. A brief description of them is annexed.
1. ITie parallel ruler. It has been proved, Geom. th. 23, that when a quadri-
lateral has each pair of opposite sides equal, they will also be parallel. Let,
then, AB = BD : hence, whatever be the angle BAC, the Une CD will be
parallel to AB. If now we suppose the
lines AB, CD, to be respectively traced -^'iP^'^-^^^ ^"^ w- ''
on two flat rectangular rulers HK, LM, ■*; k' ?'■ Ov ^ ^f" O^ ''•^^"
along the middles of them, and of the ^1"^ -^a ^\ ^aa^
same length, and then cross pieces AC, \ \
BD, fastened by means of axes to them,
as in the figure, then it is clear that the edges of these rulers will be parallel in
all positions which the rulers so united can possibly take. Whence, if the upper
edge of the upper ruler were placed along a line A'B', and the lower held firmly
whilst the upper is pushed forward to the given point P | ; then the line RS
drawn through P by the edge of the ruler in this position, will be parallel to
A'B'. In like manner, if the upper edge be still moved forward to another
point P', a line R'S' drawn along the edge of the ruler in its new position, will
also be parallel to A'B'. And so on for any number of lines.
* None of the intersecting lines or arcs need be drawn, it being sufficient to mark the several
points of intersection : and the same remark applies throughout this entire series of problems.
+ The advantage of this method is, that it requires only a single opening of the compasses, and
the entire use of it completed without intermediate operations.
X In the figure the ruler HK adjoining the ruler LM indicates the relative position of the
two rulers when the instrument is closed to be laid aside. The descriptive part of the work
refers to the upper position of HK.
374
GEOMETRY.
2. The marquois, or parallel scales.
This is merely a right-angled triangle
and a flat ruler, altogether unconnected
with each other mechanically. They are
generally of box-wood, and the scales
are variously graduated at the edges.
The triangle has no marks except one for the middle of the hypothenuse : and
its dimensions are usually the longer and shorter sides about the right angle
nine inches and three inches respectively.
In using the marquois, the longer leg GH of the triangle is laid along the line
to which another is to be drawn parallel, and whilst held in this position, the
ruler BE is placed against the hypothenuse GK. The ruler being held in this
position, the triangle is slid with the right hand, up or down, as the case may
require, till the edge GH passes through the given point, as in the position
G'H'. Then both ruler and triangle being held firmly with the left hand in this
position, the line is traced along G'H', and this is the parallel required. If there
be more parallels through more points required, again hold the ruler firmly, and
again slide the triangle up or down till the edge GH passes through a second
point, as in the position G'H", and draw the second parallel as before.
When the point K" is arrived so far in advance of the end E of the ruler as to
give an unsteadiness to the instrument, slide the ruler down before moving the
triangle, and then proceed with the triangle as before *.
I
PROBLEM IV.
To bisect a given angle ABC.
First method. Take any equal distances BA, BC, in the
sides containing the angle, and with A, C, as centres, and
any equal radii describe circles cutting in D. Then BD
being drawn will bisect the angle ABC.
For if AD, CD be drawn, the three sides of the triangle ABC are equal
each to each to those of the triangle CBD. Hence, Geom. th. 8, the angles
ABD, CBD, are equal.
Second method. Produce one of the sides AB to D, and
take BD equal to BC. Then a line BH through B, parallel
to CD will bisect the angle ABC.
For by construction BD is equal to BC, hence the angles
BDC, BCD, are equal {th. 3). But by the parallels DC,
BH, the angles ABH, ADC, are equal, and the angles
HBC, BCD, are equal {th. 12) : whence the angles ABH,
HBC, are equal f.
The U80 of this instrument is easily acquired, though it requires a little more practice than
the common panillcl ruler to use it with complete facility. On the whole, especially in respect
of its use in drawing pcqwndiculars, it is a more convenient instrument than the old parallel
ruler : and it is matter of surjirise that it lias not obtained more attention from architectural
and mechanical draughtsmen than it has yet done. Military draughtsmen, to whom time, as
well as accuracy, is an important object, seldom use any other; aud this is a good proof of its
value as a panillcl ruler.
t This latter method is the preferable one in drawings, where a parallel ruler is admitted, as
the line DC need not then be drawn at all.
PROBLEMS.
375
PROBLEM V.
Through a given point C in a given line AB to draw a perpendicular.
First method. Take equal distances CD, CE, on each
side of the given point, and with any convenient equal
radii describe arcs meeting in F. Join FC j it is the
perpendicular required.
For, conceive DF, EF, joined : then the equality of the
sides of the triangles DCF, ECF, give the angles at C
equal {th. 8), and hence CF perpendicular to AB
{def. 25) *.
Second method. With centre C and any radius
describe a circle cutting the line AB in G. Set off
the arcs GD, GE, equal to one another ; and through
C draw CF parallel to DE.
For DE is perpendicular to AB, and CF parallel to
EDf.
Third method. Take any point D with which as a
centre, and DC as radius, describe a circle, cutting
AB in E. Draw ED to meet this circle in F, and
join FC. This will be the perpendicular required.
For the angle ECF being in a semicircle is a right
angle {th. 52) J. ^
Scholium. These methods are often, in practice,
superseded by the following one.
In every case of instruments there is a rectangular
parallelogram of wood or ivory, DGHE called a scale,
having, amongst other lines, one CF drawn from C,
the middle of DE, at right angles to DE, or parallel
to the ends DG, HE. The edge of the ruler is placed * " I*" *= ^
to coincide with the given line AB, and its middle
point with the point C : then the opposite end of CF |
being marked on the paper, the scale is removed and
CF drawn.
It is also often effected by means of the triangle and ruler described in the
Scholium to Prob. 3.
* This method is applicable where there is sufficient space obtainable in the drawing on both
sides for setting off CD and CE.
i* This is adapted to the case where AB docs not admit of prolongation sufficiently beyond C,
as near the edge of the drawing.
X Tiiis metiiod is adapted to the case where AB does not admit of sufficient prolongation to
apply the first method, as when near the edge of the drawing. When C is near a comer,
this is the only method applicable. It may, indeed, be generally advantageous to the student
to practise each of these problems and modes of construction in all the comers and edges of the
drawing, as he will then see at once the circamstances which detenuine the applicability of each
method.
376
GEOMETRY.
Place the niler DEHK to coincide with the given
line AB, and holding it in this position, place the
right angle of the triangle at C. Hold the triangle
firm, (the ruler being either retained or removed,) and
draw the line FC hy its edge. ^.p c,
If there he several perpendiculars to be drawn at £ j
equal distances, this method is convenient, if HD be |
a graduated scale adapted to those distances. How-
ever, in most cases, if the divisions be already made
on AB, it will be more convenient to draw one perpendicular to AB, and
parallels to it through the given point by the parallel ruler.
draw
PROBLEM VI.
From a given point A toithout a given line BC to draw a perpendicular to BC.
First method. With centre A, and any convenient
radius AD, describe a circle cutting BC in two points
D and E ; and from D and E with any equal radii
describe circles intersecting at F. Then FA being
joined, cutting BC in G, will be the perpendicular
required.
For conceiving the lines DA, AE, DF, FE, to be
drawn, it may be proved, as before, that the angle
DAE of the isosceles triangle is bisected by AG ; and
hence AG is perpendicular to DE *.
Second method. With any two centres H and D in
BC describe circles passing through A and intersect-
ing again in F. Join AF, cutting BC in G. It will
be the perpendicular required.
For, as before, HD bisects the angle AHF of the
isosceles triangle AHF, and is therefore perpendicular
to AF ; that is, AF is perpendicular to BC f.
Third method. Draw any line AE nearly perpen-
dicular to BC by estimation, and produce it till
EF = EA. With centres A and radii AF, AE, de-
scribe circles, the former meeting BC in H, and the
latter to intersect the line drawn from A to H. With
centre D and radius DA describe a semicircle cutting
BC in G : then AG being drawn, will be the perpen-
dicular required.
For since AE = EF by construction, and that AH = AF ; and AE being
• This method can only be used wlien A is not near the edge of the drawing. When it is near
the top or iM.ttoui of the drawing, (BC supiH>s,ed horizontal,) the two triangles DAE, DFE, may
be drawn on the same bide of AB, but not with the bides of the one equal to those of the
other.
t This niclhoil is very convenient when the point A is near the margin, but not near a comer
of the drawing. The points H and D should be taken as remote as circumstances will allow, to
prevent the arcs intersecting under loo acute an angle.
PROBLEMS.
377
half AF, D is the middle of AH. Hence the semicircle passes through H, and
AGH in the semicircle is a right angle *.
Scholinm. In practice these methods are sometimes superseded by others
analogous to those spoken of in the last Scholium, especially where A is not
very remote from the line BC. The former method is more convenient than
the latter of those processes.
Adjust the scale so that the cross line CF shall coin-
cide with AB, and the side DE shall pass through A;
and draw AC by the edge DE.
Or, thus again, by the scale.
Place the ruler abed along the given line, and the
shorter side of the triangle ef(/ against the ruler : then
;he longer one will be perpendicular to AB, however
t be slid along the ruler. If the point A be not more
listant from BC, the triangle in being slid along may
be made to pass through A, and the line drawn along
it will be the perpendicular required. If, on the con- j
trary, the side fff will not reach A', draw any perpen-
jicular AH by means of it, and then through A', a
line A'H' parallel to this perpendicular. It will be
:hat required.
A very accurate rectangular ruler may be made by
loubling a piece of stiff paper, ABCD, so as to obtain
1 straight line EF ; and then carefully doubling again
in GH, 60 that GF shall coincide with GE.
PROBLEM VII.
To bisect a given straight line AB.
First method. With centres A, B, and any convenient
^qual radii, describe circles intersecting in C and D. Join
CD : then it will bisect AB in E.
For it may be shown, as before, that CE bisects the
vertical angle CB of the isosceles triangle, and hence also
it bisects the base {th. 3. Cor. 1) f.
Second method. From one extremity A draw any line
A.G, and take any equal distances AC, CD, in it. Join
DB, and draw CE parallel to it, meeting AB in E, then
E is the middle of AB.
For the line CE being parallel to the base DB of the
* This metliod is generally given in the form of" take any line All and bisect it in D," and
)0 on. The present only differs as to the mode of finding the middle point D. The process is
ipplicable, in this latter form, to the case where G falls near the corner of the drawing.
■j* It is not essential that the radii of the circles intersecting in C and of those in D should be
;qual ; but when circumstances admit, it is convenient to take them so as the compasses require
10 alteration. They need not even be on different sides of the line AB ; and hence, when AB
s near an edge of the jjictiire, it will be requisite to take these pairs of radii different, and ob-
tain two intersections C and D on the same side of AB.
378
GEOMETRY.
triangle BAD, it divides the sides AD, AB, proportionally. But AD is bisected
in C ; hence AB is bisected in E *.
Third method. From the two extremities A, B, draw
parallel lines on alternate sides of AB. Take AD, BC,
any equal lines, and join CD : it will bisect AB in E.
For, obviously, the triangles are equal, having the angles
at A and D equal to those at B and C, each to each, and
the side AD equal BC. Whence the sides AE, EB, are also equal t-
Scholium. When the common parallel ruler is used, a still better form of
construction will be : —
From A, B, draw any two pairs of parallels intersecting in C and D ; then
the diagonal CD will bisect the diagonal AB.
Fourth method. Draw any line KL parallel to
AB, and taking any point D in it, set off CD, DF,
on each side of any equal lengths. Join CA, FB,
meeting in G or CB, FA, meeting in G' : draw the
line DG or DG', cutting AB in E. Then E is the
middle of AB.
For by similar triangles, AE : EB : : CD : DF,
and CD = DF ; hence AE = EB. In the same way
for the point G' J.
Fifth method. Draw any line AF, and in it take
any three equal distances AC, CD, DH ; join HB,
and produce it till BG = BH ; join CG, cutting AB
in E. This is the middle of AB.
For, {th. 95,) AE : EB : ; GH . AC : BG . CH
::2BG. AC :2AC. BG.
Hence, since the last two terms are equal, the two
former are so too : that is, AE = EB, and AB is
bisected in E.
PROBLEM VIII.
To divide a given line AB into any given number of equal parts.
First method. Draw any straight line AK making
an angle with AB, and in it set off AC, CD, DE ....
all equal to one another (and as nearly as possible of
the estimated length of one of the required parts of
AB as can be estimated, will be convenient) : join
FB, and draw EN, DM, CL, . . . parallel to FB. The
sections of these lines with AB will be the points
sought.
* This nictliod is convenient when a parallel ruler is used, and AB is near and parallel to a
Diarpiii of the drawing.
+ Tliig method is also ronvenient where AB is not nearly parallel and adjacent to a margin, if
at the sjinic time a parallel ruler be used.
X This method is very convenient when there is a line KL parallel to AB already in the
drawing.
PROBLEMS.
379
For by parallels, AC : CD : DE : : AL : LM : MN and as the
former are all equal, the latter will be so too *.
Second method. Take any two parallel lines AC,
BD ; and in them respectively set off equal parts AE,
EF, FG .... and BH, HI, IK ; the number of
parts in each being one less than the number of parts
into which AB is to be divided. Draw EK, FI, GH
. . ., (the point E being the nearest to A, and K the
most remote from B,) meeting AB in L, M, N . . . .
Then these will be the points of division sought.
For by the parallels AC, BD, and the construction, EK, FI, GH ... are all
parallel, and hence the line AB cutting them is divided proportionally f.
Third method. Draw any line KL parallel to
AB, and in it take CD, DE, EF, FG, GH . . . aU
equal to one another, and the same number as
there are to be of parts formed of AB. Join AC,
HB, the extremes meeting in P ; and draw PD,
PE, PF, PG . . . . meeting AB in M, N, R, S
Then these will be the points of division sought.
For by the parallels, CD : DE : EF ....:: AM
: MN : NR .... and the former being all equal,
the latter are so too |.
Fourth method. Draw any line AC making an
angle (a small one better than a large one) with
AB. Set off equal distances AX, XV .... FE,
ED, such that AE is as many times AX as AB is
of the part required, and let D and F be the points
of division one on each side of E adjacent to it.
Join DB, and produce it till BH = BD; join
FH, cutting AB in G : then BG is the part of AB
required.
For, (th. 95.) AF . DH . BG = FD . HB . GA.
Hence, if we denote by m the number of parts into
which AB is to be divided, we have AE = m . AX, AF = (m — 1) AX, FD =
2AX, and DH = 2DB. Whence (m — 1) AX . 2BH . BG = 2 AX . BH .
AG or
AG = (m — 1) GB ; and therefore AG + GB = mGB : that is,
AB = jnGB, or GB is the mth part of AB §.
• This is a convenient method in practice if the parallel ruler or parallel scales are used. It,
however, can also be used when AB is near and nearly parallel to the margin of the drawing.
"t" This is a convenient process when the compasses and ruler are alone used. In that case, the
lines AC, BD, should be the sides of the equilateral triangles described on AB, one on each
side. It requires, however, that AB should not be near and nearly parallel to the margin of the
picture.
J When the line KL can be taken sufficiently long, it will be desirable to take it at least about
twice the length of AB, as otherwise the point P will fall very remote from AB. When this
cannot be done, and the line will allow of sufficiently distinct divisions, take it about or less than
half the length of AB, as in the lower position of the figure, for the same reason. When it is
possible to take it at some distance from AB, set off the divisions nearly equal to the estimated
divisions of AB, and draw AH, CB, crossing between AB and KL, as in the corresponding con-
struction of the lust problem.
§ This process is due to M. Chenou, Professor of Mathematics in the Royal College of Douay,
380
GEOMETRY.
mh method. On AB describe any parallelo-
erara \BCD, and draw the diagonals AC, BU,
fntersicting in E; draw EF parallel to AD,
meeting AB in F, and join DP, meeting AC m
G • draw GH parallel to AD, meeting AB in H,
\ ;^in nH meetinc AC in K. Continue this process as far as may be neces-
::tZ Ihe runlisf. Then AF is the half of AB, AH is the third of AB, AL
is the fourth of AB ; and so on, as far as the operations may be earned on.
Since BD is bisected in E {th. 22), and EF is parallel to the base AD of the
triande ABD, we have AF = FB.
Since AD • EF •• AB • BF, we have AD = 2EF, and hence by parallels we
have also AF : AH :: AE : AG :: DF : DG :: 3 : 2, or 2AF : AH •: 6 t 2, or
^^In a simUar manner may the truth of the succeeding divisions be proved : but
it is more fully detailed in the demonstration of Prob. 7 of the next section on
Practical Geometry in the Field.
PROBLEM IX.
To draw a third proportional to two given lines AC, BD
Take any two lines EF, EG, meeting in E ; and in
these respectively take EH = AB, and EK = EI
= BD. Draw KL parallel to IH : then EL is a
third proportional to AB and CD.
For by parallels EH : EI :: EK (= EI) : EL ; and
by construction, EH = AB, EI = EK = BD ;
whence AC : BD :: BD : EL *.
PROBLEM X.
To draw a fourth proportional to three given lines AB, CD, EF.
Take two hues GH, GI, meeting in ^
G, and in GH take GK, GM, equal to
the first and third lines AB, EF, and in
GI take (JL equal to the second hne
CI). Draw MN parallel to KL, meet-
ing GI in N. Then GN is the fourth
projwrtional to the three given lines
AB, CD, EF. The proof is similar to that of the last proposition f.
(t
and it much u»c<l l)y ilic Fri'iirh draii(rlitsiniTi. Its chief recommendation is, that it disponsei
wiih the u»c uf iKimllclit ; wliilsl its chief ohjcction arises from the time and care required to ge
llic cniii|i%)i.<'« arcuratcly set to the distance RG, so as to sli])_off tlie other points of division o
thr line AH, the fuiallrst amount of error hcing so multiplied in the process iis to create a grca
•iiffrrnirc Im-iwccm the oeveral intermediate portions and the 1 ist towards A. It is well .idaptc(
l<i thr ra»»-, where iiiil) the ii\h j.art of a given line is required without the other points of th^
line Ali l>riii;' ►oiit.'lit ; hut wiiere all are required, it is inferior in precision and simplicity, am
Ihc pirrc<liii.' melliods are preferable, es]>ecially if parallel instruments be allowed.
Thr M>iuti<iii may W varied in several different but very obvious ways; but they are al
alike in prim iple.
t T\tr MOir remaik applies to this as to the last problem.
PROBLEMS.
381
^^
X
PROBLEM XI.
To find a mean proportional between two given lines AB, BC.
First method. Let them he placed in one straight »
line, as in the figure : on AC, as diameter, describe a
semicircle, and from B draw BD perpendicular to '
AC, meeting the circle in D. Then BD is the mean \
proportional required. I
This is evident from th. 87 *. ' '
Second method. Let them be placed in a straight line
as before. With any equal radii from centres A and C
describe arcs intersecting in E and F, and let O be the
intersection of EF, AC. Make BP = BO, and from O
and P as centres, with radii equal to AO describe arcs
intersecting in D. Then DB will be the mean propor-
tional between AB and BC.
For the first part of the construction finds O the middle
of AB, and the second at the same time finds BD perpen-
dicular to AC and OD := OA. The construction is therefore identical with the
preceding one.
Scholium. The fifth method of bisecting the line AC also works well with the
subsequent operations here employed.
Third method. Let AB, BC, be placed as in the
figure. Bisect AC in O, and describe semicircles
on AC and BO, intersecting in H. Then BH is the
mean proportional required.
For by the construction, the angle OHBisin'^ ° *^"
a semicircle, and hence is a right angle, and AOH a radius of the circle AHC.
Whence BH is a tangent to AHC, and hence AB : BH :: BH : BCf.
X
PROBLEM XII.
To find any number of continued proportionals to two given lines AB, AC.
On AC, the greater of the two lines,
describe a semicircle, in which place the
less line AB, and produce both lines in-
definitely. Draw CD perpendicular to
AC, meeting AB produced in D ; draw
DE perpendicular to AD, meeting AC
produced in E ; draw EF perpendicular
to AE, meeting AD produced in F ; and
so on. Then AB, AC, AD, AE
are a series of continued proportionals.
aX
3D y
* Tliis method iinpfiei more actual work than the statement of it might lead us to expect. The
next method, which is in reality identical with this, comprises the entire operation.
+ This method, though more laborious, is sometimes convenient : viz. when one or both the
given lines are of considerable length. Its advantage in this case is that the circles do not
require so much space, nor are they so diflBcult to describe, as in the other processes.
382
GEOMETRY.
For by similar triangles ABC, ACD, ADE, .... we have
AB : AC :: AC : AD :: AD : AE :: AE : AF ::
This construction is adapted to the antecedent AB, being the less of the given
lines : when AC the greater is the antecedent, the construction will be reversed,
as is exemplified in the unlettered lines of the figure *.
PROBLEM XIII.
To divide a given line AB in extreme and mean ratio.
At one extremity B draw a perpendi-
cular, and take in it BC = 4 AB : with
centre C and radius CB describe a circle,
and join AC meeting it in D. With
centre A and radius AD describe a circle
cutting AB in E. Then AB is divided
in E in extreme and mean ratio ; or such
that BA : AE : : AE : EB.
i'or, produce AC to meet the circle again at F. Then, since AB is at right
angles to BC, therefore it is a tangent to the circle at B, and BA : AD : : AF ;
AB ; or since AD = AE, and AB = 2BC = DF, we have AB : AE : : AB +
AE : AB, and, dividendo, AB : AE : : AE : EB.
PROBLEM XIV.
(See preceding figure.)
To extend a given line AB so that AB shall be the greater segment of the whole line
divided in extreme and mean ratio; or, to extend a given line in extreme and
mean ratio.
Construct as in the last problem, but set off AF in prolongation of AB to G.
Then GB : BA : : BA : AG, and the line AB is produced in extreme and mean
ratio.
For, since BA : AE : : AE : EB, by comp. BA + AE : BA : : AE + EB :
AE ; that is, GA : BA : : BA : EA.
PROBLEM XV.
To divide a given line AB harmonically in a given ratio, M : N.
Draw any two hues through A and
B parallel to each other, and in them
take AG, AH, each equal to M, and BK
equal to N. Join GK, HK, meeting AB
in D and C : then the line AB will be
harmonically divided in C and D in the
given ratio of M to N.
For AC : CB : : AH : BK : : AG : BK
AD : DB.
V\t. ..,.ir,.„K,„ f.„„„Ic.l on il.is ,,n„ci,,k. «-as proposed by Descartes for tlie construction of a
7^LT!'' ir''"""""'^ •"" ■' ''"* "'■'■" '■'■"■''"'''* any improvement, and is mechanically
.u.d«ju.tc tu the punK.»e, though theoretically corrtxt in principle.
PROBLEMS.
383
PROBLEM XVI.
Throvgh a given point P to draw a line which shall make equal angles with two
given lines AB, CD, whose intersection K would be beyond the limits of the
picture.
Through P draw PD, PE, parallel to AB, CD,
respectively, and take in them equal distances, PF,
PG, PH, as in the figure, of which the dark lines
represent the edges of the paper. Through P draw
PL parallel to GH, and MN parallel to FH. These
will be the lines sought.
For, since PE is parallel to CD, the angle EPL is
equal to PLD, and since PD is parallel to AK, the angle DPL is equal to
EBP. But by problem iv. the angle EPL is equal to DPL ; and hence EPB
is equal to PLK.
In the same way we may prove that EMP is equal to DNP.
PROBLEM XVII.
To draw through a given point C a straight line which shall tend to the intersection
of two given lines BD, AE, hut whose point of intersection falls beyond the limits
of the drawing.
First method. Through C draw any line whatever
meeting the given lines BD, AE, in B and A, and
any other line EG parallel to it meeting AE in E. b
Draw any two parallels AD, EF, meeting BF in D
and F. Then if FG be drawn parallel to CD meet-
ing EG in G, and CG be drawn, it will tend to the intersection of BD, AE.
For, by parallels, DE : BA : : DF : DB : : DG : BC, and hence AB : BC
: : ED : DG, and the lines AE, BD, CG, will, if produced, meet in the same
point H *.
Scholium. The usual process for constructing this problem, hitherto adopted
in England, is as follows : —
Draw any line AC through the given point C and any other line EG parallel
to it. Find DG a fourth proportional to AB, BC, ED. The line CG being
drawn is that sought. It is the method of finding DG adopted above, that con-
stitutes the improvement of the process.
Second method. Draw any two parallels CA, GE, cutting
the given lines as before. Join AD, BE, meeting in F, and
through F draw QF parallel to AB or DE; and join CE
meeting it in K, and AK meeting ED in G. Then CG
will tend to the intersection of BD, AE.
For, AB : BC :: QF : FK :: ED : DG, and hence, as be-
fore, the three lines AE, BD, CG, tend to one point, H.
• This solution was first given, so far as lie is aware, by the present editor of the Course, in
the Monthly Magazine for August 1825. It is here divested of the technicalities of perspective,
the original one having been published as the solution of a problem of frequent occurrence in
drawings, and therefore given in a form exclusively adapted to that particular art. That is,
liiiwever, almost the only case in which it particularly occurs in practice. It may be remarked,
that the point D, to which the arbitrary line from A is drawn, nsay be taken anywhere in BH ;
and it is only taken in EG for the convenience of having as few marks a8 possible in the line
BH.
384
GEOMETRY.
Third method. Through C draw any line AP, and
from any point P in it draw any other line PE, and
let the intersections be as in the figure. Join AD,
BE meeting in F, and draw PFQ ; and through the
intersection K of CE and PQ draw AK meeting PE
in G. Then CG being joined, will tend to the point
H, in which AE, BD, intersect.
The truth of this construction follows from the demonstration of th. 97,
p. 339*.
PROBLEM XIX.
On a given line AB to describe a square.
First method. At A draw AC perpendicular
to AB, and take AD equal to AB. Draw DE,
BE, parallel to AB, AD : then ABED is the
square required {def. 39) f.
Second method. Produce AB till AG is equal to AB. With any radius
greater than AG or AB describe arcs intersecting in C ; join AC, and take AD
equal to AB. With centres B and D describe arcs intersecting in E, and join
DE, BE ; then ADEB is the square required X.
PROBLEM XX.
To make a rectangle whose length and breadth shall be given lines AB, CD.
At a make AE perpendicular to AB and equal to . i
CD, and draw the parallels as in the last problem. ^ ]
Or a course of constructions analogous to the
second construction of the last proposition may be
employed.
• These methods all effect the purpose very completely, and without any bye-work beyond
wliat a|i|M-ar8 in the figures. Tlic latter methods are much used by the continental draughtsmen ;
•nd the last especially, being independent of parallels, is much valued by them. As, however,
thew problems are never |)erfornied but under advantageous circumstances for drawing, it seems
to p.irtakc of affectation to reject tlic use of instruments so simple as the parallel ruler. Besides,
in < onipanson of the work which iniiiit be done in executing the process by the different methods,
it will ap|H-ar that each of the two last has more than the first; and what is of greater conse-
quence, it is work of that kind to injure tlie picture itself.
Mr. Niiholsoii invented two instruments for the purpose of praci icalli/ so\v\ng this problem,
which lie called frntriiliiie(tils, and for which he received rewards from the Society of Arts. The
jiorallil ruler d<Hi the work as effectively and rapidly by tlie first method, as the centrolinead
ilopi ; and bring a simple instrument, is less liable to derangement.
T Tliit nietbixl in convenient when instruments for perpendiculars and parallels are available.
The iniwt >unple i* the Mar<)Uois.
t This ineihod doeii not diU'er in principle from the last ; the additional work prescribed
bring only ihut neccsMry for finding tlic perpendicular AC and the parallels DE, BE, when the
ruler and couijiomc* only arc used.
PROBLEMS.
385
PROBLEM XXI.
To make a square which shall be equal to any number of given squares, viz. those
whose sides are A, B, C, D, E, .. .
Draw two lines HG, GK, at right angles to each other ;
in GH take GL equal to A, and in GK take GM equal to
B, and join ML. Then in GH take GN equal to ML, and
GP equal to C, and join PN. Again, in GH take GQ equal
to NP, and in GK take GR equal to D, and join RQ. Pro-
ceed thus till all the given lines have heen employed in the
construction. The square described upon the last, as ST,
is equal to the sum of the squares on A, B, C, D, E : as is evident from sue
cessive applications of th. 34.
PROBLEM XXII.
To describe a square eq>JMl to the difference of two squares, viz. of those on
A and B.
Draw two lines HG, GK, at right angles to each other;
in GK take GC equal to the less of the given sides, B ; and
with centre C and radius equal to A, describe an arc EF
cutting GH in D. Then (cor. th. 34) the square of DG is
the difference of the squares on A and B. xi. ^^
PROBLEM XXIII.
To describe a triangle equal to a given rectilineal figure ABCDEFG.
Having fixed upon the line, as AG,
•which is to be the base of the triangle
required, find the highest point, D, in
reference to the base (by sliding the
parallel ruler is the most ready method),
and draw line DA or DG, separating
the polygon into two polygons of equal
altitude.
Produce AG both ways. Through B draw BB' parallel to AC, meeting AG
in B', and join B'C ; through C draw CC parallel to B'D meeting AG in C',
and join CD ; and proceed thus till the highest point of the figure is attained.
Then, commencing at the other end G of AG, draw FF' parallel to GE meetingr
AG in F', and join EF'; parallel to DF' draw EE' meeting AG in E', and join
DE'. Proceed thus till the highest point is attained, as at D. Then the triangle
C'DE' is that required.
For the triangle CB'A is equal to the triangle CBA (Ih. 25), and hence the
quadrilateral DCB'A is equal to the quadrilateral DCBA. Again, the triangle
DC'B' is equal to the triangle DCB', and hence the triangle DCA' is equal to
the quadrilateral DCB'A, or to the quadrilateral DCBA. In like manner, the
triangle DAE' is equal to the pentagon ADEFG : and hence, the triangle C'DE'
is equal to the given polygon.
S^ GEOMETRY.
PROBLEM XXIV.
To construct a parallelogram equal to a given polygon ABCDEFG, and which shall
hare an angle equal to a given angle HKG (see figure to last problem).
Constrict the triangle C'DE' as in the last prohlem. Through D draw LM
parallel to AG, and make the angle GC L equal to the given angle HKG, and
draw E'M parallel to C L. Draw the diagonals LE', MC', meeting at N, and
through N draw PQ parallel to AG. Then PC'E'Q is the parallelogram
required, as is obvious by th. 26, cor. 2.
PROBLEM XXV.
Tofitid a square equal to a given parallelogram.
The side of a square is a mean proportional between the two sides of the
rectangle (th. 87) ; and constructions for the mean proportional have been
given in problem 1 1.
When the parallelogram is not a rectangle, a rectangle equal to it stands upon
the same base and between the same parallels {tk. 25), into which, therefore,
the parallelogram may be converted, and the square equal to it found as above.
PROBLEM XXVI.
To describe a parallelogram one of whose angles BAD, and one of whose .tides AE
are given, and which shall itself be equal to a given poh'gon.
Construct a parallelogram ABCD, {prob. 24,) equal to
the given polygon, and having an angle BAD equal to the
given angle : and let AE be the given side of the iiarallelo-
gram to be constructed. Join DE, and draw BF parallel
to DE meeting AD in F ; and complete the figure FGEA.
It is that sought.
For, {th. 82,) BA : AF :: AE : AD, or the sides about the angle A are re-
ciprocally proportional; and hence, (th. 81,) the parallelograms AG, AC, are
equal •.
PROBLEM XXVII.
To find the centre and radius with which a given circle or segment of a circle was
described.
First method. Take any point B in the circum-
ference, and with any radius BA describe a circle
AEG { and from the jmints A and C in which it cuts
the given circumference, and the same radius as
»»efore. describe arcs cutting AEG in E and F, G and
H. 'Hien (ill, LF, being drawn to meet in O, will
give O the centre of the circle ; and the distance at
which either of them, as GH, cuts the given circle
from O, t« the required radius.
• Tl,.. probl,-,,, n,...i frequently occurs where the given and required figures are rectangles.
PROBLEMS. 337
For, CB, BA, being drawn, GH, EF, bisect them at right angles, and hence
pass through the centre ; dem. of /A. 41, and prob. 7 *.
Second method. Draw any bne AB cutting the
circle, and set off AE, AD, BF, BC, each equal to
AB, the two former in the line, and the two latter
in the circle. Parallel to DE and CF draw AO, BO,
intersecting at O. Then O is the centre of the
circle.
For this is only bisecting the equal angles at A and
B, (prob. 4, second method,) and the bisecting lines pass through the centre.
Third method. Take any four equal distances AB, BC,
CD, DE, in succession in the circumference of the circle :
draw AB and CD meeting in G, BC and DE meeting in
H, AC and BD meeting in K, and BD, CE meeting in
L: then GK, HL, being drawn to meet in O, give this
point as the centre of the circle.
For it is easily shown that GK, HL, bisect BC, CD,
at right angles, and hence they pass through the centre.
Scholium. Any of these methods enable us to find the centre and radii of the
inscribed and circumscribed circles to a regular polygon. The chief value of
the latter two consists in their ready adaptation to this purpose ; especially the
last, which requires the use of the ruler only.
PROBLEM XXVIII.
To draw a tangent to a given circle, from a given point A in the circumference.
First method. Find the centre O, and join OA ;
draw PQ through A perpendicular to AO : it will be
the tangent sought, th. 46.
Second method. Draw any line BD through A, and
with radius BA describe the circle BCD from centre
A, cutting the circle in C, and BA produced in D ;
with centres D and C, and with equal radii describe
arcs intersecting in E ; then AE being drawn, it is the tangent required.
For, (prob. Hi. fourth method,) AE is parallel to BC; and BC is perpendicular
to the diameter through A, and hence it is a tangent at A.
Third method. Describe the arc BC from centre A, as iu the last, and draw
AQ parallel to it : then AQ is the tangent f.
Fourth method. Take any other point E in the given
circumference, with which as centre, and with radius
EA, describe a circle cutting AE produced in F, and the
given circumference in B : make the arc FG equal to
FB, and join GA, which will be the tangent required.
For, since BF is equal to FG, and E is the centre of
the circle BFG, the angles BEF, FEG, are equal ; and
because BEA, GEA, are isosceles triangles, the angle
• It is only from convenience of working, and not from matliematical necessity, that AB i»
taken equal to BC.
t In the case wliere the parallel ruler is available, this is by far the neatest method.
C C 2
388
GEOMETRY.
FEB is double of EBA. and FEG is double of GAE : the angle GAE is equal
to the angle EBA in the alternate segment. Whence AG touches the circle
in A».
PROBLEM XXIX.
To draw a tangent to a given circle BCD from a given point A without it.
First method. Find the centre O of the given
circle, join AO, and on it as diameter describe a
circle ACOC', meeting the given circle in C and C' :
then if AC, AC, be drawn, they will be tangents to
the given circle.
For since AGO is a semicircle, the angle AGO is a
right angle ; and hence AC is perpendicular to the
radius OC, and therefore a tangent. In a similar manner the other case is
proved.
Second method. With A as centre, and any radius,
describe an arc cutting the given circle in B and D ;
draw AB, AD, meeting the circle again in E and F,
and draw BF, DE, meeting in G, and through G a
line CC' parallel to BD; it will cut the given circle in
C, C' the points of contact.
Third method. Through the given point A, draw
any three lines cutting the circle in E, F, in B, D^
and in G, H, respectively: draw the diagonals BF,
ED, intersecting in L, and GF, EH, intersecting in
K : then a line drawn through KL to cut the given
circle in C and C determines the points of contact f.
" This, or some modification of this, is the only method available when the centre is
unknown or inaccessible, and the point A at or near one of the extremities of the given portion
of the circumference.
+ Tlie first of these methods is the most usually employed in this countrj'; but the others
are more convenient when tiic centre of tlie given circle is not given. The last requires the
utc of the ruler only, and can be employed when only a part of the circle is given, provided it
be not almost wholly one side of the point of contact. If, however, this last condition should
not Ik- fulfilled, we liave no alternative but to find the centre, or at least in some way find other
jx'inti in the circumference in the required region for completing the openition.
Re»i)ccting the two latter methods it may be generally remarked that they depend upon the
principle* intimated in theorems 70 and 71 of the Miscellaneous E.\ercises, pp. 346, 7, of this
vulunic. Tiiey form a branch of a system of inquiries much cultivated by tlie continental
wntir» under the name of 6V.>mWrv of l/ie Utile, and sometimes under the name of transversals.
It fonnn one of the most interesting of all the departments of geometry in reference to practical
utility. The same coiniruction that is here used (the third) is also applicable to all the conic
i^ciion.. Fur »..mc further notice of tliose subjects the reader is referred to the second volume.
The »riiin-» of Cariiot. (;ari,ier, nri;mchon, Cluislcs, Scrvois, and De Gelder, mav also be cou-
•ullcd >»uh cou»ideniblc advant.ige by the inquiring student.
PROBLEMS.
PROBLEM XXX.
From a given point A in the circumference of a given circle ABC, to cut off a
segment to contain a given angle EDF.
Draw any line AC from the given point to meet the circle
in C : with any radius DE describe the arcs EF and GK,
from centres D and C; and take GH equal to FE, and join
CH cutting ABC in B : then AB being drawn will be the
line, or the arc BCA will be that required.
For this construction makes the angles ACB, EDF, equal,
and hence fulfils the condition.
PROBLEM XXXI.
In a given circle ABC to inscribe a triangle similar to a given triangle EDH, and
having the angle which is equal to EHD at a given point A in the given circle.
From A draw a line AB to cut ofi* a segment containing
one of the other given angles, as EDH, by the last pro-
blem ; and make the angle BAC equal to the angle EHD.
Then joining BC, the triangle ABC is that sought.
For it has by construction two of its angles BAC, ACB,
equal to two of the angles EHD, EDH ; and hence the
third angles are equal and the triangles similar.
PROBLEM XXXII.
On a given line AB to describe a segment of a circle to contain a given angle
CEF.
First method. If CEF be less than a right angle,
make CED equal to it, and take ED equal to EF,
(the particular length of ED being any whatever,)
and join FD. In FD take FG equal to the given
line AB, and draw GH parallel to EF to meet ED in
H. Then with radii equal to EH, and centres A, B,
describe arcs meeting in M ; and M will be the centre of the circle sought, and
the segment ANB may be described.
If the given angle PEC be greater than a right angle, construct as before with
its adjacent acute angle CEF; and the segment AN'B will be that required.
For, suppose HK drawn parallel to GF: then HK := GF = AB ; and hence
the sides of HEK are equal each to each to those of AAIB, and consequently
the angle AMB is equal HEK. But HEK is double of ChF by construction
and AMB is double of the angle ANB in the segment. The angle in the seg-
ment is, therefore, equal to the given angle CEF.
Second method. Take any point O in EF and describe
a semicircle from O as centre, and with OE as radius,
cutting EC in C : join FC, and produce it till FG is
equal to the given line AB : through G and F draw
parallels to EF, EG, meeting in S, and join SE cutting
a OS. T
390
GEOMETRY.
FG in T : draw TV parallel to EF meeting EC in V, and VK parallel to GF
nieetini; EF in K. Then EK is the radius of the segment, with which proceed
as in the last construction *.
Third method. Take any line AL makinjr any
angle with AB, and at any point G in it make the
angle AGH equal to the given angle DEF, and
through B draw BK parallel to HG meeting AL in
K. Then a circle described through the three points
A, K, B will evidently be that which has the required
segment cut oflf by AB f.
Fourth method. Draw the perpendicular GH to
bisect the given line AB, and make the angle HGK
equal to the given angle DEF, and through B
draw BM parallel to GK, meeting GH in M.
'ITien M is the centre from which the segment is
to be described.
For join MA : then, since M bisects the base
AB. of the triangle AMB, it bisects the angle
AMB. Whence the angle AMB is double of
BMP, that is of KGP, that is of DEF ; and AMB is likewise double of ANB :
whence ANB is equal to DEF %.
Fifth method. With radius equal to
the given hne AB, and with centres
A, B, E, describe circles, the two former
mutually intersecting in G and H, and
the latter cutting CE, FE, in C and F.
Make BI equal to CF, draw AI meeting
the circle HAL in K, and draw the dia-
meter KBL. Join AL, GH, intersecting
at M : then M is the centre from which
the segment is to be described.
For, GH is by this construction a perpendicular to the middle AB of the base
of the triangle, and hence the centre of any circle through A and B is in this
line. Also, the angle BAK being equal to the given one at E, and KAL, being
an angle in a semicircle, is a right angle ; and therefore, BAK and BAM are
together equal to one right angle: but MPA being a right angle, PAM and AMP
are together equal to one right angle ; and hence AMP is equal to BAK or to
CEF. Again, .VMB is double of AMP, and therefore double of CEF; whence M
is the centre of the circle lequired, in the same manner as in the former con-
structions §.
Thi» is but an obvious variation of the last construction. With the parallel ruler, how-
ever, it i« a convenient one.
t This nictiiod anticiputcs the construction for a circle through three given points; but this
i* immaterial, a* tliat prohlcni is quite in(U-pen(lcnt of this. It has, however, when carried out,
nion- wtual work tlian eitiier of llic former two; .ind has, moreover, the disadvantage of per-
forn.iii({ the work in the important part of the drawing, and thcrehv rendering tlie paper liable to
injury.
.: I ii "I ''ji'" '""■* ^'''- '■^^'' ^^^ ^^^1 "ot requisite in the construction, are omitted, inten-
tionally, Ml the fi(jure. > -i i '
8 Tlui II c»M-ntially the sarac construction as that given by Euclid, but having all the
implird o|»rationB detached, so as to suit the particular circumstances of the problem.
PROBLEMS.
391
PROBLEM XXXIII.
Through three given points A, B, C, to describe a circle.
From B describe a circle with a radius greater than
half the distance AB or BC, and from A and C circles
with the same radii, cutting the former in D, E, and
F, G : then DE, FG, being drawn to intersect in O,
will give the centre required, and hence the circle may
be described.
For if AB, BC, be joined, DE and FG would bisect them at right angles, and
hence pass through the centre.
1. Scholium. The construction is the same when a circle is required to be
described about a given triangle ABC.
2. Scholium. Sometimes a particular case of this problem arises in practice
under the following form : —
Given the span AB and rise SR of a circular arch to describe it.
Join AR, RB, and bisect them by the perpen- .^'^uT
diculars MO, NO : then, as in the problem, O is
the centre.
The joints between the stones, or vousisoirs, are
only continuations of the radii from the centre O.
>k«
PROBLEM XXXTV.
To ijiscribe a circle in a given triangle ABC.
Produce AB both ways, and take
AE, AD, BG, BF, all equal : paral-
lel to ED and FG draw AO, BO,
intersecting in O : then O is the
centre of the inscribed circle. With
centre O and radius AO describe an
arc cutting AB in P, and with centres
A and P and equal radii describe
arcs intersecting in H. Join HO cutting AB in K : then OK is the radius
of the inscribed circle, whence the circle may be drawn.
For, draw OL, OM, perpendicular to AC and BC : then, since AO bisects the
angle CAB, and those at K and L are right angles, the two triangles AOK,
AOL, are equiangular, and have the side AO opposite two equal angles common,
they are equal in all respects, and KO is equal to OL *.
Scholium. When the given triangle is equila-
teral, and a parallel ruler not available, the follow-
ing method is useful.
Find by intersections the vertices D and L of
equilateral triangles on the sides AC, BC. Draw
AL, DB, intersecting in F, and let one of them
• Tliis method implies the use of the parallel ruler for bisecting the angle. When this is not
available, bisect the angles at A and B by means of the firet method of prob. iv. The rest
of tlic construction is the same.
392
GEOMETRY.
cut the opposite side in G. Then FG is the radius of the inscribed, and FA
that of the circumscribing, circle.
PROBLEM XXXV.
In a given circle to inscribe a square.
First method. At A with any convenient radius
describe a circle EFG, cutting EA in G ; and from
A draw AC parallel to FG, meeting the circle
again in C. From C set off CH equal to AE, and
draw CH to meet EA in K ; and through K a line
parallel to EF, meeting the circle in D and B.
Then A, B, C, D, are the angular points of the
inscribed square.
For the construction {prob. 28) gives AC a diameter; and since the arcs
E.\, CH, are equal, the angles EAC, HCA, are equal; whence AKC is isosceles;
and {prob. 6) KB is perpendicular to AC ; and the two diameters AC, BD,
are at right angles. Whence the four right-angled triangles AOB, BOC, COD,
DOA, have all their sides about the right angles equal; and, therefore, AB,
BC, CD, DA, are all equal. Also, since each of the angles ABC, BCD, CDA,
DAB, is in a semicircle, it is a right angle. The figure, therefore, is a square *.
Second method. With centre A, and any radius a
greater than that of the circle, but less than its
diameter, describe the circle KGHL, cutting the
given circle in G and H : from G and H, with the
same radius, describe arcs intersecting at E, and
join EA, cutting the given circle in C : from C,
with same radius, describe a circle meeting KGHL
m K and L, and draw KL meeting the given circle
in D and B. Then A BCD will be the square, as
before f-
PROBLEM XXXVI.
To describe a square about a gicen circle.
(See the figures of preceding problem)
Find the diagonals of the inscribed square, viz. AC, BD : through A and C
draw parallels to BD, and through D, B, draw parallels to AC. ITie intercepted
{Hjrtions of these jwrallels constitute the square required, as is obvious.
• Thi. Miliiiiun »uppo>es tliat the icntro of tlic circle is not given : but wlicn the centre is
»lrra<ly known, all wc have to do is to dnuv the two diametei-s AC, BD, at right angles to one
■nollicr for finding the angular points of the square. The construction above is only a combina-
tion of ihr ,,r<Kc«» for finding the centre with that of drawing two diameters at right angles to
out another. The use of the p.irallil ruler is, however, implied; but the next solution is by the
nilcr and ci>n-|uuM-s only.
t Thi. !• only another method of getting the rectangular diameters, drawn through tbe centre
«r the girin ct.rle. Other ver> obvious variations of these constructions might easUy be given ;
bm lhf«-. xh. n the centre is not given, and the method intimated in the preceding note when
the .cni.r .. gucn, arc .uflxicnt for all practical purposes. When, however, onlv a part of the
PROBLEMS.
PROBLEM XXXVII.
To inscribe and circumscribe circles to a given square ABCD.
Draw the diagonals AC, BD, intersecting in F,
and draw FE bisecting the angle DFC (or parallel
to AD) : then F is the common centre of the two
circles sought: FD is the radius of the circum-
scribed one, and FG of the inscribed one : and the
circles can be drawn.
PROBLEM XXXVIir.
To inscribe and circumscribe circles to a given regular polygon, of which two sides
AB, BC, and the angle ABC are constructed.
This problem is precisely the same as to describe a circle
throuj<h three given points, that is, as prob. 33 ; where
AB is equal to BC, which lessens somewhat the actual
labour of construction.
PROBLEM XXXIX.
On a given straight line AC to describe a regular pentagon.
'-^^
First method. At right angles to AC take CO
equal to half AC, and with centre O and radius OC
describe a circle : join AO, and produce it to cut the
circle in D : with A and C as centres, and radius AD,
describe arcs to intersect in E : and finally, with A,
E, C, as centres, and AC as radius, describe arcs cut-
ting in F and G respectively. Then A, C, G, E, F,
are the angular points of the pentagon.
Second method. Construct as before to find D, and
with centres A and C, and radius CD, describe arcs
intersecting in H : then H is the centre of the circle in which the angles of the
pentagon will be situated ; and by setting off AF, FE, EG, all equal to AC,
the angular points will be obtained.
As these two constructions are derived from the same process, it will be
advisable to employ one as a check upon the accuracy of the other : and as they
must flow from the same principle, their demonstrations may be conjoined in
one course of reasoning, founded on Ex. 73 of the Miscellaneous Exercises, p.
347 : but which, as it is presumed that the student has investigated for himself,
it need not be given here.
circle is given, it will be requisite to find the radius of the circle, and on the diameter passing;
through the given point taken as a diagonal to describe the required square.
394
GEOMETRY
Third method. From A and C, with radii equal to AC,
describe circles HCD, KAD, intersecting in B and D.and
draw the indefinite line BD cutting AC in L. Make LE
equal to AC, Join AE and produce it till EF is equal to
AL or LC. With centre A and radius AF describe an
arc cutting BD in G ; and from centre G with radius
equal to AC describe a circle cutting HCD in H and KAD
in K. Tlien the points A, C, K, G, H, are the angular
points of the pentagon.
The proof of this is also dependent on the same principles as the two former
constructions.
PROBLEM XL.
In a given circle ALBM to inscribe an equilateral and equiangular pentagon.
Let O be the centre, and B the position of one of
the angles. Draw the diameter BOA : from centre A
with radius AO describe the circle POQ, and parallel to
PQ draw LM and AC through O and A respectively ;
then from centre D, where BC cuts LM, describe the
circle BE, cutting LM in E. Lastly, with centre B
de.scribe the circle FEG cutting the given circle in
F, G ; and with the same radius and centres F, G
describe arcs also cutting the given circle in H and
K. Then B, F, H, K, G, are the angular points of the required pentagon *.
PROBLEM XLI.
About a given circle ABCDE to describe an equilateral and equiangular pentagon.
Find, by the last problem, the angular points A, B, C, D, E, of the regular
inscriljfd pentagon. Parallel to each side, as C, D, (or to the line which joins
the other extremity of the sides AB, AE, which meet in A,) draw lines as GH :
similar operations being performed through all the angles, viz. HK drawn
through B parallel to DE, KL through C parallel to AE, LM through D parallel
to AB, and MG through E parallel to BC, the figure GHKLM formed by them
will Im; the circumscribed pentagon required -j-.
• The principle of this simple construction was given by Ptolemy in his Almagest ; but it
ha<l never been denionstiated independently of the doctrine of proportion till about sixty years
»(fo by Mr. Bonnyca«th-, formerly Professor of Mathematics in the Royal Military Academy.
Itt proof i<i \ml cfTectcd by the same principles as the other methods.
+ The di-M-ription Wing so simple, the student is left to construct his own figure. It may
b* remarked here once for all, in reference to circumscribed regular polygons, that their
conitniclion may Ik- always effected by drawing tangents to the circle at the angular points
of the inwrilK-d ((olygon of the same numl>er of sides ; but the lines which form the circum-
•rnlxsl |io1\g<iii may always be drawn by means of parallels, without reference to their tangency
U> the rirrle. Wlicn the jMjlygon has an ixltl number of sides, draw the parallel to the most dis-
tant liilc ; and when an even numl)er, draw it parallel to the diagonal line, joining any two
KjiiiditUnl an^'ular points of the inscribed polygon. The construction for the odd number is
intlanred u in the pentagon above, and for the even number in the hexagon of the next
piublrui.
PROBLEMS.
395
PROBLEM XLir.
To describe a regular hexagon: (1) on the given line AB as side: (2) in a given
circle .- and (3) about a given circle.
1 . When the hexagon is to be constructed on the given line A B,
From centres A and B, with radii AB, describe
segments of circles BCD, ACE, intersecting in C ;
and with the same radius describe from centre C
another circle, ABD cutting the preceding segments
in E and D respectively. Draw AC, BC, to meet the
circle ABD in F and G. Then A, B, E, F, 0, D, are
the angular points of the hexagon.
For, the three arcs AD, AB, BE, are equal by con-
struction, and hence the angles DCA, ACB, BCE, are equal : and the three
triangles DCA, ACB, BCE, are equilateral triangles ; and hence the three angles
at C are equal to the three angles of one of the triangles, that is to two right
angles (th. 17); and hence again, DC, CE, are one straight line, and each of the
angles one-third of two right angles. The opposite angles to these at C are
also equal {th. 5), and hence all the angles at C are equal. The lines AB, BE,
EF, FG, GD, DA, are therefore equal, and the angles which they contain also
equal ; and the figure is, therefore, a regular hexagon.
2. When the hexagon is to be inscribed within the given circle ABE.
With the radius of the given circle, step the compasses round the circumfer-
ence of the circle : it will, by the last case, mark out the angular points of the
hexagon.
3. When the hexagon is to be circumscribed about a given circle.
Set oflF the six angular points in order A, B, C, D,
E, F, of the inscribed hexagon by the last proposi-
tion. Parallel to FB or EC draw the lines GH,
LM, through D and A : and proceed similarly for
the hnes HK, MN, parallel to DF or AC, and for
the lines GN, KL, parallel to EA or DB. Then the
six points G, H, K, L, M, N, will be the angular
points of the circumscribed hexagon.
For, join DA, which by the preceding demon-
strations will pass through the centre O of the circle.
Also, since the arcs AF, AB, are equal, the line FB is perpendicular to AD, and
hence also the lines LM and HG are perpendicular to AD. These, therefore,
are tangents to the circle at A and D. In like manner, all the other lines are
tangents, and the hexagon is circumscribed. Also KH being parallel to AC,
and HG to FB, the angle KHG is equal to FPC, which is measured by half the
sum of the arcs FEDC and BA, that is by one-third of the circumference ; and
in the same way each of the other angles at G, N, . . . is measured by one-third
of the circumference. The angles are, therefore, all equal, and the figure is a
regular hexagon.
396
GEOMETRY.
PROBLEM XLIII*.
To inscribe and circumscribe regular heptagons to a given circle.
Divide, by trial, the circumference of the circle into seven equal parts, and
these will 1)6 the angular points of the inscribed heptagon, and the points of con-
Uct of the circumscribed one f.
PROBLEM XLIV.
On a given line AB to construct a regular heptagon.
From centres A, B, with radius AB, describe
circles intersecting in L and M, and produce
AB to meet one of them (as that about B) in
C : divide by trial the circle about B into seven
equal parts, and let CD be one of these divi-
sions : about D, with radius AB, describe a
third circle cutting ADC in P and N : and
draw LM, NP, to meet in O. About O, with
radius OA or OB, describe a circle ASB. This
will pass through D, and in this set off succes-
sively the points T, S, R, Q, with distances each
equal to AB. The points thus determined are
the angular points of the heptagon described upon the given straight line AB.
Scholium.
No geometrical method (that is, by means of the ruler
and compasses) can be given for the construction of re-
gular polygons, except in very limited cases. The poly-
gons of 7, 9, 11, 13, 18, 19, 21, sides belong to this
class ; but it fortunately happens that they do not very
often occur in drawing operations. The following has
been often given by practical writers; but it is only an
approximation, and generally a very rude one. It may
serve, however, as a first step, or guess at the probable
opening of the compasses to be taken, after which the
distance may be easily corrected by the eye to any greater degree of accuracy.
Let AB be a diameter of the circle, of which it is required to find the n"" part
by a construction. From centres A and B, and with AB as radius, describe arcs
intersecting in C : divide AB into as many parts (the figure is adapted to seven,
and tlie first method of eflfecting the problem, p. 378,) as it is required by the
jjroblem to divide the circumference : through C and the second point of the
division D, draw CD, to meet the circumference in H ; then BH approximates
to the required part of the circumference. f
* The Millie procegs applies to fiijiircs of 9, 11, 13, sides; and indeed, the whole series of
riKurm not i/romelrical/i^ constnictible might have been included under one enunciation ; as
there it, with too rare exceptions to be worthy of notice, no difference whatever in the manner
of forming them. See, however, tlic sclwliiim to the next problem.
t Thi« iiieUiod was, 1 believe, invented by the elder Malton. and first published in his Roi/al
K<Mttl to (Ifiimrtrii. \ scrutinizing investigation of the degree of its approximation was given by
Dr. Henry Clurke, who proposed amendments in it; but these are also, besides being very
troiiblc»oiiic, only one degree more close in their appro.ximation.
rnuuLJiMS.
SHi
PROBLEM XLV.
On a given line AB to construct a regular octagon.
First method, by the rule and compasses. From
centres A and B, with radii equal to AB, describe
the circles CQB, DRA, intersecting in E ; from
centre E, and radius AB, describe the circle FHKG,
cutting the former circles in F and G ; from centres
F and G, and radius AB, describe arcs cutting FHKG
in H and K ; draw AH, BK, cutting the first two cir-
cles in Q and R; draw BQ, AR, and in them pro-
duced take QL, RM, each equal to AB ; and lastly, with centres L and M,
describe the circles NQC, PRD, cutting the lines AH, BK, in N and P, and
the first circles in C and D. Then A, B, D, M, P, N, L, C, are the angular
points of the octagon.
Second method, adapted to the use of the parallel
ruler. Describe the circles from centres A and B,
with radius AB intersecting in C and D ; through A
and B draw AM, BN, parallel to CD, cutting the
circles in G and H ; draw AH, BG, and parallel to
them AE, BF, to meet the circles in E and F ; through
E, F, draw EK, FL, parallel to CD, AM, or BN,
meeting BG in K, and AH in L ; and lastly, draw
KM, LN, parallel to AL, BK, meeting AG, BH, in
M and N respectively, and join MN. The figure
ABFLNMKE, is a regular octagon on AB.
The principles of these constructions are too obvious to need detail here.
PROBLEM XLVI.
To construct an isosceles triangle, whose vertical angle shall be the half or the
double of the vertical angle ACB of a given isosceles triangle ABC on the same
base AB.
1. With centre C describe the circle ABD, and with the u
same radius, and centres A, B, arcs intersecting in E ; join
EC, and produce it to meet the circle in D : then AD,
DB, being drawn, ADB is the triangle required.
For, the construction gives EC, perjiendicular to AB, and
bisecting it; hence also ADB is an isosceles triangle ; and
since ADB is an angle at the circumference, and ACB an
angle on the same arc at the centre, ADB is the half of
ACB. The conditions are, therefore, fulfilled.
2. With centres A, B, C, describe circles with radius
greater than half the side AC or BC, so that that about A
intersects that about C in E and F, and that about B in-
tersects it in G and H : then EF, GH, will intersect in the
vertex D of the triangle sought.
For, D is the centre of the circle about ACB, and hence AD, DB, are equal ;
and the angle ADB at the centre is double ACB at the circumference.
398
GEOMETRY.
PROBLEM XLVir.
On a given line GH homologous to a given side AB of a given rectilineal figure
ABCDEF, to construct a figure similar to the given one.
From one of the extremities of the line AB,
which is homologous to GH, draw lines to all
the angles of the figure; on GH construct a
triangle GKH, equiangular to ACB ; on GK a
triangle KLG, equiangular to ACD ; on GL
a triangle LMG, equiangular to ADE; on MG
a triangle GMN, equiangular to AEF ; and
continue the process as long as any triangles of the given figure remain : then
GHKLMN is the figure required.
For GKH + GKL = ACB + ACD, KLG + GLM = CDA + ADE, and
80 on {constr.) : hence the two figures are equiangular. Also BC : CA | ] HK
: KG and AC : CD ; ; GK : KL; and hence BC : CD ; ; HK : KL, or the
sides ahout the equal angles at C and H are proportional ; and the same may be
shown of all the other homologous sides of the two figures. The figure GHKLMN
is hence similar to the figure ABCDEF.
Scholia.
1. When the two sides AB, GH, are coincident, in the
manner of AB, AH, the construction becomes simpler, since
it consists merely in drawing HK parallel to BC, cutting
AC in K, KL parallel to CD, cutting AD in L, and so on
till MN is drawn parallel to EF, cutting AF in N, For, in
this case the figures are composed of similar elementary tri-
angles, and are therefore, similar to one another, as before.
