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I
«
I
I
COURSE
OF
MATHEMATICS.
IN THREE VOLUMES.
COMP08BD FOR
THE USE OF THE ROYAL MILITARY ACADEMY,
BT ORDBR or HIS LORDCHIP
THE MASTER GENERAL OF THE OM)NANC£.
BY
CHARLES HUTTON, LL.D. F.R.S.
J^TB PROFESSOR OF MATHEMATICS IN THE BOTAK.
MILITARY ACADEMY.
THE SIXTH EDITIOtr,
XHLAROID AilD CORRBCTBI^
toL. I.
LONDON:
PRINTEB FOR F, C. AND J, RIVINGTON; O. WILKXB AN9 f,
ROBINSON ; J. walker; G. ROBINSON; LACKINOTON, ALLBNf'
AND CO.; VERNOR, HOOD, AND SHARPE ; 6. KSARSLBY ;
LONGMANy HURSTy REES, ORME, AND BROWN ; CADELI AN1I
DA VIES ; J. CUTHELL ; B. CROSBY AND CO. ; J. RICHARDSON ;
J. M. RICHARDSON; BLACK^ PARRY^ AND I^^NOS^URY ; QAIA
AND CURTIS ; AND J. JOHNSON AND CO.
1811.
' * »• J
I ,
• «
\
V •■
I *
' t *
' • • • F
w >'^ *<r*
T. Dttviwp, Lombard 8tice<»
X
PHEFACE.
A SHORT and Easy Course of t&e Mathematical Sciences
h^s* long been ^considered as a desideratum for the use of
Students in the different schools of education: one that
should hold a middle rank between the more voluminous
^nd bulky collections of this kind, and the mi^e abstract
% and brief commonplace forms, of principles and. memo*
fandums.
«^ For Ung experience, in all Seminaries of Learningji ha$
j> shown, that such a work was very much wanted, and would
>l prove a gre^t and general benefit ; as, for want of it, re
4iourse has always been oblig^ to be bad to a number of
other books, by different authors; selecting a part from one
and a part from another, as seemed most suitable to thef
purpose in hand, and rejecting the other parts— a practice
which occasioned much expence and trouble, in procuring
and using such a number of odd volumes, of various forms
and modes of composition; besides wanting the benefit of
uniformity and referencei which are found in a regular series
of composition.
To remove these inconveniences, the Author of the pre
sent work has been induced, from time to time, to compose
various parts of this Course of Mathematics; which the
experience of many years' use in the Academy has enabled
him to adapt and improve to the most useful form' and
quantity, for the benefit of instruction there. And, to render
ti^t benefit more eminent and lasting, the Master General
of the Ordnance has been pleased to give it its present form,
Ky ordering it to be enlarged and printed, for the use of the
. Roayl Military Academy.
A 2 As
IT PREFACE.
As this work has been composed expressly with the inten
tion of adapting it to the purposes of academical education,
it is not designed to hold out the expectation of an entire
new ipass of inventions and discoveries: but rather to collect
^d arrange the most useful known principles of mathe
matics, disposed in a convenient practical form, demonstrated
in a plain and concis'e way, and illustrated with suitable ex
amples; rejecting whatever seemed to be matters of mere
curiosity, and retaining only such parts and branches, as have
a direct tendency and application to some useful purpose in
life or profession.
It is however expected that much that is new will be found
in many parts of these volumes; as well in the matter, as in
the arrangement and manner of demonstration! throughout
the whole work, especially in the geometry, which is" ren
dered much more easy and simple than heretofore ; and in
the conicsections, which are here treated in a manner at
once new, easy, and natural ; so much so indeed, that all the
propositions and their demonstrations, in the ellipsis, are the
very same, word for word, as those in the hyperbola, using
only, in a very few places, the word sum, for the word differ^
ence: also in many of the mechanical and philosophical parts
which follow, in the second volume. In the conic sections^
too, it may be observed, that the first theorem of each sec
tion only is proved from the cone itself, and all the rest of
the theorems are deduced from the fint, or from each other,
in a very plain and simple manner.
Besides renewing most of the rules, and introducing every
where new examples, this edition is much enlarged in several
places; particularly by extending the tables of squares and
cubes, square roots and cube roots, to 1000 numbers, which
will be found of great use in many calculations ; also by the
table of logarithms at the end of the first volume, and of lo
garithms, sines, a(nd tangents, at the end of the second
Yolume ; by the add;ition of Cardan's rules for resolving cubic
equations I
PREFACE. V
equations; with tables and rules for annuities; and many
other improvements in different parts of the work*
Though the several parts of this course of mathecnatics are
ranged in the order naturally required by such elements^ yet
students may omit any of the particulars that may be thought
the least necessary to their several purposes; or they may
study and learn various parts in a di&rent order from their
present arrangement in the book, at the discretion of the
tutor. So, for instance, all the notes at the foot of the pages
may be omitted, as well as many of the rules ; particularly
the 1st or Common Rule for the Cube Root, p. 85, may well
be omitted, being more tedious than useful. Also the chaf^
ters on Surds and Infinite Series, in the Algebra : or these
might be learned after Simple Equations. Also Compound
Interest and Annuities at the end of the Algebra. Also any
part of the Geometry, in vol. 1 ; any ojf the branches in
vol. 2, at the discretion of the preceptor. And, in any of
the parts, he may omit some of the examples, or he may
give more than are printed in the book ; or he may Tery pro
fitably vary or change them, by altering^the Aumbera oc»
casionally.— As to the quantity of writing ; the author would
recommend, that the student copy out into his fair book no
more than the chief rules which he is directed to learn off by
rote, with the. work of pne example only to' each rule, set
down at full length : omitting to set down the work of all thff
other examples, how many soever he may be directed to
work out upon his slate or waste paper. — In short, a great
deal of the business, as to the quantity and order and maimeri
must depend on the jtt4goient of the discreet and {Hmdeiit
#utor or director*
CONTENTS
OF VOLUME I.
GENERAL Pr^mmaty Prtnci^
P«f»
ARITHMETIC.
/
Ihtaiiim and Nnmeratwn • • . • .
4
Rsman Notation . • • • * •
7
AMUim
8
jfuitraetion .......
11
J/ukiptfcation • .
.13
18
^Bedmtion . . ^
2S
' Gmpaund JMMon • .
92
■ Subtracit&H ....
S<
M
^o
— Dioision . . . . «
41
Getien Bule^ or Stde of Three • . • .
44
Compound Pfopofpfion . . . . ,
49
Vtdgar Fractions
51
Beduetion of Vulgar Fractions . . . ,
54
Addition of Vulgar Fractions . . . ,
62
Subtraction of Vulgar Fractions
62
MtdttpKcaiim of Vulgar Fractions . .
63
DMsiou of Vulgar Fractions . . .
64
i?Ei& g/" 7%retf in Vulgar Fractions
65
Decimal Fractions
66
Addition of Decimak . . . ^ . ,
67
Subtraction of Decinuils
68
Multiplication of Djscimals
ib.
C0N1!ENI3.
«n
r»€»
Dwision of Decimals . • « .
70
JMuction of Decimals • •
. . 75
Mute of Three in Decimals •
T«
Dmsdecimals ...
. . TT
Iwfolution
•»
Evolution ......
«•
To extract the Square Soot
«1
To extract the Cube Root . . ,
«5
To extract any Root whatever
«8
Table qfJPawers and Roots
«0
Ratios f Propor turns, and Progressions ,
. , . 110
Arithmetical Proportion • • •
111
Geometrical Proportion
lie
Musical Proportion . • • .
i»
fiUasosh^fj or Partnership ,
ft.
Single Fellowship . • . .
ISO
Btmble Fellowship . . . .
m
Smple Interest . . • . .
IS*
Vmpound Inttrest . • « .
. . . U7
dU^gatim Medial
U9
^Hkgatwn Alternate . »  . . .
•
Jbh^/^ Position . . . . . . .
iS5
Double Position
1S7
Practical Suestions • . . <
•
140
LOGARITHMS.
•
i^nitim and Propertks vfJJDgari^nm
r . . 14S
To 'Compute Logarithms . ■ , ■
: . . 149
Description and Use cf Lagarkhim
. « . i»
Mnltiplieationby LogariU^ms .
157
iiivisumby Logarithms . .
. . .158
involution by Logarithms ,
159
iSvoluiion h/ Logarithm* . . .
kW
«m
CCKNTENTS.
ALGEBRA.
Definitions and Notation
Addition . . • .
Subtraction . .
Multiplication . .
Division . . . .
Fractions . , .
Involution *
Evolution • . .
Surds
Infinite Series
ArUhmetical Proportion .
Arithmetical Progression
PUes of S/iot or Shells
Geometrical Proportion .
Simple Equations . .
Suadratic Equations .
Cubic and Higher Powers
Simple Interest
Compound Interest  .
Annuities • . . •
GEOMETRY.
Definitions • • • . .
Axioms
, Remarks and Theorems ,
Of Ratios and ProportionS'^Definitions
Theorems
Of Planes and Solids^Definitions
Theorems . • . . ;
Problems .
Application of Algebra to Geometry
, , Problems . • . . .
Table qf Logarithms . . . •
Paftf
161
165
170
171
174
178
189
198
196
203
208
210
213
218
28(1
23d
247
256
257
260
265
271
ib.
S09
318
326
328
S43
359
360
S66
COURSE
OP
MATHEMATICS, 40.
GENERAL PRINCIPLES.
UANTITT, or Magnitude, is any thing that mil
admit of increase or decrease ; or that is capable of any sort
of calculation or mensuration : such as numbers, lines, spacey
time, motion, weight.
2, Mathematics is the science which treats of all kincls
of quantity whatever, that can be numbered or measured.—
Th^ part which treats of numbering is called Arithmetic;
and that which concerns measuring, or figured extension, is
called Geametfy^^These two, which are conversant about
multitude and magnitude, being the foundation of all the
other parts, are called Pure or Abstract Mathematics; be
cause they investigate and demonstrate the properties of ab
stract numbers ^d magnitudes of all sorts. And when th^se
two parts are applied to particular or practical subjects, they
.eonsdtute the branches or parts called Mixed Mathematics.^
Mathematics is also distinguished into Speculative and Prvnv
Jica/c viz. Speculative, when it is concerned in discovering
properties and relations ; and Practical, when applied to
practice and real use concerning physical objects*
\, Vo^L B 3.to
2 GEigERAL PRINCIPLES.
3. In Mathematics are several general terms orprinci^es;
such as. Definitions, Axioms, Propositions, Theorems, rro
blems. Lemmas, Corollaries, Scholiums, &c.
4. 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 ex
pressed in words that are common and perfectly well under
stood.
5. A Proposition is something proposed to be proved, or
something required to be done ; and is accordingly either si
Theorem or a Problem*
6. A Theorem is a demonstrative proposition; in which
some property is asserted, and the truth of it required to be
proved. Thus, when it is said that. The sum of the three
angles of any triangle is equal to two right angles, this is a
Theorem, the truth of which is demonstrated by Geometry.
—A set or collection of such Theorems constitutes a Theory.
7. 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 all the three angles of any^triangle, or to
di'aw one line perpendicular to another. A Limited Pro»
hlem is that which has but one answer or solution. An Un^
limited Problem is that which has innumerable answers.
And a Determinate Problem is that which has a certain num
ber of answers.
S. Solution of a Problem, is the resolution or answer given
to it. A Numerical or Numeral Solutiony 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.
9. A Lemma is a preparatory proposition, laid down in
order to shorten the demonstration of the main proposition
which follows it.
10. A Corollary y or Consectary^ is a consequence draWx
immediately from some proposition or other premises.
11. A Scholium is a remark or observation made by some
foregoing proposition or premises. •
12. An Axiom, or Maxim, is a selfevident proposition ;
requiring no formal demonstration to prove the truth of it ;
but is received and assented to as soon as mentioned. 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. /
13. A
GENERAL PRINCIPLES. S
Id* A Pastulatey or Petition, is sometliing required to be
donC) which is so easy and evident that no person will hesi
tate to allow it.
14. An Hypothesis is a supposition assumed to be truei in
order to argue from^ or to found upon it the reasoning and
demonstration of some proposition. ^
15. Demonstration is the collecting the several arguments
and proofs, and laying them together in proper order, to
«}iow the truth of the proposition under consideration*
16. A Direct, Positive, or Ajffirmative Demonstration, h
that which concludes with the direct and certain proof of the
proposition in hand. — This kind of Demonstration is most
satisfactory to the mind ; for which reason it is called some
times an Ostensive Demonstration.
17. An Indirect, or Negative Demonstration, is that which
shows a proposition to be true, by proving that some absur
dity would necessarily follow if the proposition advanced were
false. Thrs is also sometimes called Reductio ad Absurdum;
becausie it shows the absurdity and falsehood of all supposi
tions contrary to that contained in the proposition.
18. 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.
19. Analysis, or the Analytic Method, is the 2rt or mode
of finding out the truth of a proposition, by first supposing
the thing to be done, and then reasoning back, step by step,
till we arrive at some known truth. — ^Thi$ is also called the
Method of InventioHy or Resolution; and is that which is com
monly used in Algebra.
20. Synthesis, or the Synthetic Method, is the searching
out truth, by first laying down some simple and easy princi
jdes, and pursuing the consequences flowing from them till
we arrive at the condusion. — This is also called the Method
rfCon^sjfion; and is^ the reverse of the Analytic method, as
£bis proceeds from known principles to an unknown conclu
sion ; while the other goes in a retrograde order, from the
thing sought, considered as if it were true^ to some knpwn
principle or fact. And therefore, when any truth has been
foupd out by the Analytic method, it may be demonstrated
by a process in the contrary order, by Synthesis.
B 2 ARITH
t * 3
ARITHMETIC.
jljLRITHMETIC is the art or science of numbering ; be
ing that branch of Mathematics which treats of the nature
and properties of numbers, — ^When it treats of whole num
bers, it is called Vulgar^ or Common Arithmetic; but when of
broken numbers, or parts of numbers, it is called Frmctions.
Unity J or an Unitj is that by which every thing is called
one ; being the beginning of number ; as, one man, one baU^
one gun.
Number is either simply one, or a compound of several
units % as, one man, three men, ten men.
An Integer^ or Whoh Number^ is some certain precise
quantity ofunits ; as, one, three, ten.— These are so called as
distinguished from Fractions^ which are broken numbers, or
parts of numbers ; as, onehalf, twothirds, or threefourths.
NOTATION AND NUMERATION.
Notation, or Numeration, teaches to denote or ex
press any proposed number, either by word$ 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 intitxlucea
into Europe by the Moors, about 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, cipher, or nothing. These cha*
racters or figures were formerly all called by the general
name of C^hers ; whence it came to pass that the art of
Arithmetic was then often called Cohering. Abo the first
nine are called Significant Figures^ as distinguished from the
cipher, which is of itself quite insignificant.
Besides this value of those figures, they have also another,
which depends on the place they stand m when joined toge
ther; as in the following table :
Units
NOTATION AND NU»ffiRATION.
S 9
Si ^ '
o
CO
« »3 r^* a J rS S*3 ,2 JS «
&€. 98765432 1
98 7 65 4 3 2
9 8 7 6 5 4 3
9 B 7 6 .5 4
9 8 7 6 5
9 8 7 6
9 8 7
9  8
9
fl
Here, any figure in the first place, reckoning from right to
left, denotes pnly its own simple ralue ; 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 righthand side to other figures, it in
creases their value in the same ten>fbld proportion : thus, 5
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 halfperiods, each halfperiod con«
sisting of three figures ; the name of the first period being
units; of the second, millions; of the third, millions of
millions, or bi*millipns, contracted to billions : of the fourth,
millions of millions of millions, or trimillions, 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 thousand?.
The
6 ARITHMETIC.
The following Table contains a summarj ef the whole
docUine.
Periods* QuadrilL; Trillions; Billions; Millions; Units.
th. un. th« un, th. un. th. un. th. un.
123,456; '78»,098; 765,432; 101,234; 567,890.
Numeration is the reading of any number in words
that is proposed or set down in figures ; which will be easily
done by help of the following rule, deduced from the fore
going tablets and observations — ^viz.
Divide the figures in the proposed number, as in the sum
mary above, into periods and halfperiods ; then begin at the
lefthand side» and r^ad the figures with the names set to
them in the two foregoing tables.
EXAMPLES.
Express in WQr4s the fbllo\ving numbers ; viz^
34
96
180
304
6134
5028
15080
72003
109026
483500
2500639
7523000
13405670
47050023
309025600
4723507689
274856390000
6578600307024
Notation is the setting dgwn ii figures any number pro
posed in words ; which is done by setting down the figures
instead of the words or names belonging to them in the sum
mary above ; supplying the vacant places with ciphers where
any words do not occur.
EXAMPLES.
Set down iii figures the following numbers ;
Fiftyseven.
Two hundred eighty six.
Nine thousand two hundred and ten.
Twentyseven thousand five hundred and ninetyfour.
Six hundred and forty thousand, four hundred and eightyone.
Three millions, two hi^ndred sixty thousand, one hundred
and six.
Four
NOTATION AND NUMERATrON.
Four hundred and eight millions, two hundred and fiftyfive
thousand) one hundred and ninetytwo.
Twentyseven thousand and eight millions, ninetysix thou
sand two hundred and four.
Two hundred thousand and five hundred and fifty millions)
one hundred and ten thousand, and sixteen.
Twentyone 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; Y for Jive; X for ten; lu for fifty;
C for an hundred; D for five hundred; ^ for a thousand*
The other numbers they expressed by various repetitions and
combinations of these, aft:er the following manner :
1 =
2 =
S =
4 =
5 =
6
7
8
9
10
50
100
500
1000
2000
5000
6000
10000
50000
60000
100000
1000000
2000000
&c.
I
II
III
mi or IV
V
VI
VII
VIII
IX
X
L
C
Dor ID
M or CIO
MM
As often as any character is re
peated, so many times is its
value repeated.
A less character before a greater
diminishes its value.
A less character after a greater
increases its value.
V or IDD
VI
X or CCIOO
Lj>r IDOD
LX
C^or CCCI0D3
Mor CCCCIODDD
MM
&c.
For every 3 annexed, this be
comes 10 times as many.
For every C and O, placed one
at each end, it becomes 10
times as much.
A bar over any number in
creases it 1000 fold.
ExPLA
$ ARITHMETia
• • •
Explanation of certain Characters*
There are various characters or marks used in Arithmetic,
amd Algebra, to denote several of the operations and proposiF
tions ; the chief of which are as follow :
+ signifies/Zi/x, or addition.
—  •«■ minus, or subtraction.
X or .  multiplication.
5   division.
: :: :  proportion!
=   equality.
\/   square ropt.
J/ •  cube root, &c.
^   diffl between two numbers when it is not
known which k the greater.
Thus,
5 + Sj denotes that 3 is to be added to 5.
.6 <— 2, denotes that 2 is to be taken from 6.
'7 X 8, or 7 . S, denotes that 7 is to be multiplied by S.
5 r 4, denotes that 8 is to be divided by 4.
2 : 3 : : 4 : 6, shows that 2 is to 3 as 4 is to 6.
6 + 4 = 10, shows that the sum of 6 and 4 is equal to 10.
V^S, or 3i, denotes the square root of the number 3.
^5, or 5% denotes the cube root of the number 5.
7^, denotes that the number 1 is to be squared.
8^, denotes that the number 8 is to be cubed.
&c»
OF ADDITION.
Addition 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 as follows : ^ . ^
Set or place the numbers under each other, so that each?
figure may stand exactly under the figures of the same value,
iha^
ADDITION. ft .
thatisi 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 tens, or
if nothing remains, a cipher, and carry as many ones to the
next row as there 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 top, 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 num
ber to be proved.— rLastly, add this last found number
aad 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 probf is founded on the
plain axiom", that " The whole is equal to all its parts taken
together.*'
»
Third Method.'^Add the figures in
the uppermost line together, and find example i.
how many nines ^e contained in
their sum. — Reject those nines, and 3497 g 5
set down the remainder towards the 6512 .S 5 •
right hand directly even with the 8295 ^ 6
figures in the line, as in the annexed © —
example. rDo the same with each 18304 g 7
of the proposed lines of numbers, set ^ • ^ —
ting all these excesses of nines in a co W
himn on the righthand, as here 5, 5, 6. Then, if the excess
of 9^s in this sum, found as before, be equal to the excess
of 9's in the total sum 1 8804, the work is probably right.*
Thus, the sum of the righthand column, 5, 5, 6, is 1^, the
excess of which above 9 is 7. ' A^o the sum of the figures in
the
10 ARITHMETIC
the sum toul 18304, is 16, the excess of which above 9 is
also 7 J the same as the former^.
2.
12345
OTHER EXAMPLES.
3.
12345
4.
12345
67890
98765
43210
12345
67890
67890
9876
543
21
9
876
9087
^ \ 56
234
1012
302445
90684
•I
23610
290100
78339
11265
302445 90684 23610
* This method of proof depends on a property of* the number g,
which, except the number 3, belongs to no other digit whatever ;
namely^ that '^ any number divided by g, will leave the same re
mainder as .the sum of its figures or digits divided by 9 :** which
may be demonstrated jn this manner.
Demonstration. Let there be any number proposed, as 4658.
This, separated into its several parts, becomes 4000 + 600 + 50
+ 8. But 4000 = 4 X 1000 = 4 X (9«9 + 1) = 4 X 999 J 4.
In like manner 600 = 6 xgg + 6', and 50 = 5 X 9 + 5. There
fore the given number 4658 = 4 X 999 +446x99 + ^ +
5 X9P5 + 8=:4X999 + <5X99 + 5X9 + 4 + 6 + 5
+ 35 and 4058 rp = (4 X 999 f 6 X 99 + 5 X 9 + 4 + 6
+ 5 + 8) r 9. But 4 X 999 + 6x99 + 5 X9i8 evidently
divisible by 9, without a remainder 5 therefore if the given num
ber 4658 be divided by 9ji it will leave the same remainder as
4 + 6 + 5 + 8 divided by p. And the same, it is evident, will
hold for any other number whatever.
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 practice.
Now, from the demonstration above given, the reason of the
rule itself is evident ; for the excess of 9*s in two or more numbers
being taken separately, and the excess of 9's taken also out of the
sum of the former excesses, it is plain that this last excess must be
equal to the excess of 9's contained in the total sum of all these
numbers 5 all the parts taken together being equal to the whole.
T his rule was first given by Dr. Wallis in his Arithmetic,
published in the year l657*
Ex.
SUBTRACTION. 1 1
'Ex.5. Add 3426 J 9024; 5106; 8890; 1204, together.
Ans. 27650*
6 Add 509267; 235809; 72920; 8392; 420; 21; and 9,
together. Ans. 826838.
7. Add 2; 19; 8l7; 4298; 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
thie 24th day of November, both days included ? Ans. 224.
40. An army consisting of 52714 infantry*, or foot, 5110
horse, 6250 dragoons, 3927 lighthorse, 928 artillery, or
gunners, 1410 pioneers, 250 sappers, and 406 miners : what
is the whole number of men? Ans. 70995.
OF SUBTRACTION.
• >
SuBTKACtiON teaches to find how much one number
exceeds another, called their drfferencty 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 man«
ner 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 number, from
the figure above it, setting down the remainder below it,—
But ifthe figure in the lower line be greater than that above
it, first borrow, or add, 10 to the upper one, and then take
the lower figure from that sum, setting down the remainder,
and carrying 1 , for what was borrowed, to the next lower
figure, with which proceed as before; and so on till the
whole is fiqished.
* The whole body of foot soldiers is denoted by the word Iri"
fantry; and all those that charge on horseback by the word Cccvo/r^.
— Some authors conjecture that the terra infantry is derived fiom
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 which she engaged
^and totally routed the enemy. In honour of this events 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.
To
12
ARITHMETIC.
To psovs Subtraction.
Add the remsunder to the less number^ or dut which b
just above it; and if the sum be equal to the greater or upper
most number^ the work is right*.
1.
From 5386427
Take 2164815
£XAMFL£S.
2.
From 5386427
Take 4258792
S.
From 1234567
Take 70297S
Rem. 3222112
Rem. 1127635
Rem. 531594
Proof.5386427
Proof. 5336427
Proof. 1234567
4. From 5331806 take 5073918.
5. From 7020974 take 2766809.
6. From 8503602 take ^74271.
Ans^. 257888.
Ans. 4254165.
Ans. 79291 3 U
7. Sir Isaac Newton was bom in the year 1642, and he
died in 1 727 : how old was he at the time of his decease ?
Ans. 85 years.
8. Homer was bom 2543 years ago, and Christ 1810 years
ago: then how long before Christ was the birth of Homer ^
Ans. 733 years.
9. Noah's flood happened about the yeai^ 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 tp this present year 1810 ? Ans. 660 years.
1 1 . Gunpowder was invented in the year 1 330 : then how
long was this before the invention or printing, which was
in 1441 f Ans. 1 1 1 years.
1 2. The mariner's compass was invented in Europe in the
year 1302: then how long was that before the discovery of
America by Columbus, which happened in 1492.^
Ans. 190 years.
* The reason of this mediod of proof is evident; for if the
diflerenoe of two numbers be added to the less^ it must manifestly
siake up a sum e^ual to ibe greater.
OF
ikrLTlPLICATION.
IS
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
Mub^licand^'^The number you multiply by, or the number of
repetitions, is the Mubipiier.^^Aj^ the number found, being
the. total ambunt, is called the Productr^Aho, both the
multiplier and multiplicand are, in general, named the Termi
or Factors.
Before proceeding to any operatiops in this rule, it is ne«
cessary to learn off very perfectly the following Table, of all
the products of the first 12 numbers, commoidy called the
Multiplication Table, or sometimes Fythagoras's Table^ from
its inventor.
Multiplication Table.
1
2
2
3
4
8
5 6
7
14
8
16
9
10
11
12
4
6
10
15
20
25
30
35
12
18
18
20
22
24
S
6
9
12
16
21
24
27
30
33
36
4
.8
12
24
30
28
32
36
40
44
48
60
5
10
15
20
24
28
35
40
45
50
55
6
7
12
18
36
43
48
54
60
66
72
14
21
4^
49
56
63
70
77
84
96
108
8
16
24
32
40
48
56
64
72
72
80
B8
99
9
18
27
SO
86
40
44
45
50
55
60
54
63
81
90
10
U
12
20
60
66
70
77
80
90
100
110
120
22
24
33
88
99
110
121
132
36
48
72 1 84
96
108
120
132
144
T0
14 ARITHMETIC.
To multiply any Given Number by a Single Figure, or by am
^^ ' Number not nJf than I'i. * ^^
■ «
* Set the multipliei^ under the units figure, or righthand
pl;(ce, of the multiplicand, and draw a line below it.*— Then,
beginning at the right*hand, multiply every figxire in this by
the multiplier.^ — Count how many tens there are in the pro
duct of every single figure, and set down the remainder di
rectly 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.
To multiply by a Number consisting of Several Figures.
\ Set the multiplier below the multiplicand, placing thenv
as in Addition, namely, units under units, tens under tens, &c.
drawing a line below it. — ^Multiply the whole of the multi
plicand by each figure of the multiplier, as in the last article;
setting
^ The reason of this rule is the same as for
the process in Addition^ in which 1 is car
ried for every 10, to the next place, gra
dually as the several products are produced^
one after another, instead of setting them
all down one below each other, as in the an
nexed example.
5§78
4
■
32
280
2400
2000O
8X4
70 X 4
600 X4
5000 X 4
22712 =5678 X 4
f After having found the produce of the multiplicand by the first
figure of the multiplier, as in the former case, the multiplier is
supposed to be divided into parts, and the product is found for the
secQud 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 must be set in the place of
tensi
MULTDPLICATIOI?. IS
setting down a line of products for each figure m the multi
ptier, so as that the first figure of each line ma^c stand straight
under the figure multiplying by. — Add all the lines of pro
ducts together, in the order as they stand, and their sum will
be the answer or whole product required.
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. — *Cast all the 9's 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 9's out of the product, as also out of
tens ; or, which is the same things directly under the figure multi
plied by. And proceeding in
this manner separately with all
the ^ures of the multiplier, 1234567 the multiplicand,
it is evident that we shall mul 4567
tiply all the parts of the mul —
tipdicand by all the parts of 8641969= 7 times the mult,
the multiplier, or the whole of 7407402 r= 60 times ditto,
the multiplicand by the whole 6172835 r= 500 times ditto,
of the multiplier : therefore 493826S =4000 times ditto.
these several products being _
added together, will he equal 5638267489=4567 times ditto.
tothewhole required product; ■■ ■■
as in the example annexed.
* This method of proof is derived from the peculiar property of
the number 9, mentioned in the proof of Addition, and tha reason
for the one may serve for that of the other. Another more ampla
demonstration <^ this rule may be as follows : — Let P and Q denote
the number of 9^s in the factors to be multiplied, and a and 6 what
remain ; then 9 P+a and 9 Gl+6 will be the numbers themselves,
and their product is (9 P X 9 Gl) + (9 P X A) + (9 Gl X a) f
(a X h) i but the first three of these products are each a preiCise
number of 9's> because their factors are so, either one or both :
these therefore being cast away, there remains only a x 6; and if
the 9's also be cast out of this, the excess is the excess of 9*s in the
total product : but a and h are the excesses in the factors them
Bcjves^ and a x 6 is their product ; therefore the rule is true.,
the
16
ARITHMETIC
the whole product or answer bf the questioni reserving the
renuinders of these last two, which remainders must be eauai
when the work is right. — Ifote^ It is ccMnmon to set the lour
remainders within the four angular spaces of a crossj at in the
example below.
Third 3/^^. Multiplication b 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.
Muk. 3542
by 6196
21252
34878
3542
21252
21946232 Product.
EXAMPLES.
Proof.
or Mult. 6196
by 3542
12392
247S4
30980
18588
21946232 Proof.
OTHER EXAMPLES.
Multiply
Multiply
Multiply
Multiply
Multiply
Multiply
Multiply
Multiply
Multiply
•Multiply
Mukiply
Multiply
Multiply
Multiply
Multiply
Multiply
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
123456789
302914603
273580961
402097316
82164973
7564900
8496427
2760325
by S. Ans.
by 4. Ans.
by 5. Ans.
by 6. Ans.
by 7. Ans.
by 8. Ans.
by 9. Ans.
by 11. Ans.
by 12. Ans.
by 16. Ans.
by 23. Ans.
by 195. Ans.
by 3027. Ans.
by 579. Ans.
by 874359. Ans.
by 37072. Ans.
370370367.
493827156*
617283945.
7407407S4.
864197523.
987654S12.
1111111101.
1358024679.
1481481468.
4846633648.
6292362103.
78408976620.
248713373271.
4380077100.
7428927415293*
102330768400.
COHTIUG
MULTIPUCATION. 11
Contractions in MoLTtrucATioM.
I. Jf^im there an Ciphers in the Factors.
If the ciphers be at the righthand of the numbers ; mul
tiply the other figures only, and annex as many ciphers to
the righthand otthe whole prodocti as are in both the fac*
tors.— When the ciphers are in the middle parts of the mul*
tiplier } neglect them as before^ only taking care to place
the first figure of every line of products exactly ttnd«r th*
£gure multiplying with.
EXAUPLES.
Mult. 9001635 Mult. 3901S0400
by • 70100 by  406000
■■III ■■■■■II .^ ■ ■ » — — — .— ^
9001635 2844S284
63011445 15628816
631014613500 Products 158632482400000
S. Mukiply 81503600 by 7080. Ans. 572970308000.
4. Multiply 9030100 by 2100. Am. 18963210000.
5. Multiply 8057069 by 70050. Ans. 56439768345Q.
n. When the Multiplier is the Proiiict of two or more Numbers
in the Tabk; then
^ Multiply by each of those 'parts separately, instead pf
the whole number at once.
EXAMPLES.
1. Multiply 51307298 by 5Q^ or 7 times t.
61307298
7
359151086
' 8
2873208688
f The reason of this rule is obvious enough $ for any number
mttltif^lied by the component parts of another^ must give the same
product as if it were multiplied by that number at once Thus, in
tbe 1st example, 7 times the product of b by the giv^n number^
ifnakes 5Q times the same number, as plainly as 7 times 8 makes a6.
Vol. I. C 2.Mul
It: ARITHMETIC
2. Multrpl7 3!7045d2 by 86. Aus, .1141365512.
S. Multiply. 2S753804 by 72. Ans 2142273888*
4. Multiply 71283^8 by 96. Ans. 684i52:3326.
^. Ai*hiply 160430800 by 10$, Aiis. 11^26526 \QO.
,«, Multiply 61835720 by 1320. Ans. 81623150400.
. 7* There was aa army composed of 104 "^ battalions, each
consisting q£ ^00 men j what was the number of men con*
tained in the whole? Ans. ^2000* /
•. 9f A conToy of ammnnkion f bread, consisting of 250^
waggonst and each waggon containing 320^ loaves, hzvinfr^
been intercepted and taken by the enemy ^ what is the num
ber of loaves lostA Ans. 80000.
tsam
OF DIVISION.
' Division is ,a kind of compendious ^lethod of Subtrac*
tion, teaching to find how often one number is contained in
another, or may be taken out of it : wiiidx is the same thing.
The number to be divide is called the Dividend. —
The number to divide by, is the Z>;wflr.4r— And the number
of times the c^vi^nd contain^, the divisor, is called the Quo*
tient. — iSometiiqes thfire b g Rnnaitider left^ after the division
i^^nished.
The usual manner of placing the terms> is, the dividend in
the. middle^ haying.the cU visor en the left hand, and the<}uo*
tient on the right, each separated by a curve line j as, ta
divide 12 Iq^ 4, the quotient is 3,
• Dividend ' ' '  l2
Divisor 4) 12 f 3 Quotient; 4 subtr,
showing that the number 4 \& 3 times —
contained i9 12, t)E may be S times 8
subtracted out of it, as m the ipargin. 4 subtr*
X RuU. — Having placec^ the divisor —
before the dividend, as above .direct 4
ed, find how often the 'divisor is con 4 subtr*
tained in as many figures of the divi —
dend as are just necessary, and place the
number on the right in the quotient. —
Mul
*yi^^— W^ I ■ ■■ I H ■ II ,1 ^ ■ ■ I I III I I ■■ fcl^M^I—— *^^ll ■
* A battalion is a body of foot, censisting of 500, or 600, or 700
men, more or less.
f The amitoufiitioii bread, is that which b provWedfor, ami ifi^
'triboted to, the soldiers j t!^ u^u^ allowanee being a leaf of ^
pounds to evety soldier, once in 4 days. ' ' ■>
 Xlta this way the dividend b resolved into parts, end by trial i*
*   ■ '•  ibdnd
Multiply the divisor by this number, and set the product
tinder the figures of the dividend before^mentioned.**Sub«
tract this product froro that part of the dividend undi^ which
it stands, and bring down the n^xt figure of the dividend, ox*
more if necessary, to join on the right of the temainder.—Di*
vide this ntunber, scr increased, in the same n^aiinera;^ before 
and so on till all the figures are brought down and used.
•
If. B, If it be necessary to bring down more figures than
one to any remainder, in order' to make it as large as the
divisor, or larger, a cipher must be set in the quotient for
every Bf^ate so brou^t down more than one.
To PRQvt Division*
* Multiply the quotient by the divisor; to this product
add the jpemainder, if there be any ; then the sum will be
equal to the dividend when the work is right*
found how often the divisor is contsllned in each of those patt8> one
idler another^ arranging the several figureS'Of the qaotieutone after
another^ into one numben
When there is no remainder to a division, the^ quotient is the
whole and perfect answer to the question. But when there is a re
mainder^ it goiea 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 ^ time iQore) if the 4th part of the divisor^ it will go. one
fourth of a dq^e more ; and so on. Therefore, to complete the
quotient, set the remainder at the end of it, above a small line, and
toe divisor below it> thus forming a fractional part of the whole
quotient.
* This method of proof is plain, eoough : for since the quotient
is the nimiber of times the dividend contains the divisor, the quo
tieni multiplied by the divisor must evidendy be equal to the
dividend.
There are also several other methods sometimes used for proving
IHvision, some of the most useful of which are as follow :
Stccftd Me/W.*Snbtract the remainder from the dividend^ end
4ivide what is left by the quotient ; so shall the nei^ quotient £ro9i,
this last divisi9n be equal to the former diviaor, when the wofk is
tight.
T^r4 Method, — Add together the remainder and all die pro
ducts of the several quotient figures by the divisor^ according to the
order is which they stand ia the work $ and the sum will be
^ual to the dividend wl^en the work is right.