2. When GH lies in the same line with AB,
it is only requisite to draw GK, GL, GM, GN,
parallel to AC, AD, AE, AF, respectively; then
HK para'lel to BC, KL parallel to CD, and so
on till we arrive at MN, parallel to EF, and join
NG. The same similarity of figure as in the
preceding construction obviously takes place.
3. When the homologous sides AB, GH,
are parallel, draw AG, BH (or AH, BG),
meeting in P : then draw HK parallel to BC,
meeting CV in K, KL parallel to CD, meeting
DP in L, and so on. If the line GH be very
distant from AB. or very nearly equal to it, then as the point P would fall at an
inconvenient distance on the drawing, the point P may be taken between the
lines, by joining BG and AH. The remaining part of the construction would
be almost identical with that here given, but the order would be reversed.
4. Since ccjniangular triangles have the sides about the equal angles propor-
tional, and Iriannk's in general are most easily constructed by means of their
■ides, the proportional compasses are convenient in the general construction of
this problem.
nvwuijCi-yio.
HifV
PROBLEM XLVIII.
To draw a complex figure similar to another figure, on the same or different scales,
by means of squares
Surround the given figure by a square or a rectangle of convenient size, and
divide it by pencil lines, intersecting perpendicularly, into squares, as small as
may !)e deemed necessary. Generally, the more irregular the contour of the
figure, or the more numerous the sinuosities or subdivisions, the more numerous
the squares should be.
Then draw another square or rectangle, having its sides either equal to the
former, or greater or less in the assigned proportion, and divide this figure into
as many squares as there are in the original figure. Draw in every square of the
new figure, right lines or curved to agree with what is contained in the corres-
ponding square of the original figure ; and this, if carefully done, will give a
correct copy of the complex diagram proposed.
The pentagroph is also often used for the same purpose ; but as there are great
practical difficulties attached to it (especially its deficiency in easy, and conse-
quently certain, motion), that it is not so valuable as its theoretical principles
would lead us to anticipate. Dr. Wallace has obviated this and most other incon-
veniences by his eidograph : still its expense and liability to derangement have
greatly militated, and perhaps ever will, against its introduction jnto general use.
It is, however, but just to remark that Dr. Wallace's instrument is, in reality,
but another, though much improved, form of Scheiner's pentagraph. Another
form, somewhat intermediate to the common pentagraph and the most improved
form of the eidograph, was exhibited to the Society of Arts in Scotland a few
years ago, and which appears on the whole to answer the purpose of copying as
well as the eidograph, and to be considerably less expensive.
400
GEOMETRY.
The chord
PROBLEM XLIX.
To draw a straight line nearly equal to the circumference of a given circle.
First method. Let AE be the diameter and C the
centre of the circle, and let the semicircle be de-
scribed upon AE. Set off AB, ED, each equal to
radius AC. With centres A and E, and distances
respectively equal to AD, EB, describe arcs inter-
secting in F. Then with centre B and radius BF
describe a circle cutting the circumference (on the side D) in G.
AG is nearly equal to the quadrant of the circle.
For AF = AD = 2 sin 60o = ^3, and CF = n/AF^ — AC- = ^2 ; and
FH = CF — CH = ^{2 a/2 — ^/ 3)} Again, from the triangle BHF, we
have BG = BF = 'Z BH^ + HF^ = v/ 3 — V6; hence sin ^BG = i s/s-VQ,
cos ^ BG = i n/ 1 + a/6, sin A AB = a. cos i AB ^ § ^/3. Therefore 2AG =
4 sin ^ ABG = 4 sin {^AB + h BG) = 4 sin i AB cos 4 BG + 4 sin i BG cos i AB
= Vi + a/6~+ n/ 9 — 3^ 6 = 3-142399 .. . .But to the same decimal extent the
true value of the circumference is 3*14 159; whence the degree of approximation
is sufficiently close for most practical constructions.
This method, which may be performed by the aid of the compasses only, was
invented by Mascheroni.
Second method. Let AB be the diameter of a circle,
and C its centre. Draw an indefinite tangent at the
point A, and a radius CD parallel to this tangent.
Set off the radius DC towards A terminating in F,
and draw CF to meet the tangent in E ; and take ^
upon this tangent from E on the side of A, EG = 3CD, the straight line BG
is nearly equal to the semicircle.
For, AE = tan 30° = ^ a/3 : hence AG = EG — AE = 3 — | V 3 =
4(9 — a/3), and BG = ^ABM^AG^ = ^4 -j- U9 — \/3)2 = j -^6(20 — 3 a/3)
= 31415334.
This method, which is by an anonymous German author, gives a closer approxi-
mation than that of Mascheroni.
Third method. From any point
A in an indefinite straight line
draw a perpendicular AB equal _ \
to the given radius. Set off "^
three limes this radius from A to
D, and draw BD. At the first of these divisions C of AD, draw the perpen-
dicular CE. Set off DE in the prolongation of AD to F. Prolong AF beyond
Its extremities A and F by the lines AH and FG equal to radius AB. Take
FK = (i AB -I- H AB) = i AB ; and make AL = I AH. Then KL is
nearly equal to the circumference of the circle whose radius is AB.
For KL = AL -h AD -h FG + FD = * + 3 + ? + DE = — ^ + ? BD
5 '8 40 ' 3
= 'f + ? vaOTTaD-. = 52i±i!u^ = 3.,„5
40
120
925534.
Tins method, which is remarkable for its extent of approximation, being true
to Mix place*, was invented by M. Pioche, an eminent statuary of Metz.
X i\,\yiJUi^i.fM.K>,
^Ul
Fourth method. Take a circle
whose diameter BA = D, and
from the point B, in which the
circle touches the indefinite Hne
BR, take Ba = iD ; and set off D
three times from the point B to D",
and aR = 15aB. From a draw the perpendicular am := BA = D, and draw
Rm, cutting AB in c; and finally draw D' c. This will be nearly equal to the
circumference, whose diameter is AB.
For by similar triangles, RBc, emn give mn : nc :: ma : aTX ; whence
nc . ma
mn = Ac =
aB.D D „, „ „, ^ ^ D
— ,v- = .— ,T- ^ — • Then Be = BA — Ac = D —
all 15aB 13 15
14D
]5
Dv/2-221
15
Again, by the right-angled triangle D"cB, D"c = D /( - ) + 9 =
= 3-141S.
This method is by M. Qiietelet of Brussels, and though not extremely ap-
proximative, yet, being easy of a;:plication, is very convenient in practice.
Fifth method. Upon the circumference, whose centre is O, and
radius OB = 1, take the arc BC = 30° (which is found by the
rules and compasses), draw the tangent BC, and by the other
extremity A of the diameter BO, draw the indefinite tangent
AD, upon which set off AD equal to three times the radius OB.
Through C draw CE, parallel to BA, and join CD, which will
represent very nearly the semicircle to radius OB.
For, the right-angled triangle DCE gives DC ^ \/DE'. -f- EC^ =
^ - Hence DC
V(DA
CB)'' + CE* ; and BC = tan 30o = ~, and CE = 2.
v3
=y(-->3y+-
3 14153.
This method is by M. De Gelder of Leyden, and is also convenient, from the
simplicity of the work required.*
PROBLEM L.
To measure an angle by means of a pair of compasses only.
This will be easily comprehended by giving a single example. The method,
in fact, consists in measuring an arc or angle proposed with a pair of com-
passes, without any scale whatever, except an undivided semicircle. Produce
one of the sides of the angle backwards, and then with a pair of accurate com-
passes describe as large a semicircle as possible, from the angular point as a
centre, cutting the sides of the proposed angle, and thus intercepting a part of
the semicircle. This intercepted part is accurately taken between the points of
the comjjas^es, and stepped upon the arc of the semicircle, to ascertain how
often it is contained in it ; and the remainder, if, as usual, there be one, marked ;
then take ihis remainder in the compasses, and in like manner find how often it
* Tlicse methods of construction, almost unknown in tliis country, were first collected
togc'.lier nnd publislicJ by the editor in Leybounrs Mathematical Repositor)- a few years ago.
VOL. I. D d
402
GEOMETRY.
is contained in the last of the integral parts of the first arc, with again some
remainder ; find in like manner how often this last remainder is contained in the
fonner • and so on continually till the remainder becomes too small to be taken
and applied as a measure. By this means we obtain a series of quotients, or frac-
tional parts, one of another, which being properly reduced into one fraction,
give the ratio of the first arc to the semicircle, or of the proposed angle to two
right angles, or 180°, and consequently that angle itself nearly in degrees and
minutes.
Thus, suppose the angle BAG be proposed
to be measured. Produce BA out towards /;
and from the centre A describe the semicircle
ahcf, in which ab is the measure of the pro-
posed angle. Take ab in the compasses, and
apply it four times on the semicircle as at
b, c, d, and e; then take the remainder /e,
and apply it back upon ed, which is but once, viz. at g; again take the remainder
gd, and apply it five times on ge, as at h, i, k, I, and mj lastly, take the remainder
me, and it is contained just two times in ml. Hence the series of quotients is
4, 1, 5, 2; consequently the fourth or last arc em is ^ the third ml or gd, and
therefore the third arc gd is ^, or -^ of the second arc ef; therefore again this
second arc ff is — or \\ of the first arc ab ; and consequently this first arc ab
is — or il of the whole semicircle af. But ^ of 180° are 377 degrees, or
37° 8' 5 nearly, which therefore is the measure of the angle sought. When
the operation is. carefully performed, this angle may be obtained within two or
three minutes of the truth.
In fact, the series of fractions forms a continued fraction. Thus, in the
example above, the continued fraction, and its reduction, will be as follows : —
1 1 J._] _l__i__lj.
4 + 1 + 5i ~" 4 + Iff ~4iJ ~ 63'
the quotients being the successive denominators, and 1 always for each nu-
merator.
Scholium.
This method is due to De Lagny, and a corresponding process has been ap-
plied by Adams to finding what portion any given line is of another given line,
Tlie problems are so precisely 'alike, that any specific detail is altogether un-
necessary. Mr. Sankey has also applied the metall'c cycloid to the measure-
ment of circular arcs : but of course its determinations are only rough approxi-
mations, though their being obtained with great facility, and near enough for
most practical purposes, is a recommendation to its familiar usage.
PROBLEM LI.
To find the diameter of any solid sphere, as a ball or shell.
From any point P on the surface of the
given sphere, and with any convenient radius
(about the estimated chord of 60° will gene-
rally be l)est for accuracy) describe a circle
ABC, and in it take any three points (nearly
equidistant by estimation) A, B, C ; and on
I'KACriUAL. UEUAlIiTRY IJV THE FIELD. 403
paper describe a triangle abc whose sides are equal to those of ABC : about abc
describe a circle abed, and draw its diameter bd : with centres b, d, and radii
equal to the linear distance of P from A, B, C, describe arcs intersecting in /:
and lastly, about hfd describe the circle bfd. Its diameter will be equal to that
of the given sphere.
For let PO be drawn perpendicular to the plane of the circle ABC meeting it
in O : then, being produced to Q, it will be a diameter of the sphere. If also
BD be a diameter of the circle ABC, it is equal to the circle described about
abc, since circles described about equal triangles are equal.
Again, if PD, PB, be joined, the three sides of the triangle DPB are equal to
the three sides of the triangle bfd, each to each : and hence the circles about
them are equal, and their diameters also equal. But PQ is the diameter of the
sphere, and hence of the circle DPQ, the plane of which passes through PQ :
and hence, again, the diameter of the circle bfd which is equal to that of the circle
DPQ is also equal to the diameter of the given sphere.
Scholium.
This problem, which has often been proposed as a new one, owing to its not
being inserted in books which are generally consulted, is yet as old, at least, as
the first century b.c. : for it is found in the Spherics of Theodosius, and con-
structed almost exactly in the same way as above.
PRACTICAL GEOMETRY IN THE FIELD.
The absence of instruments in cases of exigence renders it of much importance
to be possessed of means of determining approximately the positions of certain
inaccessible points with regard to others that are accessible, by having recourse
only to lineal measurements made in accessible places. A few of the more useful
problems of this nature are annexed ; but the want of sufficient space prevents
the insertion of a greater variety. ♦
All the solutions here given are effected by means of staves set up at particular
and specified stations, together with the use- of the chain or other lineal measure ;
as the determinations are practically made with greater certainty by this than by
any of the means usually employed. So far, however, as other modes are con-
cerned, there are solutions of the main part of these problems, though not speci-
fied in reference to this use, to be found amongst the problems in Practical
Geometry already given.
PROBLEM I.
To continue a straight line on the ground, the two determining points. A, B, of
which are given, there being no visual obstacles intervening.
Fix upright staves* at A and B, and walk as nearly as
you can judge in the required prolongation to any point
C' ; which we will suppose to be a little to the right or X"
left of the actual prolongation of AB. This will be known
* It is important in all these problems tha^ the staves be placed as nearly peifectly vertical to
the horizon as possible.
Dd2
40\> PRACTICAL GEOMETRY IN THE FIELD.
by the visual lines C'A, C'B, not coinciding. If the left side of the staves be
visible, we are then on the left of the line AB, and if the right, we are on the
right of AB. Move slowly towards the line, till the visual lines from A and B
coincide, as at C; then C will be in the prolongation of AB. If we wish to
place a mark as at II between two points A, B, in lineation, v,'e must first lineate
to C beyond one of them ; and then by C, B, lineate H by a subsequent, and
similar, process.
Scholium.
It will often in the following problems be necessary to find the intersection of
two lineations ; and though the process is very simple, it may be well to explain
it in such a way as to be effected with the least possible trouble.
Case I. When the lineations AB, CD, do not intersect within the figure
ABCD, bounded by lines joining each of the stations to the adjacent ones.
Let the point H be taken in lineation of AB
beyond the probable point of intersection of
AB, CD ; then if the observer walk from H
towards B in lineation, till the points C, D,
also appear in lineation, to E, the point E will
be that required.
Case II. When the intersection G of AC, BD, are required. Set up two
auxiliary marks, K and F, in lineation of AC and BD respectively; then l)y
means of B and F let the observer walk in lineation of DB, till he arrives in
lipeation of AC, as indicated by the marks C, K. The point G, where this
occurs is the intersection of AC, BD.
Many obvious facilities may be brought into the operation when there is more
than one observer ; but these solutions are adapted to the most unfavourable
case as to assistance.
PUOBLEM II.
To find the lineation of ttio points B, C, when obstacles intervene G, which render
the points B, C, invisible.
First method. Take any point A without the
line BC. from wliich B and C can be seen ; also
points c and b in them ; find the intersection D
of lib and Cc, and then the intersection of cb
and AD. Calcuhite the di tance bd from the
formula obtained below, and measure that dis-
tance in lineaticm of 6c.- the j-oint d will be in
lineation with BC. ^
Produce AD to meet BC in a.- then (th. 07) we have cE : Eb :: cd : db, or
rE - Eb lEby.cd- db c= cb) : bd. Hence bd = -^^^^j ; and all the
cE — Eb
lines on the right side of the equation are measurable, and hence bd can be com-
puted.
Scholium.
Slu.uM a second point be required for the purpose of continuing the lineation,
we may rcprat llie process with another, or with the same triangle, ABC. The
latter will be the better method, and is thus performe.l :-
KeUmw.g one of the marks c, change the other from b to b' : then find the
PROBLEMS.
405
point D', the point E', and finally the distance b'd' from the above formula with
the new values of tlie several parts of the line cb'.
Second method, without any calculation. Take
any point A, and in the lineations AB, AC, take
any points c and b : find the intersection D of the
lineations B6, Ce ; take any point G in AB, and
find the intersection E of Gb, AD, and the inter-
section F, of cE, AC : then the intersection H of
be, FG is in lineation with B and C.
For, produce AD to meet BC in a: then (Ih. 97) the lines cb and GF divide
BC in prolongation in the same ratio that BC is divided in a. Hence they cut
BC in the same point, and each other in lineation with B and C*.
PROBLEM III.
Through a given point B to lineale towards the invisible intersection H of two
yiven lineations FG, be. (Fig. prob. 2, second method.)
Take any point A and mark F, b, in any lineation through A, and the points
G, c, in AB : find the intersection E of Gb, Fc, and D the intersection of AE,
B6 ; then cD, Ab intersect in a point C, which is in the same line with the given
jjoint B and the invisible point H.
Tliis depends on the same principle as the last, and is proved in the same way.
PROBLEM IV.
To find the length of a line AP, inaccessible at one extremity
First method, when one end of the line is accessible. Take
any convenient station B on the ground in lineation with
A, P, and a station R out of that line; prolong BR to any
convenient point C: then marking the point Q where the lines
RP, AC, intersect, we shall find AP by the equation
QARCAB '
QA.RC — QCRB"
For, {th. 95) PA : PB :: QA.RC : QC.RB; or div°.
PA : PB - PA : : QA.RC : QC.RB - QA.RC,
■whence the theorem follows.
Second Method. Take any two points, B, C, lineating
with A, and any point F in AP : find the intersection E
of BF, CP ; the intersection D of BP, CF; and the in-
tersection G of DE, AP : and measure GF, FA. Then
AF.AG
PA =
AF-FG"
• On the same principle we may resolve the following problem, which at first sight may
appear altogether impossible.
The direct wns of two lineg rchose point of concourse is invigilile, and tlie directions of two others
in like circuni.'.iances, are <jiven : to find a point in lineation of the two invidUe points of concourse.
Let Ac, 6D, be two lines whose point of concourse B is invisible, and cD, Aft, two others
whose conconrse, C, is also invisible : to find a point d in lineation of the invisible points B, C.
Produce the four lines backwards till they form the q\iadrilateral Acdli; draw the diagonals
meeting in E, and find Ld by the equation of the text : then d is in lineation of B, C.
Tiiis problem is one of frequent occurrence in practical lineation, on account of visual
obstacles occurring to prevent B and C from being seen from any point in lineation with them.
406 PRACTICAL GEOMETRY IN THE FIELD.
For ith 95) PA : PG :: AF : FG, or PA - PG : PA :: AF - FG : AF,
AP-PG .^ _ FA_AG
or finally, PA = ^^p^^FG ~ AF-FG"
One or other of these methods, where the relative positions of the fundamental
points, from their entirely arbitrary character, admit of indefinite variation, will
apply to all cases whatever.
PROBLEM V.
Through an accessible point Q to draw a line parallel to the accessible line AB.
Take any three convenient equidistant points A, P, B, in
the given line AB ; take any point C in the lineation AQ,
and find the intersection D of CP, BQ ; and then the inter-
section R of BC, AD : the lineation QR will be parallel to
the lineation AB.
For, (th. 96) AQ : QC : : AP.BR : PB.RC, and AP = PB (Jy constr.) : hence
AQ : QC : : BR : RC, and {tk. 82) the Hne QR is parallel to AB.
When the line AB is inaccessible, the solution will be eflfected hy prob. 9.
PROBLEM VI.
Tlirough a given point A to lineate parallel to two given parallel lines
QR and ab.
Through A lineate AC, cutting the
parallels in Q and a, and from any point
C in it lineate Cb, cutting the same two
parallels in R and b .- find the intersec-
tion d of the lines iQ, aR; then the in-
tersection D of Cd, RA ; and lastly the
intersection B of QD, Cb. The line AB
will be parallel to each of the lines QR, ab.
For by a converse course of reasoning to that employed in the last problem, it
may be shown that CDd would bisect ab and QR ; and by similar reasoning to
the last, that AB is parallel to either QR or ab, and hence to both of them.
Scholia.
Tliis process is often applicable under circumstances where the last cannot be
applied, arising from the intervention of practical obstacles. For instance, the
lines might not be traceable so as to get the parallel through A to the line ab ;
whilst it might be jiossible (by the preceding problem) to trace a parallel to ab
through Q, and thence by this to also trace a parallel to ab. This latter part,
too, where two parallels already exist, is attended with some practical conve-
nience, of which the preceding is destitute; as it requires not even the use of the
chain : and if Q, R, b, are remarkable points, we may even solve the problem if
they be inaccessible.
Two parallel lineations may be traced upon the ground by the following con-
siderations : —
1. The lineations of a very distant olijcct will be sensibly parallel.
2. Tlie shadows of two upright staves taken at the same time are parallel ; as
are likewise the lineations of any star taken at the same time.
PROBLEMS.
407
PROBLEM VII.
To divide a given line AB into two equal parts without measurement.
{Fig. prob. 5.)
LiNEATE any parallel QR to AB, and from the extremities A, B, of the given
line, the lineations AC, BC, to any point C, without or between the parallels :
find the intersection D of AR, BQ : then CD, being produced if necessary, will
bisect AB in P.
This rests on reasoning the converse of that employed in the demonstration of
prob. 5.
PROBLEM VIII.
To cut off any part successively of a given lineation AB without the use of measures,
supposing a line MN already drawn parallel to AB.
(1). Take any point P, and find M, N, the
intersections of PB, PA, with the lineation
MN ; and mark the point L of intersection
of AM, BN : then PL will cut off AC, one
half the lineation AB, or AC = i AB.
This is founded on the reasoning of
prob. 5.
(2). Find K the intersection of AM, CN;
then PK will cut oflF AD = J AB.
For BC : CD :: BA : AD; or
AB — AC : AC — AD : : AB : AD ; whence AD = ^ AB.
(3.) Find I the intersection of AM, DN ; then PI will cut oflF AE = i AB.
For, as before, we have CD : DE : : CA : AE ; and hence
CA — AD : DA — AE : : CA : AE, or AE = i AB.
(4.) Find H the intersection of AM, NE; then PH will cut off AF = J AB.
For again, as in the former cases, DE : EF : : DA : AF; and hence
DA — AE : AE - EF : : EA : AF, or AF = ^ AB.
We may thus proceed successively to finding the n"" part of AB.
The demonstrations here given apply equally to prob. 7,fflh method, of the
Practical Geometry : the two problems being in fact identical.*
* The following metbods of constructing the corresponding problem in the Practical Geometry
■were accidentally omitted in their proper place : and are inserted here under the belief that they
■will be found quite as simple as any of the constructions there given. See p. 379, prob. 8.
To divide a given line AB into any number of equal parts.
Fifth method. Draw any line AC, and take n equal portions AD,
DE, .... GH; draw the line HZ through B, and from centre A describe
the circle HK cutting HZ in K ; set off AL, LM, ... PK, each equal to
AD, and which will be n in number; lastly, join DL, EM, .... GP cut-
ting AB in Q, R, .... T : then AB will be divided into n equal parts.
Sirth mciJuxl. Through A and B (the reader can sketch the figure)
draw parallels AK, BC, and in AK take n parts each equal to BC, viz.
AD, DE, .... and draw CD, CE, ... cutting AB in d, e, ... : then Bd,
Be, ... will be respectively the half, the third, etc. parts of AB.
408
PRACTICAL GEOMETRY IN THE FIELD.
PROBLEM IX.
Through a given accessible point A to Uneate parallel to a line, of which any tico
points B, C, are visible, but not accessible.
First method. Take any point D in lineation of A, C, and u ^ ^
E any other convenient point : through A lineate AF pa-
rallel to the line (a part at least of) which is accessible,
meeting ED in F : and through F lineate FG parallel to V" /yj
EB, intersecting BD in G: then GA will be parallel to BC. i:\.'^y"T>
For by similar triangles CED, AFD, and BED, GFD, we
have CE : AF : : ED : DF : : BE : FG, or CE : EB : : AF : FG ; and hence
the triangles FGA, EBC, are similar, and the angles EGA, EBC, equal. Also
the parts FGD, EBD, of these are equal: and hence AGD, CBD, are equal,
and AG is parallel to BC *.
Second method. Take any points D, i,- b r
E, in AB, AC, and find P the intersec-
tion of EC, DB; and the intersection F
of AP, ED : in AF take FG equal to
FA, and in FD take FH equal to FE :
find the intersection K of DB, GH, and
l)roduce GH, to M, till K.M is equal to
KG : then AM will be parallel to BC.
For, since AF, FE, are equal to GF,
FH, and the angles EFA, HFG, are equal, AE is also equal to GH; and the
remaining angles of the two triangles are equal, viz. : FGH to FAE, and FHG
to AEF. Also, since in the triangles LFA, KFG, the angles LFA, FAL, are
equal to the angles KFG, FGK, each to each, and the sides AF, FG, adjacent to
the equal angles are equal; therefore, the sides LF, FK, are equal, and the line
LK is bisected in F.
Again, produce DE to meet BC in N, and join BE : then {th. 97) the line
DE is harmonically divided in F and N : and the line LK which cuts the three
liarmonical sectors BE, BF, BD, in L, F, K, and is bisected in F, is parallel to
the fourth sector NBC {conv of th. 99.)
Lastly, since FK bisects the sides AG, GM, of the triangle GAM, it is parallel
to AM ; and hence AM is parallel to BC.
PROBLEM X.
To bisect an accessible angle BAC.
Take any two points B, D, in AB, and the distances AE,
AC, equal to AD, AB, respectively: the intersection F of BE,
IH', will be in the lineation which bisects the angle BAC.
For. BA, AE, are equal to CA, AD, (constr.,) and the
nngle at A common : hence the angles AEF, ADF, are
• qual ; and likewise the two DBF, ECF. But since ADF,
AEF. are equal, the angles FDB, FEC, are equal : and
hence, in the two triangles CEF, BDF, there are two angles
' ll^*^' »'i!'' I- *" "''"^" ^'"" '"'" *''*'""' "'■J'''-"" "' ^' ^''^ '" l'n"''on with B and C, then the
•wllcl. AF, FCi.n.ay be dmwn, scntibly accurate, bv the remark in tlie note on prob. 6,
ft:
P
PROBLEMS.
409
of the one equal to two angles of the other, and the sides adjacent to the equal
angles equal ; therefore DF is equal to FE.
Again, the sides DA, AF, are equal to EA, AF, and the bases DF, FE, also
equal ; therefore, the angles DAF, EAF, are also equal, or BAG is bisected by
the lineation AF.
Scholium.
Should any obstacle interpose to prevent our seeing A from F, we may take
two other points G and H equidistant from A, and find the intersection K of
CG, BH : then KF will be the bisecting lineation.
PROBLEM XI.
To bisect an inaccessible angle BAC.
Take any points B, C, in the sides of the angle or their
lineations, and join BC ; bisect the accessible angles BCF,
CBG ; then H, their point of intersection, will be in the
lineation bisecting the angle BAC.
If any obstacle prevent A being visible at H, bisect the
angles ACB, ABC, by lineations meeting in K : then HK is
the bisecting lineation *.
For K and H are the centres of the inscribed and escribed
circles, and hence both HA and KA bisect the angle BAC
{th. 99, schol. 6.)
PROBLEM Xir.
To draw a perpendicular from a point A in a given lineation BC.
Take any point B in BC, and lineate BD in any
direction through B : make BD equal to BA, and AC
and DE each equal to AD : in CE take CG, GF, each
equal to AB : then AF will be the perpendicular
required.
For, join GA. Then, since AC is equal to DE,
and AB, BD, parts of them are equal, the remaining parts EB, BC, are also
equal. Hence the angles ABD, EBC, being equal, and AB : BD : : EB : BC,
the triangles are similar, and the angles ECB, BAD, equal.
Again, in the triangles AGC, ABD, we have GC, CA, equal to BA, AD, and
the angle GCA equal to the angle BAD : hence AG is equal to BD ; and hence,
again, the triangle AGC is isosceles, having AG equal to GC. Whence G is the
centre of a semicircle passing through F, A, C ; and the angle FAC, in it, is a
right angle.
PROBLEM XIII.
To lineate perpendicularly to a given lineation BC from a point A withotU it ;
all being accessible.
First method. Take any three equidistant points
B, D, C, in the given lineation, and measure the dis-
tances AC, CB, BA : then the distance of the foot P
of the perpendicular AP from D is found from the
• When the point K would fall too near to A, to render the operation by means of it prac-
ticable, we may either re])cat the first part of the process with new points C, B', instead of C, B :
or we mav employ prob. 2 for finding the lineation of a point to the inWsible intersection of
the given lines AB, AC.
410
PRACTICAL GEOMETRY IN THE FIELD.
(CA + AB) (CA - AB) , . , , . ^ ^ . .
equation DP = ^jVq , which being set off gives the point
D. Tliis is evident from th. 35, Plane Geometry, p. 311.
Second method. Take from any point D in the
line BC, four equal lines DB, DE, DF, DG, the
two former in the line BC, and the two latter in any
directions whatever: find the intersections K, H, of
BG, EF, and BF, EG : lineate HK cutting BC in
Q, and in it take QL equal to QH : lineate HA cut-
ting BC in M : and in ML take MN equal to MA.
The lineation AN is perpendicular to BC.
For since DB, DE, DF, DG. are all equal, the points B, E, F, G, are in the
circumference of a circle, of which D is the centre and BE the diameter : hence
BG and EF are perpendicular to the sides of the triangle HBE, and therefore,
also, HK is perpendicular to the base BE (/A. 97, schol. 6.)
Again, since HQ is equal to QL, and MQ common to the two right-angled
triangles MQH, MQL, the side HM is equal to LM ; and (cotistr.) MA is equal
to MN. Hence AM : MN : : HM : ML, and AN parallel to HL ; that is,
AN perpendicular to the lineation BC.
PROBLEM XIV.
To lineate perpendicularly to an accessible lineation BC from an inaccessible point
A without it.
Find by either of the preceding constructions the per-
pendicular BH to the hneation BC, and make BE equal
to BH : find D the intersection of AH, BC; G the in-
tersection of AE, BC ; and, lastly, the intersection F of
DE, HG : then FA will be perpendicular to BC.
For, the right-angled triangles DBF, DBH, have
their sides HB, BD, equal to EB, BD, and therefore
HD, DE, are equal, as are also the angles HDB, EDB.
Again, the triangles GHD, GED, have the sides GD, DH, equal to GD, DE,
and their included angles equal ; hence the angles DHG, DEG, are also equal.
Again, since the triangles DHF, DEA, have the sides HD, DE, equal, the
angles DHF, DEA, equal, and the angle HDE common, the sides DA, DF, are
equal. Lastly, since AD : DF : : HD : DE, AF is parallel to HE, and there-
fore perpendicular to BC.
PROBLEM XV.
From an accessible point A to find a line perpendicular to an inaccessible line, two
points B, C, of which are visible.
First solution. Form the inaccessible points B, C, lineate the
p«r|H«ndiculars to the lines AB, AC, {prob. 14,) to intersect in
G: lh«-n Ml will he perpendicular to BC.
The proof of this consists in showing, that if two lines drawn
from the angles of a triangle be perj)endicular to the opposite
BKles the line from the third angle through the point of intersection will be per-
pendicular to the third side (conr. of th. 97, schol. 2.)
PROBLEMS. 41 1
Second solution. Through the point A draw AH parallel to the inaccessible
line BC (prob. 9), and from A in AH draw AD perpendicular to it {prob. 13).
llien AD will be perpendicular to BC.
PROBLEM XVr.
To find the length of a Uneation BC which is inaccessible at both extremities.
{See fig. prob. 9, second solution.)
First method. Take any point A, and proceed as in the second solution of
prob. 9, except the final one of finding M : find Q the intersection of KL, AC ;
and R, S, the intersections of BF with EC, CA : lastly, measure FS, FR, and
^T- 1 x^/-. QF.SR
QF: then BC = -^^--^.
For, by the reasoning of the second solution of prob. 3, we have at once
BS : SF : : RS : SF — FR; and by the similar triangles, BSC, ESQ, we have
BS : SF : : BC : FQ ; and hence, also, SF — FR : RS : : FQ : BC, which
gives the value stated above.
Second method. Let HK be the inaccessible line :
take any point A: and in the lineations AH andAK.any
points C and B : find G the intersection of CK, BH,
and D, the intersection of CB, AG : find E the inter-
section of KD, AH, and F that of HD, AK : and,
finally, measure the sides of the triangle ABC, and the
parts AE, AF. The distance HK is found from the
following calculation.
The lines AH, AK, are harmonically divided in their several points of section :
and hence, reasoning as before, we have, putting a, b, c, for the sides of the
triangle ABC, and putting m and n for the specified quotients : —
.„ AB.AF , ,„ AC.AE ,
^^ = aF=FB = "*'' ^"'^ ^^ = AE-EC = "*'
cos CAB = ~ °^ t^f "^ ^> and HK = a/AH- - 2AH.AK cos A + AK=;
which, upon substitution, gives the following formula for HK,
HK ^ ^/m (jn — n) c^ ^ mna^ + n (n — m) P.
Scholium.
1. Though the calculation of the latter method of solution is longer than that
of the preceding, yet as its field-work is shorter and less liable to error, it seems
to deserve the preference.
2. We may also, generally, take AC equal to AB, in which case the formula
becomes simplified, viz. HK = \/ (m — n)- c^ + mna^.
3. If by any means we can obtain a moveable equilateral triangle (as by join-
ing three equal rods or staves) we may move along AH, with one side of it
lineating with AH, till the other lineates with K. In this case, a, b, c, are all
equal ; and we have HK := a ^/m"^ — mn + n^ = a ^/(m — n)* -)- mn.
4. A better method, perhaps, where it can be applied than either of the pre-
ceding is annexed.
4lJ
PRACTICAL GEOMETRY IN THE FIELD.
Third method. Let QR be the inaccessible line, and B
any convenient point: in the hnealions BQ, BR, take
BA, AE, BC, CF, equal to one another : find the inter-
sections G of AF, CE ; D of BG, EF ; O of AD, QC ;
and P of CI), AR : and, finally, measure the sides of the
tiian 'le DOP. Then calculate QR from the equation
1 ORADl
^ ~ OD.DP
For, by similar triangles ODC, CQB, and DAP, ABR, DO : DC : : BC : QB,
DA : DP :: BR : BA; and since DA = DC = BC = BA, we have also,
DO : DP : : BR : QB ; hence the triangles DOP, BRQ, are similar.
Again, by the simdar triangles ODP, RBQ, we have OD : RB : : OP : RQ
:: OD.DP : RB.DP; and DA : DP : : BR : AD, or RB.DP = DA^ hence
OP.AD^^
we have OD.DP : AD- : : OP : RQ =
OD.DP
PIICBLEM XVII.
To show the use of the equality of the angles of incidence and refection in the
det(.rm' nation of altitudes of trees, buildings, etc.
When- a ray of light falls upon a smooth or polished plane, as of quiescent
water, or quicksilver, or a mirror, it is so reflected that the angle which it
makes with the said plane after reflection is equal to the angle which it
made before reflection. Of these angles, the first is usually called the angle of
incidence, the other the angle of reflection. Sometimes the angles made re-
spectively with the perpendicular to the plane of reflection receive those names.
In either case, however, the practical application is the same.
Suppose AB in the anne.xed figure to
be a tower, whose altitude is required,
and which stands on a horizontal plane
AD. At a convenient point C place a
vessel of water or mercury ; and when y'
the surface is quite smooth and quies- /'
cent, recede a httle way from it in the /
continuation of the line AC, until with £
your eye as at E you see the top of the d?;>~ ,/' ^,^
building, or the point whose height you DC A ~
wish to ascertain, reflected from the sur-
face. Then, by the similar triangles CDE, CAB, we have CD : DE : : CA : AB.
Thus, if CD be C feet, DE = 5\, AC = GO, then will AB = 55 fe.% the height
required.
Scholium.
A convtidcra]>!e number of problems analogous to those which are solved in
this chapter might easily Ije given : but as space could not be afibrded for treat-
ing them with adt(|u,ite exi)ansion, this course has been confined to those of the
Rreatfst iinixTtaiice, and of the most frequent occurrence in the field. Some of
tliein. wiih the aid of the sextant, will, however, be resumed in the second
vol urn.-.
APPLICATION OF ALGEBRA
GEOMETRY.
Algebra is essentially a numerical system ; and in every case to which
we apply it, we must be able to express the relations between the objects under
consideration numerically. In the application of algebra to geometry, this is
effected by considering every magnitude concerned in the investigation as con-
taining some number of times another magnitude of the same species, which
other magnitude may be called the standard-unit of the system. Thus, a foot, a
yard, a mile, are each some number of times the length of an inch taken as the
standard-unit; a pole or an acre are some number of times the square foot or
the square yard taken as standard-units ; the mass of the earth is some number
of times the magnitude of a cubic foot, a cubic yard, or a cubic mile, taken
respectively as the standard-units. The same mode of estimation may be applied
to angular space, where any angles may be considered as the repetitions, each a
certain number of times, of any assigned angle taken as the standard-unit of
angular measure. The present chapter will not, however, include any problem
relating to angles, as that subject properly belongs to the next treatise, Tiu-
GoxoMETRV, immediately following it.
It will be obvious from this statement, that the standard-unit for each species
of magnitude must be of the same species as the magnitude which it measures ;
and that, though arbitrary in the outset as to its own magnitude, it must be
kept constantly the same for all the quantities concerned in the same problem.
In all expressions involving general symbols, this is presumed in the notation ;
but in the actual reduction to numbers, j)articular care must be taken to reduce
all the numbers concerned to the same denomination in the final result. {See
prob. \, p. 415 ,)
It greatly conduces to convenience, though it is not absolutely essential to the
nature of the system, to take as the standard-unit of surface, the square de-
scribed upon the linear unit, and as the unit of volume, the cube whose edge is
the linear unit. In practice, therefore, the units of surface and volume are ulti-
mately referred to the linear unit.
In general symbols it is not necessary to specify the actual magnitude of the
unit, as whatever magnitude we conceive it to be, the quantities measured by it
are expressed in numbers which have among';t themselves the same ratios, and
the final result is given in terms of the same unit. The symbols a, b, c, d . . . .
X, y, z, then only express the number of times which the raagn.tudes they de-
signate contain the standard-unit by which they are severally measured.
Geometry is not conversant with magnitude only : it also treats oi form and
position. The first of these implies the consideration of angles, and hence does
not fall under our present discussion Position may be determined several
ways, most of which include a consideration of angular magnitude ; but there is
one mode which may be considered essentially linear ; and it enters extensively
into every view which can be taken of the app'ication of algebra to geometry.
414 APPLICATION OF ALGEBRA
Suppose we had given the expression a + b — c ,«.
_]- J ^ e — y, where each of these letters expresses I
a certain number of standard-units of length. We -i" ? c 'qa^ -n t.-t
have first to draw the indefinite line X'X, and to take I
the point O in it as the origin from which we set off _^>
the lines in question. Let, also, the positive values
tend towards X, then the negative ones, whose effect is to diminish the sum of
those already set off, will be drawn in the contrary direction, or towards X'. So
far our selection is perfectly arbitrary, (except previous data shall have fixed it,)
both as to the position of the line X'X, the point O from which we are to mea-
sure, and the side of 0 which shall be estimated as the direction of positive
measurement. The standard-unit, too, (under the same exception,) is also
arbitrary.
Having determined upon all these conditions, there is no longer any thing
arbitrary, or assumable at pleasure, in the problem ; and the problem itself, as a
geometrical one, is not merely to find the length of the line, but the position of
the final point resulting from the following operation, with respect to the
origin O.
From O set off OA = 4- a, towards X pos. side.
A AB = + 6, X
B BC = — c, X' neg. . . .
C CD = + rf, X pos. . . .
D DE = + e, X
E EF = - /, X' neg. . . .
Then OF is the length of the line to be constructed, and F falling (in the case
here supposed) on the negative side of O, shows that the negative values exceed
the positive by as many standard-units as there are in OF.
It is thus that position combines itself with the simple consideration of magni-
tude in the application of algebra to geometry ; and there are very few cases in
which it cannot be distinctly traced, though in many of them our attention may
not be specifically called to it by the forms of the results.
If we draw a hne YY' perpendicular to XX' through the origin O, we may
clearly in the same manner construct any given expression in the same way upon
this Hne as we did upon X'X.*
In the next place, if we wish to express the po-
sition of a point with respect to two given lines at
right angles to each other; it is sufficient if we
can express its distance from each of them, and
likewise on which side of each of them it lies. This
will be done if we attend to the observations already
made respecting -|- and — as signs of position to
the right or left of O, and above or below O.
Taking, then, as before, OX, OY, for the lines which express positive direc-
tions from O, and OX', OY', for those which express negative directions, let the
distance of the point P from the line YY' be + a, and the distance of P from
X'X be + b. Then the values + a, + b will express the point being at P, ;
Pf
B Pi
1
t 1
0 -« Ia
n
1 -
t 1
p»
In fixing the position of these lines X'X and YY', if we commence from no preceding
condili.>n», it i» most usual, ihoujrli rendered so from custom rather than from motive, to draw
thcni hoiuoi.tal an.l vcrlital, and to take the positive values to the right, and upwards from O,
u wc have hm done.
— a, -\- b, that it is at Pj ; — a, — b, that it is at Pj ; and + a, — b, that it is
at P,.
The lines X'X, YY', are called axes of co-ordinates, and the two distances
from thera are called the co-ordinates of the point. As, however, the suhject of
co-ordinates as a system will not be entered upon till we come to the second
volume, this brief notice of the nature of the notation for position will be suffi-
cient in the present chapter.
In respect of notation, the early letters of the alphabet in this section are
mostly used to designate known or given magnitudes; and the latter, unknown
ones. Of the unknowns, lines drawn horizontally are denoted by x, and those
vertically by y ; as, for instance, the unknown base and perpendicular of a right-
angled triangle. When the sides of a triangle are given, they are denoted by
a, b, c, the side a being opposite to the angle A, and so on of the others. Some-
times the initial letter of a name is used, asp for a given perpendicular, and *for
semiperimeter, or h [a + b -{- c) ; but these, though matters of usual practice,
need not be dwelt on here.
The conditions of a problem, /orma//y given, are seldom so many as to furnish
as many equations as there are unknowns ; but there are always as many geo-
metrically-demonstrated properties implied (and which must be taken from geo-
metry), as will complete that number of equations if it be properly proposed.
All the results may be constructed were it an object to do so ; but as this is
never required, it is unnecessary to say anything on the subject here. Most of
the older works on this subject contain instructions on this head, but the con-
struction is now a mere object of curiosity, and therefore not worthy to arrest
the student's progress in the present stage of his studies.
PROBLEMS.
Prob. I. In an equilateral triangle, having given the lengths of the three per-
pendiculars drawn from a certain point within it to the three sides, to determine the
sides.
Let ABC be the equilateral triangle, and DE, DF, DG,
the perpendiculars from the point D upon the sides respec-
tively. Denote these perpendiculars by a, b, c, in order, and
the side of the triangle ABC by 2x. Then if the perpen-
dicular CH be drawn, CH = ^AC — AW = ^4x^ - a^
= x-v/3.
Now (/A. 81. cor. 2) we have triangle ADB = JAB . GD
= h ex. Similarly, triangle BDC = ^ax, triangle CDA =:
ibx',and triangle ACB = ^AB . CH = a^V3. Also BDC -|- CDA + ADB =
ABC : that is, in symbols x' V^ = (a + b + c) x, or x = - — -— — -^ which
is half the side of the triangle sought.
Suppose a = 3 ft, 6 = 9 in, and c = 2 yds 1 ft 6 in. Then, in ft, we have
3 + 1 + 7* 45 . , 15
^= — 73— ==4-75'°'"''°y*^^'^=Wl'
Cor. From the resulting equation we have x^3 = a + b + c ; and again,
CH = x^/3. Hence CH = a + 6 + c, or the whole perpendicular CH is equal
to the three smaller perpendiculars from D upon the sides, wherever the point
D is taken within the triangle. Had the point D been taken without the triangle,
the perpendicular upon the side which subtends the angle wthin which the point
416
APPLICATION OF ALGEBRA
lies would have become negative. Tims, had it lain without the triangle, but
between the sides AB, AC, produced, then CH = DF + DG — DE.
Prob, II. a maypole was broken by the wind, and its top struck the ground
twenty feet from the base, and being repaired was broken a second time Jive feet
lower, and its top struck the ground ten feet farther from the base. What was the
height of the maypole F
Let AB be the unbroken maypole, C and H the points in
which it was successively broken, and D and F the cor-
responding points at which the top B struck the ground. 1
Then CAD and HAF are right-angled triangles.
Put BC = CD = a-, CA = y. AD = a, and AF = b, and J"
CH = c. Then AB = x + y, BH = HF = a; 4- c, and ^<^ T
HA = y — c. Then in the triangles CDA, FHA, we have ^ i>^ i
J/2 _|_ a2 __ jj2^ j^jj(] (y — p-j2 _J. J2 -— (j; ^ p-j2_
Expand the second and subtract the first equation from it, and we have, finally,
b'^ — ri'^
a; + y = — = 50 feet, the height required.
Prob. III. A statue eighty feet high stands on a pedestal fifty feet high, and
to a spectator on the horizontal plane they subte7id equal angles ; required the dis-
tance of the observer from the base, the height cf the eye being five feet.
Let AB = a the height of the pedestal ;
BC = b the heigl:t of the statue ;
DE := c the height of the eye from the ground ; and
D.\= EF ^ X, the distance of the observer from
the base.
Tlien EC-' = EF= + FC- = x- -\- {a + b — cf, and
EA- = EF -f ED2 = x- + C-.
But {th. 83) EC2 : EA2 : : CB- : BA- ; in which inserting the preceding
values of these lines, we have, after easy reductions,
=±y
a (a — c)- -f b {a- — c-)
b — a
the double sign merely indicating that x may be measured either way, from A
towards D, or from D- towards A.
Inserting the given values of a, b, c, we have x = +5 v/399 = 99-874922 feet.
Proh. IV. Given the three sides, a, b, c, of a triangle, to find :—
(. 1 ). The three per pendicidars from the angles upon the opposite sides :
(2). The area of the triangle :
(3). The rhdius of the circumscribing circle ;
(4). The radius ff the inscribed circle ;
(5). The radii of the escribed circles.
Ix.t ABC be the triangle, and Bli a perpendicular from B
to the opi^o^te side AC. I^t a, b, c, denote the sides oppo-
site to the angles A, B. C, respectivelv. and /)„ p.^, p^, the
perpendiculars from A, B, C, viz. AD, BE, CF; A the area
of the triangle ; R the radius of the circumscribing circU ;
r that of the inscribed circle ; and r„ r„ r^, tie radii of the
three escribed circles which touch the sides a, b, c, e.xter-
naliy.
TU UEUMiiTKlC. 417
(1). The perpendiculars. By th. 37 we have BC^ = BA" + AC^ — 2CA.AE
,„ BA2 + AC2 — Bt;» _ a2 + 62 + c» , .
°'"^^= 2CA = 26 • ^«^'"'
BE« = BA2 - AE^ = c= - 1""'"^^"^ I'
= 46V— ^— a2 + 62 + c^p
462
_ [2bc + (—a^+P+c-)] {2bc—i—a^+¥+c^
~ ' 462
_ {- a' + (b + c)^ {a^-(b-vY]
(a + 6 + c)(-
462
-a + b + c)(a-
-6+c)(a+6-
-c)
a+b+c -
2
462
-a + 6 + c a-
2
-b + c
2
a + 6-
2
-c
i6-^
„^a + 64-c . —a + b + c a + b + c
Put — ^-— - — - = s : then ■ = — ^ a = « — a;
2*22 '
,.., , a — 6 + c , . a + b — c
and similarly, = « — 6, and = s — c.
Making these substitutions, we have the equation at once converted into
TJT?2 2 jy (S — a) (S — 6) (S — C) A ■ I t
BE^ = p^^ = — ^ ^ i and similarly,
4 "
5 s (* — a) (s — 6) (s — c) J , s (s — a) (s — b) (s — c)
i'.' = ^—. , and p/ = ^ .
(2). The area. This by th. 81, cor. 2, is 4 AC.BE = A. But putting in
this the value of BE, obtained in the last case, we have,
.5 , , , * (s — a) (s — 6) (s — c) , . , , V / .
a2 = i 62^ ^j^ fp =sis — a)(s — b){s — c).
(3). The radius of the circumscribing circle. Let the diameter BG be drawn :
then (th. 63, Geom.)
^ _ AB.BC ^a6c
~ 2BE ~ ^s(s'^l0(s—^)Js — c)
(4). The radius of the inscribed circle. Let ABC be
the triangle, O the centre of the inscribed circle, BD
the perpendicular from B upon AC, and E, F, G, the
points of contact. Then, th. 81, cor. 2, we have trian.
BOA = i re; trian. BOC = ^ ra; trian. COA = ^ rb;
and trian." ABC = h bp,. But BOA + AOC + COB
= ABC ; or in symbols.
r{a + b + c)= bp„ and hence r = *{-^^-= A' - «) (^ - b) js — c).
(5.) TTie radii of the escribed circles. Let O' be the centre of the escribed
circle, touching AC exteriorly, and E', F', G', the points of contact. Then
ABC = ABO' + CBO' - AO'C ; or in symbols, as before,
r» zz r ^^ ~~ "li^ ^' ; and in a similar manner we obtain
V s — b
r.= A(.-6U.-c) ^^^ Is^-ans-b)
V s — a ^V s — c
VOL. I. EC
418
APPLICATION OF ALGEBRA
Cor. 1. By multiplying the values of the four radii of the circles of
contact together, we have r r, r^ r3*= « {s — a) (s — b) (s — c) = A-, a
remarkable theorem discovered by LhuilUer.
Cor. 2. Also, taking their reciprocals, we get = - H \- - .
r Ti r, Tg
Cor. 3. By multiplying together the values of R and r, we get 2Rr =
" ,- — ; and similarly, 2Rri = , , , — , and so on.
a + b + c' ^ —a + b + c
Many other curious properties may be seen in the supplement to the " Ladies'
Diary" for 1835 and 1836, by the editor of this work.
Prob. V. Given the radius r of a circle to find the sides of the inscribed and
circumscribed pentagons and decagons.
I. The inscribed pentagon. Let ADBCE be
the pentagon inscribed in the circle, (prob. 40,
or Euc. iv. 11), and let O be the centre of the
circumscribing circle. Join AB, AC, and draw
AF perpendicular to BC. Then by known pro-
pertiesAC is bisected in G, and the line A F passes
through the centre 0 of the circle ; and likewise,
AB : BC : : BC : AB — BC.
Put BC = -Ix, or BG = x, BA = r/, and BF
= r. Then the preceding proportion becomes
y : 2x :: 2x I y — 2x, and hence we have,
y =(1 + V5) X.
Now we have OG = A^r^ — x^, AG = \^y'- — ar, and hence AG = AO -f
OG, gives A^y- — oc' = r + \/r"- — x'-, or squaring y- — 2r- = 2r \/f- — x^,
and squaring again, y* — 4r-y- -f 4r-x- = 0. Also we have seen that y =
(1 + i/5)x, which, inserted and the expression reduced, gives for the side of the
inscribed pentagon, 2j; = i r '^lO — 2 %/5.
Cor. I. y =zx{l + ^,/5) = 5 r a/10 -f 2^y5 ; a value which will be useful in
the other parts of the problem, though no part of the quaesita in any one.
Cor. 2. OG = ^/f'-—lc-= N/r- — |(5 — ^^5) r" = i r (I + ^/5).
2. TTie inscribed decagon. Join BF : then, since AF bisects the line BC at
right angles, it bisects the arc BFC in F, and hence BF is the side of the in-
scribed decagon.
But ABF being a right angle, since it is inscribed in a semicircle, we have
BP = FA- — AB^ : or adopting the preceding notation, and denoting as above
BF by z, we have it converted into
^* = 4r* — y^ = 4;^ — A (10 + 2 V 5)r- = § (3 — V 5)r* ; or extracting
^ = i Kn'' 5 — 1), as the length of the side of the inscribed decagon.
3. The circumscribing pentagon. The inscribed and circumscribing pentagons
bcmg regular, are similar figures, and their sides are as the perpendiculars from
the centre upon the sides. That is, if PK be a side of the circumscribing pen-
tajfon, OG : OB : : BC : PK ; which put into symbols, gives the value.
PK = ^^Bl^t; _ r.Ars/ 10 — 2^5
OG
i r n/ i (3 -h V 5)
= 2r \/ 5 — 2 ^ 5.
4. The circumscribing decagon. Let QR be one of the sides ; and draw OH
pfr|>endicular to BF, which it bisects in H. Also, by similar triangles, ABF,
OHF, we have 0H = iAB = 4y = ir V 10 + 2 ^5. Also, as in the last
case, OH : OF : : BF : QR, and hence,
OF^]BF_^^^ir0y5-J^_ A -2^/5
'^"~ OH -Jr^/lO+~2v^5-^ ^ 5 •
Further problems for exercise.
6. In a right-angled triangle, having given the base (3), and the diflference
between the hypotlienuse and perpendicular (1) ; to find these sides.
7. In a right-angled triangle, having given the hypothenuse (5), and the dif-
ference between the base and perpendicular (1); to determine these sides.
8. Having given the area, or measure of the space, of a rectangle, inscribed
in a given triangle ; to determine the sides of the rectangle.
9. In a triangle, having given the ratio of the two sides, together with the
segments of the base, made by a perpendicular from the vertical angle; to
determine the sides of the triangle,
10. In a triangle, having given the base, the sum of the other two sides, and
the length of a line drawn from the vertical angle to the middle of the base; to
find the sides of the triangle.
11. In a triangle, having given the two sides about the vertical angle, with the
line bisecting that angle, and terminating in the base ; to find the base.
12. To determine a right-angled triangle; having given the lengths of the
lines drawn from the acute angles, to the middle of the opposite sides.
13. To determine a right-angled triangle; having given the perimeter, and
the radius of its inscribed circle,
14. To determine a triangle; having given the base, the perpendicular, and
the ratio of the two sides.
15. To determine a right angled triangle; having given the h)'pothenuse, and
the side of the inscribed square.
16. To determine the radii of three equal circles, described in a given circle, to
touch each other and also the circumference of the given circle.
17- In a right-angled triangle, having given the perimeter, or sum of all the
sides, and the perpendicular let fall from the right angle on the hypothenuse ; to
determine the sides of the triangle.
18. To determine a right-angled triangle; having given the hypothenuse, and
the difference of two lines drawn from the acute angles to the centre of the in-
scribed circle.
19. To determine a right-angled triangle ; having given the side of the in-
scribed square, and the radius of the inscribed circle.
20. To determine a right-angled triangle; having given the hypothenuse, and
the radius of the inscribed circle.
21. To divide a line of ten inches in extreme and mean ratio.
22. To add to it a segment, such that the rectangle under the whole hne thus
increased, and the part of it increased, shall be to the square X)f the difference
of the two segments into which the line is now divided, as 12 : 5.
23. A circle AFB and a point D are given, the distance DC of the point from
the centre C of the circle being b, and the radius r ; to draw through D a line
EF, terminated both ways by the circle in E and F, so that its length shall be 2a.
24. Given the adjacent sides a, b, and the diagonal, c, of a parallelogram, to
find the other diagonal.
25. Given the chords of two arcs of a given circle, to find the chord of their
sum, and the chord of their difference.
£ e 2
420 APPLICATION OF ALGEBRA TO GEOMETRY.
26. To divide the base, a, of a triangle into two segments proportional to the
sides, b, c.
2". Two circles being given which touch one another inwardly; to describe
a third circle that shall touch both the former, and also the right line passing
throuuh their centres.
28. Having given the lengths of two chords which intersect at right angles,
and the distance of their point of intersection from the centre; to find the di-
ameter of the circle.
29. Given, to determine the area of the triangle, and the lengths of its sides,
the three perpendiculars from the angles upon the opposite sides.
30. If a, b, c, d, taken in order, be the sides of a quadrilateral inscribed in a
circle, and x, the diagonal, joining the extremities of the sides, a, d, or b, c, and
y the other diagonal ; it is required to show that
xy = ac -{- bd, X : y : : ad -\- be : ab + cd,
and thence to find x and y, together with the area, and the radius of the circle
circumscribing the quadrilateral.
31. Supposing the town A to be 30 miles from B, B 25 miles from C, and C
20 miles from A ; if a house be erected to be equally distant from each of those
towns, what will its distance from them be ? Ans. 15'11856 miles.
32. The chords of three arches completing a semicircle being given, 3, 4, and
5 respectively; required the diameter. Ans. 8'03581.
Theorems for Exercise.
1. Show that r, r„ + r^r3 + r^ r, = s^, and r, + r^ + Tg = 4 R + r.
1.1,1 1,1,1 , R r, r„ r,
+ ^ = :r + T + -; and -- =
Pi Pi P3 »•. r^ r, 2r p^p^Pa
3. Establish the relation i R r = = .
PiPi + P2P3 + PzPi
4. Prove that o, = -^^x^^-, and - + - _ i = i.
^■2 + r, p^ p^ p^ r.,
5. If a, b, c, d, be the sides of a quadrilateral, and a be parallel to c, and h the
distance of these parallels : then,
4A- (a — cy- = 2{a — c)' (.b- + cP) — (a — c)^ — (^2 _ (py
6. If a circle be inscribed in an equilateral triangle, and a triangle in this circle,
and again a circle in the triangle, and so on, ad inf. ; prove that r = r, + r, +
Tj + . . . where r, r,, r., . . are the radii of the successive circles.
7. Show that the side of a hexagon inscribed in a circle is mean proportional
between the sum and diflerence of the sides of the inscribed pentagon and
decagon.
8. If /„ /j, /j, be the lines drawn to bisect the angles of a triangle, whose
sides are a,b,c; show that ^-' ^' ^' = „^ (« + b + c) A
abc (a -}-b){b + c) (c + a)'
16 (/,< + V + V) = 9(a* + b* + c'), and
16 (/,' /,- + /,2 l^- + l^^ l^^) = 9 (a» 62 ^ 62 p2 _,. p2 ^2 )
9. If m^, m.j, wjj, be the lines drawn from the angles to bisect the opposite
■ides: then m,- + m/ + m/ = s ^a^ 4. fts ^ ^2 ^
10. If fl, 6, c, be the distances of a point from three of the rngles A, B, C, of
a wjtiare, B beiiiK the nearest or most distant angle, show that the area of the
Kjuare is i J„' + c' + n/T-!;-' («^ - b' + c') - {a' - c^f^ ; and that if the
four distances from a joint to the angles of a rectangle, be u, b, c, d, taken in
order then will a' -f t^ = 6' ^ di.
PLANE TRIGONOMETRY.
Plane Trigonometry is that particular portion of the application of algebra
to geometry, in which the angles of a triangle occiir either in the data, the
quaesita, or the intermediate equations, or in all of these.
It has been established {Euc. vi. 33, or Geom. th. 94), that in the same or
equal circles, angles at the centre are to one another in the same ratio as their
subtending arcs. It hence follows, that in any algebraic expression of the
angle in terms of a given standard angular unit, we may substitute the expres-
sion of the subtending arc of a circle in terms of a corresponding standard
circular unit. If we take the right angle as the angular unit, the quadrant
which subtends the right angle will be the corresponding circular unit : if four
right angles be the angular unit, the whole circumference will be the cor-
responding circular unit ; and so on for any other corresponding angular and
circular units. Whatever, therefore, is stated or proved with respect to the arc,
may be stated or proved with respect to the angle ; and the converse. We may,
therefore, indifferently employ the terms arc or avgle to express the inclination
of two lines to one another. In the higher departments of mathematical science,
however, it is not a matter of indifference which term we use, or which thing
we reason from ; whether it be viewed in reference to the conception of the
things, or the manner in which we reason upon them. The advantage is in all
cases in favour of the arc, but especially in all general investigations respecting
the unlimited extension of magnitude, which, without departing from the prin-
ciple of continuity, we may suppose the arc or angle to admit of. In the mere
solution of problems, however, relating to triangles or polygons, it is usual to
employ the word angle in our phraseology, notwithstanding that all our investi-
gations turn upon its measure, the arc.
Since the relation established (p. 367), already referred to, is true, whatever
may be the radius of the circle, it is obvious that our conclusions will not be
affected by changing the radius, so long as all the subtending arcs involved in
the same investigation are measured by circles having the same radius. In
other words, the radius-unit of the circle is altogether independent of the linear
unit by which the linear magnitudes of the same problem are measured. It is
found to be most generally convenient to take the radius, in calculation, as 1, and
in any case where a different radius is required in connecting by an equation any
parts of two different circles, to multiply the part of the circle belonging to the
unit radius by the new radius, to obtain the corresponding part of the new
circle. This is founded on th. 93, or Euc. xji. 2 ; or rather, on an obvious
corollary from it.
422 PLANE TRIGONOMETRY.
Facility of investigation and calculation, however, is better consulted by the use
of certain lines drawn in uniform and specific ways with relation to the subtend-
ing arcs, than by the direct introduction of the arcs themselves into the equation.
These lines are called the trigonometrical functions of the arc or angle: and they
have been calculated by methods which will be hereafter explained for very
minute divisions of the quadrant, and arranged in tables for convenience of use.
In all cases these lines are tabulated for the radius unity. These lines or func-
tions we shall now proceed to describe, and then to lay down their fundamental
geometrical relations.
I. DEFINITIONS AND NOTATION.
Let .\0B be an angle, and with the unit of length, as the radius, describe a
circle, commencing at A, and proceeding towards and through B (in all the
figures annexed passing upwards from the line OA) till it again arrives at A.
Through O draw the diameter CL perpendicular to OA, and produce OA to
meet the circle at K. From B, the intersection of AB with the circle, draw BD,
BE, perpendicular to OA, OC, (produced if necessary,) and from A and C tiraw
tangents to the circle meeting OB in F and G.
E-! o'
1. Then denoting the arc AB by a, BC is called the complement of a, and
BK the svpplement of a.
2. The semicircle ACK is usually denoted by tt, and consequently AC the
quadrant by ir. Whence \ir — a, and tt — a, are respectively the complement
and supplement of a.
3. The perpendicular BD is called the sine of o, and written for brevity
sin a.
4. The jwrtion of the tangent AF cut off by the other line OB of the angle
AOB, is called the tangent of a, and w ritten tan a.
5. The portion OF of the line OB intercepted between O and the tangent is
called the secant of a, and written sec a.
6. The distance between the sine and the arc, estimated on OA, viz. AD, is
called the versed sine of a, and written vers a.
7. The line BE ^ OD is the sine of the complement of a, or sin (W — a)
It ii called for brevity of expression, the cosine of o, and is written cos a.
8. The lines CG, OG, CE, are in like manner the tangent, secant, and versed
nine of the complement CB of the arc AB ; and are hence respectively called,
for the same reason as before, the cotangent, cosecant, and coversed sine of a ;
and written cot a, cosec n, and covers a.
'.». NN hen wc have occasion to calculate numerically the arcs concerned as mea-
uurcs of the angles, the unit by which they are commonly estimated is a degree,
of which 3G0 make up the entire circumference. Hence a semicircle contains
180, and a quadrant 90 degrees. These degrees are subdivided into minutes, of
PLANE TRIGONOMETRY.
4QS
which 60 make a degree, and each minute again into 60 seconds. All subdivi-
sions of seconds are expressed as decimals of a second. The notation for degrees,
minutes, and seconds, is, the marks °, ', ", written in the place usually assigned
in algebra to the indices of powers. Thus 37° 15' 18"-279 signifies 37 degrees,
15 minutes, and 18279 seconds*.
10. For the general investigation of theorems, angles are denoted by A, B, C,
. . . . or a, 6, c . . . . , or a, /3, y, . . . . : but in the investigation of the solutions
of general problems, not specifically relating to mere triangles, the data are ex-
pressed by a, j8, y, . . . . and the unknowns by later letters of the Greek alphabet,
as 0, 6, X, (0.
1 1 . When the investigation relates to a triangle, whether of a theorem or pro-
blem, the angles are usually denoted by A, B, C, and the sides respectively
opposite them by a, b, c.
II. RELATIONS AMONGST THE TRIGONOMETRICAL FUNCTIONS OF
A SINGLE ARC.
In the several figure's we have by right-angled triangles (Enc. i. 47, or th. 34),
and similar triangles (Euc. vi. 2, or th. 72).
Geometrical properties.
DB2 + OD" = 0B2
0A2 + AP = 0F2
0C2 + CG2 = 0G2
OA.DB
OD : DB
OE : EB
OD :0B
OE : OF
AF : AO :
OA : AF =
OC : CG =
OA : OF
OD
OC.EB
OE
OA.OB
OC : OG =
OC : CG =
OD
F0.0£
EO
AO.OC
AF
AD = AO - OD
CE = OC - OE
Corresponding equations.
sin- a + cos^ a =■ 1
1 -j- tan" a = sec^ a
1 + cot^ a = cosec^ a . .,
sin a
tan a =
cot a =
cos q.
cos a
sm a
1
cos a
_1
sin a
1
tan a
vers a = I — cos a . .
covers a= 1 — sin o .
sec a ^
cosec a =
cot a =
. (1)
. (2)
.(3)
. (4)
. (5)
. (6)
.(7)
.(8)
. (9)
(10)
By the established principle respecting the signification of + and — as signs
of geometrical position, (see p. 414,) we shall be able to determine in which of
the four successive quadrants the pomt B is situated, from the value of any two
of the functions being given with their proper signs prefixed.
• The French division of the quadrant is into 100 parts or grades, each grade into 100
centimes, and so on, each part heing siiccessively divided into 100 of the next. This has the
advantage of rendering tlie work entirely decimal : but its disadvantage is, that it requires all
preceding tables and astronomical observations to be transformed to suit this division — a work
of too much labour to encourage even a hope of it ever being performed. It would also render
a great number of excellent instniments almost useless. Tlie conversion in any given case is
easily performed by the equation E =: -fg F, or F =: '9 E ; where E signifies any given number
of English degrees, and F the corresponding of French grades : or again, in this form, n degrees
= (n -f- ^n) grades, and n grades := (n — -^n) degrees.
424
PLANE TRIGONOMETRY.
In order to give unity to the entire system, let all the functions of arcs less
than a quadrant, that is, of AB in the first figure, be taken positive. Tlien the
sine is positive whilst it lies above the line KA, and consequently negative when
below. The cosine is positive whilst it lies to the right of CL, and negative
when to the left. The tangent is positive when it is above the point A, and
negative when below. The cotangent is positive when to the right of C, and
negative when to the left. The secant is positive when the line proceeds from O
through B to meet the tangent, and negative when it proceeds from B through
O to meet the tangent ; or in other words, it is positive when B hes between O
and F, and negative when B is on the opposite side of O from F. The cosecant
is positive when B is between O and G, and negative when on the opposite of O
from G. The versed sine and coversed sine are -f in all the quadrants, since
they do not change the direction in which they are estimated. All these relations
are consistent with those in the preceding section, and the sine and cosine being
given with their proper signs, the same conclusions may be obtained by means
of those relations.
It also appears from this, that if we take the arc negatively, or proceeding from
A in a contrarj' direction round the circle from that assumed as positive, the
several functions will be the same as those of 2;r — a. Hence
sin (—a) = sin (2ir—a) = — sin a cos (—a) = COS (2^ — a) := + COS a
tan ( — a) =: tan (27r — a) = — tan a cot ( — a) = cot (2jr — a) = — cot a
sec ( — a) =z sec (2ir—a) = + sec o, cosec (—a) = cosec (27r — a) =: — cosec a
vers (—a) = vers(2n-— a) = + vers a covers (—a) =covers (2:r— a) = +covers a
The fifth and subsequent quadrants, being as to position only repetitions of
the first four, the signs of position of the several repetitions will be identical
with those of the first four quadrants taken in order. It will therefore be neces-
sary to tabulate only the first four.
sm a
1st, 5th quadts. +
2nd. 4th +
3rd, 6th —
4th, 8th . . —
cos a
tan a
cot a
sec a
cosec a
+
+
+
+
+
—
—
—
—
+
—
+
+
—
+
—
—
+
—
vers a and covers a
+
+
+
+
^For the particular values of these functions at the particular values of a, viz.
0 , 90°, 180°, 270°, 360°, we have, by a simple observation of the several
figures, taking into account their known geometrical properties : —
when a z=o°
when a
=z
90°
when a
—
180°
when a
270°
sin a = 0
sin a
=
1
sin a
=
0
sin a
—
— 1
cos a = 1
cos a
=
0
cos a
=
— 1
cos a
-_
0
tan a = 0
cot a = i
0
tan a
cot a
=
1
6
0
tan a
cot a
=
0
1
0
tan a
cot a
=
1
"o
0
sec a = 1
1
sec a
:^
— 1
1
0
— 1
cosec a =
0
sec a
cosec a
0
1
cosec a
=
1
0
sec a
cosec a
^-
vers a = 0
vers a
=
1
vers a
=
2
versin a
_.
1
covers a = l
covers a
=
0
covers a
=
I
covers a
=
2
and the values at the end of the fourth, fifth, and successive quadrants, repe-
titions of these severally.
We might also here find the values for some specific angles, as of 30°, 45°.
<>o ; l>ut a more convenient place will occur hereafter. {Chap, vi.)
PLANE TRIGONOMETRY.
4.?5
III. ON FUNCTIONS OF TWO ARCS.
To find the sine and cosine of the sum and difference of two arcs, in terms of the
sines and cosines of the arcs themselves.
Let AB, BC, be two arcs ; draw BD, CE,
perpendicular to OA, and CF perpendicular to
OB. Then BD, CF, are the sines, and OD,
OF, are the cosines of the arcs AB, BC; and
CE is the sine, and, OE the cosine of their sum
in fig. 1, and of their difference in fig. 2. Draw '• '^ "" -^ » uu *• -»•
FH perpendicular to AO, and FG perpendicular to CE.
In both figures, the triangles OFH, OBD, CGF, are similar; and therefore
HF(=GE) : FO :: DB : BO; or GE.OB = OF.BD
GC : CF :: DO : OB; or CG.OB =CF.OD
HO : OF :: DO : OB; or HO.OB = FO.OD
GF(= HE) : FC :: DB : BO; or HE.OB = CF.BD
By adding and subtracting the first pair of these, we get the following forms,
CE . OB = OF . BD + CF . OD in fig. 1.
CE . OB = OF . BD — CF . OD in fig. 2.
Similarly, subtracting and adding the last pair, we get the following forms,
OE . OB = FO . OD — CF . BD in/<7. 1.
OE . OB = FO . OD + CF . BD in fig. 2.
Put AB = a, and BC = /3 : then, writing the results just obtained in terms
of a and /3, and recollecting that the radius OB = 1, we have at once
sin (o + /3) = sin a cos j3 + cos a sin /3 (1)
cos (a + jS) = cos a COS (3 — sin a sin /3 (2)
sin (a — /3) = sin a cos /3 — cos a sin |3 (3)
cos (a — /3") = cos a cos /3 + sin a sin /3 (4)
Combining these by addition and subtraction we obtain the following values :
sin (a + /3) + sin (a — /3) = 2 sin a cos /3 (5)
sin (a + /3) — sin (a — /3) = 2 cos a sin /3 (6)
cos (a + /3) + cos (a — y3) = 2 cos a cos ^ (7)
cos (a — /3) — cos (a + /3) = 2 sin a sin /3 (8)
Again, a = i (a + /:■!) + Ka - /3) and /3 = i (a + /3) — i (a — ^). Hence
applying formula (1, 3,) to these values of a, /3, we get,
sin a = sin i (a -f /3) cos 2 (a — /3) + cos ^ (a + /3) sin ^ (a — /3) . ,
sin /3 = sin i (a + /3) cos ^ (a — /3) — cos 4 (a + /3) sin i (a — /3) .
By adding and subtracting these two equations, there results at once,
sin a + sin /3 = 2 sin n« + /3j cos J (a — /3) (1 1)
sin a — sin /3 = 2 cos 4 (a + /3) sin i (a — /3) (12)
Also applying (2, 4,) to the values of a, /3, as before,
cos a = cos I (a + /3) cos ^ (a — /3) — sin i (a + /3) sin i (a — /3) .
cos /3 = cos § (a -I- j3) cos i (a — /3) + sin i (a + /3) sin J (a — /3) .
And by addition and subtraction of these we obtain,
cos a + cos /3 = 2 cos § (a + /3) cos 4 (a — /3) (15)
cos j3 — cos o = 2 sin i (a + /3) sin 4 (a — /3) (16)
Dividing the results already obtained, as indicated on the left side of the page,
ne get successively.
(9)
(10)
(13)
(14)
426 PLANE TRIGONOMETRY.
(11) sin g + sin /3 _ sin * (a + ft) cos j (a — 1^) _ tan ^ (a + ^ .
(12) — sin a - sin /3 ~ cos a (a + ft) sin 4 (a — /3) tan J (a - /3) ' ' * "
(15) cos a+cosft _ cos ^ (a + /3) cos ^ (a — /?) _ cot ^ (a + /3) ,^gx
(ley ~ cos /3 — cos a ~" sin 4 (a + /3) sin l{a — ft) cot ^ {a — ft)
(11) sin a + sin ft _ sin 4 (« + /3) cos * (« — /3) ^ ^^^^ , ^^ ^ y^^ _ _ ^jg^
(15) ~ cos a + cos /3 cos A (a + /3) cos i (a — p)
(12) _ sin g - sin ^ ^ cos 4 (oj^^) jin^ 4 («^ /3) ^ ^„, 4 (^ -t- /^) .... (20)
(16) cos /3 — cos a sin i (g + ^) sin * (a — /3)
Atrain, since these equations are true for all values of o and ft, they are true
when any given values of a and ft are doubled ; that is, when 2a and 2ft are
wTitten for a and /3. This substitution beinsf made in the equations we find,
sin 2g + sin 2/3 = 2 sin (g + jS) cos (a — p) (21)
sin 2a — sin 2/3 = 2 cos (g + /3) sin (g — /3) (22)
cos 2g + cos 2/3 = 2 cos (g + ;3) cos (a — /3) (23)
cos 2/3 — cos 2g = 2 sin {a + ft) sm {a — ft) (24)
(21) _ sin 2a + sin 2/3_tan (a + ft) ,^^.
(22) ~ sin 2a — sin 2/3 ~ tan (a — /3) "
(23) cos 2a + cos 2/3 ^ / , on i. /■ o\ rnc\
y=COs2a-COS2g=^"^^° + ^^'^"^^"-^^ ^''^
(_21)^Mn2a + sin2/3^^^
(23) cos 2a + cos 2,3
(22)^ sin 2a -sin 2/3^^^
(24) cos 2/3 — cos 2a ' '^ ^ ^
The formulae here given enable us, in connection with those of chap. II., to
obtain any of the other functions of the sum and difference of two arcs, as well
as some useful formulae relating to two arcs which have specified relations. Also
by supposing one of the arcs to be itself the sura or difference of two arcs, some
useful results are obtained ; though from the complexity of the formulae this
inquiry has never been carried to great extent. Indeed, transformation is the
main object of all these researches ; inasmuch as they enable us to put the ex-
pression under different forms, better adapted to facilitate investigations and
calculations, than they appear originally in the proposition. The diversity of
forms that may be given would evidently be verj' great ; but in an elementary
work only those which are of frequent utility can possibly find a place.
♦ /■ I ^ sin (g + /3) sin a cos ft + cos a sin ft tan a + tan ft , ^
tan (a + /3) = — -L^ = !— nr — =z — =r — = — (29)
— cos (a + ft) cos a cos /3 + sm g sin ft 1 -(- tan a tan ft
cot (g + |3) — ^"^ ^" ± /^) _ cos g cos /3 + sin a sm ft _ cot a cot /3 + 1
~ sin (g -h /3) sin a cos /3 + cos a cos ft cot ft + cot a
tan g + tan /3 = ^'" « ^os /3 + cos a sin ft ^ ^i" ^« + /^ ,3.^
cos n cos /3 cos a cos /3 ^
cota + cot/3 = c»^ « sin /3 + sing cos /3 ^ sin (g + ft)
— sm ojin ft — sin a sin /3 ^ ^
tana_4. tan ft _ sin a cos /3 + cos a sin ft _ sin (g 4! ft)
tan g + tan /3 ~ sin a cos ft + cos a sin ft ~ *" sin {a + ft) ^^^^
cot g + cot /3 _ cos a sin ft + cos /3 sin a _ sin (a ^ ft)
cot a + cot /3 ~ cos a sin .d + cos ft sin a ~ ~ sin (a + ft) ^^*^
Again we have the following very useful transformations,
sm (a + ft) sin (g — /3) = (sin g cos jS + cos a sin ft) (sin g cos /3 — cos g sin ft)
— sin- g cos"' ft — cos2 « sin^ /3 = sin^ g — sin^ ft (35)
cos (g + ft) COS (a — /3) = cos2 g cos= ft - sin^ a sin" /3 = cos= g - siu2/3 (36)
PLANE TRIGONOMETRY. 427
IV. PARTICULAR RELATIONS AMONGST THE ARCS.
1. Let a = «/3. Then we have a + j3 = (ra + 1) jS. Whence
sin (w + 1) /8 = sin n/3 cos j8 + cos ?3/3 sin j3 (1)
cos (w + ]) jS = cos nj3 cos /3 + sin n/3 sin j8 (2)
Adding and subtracting the two forms of (1) we have,
sin (ra + 1) /3 + sin (w — 1) jS = 2 sin n/3 cos j8 (3)
sin (n + 1) i3 — sin (» — 1) /3 = 2 cos nj8 sin j3 (4)
Similarly from equation (2) we get,
cos (» — 1) /3 + cos (n + 1) j8 ^ 2 cos w/3 cos j8 (5)
cos (ra — 1) j8 — cos (71 + 1) /3 = 2 sin n/S sin j8 (6)
2. Let a = (3, orn = 1. Then a -f- /3 = 2a; and we get,
sin 2a = 2 sin a cos a (7)
cos 2a = cos^ a — sin^ a (8)
But (8) may be changed into (1 — sin^ o) — sin^ a or cos^ a — (1 — cos^ a), and
hence,
, -, • 9 t- • o 1 — cos 2a . .
cos 2a =: 1 — 2 sm^ a ; whence sm- a = .... (9)
1 ~i" cos 2a
cos 2a = 2 cos^ a — 1 ; whence cos^ a = • (10)
(9) . sin- a , 2 1 —cos 2a ,,,^
— — gives 5— = tan'' a = :; — ; (11)
(10) ^ cos2 a 1 + cos 2a ^
And the same holds good if for a we write ^a in the three last formulae, giving,
sin2 Aa - ^ ~ ^"^ ° cos2 ia - ^ + ^"^ " tan2 la — ^ ~ ^"^ " (12^
sm^a- - , cos 2«- 2 '^^^ 2"- 1 + cosa '"^^^^
But we may proceed differently, and get another pair of useful forms for tan Ja;
thus, put /3 = ^a, then a — (i = ^a.
sin (a — ^a) = sin Ja = sin a cos ^a — cos a sin ^a : whence,
V • , • . 1 sill « , .
(1 + cos a) sm ^a ^ sm a cos ^a, or tan ha = . — ; (13)
' -^ ^ ' -= 1 _j_ cos a
cos (a — ^«) = cos \a = COS a COS \a + sin a sin \a : whence,
1 — cos a.
(1 — cos a) COS |a = sin \a sin a, or tan \a = ^ (14)
'^ sm a ^ '^
Two other formulae may be thus obtained which are often useful. We have
sin^ a + cos' a = 1, and 2 sin a cos a = sin la. Hence, by addition and sub-
traction,
sin' a + 2 sin a cos a + cos^ a = 1 + sin 2a.
sin" a — 2 sin a cos a + cos' a =: 1 — sin 2a.
Extracting the roots, and adding and subtracting the results,
sin a = i {v'l + sin 2a + -/l — sin 2a} (15)
cos a = i f \/l + sin 2a + -\/l — sin 2o| (16)
Many other formulae respecting double and half arcs are easily obtained ; but as
they are not of frequent use in elementary study, they are left to the student's
choice to pursue or not.
3. Multiple arcs are generally most elegantly expanded by Demoivre's theorem,
hereafter to be given ; but as for small multiples, they frequently occur in early
stages of trigonometry. One method, that of successive deduction, is indicated,
rather by example than precept, but sufficient for the present purpose. It will
be kept in mind that sin 2a and cos 2a have been found in (7, 8).
428 PLANE TRIGONOMETRY.
Then sin 3a = sin (2a + a) = sin 2a cos a + cos 2a sin a
= 2 sin a (1 — sin' a) + sin a (1 — 2 sin' a) = 3 sin a - 4 sin' a.
cos 3a = cos (2 a + a) = cos a COS 2a - sin 2a sin a = — 3 COS a + 4 COs' a.
Similarly sin 4a = sin (3a + a), cos 4a = COS (3a + a), and so on.
V. THE EXPRESSIONS FOR SIN (a ± )3) AND COS (a ± /3) WHEN
a IS SOME WHOLE NUMBER OF QUADRANTS.
Let a = - ; then, sin ^ = 1, and cos 'J = 0 ; and we have,
2 2 2
sin i- — p\ z= sin | cos /3 — cos | sin /3 = + cos /3,
\~ — p] = cos - cos /3 + sin ^ sin /3 = + sin (i,
[2 } 2 2
|- + /sl = sin I cos /3 + cos ^ sin /3 = + cos/3,
cos |- + /^l = '^os I cos /3 — sin | sin /3 = — sin /3.
Let a = TT : then sin ir = 0 and cos ir = — 1 ; and we have,
sin (tt — /3) = sin x cos /3 — cos tt sin /3 = + sin /3
cos (it — /3) = cos TT cos /3 + sin tt sin /3 = — cos /3
sin (tt + /3) = sin tt cos /3 + cos jr sin /3 = — sin /3
cos (ir + /3) = cos TT cos /3 — sin tt sin j3 = — cos /3
Let a = — ; then sin — = — 1, and cos -^ = 0 ; and we have,
2 2 2
sin I — — /? [ = sin — cos /3 — cos -^ sin /3 = — cos j3
{3t _ ) 37r - , . 3;r . . -
/3 > = cos — cos /3 + Sin — sin p =: — sin ;3
sin ] - -|- /3 [ = sin -^ cos /3 + cos — ^ sin /3 = — cos /3
^ 2 J 2 .^
cos
sin
If + ^l
3n- . 33r . , • o
COS — cos /3 — Sin — sin /3 = + sin /3
Let a = 27r : then sin 27r = 0, and cos 27r = 1 ; and we have similarly,
sin (2x — /3) = sin 27r cos jS — cos 2x sin /3 = — sin /3
cos (27r — /3) = cos 2ff cos /3 + sin 27r sin /3 = + cos /3
sin (2jr + /3) ^ sin 27r cos /3 + cos 2-ir sin /3 = + sin /3
cos (2flr + /3) = cos 2;r COS /3 — sin 27r sin /3 = + cos /3
and for all additions of quadrants these values will be repeated in the same
order. The same results might have been inferred by combining the four figures
at p. 422 in one, and reasoning from known geometrical relations.
'Hie tangents and co-tangents might also be inferred from the figure in the
same way ; or they might be obtained from these results by the equation
tan y = , and cot v = . ^. The secant and cosecant also being the re-
cos 7 ' sm -y
ciprocals of the cosine and sine have the same signs as those functions.
Some interesting discussions of the signs connected with these functions may
be found in Professor Young's Mathematical Dissertations, p. 8.
PLANE TRIGONOMETRY. 429
VI. THE VALUES OF THE TRIGONOMETRICAL FUNCTIONS OF
CERTAIN ARCS.
1. To find cos 30°, sin 30°, and tan 30°. If a = 30° we have cos 3a = cos
90° ^ 0. Whence taking the value of cos 3a from p. 428, we have,
4 cos^ 30° — 3 cos 30° = cos 90° = 0. Whence cos= 30° = J, or cos 30 = i ^Z,
and sin^ 30° = 1 — cos^ 30° = 1 — J = i ; or sin 30° = \. Whence also
tan 30°= ''\1%75 = W3.
cos 30° V 3
2. Find sin 45°, cos 45°, and tan 45°. As before, cos 2.45° = cos 90° = 0;
and hence 2 cos^ 45° — 1 = 0; or cos^ 45° = \, and sin^ 45° = 1 — cos^ 45° = § ;
and we have cos 45° = ^ v' 2, sin 45° = ^ v^ 2, and tan 45° = 1.
3. Find sin 60°, cos 60°, tan 60°.
These may be inferred from (1), since the sine, cosine, and tangent of 30°»
are the cosine, sine, and cotangent of 60° ; but for illustration of the method of
proceeding, the investigation is annexed independently of 30°.
sin 3.60° = sin 180° = 0; whence, as before, 3 sin 60° — 4 sin' 60° = 0,
or sin^ 60° = f , cos^ 60° = 1 — f = i ; and sin 60° = ^ v^3, cos 60° = \,
and tan 60° = V 3.
Having obtained the functions of these arcs, (the final e.\pressions for which
are the simplest that occur throughout the quadrant, for any arcs,) we can con-
tinually obtain their halves or doubles : their halves by the resolution of
quadratic equations, and their doubles by squaring certain functions of the sine
and cosines already obtained. We can also obtain expressions containing the
functions of the third part of an arc by the resolution of cubic equations, and so
on to any extent. Two examples are annexed, to find the functions of 15° and
10°, the half and the third parts of 30°.
cos 30° = cos 2.15° = 2 cos- 15° — 1 = ^ -v/ 3 ; hence
cos^ 15° = i (2 + V 3), and sin^ 15° = 1 — cos^ 15° = i (2 — >^ 3).
Hence, extracting the roots, we have the following expressions of value :
cos 1.50= i{Vl-\- N/i},sin \b° = ^{Vi— V^}. tan 15° = 2 — ^/ 3.
Again, for sin 10° we have sin 3.10° = — 4 sin^ 10° + 3 sin 10° : hence
4 sin^lO° — 3 sin 10° + i = 0, and by Cardan's formula, we have,
sin 10° = iv/- h + h V'^^s + i v/- h-h n/"^^.
This expression taking an imaginary form, indicates that all three roots are
real, whilst neither of them can be exhibited in a real form by such a process. The
same circumstance happens universally in obtaining the sine or cosine of an arc,
by supposing it a third part of an arc whose sine or cosine are given, except
when that given sine or cosine is 0. The method of trisection, therefore, is in-
applicable to the finding of useful expressions * for these functions ; but it gives
an opportunity of making a remark which wiW be further expanded in the second
volume.
We have sin 30° = sin 390° = sin 750° = \; and hence the problem which is
virtually put into equations, has in reality three different cases, according as we
suppose these three angles to be trisected. Hence the roots are sin 10°, sin 130°,
and sin 250°, all which are real, and answer to the real roots of the equation
before found. We might, hence, have anticipated this result : and, indeed, the
• However, in all cases the values of the roots can be readily calculated by Homer's method;
and as the same reasoning will apply to every section of a given arc, it is quite clear that we
can always actually compute any function of any given part of an arc or its angle, when we are
in possession of the value of any one of its trigonometrical functions.
450 PLANE TRIGONOMETRY.
double values of the radicals in the solution of the other problems indicate the
same kind of circumstance, viz. two values of the sine, cosine, and tangent
sought ; which, on the same principle, were indications of the sine, cosine, and
tangent of 5 a and tt -\- h a.
The surd values of the sines, and hence of the cosines, and the tangents which
may be obtained from them, are given in the Introduction to Hutton's Tables,
p. xxxix. for every third degree of the quadrant. The deduction of these will
furnish sufficient exercise to the student.
VII. THE CALCULATION OF TRIGONOMETRICAL FUNCTIONS.
These functions can be expressed in a series of positive integer powers of the
arc itself, and the coefficients of the series determined ; and conversely, the
arc can be expressed in a series of positive integer powers of any one of these
functions. These series may be found either by indeterminate coefficients, or by
the differential and integral calculus. The former method, however, is laborious;
and the latter implies a degree of acquirement beyond our present progress.
Hence, we shall adopt a more simple method of proceeding in this place, leaving
the deduction of the series in question for its proper analytical position in the
Course.
The method is founded on the principle, that in very small arcs the sine varies
very nearly as the arc itself. For let a be a minute arc, and /3 one more minute,
by which a is increased. Tlien sin (a + /3) = sin a cos /3 + cos a sin /3 ; and
since a and /3 are minute arcs, cos a and cos ji are very nearly equal to unity.
Hence, taking them actually as unity, we get sin (a + /3) = sin a -f- sin j3, and
the arc, therefore, increases nearly as the sine, when these arcs are very small.
Now, by IV. 15, we have sin a = § {^1 + sin 2a — ^/l — sin 2a}-
Tf .u . n 1 ^ . ^^-> 30° 30° 30° ,
If, then, we put 2a successively equal to 30°, — , -^2 ••• • ^ n ^^^ compute
30° 225
the several sines, we at last arrive at sin — rr = sin — ^. 1' = '000255625, and
2'^ 256
1 . ,, 256 . 30^
hence s^" 1 = ^ si" ^ = -0002908582 nearly.
The cosine, tangent, or any other function of 1' can now be obtained, as
cos r = n/i — sm ^T' = -OgQQQOgsrr, and tan 1' = ?i^ = -0002908882.
cos 1'
whence so far as the first ten decimals, there is no difference between the sine
and tangent of l'.
.\gain, from (IV. 3,) we have sin (n + I ) /3 = 2 sin n/5 cos /3 — sin (n — 1) /3 :
and if we put n = 1, 2, 3, . . . . 1 799, and /3 = 1', we shall be able to calculate
the sines of all angles from 0^ to 30% for every minute of a degree; and, con-
currently, all the other trigonometrical functions of the same arcs.
Ti) calculate those from 30= to 45°, we may use the formula thus obtained :—
sin (30 -f /3) + sin (30°- ,3) = 2 sin 30° cos (3 = cos (3, from which,
Bm (30^ + /3) = cos /3 — sin (30°— /3)
and making /3 successively equal to I', 2', 3', . . . . 899', we shall obtain the sines,
and ll.ence the other functions of the arcs from 30^ to 45° inclusively.
Also, ihe sine of any arc is the cosine of its complement ; and hence, as we
have computed all the complementary functions, we have the direct functions of
all arcs from 45 to 90 , and the functions of the entire quadrant are computed.
1 he functions of arcs greater than 90° are at once obtained from the equations
cAa/7/fr \., p. 423.
PLANE TRIGONOMETRY. 431
VIII. THE CONSTRUCTION AND USE OF THE TABLES OF TRIGO-
NOMETRICAL FUNCTIONS.
1. Since by the preceding method we can calculate the sines to radius 1, of
all the angles from 1' up to 90°, we may suppose them prepared for tabulation ;
and thence also by means of the relations deduced in chapter II. all the other
functions. In Ilutton's tables they are computed to seven decimal places. On
each page are given the values of all the functions, sin, cos, tan, cot, sec, cosec,
versin, and coversin, of all the minutes from p degrees to (/) + 1) degrees inclu-
sive. The number^, if under 45°, is found at the head of the page to the left;
and if 45° or upwards to 90°, at the bottom of the page to the right. The
minutes, if p be less than 45°, are numbered from the top 0', to 60' at the bot-
tom, the numbers being the left column of the page; but if 45° or upwards,
they range from the bottom 0' to the top 60', and constitute the right column of
the page. The name of each column of functions is placed at the top or bottom
asp is less or greater than 45°.
It will also appear that the degrees at the top and the minutes at the left side,
together with the degrees at the bottom and the minutes at the right side, of any
horizontal column, together make 90° ; or in other words, that any given func-
tion of a given arc is the complementary function of the complement of that arc
in the structure of the tables. Thus sin 9° 10' = -1593069 = cos 80° 50' (^ee
page 2S6 of the Tables), and so of the other functions. This is an arrangement
depending on the equation sin a = cos (90 — a), and reduces the table to half
the dimensions it would otherwise require to carry the functions up to 90°.
These natural sines, natural cosines, etc. are always placed on the left page
whenever we open the tables, and headed "natural sines, &c." The differ-
ences between the sines and between the cosines of each two consecutive arcs
differing by 1', are placed in columns and adjacent to them, marked "differ-
ences;" thus sin 9° 10' — sin 9° 9' = -1593069 — -1590197 = 0002872, the
effective figures 2872 of which is found on a line lying horizontally between sin
9° 9' and 9° 10'; or again cos 9° 9'— cos 9° 10' = -9871827 — -9871363 =
•0000464, and the effective figures 464 are put down horizontally.
Again, let a be any number of degrees and minutes; then since covers a = 1
— sin a, we have covers a — covers (a + 1) = (1 — sin a) — [I — sin (a-fl')]
=: sin (a -1- 1') — sin a. The differences between two consecutive coversines is
equal to the difference between the sines of the same angles. The coversines of
the angles are therefore put down on the opposite side of the column of differ-
ences from the sines, the same difference applying to each of the columns. For
the same reason the versines are placed on the opposite side of the column of
differences from the cosines. No other remark remains to be made on the table
of natural functions.
The table of " log sines, &c." on the right hand page is formed by takinpf
the logarithms of the numbers on the opposite page. Thus log sin 9° 10' :=
log 1593069 = 1-2022345, and so of all the rest. However, to avoid the
negative indices in the logarithms, which would create great difficulties in print-
ing and much liability to mistakes in calculation, 10 is added to all the loga-
rithms of the sines, etc. throughout the entire tables. Hence tabular sin a =
10 + log sin a, and hence tab. sin 9° 10' = 10 + 1-2022345 = 9 2022345 ; and
similarly with all the other functions and values of a.
The succession of columns in the two tables is different. In the table of
4S2 PLANE TRIGONOMETRY.
natural functions, the sines and coversines have the same diflPerences ; and for
this reason, the sines and coversines are placed in succession with their differ-
ence-column between them ; and the cosines and versines in the same manner.
In the logarithms of these functions other relations exist, bringing together the
sine and cosecant, the cosine and secant, and the tangent and cotangent.
For cosec a =■—. , and hence log cosec a = — log sin a ; and in a similar
sm o
manner log cosec (a + I') = — log sin (a + 1); hence ^10 -|- log cosec a^
— pO -1- log cosec (a -I- l')| = {10 + log sin (a -f- 1')] — JlO -f- log sin a]
or tab cosec a — tab cosec (^a -\- I') = tab sin (a + O — tab sin a, whence
the sines and cosecants are brought together, with the column of common differ-
ences intervening. The relations sec a = and cot a := , give rise to
" cos a tan a
a corresponding arrangement respecting these functions.
As the sines and tangents at the commencement of the table, and the cosines and
cotangents at the end, vary very rapidly, the tabular functions for every second
of the first two degrees are given at pp. 238 — 267. The structure wiU be evident
on inspection.
2. Having explained the construction of the tables, their usage is next to be
described.
To take out the sines, cosines, etc. or their tabular logarithms to degrees and
minutes not greater than 90" is the immediate and first application of the tables,
and the method is obvious from the construction of them already explained.
When the function is that of an angle greater than 90° we must have recourse
to the results obtained at p. 428, chapter V.
Now sin a = sin(7r — a) and cos a = — cos (tt — a); hence
sin a sin (tt — a) ^ ,
tan a= = — ~ ] ,=: — tan (tt — a)
cos a cos (tt — a)
cos a COS (tt — a)
cot a = '. = — -. ; ^ = — cot (it — a)
sm a sm (- — a)
sec a = = ; ^ = — sec (tt — a)
COS a cos (tt — a)
cosec a = -. = -I — : — ^ 4- cosec (tt — a)
sm a sm (tt — a)
vers a = 1 — cos a — 1 -f cos (tt — a)
coversa =1 — sin a =: 1 — sin (tt — a)
Hence if a be greater than 90' and less than 180°, the trigonometrical func-
tions of TT — a may be substituted for them, subjected to the changes of sign
indicated by the above equations. Thus if a = 96° 10' we have sin 96° 10' =
sin 83 50', cos 96 10' = — cos 83' 50', and so on; all which values, signs ex-
cepted, fall amongst the tabulated numbers.
In the same manner, if the arc be greater than 180° and less than 270% the
corresponding vahies of the functions of o in terms of those of a — tt can be
a-M;;ned ; and so on to any number,/), of complete quadrants above or below the
given value of a. "Whence the functions of any arc, however large, can be ex-
pressed in functions of an arc not greater than 90", and therefore the tables t6
IK)° suffice for all the jiurposes of angular calculation.
The only thing remaining is the calculation of the interpolated values of the
functions for any number of seconds intermediate between two consecutive
minutes in the tables, and conversely the number of seconds corresponding to
the excess of a given function over the less of those for two consecutive minutes
USAGE OF THE TABLES. 433
in the tables. This is effected on the principle, that (except in extreme cases)
the difference of two arcs, being less than 1', the difference of any of their trigo-
nometrical functions is as the difference of the arcs themselves. This is suffi-
ciently apparent as to fact, from the very slight changes of the differences
between each tabular function and that of its consecutive minutes in successive
steps of the increase or decrease of the arc. Thus,
nat sin 9
sinl5°0'= -2588190
sin 15 1 = -2591000
sin 15 2 = -2593810
sin 15 3 = -2596619
A nat sin 9
-0002810
•0002810
-0002809
A^ nat sin 9
•0000000
— 0000001
And the same is true for any other part of the table, e.xcept at its extreme limits,
and for all trigonometrical functions of the arc, both natural and logarithmic.
It will, however, conduce to clearness to show under what circumstances this
law holds good, at the same time that a general mode of proving the truth of the
assertion is pointed out. Now we have, nearly,
sin (0+p'O — sin 0 2cos(6+^ p") s\n ^p'' sinip" sin/j"
sin («+ 60") — sine "" 2 cos (0+| 60") sin ^ 60" "~ smi 60^ ~ sin 60"
except 9 be very small. In this case, therefore, the difference of the sines is
sensibly as the difference of the arcs within the assigned limits ; and in a similar
manner it may be shown for the other natural functions.
Again, for the logarithmic functions, it has already been shown (p. 255) that
for small differences of the numbers, the differences of the logarithms are as the
differences of the numbers. Hence in all the functions, except near the begin-
ning and end of the quadrant, dif for 60" : dif iorp" : : 60" : p". Hence for
corrections generally we have, according as the difference of the arcs or the dif-
ference of the functions is given, respectively,
JT r -/ »" dif for 60" , „ 60" dif for p"
dif for;, = ^, andp" = ^-^ ^^^ g/ •
The differences for 60" are those before spoken of, and are found by actual
subtraction ; but generally tabulated to save the trouble of the subtraction
in each case. When arcs between 0° and 2°, or between 88° and 92°, or 178°
and 182°, etc. are concerned, it will be necessary to use the corresponding tables,
and work for decimals of seconds to one or two places at least.*
It is necessary to bear in mind that the sine, tangent, and secant (under 90°
for which the tables are constructed) increase as the arc increases ; whilst the
* It often happens tliat we have to use one function of an angle where another has been pre-
viously calculated, as, for instance, given tan 6 to find sec 6, or given sin x to fi"*! cos )(. The
direct mode of proceeding is, of course, to find 6 or x to seconds from the given values of tan 0
or sin x, and thence to find the values of sec 6 or cos x as above explained. This, however, is
not necessary ; for the correction of the angle from the tangent, and the correction of the secant
from tlie angle, involve six proportional terms, of which the middle ones are supci-fluous. Thus,
if d be the difference between tan 6 and the next less tabular tangent, and D the difference
for 60" ; and if D, be the tabular difference of the secant on the same horizontal line, and d^ the
corresponding correction of the secant ; and finally, if p" be the seconds con-csponding to the
correction of the tangent : —
D ; rf, : ; 60" ; p", and 60" ; /)"; ; D, ; rf, ; and, hence, D ;(/;", D, ; rf, =: -j— =: correction of
the secant. Tiie same holds with respect to any other functions, and in fact the correction ia
given rather more accurately, as well as with less work, than by finding the intermediate quan-
tity, p".
VOL. I, r f
434 PLANE TRIGOXOMETRY.
cosine, cotangent, and cosecant decrease as the arc increases. This will require
the corrections connected with sin a, tana, sec a, to he added, and those con-
nected with cos a, cotrt, coseca, are to be subtracted, whether arcs or their
functions be sought from the tables.*
For instance, given sin x = 9'8265832, to find cos x- (See note, p. 433 J
Given sin x = 9'8265832 ; and Cp- 353 of the tables, seventh edition).
next less sine = 9-8264910, and the corresponding cosine = 9 8702756
Hence d = 922 correction for cos x = — 754
But D = 1397 and corrected cos X =98702002
andD, = 1143
. , rfD, 922.1143 ^,, V I,- . V . 1
whence correction for cos x = -jt- = ,307 — = ' ^*» ^'bich is to be taken — .
EXAMPLES.
1. Find the tabular cosine of 28° 10' 15".
Here we have (from the tables, p. 325,) cos 28° 10' = 9-9452609, and
15diffor60" 677
ppl5" = ^ = — — ; or pp 15" = — 169
cos 28° 10' 15" = 99452440
2. Find uat tan 212= 15 IS" and tab sin iGg"" 18' 45".
sin 212° 15' 18" sin (180=-(-32° 1 5' 18")
Here tan 212° 15' 18" =
cos 212 15 18 cos (180 +32= 15' 18")
— sin 32° 15' 18''
= tan 32° 15' 18".
— cos32=15'18"
XT , ..o^^-.,^ J ,^„ ISdif. 60" 18.4068
Hence nat tan 32M5' = -6309530, and pp 18 " = = — ^^-—
or pp 18 " = +1220
-6310750 = nat tan 212° 15' 18"
Also sin 169° 18' 45 " = sin (180° — 169' 18' 45") = sin 10° 41' 15".
Hence tab sin 10° 41' = 9"2680647, and pp 15" = ^' ' , or,
00
pp 15" = + 1673
9-2682320 = tab sin 169° 18' 45"
3. Find G from n sin 0 = -1625946, and x from tab cosec x = 10-1653829.
Looking in the tables for the next less nat sin and next less tab cosec we have
given nat sin 0 = -1625946
nat sin 9' 21' = -1624650
., „ 60" , 1296 60". 1296
Also p" = ..^ ^ „ — = -„„ = 27"
■^ dif. 60" 2870
1296
and hence 0 = 9° 21' 27".
Again, given tab cosec x = 10- 1653829
tab cosec 43° 7' = 101652703
60". 1126
Also/'=- j3^g =-50 ;
1126
and hence 0 = 43° 7' 0" — 50" = 43° 6' 10".
4. Find tab cos SO'' 30' 35", tab cos 157° 10' 18", tab cot 196° 10' 18", and
tab cosec 325° 10' 15".
5. Find nat cos 57° 18' 15", nat cot 59° 59' 59", and nat sec 525° 15' 58".
It i* iiiiial to write nsin, n tan, iifcc, e/c. instead of natiiral sine, etc.; and for tab
»in, Ub t*n. ric, ►iinplv f.in, tan, ric. Also to place the work of finding the parts in any vacant
•pwc apart fn)m tlic grniral working fomuila;.
EXPANSION OF SINES AND COSINES. 435
6. (II, p. 34.) Find the values, natural and tabular, of the following expres-
sions:—
(1.) sin 1° 5' 10'' sin 91° 4' 15" sin 196° 10' 18" sin 300° 10' 15".
(2.) tan 18" tan 108° tan 196° tan 271° tan 305° tan 375° tan 400°.
sin (—18°) sin 367° cos 95° cos (—195°) tan 300° 10' 16"
(3.)
(4.)
(5.)
cos 18° cos (—367°) sin (—95°) cos 196° tan (—300° 10' 16)'
sin 270° 10' 16" sin 175° 0' 16" — sin 536° 10' 15" cos 100°
cos 17° 18' 16" — sin (— 100° 15' 16") '
sin 100° 15' 18" sin 375° 18' 16'' — cos 92° 0' 16" cos 325°
cos 100° 15' 16" — cos 460° 15' 52"
IX. TO EXPAND SINJ7 AND COSa; IN TERMS OF X.
Assume sin a? = o + br -\- ex' + dx + . . . .
and cos x =: a^ + 6,a/ ' -|- c,a;'''^ + £?,j? ' + ....
the several indices and coefficients being yet unknown.
1. Since for any given value of x there can be only a single determinate value
of sin X or cos x, these sines can contain no fractional indices. For if any one of
them do contain a fractional index, it indicates multiple values of the term in
which it appears, and therefore also multiple values of the sine and cosine them-
selves; that is, a determinate quantity has several different values, which is
impossible. Hence all the indices of both series are integers.
2. These series can contain no negative indices. For if they can, let them be
d and d,. Now as the expansion is independent of the value of x, it is true for
every value, and hence for x = 0. Now we have seen (p. 424) that sin 0 = 0,
and cos 0=1; whence we shall have in this case,
, /3 yd 6
0 = a + bo + cO -1 — ^ + eO +
0
= a + 0 4- 0 + infinity + 0 +
which is impossible. Hence the series for sin x contains no negative indices.
Nor are there any in the series for cos x, since in this case, also, we should have
1 = a, + 0 -H 0 + infinity + 0 + . . , .
which is, again, impossible.
The indices, therefore, of both series are positive integers.
3. Since sin a? =: a -f- bx + ex'' -f- dx'^ + • ••
— sin a; = — a — bx — ex"" — dx — ... and (/). 424)
sin i—x)= a + b (,— xf + c (— xY + d (— a?)* + . . .
But (424), sin ( — x) = — sin x : hence the two series for these must be
identical : and this can, obviously, only be the case when a = 0 and /3, y, 5, . . .
are odd numbers. Hence we may represent, with all possible generality, the
series by sin x =z hi x + h^ x^ + h,^ x^ + . . . . (a)
Again, cos a? = o, -f- 6a/^' + Cix''^ -f- d^x^^ + . . . .
and cos (— x) = a, + A, (— xf^ + c, (— a?)^» + rf, (— x)^^ + "
But (424), cos ( — a;) = cos x ; and hence the two series just given are iden-
tical. Now this gives a, = 1, and requires that all the indices j3i, yj, 5i, ...
shall be even numbers. Hence representing a, by h^, we may generally write
the series as follows :
cos X = h^° + hx"- + h^x* + .... (6)
Ff 2
4S6 PLANE TRIGONOMETRY.
4, Put X = y + !:■■ then by the theorems (III, 1. 2), p. 425, we have
sinx = siny cosz + cosy sin r, and cos (y + z) = cosy cos 2- — sin y sin z..{c)
But sin X = sin (y + r) = A,(y + z) + ^sCy + •^)' + *5(y + «)*+....
and cos x = cos (y + z) = K{y + z)°+ h.,{y + z)"^ + A,(y + z)^ + ....
Next insert the expansions (a), (6), in the equations (c) : then we shall have
Bmx = (h,y + h.y + ...){Kz'>+ Kz^^-) -\- ihz + hz^ + ...){h^y'' + hy'- + ...) + ...
cosx = (h^'>+ h.j/- + ..) (hoz''+ h,z'- + ...) - {k,y+fi.y4-...) {h,z + h^z^ + ...)+...
Expand the first pair of series by the binomial theorem, and arrange the
results according to powers of y ; and multiply out the factors of the latter pair,
and arrange them in the same manner. Then since the two series for sin x are
but difTerent forms for the same function of y, the coefficients of the like powers
of y must be equal, each to each ; and, in the same manner, the coefficients of
the like powers of y in the series for cos x must be equal each to each. It will
be sufficient for our present purpose, to consider the coefficients of the first
power of y only in the values of sin x and cos x ; and these give the equations
A, + 3A,Z' + 5*52^ + 7M^ + ... = A, + A,^,22 ^ J^^J^^^^ ^ |^^}^^^6 ^ ,._
2Aj2 + 4^^^=* + GA^z^ + Bh^z' + ... = — h^^z — h^li^ — h^h^z^ — h^h^z'^ — ...
Now we have already seen that Jiq ^ 1, and we must find h^. In the expres-
sion for the sine we have = h.-\-h-jC'-\-'h.x^-\- ; and as x decreases, ^^
X X
approximates continually towards 1 as its limit, and hence when
X = 0, we have 1 = A, + Jijd" + hX)^ + ..., or h^ = 1.
Again, these being expansions, the coefficients of the like powers of z are
equal in each of them. Taking those oi z, z-, z^ in succession, we have
1
2hj = — h,^, or h^ = — ^
3^3 = + h^h.,, or A3 = — Y^
Ah^ = — A,A3,orA^ =
5A, = + h,h„ or h = 2.3.4.5
6^6 = — A1A3, or Ag = —
2.3.4.5.6
and so on to any required extent.
2.3.4
Hence, finally, we have the series for the sine and cosine converted into *
x^ , ifi x'^
cos X = I
1.2.3.4.5.6
III the same way we m.iy obtain series for the t.ingent and cotangent :
.>' ,r5
cin .r '' ~ I. •_'.;{ 1" 1 -.'.a 4 ,5 — • • •
"" 1.-/ "^ 1. -2.3.4
anil iiiiiltiplying out and equating the coefficients of the like powers of j-, we obtain,
sin X ■J.r' 2-*.r^
ci;i:; = »»"'- = ^+ T.2j+ij^o + ---'
u- -t , <""s .r 1 2,r •'•'J'^
.Sinularlv, . = rot .r = - ^J^ L
■ >»"' -^ ■r l.-2.:i 1. -2.3.4.5 ~
By reverting these several scries (sec p. -27-2), the arc itself may he found in terms of sinjr,
co»j, tanj, or cot .r: hut as these expressions are of little use in our present stage, and they
can be obtaiiic.l more simply by other processes, they will not be further discussed till the second
volume. Tlie cnlirc iiivcsti^'ation of all scries of this nature, is best effected by means of the
inM/ru/ ndrutus , but as the scries in the te.\t were essential to elenieutaiy trigonometry, it was
deemed adviublc to give investigations of them here by the method of indeterminate coefficients.
FORMULA OF EULER AND DEMOIVRE. 437
X. Euler''s and Demoivre's theorems.
1. To prove Euler's theorem, viz. that
cos e = ^ e'^~' + e-^^/-^ } , and sin 9 = ^- L^ | eV^_ ,-V=rj
By (p. 251) e'=l-f-4. *^J--fl.L ^ + ..
^ ^ 1 ^ 1.2 ^ 1.2.3 ^ 1.2.3.4 ^
In this, substitute successively Oj—\, and — 9^—1 for a: : then we shall
have
/n/~1 ^ 1 + ^^ _ il _ ^V-1 , 0' ,
1 1.2 1.2.3 "*" 1.2.3.4 "T ••••
1 12"^ 1.2.3 "^ 1.2.3.4
Whence, by addition, subtraction, and division by 2 and 2 n/— 1, we shall have
1 ( e^/-l , -ev-1 1,6',^ o
2 I J 1.2 ^ 1.2.3.4
•2^—1 (.
^ev^i_^-ev-i) _ _^_^
^~iV j -" 1:2:3 "^ iXsXs '*^''''-
2. To prove Demoivre's theorem, viz. that for all values ofn,
(cos 9 + V— 1 sin 0)» = cos n0 + V— 1 sin n9.
By addition or subtraction of the two preceding results, we have at once
e^ '^~' = cos 0+ \/— 1 sin 9; and as this is independent of the value of 9,
we also have e = cos nO +,/ — 1 sin n9, whatever n may be. Hence
we have universally
(cos 9 + v^— 1 sin 9Y=\e j = « = cos n9 ± ^— I sin n9.
XI. APPLICATIONS OF THE THEOREMS OF EuLER AND DeMOIVRE.
1. To expand cos n9 and sin n9 in terms of sin 9 and cos 9.
For cos n9 + ,^ — 1 sin n9 := (cos 9 + V—l sin 0)« and
cosn0 — -v/ — 1 sinn0 ^ (cos0 — v' — 1 8in0)n ;
and by addition and subtraction we have
2 cos n9 = (cos 9 + ^/—l sin 9)" + (cos 9 — ^/^ sin 9)'*
2 ^^—[ sin 710 = (cos 0 + V^ sin 0)'» — (cos 0 — V'-^ sin 0)».
Expand the second sides of these by the binomial theorem : then in the former
all the odd powers of ^ — 1 sin 0 will mutually cancel, and in the latter all the
even powers. This being done, and the equations divided by 2 and 2/^^ — 1
respectively, we get
cos«0=cos"0-"-^p^ cos"-^© sin^0 + "^"~\^ ^^^^^^^""^ W-^esin^g-...
sin n0 = n cos «-'0 sin 0 — "^"~^^^"~'^^ cos »-^ 9 sin ^0 + ...
These may, obviously, be reduced to expressions containing only cosines and
sines respectively, by means of II. I, p. 423.
438 PLANE TRIGONOMETRY.
2. We may also obtain, in a similar manner, the tangent of a multiple arc. Thus,
sinnfl 1 (cos 0 +^A^ sine}"— {cos 0—-vA^ sine) «
*^ "^ ~ cosnO ~ V—^ {cos 0 + V—l sin O} « + {cos 9 — V^ sin 0} "
1 {l +V'^ tane}"— {l — v/^ tan0}«
~~ s/^ {l + v^^ tan 9} " + {l —^'—l tan 0} «
n tan 0 - ^^ - - 3- tan30+ 1^.2.3,^.5 tan-
~ 1 _ "1"- li tan-^0 + .(n-1) (n-2) (.-3) ^^^,^ _ ^ ^ _ ^
1.2 1.2.3.4
3, To find sin" 0 and cos" 0 in /erms of the sines and cosines of multiples of 9.
By Euler's theorem we have cos« 0 = -^ < e ^ + e ^ fjor,
expanding and bringing together the terms equidistant from the extremes of the
series, we get
cos-0=|„{{e«^^/-^ + e-"^^/=^] + «^(«-^)^^/~^ + e'^""^) ^v'-l H . • •
= i,|cosn0 + YCOS(n-2)0+^y^cos(n— 4)0+ ....|
sin"0= — \ e ^"~ — e~ '^~ \ , and proceeding as before, we have
(2v/-l/( )
= L=^ I sin n0 — 7 sin (n— 2)0 + " ,~"^^ sin (n— 4) 0 + |
(2^/— 1)» ( 1 i • 2 J
where the signs depend on {s^ — 1)", and are easily assigned, and need no remark.
For a complete discussion of multiple and subraultiple arcs, the reader may, how-
ever, be referred to Poinsot, Recherches sur I' Analyse des Sections Angulaires.
4. To express the value of the arc 0 in terms of tan 9.
2e. yUl — g^^~^ _ cos 0 + ^'^l sin 0 _ 1 + ^/^ tan 0 .
Here, C-" "" ^-^^-1 ~ cos 0 —^'^ sin 0 1— v/^tan0'
and taking logc of both sides, we have, as at p. 250, after dividing by 2v''— 1, the
following series : 0 = tan 0 tan^ 0 + - tan^ 0 — ... .
3 5
This theorem, which is by foreign writers attributed to Leibnitz was dis-
covered and published originally by James Gregory, Professor of Mathematics
in the College of St. Andrew's. Applications and modifications of it will be found
in the Mensuration in this volume.
XII. SUBSIDIARY ANGLES.
Tins term is apj)lied to those angles which it is either necessary or convenient
to calculate, as intermediary steps between the data and the final solution of a
problem. Their use is not, however, confined to cases where angles naturally
enter into the inquiry : though their chief advantage occurs in the solution of
trij<onometrical problems.
It is not usual to designate by this term those angles which inevitably arise as
mtermt diate subjects of calculation between the data and the solution : but only
§uch as facilitate the calculation of those which are necessary, whilst that calcu-
lation might have been effected, though more laboriously, without the introduc-
SUBSIDIARY ANGLES. 439
tion of these subsidiary angles. For instance, suppose we have the equation
c = \/a^ — 2ab cos 0 + 6^: we may compute c by natural numbers, without
any contrivance whatever beyond the operations indicated by the symbols : but
as this is very laborious, an intermediate angle may be found, which, by the aid
of logarithms, will render the operation much more concise. The angle so em-
ployed is, then, according to this description, a subsidiary angle. A few exam-
ples of the contrivances to be employed, are annexed; and others will occur in
the places where they can be advantageously used.
1. Let X = ^/d^ — b'^ be given for solution by a subsidiary angle. It may be
written x=za / 1 -; and finding sin 0 = , we have x = a \/ 1 — sin *0
'>J a- a
= a cos 0. The value of sin 0 may be computed by logs, and hence we have
log X =: log a + log sin 9 = log a -|- tab sin 9 — 10.
Suppose, for instance, the expression had been x ^ >v/(32-965)^ — (27682)^:
^, . . 27682 , ,
then sm 9 = „r-;,^ ; and we have
32965
10 + log 2 7682 = 10-442197'5
log 32-965 = 1-5180531
8-9241444 = sin 4° 49' nearly.
and cos 4° 49' = 9-9984636
log 32-965 = 1-5180531
11-5165167 = 10 + log 32-8486; or
X = 32-8486 nearly.
2. If a? := \/a- + b^, we may write it x ^ a / 1. + — ; and taking - =r
tan 9, we have x ■=:■ a i^l + tan^ & = a sec 9. This may be calculated simi-
larly to the last *.
3. Suppose X ■= a/ a^ — lab cos C + 6" : then it admits of the following
transformations :
X = V'a^-2abcosC + b^= (a + b) Jl - '"\[,'^^'°^
X = -v/a^' — 2ab cosC + b- = {a — b) ^ I +
, , I.X /, 4a6 cos C^i
= (a — 6) ^1
2a6(l — cosC)
(a - bf
. I . 4a6sinHC
= («-&) V^ + {a-bf •
In the former case take cos 0=—^- J^^,^ \^ tte latter tan x= ^ ^—^ . <Jc^:
then, a? = (a + 6) sin 9 in the former, and x ^=^{a — h) sec x in the latter
case : and both final solutions will give the correct values of x to the extent
that the logarithmic tables enable us to carry the computations.
• It will be obvious that we might have taken cos 6 = - in the first exsunple, and cot 6 = -
in this. The same numerical results would have been finally obtained, as the student will see
on working out any given numerical examples.
440 PLANE TRIGONOMETRY.
4 The coefficient ~ . often occurs in trigonometrical calculations, whilst
a ^ 0
only log a and log b are given. To avoid going to the tables to find a and h,
and thence the above quotient, a subsidiary angle is generally used. The pro-
cess is as follows : —
_ b
a — b^ a , .rb „ , a — b 1 — cos 20 ^ ,„
= J, where if - = cos20, we have — —-r = :, , „„ = tau-0,
a ■\- b b a a + 6 1 + cos 29
a
and as tab cos 29 = log 6 — log a + 10 ; whence tan 9 can be found from the
logarithms without first finding the numbers a and b themselves.