C 2 EXAM
2«
ARrrHMETIC.
EXAMPLES.
I. Quot.
3} 1234567 (411522
12
mult. S
2. Quot.
37 ) 12345678 (333666
111 37
3
3
1234566
add 1
4
S 1
1234567
9 •■
15
15
Proof.
6
7
6
Rem. ' 1
\
2335662
1000998
rem. 36
124
135 .
1 1 1 1234567$ ^
246 Proof.
222
247
222
258
222
Rem. 36
3. Divide 73146085 by 4.
4. Divide 5317986027 by 7.
5. Divide 5701^6382 by 12.
6. Divide 74638105
7* Divide 137896254
8. Divide 35821649
9. Divide 72091365
Ans. 1 828652 Ij^.
Ans. 7597122894*.
Ans. 47516365^.
Ans.20l7246yy.
Ans. 1421610^.
Ans. 46886fJ^.
Ans. I386I3WT.
Ans. 80496U?^
by 37.
by 97.
by 764.
by 5201.
10. Divide 4631064283 ^ 57606.
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 dbtance of 378 miles from the head
quartersi receive orders to join their corps in 18 days : what
number of miles must they march eadi dajr to obey their
orders? Ans. 21.
13. The annual revenue of a gentleman being 88330/;
kow ^uch per day is that equivalent to, there being 365 days
in the year ? Ans. 104A
CoNTHACTlONS IN DIVISION.
. There are certain contractions in Division,, by which the
operation in particular cases may be performed in a shorter
jmanner : as follows : . .
I. Divp
DIVISION.
«
!• jPivision if any Small Number^ not greater tlian 12, vaxj
be expeditioualy performedi bj multiplying and subtracting
mentallyy omitting to set down the work, except only the
quotient inunediately below the dividend.
3) 56103961
Quot. 18701320J.
EXAMPLES.
4) 52619675
5) 1379192^
6) 38^72940 7 > 81396627 3 ) 23718920
9 ) 439819^2 1 1 ) S76 14230 J2 ) 27980373
«■
»■ ■ ' U
n. * When ethers are annexed to the Divisor i cut off those
ciphers from tt, and cut off&e same number of figures from
the right*hand of the dividend; then divide with the remain
ing 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.
EXlMPi.ES.
1. Pivide 3704196 by 20. 2. Divide 31086901 by 7100.
2,0) 370419,6 71,00)310869,01 (4378ff
284
Ol
Quot. 185209ii
268
213
556
599
568
SI
■y^
3. Divide
* This method is only to avoid a needless repetition of ciphers^
which wQuld happen in the coQimoa V9y. And ihe truth of the
principle
B2 AlUTHMETICt
3. Divide 7380964 by 23000^ Am. ilpii^^^
4. Divide fli04109 fay 580a Am. 397if^.
nif W7>en tie Dhuistnr is the ex^ct Product rf two or moh
jf the small Numbers not grettter than 12 : * Divide by eadl
of those numbers Separately^ iiutead of the whole divisor sit
pnce.
N. B» There are commonly several remainderi in work*
ing by this rule, one to each division', and to find the true w
whole remainder, the same as if the division had been perr
formed ^U at opce, proceed as follows; Multifdy the. last
remainder by thf preceding divisor, or last but one and to
the product add the preceding remainder; multiply this sum
by the n^xt .preceding divisor, and to the product add the
nex^ preceding remainder i and so pn, till you have gone
backward through all the divisors ^4 remainders to the firsts
As in the example following :
EXAMPILES.
1. I^ivide 31046895 l^y 56 or 7 times d.
7 ) 810^835 6 the iastremf
tsxdttf 7 precpd^ iHvisof^
p ) 44352621 iirst rem.
; 48
554401 — fi second ren^. ^lid> 1 the I'st rem*
Ans. 55440744 43 'whole rem.
TS
principle on which it is foande^, is evident ; for, pi^ti^Dg off the
fame number of ciphers^ pr figures, from each> is the same at
dividing each of them by 10, or lOO, or 1000, &c. according to
the number of ciphers cut off; and it is e'^rdent, that as ofien as
f be whole divisor is cpn^ined in the whole dividend, so often must
^ny part of the former be containcji in a like part of the latter.
*' This fdlows from the second contraction in Multiplication^
being only the cpnverse of it; for the half of the third part of any
thing, is evidently the same as the ^ixth part of the whole ; and
00 of ^ny other numbers.— The reaspn of the method of finding
the whole remainder from the .several particular ones, will best
appear from the nature of Vulgar Fraptipps. Thus^ in the firs^
example abpvfs, the first remainder being 1, when the divisor i^
7 makes j ; this must be added to the second remainder, 6^^
making 6f to the flivisor 8^ or to be divided by 8. But €f =:
r— ^ = yi and thi^ divided by 8 gives ^^=:^..
2. P^vide
r
«. Divide 7014596 by 72. Ans. 97424f^
3. Qisrid^ .5 130652 by .132. . . Ans^ SSiea^ZyV
4. Divide 83016572 hj 2ii9. Ans. 945502^.
. IV. Common Divisign may be pftfirmed more condse/jj
by omitting the several products, and setting down only the
remainders; namely, x multiply the divisor by the quotient
figures as before, and*^ without setting down the product^
subtract each figure of it from the dividend, as it is produc^ds
mlvrays remembering to carry as many to tbft next figure as
wbre borrowed before.
[ EXAICPIES*
U Divide 3104679 liy $33.
«33 ) 3 104679 ( 3727^V»
6056
2257
5919
88
«. Divide 79165288 by 238* . Ans^ 332627^^^.
8. Divide 291S7062 by 5317. . . Ans. 5479mf.
4. Divide 62015735 by 7803  Ansu 7?47ff
OF REDUCTION.
Reduction is the changing of numbers firofcn oot name
or denomination to another, Hithout altering their valqe.—
This is chiefly concerned in reducing' money^ weights^ and
measures. .
When tilt numfiers are to be reduced from a higher name
to a lovrer, it is called Reduetkn DescenSngi mst wheii^
contrarywi^, from a lower name to a higher, it is Redudkn
AscenHngm
Before proceeding to the rules and qtresttc^fs of Rednctim,
it will be proper to set down the usual Tables of mdns^^
vireights, and measures^ whick ar^ as follow : \
bf
f4 ARITHMETIC
(y MONET, WEIGHTS, amd MEAStTRES.
Tables of Moi^et*,
. 3 Farthings
4 Farthings
)2 Penc9
20 ShilUngs
=s I Halfpenny f
^* 1 Penny d
«; I ShilUng /
= 1 Pound £
4
48
S60
S5r 1 /
== 12 =^ 1
=; 240 = 20
£
= I
PENCE TABLE.
fHltLINGS TABLE,
1
d
i d
S
l/
1
20
is 1 8
1
is
12
•
30
— 2 6
2
—
24
40
^3 4
3
«*
36
30
 4 2
4
«.»
48
60
— 5
5
—
60
1
70
— 5 10
6
.^
72
'
80
— 6 8
7
«^
84
i
90
—7 6
8
.
96
'
100
— 8 4
9
^^
lOS
HO
— 9 a
10
~
120
1
120
, JO
11
—
132
1
Trot
Gold.
* £ denotes ponndi , « shillings, and d denotes pence.
^ denotes i farthing, or one quarter of any thing.
\ denotes a halibenny^ or the half of any thing.
^ denotes 3 farthings, or tliree <uarter8 of aoy thing.
The full weight and value of the English gold and silver coin,
i^ as here below :
Weigh.
dwt gr
a 16
1 8^
I The usual value of gold is nearly 4/ an ounce^ or 2<i a grain 
an4 that cf ^silver is oearlv ^s an ounce. Alsoj the value pf any
quantity of gold^ is to the value of the s^me weight of standard
iiiver^ nearly as t. 5 to I « or more nearly as 15 and 11 4th to I.
Pure gold, free from mixture with other metals, usually called
\ $Dt gol4» U of so puie 9 Datur^> that it wiU endure the fire
•' vitfeQut
Vahe.
£ » d
1 1
10 d
A Guinea
Halfguinea
Seven Shillings 7
QuarterguineaQ ^ 3
SlLVEH.
Value.
Weight.
9 d
dwt gr
A Crown
5
19 84
Halfcrowa
% 6
9 l^i
ShiUing .
1 O
3 21
Sixpence
Q 6
. 1 ^^i
24= 1 oz
480=r 20= 1 A
5760=240 = 12= r
TA5US5 OF W^QHTS. 15
Trov Weight*.
Grains   marked ^r \ gr dnvt
24 Graim make i Pennyweight dwt
20 Pennyweights 1 Ounce «z
1 2 Ounces 1 Pound lb
By this weight are weighed Gold, Silyer, and Jewels*
Apothecahies' Wjeight.
Grains   marked gr
^0 Grains make 1 Scruple sc or 9
3 Scruples 1 Dram dr or 3
.8 Dr^mi. 1 Ounce w; or t
12 Ounces 1 Pbund lb or jl,
" gr sc •
20 = 1 ir
60 ass 3 = 1 «5
480 = 24 = 8 = 1 lb
. 5760 = 288 = 96 = 12. =3 1
This is the same as Troy weight, only having some dif
ferent divisions. Apothecaries make use of this weight in
compounding th^ir M^icines i but they buy and sell their
Drugs by Avoirdupois weight.
AvoiR
widiout wasting, though it be kept contigually melted. But silver,
not having the purity of gold^ will not endure the fire like it : yet
£ne silver will waste but a very little by being in the fire any
moderate timej 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 very 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 wnen they
are allayed, or mixed and hardened with copper or brass. And
though most nations dlifer, more or less, in the quantity of such
allay, as well asHn the samis place at different times, yet in £ng
land the standard for gold and silver coin has been for a long time
as follows — viz. That 2*JL parts, of fine gold, and 2 parts of copper,
being melted together, shall be esteemed the true standard for gold
coin : And that 1 1 ounces and 2 pennyweights of fine silver, and
. 1 8 pennyweights of cop)er^ being melted together, is esteemed
the true standard for silver coin, called Sterling silver.
♦ The original of all weights used in England, was a grain or
com of wheat, gathered o^t of the middte of the ear, and, being
W^Jl dri^d^ 32 of them were to m^e one pennyweight, 20 penny
weights
/
t6 ARTHM£71C.
Atoiildupois Weight.
DraiQS
•

naarked (ir
16 Drains
make
1 Ounce
f
•  •
OZ
16 Ounces ^
• •
1 Pound
• • •
a
t9 Pounds 
. •
1 Quarter
1 Hundred
•  .
9^
4 Quarters 
2(X Hundred 1
. .
1 Weight ^
nvt
RTeight
iTon
p»
 
Jr '
'OZ
^
16 =
1
a
256 =
16 =
= 1
i^
1
7l6« =
448 =
= 28 =s
1
cwt
28^72 =
1792 =
s H2 =
4
s 1
tm
573440 3e
35840 =
= 2240 8=
80
= 20 s=
1
By this weight are weighed all things of a coarse or drossy
nature^ as Com, Bread, Butter, Cheese, Flesh, Grocery
Wares, and somt liquids ( also all Metak, except Silver and
Cold..
OZ dtift gr
N§if^ that la AToirdupcns at 14 U 154 Troy,
las   =s 18 54
Ur   «= 1 34
LoN^ Measure.
*
•
3 Barley*coms
make 1 Inch
*>
Li
12 Inches
«
1 Foot
»
' Ft
3 Feet
•
1 Yard
•
rd
6 Feet
■■
1 Fathom /
•■•
Fth
S Yards and a
half
1 Pole or Rod
«»
PI
4b Poles
*
1 Furlong 
J
Fur
S Furlongs 
«•
1 Mile
•
Mile
3 Miles
m
1 League 
•
Lea
694 ^1^ nearly 
1 Degree 
.
Dfg at ".
•
wdghts one ounqe> and 1^ ounces one pound. But in later times,
it was thought sufficient to divide Hhe same pennyweight into
24 equal parts, still called grains, being the least weight now in
common use ; and from tb^ce the vest are computed, as in the
Tables above.
In
TABUS «fM&A9C7K£8. ft
In
2%
12
SSI
1
rd

36'
sa
3
s=
1
fl
■
198
=r
1<54
' zs
5i
=:
1
Fur
^7920
=:
660
=:
220
ss
40
= I
13960
ss
6230
^^
1760
ss
320
s: 8
ClOTH MB4SyKB.
Mik
2 Inches and a ijuart^r make 1 Nail < * Nl
*4 Nails r  sr 1. Quarter of ^ Tard Or
3 'Quarters .r  » 1 EU Flemish  £F
4 Quarters  ^ ^ 1 Yard  * Td ,
5 Qu;qpters r «. * 1 Ell English  EE
4 Quarters 14 Inch  1 £11 Scotch ^ ^ £S
»
144 Squarefochetmake 1 Sq Foot  R
9 Square Feet r 1 SqTard  rd
90i Square Tafds  1 Sq Pol^  Pok
40 Sq^are Poles  1 Rood * Rd
4 Roods pi Acw f jfcr '
SqLic ^Ft
; 144 « i SqTd
1296 = 9=1 SqPi
39204 .s 272^ s 30f r? 1 Jtd
1568160 s: 1O890 b 1210 s 4Q « i >/rr
6272640 = 4^^6Q b 4840 == 160 » 4 ^ i ,
By diis measure^ Land, and Huslaadme& and Oardenen?
irork are measured; also Artificers' work, ancl^ ^s Board,
Glass, Pavements, PlaMering, Wainscoting, Tilj9g, Floorilig,
and every dimension pf laigth smd breadth only;
. When three dimensions arjs concemed namely, length,
breadthy aapd depth or thkkness, it is called cubic or solid
measure, which is used to measure fllmbery Stone» &c.
The cubic or s^lid Foot, which is 12 inches ia length and
)>readth and thickness> contains 1728 cubic or sblici inches,
u^d {27 solid feet make one solid yard.
29
ARnHMETia
DrY) or Cork Measchb.
1 Quart
< 1 Gallon
 1 Peck
f Pifits make
2 Quart*  1 Potde
2 Pottles
2 Gallons
4 Pecks
5 Bashels
S Quarters
2 Weys
■■ « M
Bushel
Quarter
1
1
1 Wey» Load> or Ton
1 Last 
Gai
Pec
Bu
Wej
JLaU
Pts
' %
16
64
512
2560
5120
Gal
1
2 =x
8 =
64 =
S20 =
640 =:
PiC
1
4
32
160
S20
1
40
80
Or
J
5
10
: 1 ZdRlf
: 2 =: I
By this are measured all dry wares> aS) Corn> Seeds, Rooft%
Fruits, Salt, Coals, Sand, Oysters, &c.
The standi Gsdlon drymeasur^ cpntauis268 cubic ot
solid inches, and the Cdm or Winchester husl^el 21 50 cubic
inches j for the dimensions of the Winchester bushel, by the
Statute, are 8 inches deep, and 1 8^ iiv;hei wide or in diameter.
But the Coal bushel must be 19^ inches in dlanneter; and 36
bushels, heaped up, make a London chaldron of coals> the
weight of which is ^ 1 5 6 lb Avoirdupois*
AhZ and Beeil Measure ^
2 Pints make
4 Quarts  ••
36 Gallons ^. w
1 Barrel aud a half
f Barrels m
2 Hogsheads ^
2 Butts; <* r
Pts Of
2 =; I Gat
8 =; 4 =?, 1
28^ = 144 =: 36
482 23 2li5 = 54
864 2= 432 = 108
1 Quart
1 Gallon
1 Barrel
1 Hogshead
\ Puncheon
1 Butt
\ Tw 
Gal
Bar
Hhd
^ Pim
fiuu
Tut\
Bar
: 1
mi
1 Sm
2 i= 1
NitCji The Ale Gallon contains ^82 cubic or soSd Lichee.
Win*
TABLES ot MEASURES amo TIME.
2»
Wine Measure.
••
2 Pints make
1 Quart
Qt
4 Quarts  *•
1 Gallon
Gal
42 Gallons  . >
1 Tierce
Tter
63 Galkms or it Tierces
1 Hogshead 
Hhd
2 Tierces 
1 Puncheon 
Pun.
2 Hotheads
1 Pipe or Butt
Pi
2 Pipes or 4Hhds
I Tun 
Tun
Pti Qt
V
2 = i Go/
« = 4= 1 :
tter
9J6 = 168 r: 42 =
I Hid
504 = 252;= 63 =
lt= I P«/»
672 = 336 = ^4 =
2 = lt= 1 Pi
1
1008 :;= 504 = 126 =
3=2= lt= I
Tim
2016 =: 1008 = 252 =
6=4=3 =a
= 1
/
.Notit By this are measured all Wines, Spirits, Strong
Waters, Cyder, Mead, Perry, Vinegar, Oil, Honey, &c.
The Wine Gallon contains 231 cubic or solid inches.
And it is remarkable, that the Wine and Ale Gallons have the
same propprtipn to each other, as the Troy and Avoirdupois
Pounds hare^ that is, as one Pound Troy is to one P4!>und
Avoirdupois^ so is one Wine Gallon to one Ale Gallon*
Of TIMiE.
60 Seconds or 60'' make
•
1 Minute 
Mot'
^ Minutes

1 Hour
Hr
24 Hours 

i Day
Da*
7 Days *  

1 Week .
ni
4 Weeks   
•
1 Month 
M,
13 Months I Day 6 Hours,)
or 365 Days 6 Hours )
1 Julian: Year
Tr
4
■ Sec Min
,
60 =: 1
Hr
,
. 3600 = 60 =
I
■ ^"^ „.
•
S6400 = 1440 =
:■ 24.
=s 4 IVk
*•
604800 =: . lOOSO =s.
168
= 7 = 1
M» ■ .
1 2419200 =: 4.0320 =
672
= 28 = 4 .:
= 1 !
31557600 == 5250QQ ^
8766
= 365t;=i
llrfar.
V
Or
(
y
m ARITHBIEnC*
Wk Da Hr Mo Da Hr
Or 52 I 6 = 13 1 6 = 1 Julianrear
Da Hr M Sec
But 365 6 48 48 =r 1 Sflcir rear!
ItULES FOR REDUCTION.