„,, , b ,,11 ijO — b 1 — tan v
Or had we taken - = tan y, we should have had — --7 = , — -— — ^ =
a ^ a 4- 6 1 + tan X
cos Y — sin Y ^ , .. V
. - = tan (45° — y).
cos X + sm X
This latter is, perhaps, the better mode, when the fraction itself is the quan-
tity sought, and the former when its square root is required.
5. The expression a; = a sin A + fi cos A is one of frequent occurrence. Put
- = tan 9 ; then x = a (sin A + - cos A) =: a (sin A + cos A tan 0),
= — — , - sin A cos 6 + cos A sin ei = a sec 6 sin (A+0),
cos 9 { — ) —
which is in an entirely logarithmic form.
6. Let the roots of x- — px — 5 = 0 be calculated by a subsidiary angle.
Hare .=e±Ji+, = V, {^^ ± V|T7j . Take J^^ =cot,;
COS 0 ^T' 1
then X =^q Jcot 0 + cosec d\ =Jq. . — =— >yo. tan \9,OTjq.coXh9.
4o
Had the equation been x^ — px -\- q := 0, assume sin^0 =: -.,; then the roots
would have been x ■=. p cos-^Q, ox p sin-iS.
7. The real root of the cubic equation x^ — jj: — r = 0 is capable of being
put under the form
I 27r^
Let cosec =0 = - ^ -. then substituting, we obtain
X — A f ^ /FhTcosI 3 lJZr^^\ fq r ..
' - V 3 1 V '.VnT- ■^J~^9-] = Jl (V cot i 0 + V tan .^01
Again, find tan ^ = V tan ^ 0 ; and the final value wiU be expressed by
If the equation belong to the irreducible case, or 4g3 be greater than 27r^', let
» = aco80: thencos30=-*f_3^, or ^3 - 3-^% _ «-%os 30 = 0. To
a a 44
make this coincide with the given equation, we shall have g = ^', r =
a' cos 30
~ :( , from which a and 0 are determinable ; and thence a; = a cos 0 may be
found.
CHANGING THE RADIUS. 441
It might, also, have been solved by making a? = cos | -;- + 0 [ , or
X = cos I + ^[ ; and hence the three roots of the given equations may be
found : which in this case are all real.
8, The following expression, not involving angles at all, occurs in physical
astronomy, where e is always less than 1, viz. :
P = (1 + e.) (1 + e,) (1 + es) (1 + e,) . . . . (I + e.), where
— LlL^ ^ ~^^ _ I — \/I— gi' _ 1 — a/1 — e^^,
' ~ 1 + a/1 — e^' ^* ~ 1 + s/T^^f ■ ■ ■ ^" ~ 1 4- ^1 _ e\-x
Put e = sin 0, then a/1 — 63=: cos 0, and we get successively,
1_^^L^= LzL^5!i| = tanH0; and 1 + e, = 1 + tan^^S = sec^^e.
1 4. ^1 _ e2 1 + cos 0
Also, since the numerators are in aU cases less than the denominators, tan ^^9,
and therefore tan^Q is less than 1 : the process may, therefore, be continued
without limit, by making tanJ9 = sin9,, and thence finding sec^^0„ and then
again sec-i02, and so on. The e.vpression is, therefore, reducible to P :=
sec-^9 sec-^9, sec^^Q^ ' which is adapted to the application of logs at once :
and this, perhaps, is the only way in which the calculation could be practically
effected without extreme labour.
XIII. CH.\NGING THE RADIUS IN TRIGONOMETRICAL EQUATIONS.
In all the preceding investigations, the radius has been assumed as unity : but
inquiries sometimes occur in which it becomes necessary to transform trigono-
metrical equations formed on this hypothesis into others where the radius is some
different quantity r; and, conversely, from equations to radius r into others
to radius 1.
This is effected at once, in equations involving only direct functions of the arcs,
by rendering all the terms homogeneous ; and since, as a general principle, the
terms can only cease to be homogeneous by some of the linear factors becoming
unity. The rule, therefore, will be, to restore the general value r of the radius
in all the terms, so as to render them homogeneous. Thus to radius 1, we have
sin 39 = 3 sin 9 — 4 8in-^0 ; and to render it homogeneous, we must have
r^ sin 30 ■= Sr^ sin 0 — 4 sin'0, there being now three linear factors in each
term.
In the case of inverse functions, where the occasion for this change most fre-
quently occurs, and where the mode of proceeding is less obvious, the change is
still easily made.
Since arcs subtending the same angle are to one another as the radii to which
they are described {th. 94, p. 337), if we denote by 0i and B^ the arcs subtend-
ing an angle to the radii 1 and r, we have 0, : 1 : : 0, : r ; whence 0^ = ^^i and
9. =1 — . Making these substitutions according as the expression in radius r or
r
radius 1 is given, the transformation will be complete. The reduction from
radius r to radius 1, is that which almost always occurs; and it is perpetually
required in calculating the numerical values of integrals, as they are thus
reduced to the use of the common tables.
442 PLANE TRIGONOMETRY.
XIV. ON INVERSE NOTATION AND OPERATIONS.
Taking 1 as the base of all number, x^ or x signifies 1 multiplied by x, and
ar' or signifies 1 divided by x : that is, x^ and x~''- indicate inverse operations,
X
or operations such that they mutually destroy each other. In analogy to this
idea, Sir John Herschel proposed a notation for the general expression of inverse
operations, which, from its great convenience, has been generally adopted in this
countrv, and of which a very brief explanation is annexed.
Whatever function any one quantity be of another, the inverse function is
expressed by the index, ~', written after the symbol indicating the function.
Thus, in the case above, a;' becomes x~^ ; in log h := b, we have log~^ log'A :=
log~'i ; and since log~^ neutralises log ^ or log, it becomes h = log~^6. The
same applies to trigonometrical and all other functions ; as, for instance, if sin
0 = K, then sin"' sin'0 = 6 = sin~'ic ; or if 0 ^ tan~V, we have tan 0 = e ; and
80 on *.
Though the value of this notation is most obvious in the integral calculus, it
is not destitute of utility even in elementary trigonometry, as the following
examples will show.
1. Let tan a = t, tan a, = ti, tan a., = ti, ... tan a, = t,: then
, . tana — tana, t — f, , ,
smce tan (a — a,) = :; — ; = ; hence we nave
'■^ 1 + tan a tan a, ] + tt^
tan~' ' = a — a^ = tan~7 — tan"'/,.
Employing the same notation, we have the following equations in succession :
tan-'f — tan-'/, = tan"' ^ ~ ^' ,
1 + «. '
tan-i/, — tan-V, = tan"' ' ~ *
tan-'/._, — tan-'/. = tan"'
1 + /,/, '
/._. — /.
1 + ^_. t.
and by addition of all these results, the remarkable formula
Un-'/ - tan-'/. = tan-'/-=- j-' + tan-' ^-^^^' + .... tan"' -^--' ~ ^•- .
1 + //.^ 1 +/,/./ l+t,_,t.
2. Given the equations 9 + <p = a and tan 0 = m tan ^ to find ^ and 9.
From the second 0 = tan"' (m tan ^), which inserted in the first, gives
tan-' (m tan ^) 4-^=0, or taking the tangents it becomes "* ^^9+ — "9^ _ ^^^
l—m tan^ tan f
or, m tan a tan =0 + (m + 1) tan <{, = tan a.
From this, tan <p, and hence tan 0 = m tan f, will be found by the solution of
a quadratic equation : thus giving,
tan 6 = — ^"'-+AI± "^'*'" tanVfJ^+Tyi _(m-f- 1)+ ^/4m tan^a + Cm + l)
V n A ' " 9 tun U — — — - ■
2m tan a 2 tan a
• Ikforc this notation w-as invented, tlic circumlocutions, d — arc wliose sine is k. 6 = arc
who.* Ungcnt is k, r/r. were obliged to be used to express the equations 6 = sin-'*:, 0 =
Un -'«. rtc. Many foreign writeis still adhere to the old notation, which gives their books an
exlreincly awkward appearance, and renders them much more difficult to read.
INVERSE OPERATIONS. 443
EXAMPLES FOB PRACTICE.
1. Show that cosec~^\/50 + cosec~^v'65 = cosec"' — - — ,
/65
and cosec~^,v/10 + cosec~' v^26 = cosec"'— .
2. Show that tan-i - + tan"' - + tan-^- + tan"'- = 45°.
3 J 7 o
3. Solve the equations <p -\- 6 = a, and cos <p : cos 6 :: m : n.
t
4. Show that if 2 tan~'/ = sin~'2s, then s =
1 + f^'
5. Resolve the equation vers"' - — vers"' — ^ vers"' (1 — b\
^ a a
6. Find <p and 9 from sin <p = »/2 sin 9 and tan 0 = ^3 tan Q.
7. Estabhsh the following equalities amongst arcs : —
(1.) sm ' 2 + «"^ 2 ~ 2"
3 9 1
(2.) COS"' - + COS"' -r -)- COS"' ^ = T.
(3.)cot"'i + cot"'l-cot-'^-i-^3=|.
(4.) cosec"'VlO -j- cosec"';y26 + cosec"'^/50 + cosec"'»/65 = — .
4
(5.) 2 tan"' J + tan"' 1 = 2 tan"' ^ + tan"' 1 + 2 tan"' -^ =- .
3 7 8 7 5 4
(6.) tan"' 1 - tan"' 1 + 4 tan"' ^ = |.
8. When tan"' {x + 1) = 3 tan"' (x — 1), show that x = ± a/2.
10. Given cos ' + cos"' --r = + - to find x*.
2ax — 2ax — 3
MISCELLANEOUS EXAMPLES FOR EXERCISES ON ARCS.
1. Investigate the equation tan a + sec a = tan (45° + ^a).
, oi. 1 ^ 1 + tan^e , 1 _|_ tan^e sec^e
2. Show that sec 20 = — 5-, and cosec 20 = - . „ := -.
1 — tan^e' 2 tan 0 2 tan 0
3. If a; = cos 0 and y = sin 0, find 0 from Axy ^ B^^ + Ca;^ + D»
4. Prove that tan-^ — sin^X = tan^x sin^x-
5. If tan A = ^/n, show that sin (p = / — - — and cos 6 = /, — . — :
'^ '^^/l + n '^Vl+n
and assign 0 when n is 1, 2, 3, 4, and -, -, -, -, in succession.
2 3 4 5
• This equation expresses trigonometrical] y the problem : — given the three radii a, 6, c, of
concentric circles to find the side J, of an equilateral triangle which shall have its angular points
in the three circumferences.
From the Same equation is also deduced the remarkable theorem : if on the sides of any
triangle, equilatenil triangles be described exteriorly, the lines drawn from the vertex of each to
the opposite angles of the first triangle will be equal to each other, and intersect in the same
point.
444 PLANE TRIGONOMETRY.
6. Find x and 0 from sin x = » cos x and sec 0 = m tan 9, both generally,
and when m and n are 56 and 10 respectively *.
7. (II. 25.) Demonstrate the assigned values of sin A, cos A, and tan A
given in Hutton's Tables, pages 362, 363.
8. (II. 10, 1 1 ) Show that sin^e + vers-0 = 2 vers 0 ; and sink's = 4 vers 19.
9. (II. 12.) Prove that sec x = tan x + tan A (90 — x). for all values of x-
10. Estabhsh the surd values of the sines and cosines, p. xxxix of Mutton's
Tables.
11. (II. 18.) Show that sec 60° = 2 tan 45°, and tan 45° sec 60° = secHo°.
12. (II. 21, 22.) Solve the equations tan ^ + cot 0 = 4, cot ^ + tan ^ = 2n,
and sin 0 + cos <p =z a ; and show the limits of possibility.
13. (II. 9.) If 0 + 0 = 60 and x + w = 90, show that we shall have
sin 0 — sin 0 ,„ , sin x — sin w
-■^-r7 7W = ^^3, and . Z . r = ^2.
sin i{<p — 9) sm i (x — w)
14. (II. 25, 3.) Show that cos'^m — sin'*m = cos 2m, and that
sin-ni sin^Ji + cos-m cos'n + cos-??i sin% + cos% sin% = 1.
TT . , T.T, , T , cos a , cos b C t , ^,\l
15. (II. 30.) \V hen tan^x=cos 0 sec a, then — -. = •< cos'^a + cos'*o > \
cos a? sm a? L J *
16. (II. 42.) Determine the arc which is a third of the arc whose sine is s, or
find 9 from sin 39 = s; and show the meaning of the three answers.
17. (II. 43.) Prove that tan"'- -f tan"' - = tan"' ^ + 2 tan"' -=45°,andthat
_i_ ^
tan""' a + tan~'6 = tan~' — j:^^—..
— } -\- ab
18. (II. 47.) If a, b, a, be any three angles, prove the equality, tan a tan b tan c
— sin (a + & + c) + sin (— a + 6 + c) + sin (a — b + c) + sin (a + i — c)_
cos (a + 6 + c) + cos ( — a + b + c) -}- cos (a — b + c) + cos (a + 6 — c)
,« ,iT ^^ , s Tf cos (t) tn + n .sin* m + n.,, . ,
19. (II. 26. 1.) If ^= — — and ^^ = —i^ find the expressions for
cos 0 m — n sm 9 V2»in
0 and 9, and ascertain whether they be real or not.
20. If a, /3 denote two given arcs, show that in the equation sec a sec <b +
tan a tan f = sec /3, we shall have
sec 0 = sec a sec |3 — tan a tan /3, and tan 0 = sec a tan j3 — tan a sec /3.
21. Prove that tan 9° = 1 + V^ — v's + 2^5, and assign the arc ^from
the equation tan 0 = 1 + v^5 + ^5 + 2,y5.
22. If sin (p + cos 9 = a, and cos 0 + sin 9 = b, show that
a -{- bn
a + bn . b + an , / 4
tan 0 = , and tan 0 = — ^-^, where n = / , , ,„
b — an a — bn \/ a^ + b^
23. Find the value of 9 in the equation a^ — x = 2, where
— 1
X ^ ^ i
8ec"9 + V sec"0 + a/ sec^0 + . . . ad inf.
24. Resolve the equation, cot 0 tan 20 — tan 0 cot 20 = 2.
• The quiHiioiis, .IS far as (C), stand in tlic same order as in the last edition of vol. i, and
oicii(>icd p. 40.i. Tliose which follow arc mainly from vol. ii. p. 34 — 42 of the last edition,
and though luhjcctcd to a new arrangement here, the old numbers are still retained in paren-
lhc«i» for more nadv reference to another work dependent on the former arrangement.
PROPERTIES OF PLANE TRIANGLES.
445
2a?' cos 2o -(- I = 0, are + a±l, and
25. In sin n9 = = and cos m9 = — — , what is the relation
2a/- 1 2
between m and n ?
26. Show that the four roots of x*
assign the value of a.
cot^0 4- 1
27. If —^a r ^^ cosec^^ + cot'^, find the relation between 0 and 9.
r.o OL xi . . 3 sin rb — sin 3* + 3 sin 6 cos 26 — sin 36 cos 2*
28. Show that tan 6 = ^—^ ~ i.
3 cos (p + cos 30 — 3 cos <p cos 20 — cos 30 cos 20
29. If a + /3 4- 7 + ^ = 2'''» prove that 1 + tan « tan (3 tan y tan S =
tan a tan /3 + tan a tan y + tan a tan S + tan /3 tan y + tan /3 tan S + tan y tan S.
30. If c = ch 0, c, = ch iff) Cj = chla, ... .; then
sin rt = c — c Ci Cj + c Ci c^ C3 c^ — c c, c^ Cj c^ Cj Cg + • • • •
cos 0 = 1 — c Cj -\- c Ci c^Cj — c c, Cj C3 e^ C5 + . . . .
31. If ^tan (45 — -^0) =: tan 0, then it is required to prove that
V^tan e + sec9 + V'tan ^ — sec 0 = 2 cot 20.
32. Prove Euler's series, a = sin a sec la sec ^o sec la .... ad inf.
33. Given cos 0 -|- cos x = a,, and cos 50 + cos 5x = b, to find 0 and x-
34. Given 0 + x = a and sin x cos 0 = sin 0 cos Xj to find 0 and x-
XV. THE PROPERTIES OF PLANE TRIANGLES.
7. The right-angled triangle.
Let ABC be a triangle right-angled at B ; with centre
A and the unit-radius describe the arc DE, and draw
DF perpendicular to AB. Then DF is the tangent of
the arc DE or angle A ; and AF is the secant.
Also, since the triangles ADF, ABC are similar, we
have .A I)
AB : BC : : AD : DF ; that is, AB . DF = BC.AD,
AB : AC : : AD : AF; that is, AB . AF = ACAD.
But these in trigonometrical symbols become, since AD = 1,
BC = AB tan A . . . . (1) , AB = BC cot A . . . . (3)
AC = AB sec A .... (2) \ AB = AC cos A . . . . (4)
Also from (3, 4) we get
BC cos A
AC cos A = BC cot A :
sin A
or BC = AC sin A
(5)
But C 4- A ^ hir, or C is the complement of A ; and substituting this value in
(1. . . .5) we have the remaining equations
BC = AB cot C (6) I AB = BC tan C . . . . (8)
AC = ABcosec C....(7) | AB = AC sin C . . . . (9)
BC=ACcosACB (10)
When, therefore, of three parts of a right-angled triangle (except all three
be sides) any two are given, the third can be found from these equations. When
the three sides are concerned, the most generally commodious rule is that fur-
nished by Euc. i. 47, or Theor. 34 Geom. The solution, however, may, even in
this case, be eflTected by the equations given above.
44<5
PLANE TRIGONOMETRY.
EXAMPLES.
1. In the right-angled triangle ABC right-angled at
C, there are given the side BC = 379"628 and the angle
BAC = 39° 26' 15" : to find the other parts.
Here A + B = 90°, or B = 90= — A = 50° 33' 45".
Also AC = BC tan B, and AB = BC sec B. The work
will be
log 379-62 = 2-5793491
pp 8 = 92
tan 50= 33' = 10'0846678
pp 45" = 1931
A
log
37
9-62
=
2-5793491
8
=
92
sec
50=
33'
=
10-1969496
PP
45"
=
1152
log AC = 2-6642192 logAB = 2 7764231
Hence AC =: 4615504 and AB := 597-6171 nearly.
The advantage of drawing out the forms for the entire operation cannot be
too much insisted on. In the present very simple example, log BC occurs
twice, and therefore may be written down at the same stage of the work : and
tan B, sec B, occur at the same opening of the tables ; and therefore may be
taken out in another stage with one single reference to the book. In taking the
logs of numbers, the corrections are set down by inspection ; and the differences
for one minute may be registered at the time of taking out the log functions to
the nearest minute, and the calculation of the correction made and entered in
its place afterwards. The method of making these corrections has already
been explained (p. 434).
2. Given BA = 402-015 and B = 56° 7' 18" to find the other parts.
Here A = 9j° _ B = 33° 52' 42", BC = AB cosB, and AC = AB sin B.
log 402-01 = 2-6042369 I log 402-01 = 26042369
pp 5 = 54 pp 5 = 54
cos 56° 7' = 9-746-2477 sin 56= 7' = 9-9191694
pp 18" = —565 pp 18" = 254
2-3504900 log AC = 2-5234371
log BC = 2-3504335 |
Hence BC = 224-0957 and AC = 333-7621 nearly.
In this example as the correction for cos B is negative, the positive terms are
added, giving 2-3504900, and 565 subtracted downwards, leaving the corrected
log BC as the remainder. When there are several subtractive corrections, it will
be more convenient to add the positive terras together and then the negative
together, and take the result of the two sums, as in the example above.
3. Given AB = 501-625 and AC =437- 128 to find the other parts.
/AB2 — AC2, or BC = AB cos B.
Here sin B = ^'Jl, A = 90° - B, BC =
AB
log 437-12 = 2 6406007
PI» 8 = 79
Adding 10, 126406086
2-7003792
log 501-62 = 2-7003748
pp 5 = 44
2-7003792
sin B = 99402294
or B = C0° 37' 28", and hence, A = 29° 22' 32'
PROPERTIES OF PLANE TRIANGLES. 447
Then to find BC we have BC = «y(AB f AC) (AB — AC), and hence
By the question, AB = 501 625
and AC = 437-128
log (AB + AC) = log 938753 = 2-9725514
log (AB — AC) = log 64 497 = 1-8095395
2 4-7820909
log BC = 2-3910455, or BC = 2460626
Or again, by the equation BC = AB cos B = AB sin A we have
log 501-625 = 2-7003792
cos 60° 37' = 9-6907721
pp 28" = —1048
2 3911513
log BC = 2-3910465
or log 501-625 = 2-7003792
sin 29° 22' = 9'6905476
pp 32" = 1197
2-3910465
or BC — 2460631.
The former process is generally the more accurate, but the latter is the less
laborious method, as log AC had already been found in the preceding part
of the solution, and the tables were open to cos B in finding its value from
CA
sin B=: . *„. Also.it conduces to convenience, as the work shows, to use the direct
Ad
functions sin, tan, sec, in preference to the complementary ones, cos, cot, cosec.
4. Given AC = 299015, BC = 325162, to find the other parts.
AC
tan B = ^, A = 90° — B, and AB = BC sec B.
log 29901 = 2-4756857
pp 5 = 73
logAC+10= 12-4756930
2-5120998
325-16 = 2-5120971
pp 2 = 27
BC = 2-5120998
99635932 = log tan 42° 36' 5",
Hence B = 42° 36' 5" and A = 47° 23' 55".
Then AB = BC sec B = 325-162 sec 42° 36' 5" ; and hence
log 325-162 = 2-5120998
sec 42=36' = 10-1330649
pp 5' = 97
log AB = 2-6451744, or AB = 441-7478.
5. Given AB = -6293 5, and ABC = 50° 10' 33", to find the other parts.
6. Given BC = 35826 and CA = 286-325, to find the rest.
7. Given log BC = 3-1296578, and log BA = 3-2965782, to find the rest.
8. Given AC = 5, BC = 6, to find AB and the angles.
9. Given AC = 162, B.\C = 53° 7' 48", to find the rest.
10. Given AB = 25 and BC = 24, to find the angles.
II. Oblique-angled triangles.
1. Let ABC be a triangle, and from the
angle C draw the perpendicular CD to the
base AB. Denote the angles by A, B, C, and
the sides respectively opposite them by a, b, c.
By the right-angled triangles ACD, BCD we
liaveAB = AD-fDB, orc = flcosB+ficosA. * '"''
Forming similar equations with respect to the other sides, we get
44.8
PLANE TRIGONOMETRY.
a = 6 cos C + c cos B ; from which a^ = ab cos C -\- ac cos B,
6 = a cos C + c cos A ; 6^ = aft cos C + 5c cos A,
c = a cos B + 6 cos A ; c^ = ac cos B + 6c cos A.
Subtract each of these equations from the sum of the other two ; then we get
7o o » — «■ + b' 4- c-
26c cos A = — a- + 6- + c2 ; or cos A = -
2ac cos B = a2 — 62 + c- ; or cos B =
or cos C =
2bc
— b- + c'
lac
+ 6- - c2
2ab
(1)
2ab cos C = a^ Jrb- — c^
These expressions are convenient for calculating the cosines of the angles of
a triantrle when the sides are given in small numbers, or in large numbers which
have the same ratio with three small numbers. They are essential when a, b, c,
are themselves the square roots of numbers not exact squares.
2. Since sin A = \/l — cos ^A, if we substitute for cos A its value just
found, and proceed similarly for cos B, cos C, we shall have
26c sin A = V{2a^^ + 26'c^ + 2aPc^ - a* - ¥ - c*] 1
2ac sin B = ^/[2a^6•- + 2b'-c^ + 2aV _ a* — 6* — c^j V (2)
2a6sin C = V{-2a-b^ + 26-c2 + 2aV — a* — b* - c*] J
From the equality of the right-hand sides of these equations we get
26c sin A = 2ac sin B = 2a6 sin C ;
and from these three, taken two and two, we obtain
sin A sin B sin C
a 6 c
which is usually expressed in words by saying that the sides of a triangle are
proportional to the sines of their opposite angles; and often written
sin A : sin B : sin C::a:6:c (4)
The quantity under the radical in (2) is rarely adapted to convenient calcula-
tion, and never, excejjt when also the equations in (I) are also adapted ; and
as (1) is less laborious than (2), the latter is never used for that purpose, but
merely a step in general investigation. It is, in fact, the expression for the pro-
duct of one side of a triangle and perpendicular upon it from the opposite angle.
We shall presently find another form for it.
.3. Resume the equations (1) : then if a -|- 6 + c = 2s
sin-A= i(l-cosA) =i f 1 - -a^ + b^ + c^] ^«^ " (^ - c?
(. 2oc )
26c
(« — 6 -f c) (a -h 6 — c)
46c
cob^iA = i (I -i- cos A) = h\ 1 +
A) = ^-{
— a- +
4.bc
(s -b) (s — c)
be
-a- + (b + cY
_(-a + b+ c){a + b + c)
From these two last results tan -^A ::^
s (s — a)
Performing similar operations for B and C, we get the following equations : —
As-b) (s-c)
""^^=7 6c
smiB= /("-") i^-c)
. A8—a){s—b)
""i^=7 — ^r-
s (s — a)
6c~
, „ ,s (s — 6)
cos 4B= / -^ ^
-V ac
_ /s is—c)
ab
cos AA=r /'—
jc=y
tan
tan
(s—b) (.s-c) 1
1A_ /(s-6) {s-
AB_ /(s-g) js-c)
' V s^s-b)
is— a) (s—b)
>...(5)
4c=y
tan iC=
s (s—c)
J'^t'chl /iiH>/c
Fiitr T(Ut<^ 44f/,
//,/,/ /ioo/c
-\ 1 <■>
4 i'o
C •-' t
.". if-'
.1 <>u
o 7<>
(| oo
«.to
„ n
// /./i.,///rr,/
7 |0
1 TO
'.'M
-II.
/'Vr/t/ /iftft/,:.
3,
////// //om ///(■ ton-f/o/'/n; ^'/</(/ 7>ot>A- .
n
If
r oi
u\
V/»
m
l\'/i,>/r ('miliiit J<»;','. ? .'l«> .
PROPERTIES OF PLANE TRIANGLES. 449
When it is required to compute all the angles of the triangle, the third set of
formulas is the most convenient, as there are only four logarithms and three
arithmetical complements to be taken out. In each of the others there are
seven required to obtain the three angles. A still more convenient form, how-
ever, is the following, as it requires only four logarithms : —
_ J(s — a) (s — b) (^s — c) . , „ ,
Put r = /- : then we shall have for the tangents,
s—c
tan hA = , tan |B = — 7, and tan AC = .... (6)
s—a s—b ' s—c
le equations marked (3) : then, since r =: • — w
^ o sm B
sin A — sin B tan ^(A — B) tan ^ (A — B)
4. Resume the equations marked (3) : then, since r = • — « . we shall have
^ o sm B
(7)
a + 6 ~ sinA + sinB taaKA + B) cot ^C
and similar expressions for each of the other pairs of sides in succession.
Or again, from the same equations
a + b sin A -H sin B sin A + sin B cos §(A — B) "i
~c" ~ sinC "~ "sTu (A + B) ~ cos KA + B) 1
a — b sin A — sin B sin A — sin B sin ^(A — B) | * * " ^ '
c~ ~ sin C ~ sin (A + B) ~ sin ^(A + B) J
These equations may be put in a convenient form for the calculation of the
side c when the sides a, b, and angle C are given : viz.
c = (a — b) cos ^C cosec ^(A — B), and c = {a + b) sin ^C sec ^(A — B)
and the two values of c thus computed will be a mutual check upon the accuracy
of the work, by which each was obtained.
5. Returning to equation (1) we have in succession,
c^ = a^—2ab cos C + b^ = {a + bf — 2ab (1 + cos C)
r . A^2f1 4ab COS^C\
n . n 2 Vab COS ^C . • 9/, , 4a6cos*iC , ,
Put COS 9 = j-^^— ; then sm^O = 1 — —, — r-rri— and we have
a -\- b {a + by
c- = (a 4- 6)2 sin=0, or c = (a + 6) sin 0 (9)
Or, again, we may write the equation in the following successive forms,
c^={a- by + Qab (1 - cos C) = (a - b)^ fl + ^°^ ^°'^,^ 1
-- IJab sin iC ^, , , -
Put tan Y =■ —^ H=- , then we have as before
'^ a — b
c^ = {a — b)- sec^x* or c = (a — 6) sec x (10)
This method is much used by the continental mathematicians : but it is not
so convenient for use as those in (7, 8).
6. Multiplying together the values of sin ^A, cos JA found in (5), we have
„ . ,. ,. . . 2^ sis— a) {s — b) (s — c)
2 sm iA cos ^A = sm A = -^ — ^ —^ -, or
be sin A = 2 V* (* — a) (s — b)(s—c) (11)
and similar expressions for sin B, and sin C.
Hence, comparing this with equation (2) we have the expressions there given
in another form, and adapted to logarithmic use.
7. Since, in the triangle ABC we have CD = b sin A, we get from (11)
CD = p^ = ?^^-^^ ^^-^^^'-'} (12)
and similar values for the perpendiculars p„ j)„ from B and A.
VOL. r. G g
450
THE NUMERICAL SOLUTION OF
8. Again, let CD be the perpendicular from C upon
c as before. Then,
CD cot B + CD cot A = c ; or
SAB A. S
CD =
c sin A sin B c sin A sin B
(12)
cot B + cot A sin (A+B) sin C
This might have been derived, but not quite so briefly, from equation(3) : and then
BD=CD cot B=
c sin A cos B
and DA = CD cotA =
c cos A sin B
...(13)
sinC ' sinC
These forraulse are often useful, both in the determination of the heights or
distances of objects, and in mensuration.
9. Let ABCDE be any polygon, and P a point in the same plane, or not,
and situated either within or without the polygon : draw lines from P to all the
angular points of the figure.
Put PA = a
PB = i
PC = c
PAN = a, I PAB = a
PBA = /?, ! PBC = jS
PCB = 7. ! PCD = y
Then, a sin a^ = n sin v
b sin 01 = a sin a
c sin y, = b sin /3
d sin ^1 = c sin y
m sin Hi = I sinX
n sin V, = TH sin /x
and multiplying these together we get, since the linear factors mutually cancel
sin a, sin/3, sinyi. .. .sin/i, sin v, = sin a sin/3 sin y ... sin ^ sin v .... (14)
XVII. THE NUMERICAL SOLUTION OF PLANE TRIANGLES.
When only the parts of the triangle itself are concerned in the inquiry, either
as given or required, there are three general cases, the formulae necessary for
the solution of which have already been given.
1 . When a side and its opposite angle are amongst the data.
2. When two sides and their included angle form the data.
3. When the three sides form the data.
CASE I.
When a side and its opposite angle are amongst the data.
This divides itself into two subordinate cases, according as the third datura
is an angle or a side.
1. Let the third datum be an angle. Then (eq. 3, p. 448.) if ABC be the
triangle, and we have given a, A, B; then C = ^ — (A + B)
, _ a sin B . _
~ sin A~ ~ " ^'" *^°^^^ ^'
__ a sin C
^ ~ sin A' ~ ° ^'° ^^ "^ ^^ ^°^^^ ^'
PLANE TRIANGLES.
451
Ex. Let BC = 305 296, B = 51° 15' 35", and C = 37° 21' 25" be given.
Here A = 180o — B — C = 91° 23', and sin A = sin 91° 23' = am 88= 37''
Also, b =
a sin B
a sin C
sin A sin A
FK. \
fin ijutuu luu
log 305-29 = 2-4847126
log 305-29
=
2-4847126
pp 6 = 85
PP 6
=
85
sin 51° 15' = 9-8920303
sin 37° 21'
=
9-7829614
pp 35" = 591
pp 25"
=
690
sec 88° 37' = 100001266
cosec 88° 37'
logc
z
10-0001266
log b = 2-3769371
2 2678781
or, b = 238-1974.
or, c
=
185 3011.
2. Let the third datum be a side. Then if we have given a, b, A to find the
other parts, we shall have sin B ^ ; in which case two of the angles
■will become known, and we can proceed as in the former sub-case.
It is necessary, however, to observe that as sin B = sin (tt — B) we have no
reason to prefer one value of B to the other : or as it is usually expressed, the
solution is double. There are, in fact, two different triangles which will fulfil
the given conditions, and yet two of whose quaesita are different from each
other. This is altogether analogous to the double solutions of questions in
algebra which give rise to quadratic equations ; and, indeed, if this example
were solved algebraically for the third side, we should find (as may be easily
shown) that its value was given by a quadratic equation.
To explain this more clearly, let us consider the geometrical problem, in
which are given two sides AC, CB, and the angle CAB opposite to one of them
to construct the triangle.
Constr. With centre C and distance CB describe
a circle, cutting A B in B, and B'. Join CB,CB' : then
obviously either of the triangles ABC or AB'C fulfils
all the conditions, and may be taken as that required.
Also, since BC = BC, the angle CBH' = CB'B ;
but CBA is the supplement of CBB' and hence also of CB'B. This accords
entirely with the inference drawn above, that the two values of the angle B
are supplementary, the one of the other. It will also follow that the angle C
will be double according to the values of B, and likewise that the side C will
be double from the same cause, viz. AB and AB'.
In the preceding investigation of the ambiguous case, our reasonings have
turned upon the assumption that the less of the given sides is opposite to the
given angle. Suppose now that the greater of the given sides is opposite to the
given angle, and construct as before:
then since CB is greater than CA, CB'
will also be greater than CA, and the
line CB' will lie on the opposite side
of CA from CB, and the triangle
formed by the lines CA, AB, CB'
(produced where necessary for the formation of the triangle) will be ACB',
having the angle CAB' the supplement of CAB instead of equal to it. Two
triangles, therefore, fulfilling all the conditions can only be formed when the
less of the given sides is opposite to the given angle. This, therefore, is the
only ambiguous case.
Gg2
452
THE NUMERICAL SOLUTION OF
Ex. 1. Given c = 195-265, b = 203-162, and B = 45° 0' 55" to find the rest.
Here since the greater side is opposite to the given angle, the problem is
not ambiguous. The formulae of solution are
. ^ c sin B , i • * Ti
sm L = — 1 — > and a = o sm A cosec r>.
log 195-265 = 2-2906244
sin 45° 0' 55" = 98496008
ac log 203-162 = 76921575
log 203-162 = 2-3078425
sin 92° 9' 23" = 9-9996923
cosec 45° 0' 55" = 10-1503992
]og'a= 2-4579340
whence a= 2870350
sin C = 9 8323827
or, C = 42° 49' 4-2"
and hence, A = 92° 9' 23"
Ex. 2. Given a = 350-169, b = 236 291, and B = 38° 39' 15" to find A, B, r.
In this, since the less side is opposite to the given angle, the solution is double,
or belongs to the ambiguous case. The formulae are
— , C =180° — (A + B), and c = a sin C cosec A.
log 350169 = 2-5442777
ac log 236-291 = 76265529
sin 38° 39' 15" = 97956146
sin A :=
sin A = 9-9664452.
Hence A = 67° 45' 58"
B = 38 39 15
A + B = 106 25 13
C = 73 34 47
log 350-J69 = 2-5442777
sin 75° 34' 47" = 99819155
cosec 67 45 58 = 10-0335548
and A'
=
112°
14'
2"
B
A' + B
—
38
39
15
150
53
17
C
=
29
6
43
log 350 169 = 2-5442777
sin 29° 6' 43" = 9-6870985
cosec 112 14 2 = 100335548
logc= 2-5597480 I log c' = 2-2649310
Hence, we have the side AB = 362-8674, and the side A'B = 184048 0
EXAMPLES FOR PRACTICE.
1. Given A = 37° 20', a = 232 and c = 345 yards, to find b, B, C.
2. Given c = 365, A = 57° 12', B = 24° 45', to find a, b, C.
3. Given the following parts of the triangles, to find the remaining ones.
a
b
C
A
(1)
197
• . • •
237
(2)
310
• • • •
(3)
....
305
217
(4)
516
329
• • • •
(5)
232-193
345-261
• • • •
142° 29' 10"
f-..
203162
195-265
92° 9' 23"
B
62° 9' 0"
45 0 0
37 10 15
C
90° 0' 0"
41 13 10
4. Given a = 10, 6= 12, and A = 37° 15' to find the remaining parts without
the employment of logarithms.
5. Given a + b: a-b'.: 15 : 2 and 6 + c : b — c :: 14 : 3 to find the
anKles of the triangle.
6. In the same triangle, if 3a + 46- 6c = 1786, find the sides of the
triangle.
PLANE TRIANGLES.
45.3
CASE 11.
When two sides, a, b, and their included angle C are given.
First method. By th. 7. p. 449, find tan i (A — B) = ^^ cot ^C : then
B) = A, and i (A + B) — i (A - B) = B ;
we shall have ^ (A + B) + ^ (A
and the third side by Case I.
When the lojjs of a and b are given, as is generally the case when the sides
have been found by a previous investigation, employ the method explained at
p. 440 for finding tan ^ (A — B) and complete the solution as above.
Second method. When only the third side is required, employ equations (8)
p. 449, and the subsequent formula there given. This is rather shorter than the
other, though it has not been much used from not being so generally known.
Third method. Under the same circumstances employ equations (9> 10), p. 449;
which also furnishes a good mode of solution.
Fourth method. Employ the equations c = x/a^ — 2ab cos C + b^,
6 sin C a sin C
sm B = —- ;: , - — -- and sin A = . „ ^ . /^—r-co-
Va^ — 2ab cos C + b^ */a^ — lab cos 0 + 6^
The first of these methods was invented by Vieta : the second by Thacker; and
it was published in 1743, but no attention appears to have been given to it till
Dr. Gregory, in a former edition of this work, gave it a practical form. Dr.
Wallace has also discussed it in the Edinb. Trans, vol. x. The author of the
third I am not able to assign, but it is that most commonly employed by the con-
tinental mathematicians. The fourth is not adapted to calculation, and its only
use is, in the transformation of trigonometrical impressions.
Ex. 1. Given a = 16-9584, b = ir96l3, C = 60° 43' 36".
Here, tan ^A — B)
For the coef of cot ^C
a = 16-9584
b = 11-9613
a — b
a~+~b
First method.
cot ^C ; and c = a sin C cosec A = 6 sin C cosec B.
a + b
a-b
= 28-9197
= 4-9971
For the value of tan ^CA — Bj
log 4-9971 = 0-6987180
aclog28-9197 = 85388062
cot 30° 21' 48" = 10-2322235
tan i(A— B) =
For A, B, separately.
^(A+B)=59°38'12"
i(A— B)=16 26 0 23
For c =
a sinC
sin A
log 16-9584 = 1-2293848
sin 60° 43' 36" = 9 9406644
cosec 76 4 12-23= 10-0129638
9-4697477
For c =
A=76 4 12-23
B=43 12 11-77
b sin C
sin B
log 11-9613 = 10777784
sin 60° 43' 36" = 9-9406644
cosec 43 12 11-77= 10-1645703
1-1830131
1-1830130
From both of which we have c = 15 24098 nearly.
Second method.
In this, having found tan ^(A — B), as before, we have the following process :
c = {a + b) cos KA + B) sec i(A — B).
log 28-9197 = 1-4611938
cos 59° 38' 12" = 97037054
sec 16 26 0-23= 10 0181139
c = {a—b) sin i(A-|-B) cosec i(A— B).
log 4-9971 = 0-6987180
sin 59° 38' 12" = 9*9359289
cosec 16 26 0-23 = 105483662
1-1830131 1-1830131
which give the same values of c as were obtained in the former solution.
454
THE NUMERICAL SOLUTION OF
Third method.
We have cos 9 = ^^^l^*i^i^, and c = (.a+b) sin 0, for one solution : and
a -\- 0
tan V = ^^"^ "'" ^-^, and c = {a - b) sec x, for the other.
^ a — b __
For log 2'^ab or to^r v 4aJ.
log 16-9584 = 1-2293848
log 11-9613 = 1-0777784
log 4. = 0-6020600
2-9092232
cos Q =
log 2^ab
2v^cos iC
= 1-4546116
log2v^c6 = 1-4546116
cos 30° 21' 48" = 99359289
ac log 28-9197 = 8-5388062
cos 9 = 9-9293467
c ■= {a ■\- b) sin 0
log 28 9197 = 1-4611938
sin 31° 48' 13"-32 = 9-7218194
2 v^ sin IC
tan Y = = — .
-^ a — b
log 2^/06 = 1-4546116
sin 30° 21' 48" = 97037055
aclog 4-9971 = 9-3012820
tan X = 10-4595991
c z=z {a — b) sec x-
log 4-9971 = 06987180
sec 70° 51' 37"11 = 10 4842951
logc= 1-1830132 log c = 11830131
which are very nearly the same logarithms as those found by the other modes of
solution.
Scholium.
When it can be done, it is desirable to obtain check-solutions. In this case
it can be done in e?ch of the methods ; but where the operation is one of con-
sequence, it will always be better to employ the first and third, or the second and
third methods, in preference to the checks by the same method. If, however,
the same method be employed for the solution and its check, the third is the
best ; and perhaps shorter altogether than the separate application of the first
and second, as checks upon each other.
The first solution requires ten openings of the tables, the second five, and
the third eleven, for the solution and its check.
Ex. 2. Let the logs of the given sides be 2-2293848 and 2-0777784, and
C = 60° 43' 36" to resolve the triangle.
■--
a 1 — tan Y , r. . ■, b
■ r = , — ; = cot (45° + y), where tan x = "•
6 1 + tan X ^ a
Here
a-b
a + b
It may surprise the student to be informed tliat it is always much easier to observe angles
to any given degfce of accuracy than to measure lines : but s\ich, owing to a variety of causes
hereafter to be explained, is ttie case. The consequence is, that after the careful measurement
of a iinglf base, all tlic subsequent distances are found by calculation from that base and ob-
served angles. Tliis renders it e^^ntial to check every calculation, either by a different process,
or by liaving the omputations performed by two i)ei-sons independently of each other. Few
perwn* arc capalde of sucii steadiness of attention to numerical operation, as to be entirely
certain of avoiiiiiiR mistakes ; and no one would tnist to a single computation where a slight
error at any one of perhaps ten thousand steps would vitiate the entire result, and the entire
Ubour beato««J upon the work be thrown away. The practice of checking solutions should,
th«r«fore, be commenced early in every course of study of this nature.
PLANE TRIANGLES. 455
b
tan Y = -.
* a
log b = 120777784
log a = 2 2293848
tan X = 9-8483936
or, X = 35° 11'47''75
tan i(A — B) = ''—4 cot ^C.
a + 0
cot 80° 11'47"75 = 9-2375241
cot 30 21 48 = 10 2322235
tan i[A — B) = 9-4697476
very nearly as before found.
Then having the logs of the sides given, and all the angles, the solution is
completed as in the first method.
EXAMPLES FOR EXERCISE.
1. Given a = 34502»6 = 174'07, C = 37° 20' 30", to resolve the triangle.
2. Given c = 365 ,b= 154-33, A = 57° 12' 10" to resolve the triangle.
3. Given a = 112 , 6 = 120, C = 57° 58' 39", to resolve the triangle.
4. logo -f- logft = 5*1693765, logo — log 6 = '7629876, together with
n tan C = 1-8656729, to find the remaining parts of the triangle.
5. If a : 6 : : 3 : 4, cos C = O, and the perpendicular from C to c be 100,
what are the sides, and other angles of the triangle ?
CASE III.
When the sides a, b, c, are given, to find the angles A, B, C.
First method. By equation 1, p. 448, we have the solution from
_ a2 ^ 62 _,_ ^ a2 _ js ^_ c2 a''-rb^ — c^
cosA= -J — , cos 6= , cos C = -—. .
26c * - 2ca 2ab
Second method. Find s =: ^(a + b -\- c), and form s — a, s — b, s — c : then
^ \/ s (s— a) ' ^ V « (s—b) ' ^ V « (s—c)
Third method. Find r = ^/^"^"^ ^:Lr:_) i^Zf^ . then (eq. 6, p. 449.)
tan 4A=
s — a
The first of these methods is adapted to the case where a, b, c are small num-
bers, or are in the ratio of small numbers. The second (which was discovered
about 200 years ago by William Purser of Dublin) is well adapted, like the first,
to the case where either one or all the angles are required ; the use of the tan-
gents is, however, preferable to that of the sines and cosines of ^A, ^B, ^C,
given in the same place, as it requires the use of fewer logarithms. The third is
an elegant modification of the second, and was first proposed by Dr. Wallace.
In every case, to secure accuracy, compute the three angles or semiangles as
the case may require : then if the work be correct, their sum will be 180° in the
first, and 90° in the second and third methods.
Ex. 1. Let a = 3, J = 5, c = 7 : find A, B, C.
Here the first method applies, and gives successively,
q2 I k2 _|_72 IQ
cosA= ^ V Z = ;- = -9285714 = cos 21°47'12"i
2.5.7 14
32 k2 I 72 11
cosB= ^^ = — = -7857142 = cos 38 12 47i
2.3.7 14
32 _|_ 52 72 I
cosC= — —= =—-5000000 = cos 120 0 0
2.3.5 2
Proof of the work, A + B + C= 180 0 0
456 THE NUMERICAL SOLUTION OF PLANE TRIANGLES.
Ex. 2. Given n = 329S6, h = 43628, c = 62984 to find the angles A, B, C.
Here the second method may be used.
For the factors.
a =:
b -
c =
s — a=.
s-h-
s — c =
32986
43628
62984
139598
69799
36813
26171
6815
For tan ^ A.
log 26171 = 4-41 78203
log 6815 = 3-8334659
aclog 69799 = 5-1561508
aclog 36813 = 54339988
18-8414358
For tan ^B.
log 36813 = 4-5660012
log 6815 = 3-8334659
aclog 69799 = 51561508
aclog 26171 = 5-5821797
2 1191377976
tan 20° 20' 4" = 9-5688988
or B =40° 40' 8", nearly.
tan 14° 45' 35"i = 9-4207179
Hence A= 29° 31' 11", nearly.
For tan iC.
log 26171 = 4-4178203
log 36813 = 4 5660012
aclog 69799 = 5-1561508
aclog 6815= 6-1665341
2 I -20 3065064
tan 54° 54' 20"§ = 10-1532532
or C = 109° 48' 41'', nearly.
By the third method, r = A£_^lii_Ji£_?\ and then tan ^A = -^,
and 80 on. The work will, then, stand as follows :
log 36813 = 4-5660012
log 26171 = 4-4178203
log 6815 = 3-8334659
aclog 69799 = 5 156 1508
2 i 79734382
10+ logr = 13-9867191
log r — log (s— a) = 9-4207179 = tan 14° 45' 35 'i
log r — log {s—b) = 9-5688988 = tan 20 20 4
log r — log is—c) = 10-1532532 = tan 54 54 20 J
Whence A, B, C are found by doubling these results respectively.
EXAMPLES FOR EXERCISE.
1. Given a = 17407, b = 232, c = 345 to find A, B, C.
2. Given a = 30986, b = 154-33, c = 365, to resolve the triangle.
3. Given a = 1 12, b= 1 1265, c = 120, to find the angles.
4. (Jiven a = 3, 6=4, c = 5, to find the angles.
5. Given a -I- 6 = 48-2106, 6 + 0 = 601250, and c + o = 44-4114, to find
the sides and angles.
G. Given 24S04, 57876, 74412 to find the angles by the three methods.
7. Two sides of four triangles are the same, viz. 35 and 36 ; but the third
kidea are 7099, 7098, 7097, and 7096 respectively : find the angles of these
four triangles.
8. Two sides of three triangles are 315965 and 315966, and the third sides are
retpectively 20, 21, and 22 : required the angles of each triangle.
OF HEIGHTS AND DISTANCES. 457
XVIII. THE APPLICATION OF TRIGONOMETRY TO THE DETERMI-
NATION OF THE HEIGHTS AND DISTANCES OF OBJECTS.
A line is said, in technical language, to be measured, and an angle to be ob-
served. It is not, however, intended here to enter into a description either of the
instruments employed or the methods of using them : but a succinct account of
both will be given in the Geodesy in the second volume. A few specimens of
the class of calculations by which the solutions are obtained are given at the
commencement of this chapter, and a series of unsolved questions for the
exercise of the student is annexed.
EXAMPLES.
1 . From the foot B of a hill, the lower part of whose inclination was different
from that of the upper, the elevation of A the summit was observed to be 32° 10' 15",
and the inclination of the slope of the lower part to be 15° 10' 45"; and at the
upper end C of the slope the elevation of the summit was 57° 10' 30".- what was
the height of the hill, supposing the distance BC to be 952"57 yards ?
Let BC = a, ABD = /3, CBD = a, and ACH = y. Then ^
KCB = CBD = a, ACK = TT — y, or ACB = tt — y -|- a;
also BAC = TT — ABC — BCA = tt — (/3 - a) — (tt — y + «)
= y — /3. We have, therefore,
„ . BC sin BCA a sin (y—a) , , ,
BA = — : — rr-— ^ — = ■ ■ , '' ; and hence we have
sm BAC sm (y— /S)
AT^ AD • ADT-k « slu j3 siu (y — a)
AD = AB sm ABD = . ^' .
sm (y— ^)
In the example we have a = 952-57, /3 = 32° 10' 15", a = 15° 10' 45", and y =
57° 10' 30"; hence the calculation will be as follows : —
log 952-57 = 2-9788969 I y—a = 41° 59' 45"
sin 32° 10' 15"= 9-7262751 I y— j3 = 25 0 15
sin 41 59 45 = 9-8254758
cosec 25 0 15 = 10-3739840
log 802-8451 = 2-9046318 ; or AD = 802-8451 yds.
2. TTie elevations of two mountains in the same line with the observer are 10°
and 20° ; but upon approaching four miles nearer they have both an elevation
of 40° : it is required to find their heights, their distance apart, and the distances
of the two points of observation from each.
Let CE, DF be the heights of the two mountains, and
A, B the places of observation. Put, generally, AB ^ o,
CAF = a, DAF = a., DBF = /3 : then, by Case I. pi. tt. y^c
BD= -: — ^ "' and BC = ." ,^ ° ^ : hence we have
sm03— a,) sm (fi—a)
CE= BC sin/3 = °j" : ^'"f DF = DB sin /8 =«jj"«. ^i°f
sin (^ — a)
TIT7 T.i-1 r, a sin a cos /3
BE= BC cosyS = -,
. ,^ , BF = BDco8/3 =
sm (/3— a) ^
■r.n -nr nt-i o f Sin a, sin a )
EF = BF— BE = a cos /3 \ -. — ~~,— -. — jx ^, }
(sm (iS— a,) sm 03— a)j
a cos/3 Jsin a, sin (/3 — a) — sin a sin (/3 — a,) J
sin (J3 — a) sin (/3 — oi)
asin/3 cos^ sin (n,— a) a sin 2/3 sin (a, — a)
sin (/3— a,)
asinai cos/3
sin (^— a,)
sin (/3— a) sin (/3— a,) 2 sin (fi—a) sin(/3 — a,)
458
PLANE TRIGONOMETRY.
For CE.
log 4 = 0- 6020600
sin 10° = 9-2396r02
sin 40° = 9-8080675
cosec 30° = 10-3010300
For DF.
log 4 = 0-6020600
sin 20° = 9-5340517
sin 40° = 9-8080675
cosec 20° = 10-4659483
logCE = 1-9508277
and CE = 892951.
log DF =
and DF =
0-4101275
1-57115.
For EF.
log 4 = 0-6020600
sin 80° = 9 9933515
sin 10° = 92396702
cosec 30° = 10-3010300
cosec 20° = 10-4659483
log 2EF = 0-6020600
and EF = 2.
3. TTie side of a hill forms an inclined plane whose angle of inclination is known :
required the direction in which a rail or other road must run along the side of the
hill, in order that it may have an ascent of 1 in every nfeet.
In the annexed figure, let ADB be in the horizontal
plane, CD a line on the side or slope of the hill perpen-
dicular to AD, and DB a horizontal line peqjendicular
to AD, to which, as well as to AB, BC is perpendicular,
and AC the rail-road ; its projection on the horizontal
plane will be AB, and the angle BAG, therefore, will
determine the angle. Now
BD ^^„ BD ^.„ CB BD DB sin BAD
sin BAD= o-^ , cot CDB = ^p,, tan CAB = g^ = WT -^ nn —
BA'
BC cot CDB '
BC BA — BA
or sin BAD = tan CAB cot CDB.
For example let CDB = 10° and n= 100, or the ascent 1 foot in 100 : then
sin^CAB (-01)2
tan« CAB =
1— sin- CAB'
1— C-oi)2'
or tan CAB = -Ol^/l + 'OOOl = -0100005 very nearly. Hence, finally,
sin BAD = -0100005 cot 10° = sin 3° 15' 5', the horizontal angle required.
4. Going along a horizontal straight road, I wished to find the height of a tower
on a hill, and for this purpose measured two distances of 75 and 80 yds; and from
each of the three points found the elevations of the tower above the horizon to be
respectively 52° 18', 68° 15', and 67° 10' : find from these observations the height
of the tower above the horizon.
Let A, B, C, be the three stations at which the
elevations were observed, and FE be perpendicular
to the horizontal plane. Put AB = a, BC = c,
and denote the elevations FAE, FBE, FCE, at
A. B, C, by a, (i, y, respectively. Then since
EBA + EBC = n, cos EBA + cos EBC = 0.
But cos EBC =
cos EBA =
EB- -I- BC^" - CE2
2EB.BC
+ BA" — AE2
EB=
, and
aKR UA ' which being inserted in the preceding
equation, we shall have BC.AE^ + AB.CE' — AC.BE* = AC.CB.BA.
I'ut FE = X, then AE = x cot a, BE = x cot /3, CE = x coX y. hence
cr cot'a — (a -|- c) ar cot*f3 -|- ax^ cot-y =i ac {a + c), or again,
x^{c (cot'a — cot^^) -I- a (cot'-'y — cot"/3)] =z ac {a + c). But
cot'a - cot^^ = (cot a + cot ^) (cot a - cot /3) = ^Jl^Stt^l^^^Hfl
' '^^ Sin^a sin2/3
CO.', - C0../3 = (cot , + cot « (cot, - cot m = ""'^ij^;:;.^"''
PROBLEMS ON HEIGHTS AND DISTANCES.
459
Hence the coefBcient of x* may be written in the following form.
a sin 03 + 7) sin (j3 — 7)
1 +
c sin (j3 + a) sin (j3 — a) sin'y
— a) sm'y "i
— y) sin'a j '
sin 2/3 sin *y (." ' a sin 03 + 7) sin (/3
rr 1 w *u- c J * "fl c sin 03 + a) sin (/3 — a) sin '7 .
To calculate this, find tan-0 = — r-^^— — [-^- ,r> [—■-', '• then
a sm (/3 + 7) sm (/3 — 7) sm *a
„ c(a + c) sin^ /3 sin^ 7 cos^ 0
~ sin (/3 + 7) sin 03 — 7)
In the particular example we have the elementary data as follows : —
a = 67° 10'
/3 = 68 15
18
-« = Ifi = -9375
a 16
(c + a)e= 11625 7 = 52
log -9375 = 1-9719713
sin 135° 25' = 9'8463036
sin 15= 8-2766136
cosec 120 33 = 10-0649031
cosec 15 57 = 10-5609858
/3 + a = 135=
13 — a= 1
25'
5
/3 + 7 = 120° 33'
/3 — 7= 15 57
187207774
93603887
sin 52 18 = 9"8982992
cosec 67 10 = 100354398
tan 9 = 92941277
log 11625= 4-0653930
cosec 120° 33'= 10-0649031
cosec 15 57 = 10-5609858
2 4-6912819
sin 68 15 :=
sin 52 18 =
cos 9 =
logFE =
or, FE =
2-3456410
9-9679267
9-8932992
99917448
2-2036117
159-8129.
5. From the following observations made at the
extremities of a known base AB, it is required to
find the distance between two inaccessible objects
H, M: viz. AB = 900 625 yds, and the angles
HAB, MAB, MBA, HBA, respectively 93° 10' 15",
61° 15' 25", 84° 15' 30", and 52° 10' 50".
Denote the given quantities as follows :
A P _ I HAB=a I MAB=a, I AMH=9 I HAM=a— a, 1 -,_ , ^
AB _ a I jiBA=/3 I MBA=)3. | BHM=w | HBM=/3i-/3 j ''-"•"•"^ '^•
Then, th. 14, p. 450,
sin 9 sin 8 sin (o — a,) sin (a, + 0i) i • , , a\
-■ = • ■ ,, ' ■ / I ox = * sill («» + /3),
sm w sm a, sm 03i — p) sm (a + /3)
or sin 0 = A sin (a, + j8) sin w.
But sin0 = sin (a, + fl — ui) ■=. sin (a, + /3) cos w — cos (a, + S) si" ".
Equate these values of sines: then cotw =: A + cot (a, + i3), from which w
becomes known, and thence also 6 =: a, + /3 — w can be found.
FinaUy, HM = M^i° (/3. - ^) ^ a^iuo^n (3. -^^ ^^
■' sm w sm (a, + Pi) sm w
„- . HA sin (a — a,) a sin /3 sin (o — oi)
sm 9 sm (a + |3) sm 9
from either of which equations HM may be computed.
Scholium.
Had the distance HM been known, and the ground intervening between A
and B been inconvenient to measure, this might have been found with equal
ease, since the formulae of relation would in both cases have been the same, and
we should only be required to resolve the final formulae for AB instead of HM,
4.go PLANE TRIGONOMETRY.
. HMsuKgdljglliJIii^ = HM sin (g + /3) sin 9^ -n, roblemdoes,
giving AB - -^JiTa, sin (i3,-/3) sin /3 sin {a - a.) ^
in fact, appear more frequently under this form in actual surveys, than under
that given in the question : whilst the ordinary mode of solution becomes ex-
ceedingly inconvenient under these circumstances.
a = 93° 10' 15"
/3 = 52 10 50
a, = 61
/3,= 84
sin
sin 31
sin 145
cosec 61
cosec 32
cosec 145
cosec 113
15 25
15 30
52° 10' 50"=
54 50 =
a— ai= 31° 54' 50"
/3,— /3 =32 4 40
a + /3 = 145 21 5
a,+ /3, = 145 30 55
9-89759S0
97231634
30 55 = 97529595
15 25 = 10'0571069
4 40 = 10-2748483
21 5 = 10-2452375
26 15 = 10-0373965
\ogh= 1-9883101
a, + /3 = 113° 26' 15"
cot (ai + /3) = — tan 23° 26' 15"
a = 900-625.
h = -9734420
— tan 23° 26' 15" = — "4335159
cot w =
or w =
cosec 61° 38' 3" =
cosec 145 30 55 =
sin 61 15 25 =
sin 32 4 40 =
log 900-625 =
logHM
Hence the distance HM is 841-7416 yds.