^ trim tie Numben are to he reduced from a Higher Denond^ '
nation to a Lower :
Multiply the number m the highest denomination by as
many as of the next lower make an integer, or I, in that
higher ; to this jH'oduct add the number, if any, which was
in this lower denominatioir before, a^ set down the aihonnt.
Reduce this amount in like manner, by multiplying^ it by
as many as of the next lower make an integer of this, taking
in the odd parts of this lower, as before. And so pnroceed
through all the denominations to the lowest ; so shall die
number last found be the value of all the numbers which,
were in the higher denominations, taken together*.
EXAMPLE.
1. In 1234/ Idx 7i^ how many farthings I
lid
1234 15 7
20
24695 Shillings
12
296347 Pence
4
^ Answer 1185388 Farthings.
* The reason of this ru)e is very evident; for potmda are
brought i^to shillings by maltiplying them by 20 ; ^hilltngs into
pence« by multipljrtig them bj 12; and pence into ^rthings^ by
multiplying by 4 5 and tKe reverse of this nile by Division. — And
the same^ it is evident^ will be true in the reduction of numbers
consisting of any denominations whatever.
ll.When
RULES foil REDtJCnCMJ. St
IL When tie Number $ are to be reduced from a Lower .Deftomt"
nation to a Higher :
m
Divide the given number by as many as of that denomt'*
nation make 1 of the next higher, and sec down what
nemainsy as well as the quotient.
Divide the quotient hj as many as 6f this denominatbn
make 1 of the nett. higher } setting down the new quotient,
and remainder, as before. /
Proceed in the same manner through all tike denomina*
tions, to the highest ; and the quotient last fimnd, ti^ether
with the several remainders, if any, will be of the same vahif
s the first number proposed.
V
EXAMPLES.
2. Reduce 1185388 farthings into pounds, shillings, and
pence.
4) 11853*8
1*1 I I
12) 296347^
■ t ■' I !■ *
2,0 ) 9469,5 /^7rf
1S^4/ 14^ *!d
< ■ »»■»»
3. Reduce 24/ to farthinjgs. Ans. 23(Hai
4> Reduce 3375^7 farthings to pounds, &c.
Ans. S6U I3i 6^.
5. Kow many futhings^nre in 3€ guineas? Ans. 36S89«
€. In 36288 farthings how many guineas I Ans. 36.
7. In 59 lb 13 dwts 5^ how many gnttis ? Ans. 340157*
8. In 8012131 grains how many pounds, &c.?
Ans. I390ib 11 oz 18dwt t9gr
9. In 35'ton I7cwt I qr 23 lb 7oz i3dr how many drams?
Ans. 20571005.
10. How many barleycornSx will rqadb round the earth,
supposing it^ according to the best calculations, to be 25000
miles ? * Ans. 4752000000.
11. How inany seconds are in a solar year, or 365 days
5 hrs 49 min 48^ sec ? Ans. 3 1 556928.
12. In a lunar month, or 29 ds 12 hrs 44 min 3 sec, how
many second 2 Ans. 2551443.
COM.
St
ARTTHMETia
COMPOUND ADDITION,
CoMPOiTND ApDiTiON shows how to add or collect seTeral
numbers of different denominations into one sum.
RijL£.*Place the numbers so» that those of the same de^*
nomination may stand directly under eauch other, and draw z
line below them» Add up the figures in the lowest denomi
nadon, and find, by Reduction^ how many units, or ones, of
the next higher denomination are contained in their $um.~»
Set down the remainder below its proper column, and cany
those units or ones to the next denomination, which add up
in the same manner as be£bre«r*Proceed thus through all the
denominations, to the highest, whose sum, together with tht
several remainders, will give the answer sought.
The method of proof is the same as in Simple Addition.
EXAMPLES or MONET»
I.
2.
s.
4.
/ J
d
/ X
d
/ / d
/
/
rf
1 13
3
14 1
5
15 11 10
J3
14
8
a 5
lot
8 19
2t
3 14 6
5
10
H
6 18
7
7 8
It
23 6 2
93
11
6
2
H
21 2
9
14 d 4t
7
5
4^
3
7 16
8f
15 6 4
IS
2
S
n 15
4i
4
3
V
6 1^ 9^
18
7
%9 15
H
'
•
•
32 2
. •
39 15
H
/
. 5.
6.
. 7.
8
/ /
d .
/ s
d
/ s i
/
J
d
H
n
37 15
s
613 2t
472
15
3
8 15
3
14 12
H
7 16 8
9
2
2t
62 4
7 ■
17 14
9
29 13 10 J
27
12
6t
4 IT
&
23 10
9t
12 16 2
370
16
2t
23
H
8 6
1 b^
13
7
4
6 6
7
14
H
24 13
6
10
■5t
SI
lOj
54 £
ii
5 10.
30
111
»
1
\
•
EXAH<^
/
COMPOUND AjDOmON.
' aa
Exam. 9. A nobleman, going cmt of tovn, is informed by
his steward, that his butcher's bill comes to 197/ ISj '\\d: his
baker's to 59/ .'J/ 2^; his brewer's to 85/; his winemer
chant's to 103/ 13x; to his cornchandler is due 75/ %d; \o
his tallowchandler and cheesemonger, 27 i IBs 11</; and
to his tailor 55i Ss B^d; also for rent, servants' wages,, aqd
other charges, 127/ 3/; 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! 830/14/ 6^.
10. The strength of a regiment of foot, of 10 compani^,
and the amount of their subsistence*, for a month bf 30 days,
:flu:cording to l)he anaeKed Table, are r^uired i
Numb.
Rank.
Sul^istence for
 a Montih.
* / #
d
1
Colonel
27
1
1
7
11
Lieutenant Colond
IVfajor
Captains
Lieutenants
19 10
17 5
78 15
57 13
*
9
1
\
1
JEnsigns
Chaplain
Adjutant
QuarterMaster
40 10
7 10
4 10
, 5 5
1
1
30
30
2P
2
Surgeon
Surgeon's ]\{ate
Seijeants
Corporals
Pruipmers
Kfes
4 10
4 10
45
30
20
2
390
Private Men
292 10
5Qir
Total
656 iO
■*■**•
* Subsistence Money, is the money paid to the soldiers weekly
which is short of their full pay, because their clothes^ accoutre*
ments, ^c. are to be accounted for. It is likewise, the money ad
vanced to officers till their account are made op*J^<^^ is com
*monly once a year, when they are paid their arrearir^lhe follo)w
ing Table shows the full pay and subsistence of each rank on the
^English establishment .
VoLiL
D
DAILT
34
ARITHMETIC.
•
•
o o
CO ^ O o c
> 00
'•
n
»* p^
"2
£
O) QD
* (o t^ lo i*
»o »o "O
k4
C3
g
•H •
H
^^ »«^
O
It »
^ o o o o
t
oooo
o ^
O CO o o o
<oo
•£
'3S
9
o •
CO ^ ^ ^ M
, 121.^
1 c/>
»H »«
o o o o o
.000
b:.
O GO
CO 00
Cfl
rt
E
o.
1 1
1 12 2 1^
00 1 c» »o
O O O
» 00
•
o ^
CO co<o o ^
1 00
1
en
'J
9
^4
•H ^2 :2 1 ^
' ^ 1 0)«0
•
<^ 
H ^
^ O O c
> .0 00
ft
b3
•
o o
oooo O
y
•
00
£.
CO H
O "t O H 1 H
11*1
Lb
•TS
—1 rl
•I ^i* '1 ^
U.
S
g
c
3
o
(s«
^ H
p ph o o c
q
•
O CO
CO O ' <N ^
c
^
a
4><
in
1
t^ CO
O QD C^ X 11^
1 l'°l
CO
y5
w i
»H f—
oooo o
c
•H lO (X) 00 00 p
) 00 10
1 »o Ci 1 »o
u
 o
O O O O CO
1
00
C
'
>
it
o rz
ooooo oooo
<
^ •* O t^ "o 1 o t^'o *^ 1 1 1
12 l^
>
O w
c<i«^oo oooo
'o
<
u .*
o o
CO iSiO O O C
1 * 9
e
1— 1
2 I« 1
4 ^^ 1
Q^
"^ fH
O OOOO o
1 00 00,
o o
O OOOO
00.
2^ ?: ^
12^ 1
ri »f5 O C^C» I •
1 CS 1 CO GO
""^^^S
mn •^
^ OOOO
00

• "'N . .
'
>•••,•••
§ : ;
k 4
> • • • t^D • * •
> . , , u • • •
<
1
> <
» 1
> 4
1
* '<
, . / 3 . . .
. . S3 gco . . .
fc S S • h^^ •
•
3 c
13 "oJ <i> r
> O o ^ "r pj O • • ^
^ »L C t <^ CO fe
5 o«Jc
U U — cs
1 « T3 ,^ ,«^ S ^o .9 a 3
► 2 (3 eg pq vj
I °
o
•>
01
e
o
'5
a
s
o
;>
o
"2
o
so
c
o
a
■i
'^ a
^ S
•»••■
Cfa o
S w
Of "
1*^
3f^
.J
.tXAMfLSS
COMPOUND ADPITION. S3
EXAMPLES OF WEIGHTS, MEASURES, t(e.
'
TROY
WEIGfiT.
APOTHECARIES* WEIGHT.
f
K
2.
3.
4.
lb
oz dwt
oz dwt gr
lb
OZ
dr sc
oz dr sc gr
17
3 15
37 9 3
3
5
7 2
3 5 1 17
, 7
9 4
9 5 3
13
,?
3
7 3 2 5
O 10 7
8 12 12
19
6 2
16 7 12
^ 9
5
17 7 8
9
1 2
7 3 2 9
i76
2 17
5 9
36
3
5
4 1 2 18
i23 11 12
3 19
5
8
6 1
36 4 1 14
AVOI«l>UPOIS WEIGHT.
T.ONO MEASURE.
5. ,
6.
7.
. 8.
lb
oz dr '
♦ cwt qr
lb
1
mis fiir pis
yds feetinc
• 17
10 13
15 2
15
29
3 14
127 1 5 •
5
14 8
6 3
24
19
6 ^9
12 2 9
12
9 18
9 1
14
7
24
10 10
27
1 6
9 1
17
9
1^37
54 1 11
4
10 2
6
7
3
5 2 7
6
14 10
3
3
4
5 9
23 5.

CLOTH
MEASURE.
LAND 1
MEASURE.
9.
10.
11..
12.
yds
qr nls
el en qrs
nls
ac
ro p
ac ro p
26
3 1
270 1
225
3 37
19 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
3
3
2
4
2 9
53 34
9
1
10 1
42
1 19
7 2 16
55
3 1
4 4
1
7
6
75 23
WINK
E^EASURB.
—

ALE and BEER MEASURE.
13.
14
•
15.
16.
t
hdsgal
hds gal pts
hds
gal pts
hds gal pts
13
3 15
15 61
5
17
37 3
29 43 5
8
1 37
* 17 14
13
9
10 \5
12 19 7
14
1 20
29 23
7
3
6 2
14 16 6
25
12
3 15^
1
5
14
6 8 1
. 3
1 9
16 8
12
9 6.
. .57 13 4
72
3 21
4 3.6 '
6
8 42 4
5 6
D2 COM
■"S6 ARfTHMETlC.
COMPOUND SUBTRACTION.
«
Compound Subtraction shows how to find the difl&r
0nce between any two numbers of different deaomin^tio^Sr
To perform which^ ob^'erve the following Rule :
* Place the less number below the greater, so that the
parts of the same denomination may stand directly tinder
each other ; and draw a Kne below them. — Begin at the
righthand, and subtract each number ,or^part m the lower
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 as many to the upper number
is make 1 of the next higher denominatibn v 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 ity as befofe; and so proceed till* the
whole is finished. ' Then the several remainders, taken to^
jcther, will be the whole dffference sought.
The method of proof is the same as in Simple Subtraction.
iEXAMPLES OF MONET.
I.  2. 3. . 4. ^
I s i I s d I s d t s d
From 79 17 8. 103 3 2J SI 10 J I 254, 12
Take 35 12 4f 71 12 5^ 29 13 S^ 37 9 4^
Rem. 44 5 4Jr 31 10 8^
■ ' ' ■ *
Proof 79 17 8 103 ^ 2^
5. Wh^t ia the difFerence between 73/ 5^/ and 19/13/ lOrff
Ans. 53/ 6* 7^.
* The redson of this Rule wiM easily appear from what* has hcenc
said in Simple Subtraction; for the borrowing depends on the
same principle, and is only different as Hie;numberstoi>e^b<'
Uacted: are of difiete&t deApmination^.
COMPOUND SUBTRACTION.
93
Ex 6. A lends to B 100/, how much, is B. in debt afier^
has taken goods of him to the amount of 73/ 12/ 4Jr//*
Ans. 26/ Is 7i(L
, 7, Suppose' that my rent for half a year is 20/ I2/9.an<l
ihat I haye laid out for the landtax 14« 6di and for. severs^
repJlirs 1/3* 3 J^, what have I tppay of my halfyear's rent ?
Ans. 'l8/ 14j 2id.
8* A trader^ failing, owes to A 35/ Is 6d, to B 91/ 13/ id^
to C 53/ 7id, to D 87/ 5s, and to E 11 1/ 3/ 5^ When
this happened, he had by him* in c^sh 23/ 7/ Sd^ in wares
53/ 1 1/ lOj^, in household furniture 63/ 171 7^, and in re
coverable book^debts 25/ 7/ 5d. What will his creditors lose
by him, suppose these things delivered to them ?
. Ans. 2121. 5s Sid.
EICAMPLES OF WEIGHTS, MEASURES, iS^C.
TROV WfilGUT.
1.
lb ozdwtgr
Prom 9 2 12 10
Take 5 4 6 17
2.
lb oz dwt gr
7 10 4 r7
3 7 16 12
Al'OTHECARIES WfilOHT
3.
lb oz dr scr gr
73 4 7 14
29 5 3 4 19
Rem.
Proof
AVOIRDUPOIS WEIOHT.
LONG MEASURE.
4.
5. .
;6
7.
C qrs lb
lb oz dr
m fu pi
yd ft in
From 5 0.l7
71 5 9
14 3 17
96 4
Take 2 3 10
17 9 18
7 6 U
72 2,9
Rem.
Prbof
^LOTH MEASURE.
8. . 9.
yd qr nl yd qr nl
From 17 2 1 9 2
T^e 9 2 7 2 1
LAND MEA817R£*
10. 11.
ac ro p
17 1 14
16 2 8
ax: ro p
57 1 16
22 3 2d
Rem.
Proof
WINE
38
ARITHMETIC.
WINE
MEASURE.
ALE and BE
12.
t hdgal
From 17 2 23
Take 9 1 36
13.
hdgal pt
5 4
2 12^ 6
14.
hd gal pt
14 29 3
9 35 7
Rem.
Proof
IS.
hd gal
71 16
19 7
pt
5
1

^
DRY MEASURE.
From
Take
la
9
6
16. 17.
qr bu bu gal pt
4 7 13 7 1
3 5 9 2 7
TIME.
Hem.
Proof
r*~
LP ™ ' *
18. 19.
mo we da ds hrsmin
71 2 5 114 17 26
17 1 6 72 10 37
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 f^ce of the bastion? Ans. 89 yds 1 ft S iq.
"TT"
COMPOUND MU^LTIPLICATION,
Compound Multiplication shows how to find the
amount of any given number of different denominations re^
peated a certain proposed number of times \ which is per**
formed by the following rule.
Set the multiplier under the lowest denomination of th^
multiplicand, and drjiw a line below it. — Multiply the num
ber 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 n^xt higher denomination are in thi^
, , f^mowit,
COMPOUND MULTIPLICATION. 39
amount, which carry in Hke manner to the next product,
settiiig down the overplus.*^Proceed thus to the highest de
nomination proposed : so shall the kst product, with the se«
Verat remainders, taken as one compound number, be the
whole amount required. — ^The method of Proof, and the
reason of the Rule, are the same as in Simple Multiplication.
EXAMPLES OF MONEY.
i. To find the amount of 8 lb of Tea, at 5s 8 Id per lb.
i d
• 8
jf 2 5 8 Answer.
/ / d
2. 4 lb of Tea, at Is Sd per lb. Ans. 110 8
3. 6 lb of Butter, at 9^4 per lb. Ans. 4 9
4. YlbofTobacco, at li 84rfperlb, Ans. 11 11^
5. 9 stone of Beef, at 2^7 trf per St. Ans. 1 10
6. 10 cwt of Cheese, at 2/ i7« KWper cwt. Ans. 28 1 8 4
7. 12 cwt of Sugar, at 3/7/ 4rf per cwt. Ans. 40 8
CONTRACTIONS.
I. If the multiplier exceed 12, multiply successively by its
component parts, instead of the whole number at once.
EXAMPLES.
1 . 15 cwt of Cheese, at 17^ 6d per cwt»
/ / V
17 6
S
2 12 6
5
13 2 6 Answer.
/ X d
2. 20 cwt of Hops, at 4/ Is 2d per cwt. Ans. ^7 3 4
3. 24 tons of Hay, at 3/ 7/ 6d per ton. Ans. 810 O
4. 45 ells of cloth, at is 6d per ell. . Ans. 3 7 6
Ex. 5.
4a ARITHMETRi
t
Ex. 5. 63 gallon J of Oil, at 2/ %d per galh Aas* 7
$. 70 iKirrels of Ale, at l/^j per t^urel. Ans. 84
*7. 84 quarters of Oats, at 1/ 12/ 8//per qr. Ans* 187
8. 96 quarters of Barley,at l/3x4rf)per qr. Am. 112
9. 120 days' Wages, at bs 9^ per day. . Ans; 34
10. 144 reamsof Paper, at 13/ 4^^ per ream. Ans. 96
II. If the multiplier cannot be exactly produced by the
multiplication of simj^^ numbers, take the nearest number to
it, either greater or less, which can b^ so produced,' aAmuI^
tipiy by its parts, as before. — ^Then multiply the giveii mul
tiplicand by the difference between this assumed number and
the multiplier, and add the product tq that before found,
>rhen the assumed number is less than the multiplier^ bul^
subtract the same wh^n it \& greater.
/
4
1
9
41
10
G
EXAMPLES.
f.'
■?^
y?irds of Cloth, at
/
•
3/ Ot^ per yard.
/ d
3
f
15
H
,
5
f .
3
16
H
#
§
3
oi .
19
li ■Answer.