•5399261
61° 38' 3"
10-0555509
10-2470405
99428931
97251517
2-9545441
2-9251803
6. Three objects A, B, C, whose relative positions to each other are known,
are observed from a point S to subtend the angles ASB, BSC, equal respectively
to y and a : and it is required to find the distance of S from each of the three
objects A, B, C.
"Whichever three parts of the triangle ABC are given so as to render it deter-
minate, the other three can be found, and hence we may consider all the parts as
actually known. According to the position of the point S with respect to the
triangle, several different cases will arise, as in the annexed figures : but the
same general investigation serves for all of them, and only requiring a slight
modification suggested by the several figures in the form of the results. The
following investigation is immediately applicable to the second figure.
Put BAS = u ; SCB = 0 ; a,b, c, and A, B, C, the sides and angles of the
given triangle ; and B — {a+ y) =S. Then, th. 14, p. 450, we have
sin 0 sin C sin a csina . csinasinw
= . . . = . ,orsin0= : .
sm w smAsmy a sin y a sin y
Also, w-|-0-f-n+y-fA+C=7r, or0 = B— (o-l-y) — <o = S ~ u, or
■in 0 = sin (o — w) = sin S cos w — cos £ sin w. Then equating the two values
of sill 0, and reducing, we readily obtain,
c sin a cosec S ,
cot w = : + cot S.
a 8in y
PROBLEMS ON HEIGHTS AND DISTANCES.
4^1
From this w becomes known, and thence 6 = 1 — u, and thence again SBA ^
V — y — «o, and SBC.= tt — a — 9; and the solution is completed by the
first case of plane triangles.
The only alteration under different circumstances, will be in the value of S.
However, for the purposes of calculation, this formula may be a little varied :
for we have,
, . c sin a cosec 5 . , f, ,' csina )
cot w = cot 5 H . = cot ^ \l H : A.
a sm 7 (. a sm y cos S)
be +, find tan*x =
If
a sin y cos S
c sin a
a sin y cos S '
c sin a
and
if - T """ " , be — , find cos^* = —
a sm y cos S a sm y cos S
Then in the former case we have cot w ^ cot ^ sec x,
and in the latter cot w =: cot S sin 0.
The forms for the work, with blank spaces for the numbers, are added here.
c sin a cosec S . . ^ i . « f , c sin a ")
cot W
logc
ac log a
sin a
=
h c
asmy
cosec S
cosec y
log A
hence h
found by addition.
n cot 5
C0t(
logc
ac log a
sin a
sec^
cosec y
X or CCS f
X or sin (p
cot c
log cot ftJ
hence w
2
I
a sm y cos 0 ,
founc
by addition.
tan
spr
found
o
by addition.
n cot m = found by addition,
hence o) := ° ' '
As numerical examples, take questions 26 and 27, of the following unsolved
questions, the solutions being completed by Case 1, Plane Trigonometry.
7. Given the angles of elevation of the stmmit of a hill or building, observed
from three positions whose distances from each other are knoicn, to find the height.
Let A, B, C, be the three stations, and D the top of the
hill or building ; and draw DE perpendicular to the plane
ABC, and join AE, BE, CE, AD, BD, CD. Denote the
sides of the triangle ABC as usual by a, b, c ; the angles
of elevation of D from A, B, C, by a, /3, y ; the distances
AE, BE, CE, by x, y, z, and the perpendicular DE
by «.
Then a? = u cot a = mu, y = u cot j3 := n«, r = « cot y = pu.
cosAEB =
_x' + y^ — c' _(m^ + n^)u^ — (^
-f
cos BEC =
cosCEA =
2xy
-\- z" — a-
2mnu'
(n« +p^u? — a^
2yz Inpu^
+ j^ — 6^ _ (jo-+w»^M-— &"
2xz 2pmu^
(1)
(2)
(3)
4452 PLANE TRIGONOMETRY.
But AEB + BEG -f CEA = 27r, and hence we obtain the relation*
1 _ cos'AEB — cos^BEC — cos^CEA + 2 cos AEB cos BEG cos CEA = 0 :
and, inserting in this, the preceding values of these cosines, we get the equation
J + („5_/) [n^—m^)bnu* — ] +im-+p-) a-c-—n-b* u2+a=6V=0 ... (4)
[ + (p^-m-)(p'-Tv)c") \.+(n'-+p^b^c'-m^a*)
which is a quadratic equation in terms of u-, and hence the solution is analy-
tically effected.
The coefficients of this equation may be reduced in different ways, of which
two are here annexed.
1. In the coefficient of u* write for c^ its value, a- — 2ab cos C + ¥, and in
that of U-, write for a^ b^, c^, their values in corresponding forms. Then the
equation is reduced to
f (m^ -fy a^ +y -pr- bl\u*-2abc\ +^b cos b]u^ + a^iV = 0.
|_ 2ab (m- - /) (n^ _ ;,-) cos G j ( [j] ^,^ ^^^ (. j
The former of these is the square of the side Ci of the triangle, whose opposite
angle is C, and its including sides are (m' — p-^ a, and (n^ — p^) b respectively.
This, therefore, can be easily computed by the second case of plane triangles.
The terms of the second coefficient can also be computed by subsidiary angles.
2. The equation may be easily reduced to the form below :
Put nh := ma, and nk = pc : then, since y:^nu = u cot /3, the equation becomes
Assume now the relation between h and k by means of the computable angle
6, from the equation k- = h^ — 2bh cos0 -f- b-, and for e^ put its value
a' — 2ab cos C + 6' : then the coefficient of y* is transformed into
^{{0" — K^f — 4aA (o2 + h^) cos G cos 0 + ^a^¥ (cos^G + cos 20)], or into
b'^{{a- — 2ah cos G cos 0 + A-)^ — (2aA sin G sin 6f\.
In the same manner the coefficient of y^ is changed into
— 2a"b-cr ^a^ — 2ah cos G cos 0 + h-l.
Hence, transposing {2ah sin G sin 0)-, the equation is transformed into
\-2aAcosGcos0l J'-2«-'^l-2aAcosGcos0ly+°^ = l2«AsinGsin0| y*
both sides of which are complete squares : and extracting the root, we get
Ja» — 2ah cos G cos 0 + F] j/= — a» c^ = + 2ah sin G sin 0 y», or
\a^ — 2ah (cos G cos 0 + sin G sin 0) + h^y^ = a^(^, or again,
[a^ — 2ah cos (C qf 0) + h^y- = a^c^.
Hence « = y Un /3 = --=^^J^J^^
, - i^,^ =- ; the double sign prefixed
Va* — 2ah cos (C + 0) + A-
to the numerator indicating two inverted symmetrical positions of the point D,
For. if «^ -I- fl 4. X = 2ir, we have cos (<^ + 0) = cos (2ir — x), or again, cos tpcosd —
eof X — tin .^ kin 6; or squaring both sides, and writing 1 — cos»<^ and 1 — cos»e for sin*<^
»nd »in'e, wc obtain, after slight reduction, the formula in question. The same form of result
•J» tn%et from roi (<^ -|- 6) = cos x, as will be seen from performing the operation : and hence
tb« tk«ortm U true, whether the point E be within or without the triangle ABC.
PROBLEMS ON HEIGHTS AND DISTANCES. 4<;3
one above and the other below the horizon of the places of observation : and the
double sign in the denominator indicating two different positions of the point E.
It may be shown, that one point will lie within the triangle ABC, and the other
without it, and thus all ambiguity will be removed from the solution *.
Further examples for exercise.
8. From the edge of a ditch, of 36 feet wide, surrounding a fort, the angle of
elevation of the top of the wall was found to be 62° 40' : required the height of
the wall, and the length of a ladder to reach from my station to the top of it.
Ans. height of wall = 69649272, ladder = 78-402942ft.
9. Required the length of a shoar, which strutting lift, from the upright of
a building, will support a jamb 23ft lOin from the ground. Ans. 26'249ft.
10. A ladder, 40ft long, can be so placed, that it shall reach a window 33ft
from the ground, on one side of the street; and by turning it over without
moving the foot out of its place, it will do the same by a window 21 ft high, on
the other side : required the breadth of the street. Ans. 566493981ft.
1 1 . A maypole, whose top was broken off by a blast of wind, struck the
ground 15ft from the foot of the pole : what was the height of the whole may-
pole, the broken piece measuring 39ft in length ? Ans. 75ft.
12. At 170ft distance from the bottom of a tower, the angle of its elevation
was found to be 52° 30' : required the altitude of the tower. Ans. 221"55ft.
13. From the top of a tower, by the sea-side, of 143ft high, it was observed
that the angle of depression of a ship's bottom, then at anchor, was 35° : what
was the ship's distance from the bottom of the wall? Ans. 204'227lft.
14. What is the perpendicular height of a hill ; its angle of elevation, taken
at the bottom of it, being 46°, and 200yds farther off, on a level with the
bottom, the angle being 31° ? Ans. 286.2906yds.
15. Wanting to know the height of an inaccessible tower: at the least dis-
tance from it, on the same horizontal plane, I took its angle of elevation equal
to 58° ; then going 300ft directly from it, found the angle there to be only
32° : required its height, and my distance from it at the first station.
Ans. height = 307-5456, distance = 192-162.
16. Being on a horizontal plane, and wanting to know the height of a tower
placed on the top of an inaccessible hill, I took the angle of elevation of the
top of the hill 40°, and of the top of the tower 51°; then measuring in a line
directly from it to the distance of 200ft, I found the elevation to the top of the
tower to be 33° 45' : what is the height of the tower ? Ans. 93-33149ft.
* The fourth problem may be included in this, viz. ■when the three stations are in one line. For
then b zz a -{- c and C = ir ; whence 6* — c* — o* i= 2ac, c' — a* — 6* = — 2ab, and
a' — b^ — c* := — 2I)C : and the fundamental equation becomes a complete square, and equiva-
lent to {a cot^ — {a -\- c) cot'^ -f- c cot'y | u'^ = ac {a -\- c), as found at p. 459.
Again, if the distances a, e, be equal, -we get Jcot'a — 2 cot'/3 -|- cot*7 J m' = 2a'*, either by
substituting in the last, or in the fundamental, equation.
The following is the process indicated by the investigation for the solution of the problem.
1. Find A and k from Ana cot a tan /3, and k -=.0 cot y tan /3.
2. Find C and e from c* = a» — 2a6 cos C + t', and /t» = A« — 2ah cos 6 -{- a».
3. Find c, from c,^ = a» — 2 aA cos (C T 6) -j- A» ; and finally
4. Find » from u = ± -.
c,.
Corresponding steps will be applicable to the case of the three stations in one line, though the
solution will not be simpler than that already given, p. 459.
4^ PLANE TRIGONOMETRY.
17. From a window near the bottom of a house, which was on a level with
the bottom of a steeple, I observed the angle of elevation of the top of the steeple
to be 40°; then from another window, I8ft above the former, the elevation was
37° 30' : required the height and distance of the steeple, and a general formula
of solution. Ans. height = 210-436, distance = 250792 .
1 8. Wanting to know the height of, and my distance from, an object on the
other side of a river, which was on a level with the place where I stood, close to
the side of the river ; and not ha\ang room to measure backward, in the same
line, because of the immediate rise of the bank, I placed a mark where I stood,
and measured, in a direction from the object, up the ascending ground, to the
distance of 264ft, where it was evident that I was above the level of the top of
the object ; there the angles of depression were found to be, viz. of the mark
left at the river's side 42°, of the bottom of the object 27°, and of its top 19°
Required the height of the object, and the distance of the mark from its bottom.
Ans. height = 57-2734, distance = 150-5058.
19. If the height of the Peak of Teneriflfe be 2^ miles, and the angle taken
at the top of it, as formed between a plumb-line and a line conceived to touch
the earth in the horizon, or farthest visible point, be 88° 2' ; it is required from
these measures to determine the magnitude of the whole earth, and the utmost
distance that can be seen on its surface from the top of the mountain, supposing
the form of the earth to be perfectly spherical.
Ans. greatest visible dist. = 135-943, diam. = 79l7"85 miles.
20. Two ships of war, intending to cannonade a fort, are, by the shallowness
of the water, kept so far from it, that they suspect their guns cannot reach it
with effect. In order, therefore, to ascertain their distance, they separate from
each other a quarter of a mile, or 440 yds ; then each ship observes the angle
which the other ship and the fort subtends, which angles are 83° 45' and 85° 15' :
what is the distance between each ship and the fort ?
Ans. 2292-266 and 2298-051yds respectively.
21. "Wanting to know the breadth of a river, I measured a base of 500yds
in a straight line close by one side of it ; and at each end of this line I found the
angles subtended by the other end and a tree, close to the bank on the other
side of the river, to be 53° and 79° 12': what was the perpendicular breadth of
the river ? Ans. 529"4847yds.
22. Wanting to know the extent of a piece of water, or distance between two
headlands, I measured from each of them to a certain point inland, and found
the two distances to be respectively 735 and 840 yds ; also the horizontal angle
subtended between these two lines was 55° 40' : what was the distance of the two
headlands? Ans. 741-2085yds.
23. A point of land was observed, by a ship at sea, to bear east-by-south ;
and after sailing north-east 1 2 miles, it was found to bear south-east-by-east :
it is required to determine the position of that headland, and the ship's distance
from it at the last observation. Ans. 2607282 miles.
24. Wanting to know the distance between a house and a mill, which were
seen at a distance on the other side of a river, I measured a base line along the
Bide where I was, of 600 yds, and at each end of it took the angles subtended by
the other end and the house and mill, which were as follow, viz. at one end the
annles were 58° 20' and 95° 20', and at the other end the* corresponding angles
were 5.1° 30' and 98° 45' : what was their distance ? Ans. 959 604yds.
25. Wanting to know my distance from an inaccessible object O, on the
other side of a river, and having only a chain for measuring distances,
I chose two stations, A and B, 500yds asunder, and measured in the direction
HEIGHTS AND DISTANCES. 465
from the object O, the lines AC and BD each equal to 100 yds; also the diago-
nals AD.BC equal to 550, 560 yds respectively: what was the distance of the
object O from each station A and B ? Ans. AO = 536 441, BO = 500237.
26. In a besieged garrison are three remarkable objects, A, B, C, the dis-
tances of which from each other are discovered by means of a map of the place,
to be as follow, AB = 266i, AC = 530, and BC = 327^ yds. Now, having
to erect a battery against it, at a certain spot without the place, and it
being necessary to know whether my distances from the three objects be such,
as that they may from thence be battered with effect, I observed the horizontal
angles subtended by these objects from the station S, and found them to
be ASB = 13° 30', and BSC = 29° 50'. Required the three distances, SA,
SB, SC; the object B being situated nearest to me, and between the two
others A and C. Ans. SA = 757-1407, SB = 537-1028, SC = 654-0996.
27. -Required the distances as in the last example, when the object B is
the farthest from ray station, but still seen between the two others as to
angular position ; and those angles being ASB = 33° 45', and BSC = 22° 30',
aTso the three distances, AB = 600, AC = 800, BC = 400 yds respectively.
Ans. SA = 710-195, SB = 1042*545, SC = 934-29.
28. U CB (Jig. I. p. 422.) represent a portion of the earth's surface, and C
the point where a levelling instrument is placed, then DG will be the differ-
ence between the true and the apparent level ; and it is required to show that,
for distances not exceeding 5 or 6 miles measured on the earth's surface, DG,
estimated in feet, is nearly equal to § CD^, taken in miles.
29. On the opposite bank of a river to that on which I stood, is a tower,
known to be 216 feet high, and with a pocket sextant I ascertained the vertical
angle subtended by the tower's height to be 47° 56'. Required the distance,
across the river, from the place where I stood, to the bottom of the tower ; sup-
posing my eye to be 5 feet above the horizontal plane which passes through it.
Ans. 200-22ft.
30. In the valley of Chamouni three positions. A, B, C, were selected, in a
straight horizontal line, such that AB = 80, and BC = 75yd3. Three remark-
able points. A', B', C, on the side of the Jura, were also chosen to be observed.
The angles of elevation of A', as seen from A, B, and C, were 67° 10', 6S° 15',
and 52° 18'; those of B', as seen from the same points, were 72° 18', 78° 15',
and 70° 10'; and finally, those of C were 60° 5', 61° 10, and 58° 5' respectively.
It is required from these observations to find the heights of A', B', C, from the
horizon of the line of observation.
Ans. A'A" = 159-8134, B'B" = 286-3938, C'C' = 323-38G0.
31. (II. 1, p. 48.) An obstacle prevented my measuring the part BC of a line
AD, and a point E was selected from which the angles subtended by the seg-
ments AB, BC, CD, were a, /3, y, respectively, and the two accessible segments
AB and CD were found to be a and c respectively : from which data it is
required to find the length of the line AD.
A Tjr- ■ t At . I ^ . I N ac sin (a -I- /3) sin (/3 -H y)
Ans. BC = X, 18 found from (x -|- o) (a: + c) = ^^ — -. — ^~. — — .
sm a sia y
32. (II. 2.) Being on the opposite side of a river from two steeples O and W,
which I knew from a previous sur\'ey to be at the distance of 6954 yds, and
wishing to know the distance between two other objects on the side on which I
stood, but which the irregularity of the ground prevented my measuring, I took
the following horizontal angles, OAW = 85° 46', BAW = 23° 56', OBW =
31° 48', and OBA = 6b° 2'. What was the length of AB ?
VOL. I. H h
4^ PLANE TRIGONOMETRY.
33. (11. 4.) A person ■vvalking from C to D on a straight horizontal road,
can see a tower on the summit of the hill A at every point except E, where he
can just see the top of the tower over the hill B. He then measures a base EC
of 150 yds, and at C observes ihe elevation of A to be 5<j° 18' 15'' ; and. he also
finds that ACB = 10° 12' 20", ACE = 69° 18' 30", and AEC = 108° 12' 15".
From these observations, the horizontal distance of the hills from each other,
and from the places of observation, C and E are to be found.
34. (II. 5.) From three positions A, B, C, in the same horizontal plane
whose distances were AB = 150-25, BC = 179C9, AC = 20536, the eleva-
tions of the top of a tower on a hill were obser^-ed to be 6° 10' 55", 7° 18' 3", and
6° 58' 58" respectively : whilst the elevation of the bottom of the tower from A
was 6° 2' 58": and from these data the height of the tower is required.
35. (II. 6.) Four points A, B, C, D, are accessible, and three M, N, P,
inaccessible, but are to be found from the following observations :
AB = 815
BC = 670
CD = C60
ABC = 49° 54'
BCD=: 73 57
AMB = 80° 8'
CNP = 98° 44
BMN = 24 55
CRN = 29 13
CxNM = 124 16
CPD = 51 19
36. A tree growing on the side of a hill which rises due north at an angle of
30°, had the upper part blown off 12 feet from the ground by a gale from W.S.W:
now supposing the tree to stand perpendicularly to the horizon, and the top
(before the other part was wholly separated from the tree) to strike the ground
40 feet from the bottom, what was its original height ? Ans, 51 •204ft.
37. Passing along a straight and level road, near a very lofty tower on the
same horizontal plane with the road, I wished to know its height : but having
no instrument for taking othpr than vertical angles, I proceeded thus : at a con-
venient point (A) on the road, I observed the angle of elevation of the top of the
tower to be 30° 40', and 60 yds farther on the road the elevation was found to be
40° 33'; at the end of another 60 yards, I was prevented, by a high wall, from
taking the elevation, and therefore I measured 12 yds still farther, and found the
angle of elevation to be 50° 23'. From these data it is required to find the height
of the tower, and its horizontal distance from each of the stations.
Ans. the height = 94-835, and the distances 159-087, 110-8414,
and 78507, from the stations.
38. A person in a balloon observed the angle between two places A, B, bear-
ing N. and S. of each other, and a miles apart to be a°, and from B, which bore
due east of him, to a point directly under him, to be /3° : show that his altitude
is expressed by a cot a° cos/3°.
39. A tower a feet high stands in the centre of a field whose form is an equi-
lateral triangle, and each side subtends an angle of 2a : find the side of the field.
40. Walking along a horizontal road I observed the elevation of a tower to
be 20°, and the angular distance of the top of the tower and an object on the
road to be 30° ; also the nearest distance of the tower from the road was 2<)0ft :
find its height. Ans. 187-57534.
41. From a station A, the angle subtended by two objects B and C was 2a,
and at B and C the angles subtended by A and a fourth point D were right
angles ; also the distances AB and AC were b and c : show that if 29 be the dif-
ference of the angles BAD, DAC, then tan 9 = 5-^^ tan a, and the distance
AU = 6 sec {a — e)-=ic sec (a + 0).
42. AB is an obelisk on a hill BHG, and there is no horizontal ground in
front of it : on the opposite hill, I therefore measured l60ft in the same vertical
HEIGHTS AND DISTANCES. 467
plane with the castle from C upwards to E : the elevations of A and B from C
were 47° 27' and 45° 17', and that of A from E was 46° 20', whilst the inclina-
tion of CE to the horizon was 10° 10'. Find the height of the obelisk.
Ans. 367-851ft.
43. From a point on a level with the bottom of a flagstaflf its elevation was
23° 8' 15", and from another point 18 296 1ft higher, the angle subtended by
the flag-staff was 23° 15' : required its height and distance.
44. A person on the mast-head at S, 105*6ft above the level of the sea, just
sees over the earth's surface at P the top of a cliff T known to be 660 ft high :
tiow far was the ship from the cliff, the earth's diameter being 7800 miles ?
Ans. 35"05Imiles. •
45. A church O is to be built for the accommodation of three villages A, B, C,
ivhose distances asunder are BC = 2 26, CA = ri4, and AB = 1-58 miles:
3ut as they contribute unequally to the expense, their distances are to be to one
mother in the ratio AO : BO : CO :: 5 : 12 : 9: what is its distance from
iach ? Ans. If O be tcithin ABC, AO = -54699, BO = 1-31278, CO = -98459,
wilhout A0 = -99486. BO = 2-38768, CO = 1 79076.
46. Three stations of the trigonometrical survey of Britain can be seen from the
Eddystone lighthouse, viz. Kit's Hill, Carraton Hill, and Butterton Hill, (which
lenote by A, B, C respectively, arid the lighthouse by D) : and it appears from
;he survey that AB = 33427ft, BC = 131576ft, and CA = 100969ft; and
ikewise that from D, BC subtended an angle of 64° 1' 48", AC an angle of
18° 45' 53", and AB an angle of 15° 15' 55"; from which it is required to deter-
nine the distance of the Eddystone from each of the stations.
Ans. AD= 123411ft, BD = 126896ft, and CD = 121123ft.
47. In the French trigonometrical survey, three stations, Villers Bretonneux
A), Vignacourt (B), and Bourdon (C), and a station (D) within the triangle, were
aken, and the following data obtained: viz. logBC = 42734544, ABC =:
l9°4'13",ACB = 3I°49'57"-8, ADC = 130°44'l6"-5,andADB = 60°31'53"'8:
0 find the distances of D from each of the stations.
Ans. AD = 8064 61, BD=: 11124-25, CD = 7733-49.
48. In the trigonometrical survey of Scotland the three stations High Pike
A), Cross-fell (B), and Crif-fell (C) were observed to subtend angles from Hel-
rellin (D) as follows ; BDC = 100° 17' 45''-25, ADC = 32° 39' 57' 25, BDA =
i7° 37' 48" ; whilst the previously determined distances of the three stations
vere, BC = 255886- ift, AC = 147733-5ft, and AB = 120904 9ft : it is required
0 find their respective distances from Helvellin.
Ans. AD = 65724-6, BD = 129531-8, and CD = 198738-6ft.
49. At the commencement of the trigonometrical survey, a base line BC of
17404 2 ft was measured on Hounslow Heath, between Hampton Poorhouse
C) and King's Arbour (B) ; and from both these stations. Hanger- Hill Tower
A) and St. Ann's Hill A' (on opposite sides of BC) were visible, and the fol-
owing angles were observed : ABC = 70° 1' 47", ACB = 67° 55' 39", A'CB =
il°26'35" 5, and A'BC = 74° 14' 35", from which to find the distance between
llanger Hill Tower and St. Ann's Hill. Ans. 68896ft.
50. To find the distance of Inchkeith Lighthouse (A) and the spire of North
jeilh Church (A*), the following observations were made at the Edinburgh
Dbservatory (C) and Beincleuch (B) ; viz. BC = 146314 ft. BCA = 73° 16' 28"-5,
3CA' = 55° 38' 41"-1, CBA = 11° 53' 56", CBA' = 2° 34' 2 '"2 ; to find AA',
)oth points being on the same side of BC. Ans. 23045-53ft.
Hh2
4^ PLANE TRIGONOMETRY.
XIX. MISCELLANEOUS EXAMPLES FOR EXERCISES ON TRIANGLES.
1. (I. 7.) Let a, b, c denote the sides of a plane triangle, to find C, when
t^ = a" ± ab + b\ c^ = a^ ± iab + b\ and when (^ = a' ± - ab + b\
2. (IL 6.) Tlie three sides of a plane triangle are three consecutive terms in
the series of integers, and the greatest angle is double the least : find the sides
and angles.
3. (H. 14.) How must three trees A, B, C be planted so that the angle A may
be double of the angle B, and the angle B double the angle C ; and that a line
of 400 yards may go round them ?
4. (H. 15.) The sines of the three angles of a triangle are as the numbers 17,
15, 8, and the perimeter is I60 ; and it is required to find the sides, the angles,
the perpendiculars, and the lines bisecting the sides.
5. (11. 16 ) The logs of two sides are ■2-2407293 and 2-53~S191, and the in-
cluded angle is 37° 20' 1". Determine the other side without finding the angles.
6. (IL 17.) The sides of a triangle are to one another as the fractions 3, ^, 3:
what are the angles ? and under what angles do the lines from the angles to
the middles of the sides intersect ?
7. (IL 24.) If 2A denote the product of the two sides of a triangle about the
right angle C, then it is required to show that
b = x^2A cot A, a = v^2A tan A, and c =1 2 \/a cosec 2A.
8. (II. 26, 2.) If c be the hypothenuse of a right-angled triangle, then
tab tan ^ A = 10 — ^ Jlog (c + 6) — log (c — 6)|.
9. (IL 27.) Two lines ff and h are given, and a triangle, whose sides a, b, e
are respectively the arithmetical, geometrical, and harmonical means between
them, is to be constructed : find its angles, and exemplify it when g = 4A.
10. (IL 28.) From the vertex A of a triangle a line is drawn to cut the base
a in segments which have the triplicate ratio of 6 to c : find the angles which it
makes with its three sides, and show what ratio between b and c will cause the
line so drawn to make a right angle with a.
11. (II. 29.) Given the radius of the inscribed circle and the angles at the
base of a plane triangle, to find the sides and the radius of the circumscribing
circle.
12 (II. 30.) If R, r, be the radii of the circumscribed and inscribed circles,
and d the distance of their centres, then d = ^/R2 — 2 Rr"; and, if the four
circles be described, which touch the three sides of the triangle, and rf„ d^, d^,
be the distances of their centres from the centre of the circumscribing circle, then
d' + dr + d,- + d,' = 12R2.
13. (II 31.) If a right pyramid SABCD on a square base ABCD be cut by
any plane, and a, b, c, d be the distances from the vertex at which the edges
taken in order are cut by the plane : then it is required to prove that
1 1 _ 1 1
a "^ c ~ 6 ■*■ d"
U. (II. 32.) Given the perimeter, 2*, the difference, A — B, of the angles at
the base, and the perpendicular /j, from C to c, to find the angles of the triangle.
15. (II. 33.) The four sides of a quadrilateral, a, b, c, d, inscriptible in a
circle, arc given, to find the diagonals, the angles under which they intersect,
and the radius of the circumscribing circle.
GENERAL PROPOSITIONS. 469
16. (II. 34.) Let A, B, C be the angles of a triangle, a, b, c its sides, and R, r
the radii of its circumscribing and inscribed circles, and f*„ r,, r^ the radii of the
^scribed circles, and p^, p,, p^ the perpendiculars from A, B, C on c, 6, c; then,
(1) tan A + tan B + tan C = tan A tan B tan C.
(2) cot A cot B + cot B cot C + cot C cot A = 1.
(3) sin A + sin B : sin C : : a + 6 : c.
(4) sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C.
(5) tan ^A tan ^B -f tan iB tan ^C + tan ^C tan iA = I.
(6) cot ^A + cot iB + cot iC = cot ^A cot ^B cot ^C.
(7) Find r, r„ r^, r^ in terms of R and A, B, C.
i^ sin C -|- c^ ^in B a^ sin C + c^ sin A a'^ sin B + i* sin A
(8) p, = J _^- ,p,= ^^-p^ ,p,= ^^ .
/«s 1,1.1 1,1,1
(9) - + — H =- -\ h — .
fi r, Tj p, p^ p3
17. (11. 35.) Given the angles A, B, C and circumscribing radius R, to find
the three lines from the angles to the middles of the opposite sides, and likewise
the three perpendiculars.
18. (II. 38.) Three of the angles of a quadrilateral figure circumscribing a
circle whose radius is 10, are 29° 15' 10", 87° 15' 12", and 105° 15' 18": what
are its sides ?
19. (II. 39.) Find the remaining parts of the triangles where are given :
(1) b,e, B + C I (3) A, a, 6 + c I (5) A, a, be
(2) B,C, b ±c I (4) A, c, a ± 6 | (6) A, B, a + * + c.
20. (II. 40.) Adopting the notation of Ex. 16, there are given,
(1) C, c, Pa to find A, B, and a, b.
(2) a + b + c, C and ^ab sin C to find c the remaining parts.
(3) a, b, c, to find the segments of c by a line bisecting C, the seg-
ments of C by a line bisecting c and p^, p^, pj.
(4) /7,, p.^, pj to find a, b, c and A, B, C.
(5) a, b, C to find 7)3, p^, and ;j,.
(C) a -\- b + c, B, and the equation b^ = ac to find A and C.
(7) a -\- b + c, r , - io find the sides, angles, and perpendiculars.
21. (II. 41.) If a, b, c, be the chords of three arcs which together form a
semicircle, what is the radius of it ?
22. (II. 45.) Prove that in any triangle :
1 + sec A ^ (1.) — sin'A
(1) tanC= ^
tan A — cosec A^ / ( t ) — sin^A
(2) 2a = ^c^ + 1p.^c cot § C + //c* — Ip^c tan \ C.
r, o. — c COS B
(3) cos C = ■ — .
V o* — lac cos B + c*
(4) cosC = I sin^B + cosBV^i -(^^ sin^B.
23. (II. 46.) If a, 0, and a„ /3i be any two angles of two triangles, then
sin j3, sinj8 \ ^_ ( sino^ sing 1 2^4jsin34(a^a)— 8in»|(/3i— ^ J
r(ar+/3,)~8in(a+/3)j lsin(a,+/3,) 8in(a+/3)j sin (a, + /S^ sin (a+^)
24. From the angles A, B, C of a triangle draw the perpendiculars to the
470
PLANE TRIGONOMETRY.
opposite sides, meeting ihem in D, E, F respectively, and form the triangle DEF:
then aDEF = 2 aABC . cos A cos B cos C.
25. Prove the following properties of a right-angled triangle, C being the
right-angle.
sinviA = '-^ sin 2A = ,^^, ; sin (45°±A) = -;;^^ sin (A-B)
tan'iA =
c-Jb
c+ b
cos 2A =
tan2A
2ab
b^'-Ta"
b- — a^
2ab
cos (4 5°+ A) =
b±a.
c\'-2
b + a'
c^/2
b + a
a^—b^
cos (A— B) =
c-
2ab
cr
a^-b^
^,^^1 tan(450±A)=^^ tan (A-B) = ^^^
26. In any isosceles triangle, C being the angle included by the equal sides :
cos A = -, vers C = -,, r = ^V ^^TT' "°^ ^ = Via^'^^^-
27. Prove that in any plane triangle the following equations are true : viz.
sin 4(A — B) = cos iC
cot ^A -f cotiC _
cos KA - B) =
sin (A - B) = '
28. Also, that,
tr = (a+b)-sin-iC + ia — b)Uos^C =
c
cot ^B + cot^C
a + b . ^^
sm iC
c
tan iA — tan 4 B
tan ^A + tan iB
a^-b- . „
s — 6in u
c-
cot JB + tan 4 A
cot |B — tan^A
b
a
a-b
c
a + b
c
(a 4- by- sin^iC - (a — b^ cos'^C
and that c ^
cos (A — B)
_ (0 -f c) tan ^B -f (a — c) cot JB _ a
2 tan i;A^B — ^C) ~ cos B + sin B cot C"
29. R and r being the radii of the circumscribed and inscribed circles, then,
s abc tan 4(A -f B — C)
sin A + sin B + sin C
r = 5 tan ^A tan 4B tan AC =
; and
a^ + b- — (^
abc (sin A + sin B + sin C)
45^
30. If Pj be the perpendicular from C upon c, it is required to prove that
2s a- sin B -\- b^ sin A c cos(A — B) — c cos(A + B)
^*^ cotiA + cot^B ~ a + b ~ 2sinC
and its distance from the middle of c will be expressed by the following :
b-) sin (A + B)
-<«rirt*f>
c sin (.\ — B) c tan A — tan B , (a-
2sinC 'itlmAT-Man^B'
c sin C
31. Given AB = 100, A = 60°, B = 45°, to find BC, CA without the aid of
any tables. Ans. BC = 50 s/6[s/3 — 1}, CA = 100 {^3 — I].
32. The two sides of a triangle are as 3 to 5, and one of the angles at the
base three times the other : what are the angles ?
Ans. 35° 15' 54", 105° 47' 42", and 38° 56' 24".
33. In a triangle, ABC, there are given A = 80°, a = 400, and b + c = 600 ;
to find the other parts.
34. If a = 3 + v/2, 6 = 3 — ^/2, and c = 4 : then sin C = ?v/10-
35. Show that in any plane triangle ? : * ~ ^ : : sin^^A : sin'^^B.
36. If lines be drawn from the angles of a triangle to a point within it so as
to make equal angles with each other, their sum will be equal to
Va^ — 2ab cos (A + 60°) + b'K
471
MENSURATION.
In the previous applications of algel)raic methods to geometry, whether in
their simple or trigometrical forms, our object has ultimately been to find either
a line or an angle from certain data furnished by the problem. The object of
that branch upon which we now enter is to find the lengths of curve lines, are
the areas or volumes of given superficial or solid figures. To cflTect this purpose
to any great extent requires the use of the differential and integral calculus :
but in the few cases which will here be considered, it can be accomplished with-
out that end.
All lines are transformed either entirely or approximately into straight lines
expressed in terms of the linear unit; all surfaces into rectangles whose adja-
cent sides are expressed in terms of the same linear unit ; and all solids into
parallelopipedons, whose three adjacent edges are expressed also in the same
terms. See Application of Algebra to Geometry, p. 413.
In this work the general investigations will be first given altogether for figures
which belong to the same class; and then examples to each in precisely the
same order as in the previous edition.
I. THE AREAS AND LENGTHS OF PLANE FIGURES.
1. A rectangle whose sides are given.
Let AB, AD be two adjacent sides of the rectangle, and
AG, AE each equal to the unit of the scale by which the
sides are measured. Let a and b be the number of times to
which the sides AB, AD respectively contain AG or AE.
Complete the square AEFG, and this will be the superficial unit by which AC is
measured, or calculated.
Now the parallelograms AF, AC being equiangular, they are to one another
in a ratio compounded of the ratios of their sides. That is,
par" AC : par" AF : : AB.AD : AG.AE ; or alternately
par" AC : AB.AD : : par- AF : AG.AE.
But par" AF = AG.AE = 1, and hence par" AC = AB.AD = ab.
The area is, hence, expressed by the product of any two adjacent sides.
2. mien two adjacent sides of a parallelogram and their included angle are given.
Let AB, AD, and the angle A be given. Draw
the perpendiculars AE, BG, DF. Then the paral-
lelogram AC is equal to the rectangle AF. Whence
parallelogram AC = AD.AE = AD.BG = AD.AB
sin A = ab sin A, where a and b denote AD and AB
respectively.
Hence the area is expressed by the continued product of the two sides and
the sine of their included angle.
Cor. 1 . It is obvious from what is here shown, that when the one side and its
distance from the opposite side are given, the area is their product, viz. AD.BG.
Cor. 2. When the base and perpendicular of a triangle are given, the area
is half their product : for the triangle ABD is half the parallelogram AC.
472
MENSURATION.
Cor. 3. When two sides and the included angle of a triangle are given, as
AB, AD, and A, we shall have, since trian ABD = ^ parallelogram AC,
ABD = iAB.ADsinA.
Cor. 4. When in the triangle ABC, the angle C = ^, we have area=: Ja6*.
3. IVhen two of the opposite sides are parallel but not equal (the trapezoid), and
there are given those two sides with one of the other sides, and its inclination to
either of the parallels.
Let ABCD be the trapezoid, the sides AD, BC which are n ^ „
parallel, being given, as likewise the side AB and angle A.
Draw CH parallel to AB, bisect HD in G and draw GF
parallel to AB. Then GD = GH = CF. Hence the trian-
gles GKD, CKF, having one side GD equal to one side CF,
and the angles at the extremities of these sides equal each to each, the triangles
are equal. Consequently adding ABCKG to each, the trapezoid ABCD is equal
to the parallelogram ABFG. Also BF + AG = 2AG = BC -|- AD, or AG
= i(.\D + BC), that is, to half the sum of the parallel sides.
Whence, ABCD = ABFG = AB.AG sin A = iAB (BC + AD) sin A.
Cor. 1 . If the two opposite sides AD, BC and breadth BE be given, then the
area is pE (AD + BC).
4. When the two diagonals of a trapezium and their angle of intersection are given.
Let ABCD be the quadrilateral figure or trapezium,
of which the diagonals AC, BD, and their angle K of
intersection are given.
Through A, B, C, D draw lines parallel to the
diagonals : then they will together form a parallelo-
gram, EFGH whose angles are equal to the angles at
K ; and whose area is double that of the trapezium ABCD.
But EFGH = HE.HG sin H = BD.AC sin K ; whence', finally, we hare
ABCD = iBD.AC sin K.
5. When one diagonal and perpendiculars from the other angle are given.
Let ABCD be the trapezium, AC the given diagonals, and BE, DF the given
perpendiculars from the other angles B and D upon AC.
'The following, amongst many other expressions that micht be given, for the area of a triangle
■n t.m.s of a.fTerent parts of the figure, possess an analytical interest to the inquiring mind.
1 ticir invcMigation will form good exercises for the student
l<i/> till C
■ , »iii^H sinC
*" Mir.\"~
, , "in lUiiiC
K* — (t) un JA
Ejprcsmmtfor the area of the triangU.
•J\^s — a)bc\ sin i A
^\(s — b){s~i)bc} cos J-V
iV^'''''<■'sin A sin B sin C I
2aAc
iT^-t-f^ cosiAcosiBcosJC
|(a2-)-t»_c»)sinC
4ian^(A-|-B — C)
4(cot A -f col B -(- cot C)
abe'
■2(a^ — 6^)
^siu(A— B)
Ka*_ti) «Jl.^j!iJl 1 ^ y4(a< -I- &« -f r«) — («» -|- i»4. (.8)8 |
•»nU — B> \~AmJ cotU-l-cot^B-l-cot'C 's/ls{s—a){s—h)(s—c)\
RECTILINEAR FIGURES. 473
Through B, D, draw GH, LK parallel to AC, and
through A, C draw GL, HK perpendicular to AC and
therefore parallel to BE and DF. Then the rectangle
LH is double the trapezium ABCD ; and LG = BE
+ FD, and LK = AC. Whence
trapezABCD = iparLH = ^LK.LG = iAC(BE + FD).
Scholium. If a, b, c, d be the four sides of a trapezoid, which has the oppo-
site sides a and c parallel ; and if c — a = S : then the area of the figure is
?JLl^(i + d+ S) C-b + d +S) (b-d+ d)(b + d -S)
If the quadrilateral be inscriptible in a circle, 2s=:a + b + c-\-d, and
area = //(s — a) {s — b) (s — c) (s — d).
These theorems are left for the student's investigation by means of the trigo-
nometrical calculus. \
6. When the three sides of a triangle are given to find the area.
Let them be as usual denoted by a, b, c. Then, we have, from p. 448, eq. 1 1.
sm C = '—^ '— ; and hence
ab
area = ^ab sin C ^ \/s (s — a) {s — b) (s — c),
the same result as before found, p. 417, by a different method, but founded on
the same principle. •
When 6 = c, or the triangle isosceles, this becomes
area = ia ^{2b + a) t,2b — o)
and when a = 6 = e, or the triangle equilateral, area = ^ar^ys.
Also, when the sides are expressed by the radicals of the second degree,
;y/a, .yb, \/c, the form most convenient for computation will be
area = i ^2{ab + be + ca) — (a" +'b^+~c'^
The area of a triangle may also be computed by a different formula, thus : put
b + c = h, b — c^k, then area = Ja/(A"^ — a'^'i {a^ — k'^), and these again
may be separated into factors, giving the convenient equation
area = J ^{h + a) (A — a) [a + k) {a — k).
7. Given two. angles of a triangle and the included side.
At p. 450, eq. 12, we found the expression for the perpendicular on c to be
/3 = c sin A sin B cosec C ; and hence, area = ^pc = ^cr sin A sin B cosec C.
In all cases where sufficient data is given, the problem can be reduced to one
or other of the forms laid down, and the area thence found : and in all cases of
trapezia, the same remark applies. When we come to figures of more than
four sides, it will generally be better to reduce them into triangles or trapezia,
or into figures of both kinds, as the data may suggest, and take the sum of all
the areas of these partial figures for that of the given one.
When the object to be measured is small, it is more easy to obtain the several
lines and perpendiculars accurately than it is to obtain angles ; as in boards,
walls, and artificial objects generally. In larger ones, as fields or the assemblage
of fields constituting a farm or estate, it is easier to obtain angles than lines ;
and in this case it will always be better to measure one side very carefully, and
474
MENSURATION.
obtain all the other requisite data, in angular measures. In this latter case, we
reduce the figure to a series of triangles; and in the former to a combination of
triangles and quadrilaterals, or often more conveniently to a series of triangles
only. All the cases that are likely to occur have been already given ; and any
other that may incidentally present itself can easily be reduced to some of these
by the principles of trigonometry; and, indeed, the difficulty can often be evaded
altogether by a little difference of arrangement.
8. Jn a circle of given radius, a regular polygon of n sides is inscribed, and it
is required to find the perimeter and area of the polygon.
Let AE or EP = R be the given radius, and AP one of
the equal sides. Then since all the sides are equal, the
angles which they subtend at the centre are all equal, and
each of them to the angle AEP. But they are all together
equal to 2n-, and hence AEP = — . Draw the perpendicular
ER to AP, and it will bisect the angle AEP. Hence AER
?r
= ; and by right-angled triangles AQ = AE sin AER =
R sin - , and AP = 2R sin -. Whence the perimeter = 2raR sin .
n n "^ n
Again for the area, we have n triangles each equal to AEP, and expressed by
nAQ.QE = «R sin - . R cos - =-R-sin — = area,
n n 2 n
9. About a given circle a polygon of n sides is described, whose perimeter and
area are required.
In the preceding figure, let AE = EP = r, the radius of the given circle, and
ST be one of the n sides of the polygon. Then, reasoning as in the preceding
instance, we have
ST = 2r tan , and the perimeter = 2nr tan - .
« n
And the triangle EST = SR.RE = r^ tan -, and that of the entire polygon is
nr^ tan .
10. Given one side of a regular polygon of a sides, to find its area, and the radii
of the inscribed and circumscribed circles.
Ut AB, BC, CD . . . be the sides of the polygon,
each of which is denoted by 2a.
Then BOF = ^, and FO = BF cot BOF, and BO =
BF cosec BOF. That is, FO = a cot "", and BO =
n
a cosec ^, which are the radii respectively of the in-
scribed and circumscribing circles.
For the area, we have n.BF.FO = na^ cot '^.
THE CIRCLE. 75
11. To calculate the perimeter of a circle of a given radius.
(See fig. in p. 474, No. 8.)
Let ST, AP, be one of the sides of the regular circumscribed and inscribed
polygons respectively of n sides. Then the perimeter of the circle is less than
the perimeter of the former, but greater than that of the latter polygon. Also,
as the value of n is increased, the perimeter of the circumscribed polygon is
continually diminished, whilst that of the inscribed one is continually increased :
and the perimeter of the circle is the limit towards which they both tend, the one
by its continual diminution, and the other by its continual enlargement. For
the one may be diminished so as to differ from that of the circle by a quantity less
than the least assignable, whilst it can never become less than that of the circle :
and on the other hand, the other perimeter may be so increased, by increasing
«, as to differ from that of the circle by a quantity less than the least assignable,
whilst it can never become greater than that of the circle. The perimeter of the
circle, then, is a limit between the perimeters of the two polygons; and, there-
fore, to the same number of decimal places as the two polygons agree for any
value of n, to that extent the perimeter of the circle is also obtained.
Let us suppose, then, that n = 10800, in which case the angle AEP = 2';
and AP = 2r sin 1', and ST = 2r tan 1' ; and the inscribed and circumscribed
perimeters are, putting 2r = d, lOSOOd sin 1', and \OSW)d tan 1' respectively.
We have already (p. 430) computed sin 1' and tan 1' to ten decimal places, and
found them coincident to that extent : but in fact they only begin to differ in the
fifteenth decimal place, and we have the following functions to radius 1 ; viz.
sin 1' = -000290888208664, and tan 1' = -000290888208668.
Multiplying these by 2l600r or 10800(f respectively, and we shall obtain
circum. perim. = 31415926536161 ... ; inscr. perim. = 3-1415926535767 ...
These agree to nine places of figures, and hence the perimeter of the circle,
whose value is intermediate to them, is accurately found to the same extent *.
* It may be desirable to point out other methods of computing the circumference of the
circle : but as they mainly rest either upon the use of the imaginary symbol, or upon the inte-
gral calculus, the obvious principle above employed has been thought better adapted to the text.
1. By Elders theorem (p. 437) we have tan a = — =-^rr . ^^^^.
^ xr » cosa ./ 1 '
_ ^^ ,W-J-fi
9 _ /ZZT 1 4" n/ — 1 tan a
ore ^ =' — — : whence, takmg log, of both sides, we have the equation
1 — ^ — 1 tan a
2a^ — \ =log. /I + s/ — \ tanaj —log, /l — ^— 1 tanoj.
which being expanded, the equal terms with contrary signs cancelled, and the whole divided by
2^ — 1, we have o = tan a — J tan^a -h J tan'a — \ tan'a -{- ... ad inf.
This, from its discoverer, is called Gregory's series, and it is the foundation of almost every
effective method usually employed. It is not, however, in its simple form, well adapted to use,
on account of its slow convergency ; as it requires a great number of terms to be computed
for arriving at a moderate degree of approximation.
When a — 45°, tan o = 1, and we have the following expression for the value of 45°, viz.
1 -1 Ij-l-lj. -1.4. 2.2
^ir_l — g-r^ 7 + -- 1.3 T" 6.7 T" 9.11 -T —
When
476 MENSURATION.
This number might have been found correctly to a greater number of places,
if we had taken AEP smaller, (that is, n greater,) as 2" for instance : and it has '
been actually computed to 128 decimal places by Delagny, and in a MS. at
Oxford to 140 places. {See Mutton's Dictionary, Art. Circle.) It is generally
represented by v, the initial letter of the Greek term perimeter. The same symbol
has been used in the trigonometry to denote 180°, or a semicircle to radius 1 ;
and which, since the perimeters of circles are as their diameters, will properly
represent the circle to diameter 1 or radius ^.
12. To find the area of a circle, whose diameter is given.
Reasoning as in the last case, the area of the circle is the limit between
the areas of the inscribed and circumscribed polygon of 2" sides when n is con- \
tinually increased. Now circumscribed polygon = nR- tan -, and inscribed
1 . 27r
polygon = ^ nr^ sin — : and R, r, approach to equality as n is increased, whilst
sin — and tan - approach to ^ and - respectively under the same circum-
n n '^'^ n n ^ •'
stances ; and at their limits these three equalities actually take place. Hence,
•"• 1
When o =; 30° = . we have tan a =: —rr,, and hence by actual substitution we have
a ~s/A 3.3^3.3-^ 7.33 ^••" J
This latter is called Hallet/'s series, but it, like the former, converges but verv slowly.
I + I
2. Let tan o = .-r and tan /3 =: - : then tan (a 4- /3) = — -7-7 = 1, or a + fl = - ir.
2 -3
But we have by Gregory's series,
'' = 2-3:2i+5T*-7:2^ + --- ''"^^ = 3- "5:^^ + 0^-7X7 +
whence
-''=i={k^}-K^+i}+Hi+F}--
This is Elder's series, and converges with considerable rapidity.
15 1 1
3. Take tan a = t; then tan 2a = 1 ^ , and tan 4o =:: 1 -|- yy^ ; and hence a is greater than - it.
Take 4a = 6 + - x, or 6 ^ 4a ; : then tan 0 = ; — ^^ = jr--.
^ 4 ' 4 tan 4a -t- 1 239
Again iJ'tt = 4o — 6, and substituting these separately in Gregory's series, we have
1 _ «n 1 _i_ 1 1 ifl 1 1,1
*" - '^ l.5-X5»+ 5:3s- •■ • 1'=^"'^^= 23D -3:2393 + 5:239i- • " "
*"" 4 X.5 3.03+.5.0S ■•■ J '1.239 3.2393 +5.2395~ •• ' /
which converges with great rapidity. This series was discovered by Machin, and is called
after him. Tiic most convenient scries, however, for the actual computation of the circum-
fcrenrc. ever <li(*covercd, is the following, which was communicated to me some time ago by my
CollraKur, Mr. Itutlierfurd, —
T = 4/'_l 1-+^ L_ \ n 1 1 , \ , ri 11,1
4 l."-. _3- .5^^.5' ji ••i~\70 3'702"^-/ + \.99~3-99'''" • •/
The diviviri 70 and 9.") being ea.tily einph)yed from their roducibility into factors,, would render
the Oxford approximation an oporation of by no means extraordinary labour. The scries itself
U derived from tlic c<iuation xiv. 7. (()) ;>. 443,
•»• . -1 1 -1 1 _, 1
^- = 4Un 5-^" 70 +''^" 99-
THE CIRCLE. 477
supposing ir to represent the accurate circumference, we have rf = 2r = 2R, and
- nR*. - =: nr' - = r^jr = cP. - = area of the circle.
2 n n 4
Cor. 1 . Since the circumference of the circle is 2rir, this value of the area is
also represented by ^r.2rn, or half the product of the radius and circum-
ference.
Cor. 2. Since (Ih. 94, Geom. p. 337^ the sectors of a circle are as the arcs or
angles corresponding to them, these are found by the simple proportion,
o
360° : a° : : r^ir ; area of sector = -r-?: .r*ir.
360°
Cor. 3. The sector is also represented by - . /3, where /3 is the length of the
arc subtended by a°.
Cor. 4. The area of an annulus, or the space inclosed by the circumferences
of two concentric circles, is represented by r^n — ri^ ir = {r^ — »'i*)t =:
(r + fj) (r — ri)7r ; where r and r, are the two radii.
13. To find the area of the segment of a circle.
{See first fig. p. 474.)
If a be the arc corresponding to the segment, we shall have the following values :
segment ARP = sector AEP + triangle AEP
= »'■'"+ ir^siaa° = ^'{a + 8ina}
where the upper or lower sign is used according as the segment taken is less or
greater than a semicircle.
Scholia.
A few numerical particulars, which will greatly facilitate the actual solution of
the following and similar problems, may be advantageously put together here.
1. The ratio of the diameter to the circumference of a circle may be with dif-
ferent degrees of approximation calculated from
: 22, that of Archimedes,
: 355 Melius,
: 3' 14 16, that commonly used by artificers,
: 3'14159, that commonly used in works of science.
: 1 ; the same, in fact, as the last.
In some few cases, the approximation is carried to eight or ten decimal places :
but the necessity for this is of comparatively rare occurrence. In Hutton's
Tables, p. 360, the parts of the circle for degrees, minutes, and seconds, are put
down to seven decimals.
6*2831853
2. The length of au arc of 1° is -.- '"-, or -0174533 ...
3. The factor — = 0795775 .... is often used in our practical calculations.
47r
Other numbers relating to the circle and sphere are given on the last page of
Hutton's Tables, 7th Edition, and constructions of a line approximately equal to
the circumference have been given at pp. 400—1 of this volume.
diam.
: circ. :
7
diam.
: circ. :
113
diam.
: circ. :
1
diam.
: circ. :
1
diam.
: circ. :
•318309
^-g MENSURATION.
14. To find the area of an irregular figure by the method of equidistant ordinates.
When a figure is bounded by an irregular
outline A' B' C D' E' P G' on one side, this
method will enable us to obtain a tolerably close
approximation to the area enclosed by it and a
straight line AG. Set of AB = BC = CD =
FG, any number of equal parts. Denote each by h, and measure the
several distances AA' BB', GG' perpendicular to AG ; and denote them
severally bv a,b,c, g. Then the area of the space enclosed by the straight
lines A'A, AG, GG' and the irregular boundary A'G' will be expressed by
{
2
or
b ^c c + d d + e e +/ , /+g-( ^
-^r + T~ + 2 "^ 2 "^ 2 J '*'
, by {^±^ + i + c + <i + c +/}a.
For these are the areas of A'ABB', B'BCC, ; and if the distance h be so
ukcn as to render the collected excesses of one set of parts above the figure,
visibly equal to the collected defects of the others from the several trapezoids,
the area will be found to a visible degree of accuracy.
When the figure is composed of boundaries ^,
along parts of which straight lines of consider-
able length can be drawn without varying much
from the actual boundaries, a better method will
be to take the extremities A', B', C . . . . G' of
these lines as the angles of the trapezoids; to '^^ a c o t r&
measure the lengths AB, BC, CD, FG of
the intersections of the perpendiculars AA', BB', CC, .... GG' upon AG.
Then the figure will be, calling these distances A,, h^, .... h„
V
f+9
h..
2 ■■' ' 2 ■"" ' 2 "•' ' 2
On these principles, the areas of fields bounded by irregular fences are esti-
mated in land surveying : which will be illustrated under that head.
II. PROBLEMS ON PL.\NE SURFACES.
On the principles laid down in the preceding theorems the following series of
problems admit of solution. The theorems for solution are referred to without
specifying the rules in words, a step which is altogether unnecessary.
Problem I. To find the area of any parallelogram.
Nos. 1, 2, 3, page
1. The length of a parallelogram is 12'25 and its breadth is 8"5 : what is its
area? Ans. 12-23.8-5 = 104-125 = area.
2. The side of a square is 35 25 chs : required its area ? Ans. 124ac Ir Ip.
3. Find the aiea of a rectangular board whose length is 12^ft, and whose
brradlh is 9in. Ans. 9| ft.
4. Find the content of a piece of land in the form of a rhombus, its length
b«ing 6-2 chs, and breadth 5 45. Ans. Sac Ir 20p.
PROBLEMS ON PLANE SURFACES. 479
5. Required the number of sq. yds in a rhomboid whose length is 37ft. and
height 5ft 3in. Ans. 2l7j sq. yds.
6. The two diagonals of a parallelogram are 185 5 and 137*9, and they inter-
sect under an angle of 42° 10' 18" : what is the area ? Ans 8587'55.
7. The side of a rhombus is 18j, and one diagonal is 23.J : find the other
diagonal and the area. Ans. diag. = 29-223278, area = 343373.
8. If a, b, c be the distances of a tree from three of the angles of a square
field, show that its area is expressed by
, ,o ■ X , 1.0 u (a^ —c2) cos 4 5°+ 62 sec 4 5°
a^ — 2ab cos (45° + i>) + b\ where cosdi = ^^ .
9. The distances a, b, c, d of a point from the four angles of a rectangle are
given to find its sides and area.
10. The stretching frame of a picture is 24in by 18 ; the frame of the picture
(which is flat with a bevelled edge of ^in wide, and inclined to the picture in an
angle of 45°) extends over 2in of the canvas ; and the visible area of the picture
is equal to the visible surface of the frame : it is required to find the width of the
frame, including the bevelled edge.
Problem IL To find the area of a triangle.
Rule I. No. 2, Cor. 2. page 47 i.
1. The base of a triangle is 625 and height 520 links: what is its area?
Ans. 625.260 = l62500lks = lac 2r 20p.
2. Find the area in yds of the triangle whose base is 40 and perpendicular
30ft. Ans. 663 sq. yds.
3. Required the number of yds in a triangle whose base and height are 49
and 25Jft respectively. Ans. 68^^yds.
4. Required the area of a triangle whose base is 18ft 4in, and altitude lift
lOin. Ans. !08ft. 5^in.
Rule II. No. 2, Cor. 3, and No. 7, pages 472, 473.
1. The containing sides are 30 and 40, and included angle 28° 57': what is
the area ?
By natural numbers,
n sin 28° 57' = '4840462
i . 30.40 = 600
290-42772
By logarithms,
t sin 28° 57' = 9-6848868
log 600 = 2-7781513
log 290-4277 = 2 4630381
2. How many sq. yds are contained in the triangle, one of whose angles is
45°, and the containing sides 25 and 21|ft respectively? Ans. 20 86947.
3. Given the base of a triangle equal to 476-25yds, and the angles at the
base 27° 10' 15" and 35° 10' 18 ', to find the area and the three perpendiculars
from the angles to the opposite sides.
Ans. area 33678, and perpendiculars 141-43, 217-47, 274-33.
4. The base of a triangle is 2725 chains, the vertical angle is 57° 15', and
the difference of the angles at the base is 18° IS'. What is the area of the
triangle ? Ans. 328-972.
5. The difference of the segments into which the perpendicular divides the
base is 10, the base itsell is 50, and the vertical angle is 100°. Wiiat is the area
of the triangle ? Ans. 5 1 1 -938.
Rule III. No. 6, page 473.
1. To find the area of a triangle whose sides are 20, 30, 40.
Here 2s = 20 + 30 + 40, or s = 45, 5 — a = 25, * — 6 = 15, 5 — c = 5 ;
and hence the area is \/45.25.15.5 = 75V'15 = 290*4737.
4^ MENSURATION.
2. How many yards are there in a triangle whose sides are 30, 40, 50ft
respectively ? Ans. 66§.
3. Find the area of a field whose sides are 2569, 4900, and 5025 links respec-
tively. Ans 6lac Ir 39p.
4. The area of a triangle is 6, and two of its sides 3 and 5 : find the third
side and all the angles. Ans. 4 or 2 a/13.
5. The sides of a triangle are in arithmetical progression, their sum is 27 linear
feel, and the sum of their squares is 261 sq. ft : find the area in yds.
Ans. I a/is.
6. The area of a triangle is 1000 and its sides in the ratio 3, i, i : find the
sides. Ans. 40-074, 50-093, 66-791.
7. Find the side of an equilateral triangle whose area is 100. Ans. 15-197.
8. The sides of a triangle are a/S, ,J\, v/5 : what is the area? Ans. \-yJ\\.
9. A triangle has its sides each equal to Z^J'isJ'i : what is its area ? Ans. 20^.
10. Fmd the area of the triangle whose sides are 4, 4 + ^3 and 4 — ^/3.
Ans. 2 a/3.