/ s d
2. 29 quarters of Corfi, sit 2/ 5/ S^d pet qr. Arts. 65 12 lO^
si 53 loads of Hay, at 3/ 15/ 2^ per ibad. Ans. 199 3 10
. 4. 79bushelsofWh^at,atll/5.rf*^er bush. Ans. 45 6 lOf
5. 97 casks of Bepr, at 12/ 2^?peif cask. Ans. 59 O 2
6. 114 stone of Meat, at 1 5/ 3tf per stone. Ans. 87 5 74.
EXAMPLES OF WEIGHTS AND MEASURES.
1.
•
lb oz dwt gr
28 7 14 10
4
2.
lb oz dr sc
2 6 3 2
•
8
cwt
29
3.
qr lb oz
2 16 14
12
<^
•
■ """V ■
>
•♦►•'■
COMPOUND DIVISION. 41
4.
5,
6.
ttds fu pis
yds
yds
qfs
na
ro po
22 5 . 29
6
4
126
3
1
7
28
3 27
9
T»"
V.
tuns hhd gal pts
^0 2 26 2
3
8,
9.
we qr bu pe
mo we da ho min
24 2 5 3
172 3 5 16 49
6
10
«•■
"W"
*?iP
COMPOUND DIVISION,
CoMi^oUND Division teaches how to divide a number of
sj^veral denominations by any given number^ or into any
number of equal pjarts ; as follows :
Fla^e the divisor on the left of the dividend^ as in Simple
Division. — ^Begin at the lefthand^ and divide the number of
the highest denomination by the divisor^ setting dgwn th^
quotient in its proper place.«If there be any remainder after
this division, reduce it to the next lower denominattony
which add to the number, if any, belonging to that denomi
nation, and divide the sum by the divisor. — Set down a^a
tlfis 4dot}ent reduce its remainder to the next lower deho*
mination again^ and so on through all th^ denominations tq
Isfae last.
r
EXAMPLES OF MONBY.
I. Divide 237/ 8/ 6d by 2.
lid
2 ) 231 8 6
;^118 14 3 the Quotient.
2. Divids .
4a ARTTHMEnC
I s d t s d
^i Divide 4S2 12 \\ by 3. Ans. 144. 4 a.
3. Divide 507 3 5 by 4. Ans. 126 15 lOf
4. Divide 632 7 61 by 5. Ans. 126 9 6
5. Divide 690 14 S^J: by 6. Ans. 115 2 44
6. Divide' 705 10 2 by 7. Ans. 100 15 8^
7. Divide 760 5 6 by 8. Ans. 95 O 8:^
8. Divide 761 5 7 by 9. Ans. 84 11 %\
9. Divide 829 17 10 by 10. Ans. $2 19 ^\:
10. Divide 937 '8* 8byll. Ans. 85 4 5
11. Divide 1145 11 4^ by 12. Ans. ^^ 9 3^.
CONTRACTIONS.
\p If the divisor exceed 12, find what simple numbers,
multiplied together, will produce ity^and divide by them se>
parately, as in Simple Division, as below.
' . £XAMPJ^£5.
I. What is Cheese per cwt, if 16 cwt cost 25/ 14j ^i?
Isd
4) 25 14 8
4) 6 8 8
j^ 1 12 2 the Answer.
/
2. If 20 cwt of Tobacco come to
: 20 cWt ot lobacco come to 7 An « a in a
150/ 6/ 8i, what is that per cwt ?i ^^* ^ *
3. Divide 98/ 8j by 36. Ans. 2 14 8
4. Divide 71/ 13/ \Odhj 56. Ans. 1 5 75
5. Divide 44/ 4 j by 96. Ans. 9 2t
6. At 3J / lOx per cwt, how much per lb } Ans. 5 7^
IL If the divisor cannot be produce4 by the multiplication
of small numbeis, divide by the whole divjisor at once, after
the manner^of Long Division, as follows.
EXAU
COMPOUND DIVISION. 43
EXAMPLES.
1. Divide 59/6/ Sid by 19.
I s d I s d
19 ) 59 6 3 (3 2 51 An^.
57
2
20
46 (2
38
8
12
99 (5
95
•
4
4
.
19 (1
■
t
2,
3.
4.
5.
Divide 39 14 5 by 57.
Divide 125 4 9 by 43
Divide 542 7 10 by 97.
Divide 123 11 2t by J 27.
/ s
Ans. 13
Ans* 2 18
Ans. 5 11
Ans. 19
Hi
3
10
EXAMPLES OF WEIGHTS AND MEASURES.
1. Divide 17 lb 9 oz dwts 2 gr by 7.
Ans. 2 lb 6 oz 8 dwts 14gn
2. Divide 17 lb 5 oz 2 dr 1 scr 4 gr by 1 2.
Ans. lib 5 oz 3 dr 1 scr 12 gr.
3. Divide 178 cwt 3 qrs 14 lb by 53. Ans. 3cwt Iqr 14lb.
4. Dividie 144 mi 4 fur 2 po 1 yd 2 ft in by 39.
Ans. 3 mi 5 fur 26 po yds 2 ft 8 in.
5. Divide 534 yds 2 qrs 2 na by 47. Ans. 1 1 yds 1 qr 2iia.
6. Divide 7 1 ac 1 ro 33 po by 5 1 . Ans. 1 ac 2 ro 3 po.
1. Divide 7 tu hhds 47 gal 7 pi by 65. Ans. 27 gal. 7 pi.
8. Divide 387 la 9 qr by 72. Ans. 3 la 3 qrs 7 bu.
9. Dividij 206 ino 4 da by 26. Ans. 7 mo 3 vire 5 ds.
44 ARITHMETIC.
The golden RULE,' or RULE OF THREE. 
The Rule of Three teaches how to find a fourth pro
portional 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 use
fuhiess, it is often called the Golden Rule. This Rule is
usually considered as of two kinds, namely. Direct, and
Inverse.
The Rule of Three 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 tihie. Or when it is thus: if 6 men .dig 42 yards, how
much will 3 men dig in the same time ? Here then, less re
quires less, or 3 men will perform proportionably less work
than 6 men, in the same time. In both these cases then, the
Rule, OK the Proportion, is Direct j and the stating must be
thus. As 3 : 21 :: 6 : 42,
or thus,' As 6 : 42 1 : 3 : 21,
But the Rule of Three Inverse, is when nlore requires less,
or less reouires more. As in this i if S m^o dig a celtain
quantity or trench in 14 hours, in how many hour^ Will 6
men dig the Kke quantity ? Here it ii 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 nien perform the same ? Here less requires more, for 3
men will take more hours than 6 to perform the same work.
In both those cases then the Rule, or the Proportion, is
Inv^i^ ; and the stating must be
thus. As 6 ? 14 :: 3 : 7^
or thus. As 3 : 7 :: 6 : 14.
And in all these statings, the £:nirth term is found, hj
inukiplying the 2d and 3d terms together, and dividing the
product by the 1 st 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 th^ following ge
lieral Rule :.•
Stat^
RULE OF THREE.
*»
State the qt^stion, by setting down.in astraigbtline the
three given numbers, in the following mannery'viz. ao that
the 2d term be that ntunbea* of supposition which is of the
same kind that the answer or 4th term is to be ; 'making the
other number of supposition the 1st term, and the demanding
number the dd term, when the question is in direct propor
tion ; but contrariwise, the other number of supposition the
Sd term, and the demanding number the 1st term, when th^
question has inverse proportion.
Then, in both cases, multiply the 2d and Sd terms to*
gether, and divide the product by the Ist, which will give
the answer^ or 4th term sought, viz. of the same denomma
tion as the second term.
Nate^ If the first and third t^ms consist of diffinrent deno^
minations, reduce them both to the same : and if the second
term be a compound number, it is mostly convenient to re
duce it to the lowest denomination mentioned.^^If, after di«
vision, there be any remainder,, reduce, it to the next lower
denomination, and divide by the same divisor as before, and
the quotieilt will be of this last denominatioh. Proceed in
the same manner with all the remainders, till they be re*
duced to the lowest denomination which the second admits
of, and the several quotients taken together will be the an
swer required.
N^te also. The reason for thtf foregoing Rules will appear,
when we come to treat of the nature of Proportions.— Some
times two or more statings are necessary, which may always
be known from the nature of the question.
* EXAMPLES.
i. If 8 yards of Cloth cost 1/ 4^, what will 96 yards cost i
yds 1 s yds 1 s
As 8 : 1 4 *: : 9€ : 14 S the Answer.
20
24
96
144
216
8) 2304
2^0) 28,8*
jf 14 8 Answer*
£x. 2s
4^ ARITHMETrC%
Ex. 2. An engineer having rabed 100 yards of a certain
work in 24 days with 5 men ; how many men must he enw
jploy to finish a like quantity of work in 15 days ?
ds men ds men
As 1 5 : 5 ': : 24 : 8 Ans.
15) ^20 ( 8 Answer.
120
3» What will 72 yards of cloth cost> at the rate of 9 yards
for 51 12j ? Ans. 44/ 16/.
4. A person^s annual income being 146/; how much is
that per day ? Ans. 8/.
5. If 3 paces or common steps of a certain person be equal
to 2 yardS) how many yards wiU 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 V50 men require 22500 rations. of bread for a month;
how' many rations will a garrison of 1 200 men require ?
• Ans. 36000.
8. If 7 cwt 1 qr of sugar cost 26/ 10/ 4^; what will be the
price of 43 cwt 2 qrs ? Ans. 159/ 2s.
9. The clothing of a regiment of foot of 750 men amount
ing to 2831/ 5s; what will the clothing of a body of 3500
men amount to? Ans. 13212/ 10/;
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.
1 1 . What is the value of 6 bushels of coals, at the rate of
\l I4fs 6d the chaldron ? Ans. 5s 9d.
12. If 6352 stones of 3 feet long complete a certain quan
tity of walling ; how many stones of 2 reet long will raise a
like quantity ? Ans. 9528.
13. What must be given for a piece of silver weighing
73 lb 6 oz 15 dwts, at the rate of 55 9d per ounce ?
Ans. 253/10/0rf.
14. A garrison of 536 nien having provision for 12 months;
how long will those provisions last, if the garrison be increased
to 1 124 men ? Ans. 174 days and tttt'
15. What will be the tax upon 763/ 15x, at the rate of
35 6d per pound sterling ? Ans^ 1 33/ 1 3/ If ^.
16. A
RULE OF THREE. 47
16* A certain work being raised in 12 days, by working 4
liours each day ; how long would it hjive been In ra^ising bf
%Krorking 6 hours per day ? . Ans. 8 days,
IT. What quantity of com can I buy for 90 guineas, at the^
raje of 6s the bushel ? Ans. 39 qi^s 3 bu.
18. A person, failing in trade, owes in all 977/; at which
time he has, in money, gobds, and recoverable debts, 420/ 6s
S^d\ now supposing these things delivered to his creditors,
lidw much will they get per pound ? Ans. 8x 7 j€f.>
« 1 9. A plain of a certain extent having supplied a body of
SOOO horse with forage for 1 8 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 25/ 6d per day, one day with another ;
' ho'w much will he have saved at the year's end ?
Ans. 16^/ I2s 6d.
2i . What cost 30 pieces of lead, each weighing 1 cwt 121b,
at the rate of I6j W the cwt ? Ans. 27/ 2s 6d.
22. The governor of a besieged place having provision for
54 days, at the rate of 1^ lb ofbread ; but being d^irous to
prolong the siege to 80 days, in expectation of succour, in
that case what must the ration ofbread be ? Ans. l^lb.
23. At half a guinea per week, how long can I be boarded
for 20 pounds ? Ans. 38^?^ wks*
24. How much vrill 75 chaldrons 7 bushels of coals come
to, at the 'rate of 1/ 13i 6^ per chaldron ?
Ans, 125/ 19x0J.
25. If the pormy loaf weigh 8 ounces when the bushel of
wheat costs 7s 3rf, what ought the penny loaf to weighilhen
the wheat is at 8/ ^dP Ans. 6 oz 15t^ dr.
26» How much a year will 173 acres 2 roods 14 poles of
land give, at the rate oi Ills 8d per acre?
 > Ans. 240/ 2s 1^^.
• 27. To how much amounts 73 pieces of lead, each weigh
ing 1 cwt 3 qrs 7 lb, at 10/ 4j per fother of 1 9^ cwt?
Ans. 69/ 4j 2i/ 111 q.
/^ 28. How many yards of stuff, of 3 qrs wide, will line a
' cloak that is ^ yards in length and 3t yards wide ?
Ans. 8 yds qrs 2 nl.
'29. If 5 yards of cloth cost 14/ 2d, what must be. given for
9 pieces, containing each 21 yards 1 quarter?
^ Ans. 27/ \s \Q\d, ';
30. If a gentleman's estate be wortt 2i07/ 12/ a year j
what may he spend per day, to save 5C0/ in the year ?
Ans. 4/ Ss l^Sd.
31. Wanting
M ARTTHMETia
31. Wanting just an acre of land cut off from a piece
^which is 13^ pdes in hreadthj what length must the piece be?
/ Ans 11 po 4yds 2ft O^f in.
32. At 7s 9f «/ per yard, what is the value of a piece of
doth containing 53 ells English 1 qu. Ans. 25/ lb/ 1^.
33. If the carriage of 5 cwt 14 lb for 96 miles be 1/ 12/ 6ds
Jhow far may I have 3 cwt 1 qr carried for the same money i
Ans. 151 m 3 fur ^rrt^*
34. Bought a silver tankard, weighing 1 lb 7 oz 14 dwts i
what did it cost me at 6/ 4^ the ounce ? Ans. 6/ 4/ 9 V«
35. What is the half year's rent of 547 acres of land, at
1 5s 6d the acre ? Aas. 21 1/ 19j SJ.
36. A wall that is to be built to the height of 36 feet, was^
raised 9 £set 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 chargfe of keeping 20 horses for a
year, at the rate of 14^^ per day for each horse ?
Ans. 441/Oj lOrf.
38. If 18 elk of stuff that is i yard wide, cost Z9s6d'}
what will 50 ells, of the same goodness, cost, being yard wide^
Ans. 7/ 6s :i</.
39. How many yards of paper that is 30 inthes wide, will
hang a room that is 20 yards in circuit an^ 9 feet high?
Ans. 72 yards.
40. If a gentleman's estate be worth 384/ 16/ a year, and
the lapdtax be assessed at 2/ 9id per pound, what is his net
aftnuar income? Ans. 331/ 1/ 94^.
41 . The circumference of the earth is about 25000 miles i
at what rate per hour is a person at the middle of its sur£a€e
carried round, one whole rotation being made in 23 hours
56 minutes ? Ans. i044ry^ miles.
42. If a person drink 20 bottles of wine per month, when
It costs 8/ a gallon j how many bottles per month may he
drink, without increasing the expense, when wine costs 10/
the gallon? Ans. 16 bottles.
43. What cost 43 qrs 5 bushels of corn, at 1/ 8/ 6d the
. quarter ? Ans. 62/ 3/ S^d.
44. How many yards of canvas that is ell wide will line
50 yards of say that is 3 quarters wide? Ans. 30 yds.
45. If an ounce of gold cost 4 guineas, whaC is the valfae
of a grain? Ans. 2yT,it
46. if 3 cwt of tea cost 40/ 12/; at how much a pound
must it be retailed, to gain loi^ by the whole ? Ans.3y^^j.
CX)MPOUN»
C « 3
COMPOUND PROPORTKHf
Compound Proportion ^ows how to resolve such ques
tions as require two or more statings by Simple Proportion ;
and these may be either Direct or Inverse. • •
In these questions, there is always given an odd mimber of
terms, either five, or seven, or nine, Sec. These are distm.
guished into terms of supposition, and terms of demand,
there being always one term more of the former *than of the y^
latter, which is of the same kind with the answer sought.
The method is thus :
Set down in the middle place that term of supposition
which is of the same kind with the answer sought.— Take
one of the other terms of supposition, and one of the demand
ing terms which is of the same kind with it ; then place one
of them for a first term, and the other for a third, according
to the directions given in the Rule of Three. — Do the same
with another term of supposition, and its corresponding de
manding term ;, and so on if there be more terms of each
kind ; setting the numbers under each other which fall all on
the lefthand side of the middle term, and the same for the
others on the right^hand side. — ^Then^ to work
By several Operations, — ^Take the two upper terms and
the middle term, in the same order as they stand, for the first
RuleofThree question to be worked, whence will be found
a fourth term. Then take this fourth number, so found, for
the middle term of a second RuleofThree question, and the
next two under terms in the general stating, in the same
order as they stand, finding a fourth term for them. And so
on, as far as there are any iiumber$ in the general Stating,
making always the fourth number, resulting from each simple
stating, to be the second term in the next following one.
So shall the last resulting number be the answer to the
question.
By one Operathfi, ^Multiply together all the terms stand ^ ^t^o^^^^Ur
ing under eajch other, on the lefthand side of the middle I ^^eSL^
terni} and, in like manner, multiply together all those on the > ^^^^JgSul,
righthand side of it. Then multiply the middle term by \ ^f^iim^
jfthe latter product, and divide the result by the former pro «^ ^^^^^^^
'duct } so shall the quotient be the answer sought. ^t^ij?^'*^
Vq.. I, % l&^^AMPtES
50
ARITHMETIC.
EXAMPLES.
1. How ms^y men can complete ^ trench of 135 yards
long in 8 da^y when 16 men can dig 54 yards in 6 days ?
General Stating.
yds 54 : 16 :: 185 yds
dfiys 8 6 days
432
810
16
4860
81 men
432 ) 1 2960 ( 30 Ans. by one opcratioq.
1296 ^' '
The same hy two Operations.
1st.
As 54 : 16 :: 135 : 40
16
/
8IQ
135
54) 2160 (40
216
2d.
As 8 : 40 : : 6 : 30
6
8 ) 240 ( 30 Ans.
24
O
2. If 100/ in one year gain 5/ interest, what will be the
interest of 750/ for 7 years ? Ans. 2621 "iq^,
3. If a family of 8 persons expend 200/ in 9 months ; Jxow
much will serve a family of 1 8 people 1? months ? .
^ Ans. 300/:
4. If 2*7s be the wages of 4 men for 7 days ; what will be
the wages of 14 men tor ip days ? Ans. 6/ 15/.
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 be travel 360 miles ? Ans. 9 days.
"i
VULGAR FRACripNS, 54,
#
Ex. 6. If 120 bushels o£ com am s^rve 14 horyet SS daysi
how inany days will 94 bushels serye, 6 horses ?, . • • ^
Ans. 102^ days..,
7. If 3000 lb of Ijeef serve ^0 mta^ IS Jays ^ how jiiany
lbs win serve 120 men for 25 days ? >t)s.\ l764lb 1114.02^ '
8. If a barrel of beer be sufficient to last a family of 8 per.
sons 12 days ; how many barrels will be .drank by i6 persons,
in the space o^ year?  . ..AftSf.6P. barrels*^
9. If 180 men, infe d^ys^ of 10 hours ttcj^i, can dig a.
trench 200 yards long, 3 wide* and 2 deep Zip hpw many,
days, of 8 hours long, will 100 mth dig % treyh of 36 yards .
^Jong, 4 wide, and 3 deep ? Ans. '^""*"*
9*"
•"; c. /•■.■.■; ■••■I ,
OF VULGAR FRACTIONS,
part,
with a line bet\^een them :
_,_ 3 numerator i , . , .  «  , *•
Thus,  , > , which IS named 3»fourth§.
4 denominator J >
The Denominator, or number placed below the line, shhws
how m;^y equal ps^ts the whpl6(rqj(i^f^it^ i^divid^qiflo ;
and it represents the Divisor in D,iVjision«t^An4 tH^/^^IMfifr
tor, or numbeir set ^oye the line« shows, how m^y.^JFjtti^^ 
parts are expressed by the Fraction; l^^ing the r^n^^^^;4ff '
after division. — Also, both these i^i^Jser^ ^c, in gen^r^
named the Tjerms of the Fracjt;iort. '> » ^ r! ; »t
Fractions' are either. Proper* ImjMroper, Siajpjte>'jQf)aw
'pounds or Mixed. v . ; '^^ . f
A Proper Fraction, is when the numerator is less than the
denominator ; as, 4/ or f , or j* ^c*
y An Improper Fraction, is when the numerator is equal to,
or exceeds, the denominatoj* ; as, j.,' Qrr^,.or ^s &c. V /
' A Simple FractiQjSy is a single exprei^ion, deiaodng any
nuini>er of parts of the integer \ asi^,.er i^
A Compound Fraction, i$ the fir^Kririon of a fraction^ (r
several fractions connected with the word ^ bet W^0n theii^;
as, j^ of I, or f of ^ of 3, &c*
A Mixed Number, is composed of a whyle nipnb^ miA^
fraction together; at^.3^, or I24, &c. 
E2 Sr A whole
Hk
A Krliote* nt hit^br htunlier may bef expressed like t frftc**
tion, by writisg 1 btelo^'^it, ia a denominator ; so 8 is fs <^
4is4i&c,
A fraction denotes division ; and Its valtie is equal toliie
qaotient obtained by dividing the nuinerator by the dtno^
ininator : so' '^ is ecjud to'B, and V is eqnal to 4.
Hence tlien, if the nomerator be lesfi than the denominator,
the talne of the fraction is less tiian 1 . But if the numerator
be the samfe iH the deno^ninator;,' the fiction is just eqUa!
to 1 / And If the numerator be greater than the denraU^
natorj the ffattioji is greater than J^
tssst
REDUCTION OF VULGAR FRACTIONS.
REDi7CTX0N of Vujglt FractionS is the bringing them
out of.one form or denomination into another } commonly to
{)repagre.theip for the o>erations of Additionj Subtractioni ^c^
of vl^ch there ^^ure several cases.
1
I
4^ PROBLEM,
1 . ^ . . i • i m
ToApd the Great f St Ccmmcn Measure ofTtuo or more Numbers*
TrtE Cotirtnoti Measure of two or more numbers, is that
trulxiber which will divide them both without remainder ; so>
f is a^omtnon measure of 18 and 24 ; the quotient of the
Ibrm^ being 6, and of the latter S. And the greatest num«
ber that will do thisy is the grejttest common measure : so 6
is the greatest common measure of IS and 24 ; the quotient
t>f tWe former being S, and ci the latter 4, whi^rh will not
iHDtli diyidp furtJier.
t^ there be two numbers only; divide the greater by thft
tesfe \ l4ven divide the divisor by the remainder \ $md so on,^
dividing always the last divisor by the last remainder} t3l no
thing remains ; so shall the last diyisol^ of all be th^ greatest
fimimon measure sought*
When there are more than two numb^rs^ find the greatest
Wttkfofsti miKisure of two of thom^ as bisfore ; then do the
If^n^ for that cpmmpxi'^ineasure and another of the numbers}
an4
REDUCTION w VUWJAH FRACTIONS. It
wd io oo» through «U the uMibers; so uriU the great^t^ com
mon mtwmt last ibund be the answfr/ . .
If it happen that the conunooi measure thus found' is If
then the numbers are said to be incommensurabIp> Or uofi
liif ing any comqion measure^
BXAM^L^.
1. To find the griNitest common measure of 1908> 936,
and 630.
936) 1908 (2 So that 36 is the greatest common
1872 measure of 1808 and 9S6*
36 ) 936 ( 26 Hence 36 ) 630 ( 17
72 36
216 270
216 252
18) 36 (2
36
^p
Hence then 18 is the answer required.
2. What is the greatest common measure of 246 ind 372 ?
Ans. 6*
3. What is the greatest common measure of 324, 61 9,
and 1032/ ' A»3. 12.
CASE I.
T$ Abbrevii^ or Reduce Fractions U their Lowest Terms*
* Divide the terms of the given fraction by any numher
chat will divide .them irithput z remainder; tl^n divide these
guotients
* Tluit dtviding both the terms of the fraction by th^sam^
namt>er^ whatever it be, will give another fraction equal to the
former, is evident A^d when these divisions' are performed as
if^n as can be dobe^ or when the common divisor is the greatest
possible^ tbf», terms of the resuhing fraction must be ^he least
possible.
Note, 1. Aiqr number endlag widi aoevea ijiuQiher, or a cipher^
n divisible^ or can be divided, by 2.
2. Any number ending with ^> or 0^ is divisible by 5,
3, If
si ' Ai^ITHMETIC.
c^ucei^tits again in the same manner ; and so on, till it appears
that Uiere is no number greater than 1 which will divide
thetti} then the fraction will be in its lowest terms.
• • * . »
Or^ divide both the terms 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.
EXAMPLES.
1 . Rediuc^ a^ to its least terms.^
mU il = Tf =1 == ii the answer.^
Or thus :
216 ) 288 * { 1 Therefore 72 is the greatest common
216 taeasure; and 72; 4f =s ^ the An
swer, the same as before.
72) 216 {3
216
2. Reduce
3 . If the righthand place of any numfier be O, the whole is di
visible by 10 3 if there be two ciphers, it is divisible by 100 j if
three ciphers, by 1000 : and so on; which b only cutting offdiose
ciphers.
4. If the two righthand figures of any number be divisible by
4, the whole is divisible by 4. And if the thrqe righthand figures
be dtvisible by 8, the whde is divisible by 8. And so on.
If
5. If the sum of the digits in any number be divisible by a, or
by 9, the whole is divisible by 3, or by g.
e. If the righthand digit be even, and the sum of all the digits
be divisible by 0, then the vlhole is divisible by 6.
7. A number is divisible by 1 1 , when the sum of the l st, 3d, 5th,
. 3cc, or all the odd places, is equal to the sum of the 2d, 4tt, 6th,
&c, or of all the even places of digits.
8. If a number cannot be divided by some quantity less than the
sqliarfe root of the same, that number is a prime, o.^ cannot be di
vided by any number whatever,
g. All priine numbers, except 2 and 5, have either 1, 3> 7, org,
m the place of units ; and all other numbers are composite, or can
be divided.
10. When
REDUCTION OF VULGAR FRACTIONS. 55
t
^. Reduce f^ to its lowest terms. Ans. ^*
3. Reduce 441 1® ^^^ lowest terms. Ans. \.
4. Reduce fyv ^^ ^^^ lowest terms. Ans. I*
CASE II.
To Reduce a Mixed if umber to its Squivalent Improper Fraction.
* MuLTiPLt the integer or whole number by the deno^
minator of the fraction, and to the product add the numera
tor ; then set that sum above the denominator for the frao^
tion required*
EXAMPLES;
1 . Reduce 23 to a fraction*
23
5
115 Or,
2 (23x5)+2 in , .
rs — , the Answer.
117 ^ ^ ' .
5
2. Reduce 12 to a fraction. Ans. '^^
3. Reduce 14t^ to a fraction. Ans. Vtf .
4. Reduce 183^ to a fraction. Ans. ^.t**.
10. When numbers, with the sign of addition or subtraction be^
tween them^ are to be divided by any number, then each of those
, 4i ,, , . . rr.i 1018—4
numbers must be divided by it. Thus — i =5 + 4'2~7«
11. But if the numbers have the sign of multiplication between
% them, only one of them muSt be divided. Thus,
10X8X3 _ 10X4X3 _ 10 X4X 1 _^ 10X2X1 _20^^
6X2 "~ 6X1 ■" 2X1 " 1X1 ""T*"" \ .
* This is* no more than first multiplyitig atjuaritity by some
number, and then dividing the result l»ck again by the samd :
which it is evident does not alter the value 5 for any fraction re
presents i. divisiori of the numerator by the detiominatot".
. ' CASB
g$ ARITHMETIC
CASE III.
T9 Reduce an Improper Fraction to its Equivalent Wb^ or
Mixed Number,
* DiVfDE the numerator, by the denominariior, and the
quotient will be the whole or mixed number sought.
EXAMPLES*
1. Reduce y to its equivalent number*
Here y or 1 2 r 3 = 4, the Answer.
2. Reduce y to its equivalent number.
Here V or 15 r 7 = 2, the Answer.
3. Reduce V^^ to Its equivalent number.
Thus, 17) 749 (44^
68
69 So that ^ 3= 447V> the Answer.
68
1
4. Ileduce l^ to its equivalent number. Alls. 8.
5. Reduce ' ij* to its equivalent number. Ans. 544»
6. Reduce *^* to its equivalent number. Ans. 17147*
i
CASE IV.
To Reduce a Whole Nundfer to an Equivalent Fraction^ having
a Given Denmmnator^
f MULTIPLT tlie whole number by the given denominators
then set the product over the said denominator, and it will
form the fraction required*
/
* Tills Rule is evidently the reverse of the former ; and the
reason of it is manifest from the nature of Common Division*
f Multiplication and Division being here equally used> the re
sult must be the same as the quantity first proposed.
EXAMPLES.
REDUCTION op VULGAR FRACTIONS. S7
EXAMPLES.
I
1. Reduce 9 to a fraction whose denominator shall he 7.
• Here 9 X 7 s 63: then y is the Answer;
For V = 63 ^ 7 = 9, the Proof.
^ Reduce 12 to a fraction whose denominator shall be IS.
Aas. Vt •
3^ Reduce 27 la a fraction whose denominator shall be 1 U
Ans. Vi^.
CASE Y.
To Reduce a Compound Fraction to an Equivalent Simple One*
* Multiply ail the numerators together for a numerator,
and aU the denominators 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 bv 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 conunon,
they may be omitted, or cancelled.
£XAMPL£S.
1. Reduce ^ of  of  to a simple fraction.
, 1x2x3 6 1 , ^
**€^^:;: — :;; = :r: = rr> ^e Answer.
2x3x4 24 4'
Or, ^ —  — 2 ~ T* ^y cancelling the 2's and i*s.
^ X Q X 4? 4
* The truth of this Rule may be shown as follows : Let the
.compound fraction be  of j . Now j of f is  r 3» which is ^\j
consequently f of ^ will be ^V X 2 or ij, that is, the numerators
are muluplied togetlier, and ulso the denominators, as in the Rule.
When the compound fraction consists of more than two single
ones i having tirit reduced two of them as above, then the resulting
Action and a third will be the same as a compound fraction of t«vo
jparts^ and so on to the last of all.
2, Reduce
58 ARITHMETIC.
2. Reduce ^ of ^ of {4 to a simple fraction.
_, 2x3x10 60 13 4 , .
""'" sTI^TTT = 165 = 33 = IT' ^^^'^ ^"'^"'••.
2 X 3^ X jd 4
^^» ^ — f — TT ~ 7T> ^^ ^^°^^ ^^ before, by cancelling
the 3's, and dividing by 5's.
S. .Reduce ^ of  to a simple fraction. Ans. 54*
4. Reduce f of f of 4 to a simple fraction. Ans. J.
5. Reduce . of f of Si to a simple fraction. Ans. {•
6. Reduce 7^ of ^ of J of 4 to a simple fraction. Ans. i.
7. Reduce 2 and  of ^ to a fraction. Ans. ^.
CASE VI,
To Reduce Fractions of Different Denominators^ to Mquivatent
Fractions having a CommoH Denominator.
* Multiply each numerator by all the denominators ex
cept its own, for the new numerators : and. multiply all the
denominators together for a common denominator.
NotCy It is evident, that in this and several other (fperation^
when any of the proposed quantities are integers, or mixed
numbers, or compound fractions, they must first be reduced,
by their proper Rules, to the form of simple fractions*
EXAMPLES.
1 . Reduce 4> t> and ., to a common denominator.
1 X 3 X 4 = 12 the new numerator for 4.
2x2x4 = 16 ditto f.
3 X 2 X 3= 18 ditto f.
2 X 3 X 4 = 24 the common denominator.
Therefore the equivalent fractions are ^, J^, and i^.
Or the whole operation of multiplying may be best per^
formed mentally, only setting down the results and given
fractions thus: 4, 4, J = H> ih H = A» tV> tV, by
abbl'eviation.
2. Reduce y and ^ to fractions of a common denominator.
Ans. II, 14.
* This is evidently bo more than multiplying each numerator
And its denominator by the same quantity, and consequently th6
value of the fraction is not altered.
3. keduce
REDUCTION OF VULGAR FRACTIONS. v<9
5. Reduce •§, ^i and ^ to a conunon denominator.
Ans. ^, IJ, 1^.
4. Reduce i, 2^1 and 4 to a common denominator.
AnQ 2 5 7 8 »ao
Note 1 . When the denominators of two given fractions
have a common measure, let them be divided by it ; then
muhiply the terms of each given fraction by the quotient
arising from the other's denominator,
£/:• 2T and j^ =: rrr ^^^ tttj ^7 niuhiplying the former
5 7 by 7 and the latter by 5.
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.
J?ar* f and ^ =: ^ and ^^ by mult, the former by 2.
2
3. When more than two fractions are proposed, it is some
times convenient^ first to reduce two o£them to a common
denominator ; then these and a third ; and so on till they be
all reduced to their least conunon denominator.
£:e. I and J and I =;: I and J and  = ^^ and if and J^.
CASE vir.
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
parts in the next inferior denomination, and divide the pro
duct by the denominator. Then, if any thing remains, mul
tiply 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 *i
* The numerator of a fraction being considered as a remainder,
in Division^ and the deriomitiator ^s the divisor, this rule is of the
same nature as Compound Dtvisioni or the valuation of remainders
in th« Rule of Three, before e&plained.
EXAMPLES.
tQ
ARTTHMEnC.
EXAMPLES.
I. What is the f of 2/ 6s f"
B J the former part of the Rule,
2/ 6s
5)9 4
Ans» 1/ 16/, 9d 2\q.
2. What Is the value of ^of 1/r
By the 2d part of the Rule^
2
20
3 ) 40 ( 13x 4rf Am.
1
12
3) 12 (4</
$• Find the value of f of a pound sterling* Ans. Is 6d»
4. What is the value of f of a guinea ? Ans. 4i 8i^
Sm What is the value of ^ of a half crown i Ans^ 1/ lO^^.
Ans* 1/ 1 l{d.
Ans. 9oz 12dwts»
Ans* Iqr 71b*
Ans. 3 ro. 20 po*
Ans* 7brs 12mi&»
6. What is the value of f of 4/ \0d?
7. What is the value of * lb troy ?
8* What is the value of i ^of a cwt ?
9. What is the value of {of an acre?
iO.. What is the value of ^ of a day ?
CISE VJII.
I
Tq Reduce a Fraction from one Denomination to another.
^ 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 denomir
nator, if to a greater.
EXAMPLES.
2 . Reduce ^ of a pound to the fraction of a penny,
f X V X V == *r =^ 't""* ^^e Answer.
* This is tiie same as the Rule of Keduclion in whole numbers
from ooe denomiiiatign to another.
2. Reduce
ADDITION OF VULGAR FRACTIONS. 6fr
$• Reduce f of a penny to the fraction of a pound*
I X A X ^ = ^, the Answer.
9. Reduce i%/ to the fraction of a penny. Ans. V^«
4* Reduce q to the fribCtion of a pound. Ans. t^V?
5. Reduce y cwt to the fraction df a lb. Ans. V^
6. Re<htce f dwt to the fraction of a lb troy. Ans. j^^*
7. Reduce  crown to the fraction of a guinea.. Ans. ^^.
6« .Reduce ^ half<:rown to th« fract. of a shilling. Ans. 44*
9. Reduce 2s 6d to the fraction of a £• Ans. 4«
10. Reduce 17/ 7rf S\j to the fraction of a £.
*lmmmmiitmmmm^ttmmatagmmmmmmm*im
ADDmON OP VULGAR FRACTIONS.
tp the fractions have a common denominator; add all the
liumerators together, then place the sum over the common
denominator, and that will be the sum of the fractions
required.
* If the proposed fractional have not a common denomina
tor, they must be reduced to .one. Also compound fractions
must be reduced to simple ones, and fractions of diderent
denominations to those of the same denomination. Then
add the numerators as before. As to mixed numbers, they
may either be reduced to improper fractions, and so added
with the others ^ or else the nactional parts only added, and
the integers united afterwards.
* Before fractions are reduced to a common denominator, they
are quite dissimilar, as much as shillings and pence are, and there
fore cannot he 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 difference of their iedividuals. Whence the reason of thQ
Bule is manifest, both for Additipn and Subtraction.
When several fractions are to be collected, it is commonly best
^rst to add two of them together tuat mrst easily reduce to a com
fifwn dptiomn^^^f then ftdd their ^um and a thirds and so on.
EXAMPLES.
» \
63 ARITHMETIC.
EXAMPLES.
1. To add ^ and f together.
Here ^ + ^ = ^ = 1 4, the Answer.
• 2. To add 4 and  together.