11. On the perpendicular of an equilateral triangle whose side is a, another
equilateral triangle is described, and on the perpendicular of this another, and so
on ad inf. : it is required to find the sum of the areas of all the triangles so
described. Ans. fa->/3.
12. The sides a, b, and the area A, are given to show that
2/^ ' 2A
ao -r _L V a'v/a- + 6^ + 2A/a-62— 4A»
Problem III. To find the area of a trapezoid.
No. 3, page 472.
1. In a trapezoid the parallel sides are 750 and 1225, and their distance 1540
links : what is the area? Ans. 1 5ac Or 33p.
2. The greater and less ends of a plank are 1 5 and 1 lin, and its length 1 2ft 6in :
what is its area? Ans. IS^^ft.
3. A quadrangular field ACDB has perpendiculars CP, DQ drawn to the side
AB ; and the following measures were taken from which to find its area : AP =
110, AQ = 745, AB = 1110, CP = 352, and DQ = 595 links respectively.
Ans. 4ac Ir 5 792p.
4. Find the area of the trapezoid whose opposite parallel sides are 35 and 19;
and the angles made by the oblique sides with the parallel sides are 42° 10' 15"
and 73° 6' 20" respectively. Ans. 306-861.
5. From the triangle ABC, whose base AB is 40 and the perpendicular upon
it is 60, to cut oflf a trapezoid by a line parallel to AB which shall have the area
*®0. Ans. The breadth OP of the trapezoid is 13| nearly.
6. What length must be cut from the broader end of a board which is 13 ft
long, and whose ends are respectively 18in and 14in, to make lOft square?
Ans. 71 nearly.
7. The breadth of a ditch at the top was 72ft, at the bottom 38?, and the
■loping side 26i|and20ft; and the top and bottom horizontal: find the area
of tl.e vertical section. Ans. SSSJft.
8. 'JTie area of the section of the ditch being 154ft, and its depth 5^ft; also
the breadths at the top and bottom are as 9 to 5 : what are those breadths ?
Ans. 36 and 20ft.
PROBLEMS ON PLANE SURFACES. 4^1
Problem IV. To find the area of any trapezium.
Nos. 4, 5, page 4/2.
1. To find the area of a trapezium, one of whoso diagonals is 42 and the per-
pendiculars upon it IC and 18. Ans. i (16 + 18) 42 = 714.
2. A diagonal is 65 ft, and the perpendiculars on it are 28 and 33^ ft : how
naany yds does it contain ? Ans. 222-j;iyds.
3. In a quadrangular field ABCD, owing to obstructions there could only be
taken the following measures ; BC = 265, AD = 220, AC = 378, AE = 100,
CF := 70 yds, when DE and BF are perpendicular to AC : it is required to
construct the figure and compute the area. Ans. I7ac 2r 21p.
4. ITie two diagonals of a trapezium are 31*2956 and 62" 1598, and they inter-
sect under an angle of 105° 18' 25" : what is the area ?
5. One angle of a trapezium is 107° 18' 10", and the diagonal divides it in the
ratio of 7 to 5, whilst the sides containing it are 123'456 and 654"321, and the
greater side makes the less angle with the diagonal, the diagonal itself being
1000 : it is, from these data, required to find the areas of the two triangles into
ivhich the other diagonal divides the trapezium.
6. The four sides of a field taken in order are 25, 35, 31, and 19 poles, and
;he diagonals are equal : required the area of the field.
7. ABCD is a quadrangular field, whose sides taken in succession are AB =
15 ch 24 1, BC = 18 ch 86 1, CD = 9 ch 90 1, and DA = 11 ch 14 1 ; also the
ingles at A and C are 105° 28' and 89° 54' : find its area. Ans. I7'5l69ac.
8. The side AD of a quadrangular field ABCD was 311yds, and the angles
vere BAC = 44° 20', CAD = 41° 19', ADB = 24° 10', and DBC = 37° 4':
equired its area. Ans. 8"6531ac.
9. The sides of a quadrangular field taken in order are 26, 20, 16, and 10
)oles, and the angle contained by the longest sides is 56° : what is its area ?
Ans. lac 127-676p.
10. ABCD is a quadrilateral, whose angles B and C are right angles, as are
ikewise those formed by the diagonals BD, AC : the diagonals themselves are
fiven, d and df,, and it is required to find the area, the sides, and the remaining
ingles formed by the several lines of the figure.
1 1 . The two diagonals AC, BD, of a trapezium bisect each other ; one of the
lUgles, B, is 30°, the side BC is 100, and twice the product of the diagonals is
iqual to the square of their difference : required the sides, angles, and area of the
igure.
Problem V. To find the area of any polygon.
This will be effected if we divide it into triangles or quadrilaterals, or both,
■nd take suflBcient measures either of lines or lines and angles for the trigono-
netrical determination of all the parts requisite for the purpose. One or two
xamples of the manner of dividing the polygon will suffice to show the nature
if the problem, whilst the actual computation is left for the student.
1. Suppose a regular pentagon whose side is 170 fathoms, to be fortified, and
hat the salient angle of the bastion is 71°, and its face 47 fathoms; required
he flank and curtain, supposing the line of defence to be perpendicular to the
lank : determine, also, how many acres would be contained within the bounda-,
ies of the fortification, supposing the work completed.
Ans. Flank = 2565, curtain ~ 64 57, area = 26 051.
VOL. I. I i
AS2
PROBLEMS OX PLANE SURFACES.
2. The polygon in the annexed figure had the follow-
ing parts measured, from which to determine the area :
viz. AC = 55, FD = 52, GC = 44, Cm = 13, Bn =
18, Go = 12, Ep = 8, and Dq = 23.
Ans. area = 1878-5.
3. The sides FE, AB, BC, were found to be 15 726,
25'182, and 23629, and the angles were taken as follows: FED = 15-2'' 10',
EFG = 65° 18', FEG = 66° 28', EGD = 31° 15', CGD = 32° 18', GCD =
58° 40', AGC = 100° 5', GAC = 41° 50': it is required to find the area.
Problem VL— To find the area of a regular polygon *
Nos. 8, 9, 10, page 474.
1. Required the area of a regular pentagon, each side being 25 ft.
Ans. 1075 298356.
2. Find the area of an equilateral triangle whose side is 20 ft.
Ans. 173-2058.
3. Required the area of a hexagon whose side is 20. Ans. 103923048.
4. Find the area of an octagon whose side is 20. Ans. 193r37084.
5. Find the area of a decagon whose side is 20. Ans. 3077'68352
Pbob. Vn. — To find the circumference and diameter of a circle from one another.
■No. 11, page 475.
The approximation 1 : 31 4 159 is the number used for general scientific
purposes; but for mere round numbers 7 : 22 or 113 : 335 are used as the ratio
of the diameter to the circumference. In the same manner, circumference :
diameter :: 1 : -318309.
1. Find the circumference of a circle whose diameter is 20.
Ans. 7 : 22 : : 20 : 62f as the roughest approximation.
2. If the circumference of the earth be 24877-4 miles, what is its diameter ?
314159: 1 :: 24877-4 : 791872
355 : 113 : : 24877-4 : 7918-72
1 : -318309 : : 24877-4 : 7918-72
3. Required the circumference of a circle whose radius is 22J.
Ans. 13901547.
It ii sometimes convenient to possess the values « cot — in a table ; and likewise the value
n
of the circumscribed radius. In the latter case the tabulated value is multiplied by the side of
the polygon, and in the former by its square. The following is such a table to the duodecagon
inclusive.
Names.
Areas, or
Multi-
pliers.
Radius 01
circuni.
circle.
3 iTrigon or triangle ...I 0 43301-27 0 .5773.i03
4 Tetr»(ton or Mjunre .| 1 0000000 0 7071068
4 Pcnugon I 1-7-J04774 0ai(W),>08
6 |lcxa«oti ,... 2-.i.'»807(;'2 1()00(WIO
7 Hrptapon I 3-C339 1 '26 ' 11. 5-23ff2o
fi
^
Names.
8 Octagon
9 Nonagon
10 Decagon
1 1 I'ndecagon ..
\'2 Duodecagon
Areas, or
Multi-
pliers.
4-8284272
61818-242
7-694-2088
9-36.56415
11-1961.5-24
Radius of
circuui.
circle.
1-306.563(1
I1-46190-2-2
il-6180340
1-7747331
1-9318517
To show its u»c we may take Ex. 1, in which we have So^.] -7204774 = 1075-29837, the
I PROBLEMS ON PLANE SURFACES. 483
4. The circumference of a circle is 64*4 : what is the diameter ?
Ans. 20-49916.
5. Two chords whose sum is 21 and difference 1, and the rectangle of the seg-
lents of either is 24, cut each other at right angles : it is required to find the
ircumference of the circle, and the distance of their point of intersection from
be centre. Ans. circumf. = Stt^ 5, dist. = J^^ 29.
6. Let C be the circumference of a circle, d its diameter, and c the chord of an
, ,, , 16 ad jC — a) .
re, a : show that c := —p^ 77= — —^ nearly.
5C* — 4a (C — a) ■'
7. If a be an arc of a circle to radius l,o''=5.48'. - ^ — ~---- ^ - nearly.
237— cosa + 124co8^o '
8. Show that 3a ^ tan a + 2 sin a, very nearly, when a is small.
Problem VIIL To find the length of an arc of the circle.
Scholium, No. 13, page i77.
Multiply the number of degrees and parts expressing the arc by •0174533,
ad by the radius of the circle j or, take out the numbers corresponding to the
iven number of degrees, minutes, and seconds successively from Hutton's
abtes, p. 360; and then their sum, so multiplied, will be the length required.
1. Find the length of 30° in a circle whose radius is 9ft. Ans. 4712388.
2. To diameter 10ft find the length of 12o 10' 15". Ans. 4-248422.
3. What portion of the arc of a circle is equal to the radius ?
Ans. 57°'2957795
Problem IX. To find the area of a circle.
No. 12, and scholium, page 477.
Let r be the radius, d the diameter, c the circumference, and A the area : then
is found from any one of the following equations.
A = irr^= ir.ic = idc = cP.^ = 07957750-.
4
1. Find the area of a circle whose circumference is 31*41593.
Ans. 78-5398.
2. Find the area of a circle whose diameter is 7. Ans. 38-484501.
3. How many square yards are there in a circle whose diameter is 3jft.
Ans. 1069014.
4. Required the area of a circle whose circumference is 12ft.
Ans. 11-45916.
5. Find the area of a circle, the difference of whose diameter and circumfer-
ice is 1056 64 ft, Ans. 191105-4ft.
6. If the centre of a circle whose diameter is 20 be in the circumference of
lother whose diameter is 40, what are the areas of the three included spaces ?
Ans. 173 852, 140308, and 1116-332.
7. What is the diameter of that circle which contains an acre ? Ans. 78i yds.
8. If the area of a circle be 100, find the sides of the inscribed square, pen-
gon and hexagon.
Problem X. Tofindthe area of an annulus.
No. 12, Cor. 4, page 477.
1. The diameters of two concentric circles are 10 and 6 : required the area of
e annulus. Ans. 50*26552.
2. The bounding circles are 10 and 20 in dianoeter: what is the area of the
inulus? Ans. 235-61947.
I i 2
484 PROBLEMS ON PLANE SURFACES. I
3. The circumference of a ring is l6lin, and its width lin: required its
internal diameter and area. Ans. int. diam. = 49'248in, and area = 157-8634.
4. The radii of two concentric circles are in the ratio of 10 to 9, and the area
of the ring is 375-562feet : find the diameters of the circles.
Ans. 501624, and 45-1462.
5. Let c be the outer circumference, and b the breadth of a ring : show that
its area will be (c — bi:) b.
Problem XL To find the area of a sector of a circle.
No. 12, Cor. 2, page 477.
(1) Multiply the radius by half the length of the arc.
(2) Take ith of the product of the arc and diameter.
(3) 360° : given arc : : irr- '. area of sector.
1. Find the area of a sector of 18° to a diameter of 3ft. Ans. '35343.
2. The radius is lOft, and the arc 20ft : find the area and angular measure.
Ans. 100 and 114° 35' 29".
3. A sector of 147° 29' 18" has a radius of 25ft : what is its area ?
Ans. 804-3986.
4. Find the area of a sector whose radius is 50 and arc 56° 30'.
Ans. 1232-6387.
5. The area of a sector is 100, and the length of its arc 20 : what is the angle
of the sector? Ans. 114° 35 'J nearly.
Problem XIL To find the area of a segment of a circle.
No. 13, page 477-
(1) Find the difference or sum of the sector ha\'ing the same arc and the tri-
angle formed by the chord and the radii bounding the sector, according as the
arc is less or greater than a semicircle.
(2) Segment = ir^ ^a + sin a], which is only the same rule in another form.
1. The chord AB is 12, and the radius AE is 10: what is the area of the
segment ?
AD
sin AED, or sin AEC= -p = -6 = sin 36° 52' ll"-2.
Ah
Hence the arc ACB = 73° 44' 22''-4; and we have
360° : 73° 44' 22"-4 : : lO-. TT : 64-3504 = area of sector.
Also, for the triangle AEB, we have
ED = ^/AE^ - AD2 = ^T^^ZT^ = g,
and the area of the triangle is AD.DE = 6.8 = 48. Whence
segment ACB = sector AEB — triangle AEB = 163504.
Or again, by the formula : we have as before arc ACB = 73° 44' 22"-4 ; and
by Hutton's Tables, p. 360,
arc 73°
44'
22
^
1 -2740904
127991
1067
3
sin 73° 44' = -9599684
pp to 22 '-4 = 304
•4"
•9599988
1-2869965
1-2869965
-3269977 for less segment.
2-2469953 for greater segment.
Hence, less segment = \r^ \a — sin a} = 16349885.
and greater segment = \r^ {a + sin a\ = 112349765.
I PROBLEMS ON PLANE SURFACES. 485
2. The height CD is 18, and radius CE is 50: what is the area of the seg-
lent? Ans. 961-3532.
3. Required the area of each of the segments where the chord is 16, the
iameter of the circle being 20. Ans. 44 728 or 269'432.
Scholium. ^
It occurs in many problems that we have to find the arc of the sector from
;nowing the area, together with the radius of the circle, or some other given
ine in the circle : or in other words, to solve the equation.
0 + sin 0 = — .
— a
For this purpose no method seems so generally and easily applicable as the
nethod of trial and error. One example is annexed : viz. to find 9 from the
quation
e — sine = 20943951.
Take as a conjectural approximation 9 = 150° : then
arc 150° = 2-6179939 arc 149° = 2-6005406
— sin 150°=— -5000000 — sin 149° = —-5150381
2 1179939 20855025
too great by -0235988 too small by -0088926
Hence by Trial and Error, p. 202 of this work
-0088926 „„ _, ,
c, = = -26° = 16' nearly.
-0324914 ^
Taking next 9 = 149° 16' we have a second correction.
arc 149° 16'= 2 6051948 arc 149° 17' = 2-6054857
— sin 149 16 = — -5110431 — sin 149 17 = — -5107930
20941517 20946927
too small by -0002434 too great by -0002976
TT -0002434 , .„ ,
Hence c, = = -45 = 27" nearly; and hence agam
\J\)\J<J^ \.\J
9 = 149° 16' 27" nearly.
PaoBLEM Xin. To measure any long and irregular figure.
No. 14, page 478.
1. The breadths of an irregular figure at five equidistant places are 8-2, 7'4,
-2, 10 2, and 86, and the whole length is 39 : what is the area? Ans. 343-2.
2. The length of the figure is 84, and the six equidistant ordinates are 17-4,
06, 14-2, 16 5, 20-1, and 24-4: what is the area ? Ans. 155064.
3. The distances of seven points in the long side of a figure formed by straight
nes from the first extremity of that side were 108, 104, 25, 106, 18-5, 56-2,
nd the ordinates of the angular points were 0, 12, 18, 25, 18, 5, 15, and 0 :
'hat was the area ?
3. In the vertical section of a rampart AS is the horizontal base, and the
orizontal distance, in feet, of the several angular points of the work reckoned
n this line, together with the heights of those points above it, are ranged below,
•om which to find the area of the section and construct the figure : viz. —
distance on AS, AB = 16, BD = 18, DH = 2, HK = 3, KL = 2, LP = 12,
PS = 10.
leight above AS, BC = 12, DE = 12i, HG = 13i, KI = 13^, LO = 18,
PR =16.
486 MENSURATION OF SOLIDS.
4. Let ABC be the profile, or perpendicular
section of a breast-work, and EP that of the
ditch. Now, sujjpose the area of the section
ABC is 88 feet, the depth of the ditch RD
6 feet, ER = SO = 3 feet ; what is the breadth of the ditch at top when the
sections of the ditch and the breast-work are equal; that is, when the earth
thrown out of the ditch is sufficient to make the breast-work ?
5. And what must be the breadth of the ditch at top, the depth and width at
bottom remaining the same, when the profile of the breast-work remains the
same, and the earth, in consequence of removal, occupies xjth more space than
it did before it was taken out of the ditch ?
MENSURATION OF SOLIDS.
The cube described upon the linear unit is always taken as the unit of volume.
It will hence follow in precisely the same manner as for superficial measure,
that the volume of a rectangular prism or cylinder is the product of the nume-
rical measures of the base and altitude. For any two rectangular parallelopi-
pedons are to one another in a ratio compounded of that of their bases and that
of their altitudes ; and the bases are compounded of the ratios of their sides
containing one of the right-angles : hence they are compounded of the ratios of
the three edges of the one to the three edges of the other, each to each respec-
tively.
So far as the surfaces of solids bouiided by plane faces are concerned, the
surfaces are computed by the rules appropriate to them, as already explained in
the mensuration of planes : but the surface of the sphere and its segments or
zones will form a part of this section.
I. THE SURFACES AND VOLUMES OF FIGURES OF THREE
DIMENSIONS.
1. The volume of a rectangular parallelopipedon is expressed hy the continued
product of these edges which meet at one of the solid right-angles, as is evident
from the foregoing introductory lemarks.
2. The volume of any parallelopipcdon, whose plane angles forming one of the
solid angles are n, /3, y, and whose opposite edges are a, b, c, is expressed by
2abc ^sin <r sin (,a—a) sin (<t— /3) sin {a—y)
where 2<t = a + /3 -f- y.
Let Q be the solid angle, the three edges
AQ, BQ, CQ, of which are denoted by
a,b,c; and the three angles CQB, CQ.\.
AQB are a, (i, y. From A draw the per-
pendicular AP to the plane CQB, and
from P draw PD, PE perpendicular to
QC, un, and join AFl, AD, and PQ,
Then (GVom. Planes, th. 7), AEQ, ADQ,
are right-angles, and AQP is the inclina-
Uon of the line AQ to the plane CQB.
Denote this by B -. then we have
MENSURATION OF SOLIDS. 4^7
QP = a cos 9, QD = a cos /3, and QE := a cos y. Also,
„_.^ QD cos/8 , „__ QE cosy
cosPQC = TTTi = — ^. and cos PQB = w^ = £.
Qr cos Q QP cos Q
n i ^n/-k/^ ■ T»/-\r.\ cos /3 cosy // C08*j3\ / COs'yX
But COS a = cos (PQC + PQB) = — -~^, — ^/( 1 ^ ) ( 1 -)
COS'(t V \ COS^fl/ \ COS*0/
whence, transposing, squaring, and performing obvious reductions, it becomes at
ance
sin 0 sin n = ^^ 1 — cos-a — cos^/3 — cos^y -|- 2 cos a cos /3 cos y .
= 2 \/ sin (T sin (tr — a) sin ((t — /3) sin ((t — y).
But the parallelopipedon QR is compounded of the base QS and altitude AP,
ind hence we have
vol ^ 5e sin a . a sin 9 ^ abc sin a sin 9.
= 2abc ,^sin <r sin (<r — a) sin (it — /3) sin (er — y).
3. The volume of any prism is the product of one end into the distance of
the two parallel ends. (Geom. PL and Sol., th. 30, p. 366).
4. The volume of a cylinder is expressed by the product of its base and altitude ;
ind that of a cone by one-third of that product : as is clear from Geom. of
PI. and Sol., theorems 30, and Cor. 34.
5. The curve surface of a right-cylinder is the product of the perimeter of the
base into the length of the axis, or by 2rhir. For that surface is composed of
m infinite number of infinitely narrow rectangles, all of the same length.
When the two ends are also required in the expression, the entire surface is
expressed by 2r (r -|- A) v. For 2r'7r ^ area of the base, and 2rhir = curve
surface ; hence the whole surface 2r7r + 2rA7r = 2r (r + h) v.
6. The curve surface of a right-cone is half the product of the perimeter of
;he base into one of the slant sides or edges. The reason is similar to the last.
Cor. The curve surface of a frustum of a cone whose bounding sections
iiave a and b for radii, and their distance reckoned along the side of the cone is
i, is (a + b) dv.
7. The volumes of a cone and pyramid are one-third of the volumes of the
:ylinder of a prism respectively of the same bases and altitudes.
8. The volume of a truncated pyramid or cone is expressed by- A {a"-|-ai-)-6'J,
tvhere a and b are the areas of the two ends, and h is their distance.
For, let ABCD be the pyramid of which BCDEFG is the frus-
tum, a* the area of BCD, 6^ the area of EFG, and h the height
HI of the frustum. Denote AI by c : then by similar figures
, I "^ ** J 1 I. <**
c -\- h:= -J- , or c=. — r- and c -H « = — ;.
b a—b a—b
Again, vol frustum BDGE
= vol ABCD - vol AEFG = i a' (e -|- A) — i i^c
= 3°'5^*->«"»=^!a' + «» + »'{.
The same demonstration applies to the conic frustum.
Cor. If D, d, be the corresponding linear dimensions of the ends, m their
appropriate multiplier, (so that mD^ md!^, are their areas,) and S* the difference
of those areas : then the volume is expressed by \ mh |3Drf -|- ^| .
9. The curve surface of a sphere is equal to four times the area of one of it»
great circles.
4^ MENSURATION OF SOLIDS.
Let the hemisphere he generated by the revolution of
the quadrant FOG about one of its limiting radii FI, and
let also the tangent GB at the other extremity G describe
a cylinder of the same altitude. Cut both these by two
planes, KLL', MNN', parallel to the common base IGG'
of the cylinder and hemisphere ; which will cut the sphere
in two parallel circles KOO', MQQ', and the cylinder in
two parallel circles KLL', MNN ; and likewise any plane
GIF through the axis FI will be cut by them in the lines
KOL, MQN, and the planes of the ends of the cylinder
in FB, IG. Draw in the plane GIF, the line OP perpen-
dicular to MN, and join 01.
Now if the arc OQ be taken smaller and smaller continually, the angle formed
by the line OQ with the radius 10 may be made to differ from a right angle by
an angle less than any assignable one, whilst it can never exceed a right angle.
The right angle is, therefore, the ultimate one formed by the arc OP and the
radius 10, since in that case the arc OQ and line OQ are coincident.
In this case, then, ROQ is a triangle right-angled at O, and OP perpendicular
to its base : whence the triangle OPQ is similar to RPO, that is, to RMI and to
OKI. Wherefore by the similar triangles OPQ, OKI, (since OP = LN, and
10 = KL,)
OQ : LN : : KL : KO : : Gx.KL : •2;r.K0 : : circ LL' : circ 00'.*
or OQ.circ 00' = LN.circ LL'.
But the zone LNN'L' of the cylinder is expressed by the circumference of the
base multiplied by the altitude; and the zone OQQ'O' of the sphere is ultimately
the surface of the frustum of a cone whose side is OQ, and whose two ends are
00' and QQ', and is expressed by § OQ {00' + QQ'}. Also ultimately 00'
is equal to QQ', and hence the spherical zone is ultimately expressed by OQ.OO':
whence also ultimaiely we have the spherical zone OQQ'O' equal to the cylindri-
cal zone LNN'L'.
Again, since this is true wherever the point O is taken, it is true for every
point in the quadrant FG ; and hence all the elementary zones which compose
the surface of the hemisphere are equal to all those which compose the surface of
the cylinder, each to each. The entire surface* of the hemisphere and cylinder,
or any corresponding parts of them between parallel planes, must, therefore, be
equal, each to each ; and likewise, the entire spherical and the entire cylindrical
surfaces of the same altitude, or any corresponding parts of them, will be equal,
each to each.
Now S = cylindrical surface = 2r . 2r7r = 4r^7r = 4 area of the great circle
GG'; which is the proposition enimciated.
Cor. 1 . Whence S = 2r . 2rir = d . dir = diam x circ.
Cor. 2. Any zone or segment of a sphere is expressed by the product of its
altitude into the circumference of its great circle.
10. Putting V for the volume of a sphere and the remaining notation as before,
we shall have
V = Trrf . Jd2 = ■ ^rf, ^ ^ ^S = J 7r-2 (rf;r)3.
For the sphere is two -thirds of its circumscribing cylinder {Geom. of PI. and Sol.
By thcw arc meant the fndre lircumfercuccs of the circles wliose centres are K and M,
though only jiuru of them are actually marked in the figure.
MENSURATION OF SOLIDS.
489
th. 37); that is, two-thirds of the base of the cylinder inultiplied by its altitude;
or again in symbols V = § . J nd^ = ^d^ . wd, which is the first equation ; and
the others are simple transformations into factors expressive of the different
way by which the volume may be found from the data.
11. Denote the height, KF, of a spherical segment by h, and the radius of its
base by r, ; then the volume of the segment is expressed by J ir {3d — 2h)h^, or by
J TT (3r,2 + h^)h ; or again by J ir {2r^ ± (2r» + r,^') ^/r^^^^H the upper or
lower sign being used according as the segment is greater or less than a hemi-
sphere.
For the cones generated by AIB, QIM, are similar;
and hence,
FP : KP : : (2FI)3 : (2KI)3 : : cone AIB : cone QIM,
or d^ : {d—2hf : : 5', cPtt : cone QIM = J, tt {d—2hf
Hence cone AIB — cone QIM = j", tt [d?—{d—2hyi
= v. rr {3d'h — Qdh"" + 4^3}
But tl.e spherical segment is equal to the difference of
the cone and cylinder {Geom. PI. and Sol., th. 37) ; that is,
\=zl 7r(Ph — -^ TT [3d^h - Qdk' + 4h^
= ^ v {3d — 2h] ¥
■c"^ ' ^
'^^':~-^'^::L^^2:>^
\ V N. "^j//
o^'^^^j^Tb^
^~t-:^£
^ffi
'v.X
7 '
/ /
dI^^^^
/ j^^'^-^
^- « __— -^
■••• (I)
Again, PK2 = FK.KH, or r.^ = id—h)h; whence d = ''-L+E,
h
Substitute
this in (1), then
V = 4 TT (3r,2 + ¥)h (2)
From r,2 = {d — h)h we have A = r + //r^ — r 2, which substituted in (I) or
(2) gives
V = i TT f 2r3 + (2r2 + r.^) ^r^-r,^} (3)
Cor. A spherical zone may be considered as the difference of two segments,
the factors of Jtt in which must be separately calculated, and their diffeVence then
multiplied by Iv.
II. PROBLEMS ON SOLID FIGURES.
Prob. I. To find the surface of a prism or cylinder.
1 . Fmd the surface of a cube, the side of which is 20ft. Ans. 2400ft.
2. Find the whole surface of a triangular prism whose length is 20ft, and
each side of its ends 18in. Ans. 9r948ft.
3. Required the convex surface, and the whole surface, of a right cylinder,
the length being 20ft, and diameter of the base 2ft,
Ans. 125-664ft, and 131-947ft.
4. A rectangular cistern whose length was 3ft 2in, breadth 2ft Sin, and depth
2ft 6in, was lined with lead at 3d per lb, and the thickness of the lead such as
to weigh 71b to the foot surface : what must be paid for it ? Ans. 3/ 5s Q^d.
5. How many square feet of board are required to make a packing-case
whose length is 3ift, breadth 2ft, and depth 20in ? Ans. 32ift.
6. How many revolutions of a roller 40in in length and 30in in diameter, will
be required in rolling a lawn 80ft in length and 50ft in breadth ; the lawn being
rolled first lengthways and then across ?
490 MENSURATION OF SOLIDS.
Pbob. II. To find the surface of a right pyramid or right cone.
Nos. 5, 6, p. 487.
1. What is the whole surface of a triangular pyramid, the slant edges being
each 20ft, and each side of the base 3ft ?
2. Required the convex surface of a cone, or circular pyramid, the slant
height being 50ft, and the diameter of its base 8|ft. Ans. 667 59-
3. Find the whole surface of the same cone.
4. Find the cost of lining a circular reservoir whose diameter at the top is
40yds, at the bottom 38§yds, and whose side or slant depth is lift : the brick-
work being executed at 3s ICd per square yard. Ans. 311/ 18s 2d.
5. What quantity of canvas is required for a conical tent, eight feet high and
thirteen feet wide at the bottom ? Ans. 70g yds square, nearly.
Problem III. To find the surface of a frustum of a pyramid or cone.
No. 6 Cor. p. 487.
1, How many square feet are in the surface of the frustum of a square pyra-
mid, whose slant height is 10ft; also each side of the base or gi'eater end
being 3ft 4in, and each side of the less end 2ft 2in? Ans. 110ft.
2. Required the convex surface of the frustum of a cone, the slant height of
the frustum being 1 25ft, and the circumferences of the two ends being as 5 to 7,
and the area of the less one 2 86488ft. Ans. 90ft.
Prob. IV. To find the volume of a right prism or cylinder.
Nos. 3, 4, p. 487.
1. Find the solid content of a cube, whose side is 24 in. Ans. 13824.
2. How many cubic feet are in a block of marble, its length being 3ft 2in,
breadth 2ft Sin, and thickness 2ft 6in? Ans. 21g.
3. How many gallons of water will the cistern contain, whose dimensions are
as 19, 16, 15, and the diagonal drawn through the cistern is 58 0345 in, when
277i cubic inches are contained in one gallon? Ans. 13P53.
4. Required the solidity of a triangular prism, whose length is lOft, and the
three sides of its triangular end or base are 3, 4, 5 feet. Ans. 60ft,
5. If the depth of an oblique parallelopiped be 8 ft, and the obtuse angle at
the base be 133°, and the including sides be 10 and 15 ft: what is its volume in
cubic yards ? Ans. 257392.
G. If the mean velocity of water through a cylindrical pipe an inch and a half
in diameter be 13in per second : what quantity would it supply in 12 hours ?
Problem V. To find the volume of any pyramid or cone.
No. 7, page 487.
1. Required the solidity of a square pyramid; each side of its base being 30,
and ilH height 25. Ans. 7500.
2. Find the content of a triangular pyramid, whose height is 30, and each
•.de of the base 3. Ans. 38-971143.
MENSURATION OF SOLIDS. 491
3. Find the content of a triangular pyramid, its height being 14ft 6in, and
the three sides of its base 5, 6, 7ft. Ans. 71 '0352.
4. What is the content of a pentagonal pyramid, its height being 12ft, and
each side of its base 2ft ? Ans. 27'5276.
5. What is the content of the hexagonal pyramid, whose height is 6"4ft, and
each side of its base 6in ? Ans. 1 38564ft.
6. Required the content of a cone, its height being 10| ft, and the circum-
ference of its base 9ft. Ans. 22-56093.
7. The content of a cone is 677828, the radius of its base r43l76. Find
the slant height, and the length of its axis. Ans. 45 2086 and 453614.
Pbob lem VI. To find the volume of the frustum of a cone or pyramid*
No. 8, page 487.
1 . To find the number of solid feet in a piece of timber, whose bases are squares,
each side of the greater end being 15in, and each side of the less end 6in, and
the length 24ft. Ans. 19^.
2. Required the content of a pentagonal frustum, whose height is 5ft, each
side of the base 18in ; and each side of the top or less end 6 in.
Ans. 9-31925ft.
3. Find the content of a conic frustum, the altitude being 18, the greatest
diameter 8, and the least diameter 4. Ans. 5277888.
4. What is the solidity of the frustum of a cone, the altitude being 25, also
the circumference at the greater end being 20, and at the less end 10 ?
Ans. 464-216.
5. If a cask, which is two equal conic frustums joined together at the bases,
have its bung diameter 28in, the head diameter 20in, and length 40in ; how
many gallons of wine will it hold? Ans. 790613,
Problem VII. To find the surface of a sphere or spherical segment.
No. 9, page 488.
c"
Surface = cd z= ttcP = — = •31830Cc', and Segment = ch.
1. Find the surface of a sphere whose diameter is 7- Ans. 153-93804.
2. Find it when the circumference is 22. Ans. 154 06l56.
3. Required the surface of a globe of 24 in diameter. Ans. 1809 557368.
4. Find the area of the surface of the globe, its diameter being taken 7957J
miles and circumference 25000 : also find it from each of these data taken sepa-
rately. Ans. 198943750, 198943821, and 198943125, respectively.
* We may proceed ratlier differently -when the ends are either circles or regular polygons.
In this latter case, square one side of each polygon, and also multiply the one side hy the other;
add all these three products together ; then multiply their sum by the tabular area proper to the
polygon, and take one-third of the product from the mean area, to be multiplied by the length,
to give the solid content. Also in the case of the frustum of a cone, the ends being circles,
square the diameter or the circumference of each end, also multiply the same two dimensions
together ; then take the sum of the three products, and multiply it by the proper tabular number,
viz. by '7854 when the diameters are used, or by -07958 in using the circumferences ; then take
one-third of the product to multiply by the length, for the content.
4QO MENSURATION OF SOLIDS.
5. The axis of a sphere being 42in, what is the convex surface of a segment
whose height is 9in ? Ans. 1 18/ -5248.
6. R.-quired the convex surface of a zone whose breadth is 2ft cut from a
sphere 12Ut diameter? Ans. 78-539.
Problem VIII. To find the volume of a sphere.
No. 10, page 488.
Vol. = ldx surface = Id-c = l d^ = -OiesSc^.
6 2 o
1. Find the solid content of the globe of the earth, supposing its circum-
ference to be 25000 miles. Ans. 263750000000 miles.
2. Supposing that a cubic inch of cast iron weighs "269 of a lb avoird., what
is the weight of an iron ball of 5 04 inches diameter ?
3. The radii of a shell are R and r : show that its content is - tt > R' — r^^ .
P&OBLEH IX. To find the tolume of a spherical segment or zone.
No. 11, page 489-
Vol. = (3rf — 2h)h'. - = (3r2, + h^)h . -, where r, is the radius of the sec-
tion and h the height of the spherical segment.
1 . To find the content of a spherical segment, of 2ft in height, cut from a
sphere of 8ft diameter. Ans. 41-888ft.
2. "What is the solidity of the segment of a sphere, its height being 9, and
the diameter of its base 20? Ans. 1795'4244.
3. The radii of the faces of a zone are 12 and 9 respectively; and the thick-
ness of the zone is 4 find the volume of the zone, and the radius of the sphere
of which it forms a part.
LAND SURVEYING.
DESCRIPTION AND USE OF THE INSTRUMENTS.
1 . Of the chain. Land is measured with a chain, called Gunter's Chain, from
its inventor, the length of which is 4 poles, or 22 yards, or 66 feet. It consists
of 100 equal links ; and the length of each link is therefore ^^ of a yard, or -^5
of a foot, or as much as 7"92 inches*.
I.an<l ii estimated in acres, roodg, and perches. An acre is equal to 10 square chains, or as
much a* 10 chains in length and 1 chain in breadth. Or, in yards, it is 220 X 22i= 4840 square
var.lv Or, in poles, it is 40 X 4 = ICO square poles. Or, in links, it is 1000 X 100 = 100000
•quarc linkn : ihchc Ining all the same quantity.
AUi, an acre is divided into four parts called roods, and a rood into 40 parts called perches,
which arc »<iii»re |)oles, or the square of a pole of 5J yards long, or the square of J of a chain, or
of 23 link*, which is «2.') square links. So that the divisions of land-measure will be thus :
♦I'i') »<|turc links :z: 1 pole or perch ; 40 perches = 1 rood ; 4 roods = 1 acre.
The Irngthi of lines measured with a chain, are best set down in links as integers, every chain
in length Uing ItX) links: and not in chains and decim.-ils. Therefore, after the content is
found, it will be in square links ; then cut off five of the figures on the right-hand for decimals,
LAND SURVEYING. 493
2. 0/ the plane table. This instrument consists of a plain rectangular board, of
any convenient size : the centre of which, when used, is fixed by means of
screws to a three-legged stand, having a ball and socket, or other joint, at the
top, by means of which, when the legs are fixed on the ground, the table is
inclined in any direction *.
and the rest will be acres. These decimals are then multiplied by 4 for roods, and the decimals
of these again by 40 for perches.
Ei-am. Suppose the length of a rei^tangular piece of ground be 792 links, and its breadth 385 ;
to find the area in acres, roods, and perches.
Here we have, 792.385 =30492 = Sac Or 7-872p.
* To the table belong v.nrious parts, as follow.
1. A frame of wood, made to fit round its edges, and to be taken off, for the convenience of
putting a sheet of paper on the table. One side of this frame is usually divided into equal parts,
for drawing lines across the table, pai-ailel or perpendicular to the sides ; and the otiier side of
the frame is divided into 3G0 degrees, to a centre in the middle of the table ; by means of which
the. table may be used as a theodolite, &c.
1. A magnetic needle and compass, either screwed into the side of the table, or fixed beneath
its centre, to point out its directions, .and to be a check on the sights.
3. An inde.v, which is a brass two-foot scale, with cither a small telescope, or open sights set
perpendicularly on the ends. These sights and one edge of the index are in the same plane, and
that is called the fiducial edge of the index.
To use this instrument, take a sheet of paper which will cover it, and wet it to make it
expand ; then spread it flat on the table, pressing down the frame on the edges, to stretch it and
keep it fixed there ; and when the paper is become dry, it will, by contracting again, stretch
itself smooth and flat from any cramps and unevenness. On this paper is to be drawn the plan
or form of the ground measured.
Tiius, begin at any proper part of the groimd, and m.ake a point on a convenient part of the
paper or table, to represent that place on tlie ground ; then fix in that point one leg of the com-
passes, or a fine steel pin, and apply to it the fiducial edge of the index, moving it round till
through the siglits you perceive some remarkable object, as the corner of a field, &c. ; and from
the station point draw a line with the point of the compasses along the fiducial edge of the
index, which is called setting or taking the object : then set another object or corner, and draw
its line; do the same by another ; and so on, till as many objects are taken as may be thought
fit. Then me.isure from the station towards as many of the objects as may be necessary, but
not more, taking the requisite oflfscts to corners or crooks in the edges, l.iying tiie measures
down on tiieir respective lines on the table. Then at any convenient place measured to, fix the
table in the same position, and set the objects which appear from that place ; and so on, as before.
Thus continue till the work is finished, measuring such lines only as are necessary, and de-
termining as many as may be by intersecting lines of direction drawn from different stations.
0/ shifting the paper on the plain tul/le.
When one paper is full, and there is occasion for more, draw a line in any manner through
the farthest point of the last station line, to which the work can be conveniently laid down ;
then take the sheet off the table, and fix another on, drawing a line over it, in a part most con-
venient for the rest of the work ; then fold or cut the old sheet by the line drawn on it, applying
the edge to the line on the new sheet, and as they lie in that position, continue the List station
line on the new paper, placing on it the rest of the measure, beginning at where the old sheet left
oflf. And so on from sheet to sheet.
When the work is done, and you would fasten all the sheets together into one piece, or rough
plan, the aforesaid lines are to be accurately joined together, in the same manner as when the
lines were transferred from the old sheets to the new ones. But it is to be noted, that if the
said joining lines, on the old and new sheets, have not the same inclination to the side of the
table, the needle will not point to the original degree when the table is rectified ; and if the
needle be required to respect still the same degree of the compass, the easiest way of drawing
the line in the same position, is to draw them both pjirallel to the same sides of the table,
by means of the equal divisions marked on the other two sides.
4^4 LA^'D SURVEYLVG.
3. Of the theodolite. The theodolite is a brass circular ring, divided into 360
dej?rees, ttc. and having an index with sights, or a telescope, placed on the centre,
about which the index is moveable ; also a compass fixed to the centre, to point
out courses and check the sights ; the whole being fixed by the centre on a stand
of a convenient height for use *.
4. Of the cross. The cross consists of two pair of sights set at right angles to
each other, on a staff having a sharp point at the bottom, to fix in the groundf.
PROBLEM I.
To measure a line or distance.
To measure a line on the ground with the chain : Ha\ang provided a chain,
with ten small arrows or rods, to fix one into the ground, as a mark, at the end
of every chain ; two persons take hold of the chain, one at each end of it ; and
* In using this insti-ument, an exact account, or field-book, of all measures and things neces-
gan- to be remarked in the plan, must be kept, from which to make out the plan on returning
home from the ground.
Begin at such part of the ground, and measure in such directions as are judged most conve-
nient : taking angles or directions to objects, and measuring such distances as appear necessary
under the same restrictions as in the use of the plain table. And it is safest to fix the theodolite
in the original position at everv station, by means of fore and back objects, and the compass,
exactly as in using the plain table; registering the number of degrees cut off by the index when
directed to each object ; and, at any station, placing the index at the s-ime degree as when the
direction towards that station was taken from the last preceding one, to fix the theodolite there
in the original position.
The best method of laying down the aforesaid lines of direction, is to describe a pretty large
circle ; then quarter it, and lay on it the several numbers of degrees cut off by the index in each
direction, and drawing lines from the centre to all these marked points in the circle. Then, by
means of a parallel ruler, draw from station to station, lines pai-allel to the aforesaid lines drawn
from the centre to the respective points in the circumference.
+ The cross is very us.'ful to measure small and crooked pieces of ground. Tlie method is,
to measure a base or chief line, usually in the longest direction of the piece, from comer to
comer ; and while measuring it, finding the places where perpendiculars would fall on this line,
from the several comers and bends in the boundary of the piece, with the cross, by fixing it, by
trials, on such parts of the line, as that through one pair of the sights both ends of the line may
mppcar, and througli the otlier pair the corresponding bends or comers ; and then measuring the
lengths of the said perpendiculars.
Remarks. Besides the fore-mentioned instruments, which are most commonly used, there
arc some others ; as,
Tl)e perambulator, used for measuring roads, and other great distances, level ground, and by
the »idc» of rivers. It has a wheel of 8.|ft, or half a pole, in circumference, by the turning of
which tlic machine goes forward ; and the distance measured is pointed out by an index, which
is moved roimd by clock-work.
Ijerels. with telescoi)ic or other siglits, are used to find the level between place and place, or
how much one place is higher or lower than another. And in measuring any sloping or oblique
line, either ascending or descending, a small pocket level is useful for sho\ving how many links
for each chain are to be deducted, to reduce the line to the horizontal length.
An nfftft-gtaff is a very useful instrument, for measuring the offsets and other short distances.
It i« 10 link> in length, being divided and marked at each of the 10 links.
Ten tmall arruus, or rods of iron or wood, are used to mark the end of every chain length,
in uirawiring lines. And sometimes pickets or staves with flags, are set up as marks or objecte
of direction.
\ »nou» tcalet are also used in protracting and measuring on the plan or paper; such as plane
tcain, line of rhordu, protractor, compasses, reducing scale, parallel and perpendicular rules, &c.
Of plane uralM, there nhonld be several sizes, as a chain in 1 in, a chain in j of an inch, a chain
in 4 ui inch, kc. And of these, the best for use are those that are laid on the very edges of the
jvory ic«lc, to mark off distances without compasses.
LAND SURVEYING. 495
all the 10 arrows are taken by one of them, who goes foremost, and is called the
leader; the other being called the follower, for distinction's sake.
A picket, or station-stafF, being set up in the direction of the line to be
measured, if there do not appear some marks naturally in that direction, they
measure straight towards it, the leader fixing down an arrow at the end of every
chain, which the follower always takes up, as he comes at it, till all the ten arrows
are used. They are then all returned to the leader, to use over again. And
thus the arrows are changed from the one to the other at every 1 0 chains' length,
till the whole line is finished ; then the number of changes of the arrows shows
the number of tens, to which the follower adds the arrows he holds in his hand,
and the number of links of another chain over to the mark or end of the line.
So, if there have been 3 changes of the arrows, and the follower hold 6 arrows,
and the end of the line cut off 45 links more, the whole length of the line is set
down in Hnks thus, 3645.
When the ground is not level, but either ascending or descending ; at every
chain's length, lay the offset-staff, or link-staff, down in the slope of the chain, on
which lay the small pocket level, to show how many links or parts the slope line
is longer than the true level one ; then draw the chain forward so many links or
parts, which reduces the line to the horizontal direction.
PROBLEM II.
To take angles and bearings.
Let B and C be two objects, or two pickets set up perpen- j^
dicular ; and let it be required to take their bearings, or the /"
angles formed between them at any station. /
1 . With the plain table.
The table being covered with a paper, and fixed on its stand ; place it at the
station A, and fix a fine pin, or a foot of the compasses, in a proper point of the
paper, to represent the place A : close by the side of this pin lay the fiducial
edge of the index, and turn it about, still touching the pin, till one object B can
be seen through the sights : then by the fiducial edge of the index draw a line.
In the same manner draw another line in the direction of the other object C;
and it is done.
2. With the theodolite.
Direct the fixed sights along one of the lines, as AB, by turning the instrument
about till the mark B is seen through these sights ; and there screw the instru-
ment fast. Then turn the moveable index round, till through its sights the other
mark C is seen. Then the degrees cut by the index, on the graduated limb or
ring of the instrument, show the quantity of the angle.
3. With the magnetic needle and compass.
Turn the instrument or compass so, that the north end of the needle point to
the fleur-de-lis. Direct the sights to one mark as B, and note the degrees cut
by the needle : direct the sights to the other mark C, and note again the degrees
cut by the needle. Then, their sum or difference, as the case may be, will give
the quantity of the angle BAG.
4. By measurement with the chain.
Measure one chain length, or any other length, along both directions, as to B
and C. Then measure the distance BC, and it is done. This is easily trans-
406
LAND SURVEYLXG.
ferred to paper, by making a triangle ABC with these three lengths, and then
measuring the angle A.
PROBLEM III.
To survey a triangular field ABC.
1. By the chain. Having set up marks at the corners,
which is to be done in all cases where there are not marks
naturally ; measure with the chain from A to P, where a
pen)endicular would fall from the angle C, and set up a
mark at P, noting down the distance A P. Then complete v P b
the distance AB, by measuring from P to B. Having set down this measure,
return to P, and measure the perpendicular PC. And thus, having the base and
perpendicular, the area from them is easily found. Or, having the place P of
the perpendicular, the triangle is easily constructed.
Or, when practicable, measure all the three sides with the chain, and note
them down. From which the content is easily found, or the figure is con-
structed.
Ex. Suppose AP = 79i, AB = 1321, and PC = 826, to find the area.
2. By taking some of the angles. Measure two sides AB, AC, and the angle
A between them. Or measure one side AB, and the two adjacent angles A and
B. From either of these ways the figure is easily planned ; then by measuring
the perpendicular CP on the plan, and multiplying it by half AB, the content is
found.
PROBLEM IV.
To measure a four-sided field.
1. By the chain. Measure along one of the diagonals, as
AC ; and either the two perpendiculars DE, BF, as in the
last problem; or else the sides AB, BC, CD, DA. From a
either of these the figure may be planned and computed as
before directed. D
Ex. The following measures were taken, AE = 214, AF = 362, AC = 592,
DE = 210, BF = 306.
2. Otherwise, by the chain. Measure, on the longest
side, the distances AP, AQ, AB ; and the perpendiculars
PC, QD. For example. AP= 1 10, AQ = 745, AB =
1110, and PC = 352, QD = 595. ,
3. By taking some of the angles. Measure the diagonal a. P Q B
AC (see the last fig. but one), and the angles CAB, CAD, ACB, ACD. Or
measure the four sides, and any one of the angles, as BAD.
'ITius : AC=591, CAB=37°20',CAD=41°15', ACB=72°25', ACD=54°40'.
Or thus: AB=486, BC=394, CD=410, DA=462, BAD=78° 35'.
PROBLEM V.
To survey any field by the chain only.
First method. Having set up marks at the corners,
where necessary, of the proposed field ABCDEFG,
walk over the ground, and consider how it can best be
divided into triangles and tra])ezium8 ; and measure
them separately, as in the last two problems. Thus,
the foUowmg figure is divided into the two trapeziums
LAND SURVEYING.
497
ABCG, GDEF, and the triangle GCD. Then, in the first trapezium, beginning
it A, measure the diagonal AC, and the two perpendiculars Gm, Bn. Then
he base GC, and the perpendicular Dg. Lastly, the diagonal DF, and the two
perpendiculars pE, oG. All which measures \vrite against the corresponding
parts of a rough figure drawn to resemble the figure surveyed, or set them
Jown in any other form you choose.
Thus : Am=135, An = 410, AC = 550 ; Cq = 152, CG = 440; Fo = 237,
F;j = 288, FD = 520 : mG = 130, nB = 180; gD = 230 ; oG = 1 20, ;)E = SO.
Or thus : Measure all the sides AB, BC, CD, DE, EF, FG, GA; and the
3iagonals AC, GD, GD, DF.
Second method. Many pieces of land may be very
|well surveyed, by measuring any base line, either
within or without them, with the perpendiculars let
fall on it from every corner. For they are by those
means divided into several triangles and trapezoids,
all whose parallel sides are perpendicular to the base
line ; and the sum of these triangles and trapeziums
will be equal to the figure proposed if the base line fall within it ; if not, the
sum of the parts which are without, being taken from the sum of the whole
■which are both within and without, will leave the area of the figure proposed.
In pieces that are not very large, it will be suflSciently exact to find the points,
in the base line, where the several perpendiculars will fall, by means of the cross,
or even by judging by the eye only, and from thence measuring to the corners
for the lengths of the perpendiculars. And it will be most convenient to draw
the line so as that all the perpendiculars may fall within the figure.
Thus, in the annexed figure, beginning at A, and measuring along the line
AG, the distances and perpendiculars on the right and left are as below.
Ai = 315, Ac = 440, Ad = 585, Ae = 610, A/= 990, AG = 1020, bB = 350,
cC= 70, dD = 320, eE = 50, /F = 470, 0.
PROBLEM VI.
To measure the offsets.
Let Ahiklmn be a crooked hedge, a brook, or
other irregular boimdary. From A measure in
a straight direction along the side of it to B.
And in measuring along this line AB, observe"
when you are directly opposite any bends or
corners of the boundary, as at c, d, e, ; and from these measure the per-
pendicular oflfsets, ch, di, . . . ., with the ofl!setstafF, if they are not very large,
otherwise with the chain itself; and the work is done. The register, or field-
book, may be as follows : —
OflFs. left.
Base line AB.
0
ch 62
O A
45 Ac
di 84
220 Ad
ek 70
340 Ae
fl 98
gm 57
Btt 91
510 A/
634 A^r
785 AB
VOL. I.
^jjg LAND SURVEYING.
PROBLEM VII.
To survey any field with the plain table.
1 From one station. Place the table at any angle, as C, from
which all the other angles, or marks set up. can be seen; turn
the table about till the needle point to the fleur-de-lis ; and
there screw it fast. Make a point for C on the paper on the
table and lay the edge of the index to C, turning it about C till
through the sights you see the mark D ; and by the edge of the mdex draw a
dr>- or" obscure line : then measure the distance CD, and lay that distance down
on the line CD. Then turn the index about the point C, till the mark E be
seen through the sights, by which draw a line, and measure the distance to E,
laying it on the line from C to E. In like manner determine the positions of CA
and CB, by turning the sights successively to A and B ; and lay the lengths of
those hnes down. Then connect the points, by drawing the black lines CD, DE,
EA, AB, BC, for the boundaries of the field.
2. From a station within the field. TN'hen all the other parts cannot
be seen from one angle, choose some place O within, or even with-
out, if more convenient, from which the other parts can be seen.
Plant the table at O, then fix it with the needle north, and mark
the point O on it. Apply the index successively to O, turning it ^ ^
round with the sights to each angle, A, B, C, D, E, drawing dry lines to them
by the edge of the index ; then measuring the distances OA, OB, &c. and laying
them down on those lines. Lastly, draw the boundaries, AB, BC, CD, DE,
EA.
3. By going round the figure. When the figure is a wood, or water, or when
from some other obstruction you cannot measure lines across it ; begin at any
point A, and measure around it either within or without the figure, and draw
the directions of all the sides thus : place the table at A ; turn it with the needle
to the north or fleur-de-lis: fix it, and mark the point A. Apply the index
to A, turning it till you can see the point E, and there draw a line : then the
point B, and there draw a line : then measure these lines, and lay them down
from A to E and B. Next move the table to B, lay the index along the line AB,
and turn the table about till you can see the mark A, and screw fast the table ;
in which position also the needle will again point to the fleur-de-lis, as it will do
indeed at every station when the table is in the right position. Here turn the
index about B till through the sights you see the mark C ; there draw a line,
measure BC, and lay the distance on that line after you have set down the table
at C. Turn it then again into its proper position, and in like manner find the
next line CD. And so on, quite around by E, to A again. Then the proof of
the work will be the joining at A ; for if the work be all right, the last direction
E.\ on the ground, will pass exactly through the point A on the paper ; and the
measured distance will also reach exactly to A. If these do not coincide, or
nearly so, some error has been committed, and the work must be examined
over again.
PROBLEM VIII.
To surrey afield with the theodolite.
1. From one point or station. When all the angles can be seen from one point,
as the angle C (first fig. to last prob.), place the instrument at C, and turn it
aliout, till through the fixed sights you see the mark B, and there fix it. Then
LAND SURVEYING.
49a
jra the moveable index about till the mark A be seen through the sights, and
Ota the degrees cut on the instrument. Next turn the index successively to E
nd D, noting the degrees cut off at each ; which gives all the angles BCA, BCE,
iCD. Lastly, measure the lines CB, CA, CE, CD ; and enter the measures in
field-book, or rather, against the corresponding parts of a rough figure drawn
y guess to resemble the field.
2. From a point within or without. Place the instrument at O (last fig.), and
lurn it about till the fixed sights point to any object, as A ; and there screw it
ast. 'ITien turn the moveable index round till the sights point successively to
he other points E, D, C, B, noting the degrees cut oflF at each of them ; which
jives all the angles round the point O. Lastly, measure the distances OA, OB,
3C, OD, OE, noting them down as before, and the work is done.
3. By going round the field. By me£isuring round, either
vithin or without the field, proceed thus. Having set up
narks at B, C, near the corners as usual, plant the instrument
it any point A, and turn it till the fixed index be in the direc-
ion AB, and there screw it fast : then turn the moveable index
.0 the direction AC ; and the degrees cut oflf will be the angle A. Measure the
ine AB, and plant the instrument at B, and there in the same manner observe
he angle A. Then measure BC, and observe the angle C. Then measure the
iistance CD, and take the angle D. Then measure DE, and take the arigle E.
rhen measure EF, and take the angle F. And lastly, measure the distance FA.
To prove the work ; add all the inward angles, A, B, C, etc. together ; for
vhen the work is right, their sum will be equal to twice as many right angles as
;he figure has sides, wanting 4 right angles. But when there is an angle, as F,
hat bends inwards, and you measure the external angle, which is less than two
Tght angles, subtract it from 4 right angles, or 360 degrees, to give the internal
mgle greater than a semicircle or 180 degrees.
4. Otherwise, instead of observing the internal angles, we may take the exter-
lal angles, formed without the figmre by producing the sides farther out. And
n this case, when the work is right, their sum altogether will be equal to 360
iegrees. But when one of them, as F, runs inwards, subtract it from the sum
)f the rest, to leave 360 degrees.
PROBLEM IX.
To survey afield with crooked hedges, etc.
With any of the instruments, measure the lengths and positions of imaginary
ines running as near the sides of the field as you can ; and, in going along them,
neasure the offsets in the manner before taught ; then you will have the plan on
;he paper in using the plain table, drawing the crooked hedges through the ends
)f the offsets ; but in surveying with the theodolite, or other instrument, set
iown the measures properly in a field-book, or memorandum-book, and plan
;hem after returning from the field, by laying down all the lines and angles.
Kk 2
500 . LAND SURVEYING.
So, in surveying the piece ABCDE, set up marks, a, b, c, d, dividing it so as
to have as few sides as may be. Then begin at any station, a, and measure the
lines ab, be, cd, da, taking their positions, or the angles, a, b, c, d ; and, in
going along the lines, measure all the offsets, as at m, n, o, p, &c. along every
station-line.
And this is done either within the field, or without, as may be most con-
venient. When there are obstructions within, as wood, water, hills, or build-
ings, then measure without, as in the next following figure.
PROBLEM X.
To survey afield, or small estate, by two stations.
This is performed by choosing two
stations from which all the marks and
objects can be seen ; then measuring the
distance between the stations, and at
each station taking the angles formed by
every object from the station line or dis-
tance.
The two stations may be taken either
within the bounds, or in one of the sides,
or in the direciion of two of the objects,
or quite at a distance and without the bounds of the objects or part to be sur-
veyed.
In this manner, not only grounds may be surveyed, without even entering
them, but a map may be taken of the principal parts of a county, or the
chief places of a town, or any part of a river or coast surveyed, or any other
inaccessible objects; by taking two stations, on two towers, or two hills, or
such-like.
PROBLEM XI.
To survey a large estate.
If the estate be very large, and contain a great number of fields, it cannot
well be done by surveying all the fields singly, and then putting them together ;
nor can it be done by taking all the angles and boundaries that enclose it. For
in these cases, any small errors will be so much increased, as to render it very
much distorted. But proceed as below,
1. Walk over the estate two or three times, in order to get a perfect idea of
it, or till you can keep the figure of it pretty well in mind. And to help your
memory, draw an eye-draught of it on paper, at least of the principal parts of it,
to guide you ; setting the names within the fields in that draught.
2. Choose two or more eminent places in the estate, for stations, from which
all the principal parts of it can be seen : selecting these stations as far distant
from one another as convenient.
3. Take such angles, between the stations, as you think necessary, and
measure the distances from station to station, always in a right line : these things
mu«t be done, till you get as many angles and lines as are sufficient for deter-
mining all the points of station. And in measuring any of these station-dis-
tances, mark accurately where these lines meet with any hedges, ditches, roads.
LAND SURVEYING. 501
anes, paths, rivulets, &c. ; and where any remarkable object is placed, by mea-
uring its distance from the station- line ; and where a perpendicular from it cuts
hat line. And thus as you go along any main station-line, take offsets to the
nds of all hedges, and to any pond, house, mill, bridge, &c. noting every thing
lown that is remark&b?e.
4. As to the inner parts of the estate, they must be determined, in like
nanner, by new station-lines ; for, after the main stations are determined, and
svery thing adjoining to them, then the estate must be subdivided into two or
hree parts by new station-hnes ; taking inner stations at proper places, where
r^ou can have the best view. Measure these station-lines as you did the first,
md all their intersections with hedges, and offsets to such objects as appear.
Then proceed to survey the adjoining fields, by taking the angles that the sides
Bake with the station-line, at the intersections, and measuring the distances to
jach corner, from the intersections. For the station-lines will be the bases to
Ul the future operations ; the situation of all parts being entirely dependent on
them ; and therefore they should be taken of as great length as possible ; and it
IS best for them to run along some of the hedges or boundaries of one or more
fields, or to pass through some of their angles. All things being determined for
these stations, you must take more inner stations, and continue to divide and
subdivide till at last you come to single fields : repeating the same work for the
inner stations as for the outer ones, till all is done ; and close the work as often
as you can, and in as few lines as possible.
5. An estate may be so situated that the whole cannot be surveyed together ;
because one part of the estate cannot be seen from another. In this case, you
may divide it into three or four parts, and survey the parts separately, as if they
were lands belonging to different persons ; and at last join them together.
6. As it is necessary to protract or lay down the work as you proceed in it,
you must have a scale of a due length to do it by. To get such a scale, measure
the whole length of the estate in chains ; then consider how many inches long
the map is to be ; and from these will be known how many chains you must
have in an inch j then make the scale accordingly, or choose one already made,
PROBLEM XII.
To survey a county, or large tract of land.
1. Choose two, three, or four eminent places, for stations; such as the tops
of high hills or mountains, towers, or church steeples, which may be seen from
one another ; from which most of the towns and other places of note may also
be seen ; and so as to be as far distant from one another as possible. On these
places raise beacons, or long poles, with flags of different colours flying at them,
30 as to be visible from all the other stations.
2. At all the places which you would set down in the map, plant long poles,
tvith flags at them of several colours, to distinguish the places from one another;
[ixing them on the tops of church steeples, or the tops of houses; or in the
centres of smaller towns and villages.
These marks then being set up at a convenient number of places, and such as
fnay be seen from both stations ; go to one of these stations, and, with an instru-
nent to take angles, standing at that station, take all the angles between the
Jther station and each of these marks. Then go to the other station, and take
dl the angles between the first station and each of the former marks, setting
:hem down with the others, each against its fellow with the same colour. You
50S
LAND SURVEYING.
may, if convenient, also take the angles at some third station, which may serve
to prove the work, if the three lines intersect in that point where any mark
stands. The marks must stand till the observations are finished at both stations ;
and then thev may be taken down, and set up at new places. The same opera^
tions must be performed at both stations, for these new places; and the like for
others. The instrument for taking angles must be an exceeding good one,
made on purpose with telescopic sights, and of a good length of radius.
3. And though it be not absolutely necessary to measure any distance, because,
a stationary line being laid down from any scale, all the other lines will be pro-
portional to it; yet it is better to measure some of the lines, to ascertain the
distances of places in miles, and to know how many geometrical miles there are
in any length ; as also from thence to make a scale to measure any distance in
miles. In measuring any distance, it will not be exact enough to go along the
high roads ; which, by reason of their turnings and windings, hardly ever lie in
a right line between the stations ; which must cause endless reductions, and
require great trouble to make it a right line ; for which reason it can never be
exact. But a better way is to measure in a straight line with a chain, between
station and station, over hills and dales, or level fields, and all obstacles. Only
in case of water, woods, towns, rocks, banks, &c. where we cannot pass, such
parts of the line must be measured by the methods of inaccessible distances ;
and besides, allowing for ascents and descents, when they are met with. A good
compass, that shows the bearing of the two stations, will always direct us to go
straight, when the two stations cannot be seen ; and in the progress, if we can
go straight, ofl'sets may be taken to any remarkable places, likewise noting the
intersection of the station-line with all roads, rivers, &c.
4. From all the stations, and in the whole progress, we must be very particular
in observing sea-coasts, river-mouths, towns, castles, houses, churches, mills,
trees, rocks, sands, roads, bridges, fords, ferries, woods, hills, mountains, rills,
brooks, parks, beacons, sluices, floodgates, locks, etc., and in general every
thing that is remarkable.
5. After we have done with the first and main station-lines, which command
the whole county ; we must then take inner stations, at some places already
determined ; which will divide the whole into several partitions : and from these
stations we must determine the places of as many of the remaining towns as we
can. And if any remain in that part, we must take more stations, at some
places already determined, from which we may detennine the rest ; and thus
go through all the parts of the county, taking station after station, till we have
determined the whole. And in general the station-distances must always pass
through such remarkable points as have been determined before, by the former
stations.
PROBLKM XIII.
To survey a town or city.
This may be done with any of the instruments for taking angles, but best of
all with the plain table, where every minute part is drawn while in sight. In-
•lead of the common surveying, or Gunter's chain, it will be best, for this
punicsc, to have a chain 50 feet long, divided into 50 links of one foot each, and
an offset-staiV of IG feet long.
Begin at the meeting of two or more of the principal streets, through which
you can have the longest prospects, to get the longest station-lines: there having
LAND SURVEYING.
503
fixed the instrument, draw lines of direction along those streets, using two men
as marks, or poles set in wooden pedestals, or perhaps some remarkable places
in the houses at the farther ends, as windows, doors, corners, &c. Measure
these lines with the chain, taking oftsets with tlie staff, at all corners of streets,
bendings, or windings, and to all remarkable things, as churches, markets, halls,
colleges, eminent houses, etc. Tlien remove the instrument to another station,
along one of these lines ; and there repeat the same process as before. And so
on till the whole is finished.
Thus, fix the instrument at A,
and draw lines in the direction of
all the streets meeting there ; then
measure AB, noting the street on
the left at tn. At the second sta-
tion, B, draw the directions of
the streets meeting there ; and
measure from B to C, noting the
places of the streets at n and o as
you pass by them. At the third
station, C, take the direction of all the streets meeting there, and measure CD.
At D do the same, and measure DE, noting the place of the cross streets at p.
And in the same manner go through all the principal streets. This done, pro-
ceed to the smaller and intermediate streets; and lastly to the lanes, alleys,
courts, yards, and every part that it may be thought proper to represent in the
plan.
PROBLEM XIV.
To lay down the plan of any survey.
If the survey was taken with the plain table, we have a rough plan of it
already on the paper which covered the table. But if the survey was with any
other instrument, a plan of it is to be drawn from the measures that were taken
in the survey ; and first of all a rough plan on paper.
To do this, you must have a set of proper instruments, for laying down both
lines and angles, as scales of various sizes ; scales of chords, protractors, per-
pendicular and parallel rulers, etc. Diagonal scales are best for the lines, because
they extend to three figures, or chains, and links, which are 100 parts of chains.
But in using the diagonal scale, a pair of compasses must be employed, to take
ofl" the lengths of the principal lines very accurately. But a scale with a thin
edge divided is much readier for laying down the perpendicular offsets to crooked
hedges, and for marking the places of those offsets on the station-line ; which
is done at only one application of the edge of the scale to that line, and then
pricking off all at once the distances along it. Angles are to be laid down, either
with a good scale of chords, which is perhaps the most accurate way, or with a
large protractor, which is much readier when many angles are to be laid down
at one point, as they are pricked off all at once round the edge of the protractor.
In general, all lines and angles must be laid down on the plan in the same
order in which they were measured in the field, and in which they are written
in the field-book ; laying down first the angles for the position of lines, next the
lengths of the lines, with the })laces of the offsets, and then the lengths of the
offsets tlipniselves, all with dry or obscure lines ; then a black line drawn through
the extremities of all the offsets will be the edge or bounding line of the field.
After-t'.;e principal bounds and lines are laid down, and made to fit or close
504 LAND SURVEYING.
properly, proceed next to the smaller objects, till you have entered every thing
that ou^ht to appear in the plan, as, for instance, houses, brooks, trees, hills,
gates, stiles, roads, lanes, mills, bridges, and woodlands.
The north side of a map or plan is commonly placed uppermost, and a me-
ridian is somewhere drawn, with the compass or fleur-de-lis pointing north.
Also, in a vacant part, a scale of equal parts or chains is drawn, with the title
of the map in conspicuous characters, and embellished with a compartment.
Hills are shadowed, to distinguish them in the map. Colour the hedges with
different colours; represent hilly grounds by broken hills and valleys; draw
single dotted lines for foot-paths, and double ones for horse or carriage-roads.
Write the name of each field and remarkable place within it, and, if you choose,
its contents in acres, roods, and perches.
In a very large estate, or a county, draw vertical and horizontal lines through
the map, denoting the spaces between them by letters placed at the top, and
bottom, and sides, for readily finding any field or other object mentioned in a
table.
In mapping counties, and estates that have uneven grounds of hills and val-
leys, reduce all oblique lines, measured up-hill and down-hill to horizontal
straight lines, if that was not done during the survey, before they were entered
in the field-book, by making a proper allowance to shorten them; for which
purpose there is commonly a small table engraven on some of the instruments
for surveying.
PROBLEM XV.
To survey and plan by the new method.
In the former method of measuring a large estate, the accuracy depends both
on the correctness of the instruments, and on the care in taking the angles.
To avoid the errors incident to such a multitude of angles, other methods have
of late years been used by a few skilful surveyors. The most practical, expe-
ditious, and correct, seems to be the following, which is performed, without
taking angles, by measuring with the chain only.
Choose two or more eminences as grand stations, and measure a principal
base-line from one station to another; noting every hedge, brook, or other
remarkable object, as you pass by it ; measuring also such short perpendicular
hnes to the bends of hedges as may be near at hand. From the extremities of
this base-line, or from any convenient parts of the same, go off with other lines
to some remarkable object situated towards the sides of the estate, without
regarding the angles they make with the baseline or with one another; still
remembering to note every hedge, brook, or other object that you pass by. ITiese
lines, when laid down by intersections, will, with the base-line, form a grand
triangle on the estate ; several of which, if need be, being thus measured and
lajd down, you may proceed to form other smaller triangles and trapezoids on
the Hides of the former; and so on till you finish with the inclosures individually.
«y which means a kind of skeleton of the estate may first be obtained, and the
chief lines serve as the bases of such triangles and trapezoids as are necessary to
fill up all the interior i)arts.
The field-book is ruled into three columns, as usual. In the middle one are
•et down the distances on the chain line, at which any mark, offset, or other
obnervation, is made : and in the right and left-hand columns are entered the
oflwU and observations made on the right and left-hand respectively of the
LAND SURVEYING. 505
chain-line ; sketching on the sides the shape or resemblance of the fences or
boundaries.
It is of great advantage, both for brevity and perspicuity, to begin at the
bottom of the leaf, and write upwards ; denoting the crossing of fences by lines
drawn across the middle column, or only a part of such a line on the right and
left opposite the figures, to avoid confusion ; and the comers of fields, and other
remarkable turns in the fences where offsets are taken to, by lines joining in the
manner the fences do ; as will be best seen by comparing the book with the plan
annexed to the field-book, in four engraved pages, following p. 506.
The letter in the left-hand corner at the beginning of every line is the mark
or place measured from ; and that at the right-hand corner at the end is the
mark measured to. But when it is not convenient to go exactly from a mark,
the place measured from is described such a distance from one mark towards
another; and where a former mark is not measured to, the exact place is ascer-
tained by saying, turn to the right or left-hand, such a distance to such a mark,
it being always understood that those distances are taken in the chain-line.
The characters used are, f for turn to the right-hand; ") for turn to the left-
hand ; and -^ placed over an offset, to show that it is not taken at right angles
with the chain-line, but in the direction of some straight fence ; being chiefly
used when crossing their directions : which is a better way of obtaining their
true places than by offsets at right angles.
When a line is measured whose position is determined, either by former work
(as in the case of producing a given line, or measuring from one known place
or mark to another) or by itself (as in the third side of the triangle), it is called
a fast line, and a double line across the book is drawn at the conclusion of it ;
but if its position is not determined (as in the second side of the triangle), it is
called a loose line, and a single line is drawn across the book. When a line
becomes determined in position, and is afterwards continued farther, a double
line half through the book is drawn.
When a loose line is measured, it becomes absolutely necessary to measure
some other line that will determine its position. Thus, the first line ah or bh,
being the base of a triangle, is always determined ; but the position of the
second side hj does not become determined till the third side jb is measured ;
then the position of both is determined, and the triangle may be constructed.
At the beginning of a line, to fix a loose line to the mark or place measured
from, the sign of turning to the right or left-hand must be added, as at h in the
second, and j in the third line ; otherwise a stranger, when laying down the
work, may as easily construct the triangle hjb on the wrong side of the line ah
as on the right one; but this error cannot be fallen into, if the sign above named-
be carefully observed.
In choosing a line to fix a loose one, care must be taken that it does not make
a very acute or obtuse angle ; as in the triangle pBr, by the angle at B being
very obtuse, a small deviation from truth, even the breadth of a point at p or r,
would make the error at B, when constructed, very considerable ; but by con-
structing the triangle pBq, such a deviation is of no consequence.
Where the words leave off are written in the field-book, it signifies that the
taking of offsets is from thence discontinued ; and of course something is want-
ing between that and the next offset, to be afterwards determined by measuring
some other line.
The field-book for this method, and the plan drawn from it, are contained in
the four following pages, engraven on copper-plates ; answerable to which the
pupil is to draw a plan from the measures in the field-book, of a larger size, viz.
506
LAND SURVEYING.
\
to a scale of a double size will be convenient, such a scale bein^ also found oji
most instruments. In doing this, begin at the commencement of the field-book,
or bottom of the first page, and draw the first line ah in any direction at pleasure,
and then the ne.Tt two sides of the first triangle bhj by sweeping intersected
arcs ; and so all the triangles in the same manner, after each other in their order ;
and afterwards setting the perpendicular and other offsets at their proper places,
and through the ends of them drawing the bounding fences.
Note. That the field-book begins at the bottom of the first page, and reads
up to the top ; hence it goes to the bottom of the ne.<t page, and to the top ;
and thence it passes from the bottom of the third page to the top, which is the
end of the field-book. The several marks measured to or from, are here denoted
by the letters of the alphabet, first the small ones, a, b, c, d, . . . and after them
the capitals. A, B, C, D, . . . But, instead of these letters, some surveyors use
the numbers in order, 1, 2, 3, 4, &c.*
• In surveying with the plain table, a field-book is not used, as every thing is drawn on the
table immediately when it is measured. But in surveying with the theodolite, or any other
instrument, some kind of a field-book must be used, to write down in it a register or account of
all that is done and occurs relative to the sur\-ey in hand.
This book every one contrives and rules as he thinks fittest for himself. The following is a
specimen of a form which has been much used by country surveyors. It is ruled in three
columns, as below.
Here Q 1 is the first station, where the angle or bearing is 105° 25'. On the left, at 73 links
in the distance or principal line, is an ofl^set of 92: and at 610 an offset of 24 to a cross hedge.
On the right, at 0, or the beginning, an offset 25 to the comer of the field ; at 248 Brown's
boundary hedge commences ; at 610 an offset 35 ; and at 954, the end of the first line, the 0
denotes its terminating in the hedge. And so on for the other stations.
A line is drawn under the work, at the end of every station-line, to prevent confusion.
Form of this jkldAjook.
Offsets and Remarks
on the left.
Stations. Bearings
and Distances.
Offsets and Remarks
on the right.
O 1
i
lOo- 25'
RO
00
25 corner
92
73
1
24i]
1 Brown's hedge
a cross hedge 24
610
35
954
GO
©2
1
house comer 51
34
53^ 10'
120
734
21
2.0 a tree
40 a stile
a brook
3<J
f.MltjMltll Ifi
truM liciii-c IJJ
O 3
67' 20'
61
248
63.0
itlO
973
16 a spring
20 a pond
Ti.e
LAND SURVEYING. 507
PROBLEM XVI.
To compute the contents of fields.
1. Compute the contents of the figures as divided into trianf^les, or trape-
ziums, by the proper rules for these figures laid down in measuring, multiplying
the perpendiculars by the diagonals or bases, both in links, and divide by 2 ; the
quotient is acres, after having cut oflf five figures on the right for decimals.
Then bring these decimals to roods and perches, by multiplying first by 4, and
then by 40. An example of which is given in the description of the chain,
p. 462.
2. In small and separate pieces, it is usual to compute their contents from the
measures of the lines taken in surveying them, without making a correct plan
of them.
3. In pieces bounded by very crooked and winding hedges, measured by
offsets, all the parts between the offsets are most accurately measured separately
as small trapezoids.
4. Sometimes such pieces as that last mentioned are computed by finding a
mean breadth, by adding all the ofisets together, and dividing the sum by the
number of them, accounting that for one of them where the boundary meets
the station-line (which increases the number of them by 1, for the divisor,
though it does not increase the sum or quantity to be divided) ; then multiply
the length by that mean breadth.
5. But in larger pieces and whole estates, consisting of many fields, it is the
common practice to make a rough plan of the whole, and from it compute the
contents, quite independent of the measures of the lines and angles that were
taken in surveying. For then new lines are drawn in the fields on the plans, so
as to divide them into trapeziums and triangles, the bases and perpendiculars of
which are measured on the plan by means of the scale from which it was drawn,
and so multiplied together for the contents. In this way the work is very expe-
ditiously done, and sufficiently correct ; for such dimensions are taken as afford
the most easy method of calculation : and among a number of parts, thus taken
and applied to a scale, though it be likely that some of the parts will be taken a
small matter too little, and others too great, yet they will, on the whole, in all
probability, very nearly balance one another, and give a sufficiently accurate
Then the plan, on a small scale, dravn
from the above field-book, will be as in
the following figure. But the pupil may
draw a plan of 3 or 4 times the size on
his paper book. The dotted lines denote
the three measured lines, and the bhick
lines the boundaries on the right and left.
But some skilful surveyors now make
use of a different method for tlie field-
book, namely, beginning at the bottom of the page and writing upwards ; sketching also a neat
boundary on cither liand, resembling the parts near the measured lines as they pass along ; an
example of which was given in tlie new method of surveying, in the preceding pages.
In smaller surveys and measurements, a good way of setting down the work is to draw by
the eye, on a piece of paper, a figure resembling that which is to be measured : and so writing
the dimensions, as they are found, against the corresponding parts of the figure. This method
may, also, be practised to a considerable extent, even in the larger surveys.
508
LAND SURVEYING.
result. After all the fields and particular parts are thus computed separately,
and added all totrether into one sum ; calculate the whole estate independent of
the fields, bv dividing it into large and arbitrary triangles and trapeziums, and
add these also together. Then if this sum be equal to the former, or nearly so,
the work is right ; but if the sums have any considerable difference, it is wrong,
and they must be examined, and recomputed, till they nearly agree.
6. But the chief art in computing consists in finding the contents of pieces
bounded by curved or very irregular lines, or in reducing such crooked sides of
fields or boundaries to straight lines, that shall enclose the same or equal area
with those crooked sides, and so obtain the area of the curved figure by means
of the right-lined one, which will commonly be a trapezium. Now this reducing
the crooked sides to straight ones is very easy, and accurately performed in this
manner : — apply the straight edge of a thin, clear piece of lantern-horn to the
crooked line, which is to be reduced, in such a manner, that the small parts cut
off from the crooked figure by it, may be equal to those which are taken in ;
which equality of the parts included and excluded you will presently be able to
judge of very nicely by a little practice : then with a pencil, or point of a tracer,
draw a line by the straight edge of the horn. Do the same by the other sides
of the field or figure. So shall you have a straight-sided figure equal to the
curved one ; the content of which, being computed as before directed, wiU be
the content of the crooked figure proposed.
Or, instead of the straight edge of the horn, a horse-hair, or fine thread, may
be applied across the crooked sides in the same manner ; and the easiest way of
using the thread is to string a small slender bow with it, either of wire, or cane,
or whalebone, or such like slender elastic matter ; for the bow keeping it always
stretched, it can be easily and neatly applied with one hand, while the other is at
liberty to make two marks by the side of it, to draw the straight line by.
Example. Thus, let it be required to find the contents of the same figure as
in prob. 9. p. 499, to a scale of 4 chains to an inch.
Draw the 4 dotted straight lines
AB, BC, CD, DA, cutting off equal
quantities on both sides of them,
which they do as near as the eye
can judge ; so is the crooked figure
reduced to an equivalent right-lined
one of 4 sides, ABCD. Then draw
the diagonal BD, which, by apply-
ing a proper scale to it, measures
suppose 1256. Also the perpendicular, or nearest distance from A to this dia-
gonal, measures 456 ; and the distance of C from it is 428.
Then, half the sum of 456 and 428, multiplied by the diagonal 1256, ^ves
555152 square links, or 5 acres, 2 roods, 8 perches, the content of the trapezium,
or of the irregular crooked piece.
As a general example of this practice, let the contents be computed of all the
fields separately in the foregoing plan facing page 506, and, by adding the con-
tents altogether, the whole sum or content of the estate will be found nearly
equal to 103 i acres. Then, to prove the work, divide the whole plan into two
parts, by a pencil-line drawn across it any way near the middle, as from the
comer / on the right, to the corner near s on the left ; then, by computing these
two large parts separately, their sura must be nearly equal to the former sum,
when the work is all right.
'Hie content of irregular fields, farms, &c. when planned, may be readily and
LAND SURVEYING. /;09
correctly found by the process of weighing *. If the plan be not upon paper, or
fine drawing pasteboard of uniform texture, let it be transferred upon such.
Then cut the figure separately close upon its boundaries, and cut out from the
same paper or pasteboard a square of known dimensions, according to the scale
employed in drawing the plan. Weigh the two separately in an accurate balance,
and the ratio of the weight will be the same as that of the superficial contents.
If great accuracy be required, cut the plan into four portions, called 1, 2, 3, 4.
First, weigh 1 and 2 together, 3 and 4 together, and take their sum. Then
weigh 1 and 3 together, 2 and 4 together, and take their sum. Lastly, weigh
1 and 4 together, 2 and 3 together, and take their sum. The mean of the three
aggregate weights thus obtained, compared with the weight of the standard
square, will give the ratio of their surfaces very nearly.
PROBLEM XVII.
To transfer a plan to another paper.
After the rough plan is completed, and a fair one is wanted ; this may be
done by any of the following methods.
First method. Lay the rough plan on the clean paper, keeping them always
pressed flat and close together, by weights laid on them. Then, with the point
of a fine pin or pricker, prick through all the comers of the plan to be copied.
Take them asunder, and connect the pricked points on the clean paper, with
lines, and it is done. This method is only to be practised in plans of such
figures as are small and tolerably regular, or bounded by right lines.
Second method. Rub the back of the rough plan over with black-lead powder,
and lay this blacked part on the clean paper on which the plan is to be copied,
and in the proper position. Then, with the blunt point of some hard substance,
as brass, or such-like, trace over the lines of the whole plan, pressing the tracer
so much, as that the black-lead under the lines may be transferred to the clean
paper ; after which, take off the rough plan, and trace over the leaden marks
with common ink, or with Indian ink. Or, instead of blacking the rough plan,
we may keep constantly a blacked paper to lay between the plans.
Third method. This is by means of squares. This is performed by dividing
both ends and sides of the plan which is to be copied into any convenient num-
ber of equal parts, and connecting the corresponding points of division with
lines ; which will divide the plan into a number of small squares. Then divide
the paper on which the plan is to be copied into the same number of squares,
each equal to the former when the plan is to be copied of the same size, but
greater or less than the others, in the proportion in which the plan is to be
increased or diminished, when of a different size. Lastly, copy into the clean
squares the parts contained in the corresponding squares of the old plan ; and
you will have the copy, either of the same size, or greater or less in any propor-
tion. See p. 399.
Fourth method. By the instrument called a pentagraph, which also copies the
plan in any size required ; for this purpose, also. Professor Wallace's eidograph
may be advantageously employed.
Fifth method. A very neat process, at least for copying from a fair plan, is
this : procure a copying frame of glass, made in this manner ; namely, a large
* B7 a method like tbis, Dr. Long found the quantities of land and water on our globe to be
very nearly as 2 to 5. He cut up the gores of a globe for the purpose.
510 ARTIFICERS' WORK AND TIMBER MEASURE.
square of the best plate-glass, set in a broad frame of wood, which can be
raised up to any angle, when the lower side of it rests on a table. Set this frame
up to any angle before you, facl'ng a strong light ; fix the old plan and clean
nai)er together, with several pins quite around, to keep them together, the clean
paper being laid uppermost, and over the face of the plan to be copied. Lay
them, with the back of the old plan, on the glass ; namely, that part which you
intend to begin at to copy first ; and by means of the light shining through the
papers, you will very distinctly perceive every line of the plan through the clean
paper. ' In this state, then, trace all the lines on the paper with a pencil. Having
drawn that part which covers the glass, slide another part over the glass, and
copy it in the same mauner. Then another part; and so on, till the whole is
copied. Then take them asunder, and trace all the pencil lines over with a fine
pen and Indian ink, or with common ink. You may thus copy the finest plan,
without injuring it.
ARTIFICERS' WORK AND TIMBER MEASURE.
Artificbrs compute their work in diflferent ways : the chief distinctions of
which are the following : —
1. Glazing and masonry by the foot square *.
2. Painting, plaistering, paving, and paperhanging by the yard square.
3. Flooring, partitioning, roofing, and tiling by the square of 100 feet, or a
square whose side is 10 feet.
4. The removal of earth, as in forming roads and railways, the purchase of
stone, and other works on which volume is concerned, the measures are either
the cubic foot or the cubic yard.
All works, whether of superficial or solid measure, are computed by the rules
proper to the figure of the magnitude concerned, and therefore come under one
or other of the methods already explained for the mensuration of surfaces and
solids. The only peculiarity of the operations as distinct from those already laid
down, is the computation of the value of the work done or the materials sup-
plied. The particular customary allowances to be made are detailed in the notes
to the several kinds of work in which they occur.
I. BRICKLAYERS' WORK.
Brickwork is estimated at the rate of a brick and a half thick : but if a
wall be more or less than this standard thickness, it must be reduced to it, as
follows : —
Multiply the superficial content of the wall by the number of half-bricks in
the thickness, and divide the product by 3 f .
riiis is only the common mode of expressing tlie magnitude which has been previously
doipnatc-ii at llie " jqiiarc foot."
t Thi- aiinensions of a building mav be taken bv measuring half round on the outside and
h»lf roiiii.1 ..ti the iiihide : tlie sum of these two gives the compass of the wall, which, multiplied
bv the hii.'lit, gives liie content of the materials.
Chimncvi arc commonly measured as if they were solid, on account of the trouble, deducting
ARTIFICERS' WORK AND TIMBER MEASURE.
511
EXAMPLES.
1. How many rods of standard brickwork are contained in a wall whose length
or compass is 57 ft 3 in, and height 24 ft 6 in ; the wall being 2^ bricks thick ?
Ans. 8 rods, 17§ yards.
2. Required the content of a wall 62 ft 6 in long, 14 ft 8 in high, and
2i bricks thick ? Ans. 169753 yards.
3. A triangular gable is raised 17i ft high, on an end- wall whose length is
24 ft 9 in, the thickness being 2 bricks : required the content.
Ans. 32-084 yards.
4. The end-wall of a house is 28 ft 10 in long, and 55 ft 8 in high, to the
eaves ; 20 ft high is 2^ bricks thick, other 20 ft high is 2 bricks thick, and the
remaining 15 ft 8 in, is 1^ brick thick; above which is a triangular gable, of 1
brick thick, which rises 42 courses of bricks : what is the content in standard
measure ? Ans. 253'626 yards.
5. Required the number of bricks necessary to build a wall of 2J bricks
thick, the superficial area being 2346 feet.
only the vacuity from the earth to the mantle. All windows, doors, etc. are to be deducted
from the contents of the walls in which they are placed.
The dimensions of a common bare brick are, 8J inches long, 4 broad, and 2^ thick ; but, on
account of the half-inch joint of mortar, when laid in brickwork, every dimension is to be counted
half an inch more; thus making its length 9, its breadth 4J, and thickness 3 inches. Hence,
ever)' 4 courses of brickwork measure 1ft in height.
450 stock bricks weigh about a ton, and 2 hods of mortar make nearly a bushel.
The standard rod requires 4500 bricks of the usual size, including waste.
1 rod of brickwork requires 27 bushels of chalk lime, and 3 loads of road drift or sand.
Taking 4500 for the bricks employed, including waste, in a standard rod of 272 feet face, and
l.J brick thick ; the following table will serve to determine the number of bricks required in any
proposed case.
Area of
1
the face
Number of bricks required for 1, 2, 3, 4, ... feet at the respective thicknesses. |
of the
wall, in
feet.
1 brick.
ii brick.
2 bricks.
2i bricks.
3 bricks.
1
11-02.947
16-54412
22-05883
27-.i7353
33-08824
2
2205883
33088-24
44-11765
55-14707
66-17648
3
3308824
49-63-236
66-17648
82-7-2060
99-26472
4
44.11765
6617648
88-23531
110-29413
132.352.%
5
5514706
82-7-2060
110-29414
137-86766
165-441-20
6
6617648
99-26472
132-352.96
165-44121
198-52944
7
77-20589
115-80884
1.54-41180
193 01473
231-61768
8
88-23531
132-.3.5-296
176-47062
220-58827
264-70592
9
99-26472
148-89708
198-52945
24816180
297-79416
The left-hand column exhibiting the area of the face of a wall in feet, the numbers of bricks
required for 1 brick thick, 1^ brick thick, etc. are shown in the corresponding horizontal column
under the appropriate heading. For greater numbers, being 10 times, 100 times, 1000 times,
etc. the number of square feet specified in any part of the left-hand column, take 10 limes
100 times, 1000 times, etc. the niiniber given under the proper head.
Thus, 5 sq. ft of 2 bricks thick will require 110-294 bricks; 50, 1102-940; 500, 11029400;
and so on.
For much valuable and really practical information, on artificer's work, the reader may refer
to Maynard's edition ofHutton's Alensuration.
512 ARTIFICERS' WORK AND TIMBER MEASURE.
For 2000 take 1000 times the number for 2 55147"07
300 - 100 times - - for 3 8272-06
40 - 10 times - - for 4 1102*94
6 for 6 165-44
Total number of bricks required - 64687'5L
II. MASONS' WORK.
To masonry belong all sorts of stone-work ; and the measure made use of is
a foot, either superficial or solid *.
EXAMPLES.
1. Required the content of a wall, 53 ft 6 in long, 12 ft 3 in high, and 2 ft
thick. Ans. 1310| ft.
2. What is the content of a wall, the length being 24 ft 3 in, height 10 ft 9 in,
and 2 ft thick ? Ans. 521 375 ft.
3. Required the value of a marble slab, at 8s per ft ; the length being 5 ft 7 in,
and breadth 1 ft 10 in. Ans. 4Z 1* 10|d.
4. In a chimney-piece, suppose the length of the mantle and slab, each 4 ft
6 in, the breadth of both together 3 ft 2 in, the length of each jamb 4 ft 4 in,
and the breadth of both together. 1 ft 9 in. Required the superficial content ?
Ans. 21 ft 10 in.
III. CARPENTERS' AND JOINERS' WORK.
To this branch belongs all the wood-work of a house, such as flooring, parti-
tioning, roofing, etc.f.
* Walls, columns, blocks of stone or marble, etc. are measured by the cubic foot ; and pave-
ments, slabs, chimney-pieces, etc. by the superficial or square foot.
Cubic or solid measure is used for the materials, and square measure for the workmanship.
In the solid measure, the true length, breadth, and thickness are measured, and multiplied
together. In the supei-ficial, the length and breadth are taken of every part of the projection
which is seen without the general u]>right face of the building.
A ton of Portland stone is about 16 cubic feet ; of Bath stone, 17 ; of granite, 13.^ ; of marble,
at a medium, 13 cubic feet.
t I*irgc and plain articles are usually measured by the square foot or yard, etc. ; but enriched
rnouldin^'s, and some other articles, are often estimated by running or lineal measure ; and some
thing* arc mtcd by the jjiece.
In mc.-uuriug of joists, take the dimensions of one joist, considering that each end is let into
ihc wall about J of the thickness, and multiply its content by the number of them.
I'liriitums arc measured from wall to wall for one dimension, and from floor to floor, as far
M ibcy extend, for the other.
Thr nu-tisuri- of veritrriiio for cellars is found by making a string pass over the surface of the
arrh for the t.r<-;ulth, and taking the length of the cellar for the length : but in groin centering.
It it iMual to allow double measure, on account of their greater trouble.
In
SLATERS' AND TILERS' WORK. 513
EXAMPLES.
1. Required the content of a floor, 48ft 6in long, and 24ft Sin broad ?
Ans. 1 1 sq. 76^ ft.
2. A floor being 36ft Sin long, and l6ft Gin broad, how many squares are
in it ? Ans. Ssq 984ft.
3. How many squares of partitioning are there in I73ft lOin in length, and
10ft 7in in height ? Ans. 18-3973 sq.
4. What was the cost of roofing a house at 10« 6d a square; the length within
the walls being 52ft Sin, and the breadth 30ft 6in ; reckoning the roof J of
the flat? Ans. 12n2s lljrf.
5. Required the cost, at 6s per square yard, of the wainscoting of a room; the
height, including the cornice and mouldings, being 12ft 6in, and the whole
compass 83ft 8in; also the three window-shutters being each 7ft Sin by 3ft
6in, and the door 7ft by 3ft 6in, which being worked on both sides must be
reckoned work and half work. Ans. 36/ I2s 2jid.
IV. SLATERS' AND TILERS' WORK.
In this work, the content of a roof is found by multiplying the length of the
ridge by the girt from eaves to eaves ; making allowance in this girt for the
double row of slates at the bottom, or for how much one row of slates or tiles is
laid over another*.
In roofing, the dimensions, as to length, breadth, and depth, are taken as in flooring joists, and
the contents computed the same way.
In floor-boarding, multiply the length by the breadth of the room.
For stair-cases, take the breadth of all the steps, by making a line ply close over them, from
the top to the bottom ; and multiply the length of this line by the length of a step, for the
whole area. By the length of a step is meant the length of the front and the returns at the two
ends ; and by the breadth is to be understood the girts of its two outer surfaces, or the tread and
riser.
For the balustrade, take the whole length of the upper part of the hand-rail, and girt over its
end till it meet the top of the newel-post, for the one dimension ; and twice the length of the
baluster on the landing, with the girt of the hand-rail, for the other dimension.
For wainscoting, take the compass of the room for the one dimension ; and the height from
the floor to the ceiling; making the string ply close into all the mouldings, for the other.
For doors, multiply the height into the breadth, for the area. If the door be panneled on
both sides, take double its measure for the workmanship ; but if one side only be panneled, take
the area and its half for the workmanship. For the surrounding architrave, girt it about the
uppermost part for its length ; and measure over it, as far as it can be seen when the door is
open, for the breadth. Window-shutters, bases, etc. are measured in like manner.
In measuring of joiners' work, the string is made to ply close into all mouldings, and to every
part of the work over which it passes.
Note. 64 cubic feet of fir, 60 of elm, 45 of ash, 39 of oak, make each a ton, at a medium.
Battens are 7 inches, deals 9, and planks 1 1 inches wide.
* When the roof is of a true pitch, that is, forming a right angle at the top; then the breadth
of the building, with its half added, is the girt over both sides nearly.
In angles formed in a roof, running from the ridge to the eaves, when the angle bends inwards,
it is called a valley ; but when outwards, it is called a hip.
Deductions are made for chimney-shafts or \vindow-hole8.
6 inch gage, 1 square requires 760 plain tiles
7 1 660
8 1 576 Five
VOL. I. L 1
514 PAINTERS' WORK.
EXAMPLES.
1. Required the content of a slated roof, the length being 45ft gin, and the
whole girt 34ft Sin. Ans. ir44'gyds.
2. To how much amounts the tiling of a house, ,at 25s 6cl per square ; the
length being 43ft lOin, and the breadth on the flat 27ft Sin ; also the eaves
projecting l6in on each side, and the roof of tnie pitch ? Ans. 24l 9s S^d.
I
V. PLASTERERS' WORK.
Plasterers' work is of two kinds, which are measured separately; namely,
ceiling, which is plastering on laths j and rendering, which is plastering on
walls*.
EXAMPLES.
1. Find the content of a ceiling which is 43ft 3in long, and 25ft 6in broad.
Ans. 122|yds.
2. Required the cost of the ceiling of a room at lOd per yd ; the length being
21ft Sin, and the breadth 14ft lOin. Ans. ll 9s 8^d.
3. The length of a room is 18ft 6in, the breadth 12ft 3in, and height 10ft
6in; what is the amount of ceiling and rendering, the former at 8d and the
latter at 3c? per yd : allowing for the door of 7ft by 3ft Sin, and a fire-place of
5ft square ? Ans. ll I3s 3^d.
4. Required the quantity of plastering in a room, the length being 14ft 5in,
breadth 13ft 2in, and height 9ft 3in to the under side of the cornice, which
girts S^in, and projects Sin from the wall on the upper part next the ceiling ;
deducting only for a door 7ft by 4.
Ans. 53yd8 5ft 3§in of rendering, 18yds 5ft 6in of ceiling, and SQftOl^in of cornice.
VI. PAINTERS' WORK.
Painters' work is computed in square yards. Every part is measured where
the colour lies ; and the measuring line is forced into all the mouldings and
corners.
Windows are painted at so much a piece : and it is usual to allow double
measure for carved mouldings and other ornamental works.
Five hundred feet in length of laths make a bundle ; and is the quantity usually allowed to
a square of tiling.
A square of Westmoreland slates will weigh half a ton ; of Welsh rag from | of a ton to a
ton ; and a square of pantiling weighs about 7J cwt.
• The contents are estimated either by the foot or the yard, or the square of 100 feet. En-
rirhod mouldings, etc. are rated by running or lineal measure.
Deductions arc made for chimneys, doors, windows, and other apertures.
3 cwu. oflinie, 4 loads of sand, and 10 bushels of hair, are allowed to 200 yards of rendering.
1 bundle of laths, and 500 of nails, are allowed to cover 4^ square yards.
1 b«ml of cement is 5 bushels, and weighs 3 cwt. 1 rod of brickwork in cement requires
36 buiheli of cement and ."W bushels of sand.
PAVERS' WORK. 515
EXAMPLES.
1. How many yards of painting are there in a room which is 65ft 6in in com-
pass, and 12ft 4in high ? Ans. 89^Jyds.
2. The length of a room being 20ft, its breadth 14ft 6in, and height I Oft
4in : how many yards of painting are there, deducting a fire-place of 4ft by 4ft
4in, and two windows, each 6ft by 3ft 2in ? Ans. 73i'^yds.
3. Required the cost of painting a room of the following dimensions at 6rf a
yd: viz. the length 24ft 6in, the breadth 16ft 3in, and the height 12ft 9in;
the door 7ft by 3ft 6in, and the fire-place 5ft by 5ft 6in ; also the shutters
to the two windows each 7ft 9in by 3ft 6in, the breaks of the windows 8ft
6in high by ift Sin deep, and the window-cills and soflSts determinable from
the dimensions already given. Ans. 3/ 3* lOf rf.
VII. GLAZIERS' WORK.
Glaziers take their dimensions, either in feet, inches, and parts, or feet,
tenths, and hundredths ; and they compute their work in square feet *.
EXAMPLES.
1. How many square feet are there in the window which is 4"25ft long, and
275ft broad? Ans. ll§ft.
2. What will the glazing a triangular sky-light cost at lOd per foot ; the base
being 12ft 6in, and the height 6ft 9in? Ans. 1/ 15s Ifrf.
3. There is a house with three tiers of windows, three windows in each tier,
their common breadth 3ft llin: and their height are 7ft lOin, 6ft Sin, and
5ft 4in respectively. Required the expense of glazing at 14d per foot.
Ans. 13/1 Is lO^d.
4. Required the expense of glazing the windows of a house at I3d a foot; there
being three stories, and three windows in each story ; the heights of which are
respectively 7ft 9in, 6ft 6in, and 5ft 3|in, and of an oval window over the
door 1ft lOjin : also the common breadth of all the windows 3ft 9in.
Ans. 12; 5s 6d.
VIII. PAVERS' WORK.
Pavers' work is done by the square yard : and the content is found by multi-
plying the length by the breadth.
EXAMPLES.
1. What cost the paving a foot-path, at 3s 4rf a yard; the length being 35ft
4in, and breadth 8ft 3in ? Ans. 5/ 7s U^d.
* In taking the length and breadth of a window, the cross bars between the squares are in-
cluded. Windows also of round or oval forms are measured as square, measuring them to their
greatest length and breadth, on account of the waste in cutting the glass.
l1 2
5lg TIMBER MEASURING.
2. "VNTiat was the expense of paving a court, at 3s 2d per yd; the length heing
27ft lOin, and the breadth 14ft Qin ? Ans. 71 is bid.
3. What will be the expense of paving a rectangtdar court-yard, whose length
is 63ft, and breadth 45ft ; in which there is laid a foot-path of 5ft Sin broad,
running the whole length, with broad stones, at 3s a yd ; the rest being paved
with pebbles at 2s 6d a yd ? Ans. 401 5s 10^
IX. PLUMBERS' WORK.
Plumbers' work is rated at so much a pound; or else by the hundred weight
of 112 pounds*.
EXAMPLES.
1. Required the weight of the lead which is 39ft 6in long, and 3ft 3in broad,
at 8ilbs to the square foot. Ans. lOQl^lbs.
2. Find the cost of covering and guttering a roof with lead, at 18s per cwt;
the length of the roof being 43ft, and breadth, or girt over it, 32ft; the guttering
57ft long, and 2ft wide; the former 98311b, and the latter 7'373lb to the square
foot. Ans. 115? 9s Hd.
X. TIMBER MEASURING.
PROBLEM I.
To find the area, or superficial content, of a board or plank.
Multiply the length by the mean breadth, when the breadths of each end arc
equal ; but when the board is tapering, add the breadths at the two ends to-
gether, and take half the sum for the mean breadth ; or, if convenient, take the
breadth in the middle.
By the sliding rulef.
Set 12 on B to the breadth in inches on A ; then against the length in feet on
B, is the content on A, in feet and fractional parts.
• Sheet lead, used in roofiDg, guttering, &c. weighs from 61b. to 101b. to the square foot; and
pipe of an inch bore is commonly 13 or 141b. to the yard in length.
A square foot an eighth of an inch thick, weighs 7'38 or 741b. nearly; a quarter of an inch
thick l4Jlb., and so on.
t Ttic Carpenter's or Sliding Rule is an instniment much used in measuring of timber and
artificer*' works, both for Uking the dimensions, and computing the contents.
The instniuient consists of two equal pieces, each a foot in length, which are connected toge-
ther by a folding joint.
One side or faro of the rule is divided into inches, and eighths, or half-quarters. On the same
fate also arc several plane scales divided into twelfth parts by diagonal lines ; which are used in
plannmg dimensions that are Uken in feet and inches. The edge of the rule is commonly
divided decimally, or into tenths ; namely, each foot into ten equal parts again ; so that by
means of this last scale, dimensions are taken in feet, tenths, and hundredths, and multiplied as
rommon decimal number*, which is the best way. On
TIMBER MEASURING. 517
EXAMPLES.
1 . What is the value of a plank, at l^d per foot, whose length is 12ft 6in, and
mean breadth 11 in? Ans. Is bd.
2. Required the content of a board, whose length is lift 2in, and breadth
1ft lOin. Ans. 20ft 5^in.
3. What is the value of a plank, which is 12ft 9in long, and 1ft Sin broad,
at 2\d a ft ? Ans. 3« Z\d.
4. Required the value of 5 oaken planks, at Zd per ft, each of them being
17ift long; and their several breadths as follows, namely, two of IS^in in the
middle, one of H^in in the middle, and the two remaining ones, each I Sin at
the broader end, and ll^in at the narrower ? Ans. 1/ bs Q^d.
PROBLEM ir.
To find the solid content of squared or four-sided timber.
Multiply the mean breadth by the mean thickness, and the product again by
the length, for the content nearly.
By the sliding rule.
CD DC
As length : 12 or 10 : : quarter girt : solidity.
That is, as the length in feet on C, is to 12 on D, when the quarter girt is in
inches, or to 10 on D, when it is in tenths of feet ; so is the quarter girt on D,
to the content on C.
If the tree taper regularly from the one end to the other ; either take the
mean breadth and thickness in the middle, or take the dimensions at the two
ends, and half their sum -will be the mean dimensions : which multiplied as
above, will give the content nearly.
If the piece do not taper regularly, but be unequally thick in some parts and
small in others; take several different dimensions, add them all together, and
divide their sum by the number of them, for the mean dimensions.
EXAMPLES.
1. The length of a piece of timber is 18ft 6in, the breadths at the greater and
less end 1ft 6in and 1ft 3in, and the thickness at the greater and less end 1ft
3in and ift; required the content. Ans. 28ft 7in.
On the one part of the other face are four lines, marked A, B, C, D ; the two middle ones B
and C being on a slider, which runs in a groove made in the stock. The same numbers serve
for both these two middle lines, the one being above the numbers, and the other below.
These four lines are logarithmic ones, and the three A, B, C, which are all equal to one
another, are double lines, as they proceed twice over from 1 to 10. The other or lowest line,
D, is a single one, proceeding from 4 to 40. It is also called the girt-line, from its use in com-
puting the contents of trees and timber; and on it are marked \VG at 17"15, and AG at 18'95,
the wine and ale gage points, to make this instrument serve the purpose of a gaging rule.
On the other part of this face there is a table of the value of a load, or 50 cubic feet of timber,
at all prices, from 6 pence to 2 shillings a foot.
When 1 at the beginning of any line is accounted 1, then the 1 in the middle will be 10,
and the 10 at the end 100 ; but when 1 at the beginning is counted 10, then the 1 in the middle
is 100, and the 10 at the end 1000 ; and so on. And all the smaller divisions are altered pro-
portionally.
518 QUESTIONS IN MENSURATION.
2. "What is the content of the piece of timber, whose length is 24|ft, and the
mean breadth and thickness each l-04ft ? Ans. 26ift.
3. Required the content of a piece of timber, whose length is 20-38ft, and its
ends unequal squares, the side of the greater being 19gin, and the side of the
less 9jin. Ans. 29-7562ft.
4. Required the content of the piece of timber, whose length is 27"36ft ; at
the (greater end the breadth is l78ft and thickness r23ft; and at the less end
the breadth is l-04ft, and thickness OQlft. Ans. 41-278ft.
PROBLEM III.
To find the solidity of round or unsquared timber.
Multiply the square of the quarter girt, or of i of the mean circumference
by the length, for the content.
By the sliding rule.
As the length upon C : 12 or 10 upon D : : quarter girt in 12"" or 10"", on
D : content on C.
1. When the tree is tapering, take the mean dimensions as in the former pro-
blems, either by girting it in the middle, for the mean girt, or at the two ends,
and taking half the sum of the two ; or by girting it in several places, then add-
ing all the girts together, and dividing the sum by the number of them, for the
mean girt : but when the tree is very irregular, divide it into several lengths, and
find the content of each part separately.
2. This rule, which is commonly used, gives the answer about J less than the
true quantity in the tree, or nearly what the quantity would be, after the tree
is hewed square in the usual way ; so that it seems intended to make an allow-
ance for the squaring of the tree.
On this subject, however, Hutton's Mensuration, pt. v. sect. 4, may be ad-
vantageously consulted,
EXAMPLES.
1. A piece of round timber being 9ft 6in long, and its mean quarter girt
42in ; what is the content ? Ans. Il6|ft.
2. The length of a tree is 24ft, its girt at the thicker end 14ft, and at the
smaller end 2ft ; required the content. Ans. 96ft.
3. What is the content of a tree whose mean girt is S'lSft, and length 14ft
6'n ? Ans. 8-9922ft.
4. Required the content of a tree, whose length is l7ift, and which girts in
five different places as follows, namely, in the first place 9 43ft, in the second
7-92ft, in the third 6-15ft, in the fourth 474ft, and in the fifth 3-l6ft.
Ans. 42-519525.
PRACTICAL EXERCISES IN MENSURATION.
1. What difference is there between a floor 28ft long by 20 broad, and two
others, each of half the dimensions : and what do all three come to at 45* per
•quare of looft ? ^^3 ^■^f 28oft; amount 18 guineas.
2. An elm plank is 14ft 3in long, and I would have just a square yard slit off
«t ; at what distance from the edge must the line be struck ? Ans. 7;iin.
QUESTIONS IN MENSURATION. 519
3. A ceiling contains IHyds 6ft of plastering, and the room is 28ft broad;
jivhat is the length of it ? Ans. 36?ft.
4. A common joist is 7in deep, and 2^in thick ; but I want a scantling just as
I big again, that shall be Sin thick; what will the other dimension be?
I Ans. 1 1 Jin.
! 5. A wooden trough, length 102in, and depth 21in, cost me 3*2rf painting,
within, at 6d per yd : what was the width ? Ans. 27iin.
6. If my court-yard be 47ft 9in square, and I have laid a foot-path with Pur-
beck-stone, of 4ft wide, along one side of it j what will paving the rest with
flints come to at 6d per square yd ? Ans. 5/ l6s 0§rf.
7. A ladder, 36ft long, may be so placed, that it shall reach a window 30*7ft
from the ground on one side of the street ; and, by only turning it over, without-
moving the foot out of its place, it will do the same by a window 18'9ft high on
the other side ; what is the breadth of the street, and the angle of elevation of
the second window from the first ?
Ans. the street is 49-441 4ft wide ; and the elevation is 13° 25' 24".
8. The paving of a triangular court, at 18d per ft, came to 100/; the longest
of the three sides was 88ft ; required the sum of the other two equal sides ?
Ans. 106-85ft.
9. The perambulator, or surveying-wheel, is so contrived, as to turn twice in
the length of a pole, or l6ift ; required the diameter. Ans. 2*626ft.
10. In turning a one-horse chaise within a ring of a certain diameter, it was
observed, that the outer wheel made two turns, while the inner made but one :
the wheels were both 4ft high ; and, supposing them fixed at the statutable dis-
tance of 5ft asunder on the axle-tree, what was the circumference of the track
described by the outer wheel? Ans. 62"832ft.
11. What is the side of that equilateral triangle, whose area cost as much
Daving at 8d a ft, as the palisading the three sides did at a guinea a yd ?
Ans. 72-746ft.
12. A roof, which is 24ft 8in by 14ft 6in, is to be covered with lead at 8lb per
square ft; find the price at 18s per cwt. Ans. 22/ 19* lO^rf.
13. Having a rectangular marble slab, 5Sin by 27, I would have a square foot
:ut off parallel to the shorter edge ; I would then have the like quantity divided
Tom the remainder parallel to the longer side ; and this alternately repeated, till
there shall not be the quantity of a foot left ; what will be the dimensions of the
'emaining piece * ? Ans. 20"7in by 6'086.
14. Given two sides of an obtuse-angled triangle, which are 20 and 40 poles;
equired the third side, that the triangle may contain just an acre of land ?
Ans. 58-876 or 23-099.
15. How many bricks will it take to build a wall, lOft high, and 500ft long,
)f a brick and half thick, reckoning the brick 10 inches long, and four courses
0 the foot in height ? Ans. 72000.
16. How many bricks will build a square pyramid of 100ft on each side at the
)ase, and also 100ft perpendicular height, the dimensions of a brick being sup-
)08ed lOin long, Sin broad, and 3in thick ? Ans. 3840000.
17. If, from a right-angled triangle, whose base is 12, and perpendicular 16ft,
* This question may be solved neatly by an algebraical process, as may be seen in the Ladies'
)iary for lt5"33. In applying the foimulae there found, the term to stop at is that whose ordinal
umber is the number of entire feet in the slab : which in the present case is 10, since 58.*27 ^
OJft.
520 QUESTIONS IN MENSURATION.
a line be drawn parallel to the perpendicular, cutting off a triangle whose area is
24ft ; required the sides of this triangle. Ans. 6, 8, and 10.
18. If a round pillar, 7in across, have 4ft of stone in it ; of what diameter is
the column, of equal length, that contains 10 times as much ? Ans. 22136in.
19. A circular fish-pond is to be made in a garden, that shall take up half an
acre ; required the length of the cord that strikes the circle ? Ans. 274yd8.
20. When a roof is of a true pitch, the rafters are f of the breadth of the
building. Now supposing the eaves-boards to project lOin on a side, what will
the new ripping a house cost, that measures 32ft 9in long, by 22ft 9in broad on
the flat, at 15* per square ? Ans. 8/ 15s 9|d.
21. A cable, which is 3ft long and 9in in compass, weighs 22lb ; required the
weight of a fathom of a cable which measures a foot round ? Ans. 78|]b.
22. A plumber has put 28lb per square foot into a cistern, 74in and twice the
thickness of the lead long, 26in broad, and 40 deep ; he has also put three stays
across it within, l6in deep, of the same strength, aud reckons 22s per cwt for
work and materials: a mason has in return paved him a workshop, 22ft lOin
broad, with Purbeck-stone, at 7d per ft ; and upon the balance finds there is
3* 6d due to the plumber : what was the length of the workshop, supposing
sheet lead i^ of an inch thick to weigh 5-899lb per ft ? Ans. 32-282oft.
23. The girt or outside circumference of a vessel is 44in, the hoop is lin
thick, and the height of the vessel is 24in ; required its content in imperial
gallons. Ans. 9 7892 gallons.
24. If 20ft of iron railing weigh half a ton, when the bars are an inch and
quarter square ; what will 50ft come to at 3hd per lb, the bars being but | of an
inch square ? Ans. 20/ Os 2d,
25. It is required to find the thickness of the lead in a pipe, of an inch and
quarter bore, which weighs 14lb per yd in length ; the cubic foot of lead weigh-
ing 11325 ounces? Ans. •20737in.
26. Supposing the expense of paving a semicircular plot, at 2s 4d per ft, come
to 10/; what is its diameter? Ans. 14-7737ft.
27. What is the length of a chord which cuts off one-third of the area from a
circle whose diameter is 289 ? Ans. 27867 16.
28. My plumber has set me up a cistern, and, his shopbook being burnt, he
has no means of bringing in the charge, and I do not wish to take it down to
have it weighed ; but by measure he finds it contains 64^3 square feet, and that
it is precisely I of an inch in thickness. If lead was then wrought at 2/ per
fother of 19§cwt., can we from these items make out the bill, allowing 6g oz for
the weight of a cubic inch of lead ? Ans. 4/ 1 Is 2d.
29- What will the diameter of a globe be, when the solidity and superficial
content are expressed by the same number ? Ans. 6.
30. A sack, that would hold 3 bushels of corn, is 22^in broad when empty;
what will another sack contain, which, being of the same length, has twice
its breadth ? Ans. 12 bushels.
31. A carpenter is to put an oaken curb to a round well, at 8d per foot
square ; the breadth of the curb is to be 8in, and the diameter within 3§ft : what
will be the expense? Ans. 5s 9|(f.
32. A gentleman has a garden 100ft long, and 80ft broad ; and a gravel walk
i« to be made of an equal width half round it : determine both by construction
and calculation the breadth of the walk, to take up just half the ground.
Ans. 25-968ft.
33. The top of a may-pole, being broken off by a blast of wind, struck the
ground at 15fi from the foot of the pole; what was the height of the whole
may-pole, supposing the length of the broken piece to be 39ft ? Ans. 75ft.
QUESTIONS IN MENSURATION.
5^1
34. Seven men bought a grinding-stone of 60in diameter, each paying i, part
!)f the expense ; what part of the diameter must each grind down for his share i
V.lso, exhibit the solution by a geometrical construction.
Ans. the ls'4-4508, 2'* 4-8400, 3'^ 5-3535, 4t»> 6-0765, 5* 7*2079, 6* 9*3935,
7* 22-6778in.
35. A maltster has a kiln, that is ]6ft 6in square; but he wants to pull it
lown, and build a new one, that may dry three times as much * the old one ;
ivhat must be the length of its side ? Ans. •28ft 7in.
36. How many 3in cubes may be cut out of a 12in cube ? Ans. 64.
37. How long must the tether of a horse be, that will allow him to graze an
icre of ground ? Ans. 39iyds.
38. "What will the cost of painting a conical spire come to at 8d per yd ;
the height being 1 18ft, and the circuit of the base 64ft ? Ans. 14/ Os 8frf.
39. The diameter of an old standard corn bushel is 18^i», and its depth Sin;
what must be the diameter of that bushel whose depth is 7^in ? Ans. 19-1067in.
40. The ball on the top of St. Paul's church is 6ft diameter; what did gilding
it cost at 3|d per square inch ? Ans. 237/ 10s \d.
41. What will a frustum of a marble cone come to at 12« per ft ; the dia-
meter of the greater end being 4ft, that of the less end l^ft, and the length of
the slant side 8ft ? Ans. 30/ 1* lOid.
42. Divide a cone into three equal parts by sections parallel to the base, and
find the heights of the three parts, that of the whole cone being 20in.
Ans. the upper 13-867, the middle 3*605, the lower 2-528.
43. A gentleman has a bowling-green, 300ft long, and 200ft broad, which he
wishes to raise ift higher, by means of the earth to be dug out of a ditch sur-
rounding it : to what depth must the ditch be dug, supposing its breadth to be
every where 8ft ? Ans. 7iBft.
44. How high above the earth must a person be raised, that he may see 3 of
its surface : and under what angle will the earth then appear ?
Ans. to the height of the earth's diameter ; angle 38° 56' 32".
45. A cubic foot of brass is to be drawn into wire of :n,inch in diameter ; what
will the length of the wire be, allowing no loss in the metal ?
Ans. 97784-797yds, or 55mls 984-797yds.
46. Of what diameter must the bore of a cannon be, which is cast for a ball of
24lb weight, so that the diameter of the bore may be ig of an inch more than
that of the ball ? Ans. 5-647in.
47. Supposing the diameter of an iron 9lb ball to be 4in, it is required to find
the diameter of the several balls weighing 1, 2, 3, 4, 6, 12, 18, 24, 32, 36, and
42lb, and the calibre of their guns, allowing jg of the calibre, or ^g of the ball's
diameter, for windage.
Answer.
Wt.
Diameter
Calibre
Wt.
Diameter
Calibre
ball.
ball.
gun.
ball.
ball.
gun.
1
1-9230
1-9622
12
4-4026
4-4924
2
2-4228
2-4723
18
5-0397
5-1425
3
27734
2-8301
24
5-5469
5-6601
4
3-0526
3-1149
32
6-1051
6-2297
6
3-4943
3-5656
36
6-3496
6-4792
9
4-0000
4-0816
42
66844
6-8208
48. Supposing the windage of all mortars to be ^ of the calibre, and the
diameter of the hollow part of the shell to be -^ of the calibre of the mortar : it
s^?o
oi
QUESTIONS IN MENSURATION.
is required to determine the diameter and weight of the shell, and the quantity
or weight of powder requisite to fill it, for each of the several sorts of mortars,
namely, the 13, 10, 8, 5-8, and 4-6in mortar.
Answer.
Calibre
Diameter
Wt. shell
Wt. of
Wt. shell
mort.
ball.
empty.
powder.
filled.
46
4-523
8-320
0-583
1 8-903
5-8
5703
16-677
1-168
! 17-845
8
7-867
43-764
3065
i 46-829
10
9833
85-476
5-986
1 91-462
13
12783
187-791
13-151
1 200-942
49. If a heavy sphere, whose diameter is 4in, be let fall into a conical glass,
full of water, whose diameter is 5, and altitude 6in ; it is required to deter-
mine how much water will be displaced. Ans. 26-2/2 cubic in, or nearly Jfpint.