Hi = 44 + l^=^=lT4,the Answer.
3. To add 4 and 7i and 7 of  together,
4. To add* f and 4 together. Ans. If.
5. To add ^^ and f together. ; Ans. IfJ^,
6. Add y and /^ together. Aiis. ^2^.
7. What is the sum of f and ^ and ^? Ans. 1tS4*
8. What is the sum of  and ^ and 2^? Ans. S^.
9. What is the sum of 4 and ^ off, and 9,^ ? Ans. 10^^.
10. What is the sum of  of a pound and 4 of a shilling ?
Ans. '5,or 13i 10^24^.
1 1 • What is the sum of 4 of fi shilling and j^ of a penny ?
Ans. Vt ^ or Id \^g.
12. What is the sum of 4 of a pound, and . of a shilling,
and TT'of a penny ? Ans. 4^^j or Ss Id 145?*
SUBTRACTiqjf^ OF VULGAR FRACTIONS.
' Prepare the fractions the same as for Addition, when
necessary 5 then subtract the one numerator from the other,
and set the remainder over the common denominator, for the
difference of thp fractions sought.*
EXAMPLES,
1 . To find the difference between 4 and 4«
Here 4 "^ 1^ ^ 4 == t> ^^^ Answer.
2. To find the difference between 4 and 4.
T  V = T^ • If = Ty» the Answer.
3. What
MULTIPLICATION of VULGAR FRACTIONS. 63
S. what IS the difference between ^ and f^ ? Ans. ^.
4. What IS the difference between ^ and f^i Ans, y^.
5. What Is the difference bel;ween ^ and ^Zj ? Ans. rh
6. What is the diffl between 5 and ^ of 4^ ? Ans. 4,^^.
7. What is the difference between ^ of a pound, and J of
i of a shilling ? Ans. Vt ^ ^r lOx Id l^.
8. What is the difference between f of 5^ of a pound,
and I of a shilling ?  Ans. lUil or V Ss \ 1 .y.
MULTIPLICATION of VULGAR FRACTIONS.
* Reduce ^ixed numbers, if there be any, to equivalent
fractions ; then multiply all the numerators together for a
numerator, and all the denominators together for a denomi«
nater, which will give the product required.
EXAMPLES* N
1 . Required the product of ^ and ..
Here t ^ i == t% ^ t> ^*^c Answer.
Or^x J = 4x^ = i.
2. Required t}ie continued product of ^ 3, 5, and  of fi
,, ?f 13 ^ a 3 ISx 3 39
Here y x x ^ X "^ X y = ;^;3^ = = 4^, Ans.
3. Required the product of f and ■^. Ans. j^y.
4. Required the product of ^ and j^. Ans. rV.
5. Required the product of f , ^, and 44» Ans. ^*j.
* Multiplication of any thing by a fraction, implies the taking
some part or parts of the thing j it may therefore be truly expressed
by a compound fraction; which is resolved by multiplying toge
ther the numerators and the denominators.
Note, A Fraction is best inultiplied by an integer, by dividing
the denominator by it , but if it will not exactly divide^ then
multiply the numerator by it
6. Required
64 ARITHMETia
6. Kequired the product ofi, jf^nd S* An$«> 1;
7. Required the product of ^, ^, and 4^ Ans. 2^.
8. Required the product of ^ and •§ of y. Ans. H^
9. Required the product of ^, and f of 5, Ans, 20»
10. Required the product of  of j, and J. of 3f . Ans. f J^
1 1 . Required the product of 3 and 4^. . Ans, i 4444*
1 2. Required the product ef 5, > f of f, and 4. Ans. 2^.
■««»
DIVISION or VULGAR FRACTIONS.
^ Prepare the fractions as before in MuhipKcation ; then
divide the numerator bv the numeratorj and the denominator
hj the denominator, it they will exactly divide : but if not»
then invert the terms of the divisor, and multiply the dividenc)
by h, as in Multiplication*
ZXAUPtES.
1. Divide V by f .
Here V "^ 4^ = t = Ht ^7 the first method*
2, Divide^ by ,5^.
Here^^T^=:^x V=f xi = V =H
5, It is required to divide 44 ''T 4* Ans. f*
4. It is required to divide ^ by ^. Ans. ^V
5. It is required to divide \^ by ^. Ans. lf.
6. It is required to divide ^ by V • Ans. ^^
7. It is required to divide y^ by ^. Ans. f .
.8. It is required to divide 7 by ^ Ans. Jt
«
^ Divison being the 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 vyill not exactly divide^ then multiply
the denominator by it*
9. ft
RULE OF THREE m VULGAR FR A CTIONS. fi^
•9. It is required to divide V^ t)y 3. • Ans. ^
10. It is required to divide 4 by 2. Ans. ,25^'
11. It is required to divide 7 j. by 9^. Ans. ff .
1 2. It is required to divide f ttf j by f of 7f . Ans. Tfr
' * ' "
RULE OF THREE in VULGAR FRACTIONS.
Makb the necessary preparations as before directed ; then
multiply continually together, the second and third terms^
Mtkd the first with its parts inverted as in Divi^ion^ for the
answer**
Examples.
1 • If f of a yard of velvet cost  of a pound sterling ; vfhM
Will 1^ of a yard cost ?
S 2 5 S i S , , ^
y •• "5" •• li = T ^ 7^ ig =i'*^^«^^ Answer.
^i What trill S^ot bf silver cost^ at 6s 4i an ounce f
Ans. 1/ li 44rf.
3. If T^ of a ship be worth 273/2/ 6d; what are ^r of her
worth? Ans. 227/ 12/ Id.
4i What is the purchase of 1230/ b^kstbck, at 108 per
cent.? Ans. 1336/ UW.
5. What is the interest of 273/ 15/ for a year, at 3^ per
cent.? Ans. 8/17/ lli/.
.6. If I of a ship be worth 73/ 1/ 3/// what part of her is
worth 250/ iOsP Ans. j.
7. What length thust be cut off a bo^d thit is 7^ inches
broadi to tontain a square foot, or as much as another piece
of 12 inches long and 12 broad ? ' An^ I844 inched.
S, What quantity of shalloon that is ^ of a ydrd wide^ will
line 9i yards of cloth/ th^at is 24 yards widei Ans. 31^ yds.
» . ^ . . ■ .
, "* This is only multiplying the 2d and 3d terms Oogether^ and
diidding the product by the firsts as in the Rule o^^ree in whote
^umbera. >.
Vol. h T ' 9. If
66 ' ' AWtllMlETIC:
9. If'tlie penny loaf weigli 6^<5K when the prite bl
w'heat is 5/ the bushel ; what CH^t it to weigh whe^ the
wheat is 8s 6d the bushel? Ans. 4^7 oz.
10. How much in length, of a piece of land that is ll^^*
poles broad, will make an acre of land, or as much as 40
poles in length and 4 in breadth ? Ans. 13^^ poles.
11. If a courier perform a. certain journey in %5i days^
travelling 13^ hours a day ; how long would he be in per
forming the same, 'travelling only I'lru liours a day ?
Ansi 4044 days.
12. A regiment 6f soldiers, tonsisttng of 976 mw, are to
be iiew cloathed ; each coat to contain ^i yards of cloth
that is i{ yzrft wide, and lined ifith shalloon ^ yard wide :
how maily yaid$* of shalloon will line them ?
• Ans. 4531 yds 1 ^ 2 nails.
* .d' i /tiii;
DECIMAL FRACTIONS.
A Decimal Fraction, is that which has foi** its deno
minsttonaatmit (1), with as many ciphers annexed as the
humetatoi: ha^ places ; and it is usually expressed by setting
down the numerator only, with a point before it, on the left
hand. Thus, ^^ is 4, and ,^\ is •24,>and ^ittr is "074, arid
^^>5*y^^ is '00124 ; where ciphers are prefixed to make up as
many places as are ciphers in the denominator, when there
is a deficiency* of figures.
• A mixed nuipber is made up of a whole number with some
decimal iiraction, the one being separated from the .other by
a point.; Thus, 3 '25 is the same as S^y or 4^.
' Ciphers on the righthand of decimals make no alteration
in their value; for*4, or '40, or '400 are decimals having all
the same value, each being == /„, or ^. But when tliey are
placed on the lefthand, they decrease the value in a tenfold
proportion: Thus, '4 is T^y* or 4 tenths ; but '04 is oiily r^t
or 4 hunciredfhs, and '004 is only ri^j or 4 thousandths*
The 1st place of deciinals, counted from the lefthand to
wards the^rightj j.s called the place of primes, or 1 Oths ; the
2d is the place of seconds, or lOOths ; the 3d is the place of
thirdsj or lOOOths ; and so on. For, in decimals, as well ag
in whole numbers, the values of the places increase towards
the lefthand, afid decrease towards the right, both in the
same
«
same tenfold proportion ; as in the following Scale or Table
«f Notation.
CO
§ 9 .•i'.!.g,"S<'3g; iltM
to
ill I ^ •;■■*,■ .^.i I ^ ^ §
3 3 33 3 3 3 •3.3 • 3 3 3 3
ADDtTlbN'oF DEfclMALS.
S£t 't&e) ttic&bcfrs ttxubr each*oth«9 «ccdniiiig >to die faltte
.of jAac >plac€js, like as ip^wljLple nipnb^K j. jp jwrb^^ta^tp tl^
tieqijiipgl .5^para^ing poinds will stand all exacriy ynder each
other. Then, beginning at the ri;^thand, add u'p aH the
columns of numbers as in integers ; and point off as many
places, for decimals, as are in the greatest number of decimal
places in any of the litres that are addled ; or place the point
directly below all the other points.
EXAMPLES.
1. Ta add together 290146, and"3146*5, and 2109, and
; . 290 146
3146*5 :
» J > • .
2109 .. ,: . ' • ,
1416
5299*29877 ' the Sum.
Ex.2. What is the sum of ^^6, 3 9 '2 13: 72014*9, 417,
and 50S2? . ' • ' .
3. What is th^ sutti qf .7530y 16^201, «0142,;957*X3,
^•72119 and 'OSOU*:
4. What is the syrii pf 31209, 357llj 71956, 7l*4»8,
9739.215, 179, and 0027?
F2  SUBTRACTIOJI
€i ARITHMETIC
. SUBTRACTION or DECIMALS.
Place the numbers under each other according to the
value of their places^ as in the last Rule. Then^ beginning
at the righthand) subtract as in whole numbers, and point
off the decimals as in Addition.
EXAMPLES.
1. To find the difference betwfeen SlIS and 2^13«.
9r73
2138
Aixs. 89*592 the Difference.
q: Find the diff. between 19185 and 2*73. Ans. 0'81 15.
'3. To subtract 4.90U2 from 214*81. Ans* 2099085&.
4. Find the diff. between 2714 and 'QIS. Ans. 2713084.
MULTIPLICATION of DECIMALS.
/ . •
^ Place the factors^ and multiply them together the same
as if they were whole numbers.— Then point off in the pro
duct just as many places of decimals as there are decimals in
both the factors. Bufif there be not so many figures in the
product) then supply the defect by prefixing ciphers.
* The Rule will be evident from this example :«Let it be re
quired to multiply *12 by '36l ; these tumbers are equivalent to
;^ and ^t}„ 5 the product of which is rUlhs = 04332, by the
nature' (^ 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 for any other numbers.
' ' • EXAMPLES.
MULTIPLICATION of DECIMALS. 69
EXAMPLES.
1. Multiply •321096
by '2465
1605480
1926576
1284384
642192
" 1 W ' ■ I P
Ans. 0791501640 the Product.
2. Multiply 79347 by 2315. Ans. 1836'88305.
3. Multiply 63478 by 8204. Ans. .520773512
A. Multiply 385746 by 00464. Ans. 00178986144.
CONTRACTION I.
To multiply Decimal/ by 1 with any number of Ciphers^ as by lO,
or 100, or 1000, isTc.
■
This is done by only removing the decimal point so many
places farther to the righthand,* as there are ciphers in the
Qiultiplier ; and subjoining ciphers if need be.
EXAMPLES*
1. The product of 513 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 lOOOO is
CONTRACTION 11.
To Contract tie Operation^ sq as to retain only as many Decimals
in the Prodsjct,f^ may be thought Necessary^ tuhen the Product
would naturally contain several more f^lfu:es^
Set the units' place of the multiplier under that figure of
the multiplicand whose place is %\^ same as is to be retained
for the last in the product ; aind dispose of the rest of the
figures in the inverted or contrary order to what they are
usually placed in. — ^Then, in multiplying, reject all the figures
that are more to the righthand than each multiplying figure,
and set down the products, so tliat their righthand figures
may
iof
ARmiRtETlC.
may fall in a column straight below each other; but observing
to increase the first figure of every line with what would
arise from the figures omittedy in this manner, namely 1
from 5 to 14, 2 £om 15 to 24, 3 from 25 to 34, &c ; and
the sum of all the lines will be the product as required^ com<«
monly to the nearest unit.in the last figure.
EXAMPLESi
1. To multiply 27^14986 by 92.41035, so as to retain
only four j^ace^ of decimals in the product.
Contracted Way* Common Way.
2714986 » 2714986
53014^9 9241095.
24434874
542997
108599
2715
81
14
25089280
13
574930
81
4.4958
2714
986.
108599
44
542997
2
24434874
250S9280
650510
t
•
2. Multiply 480*14936 by 972416, retaining only four
decimals in the product.
3. Multiply 24903048 by 573286, retaining only five
decimals in the product.
4. Multiply 325701428 by 7218393, retaining oidythre^
decimals in the product*
DIVISION OF DECIMALS.
Divide as in whole numbers; and point off in the quo
. tfent as many places for decimals, as the dedi^i^^l places in the
dividend exceed those in the divisor*.
l^
II 1 1 1 1 1 1
♦ The reasort' of t^its. Rule is evident 5 for, since the divlsoi:
multiplied by the quotient gives the dividend, therefore the num
ber of decimar places in the dividend, is equal to those in the"di
vi^orand quotfettt, taken together, by the nature of Mnltiplfca
<fe& 5 and consequently thequoti«at itBelf must cohtdfe ds many as
i(he dividend exceeds the divisor*
Another
DIVISION ct DXCiMAJ^.
%■
Another way to know the place {gr the decimal point, is
this : The first figure of the quotient must be made to oc
cupy the same place, of integers or decimals, as doth 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 tp 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 aivi?xed ('p the divi*
dend, and the quotient carried on as far as required.
EXAMPLES. .
1.
nS) 48520998 (00272589
1292
460
1049
1599
1758
156\
2.
•2639) 2700000 (1023114
6100
8220
3030
3910
12710
2154
J. Divide 123;70536 by 54*25,
4. Divide 12 by 7854.
5. Divide 4195*68 by 100.
6. Divide 8297592 by 153.
Ans, 22802,
Ans. 15278.
' Ans. 41*9568,
Ans. 5*4232.
CONTRACTION I.
When the divisor is an integer, with any number of ci^
phers annexed : cut off those ciphers, and remove the deci.
mal 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*.
* This is no hjopc than dividing both divisor and dividend by*
the same number, either 10^ or lOO, or 1000, &c, ^c^otrdibg to<
the number of ciphers cut off, which^ leaving them in the same
proportion, does not affect the quotienf . And, in the same way,
the decimal point may be moved the same number of p^ces in
both the divisor and dividend, either to the right or left, whetheif
they have ciphers or not.
tXAMPXES^
if ARITHMETIC.
EXAMPLES,
1. Dividp 45*5 by 2100.
2100) 455 (0216, &C. V '^2
35
140
14
$. Divide 41020 by 32006.
5. Divide 953 by 2 1600,
4. Divide 61 hj 19000.
CONTRACTION IJ.
HbncEi if the divisor be 1 vith ciphers, as 10, 100, O]^
1000, &c : then the quotient will be found by merely mov
ing the decimal point in the dividend so many places farther
to the left, as the divisor has ciphers j prefixing ciphers if
need be.
3EXAM?LES.
So, 217^3 r 100 ==2173 And 419 t 10 =
And 5ll5 r 100 = And 21 — lOpO =
CONTRACTION III.
When the^e are many figures in the djvisor; or when
only a certain number of decimals arie necessary to be re
tained in the quotient ; then take only as many figures of
the divisor as ^mW be equal to the number of figures, both ^l
tegers and decimals, to be in the quotient, and find how
many times they may be contained in the first figures of the
divia^nid, as usual.
l(jpX each rjjn^aind^ b^ a ne^ dividend i^ and for ^very such
<livi4fnd, Ipve oyt one figure mpr^ on the rightrhand side of
tide divisoi: j' r^m^jmbering tQ parry for the increase ox the
figures ciit p^, as in the ^ contractipn in Multiplication.
Note. Wb^n there are not so many figures in the divisor,
as are required to be in th^ quotient, begin the operation
with all the figures, and continue it as usual till the number
of figures ill the divisor be equal to those remaining to be
found in the quoUent ; 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 tJ^e quotient will
contain six figures.
Contracted.
REDUCTION OF DECIMALS. 73
Contracted, Common.
PJ'4103,5)25O8'928,06(a7 1498
660721
13849
4608
80
6
924 103,5)2508928,06(27. 1 498
66072106
13848610
46075750
91116100
79467850
5539570
2. Divide 41092351 by 230*409, so that the quotient may
contain only four decimals. Ans. 17*8345.
3. Divide 37' J 0438 by 571396, that the quotient may
contain only five decimals. Ans. •00649.
4.Divide 913*08 by 2137*3, that the quotient may contain
fmly threie decimals.
flEDUCTION OF DECIMALS.
CASE I.
To reduce a Vulgar Fraction to ks equivalent Decimal.
«
Divide the numerator by the denominator as in Division
pf Decimals, annexing ciphers to the numerator as far as
necessary ; so shall the quotient be the decimal required.
EXAMPLES.
1. Reduce ^ to a decimal.
24=4 X 6. Thei£4) 7*
W) 1*750000.
•291666 &c.
2. Reduce ^y and ^ and \<, \o decimals.
Ans. '25, and '5, and *75.
' 3. Reduce ^ to a decimal, Ans, '625.
4. Reduce ^ to a decimal. Ans. 12,
5. Reduce ttt *o * decimal. Ans. '03^59.
6. Reduce ^:^ to a decimal. Ans.. '143155 &c.
CASE
74 . ARITHMETia
CISB II.