50. The dimensions of a sphere and cone being the same as in the last ques-
tion, and the cone only 3 full of water ; what part of the axis of the sphere is
immersed in the water? Ans. -546 parts of an inch,
51. (1) If R and r be the radii of two spheres inscribed in a cone, so that the
greater may touch the less, and that planes are drawn to touch the spheres at
their intersections with the axis of the cone : it is required to prove that the
volumes of the three cones thus cut off by the planes, and estimated from the
vertex, are respectively expressed by
2ff T^ 2ff RV2 , 27r R5
3 RcR — ryY
RV2 , 27r
(2) It is likewise required to prove that if any number of spheres are inscribed
in a cone to touch each other in succession, that they will be in geometrical
progression.
52. If a person, in an air balloon, ascend vertically from London, to such
height that he can just see Oxford in the horizon ; required his height above
the earth, supposing its circumference to be 25000 miles, and the distance
between London and Oxford 495933 miles ? Ans. nearly i^^Vjm, or 547yds 1ft.
53. In a garrison there are three remarkable objects, A, B, C, the distances of
which from one to another are known to be, AB = 213, AC = 424, and BC =
262 yds. I am desirous of knowing my position and distance at a place or
station S, from whence I observed the angle ASB = 13° 30', and the angle CSB
= 29° 50', both by geometry and trigonometry, the point S being on the same
side of AC with B. Ans. AS = 6057122, BS = 429-6814, CS = 524-2365.
54. Required the same as in the last question, when the point B is on the
other side of AC, supposing AB = 9, AC = 12, and BC = 6 furlongs ; also
the angle ASB = 33° 45', and the angle BSC = 2-2° 30'.
Ans. AS = 10-65, BS = 15-64, CS = 14-01.
55. It IS required to determine the magnitude of a cube of standard gold,
which Khali be etjual to £960000000 ; supposing a guinea to weigh Sdwts 9^grs.
Ans. 23-549ft.
56. The ditch of a fortification is lOOOft long, 9ft deep, 20ft broad at the
bottom, and 22 at the top ; how much water will fill the ditch?
Ans. 1177867gall nearly.
57. If the diameter of the earth be 7930 miles, and that of the moon 2160
QUESTIONS IN MENSURATION. 523'
lies; required tbe ratio of their surfaces, and also of their volume, supposing
lem both to be spherical.
Ans. the surfaces are as 13^ to 1 nearly; and the volumes as 49i to 1 nearly.
58. A rectangular cistern whose length, breadth, and depth, internally, were
[ft lOin, 2ft lin, and 2ft gin respectively, was rested on props at the comers,
'f 4in high; but by accident one of the props was knocked out of its place:
ow much less water would the cistern hold when it was brought with that
orner to rest upon the ground ; and how much less, still, when two adjacent
rops were removed, either those under the side or under the end ?
59. Let the section of the breast-work be as in Ex. 4, p. 486, and EO the
Teadth of the ditch at top be 20ft ; the slopes of the ditch unequal so that ER
RD : : 2 : 3 and SO : SP : : 2 : 4 ; what must be the depth of the ditch, so
hat tbe earth thrown out shall form a glacis whose height is 3ft and base OL is
4ft? Ans. 6-8ft.
60. If the area of the profile ABHC be 100ft ; and BF = 1, FH = 6, EG =
0, GR = 13, RD = 6, and ER = 3ft : what must be the breadth of the ditch
0 that its section EDPS shall be equal to the profile ABHC and OKL (the
ection of the glacis) together, when the slopes BH, KL are in the same plane,
nd the slopes ED, OP, are equal ? Ans. 25778 ft.
61. (1) The four sides of a trapezium are 6^, 15?, 12, and 9 respectively, the first
wo of these sides make a right angle : required the area of the quadrilateral.
(2) When the same four sides form a quadrilateral inscriptible in a circle,
ind its area, angles, and diagonals.
62. Find the ratio of the surfaces of the torrid zone, the two temperate, and
he two frigid zones, of the earth ; supposing the two tropics to be 23° 28' from
he equator, and the two polar circles to be 23° 28' from their respective poles.
63. A cone, whose altitude is 63, and diameter of the base 32, is to be cut,
y sections parallel to the base, into four portions of equal curved surface:
equired the respective distances from the vertex, measured on the slant side, at
(hich tbe sections are to be made.
64. The solid content of a spherical shell, is equal to that of a conic frustrum,
he areas of whose two ends are respectively equal to the e.xtcrior and interior
urve surfaces of the shell, and whose height is equal to the shell's thickness.
65. A sphere is to any circumscribing polyhedron, as the surface of the sphere
0 the surface of the polyhedron.
66. The surface of a sphere is double the curve surface of an inscribed cy-
inder whose height and diameter of the base are equal. Also, the surface of a
phere is to the curve surface of an equilateral inscribed cone, as 8 to 3.
67. If a cone be cut by a vertical section, the segment of the base cut off by
hat section, is to the corresponding segment of the same surface, as the radius of
he base to the slant side of the cone.
68. The volume of a regular octahedron inscribed in a sphere is to the cube
f the radius as 4 to 3.
69. If a, /3, y, be the angles under which any three diameters of a sphere
0 radius a intersect, and «r = ^ (a -f j3 + 7) : show that the volume of the
larallelopiped which is formed by tangent planes at the extremities of their
liameter, is expressed by 4o^ Jsin <r sin (<r— a) sin (er — /3) sin (it — 7)^ -.
70. Find the volumes of the pyramids which envelope a triaugrular and a
quare pile of balls, respectively, the side of the lower course in each being n.
71. A cannon-ball whose radius is r, may be touched by four, or by eight, or
)y twenty shells of equal radii, R, and each of which touches three of the remain-
5:24
NOTES.
1
ing ones ; or it may be touched by six equal shells of radii R„ each of which
touches four of the remaining ones ; or, again, it may be touched by twelve equal
shells of radii R,. each of which touches five of the others : it is required to
prove these contacts, and assign the several external radii R, R^, R, of these
shells in terms of r.
72. If a rectangular pile of six inch balls of 150 in length and 40 in breadth
be roofed over, the roof being in close contact with the balls : how much empty
space would be enclosed ?
NOTES.
Note I. Synthetic Division, p. 129.
A different and much more simple investigation of this process has occurred
to me since the sheet on this subject was printed off". It is as follows : —
To dinde Ax* + Bx"-' + Cx— ' + ... by ar" + fl,a;— ' + a^a;— * + . . .
Assume the quotient to be Ax"— + A,x" ' + AjX"-— « + . . . ; iu which
A„ Aj . . . are unknown : then, since quotient x divisor = dividend, let this
multiplication be made.
quot' = A + Ai + A, + A3 +
divf = 1 + ai + Oa + fla +
A + A. + A, + A3 + ....
a,A +a,A,+o,A, + OiAj +
OjA H-OjA, + ajAj + fljAj + . . .
fljA + OgA, + ajij + fljAg ...
div<« =
A+B-HC + D + ....
— a,
— QjA — a,Ai— OjAj — fliAj — ...
— th
— OjA— a^i— OjAj — fljAj —
— flj
—OjA — fljA, — CgA, — a3A3 — . . .
A+ A, + A, + A3 +
In this we have worked by detached coefficients as usual. The first part of
the operation shows the manner in which the coefficients of the dinsor, which
are known, are combined with those of the quotient, which are unknown, in form-
ing those of the dividend ; and, conversely, the second operation shows the
formation of the addends to coefficients of the dividend to form those of the
quotient, to be precisely the same as before, except that all the signs are changed.
Moreover, when A (which is the same in the dividend and quotient) is known,
we can form the diagonal column composed of — OiA, — fl,A, — OjA, — a^A, . . . ;
and thence we obtain A,, and, consequently, the next diagonal column — OiA,,
— QjA,, — fljA,, — a^A,, ; then, similarly for Aj and the next diagonal
column, as far as it is necessary to carry the process.
The change of sign of the coefficients of the divisor, it is obvious, is a conse-
quence of the converse nature of the operation of finding the coefficients of
dividend from those of the quotient and divisor, to that of finding the quotient
from the divisor and dividend. The entire result is so strictly in accordance
with the prescribed rule (p. 128) that any further detail in addition to what
ba« been given would be superfluous.
NOTES. 525
Note II. Small arcs, p. 436.
The tabular sine and tangent of a very small angle, or vice versd, which cannot
e obtained from the tables on account of the rapid variations which those func-
ous undergo at that stage, is often required to be found with great accuracy.
:liey are, however, found in the following manner, without much labour.
1. To find tab sin x when x is very small.
a^ afi
smx ^ X — - +
1.2.3 1.2.3.4.5
w
Now, as the length of an arc of 1°, or of 7-—, is '01745329 .... the third
erm of the series for this arc has no effective figure within the first ten decimal
jlaces ; and hence, h fortiori, the series for a smaller arc can have no effective
igure within the first seven decimal places. The series will be, then, effectively
'educed in this case to its first two terms, and we shall have
( a^ ) ( x' X* ){
smx = a.|l-^ = x|l-— + ^-^^1 =a.3Vcosx.
But when it is very small, cos x varies very slowly, and may be taken, for
;his purpose, with sufficient accuracy from the tables : whence the value of sin x
:an also be computed from
tab sin a; = loga? + 10 + J (tab cos a; — 10).
Let the arc x contain p seconds, or a; = Y^^r/yz-^p. ■ p" ■ then
J 80. 00. 00
log a; = log;> + log TT — log (ISO.eo^) = logjp — 5-3144251 ...
Substituting this in the preceding general formula, we have
tab sinar = log;> + 10 — 5-3144251 + J (tab cos a? — 10)
= log p + 4-6855749 — i ac tab cos x.
2. To find tab tan x when x is very small.
By the preceding we have sin a; = a; V cos x, and hence tan x = =
— — J— : wherefore tab tan a; = log a; + | (10 — tab cos a;) ; and, proceeding
t^COS x
as before, tab tan x =■ logp + 4-6855749 + 3 ac tab cos x.
3. Given tab sin x or tab tan z to find x itself.
Here, by merely reversing the former processes, we have
log p = tab sin x + 5-3144251 + J ac tab cos a; — 10,
log p = tab tana: -f 5-3144251 — § ac tab cos a? — 10:
from either of which, according to the data, the value of x is found *.
• The rules indicated by these fonnulse ■were first given, in a verbal form, by Dr. Maskelyne,
in the Introduction to Taj/tor's Tables. Many investigations of them have since been given ;
but the above, whilst they are the most generally adopted ones, are amongst the most simple.
This method of investigation itself was originally given in Woodhouse's Trigonometry.
526
NOTES.
Note III. p. 423.
The notation for powers of trigonometrical functions.
In comparing the notation for powers with that for inverse functions, it cannot
have escaped the student's attention that the same notation is used in two diflfer-
ent senses ; and a degree of confusion might arise from it, if he were not apprised
of the reason, and of the relations of the two things signified by it.
The strictly correct notation for powers is (cos a;)", (tan 0)", etc.: but the
increased space required for wTiting it in this way, as well as the additional
trouble, has caused it to be written cos'x, tan"0, etc., by some authors ; and cosa:",
tan 0", etc., by others. Now, so long as no idea of the successive trigonome-
trical functions — such as the cosine of a cosine, or the trigonometrical func-
tions of any power of an arc — was entertained, mathematicians very naturally
abridged the notation as far as possible, so as not to create doubt in the mind as
to the signification of the expression. The introduction of the notation for
inverse functions has, however, interfered with the former of these notations ;
whilst the latter has never been extensively used: and, in strict accuracy,
we should be compelled to adopt (cos x)', (tan 9)', etc., as our standard nota-
tion ; and especially, should the inquiries of mathematicians ever lead to results
involving the successive trigonometrical functions, direct and inverse, of any
expressions for the arc, in the same manner that they have already led to the
consideration of successive logarithmic functions (p. 262). Instances of this,
however, are so rare, that it would be difficult to quote one in any subject
of even a moderately elementary character. We have hence, for the present,
adhered to the old notation : though it was necessary to point out the circum-
stance for the satisfaction of the inquiring student.
It may be added, too, that sin'j: does really, in reference to the thing signified,
bear the same meaning, though founded on a different view of the subject,
as sin x ; and hence, as a fundamental principle, there is no real difference
in this stage (the utmost extent to which the notations under the two aspects
coalesce) of the inquiry, between the quantities signified.
Note IV. The angular unit taken as the arc eqvuil to radius, p. 422.
The angular unit employed by the Greeks and by all the moderns, till the
period of the French Revolution, was the ninetieth part of the quadrant : but by
the French, the quadrant was divided centesimally, the hundredth part being
the unit ; with, however, a general impression that these were more advan-
tageously considered as minor divisions of the quadrant, taken as the standard-
unit. Amongst many distinguished men of science, however, the radius of the
circle by which the angle was estimated, has been considered the most advan-
tageous standard-unit : and in some important applications of trigonometry it is
undoubtedly the case, though in reference to the ordinary ones for which tables
have already been computed, it would be altogether useless till very extensive
tables adapted to this division shall have been published. The main applica-
tions, indeed, of this unit, are to purposes for which, from particular circum-
sUnces, the existing tables are inadequate.
Denote by a and a the length and the number of degrees of the arc subtending
the angle A" ; and let the radius r contain p degrees of that circle. Then
p' : 180° : : r : rjr, and A° : 180° : : a : nr.
NOTES. 527
180°
Whence (P = — - = S7°-29S7795 = 206264"-8 nearly;
IT
^'^-r= f =57°-295m5.. = -01745329. . .A^.
In reference to this mode of estimating angles, p = 57"°2957795 ... is to be
:onsidered the unit: and the angle is then said to be estimated in circular
I Note V, pp. 451—6.
Certain conditions amongst the data of plane triangles.
In some of the examples given for solution, the student will have discovered
a certain degree of uncertainty in the results which he obtained. This arises
from the great relative variations of the trigonometrical functions compared with
those of the angles themselves, in certain parts of the tables. In actual trigo-
nometrical observations, the classes of conditions which require these computa-
tions to be employed, will, where practicable, be avoided : but numerous in-
stances arise in practice where they cannot be avoided, and hence to remove the
resulting uncertainty, other methods of solution have been devised, one or two
of which are given in this note, with one or two other particulars relating to the
subject.
1 . There are given a, b, C, where b is very small in comparison with a, to find
the remaining parts of the triangle.
Putting for cos C its value - \e^^^^~^-\- e~^'^^~^|
c2=a2-2o&cosC + 6'=a'{l +\ g^'^^^^l |l — * e-^'^^^^}
and taking log^ of both sides, and expanding, we have
or log. c = log^ a cos C — — „ cos 2C — —j .cos 3C —
ee 6e ^ 2a- 3a^
whence c can be found.
Again, tan B = y p ; or expressed in exponentials
— ^- = 5: — :p — -=^^ 7^;^ » and hence also
2BV3rr= °-^^ ^_-= \ — — ..
a -\- be 1 H — c
a
Take log^ of both sides as before, and reduce : then
h ¥ ¥
B = - sin C + — , sin2C -|- r-5 sin 3C +
a 2a^ Sa-*
tvhere B is estimated in circular measure. Or again, since sin 1" = 1'' very
Qearly, the value of B, in seconds, is
_ 6sinC fc" sin 2C b[ sinSC
a sin 1" ''" 2o2sin 1" "^ Sa^sin 1" ■"■••' *
^.^g NOTES.
1
2. Given the very acute angles A, B, and the side c to find the remaining parts.
Since A and B are very small, we have very nearly ^
sinA = A--^3;cosA = l-^3;sinB = B-^3;cosB = l-A_--
' ' (A+B)'
Hence,8inC = sin(A + B) = sinAcosB + cosAsinB = (A + B) ^j^-
c sin A cA ( . . B(2A + B)| ,
merefore, a = ,-^aTB)"-= A+B 1 ' + ~i:2:^^ i '
c sin B _ cB ( A(2B + A)|
\ "*" 1.2.3 j •
b =
sin
(A+B)""A+B
3. Gftren the two sides a, b, and the included angle C, which is very obtuse, to find
the other parts of the triangle.
Put C = 180° — a: then, since C is very obtuse, a is very small, and
cos a = 1 nearly : whence
1.2
c* = a^- 2ab cos C+ b'- = a^ + b' + 2abil - —^ = {a + bf - ab a^
Whence c = V (.a + bf — aba? = a + b — -^-—^ — nearly.
Again, sin A = — sin C = - sin a = ^ j a — ^^ | nearly ; and
c a + b
ab a^) -1
■"- - a + b\^~{a + b)^ 2j
a ( ab a?) ,
a + 6 2
_ a ( _ . ab
Inserting in sin A = - sin a, these values, we have
^"'^^ aT6 r ^ (^+*)"- 2 1 r -17273) = a+6l' + 2(^+6? " 1:273/
— g a ( , a^—ab+¥ a'^ ^
41 • A A A3 . sin^A , A • A . ?H^
Also am A = A — — — = A — — -— nearly ; or A = sm A + -3,
««_ f , a^-a&+6^ g' 1 f a ^3 «'
" a + 6\ 2(a+6)'^ ' 1.2.3J "^ (a + 6j * 1.2.3
a a ( {a — b)b a" \ ,
=«-Hr4 1 + -(^T^T:^3r^^^y-
In the same manner we have
ba ( (a — b)a a?
B =
A'- {a + bfUTsr^'^^-
a + b
These angles being understood to be in circular measure, as explained in the
preceding note.
For further information on subjects of this nature, the reader is referred to
Bonnycastle's Trigonometry, Cagnoli Trigonometrie, several of the authors on
Geodesy, especially Puissant, and to the second volume of this work.
A TABLE OF SQUARES, CUBES, AND ROOTS.
529
No.
Square.
Cube.
Sq. Root.
Cube Root
No.
Square.
Cube.
Sq. Root.
Cube Root
1
1
1
1-0000000
1-000000
73
5329
389017
8-5440037
4-179339
2
4
8
1-4142136
1-259921
74
5476
405-224
8-60-23253
4-198336
3
9
27
1-7320508
1-44-2250
75
S625
421875
8660-2540
4-217163
4
16
64
2-0000000
1-587401
76
5776
438976
8-7177979
4-2358-24
5
25
1-25
2-23C0680
1-709976
77
59-29
456533
8-7749644
4254321
6
36
216
2-4494897
1-8171-21
78
6084
474552
8-8317609
4-272659
7
49
343
2-6457513
1-91-2931
79
6241
493039
8-8881944
4-290840
8
64
512
2-8-284-271
2000000
80
6400
512000
8-944-2719
4-308869
9
81
729
30000000
2080084
81
6561
531441
90000000
4-3-26749
10
100
1000
3-16-2-2777
2154435
82
6724
551368
9-0.553a51
4344481
11
121
. 1331
3-3166-248
•2-223980
83
6889
571787
9-1104336
4-36-2071
1-2
144
1728
3-4641016
2-2894-29
84
7056
59-2704
91651514
4-379519
13
169
2197
3-605.5513
-2a51335
85
72-25
614125
9-2195445
43.%830
14
196
2744
3-7416574
2-410142
86
7396
636056
9-2736185
4-414005
15
225
3375
3-8729833
2466212
87
7569
658503
9-3-273791
4-431048
16
256
4096
40000000
2-519842
88
7744
681472
9-3808315
4-447960
17
289
4913
41-231056
2-571-282
89
7921
704969
9-4339811
4-464745
18
324
5832
4-2426407
2-6-20741
90
8100
7-29000
9-4868330 ! 4481405 |
19
361
6859
4-3588989
2-668402
91
8-281
753571
9-53939-20
4-497941
20
400
8000
4-4721360
-2-714418
92
8464
778688
9-59 16630
4-514357
21
441
9261
4-5825757
2-7589-24
93
8649
804357
9-6436508
4-530655
22
484
10648
4-6904158
2-80-2039
94
8836
830584
9-6953597
4-546836
23
529
12167
4-7958315
2-843867
95
9025
857375
9-7467943
4-562903
24
576
13824
4-8989795
2-884499
96
9-216
884736
9-7979590
4-578857
25
625
15625
50000000
2-9-24018
97
9409
912673
9-8488578
4-594701
26
676
17576
5-0990195
•2-962496
98
9604
941192
9-8994949 j 4610436
27
729
19683
5-1961524
3-000000
99
9801
970-299
9-9498744 4-6-26065
28
784
21952
5-29150-26
3-036589
100
10000
1000000
10-0000000 1 4-641589
29
841
24389
5-3851648
3-072317
101
10201
1030301
100498756 4657009
30
900
27000
5-477-2-256
3-107-232
102
10404
1061-208
100995049
4-67-23-29
31
961
29791
55677644
3141381
103
10609
109-2727
10-1488916
4-687548
32
1024
32768
5-6568542
3-174802
104
10816
1124864
101980390
4-70-2669
33
1089
35937
5-7445626
3-207534
105
110-25
11576-25
10-2469508
4-717694
34
1156
39304
5-8309519
3-239612
106
11236
1191016
10-2956301
4-73-26-23
35
1225
42875
5-9160798
3-271066
107
11449
12-25043
10-3440804
4-747459
36
1296
46656
6-0000000
3-301927
108
11664
1-259712
10-39-23048
4-76-2-203
37
1369
50653
6-08-276-25
3-33-2-?22
109
11881
12950-29
10-4403065
4-776856
38
1444
54872
6-1644140
3-361975
110
12100
1331000
10-4880885
4-791420
39
1521
59319
6-2449980
3-391211
111
1-2321
1367631
10-5356538
4-805895
40
1600
64000
6 •3-245553
3-419952
112
1-2544
14049-28
10-5830052
4-8-20-284
41
1681
68921
6-4031-242
3-448217
113
1-2769
1442897
10-63014.58
4834588
42
1764
74088
6-4807407
3-476027
114
12996
1-481544
106770783
4-848808
43
1849
79507
6-5574385
3-503398
115
13225
1520875
10-7-238053
4-86-2944
44
1936
85184
6-633-2496
3-530348
116
13456
1560896
10-7703-296
4-876999
45
2025
91125
6-708-2039
3-556893
117
13689
1601613
10-8166538
4-890973
46
2116
97336
6-7823300
3-583048
118
139-24
1643032
10-86-27805
4-904868
47
2209
103823
6-8556546
3-6088-26
119
14161
1685159
10-9087121
4-918685
48
2304
110592
6-928-2032
3-634-241
1-20
14400
1728000
10-9544512
493-24-24
49
2401
117649
70000000
3-659306
121
14641
1771561
11-0000000
4-946087
50
2500
125000
7-0710678
3-684031
122
14884
1815848
11-0453610
4-959676
51
2801
132651
7-1414-284
3-708430
1-23
15129
1860867
11-0905365
4-973190
52
2704
140608
7-21110-26
3-732511
124
15376
1906624
11-1355-287
4-986631
53
2809
148877
7-2801099
3-756-286
125
156-25
19531-25
11-1803399
5-000000
54
2916
157464
7-3484692
3-779763
126
15876
2000376
11 •-2-2497-22
5-013298
55
3025
166375
7-4161985
3-80-2952
127
161-29
2048383
11-2694-277
5-026526
56
3136
175616
7-4833148
3-8-25862
1-28
16384
2097152
11-3137085
5-039684
57
3249
185193
7-5498344
3-848501
1-29
16641
2146689
11-3578167
"505-2774
58
3364
195112
7-6157731
3-870877
130
16900
2197000
11-4017543
5-065797
59
3481
205379
7-6811457
3-89-2996
131
17161
2-248091
11-4455231
5-078753
60
3600
216000
7-7459667
3-914868
132
174-24
2299968
11-4891-253
5-091643
61
3721
226981
7-810-2497
3-936497
133
17689
2352637
11-53-256-26
5-104469
62
3844
238328
7-8740079
3-957891
134
17956
2406104
11-5758369
5-117-230
63
3969
250047
7-937-2539
3-979057
135
182-25
2460375
11-6189500
5-1-299-28
64
4096
262144
8-0000000
4000000
136
18496
2515456
11-6619038
5-14-2563
65
4-225
274625
806-22577
4-0-207-26
137
18769
2571353
11-7046999
5-155137
66
4356
287496
81240384
4041-240
138
19044
26-28072
11.7473401
5-167649
67
4489
300763
81853,5-28
4-061548
139
19321
2685619
11-7898261
5180101
68
4624
314432
8-2462113
4-081655
140
19600
2744000
11-8321596
5-19-2494
69
4761
328509
8-3066-239
4-101566
141
19881
2803-221
11-87434-22
5-2048-28
70
4900
343000
8-3666003
4-121285
142
20164
2863-288
11-9163753
5217103
71
5041
357911
8-4261498
4-140818
143
20449
2924207
11-958-2607
5-229321
72
5184
373248
8-485-2814
4160168
144
20736
2985984
120000000
5-241483
530
SQUARES, CUBES, AND ROOTS.
No.
Square.
Cube.
Sq. Root. iCubc Root
No.
Square.
Cube.
Sq. Root.
Cube Root
145
210^25
3048625
12 041.5.0461 5-2.53588
217
47089
10-218313
14-7309199
6-009-245
[4t]
21316
31 121. "56
l-208.'50460i 5-26.5637
218
475-24
10360232
14-7648-231
6018462
147
21609
3176.V23
r2-r243.5.57! 5-277632
219
47961
10503459
14-7986486
60276.50
148
21904
3-241 7i*2
r2-16,55^251
5-289572
-2-20
48400
10648000
14-83-23970
6036811
149
2'2-201
3307949
] 2-206.55.56
5-301459
-221
48841
10793861
14-8660687
6-045943
l.Wl
2'2.5<t0
3375000
12-2474487
5-313-293
■>>2
49-284
10941048
14-8.096644
6-055049
151
2^28«ll
344-2951
r2-288-2057
5-3-25074
-2-23
497-29
11089567
14-9331845
6-064127
152
•23104
.V)11808
12-3288280
5-336803
•2-24
,50176
11-2394-24
14-9666-2,05
6-073178
l,i3
23409
ai81577
1-2-36.03169
5-348481
-2-25
506-25
11390625
150000000
6-08-2-202
154
'23716
365-2-264
12-40.06736
5-360108
-2-26
51076
11543176
15033-2964
6-091199
155
•240^2.i
37-23875
12-4498996 5371685
-2-27 515-29 |
11697083
150665192
6-100170
15« -24336
37.%416
12-48.09960 5383213
•228
51984
1185-2352
15-0996689
6-109115
157 24649
3!(69893
1-2-5-29.0641
5-394691
•2-29
52441
1-2008989
15-13-27460
6-118033
1,)8 24964
3944312
12-5698051
5 4061-20
-230
5-2900
12167000
15-16.57509
6-1-269-26
\X> 2.V_>Jtl
4019679
12-6095-202
5-417501
-231
53361
.1-23-26391
15-1986842
6-135792
1 6(1 25*500
4096000
12-6491106
5-4-28835
232
53824
1-2487168
15-2315462
6-144634
161
25921
4173-281
1268a5775
5-440122
-233
54289
1-2649337
15-264.3375
6153449
10-2
26244
4-251.5-28
1-2-7279-221
5-451362
-234
54756
1-281-2904
15-297058,5
6-16-2-240
163
26569
4330747
12-7671453
5-46-2556
-235
55-2-25
1-2977875
15-3-297097
6171006
164
2»i896
4410944
12-806-2485
5 473704
-236
55696
13144-256
15-36-2-2915
6-179747
165
272^25
44921-25
12-845-2326
5-484807
237
56169
1331-2053
15-3948043
6188463
166 27556 1
4574296
12-8840987
5-495865
-238
56644
13481272
15-427-2486
6-197154
167
27889
4657463
12-9-?28480
5-506878
239
57121
13651919
15-4596-248
6-2058-22
168
28-224
4741632
1-2-.0614814
5-517848
-240
57600
138-24000
15-4919334
6-214465
169
28.161
48-26809
130000000
5-528775
-241
58081
13997521
15-5-241747
6-2-23084
170
28900
4913000
13-0384048
5-539658
-242
58564
1417-2488
15-5563492
6-231680
171
29241
5000211 I13076696S1 5-550499 |
-243
59049
14348907
15-5884573
6-240-251
172
29.584
5088448 131148770
5-561-298
244
59536
145-26784
15-6-204994
6-248800
173 29929 |
5177717 13-15-29464
557-2055
-245
600-25
147061-25
15-65-24758
6-2573-25
174
30^276
5-2680-24 131909060
5-58-2770
-246
60516
14886936
15-6843871
6-2658-27
175
3<W25
53.19375 1 13-2-287566 5-593445
-247
61009
15069-223
15-716-2336
6-274305
176
30976
.5451776 1 13-2664992 5-604079
-248
61504
15-252992
15-7480157
6-282761
177
313-29
5545-233 |13-3041347
5614672
-249
6-2001
15438-249
15-7797338
6-2911.05
178
31684
5639752
13-3416641
5-6-25-2-26
-250
62500
1 56^25000
15-8113883
6-299605
179
3-2041
57^5339
13-3790882
5-635741
-251
63001
15813-251
15-84-29795
6-307994
180
3^2400
583-2000
13-4164079
5-646216
-252
63504
16003008
15-8745079
6-316360
181
3-2761
59-29741
13-4536-240
5-656653
-253
64009
16194^277
15-9059737
6-3-24704
182
331-24
60-28568
13-4907376 5-667051
-254
64516
16387064
15-9373775
6-3330-26
183
33489
61-28487
13-5-2774.03 5-677411
-255
650-25
16581375
15-9687194
6-3413-26
184 xam
6-2-29504
13-5646600
5-687734
-256
65536
16777216
16-0000000
6-349604
18.1 34225
63316-25
13-6014705
5-698019
-257
66049
16974593
16-0312195
6-357861
186
34596
6434856 113-6381817
5-708-267
-258
66564
17173512
16-06-23784
6-366097
187
3496'9
6539^203
13-6747943
5-718479
•259
67081
17373979
16-0934769
6-374311
188
35344
6644672
13-71130.02
5-728654
•260
67600
17576000
161-245155
6-38-2504
189
35721
6751269
13-7477-271
5-738794
261
681-21
17779581
16-1554944
6-390676
190
36100
685.9000
IS-7840488
5-748897
262
68644
179847-28
16-1864141
6-3988-28
191
36481
6.067871 138-202750
5-758965
•263
69169
18191447
16-217-2747
6-406958
192
36864
7077888 13^8564065
5-768998
264
696.06
18399744
16-2480768
6-415069
193
37249
7189057 13-89-24440
5-778996
•265
70-2-25
186096-25
16-2788-206
6-4-23158
1-94
37636
7301 .-m 13 9^283883
5-788960
•266
70756
18821096
16-3095064
6-431-2-28
195
380^25
7414875 l3-964^2400
5-798890
•267
71-289
19034163
16-3401346
6-43.0277
19«i
38416
7529536
14-0000000
5-808786
•268
718-24
19-248832
16-3707055
6-447306
197
38809
7645373
14-0356688
5-818648
•269
7-2361
19465109
16-40121.05
6-45.5315
198
39-204
776^2392
140712473
5-8-28477
270
72900
19683000
16-4316767
6-463304
199
39601
7880599
14-1067360
5-838-272
271
73441
1990-2511
16-46-20776
6-471274
•200
40000
8000000
14-14-21356
5-848035
•272
73984
201-23648
16-4.0-24-2-25
6-479-2-24
•201
40401
8r20()01
14-1774469
5857766
•273
745-29
20346417
16-5-2-27116
6-487154
•202
40804
8^24^2408
14-21'2(;704
5-86-7464
-274
75076
205708-24
16-55-29454
6-495065
■203
4r209
a36.5427
14-2478068
5-877131
275
756-25
20796875
16-5831-240
6-50-2957
•204
41616
84896(>4
14-28-28.569
5-8!!6765
-276
76176
210-24576
16-613-2477
6-510830
•205
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138991832
139798359
140608000
1414-20761
14-2236648
143055667
1438778-24
1447031-25
145531576
146363183
147197952
148035889
148877000
149721-291
150568768
151419437
15-2-273304
153130375
153990656
154854153
1557^20872
156590819
157464000
158340421
1 59^2-2f M)88
160103007
160989184
1618786^25
16^2771336
163667323
164566592
165469149
166375000
167^284151
168196608
169ir2377
170031464
170953875
171879616
17^2808693
173741112
174676879
175616000
1765.58481
1775043-28
178453.547
179406144
I803<;2r25
181321496
18^>284^263
1832.504.32
184^2^_'00O9
ia5 193000
186169411
187149^248
l«tl3-'517
189119-2-24
190109375
191U«I76
-22-47-2-2051
•22-4944438
•>2-51 66605
•22-5388553
•225610283
•2-2-5831796
•22-6053091
22-6-274170
•22-6495033
•2^2-6715681
-22-6936114
•22-7156334
•22-7376340
•2^2-7596134
•22-7815715
•22-8035085
•22-8254244
•>2-8473193
•22-8691933
•22-8910463
•22-91 •28785
•22-9346899
•22-9564806
22-978-2506
•23-0000000
•23-021 7^289
230434372
•23-0651'252
•230867928
•231084400
•23^ 1300670
•231516738
•231 73-2605
-23-1948-270
-23-2163735
-23-2379001
-23-2594067
23-2808935
•23 3023604
23-3^238076
23-345^2351
23-36664-29
•23-3880311
23-4093998
•23^4307490
•23-45-20788
23-47338.92
•23^4946802
2351595^20
23-537-2046
-23-5,584380
23-57965-22
•23-6008474
23^6^?20236
•23-64;«803
•23-6643191
•23-68.54386
•23-7065392
•23-7276210
237486842
•237697^286
•237.907.545
•2.3-8117618
•23-8327506
23*537-209
-23-87467-28
23-8956063
-239165215
-2.3-.9374184
23-958-2971
239791576
24-(XM)0<MKl
7-963374
7-968627
7-973873
7-979112
7-984344
7-989570
7-994788
8-000000
8 005-205
8-010403
8015595
8-0-20779
8 0-25957
8-031129
8-036-293
8041451
8-046603
8-051748
8-056886
8-06-2018
8067143
8-07-2-262
8-077374
8-08-2480
8-087579
8-09-2672
8-0.977.59
8-10-2839
8-107913
8 11-2980
8-118041
81-23096
8-1-28145
8-133187
8138->23
8143-253
8-14^276
8-153-294
8-158305
8-163310
8 168309
8-173302
8-178-289
8-183-269
8-188-244
8-193213
8-198175
8-20313-2
8-208082
8-2130-27
8-217966
8->2-2898
8-2-278-25
8-2,3-2746
8-237661
8-24-2571
8-247474
8-25^2371
8 •257^263
8-262149
8-267029
8-271904
8-276773
8-281635
8-286493
8^^29l344
8-2961.90
8-301030
8305865
8-310694
8-315517
8-320.-J.V)
SQUARES, CUBES, AND ROOTS.
ti33
No.
Square.
Cube.
Sq. Root.
Cube Root
No.
Square.
Cube.
Sq. Root.
Cube Root
577
332929
1.92100033
24-02082431 8-325147
649
421201
273359449
25-4754784
8-657946
578
334084
193100552
24041630()
8-329954
6.50
42-2.500
274625000
2.5-4950976
8-66-2;i91
579
33.5241
194104539
2406^24188
8-;m755
651
423801
275894451
-25-5147016
8-666831
580
336400
195112000
24 0831891
8-33.9.551
652
4-25104
•277167808
-25-5342907
8-671-266
581
337561
196122941
241039416
8-344341
653
4-26409
278445077
25-5538647
8-675697
58-2
338724
197137368
24-1246762
8-;{49r26
654
4-27716
2797-26'264
-2.V57:M237
8-6801-24
583
339889
198155287
2414.53.9-29
8-353905
655
4-290-25
281011375
25-.59-29678
8-684.546
5«4
341056
199176704
-24 1660919
8-358678
656
4303;J6
•28-2300416
25-6124969
8-688963
5!i5
342225
200201625
-241867732
8-3(;3447
657
431649
28359.3393
25-63-20112
8 693376
586
343396
2012300.56
-24-2074369
8-368-209
658
43-2.%4
2fM890312
25-6515107
8-697784
587
344569
202262003
24-2-280829
8-372967
659
4;U281
286191179
25-6709953
8-702188
588
345744
203297472
-24-2487113
8-377719
660
43,5600
287496000
25-6904()52
8-706588
589.
346921
204336469
24-2693'2-22
8-38^2465
661
436921
288804781
25-7099-203
8-710983
590
348100
205379000
24-2>}99156
8387206
662
438-244
2.90117,528
2.5-7293607
8-715373
591
349281
20642.5071
-24-3104916
8-391.942
663
439569
291434-247
25-7487864
8-719760
59-2
350464
207474688
-24-3310501
8-3.96673
664
440896
292754944
25-7681975
8-7-24141
593
3.51649
208.527a57
24-351.5913
8-401398
665
44^22^25
•294079625
•25-7875939
8-72a518
594
352836
209584584
-24-3721152
8-406118
666
443556
295408296
•25-8069758
8-73-2892
595
354025
210644875
24-39-26218
8-410833
667
444889
296740963
•25-a263431
8-737260
596
355216
211708736
24-4131112
8-41.5.542
668
446224
298077632
•25-84.56960
8-7416-25
597
356409
212776173
•24-4335834
8-4-20246
669
447561
299418309
25-8650343
8-74.5985
598
357604
213847192
24-4540385
8-4-24945
670
448900
300763000
25-88435^2
8-750340
599
358801
214921799
24-4744765
8-4-29638
671
450241
302111711
25-90.36677
8-754691
600
360000
216000000
-24-4948974
8-4343-27
672
451584
303464448
25-9-2-29628
8-759038
601
361201
217081801
24-5153013
8-439010
673
45-29-29
304821217
25-94-2-2435
8-763381
60-2
362404
218167208
24-5356883
8-443688
674
454276
3061820^24
25-9615100
8-767719
603
363609
2192.56227
24-5560583
8-448360
675
45,56-25
307546875
-25 -,9807621
8-772053
604
364816
220348864
24-5764115
8-4530-28
676
456976
308915776
260000000
8-776383
605
36"6025
221445125
-24-5967478
8-4,57691
677
4583-29
310288733
26-0192-237
8-780708
606
36723(i
222545016
•24-6170673
8-462348
678
459684
311665752
26-0384331
8 785030
607
368449
223648543
24-6373700
8-467000
679
461041
313046839
26-0576284
8-789347
608
369664
224755712
24-6576560
8-471647
680
46-2400
31443-2000
260768096
8-793659
609
370881
225866529
-24-6779-254
8-476289
681
463761
315821-241
-26-0959767
8-797968
610
372100
226981000
-24-6981781
8-4809-26
682
4^124
317214568
•261 151297
8-80-?272
611
373321
228099131
24-7184142
8-485558
683
466489
318611987
26-134-2687
8-806572
61-2
374544
229220928
24-7386338
8-490185
684
467856
3-20013504
26- 1533937
8-810868
613
375769
230346397
•24-7588368
8-494806
685
469225
321419125
261725047
8815160
614
376996
231475544
24-7790234
8-499423
686
470596
3^228-28856
26-1916017
8-819447
615
378225
232608375
24-7991935
8.504035
687
471969
324-24-2703
•26-2106848
8-8^23731
616
379456
233744896
-24-8193473
8-508642
688
473344
3-2.5660672
26-2297.541
8-8-28010
617
380689
234885113
24-8394847
8-513-243
689
474721
327082769
26-2488095
8 83-2-285
618
381924
236029032
24-8596058
8-517840
690
476100
328509000
26-2678511
8-836556
619
383161
237176659
24-8797106
8-5-2-2432
691
477481
3-2993.9371
26"2868789
8-840323
620
384400
238328000
24-8997992
8-5^270I9
692
478864
331373888
•26-3058929
8-845085
621
385641
239483061
•24-9198716
8-531601
693
480-249
33281-2557
26-3248932
8-849344
622
386884
240641848
24-9399278
8-536178
694
481636
334-255384
26-3438797
8-853598
623
388129
241804367
24-9599679
8-540750
695
4830^25
335702375
26-3628527
8-857849
624
389376
242970624
•24-97999-20
8-545317
696
484416
337153536
26-3818119
8-86-2095
625
390625
244140625
25-0000000
8-549880
697
4a5809
338608873
26-4007576
8-866337
626
391876
24.5314376
2501999-20
8-554437
698
487204
340068392
26-4196896
8-870576
627
393129
246491883
•25-03.99681
8-558990
6.99
488601
34153-2099
26-4386081
8-874810
628
394384
247673152
250599282
8-563538
700
490000
343000000
-26-4575131
8-879040
629
395641
248858189
-2507987-24
8-568081
701
491401
344472101
26-4764046
8-883^266
630
396.')00
250047000
25-0998008
8-57^2619
702
492804
345948408
-26-4952826
8 887488
631
398161
251239591
25-1197134
8-577152
703
494209
347428927
26-5141472
8-891706
632
399424
252435968
-25-1396102
8-581681
704
495616
348913664
26-53-29983
8-8959-20
633
400689
253636137
25-1594913
8-586-205
705
497025
350402625
26-.55ia361
8-900130
634
401956
254840104
25-1793566
8-.590724
706
498436
351895816
26-5706605
8-904337
635
403225
256047875
25-199-2063
8-595^238
707
499849
353393-243
26-.5894716
8 908539
636
404496
2,57259456
25-2190404
8-599748
708
50r264
354894912
26-608-2694
8-91-2737
637
405769
258474a53
25-238a589
8-604^252
709
50-2681
356400829
26-6270539
891693]
638
407044
259694072
-25-2586619
8-608753
710
504100
357911000
26-6458-2.52
8-921121
639
408321
260917119
•25-2784493
8-613-248
711
505521
359425431
26-6645833
8-9-25308
640
409600
262144000
•25-298-2213
8-617739
712
506,944
3609441-28
26-6833281
8-929490
641
410881
263374721
25-3179778
8-6-2-2-225
713
508369
36-2467097
2670-20598
8-933669
642
412164
264609288
25-3377189
8-6-26706
714
509796
363994.344
26-7-207784
8-937843
643
413449
265847707
-25-3574447
8-631183
715
51^2-25
365.525875
26-7394839
8-94-2014
644
414736
267089984
•25-3771551
8-635655
716
5126.56
367061696
26-7581763
8946181
645
416025
268336125
'25-3968502
8-640T-23
717
514089
368601813
26-7768557
8-950344
646
417316
269586136
•25-4165301
8-6445a5
718
515524
370146-232
26-79.55-2-20
8954.503
647
418609
270840023
•25-4361.947
8-649044
719
516961
3716.94959
26-81417.54
89.586,58
648
419.W4 272097792
•25 4558441
8-653497
7-20
518400
373-248000
26-83-28157
8-.%-2809
534
SQUARES, CUBES, AND ROOTS.
No.
Square.
-•21
519841
7->2
521284
7-23 o->27-2ii
7-24
524176
7-25
525625
7-26
527076
7-27 i 5-2aV2.')
72« i 5-2i»0(}4
7-J<» ! 531441
730 5;j-2900
731 5343<;i
732 1 .53.j«24
733 537281)
734
538756
735
540225
736
541696
737
543169
738 oUaU
730 1 54G121
740
547600
741
549081
742
550564
743
552049
744
553536
745
555025
746
556516
747
558009
748
559504
74i»
561001
750 562500
751
564001
752
56.5504
753
567009
754
568516
755
570025
756
571536
757
573049
758
574564
759
576081
760 i 577600
761 1 579121
762
580644
763
582169
7(i4
583696
765
58.5-225
766
58675<;
767
5)58289
768
.589824
769
591361
770
592900
771
594441
772
595.984
773
.597529
774
59!*<)76
775
600625
776
602176
777
603729
778
60.5284
779
606W1
780
608400
781
6099«n
782
611524
783
61308!*
784
6146.5<J
785
616225
786
6177'W
787
619:J69
7««
6-2U944
7Ii9
622.521
7>M>
624 10(
7!»1
625681
792
627264
Cube.
Sq. Root.
Cube Root No. Square. Cube.
374805361
376367048
377933067
379503424
381078125
382657176
384240583
385828352
387420489
389017000
390617891
392223168
393832837
395446904
397065375
398688256
4W3 15553
401947272
403583419
405224000
406869021
408518488
410172407
411830784
413493625
415160936
416832723
418508992
420189749
421875000
423564751
425259008
426957777
428661064
430368875
432081216
433798093
4;i55l9512
437245479
438976000
440711081
442450728
444194947
445943744
447697125
449455096
451217663
452984832
454756609
456533000
458314011
460099648
461889917
46;i684824
4(i5484375
4672Wi576
46-.90974;J3
4709109.52
472729139
474.552000
476379541
478211768
48(i048687
481890304
4837:{<it;25
48.5587656
I 487443403
[ 48!):t(i;$872
i4!)ll(i!)(((i;(
j 49:i03!*0<M)
4949l3<i71
49679301W
26-8514432
•26-8700577
•26-8886593
•26-9072481
■26 -9258240
•26-9443872
26-96^29375
26-9814751
27-0000000
27-0ia51-22
■27-0370117
•27-0554985
27-0739727
•27 ■09-24344
■27-1108834
27-1-293199
•271477439
27-1661554
'27 -1845544
27-20-29410
'27-2213152
•27-2396769
27-2580-263
27-2763634
27-2946881
27-3130006
27-3313007
27-3495887
■27-3678644
27-3861-279
■27-4043792
■27-4-2-26184
27-4408455
•27-4590604
27-477-2633
■27-4954542
27-5136330
27-5317998
27-5499546
■27-5680975
27-586-2-284
27-6043475
27-6-2^24546
■27-6405499
27-6586334
■27-6767050
■27-6947648
■27-71-281'29
•27-7308492
27-7488739
27-7668868
27-7848880
27-80-28775
27-8-208555
27-838}}218
27-8567766
■27-874719
27-89-26514
■27-910.5715
■27-9-284801
27-9463772
27-964-2(i-29
27-9821 ;J72
28(X)0(M)00
•280178515
28 03.56915
•28-0535-203
28 071337
-28 0891438
1-2H-1(M)9;«}(;
•2J11-247-2-22
•281424946
8-966957
8-971101
8-975-241
8-979377
8-983509
8-987637
8-991762
8-995883
9-000000
9-004113
9-008-2-23
9-01-2.329
9-016431
9-0205-29
9-0-246-24
9-0-28715
9-032802
9036886
9-040965
9045042
9049114
9053183
9-057-248
9061310
9-065368
90694-22
9-073473
9077520
9-081563
9085603
9-089639
9-093672
9-097701
9- 101 7-26
9-105748
9-109767
9113782
9117793
9121801
91-25805
9129806
9-133803
9-137797
9141787
9-14.5774
9149758
9153737
9-157714
9161687
9-165656
9-1696-22
9-173585
9-177544
9-181500
9-18.5453
9-189402
9-1.93347
9-197-290
9-201-2-29
9-^205164
9-'209096
9-2130-25
9-2169.50
9-2-20873
9-2-24791
9-'228707
9-232619
9-236.5^28
9-240433
9-244335
9248^234
9252130
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
8-20
821
8-2^2
823
824
825
8^26
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
8.56
i857
858
8.59
860
861
8u2
863
864
628849
630436
6320^25
633616
635209
636804
638401
640000
641601
643-204
644809
646416
6480-25
649636
651-249
65-2864
654481
656100
657721
659344
660969
66-2596
664-225
665856
667489
669124
670761
67-2400
674041
675684
6773^29
678976
680625
68-2276
6839-29
685584
687241
688900
690561
69-2-224
693889
695556
697225
698896
700569
70-2244
703921
705600
707^281
708964
710649
71-2336
714025
715716
717409
719104
7-20801
72-2500
724201
725904
727609
7-29316
7310-25
732736
734449
736164
737881
739600
7413-21
743044
744769
746496
Sq. Root. Cube Root
498677-257
500566184
502459875
504358336
506-261573
508169592
51008-2399
512000000
513922401
515849608
5177816-27
519718464
52] 6601-25
5-23606616
525557943
5-27514112
5-294751-29
531441000
533411731
535387328
537367797
539353144
541343375
543338496
545338513
547343432
549353-259
551368000
553387661
555412248
557441767
559476224
5615156-25
563559976
565609283
567663552
569722789
571787000
573856191
575930368
578009537
580093704
582182875
584277056
586376-253
588480472
590589719
592704000
594823321
596947688
599077107
601211584
6033511-25
605495736
607645423
6098001.92
611960049
6141-25000
616295051
618470-208
620650477
6-22835864
6-25026375
6"2722-2016
6-29422793
6316-287r:
633839779
636056000
638277381
640.503928
642735647
64497^2544
28-160-2557 92560-22
28-1780056 9-259911
28-2311884 9-2715,59
28-2488938 9-275435
-28-1957444
28-21347-20
28-2665881
28-2842712
28-3019434
-28-3196045
28-337-2546
28-3548938
-28-37-25219
-28-3901391
28-4077454
28-4253408
■28-44^29253
28-4604989
28-4780617
■28-4956137
■28-5131,549
-28-5306852
28-5482048
28-5657137
■28-5832119
28-6006993
28-6181760
■28-6356421
28-6530976
28-6705424
28-6879766
28-7054002
•28-7-228132
28-7402157
•28-7576077
•28'7749891
28-7923601
28-8097-206
28-8-270706
28-8444102
28-8617394
28-8790582
28-8963666
28-9136646
•28-9309523
•28-948-2-297
28-9654967
•28-9827535
29-0000000
29-017-2363
•290344623
290516781
•29-0688837
29-0860791
29-103-2644
•29-1204396
•29-1376046
•29-1547595
•29-1719043
•29-1890390
-29-2061637
29-2232784
-29-2403830
-29-2574777
29-27456-23
29-2916370
29-3087018
29-3257566
•29-3428015
29-3598365
■29-3768616
-29-3938769
9-263797
9-267680
9-279308
9-283178
9 287044
9-290907
9-294767
9-298624
9-30-2477
9-306328
9-310175
9-314019
9-317860
9-321697
9-3-25532
9-3-29363
9-333192
9-337017
9-340839
9-344657
9-348473
9-35-2286
9-356095
9-359902
9-363705
9-367505
9-371302
9-375096
9-378887
9-382675
9 386460
9390-242
9-3.94021
9-397796
9-401569
9-405339
9-409105
9-4r2869
9-416630
9-4-20387
9-4-24142
9-427894
9-431642
9435388
9-439131
9-44-2870
9-446607
9-4.50341
9 454072
9-4.57800
9-4615-25
9-465-247
9-468966
9-47-2682
9-476396
9-480106
9-483814
9-487518
9-491^2-20
9-494919
9-498615
9-50-2308
9-505998
9-509685
9-513370
9-517051
9.5-20730
9-524406
SQUARES, CUBES, AND ROOTS.
535
No.
Square.
Cube.
Sq. Root.
Cube Root
No.
Square.
Cube.
Sq, Root.
Cube Root
«()5
74JJ225
64721462.5
29-4108823
9-.5-28079
933
870489
812166-237
30-5450487
9-771484
8()6
749956
64946 189(>
29-4278779
9-5317.50
934
8723.56
814780.504
:«)-5614l36
9-774974
867
751689
651714363
29-4448637
9,53.5417
935
874-225
817400375
;«)-.5777697
9-778462
!ifJ8
753424
653972032
29-46 183.97
9-539082
93()
87(i0.96
8200258.56-
,•50.5941171
9-781947
K()9
755161
656234909
29-478!!059
9-542744
937
877.969
8-2-26,5(;953 30-6] 04557
9-785429
870
75()900
658503000
•29-4957()-24
9 54-;403
938
879}t44
8-J.5-2!);j(i72 30 6-2(i7857
9-788.909
871
758(i4l
660776311
29-5127091
9-550059
939
881721
8-279.-{(i01.9:}0(i43HH;9
9-7.92386
87-2
760384
663054848
29-5-29()461
9-,->53712
940
883600
830584O0o!3O-6.5.941.94
9-7.9.5861
873
762129
665338617
•29-5465734
9.5,57363
941
8J1.5481
833237621 .•{0-6757233
9-79.9334
874
7C.W76
667627624
29-5634910
9.561011
942
887364
83589()W»(
;}0-()9-20185
9-802804
875
765625
669921875
29 5803989
9-564656
943
889249
838.561807
30-7083051
9 806-271
876
767376
672221376
29-5972972
9-568298
944
891136
841-232384
30-7245830
9-809736
877
769129
674526133
29-6141858
9.571938
945
8930-25
8439086-25
30-74085-23
9813199
878
770884
6768.%-152
29-6310648
9-575574
946
894916
{{46590.536
30-7571130
9-8166.59
879
772641
679151439
29 6479342
9-579208
947
896809
849-278123
30-77.33651
9-8-20117
880
774400
681472000
29-6647939
9-582840
948
898704
851.97135)2
.30-7896086
9-823572
881
776161
6}{3797841
29-6816442
9-586468
949
900601
854670349
30 8058436
9-8-27025
88-J
777924
686128968
29-6984848
9-590094
950
90-2500
857375000
30-8-2-20700
9-830476
883
779689
688465387
29-71531.59
9-593717
951
904401
860085351
.30-838-2879
9-8339-24
}i84
781456
690807104
29-7321375
9-597337
952
906304
862801408
30-8544.972
9-837369
885
783225
693154125
29-7439496
9-600955
953
908209
865.5-23177
;50-870()981
9-840813
886
784996
695506456
29-7657521
.9-604570
954
910116
868-250664
30^8868904
9-844-254
887
786769
697864103
29-782.5452
9-608182
955
9120-25
870983875
30-9030743
9-847692
888
788544
700227072
29-7993289
9-611791
956
913.936
8737-22816
30^9 19-2497
9-851128
889
790321
702595369
29-8161030
9-61.5398
957
915849
876467493
30-9354166
9-8.54562
890
792100
704969000
29-83-28()78
9-619002
958
917764
879217912
30-9515751
9-857993
891
7938}}]
707347971
29-8496-231
9-62-2603
.959
919681
881974079
30-9677251
9-861422
892
795664
709732288
29-8663690
9-6-26202
960
921600
884736000
30-9838668
9-864848
893
797449
712121957
•29-88310.56
9-629797
961
9-23521
887503681
31-0000000
9-868-272
894
799236
714516984
•29-J{998328
9-633391
962
925444
8902771-28
31-0161-248
9-871694
895
801025
716917375
29-9165506
9-636981
963
927369
893056347
31-03-22413
9-875113
896
802816
719323136
29-933-2591
9-640569
964
929-296
89.5841344
31-0483494
9-878530
897
804609
721734273
29-9499583
9-6441.54
965
931-225
898632125
31-0644491
9-881.945
898
80<)404
724150792
•29-9666481
9-647737
966
933156
9014-28696
904231063
31-0805405
.9-885357
899
808201
726572699
29-9833287
9-651317
967
935089
31-0.966-236
9-888767
900
810000
729000000
300000000
9-654894
968
937024
907039232
31-1126984
9-892175
901
811801
731432701
3001666^20
9-658468
969
938.961
909853209
31-1-287648
9-895580
902
813604
733870808
30-0333148
9-662040
970
940900
9r2673000
31-1448-230
9-898983
903
815409
736314327
30-049.9584
9665610
971
94-2841
91.5498611
31-16087-29
9.902383
904
817216
738763264
300665928
9-669176
972
944784
918330048
31-1769145
9-905782
905
819025
741217625
30-0832179
9-672740
973
946729
921167317
31-19-29479
9-909178
906
820836
743677416
30-0998339
9-676302
974
948676
924010424
31-2089731
9-912571
907
822649
746142643
30-1164407 9-679860
975
950625
926859375
31--2'249900
9-915962
908
824464
748613312
30-1330383 9683417
976
9.52576
9^2.9714176
31-2409987
9-919351
909
826281
751089429
30- 1496-269
9-686.970
977
9545-29
93^2574833
31-2569992
9-9-22738
910
828100
753571000
30- 166-2063
9-6.90.521
978
956484
935441352
31-2729915
9-926122
911
829921
756058031
301827765
9-694069
i 979
958441
938313739
31-28897.57
9-929504
912
831744
758550528
30 1993377
9-697615
1 980
960400
94119-2000
31-3049517
9-932884
913
833569
761048497
30-2158899
9-701 1.58
1 981
96-2361
944076141
31-3-2091.95
9-936-261
914
835396
763551944
30-23-24329
9-704699
982
964324
946.966168
31 -3368792
9-939636
915
837225
766060875
30-2489669
9-708-237
; 983
966289
949862087
31-35-28308
9-943009
916
839056
768575296
30-2654919
9-711772
; 984
968-256
952763904
31-3687743
9-946380
917
840889
771095213
30-2820079
9-715305
985
970225
9556716-25
31-3847097
9-949748
918
842724
773620632
30-29a5148
9-718835
986
972196
958585-256
31-4006369
9-953114
919
844561
776151559
30-3150128
9-7-2-2363
987
974169
961504803
31-416.5.561
9-956477
920
846400
778688000
30-331.5018
9-7-25888
988
.976144
9644,30-272
31-4324673
9-959839
f>21
848241
7812-29961
30-3479818
9-729411
989
978121
96736166!
31-4483704
9-963198
922
850084
783777448
30 ■36445-29
9-73-2931
990
980100
970-2.99000
31-464-2654
9-966555
923
851929
786330467
30-3809151
9-736448
99)
98-2081
973-24^2271
31-4801.525
9-969909
924
853776
788889024
•20-3973683
9-739963
992
984064
976191488
31-4960315
9-973-262
925
855625
791453125
30-41381-27
9-743476
1 993
986049
9791466.57
.31.5119025
9-976612
926
857476
794022776
30-430-2481
9-746986
994
988036
982107784
31-5-277655
9-979960
927
859329
79(i597983
30-4466747
9-750493
995
990025
985074875
31.5436-206
9983305
928
861184
799178752
30-4()30924
9-753998
996
99-2016
.98804793r
31-5594677
9-986649
929
863041
801765089
30-479.5013
9-757500
997
994009
991026973
31.5753068
9-989990
930
864900
804357000
30-4.959014
9-761000
998
996004
9.9401 1.99^.
31-.591l;«0
9-993329
931 [ 866761
8069.54491
30-5r2-29-26
9-764497
999
998001
997002998
31-6069613
9-9.96666
932 868624
809557568
30-5286750
9-767992
1000
1000000
lOOOOOOOOO
31-6227766
10000000
END OF VOL, I.
Gilbert & Rivington, Printers, St. John's Square, London.
ERRATA.
Page 160, 1. 2, note,/or /a + (« — a)d\ read |a + (« — l)d\
— 163, two place8,/br s* read s,.
— 253, three places,_/br ^ttj '"^'M' omJ-
— 380, note, 1. 10, h./or slip read step.
— 417, 1. 4. bottom, ybr F' read D'.
— 469, 1. 3,/or descnbed, read escribed.
— 471, last line,ybr A read C.
— 472, note,yor a' -f- i^ — c", read (a* -(-6* — c*) sin C, in a few of the copies.
— 478, 1. 8, bottom, after page , insert 471.
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