To find the Value of a Decimal in terms of the Inferior Den^
minationSm
Multiply the decimal by the number of parts in the
next lower denomination ; and cut off as many places for a
remainder to the righthand, as there are places in the given
decimal.
Multiply that remainder by the parts in the next 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.
Notey Tliis Operation is the same as Reduction Descending;
in whole numbers. ' •
exa'mples.
1. Required to. fin,d the value of ^775 pounds sterling*
•775
90
/ 15500
\2
d 6000 Ans. 15/ 6i:
2. What is the value of '625 shil ? Ans, 7^/.
3. What is the value of 8635/.^ Ans. 17/S24A
4. What is the value of '0125 lb troy ? Ans. 3 dwts,.^
5. What is the value of '4694 lb troy ?
Ans. 5 oz J2 dwts 15*744 gr.
6. What is the value of '625 cwt ? ^Ans. 2 qr H lb»
7. What is the value of 009^43 miles ?
^ Ans. 17 yd 1 ft 593S48 inc.
8. What is the valu^f 6875 yd? Ans. 2 qr 3 nls.
.9. What is the value of '3375 acr ? Ans. 1 rd 14 poles*
10. What is the value of '2083 hhdof wine?
Ans. 131229 gal.
iCASE
REDUCTION OF DECIMALS. 75
CASK III.
To reduce Integers or Decimals to Equivalent Decimals (f Higher
De/iominations,
Divide by the number of parts in the next higher deno
mination ; continuing the operation to as many higher deno^
minations as may be necessary, the s^me as in Reduction
Ascending of whole numbers.
EXAMPLES.
1. Reduce 1 dwt to the decimal of a pound troy*
20
12
1 dwt
0*05 oz
0004166 &c.lb^ Ans,
2. Reduce 9fl? to the decimal of a pound: Ans. .0375/.
3. Reduce 7 drams to the decimal of a pound avoird.
Ans. 02734.375^.
' 4. Reduce '26^ to the decimal of a /. Ans. '0010333 &c. /.
5. Reduce 2*15 lb to the decimal of a cwt.
Ans. 019196 + cwt.
6. Reduce 24 yards to the decimal of a. mile.
^ Ans. 013636 &c. mile.
7. Reduce *056 pole to the decimal of an acre.
Ans. 00035 ac.
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 &c. da.
10. Reduce *21 pint to thedecimalof a peck.
Ans. 013125 pec.
1 1 , Reduce 28" 1 2"' to the decimal of a minute.
Note, When there are several numbers^ to be reduced all to the
decimal of the highest :
Set the given nuihbers directly under each otherj for divi
dends, proceeding orderly from the lowest denominatiori to.
the highest. %
Opposite to each dividend, on the lefthand, set such a
number for a divisor as will bring it to the next higher name;
drawing a perpendicular line between all the divisors and
dividends.
Begin at the uppermost, and perform all the divisions:
qnly observing to set the quotient of each division^ as decimal
parts.
16 ARITHMETIC.^
parts, on the righthand of the dividend next below It } so
shall the last quotient be the decimal required.
EXAMPLES.
1. Reduce 17/ 9^rf to the decimal of a pomid* .
4
12
20
3
975
17*8125
£ 0*890625 Ans.
2. Reduce 19/ 17/ S^d to /. Ans. 1986354fl66 &c. /.
3. Reduce 15/ 6d to the decimal of a /. Ans. '775/.
4. Reduce 74^ to t^e decimal of a shilling. Ans. '6S5s.
5. Reduce 5 oz 1 2 dwts 1 6 gr to lb. Ans. *46944 &c. lb;
iSKTrnffTTT
RULE OF THREE in DECIMALS.
PiEPARfe the terms, by reducing the vulgar fractions to
decimals, and any compound numbers either to decimals of the
higher denominations, or to integers of the lower, also the
iirst and third terms to the same name : Then multiply and
tlivide as in whole numbers.
Note^ Any of the convenient Examples in the Rule of
Three or Rule of Five in Integers, or Vulgar Fractions, may
be taken as proper examples to the same rules in Decimals.
— ^The following Examplei which is the first in Vulgar Frac
tions, is wrought out here, to show the method.
Jf I of a yard of velvet cost /, what will ^^ yd cost ?
yd / yd / s d
1 = 375 .
•375 : 4 :: 3125 : 333 &c. or 6 «
•4
T — *
•375 ) 12500 ( •333333 &c.
1250 20
rf 1^1? . _
/ 666666 Sec:
12
«
Ans. 6s Sd. d 7*99999 &c.=8rf.
]>UOD£*
DUODECIMALS. 77
DUODECIMALS.
1 • f
DuobsciMALs, or Cross Multiplication, is a rule used
by workmen and artificers^ in computing the contents of
tneir works.
Dimensions are usually taken in feet, inches, and quarters ^
any parts smaller than these being neglected as of no conse
quence. And the same in multiplying them together, or
casting up the contents. The method is as follows.
Set down the two dimensions to be multiplied together,
one under the other, so ^t feet may stand under feety
inches under inches, &c. '
Mukiptyeach term in the multiplicand, beginning at the
lowest, by the feet in the multiplier, and 6et the result of
each straight under its corresponding term, observing to
carry 1 for every 12, from the inches to the feet.
In like mai^ner, multiply all the multiplicand by the inches
and parts of the multiplier, and set the result of each term
one place removed to the righthand of those in the multipli
cand; omitting, however, what is below parts of inches, only
carrying to these the proper number of units from the lowest
denomination.
Or, instead of multiplying by the Inches, take such parts
<£ the multiplicand as there are of a foot.
Then add the two lines together, after the manner of
Compound Addition, carrying 1 to the feet for 12 inches^
when these come to so many#
EXAMPLES.
J Multiply 4 f 7 inc
by 6 4
2.
Multiply Uf 9 inc.
by 4 «
27 6
59
7 4i
Ans. 29 0\
Ans. ^6 41
t
3. Multiply 4 feet 7 inches by 9 f 6 inc. Ans. 43 f. 6^ inc.
y 4, Multiply 12 f 5 inc by 6 f 8 inc. Ans. 82 9J
5. Multiply 35 f 4 mc by 12 f 3 inc. An?. 433 4 J.
6. Multiply 64 f 6 inc by 8 f 9J inc. Ans. b^o 8
INVOLUTION.
1$
AIUTHMETIC:
INVOLUTION.
Involution is the raising of Powers from tny given
number, as a root. ' ' ' '
A Power is a quantity produced bytnultiplying any given
number, called the Root, a certain number of times conti
nually by itself. Thiis, •' *
2 = g is the*0Qt, qr iBt ppwer of 2*
2x2= 4 is the 2d power, or square of'il^ ,
2x2x2= 8 is, the 3d powerj or cubeiof 2. .
2 X '2 X 2 X 2 == 16 is the 4th. power of 2, &€•
/
jAnd in this tnanner may be caktflated the followitig Table
of the first^nine powers of the fkaj 9 numbers.
TABLE of th.^ jfirst Ninb Powers of Numbers.
1
1st
4
2c
if 3d
9.
5
6
7
s
9
16
I
8
27
04
25
36
J25
40
2J6
343
64 512
_!
8l/2,9
4th
16
81
256
625
I2g6
240]
•5th j ^t1i''>ih
■t
' 1
32
64
128
243
J024
729
4096
2187
16384
3125
15625
7776
16807
409632768
0561
59019
46656
78125 .
2799^6
II 7649 823543
I
2621442097162
531441
8th
256
6561
65536
39062^
J 67961 6
^•^'••••^va
5764591
16777216
4 782969 4304672 J
4 *
512
.196.88
262144
19531 2 j:
10077696
40353607
■S.t.
1«4217
»>
387420489
The
r^
INVOLUTION. ^9
The Index or Exponent of a Powerj is the number de
noting the height or degree of that power; and it is 1 more
than the number of muUiplications used in producing the
same. So 1 is the index or exponent of the 1st power or
rooty 2 of the 2d power or square^ 3 of the third power or
cube, 4 of the 4th power, and so on.
Powers) that are to be raised, are usually denoted bj placing
the index above the root er first power.
So 2* S3 4 IS the 2d power of 2.
.' ■ 2'= 8is the. 3d power of 2. '
2;* ac 16 is tb« 4th power of 2.
540* is the 4th power of 540, &c.
When two or more powers are multiplied together, their
product is that power whose index is the sum of the expo*
nents of the faaors or powers multiplied. Or the multipH*
cation of the ppmrs, answers to the addition of >tfaje indices.
Thus, in the following powers of 2,
1st
2d
3d
4th
5th
6th
7th
8th
9th
10th
£
4
8
16
32
64
123
256
512
1024
fflp ^'
«•
2^
£♦
25
2*
2'
2«
go
gto
Here^ 4 X 4 ±=: 16, and 2 + 2 5= 4 its index;
and 8 X 16 = 128, and 3 + 4 = 7 its index;
ako 16 X 64 = 1024, and 4 + 6 = 10 its index.
eTHfiR EXAMPLES. '
1. What IS the 2d power of 45 ? . Ans. 2025.
2. What is the square of 4' 16 ? Ans*^ 17*3056.
S. What is th« 3d power of 3*5 ? Ans. 42'875.
4, What is the 5th power of '029? Ans. '00000002051 1 149.
5, What is the square off ? Ans. ^.
6. What is the 3d power of 4? Ans. fj^.
7. What is tjtie 4th ppw^r of 4: ? Ans. ^Vy*
EVOLUTION.
80 ARITHMETIC.
EVOLUTIO^f.
EvoLUTiONi 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 number^ 29
being multiplied into itself a certain number of times, wul
produce that power. . Thus, 2 is the square root or 2d root
of 4, because 2' = 2 x 2=4; and 3 is the cube root of 3d
root of 27, because 3' = 3 x 3 x 3 = 27.
Any powei: of a gfiven number or root toay be found ex
actly, namely, by multiplying the number continually inta
itself. But there are many numbers of which a proposed root
can never be exactly found. Yet, by means of decimals, we
may approximate or approach towards the root, to any de
gree of exactness.
Those roots which only approximate, are called Surd
roots; but those which can be found quite exact, 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 cfqual to 2 :
also the cube root of 8 is rational^ being eqvial to 2 ; but the
cube root of 9 is surd or irrational.
Roots are sometimes denoted by writing the charstcter 4/
before the power, with the index of the root against it.
Thus, the 3d root of 20 is expressed by ^0 5 and the square
root or 2d root of it is v'20, the index 2 being always omit
ted, when oiily the square root is designed.
When the power is expressed by several numbers, with the
sign + or ~ between them, a line is drawn from the top of
the sign over all the parts of it: thus the third root of
45 12 is ^45  12, or thus ^(4512), inclosing the
numbers in p^entheses.
But all roots are now often designed like poweti, with
fractional indices : thus, the square root of 8 is 8^, th6 cube
root of 25 is 25"^, and the 4th root q{ 45 ^ 1« is 45 1,8 J*,
•r (45 .18)^.
TO
SQUARE ROOT. M
TO EXTRACT THE SQJIARE EOOT.
* DfViDB the given number into periods of two figuret
each, by setting k point over the place of units^ mother over
the place of hundredsi and so on, over every second figure^
both to the left hand in integers, and to the right in de^
cimals.
Find the greatest square in the first period on the lefthand,
and set its root on the righthand of the given number, after
the manner of a quotient figure in Division. .
■»•»•
* The reason for separating the figures of the dividend into
periods or portions of two places each^ is^ thdt the square of any
single figure never consists of more thui two places $ the square of .
a number of two figures, of not more than four places, and so i^
So that there will be as many figures in the root as the given nUin
ber contains periods so divided or parted off.
And the reason of the several steps in the operation appears
from the algebraic form of the square of any number of terms,
whether two or three or more. Tlius,
(a + b)^ = a» + 2ab + 6* =: a* + (2a + b) b, the squareof two
tffirms ; where it appears that a is the first term of the root, and b
the second term ) also a the first divisor, and the new divisor is
2a 4 ^> or denize the first term increased by the s^copd. And
hence the manner of extractipp is thvs :
1st divisor a) a^ ^ 2ab f 6* {a + b the root*
a*
2d divisor 2a f &  2a6 + b*
b\2ab + bf
Again, for a root of three parts, a, b, c, thus :
(a + 6 + c)* =; a* + 2a6 + 6® + 2tfc + 2bc + e« =
a^ + (2a + 6) 6 f (2a + 26 + c) c, the
squaie of three terms, where a is the first term of the root, b the
second, and c the third term ; also a the first divisor, 2a f ^ the
second, and 2a ^ 2b + c the third, each consisting of the double
fif the root increased by the next term of the same. And the mode
of extraction is thus :
1st divisor a) a* + 2a6 ^ 6^ + 2ac + 2bc + c'^ {a + b +c the root.
a?
2d divisor 2a + b
b
2ab + i*
2a6 + A»
N
3d divisor 2a + 26 + c I 2flc + 26c + c*
c I 2flc + 26c + c*
Vol. I. G Subtract
sr Aum^/^tic.
»%
• ^
Subtract the square dius found from the said period, and
to the remaindet annex tNi^>^two fTg^ures of the ntext following
pQriodffbrfidivi4^. '
* Doiri>le the rootabove aaeadoned fisr a divisor ; and.ftid^
how oiften k is contained in the said dividend^ exclusive of
its righthand figure ; and set that quotient figure botl)^ in
the quotient and divisor.
Multiply the whole augmented divisor by this last quotient
figure, and subtract the p]t)duct from the said dividend,
bnnging down to it the next period of the given number,
foraaew^Uvidend. .
Rppeat tb^ same procesi o^r agaln^ viz. find another new
dsmei^. by douUipg all the figures new hund in the root (
fixmi wludi, and the last dividend, find the next figure of
the root as before ; and so on through all die periods, to the
last.
Note^ The best way of doubling the root, to form the new
divisor's, is by adding the last figure always to the last divisor,^
as appears in the following examples. — Also, aftejr the figures
belo^ng to the given number are all exhausted, the operas,
tipn may be continued into decimals at pleasure, by adding
any number of periods of ciphers, two in each period,
EXAMPLES.
i. To find the square root of 29506624.
• • • •
29506624 ( 5433 the root.
25
104
4
450
416
1083
8
3466
3249
2
[21724
21724
Note, When the rooi is to he extracted to many places of figures ^
the work may be considerably shortenedy thus'/
* • . ' *
Having proceeded in the extraction after the common me
thod> till there be found half the required number of figures
•'in
S(^tTAR£ ROCyr.
»
in the robt^ or one figure more; tb^> Ibr ib(t fM, dii^e
the last remainder by its eori^e^pohdihg divl^rrafter the ittali'
aer of the thu*d contraction in Division of Deciihdisi tluis,
2. To find the root of 2 to nine placid of fi^M^
2 ( 1*4142195^ the tooti
1 • •
^ 100
4j 96
2ii
1
400
3^1
2824
4
11900
11296
28282
2
60400
56564
v.*
28284 }
3. What
4. What
5. What
6. What
7. What
i. WlKit
9. What
10. What
11. What
12. What
id the
is the
is (he
is the
is the
is the
is the
is the
is the
is the
square
square
square
square
square
sqtiarje
square
square
square
square
S8S6 ( 1856
1008
160
19
root of 2025 ?
root of 173056 ?
root of •000729 f
root of S ?
root of 5 ? .
root of 6 ?
root of 7 ? ^
root of 10 ?
root of II ?
root of 12?
Am. 45.
Ans. 4*16.
Ans. *027.
Ans. 1732050.
Ans. 2*236068.
Ans. 2*449489.
Ans. 2*645751.
Ans. 3162277.
Ans. 3*316624.
Ans. 3*464201.
RULES tOK TB£ s'qpARE ROaTS OF VUXOAR FRACTIONS
AND MIXED NUMBERS.
FitLsT prepare all vulgar fractions, by reducing theni to
their least terins^ both for this and all other roots. Then
1. Take tbe root of the numeratoi' and of the denominator
for the respective terms of the root required. . Aiid fKb is the
best way if the den6i£iiilator be a complete power: but if it
be not, then
2. Multiply the numeraitbr and denominator together;
take tlie root of th^ product : this root being zidade the nume^
^ G 2 rater
M
ARITHMETia
xatorto the denominator of the riven fraction, or mado tl^e
denominator to the numerator of it> will form the fractio^al
root required.
That is, v^y
f^a \/ah
s^b b s/ab'
And this rule will serve, whether the root be finite or infinite.
3^ Or reduc(e the vulgar fraction to a decimal, and extracf
its root,
4. Mixed numbers may be either reduced to improper
Actions, and extracted by the first or second rule» or the
vulgar fraction may be reduced to a decimal, then joined tp
the integer, ami the root of the whole extracted.
SAMPLES.
Ans.
4»
Ans. f .
!• What is the root of 4 ?
2. What i$ the root of^l
S. What is the root of A ? Ans. 0866025.
4. What is the root of T^j. ? Ans. 0645497^
5. What is the root of n ? ^ns. 4168333;
By means of the square root also may readily be found the
4th root, or the 8th root, or the l^th root, &C that is, the
root of any power whose index is some power of the numoer
2 ; namely, by extracting so often the square root as is de*
.noted by that power of 2; that is, two extractions fbr 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 as follows :
210358000
I
( 145037237 (120431407 the 4th root,
1
24
4
110
96
22
2
45
44
285
5
1435
1425
2404 I 10372
4 9616
■▼ .
29003
3
108000 24083
87009 3
75637
72249
20591(7237 3388 ( 140t
687 980
107 17
Jlx, 2. Wh^t is th? 4th root of 97^ ^ I
n
ROOT; §5
TO EXTRACT THE CUBE ROOT.
4
I. By the Common Ruk^i
. i. Having divided the given number into periods of three
£gures each, (by setting a point over the place of units, and
also over every third figure, from thence, to t&e left hand iri
y^hole numbers, and to the right iirxtedmals), find the nearest
less cube to the first period ^ set. its root in the quotient, and
subtract the said cube ftoffl^the fifsrperiod f to the remainder
bring down the second period, and call this the resolvend.
2. To three times the square of the root, jiist found, add
three times the root itself, setting this one place more to the
right than the former, and calltliis siiiri the divisor/ Thctn
divide the resolvend, wanting the last figure, by the divisort
for the next figure of the root, which annex to the former ;
calling this last figure #, and the part of the root before
found let be called a.
3. Add alltogether these three products, ns^ely, thrice
^uare multiplied by ii^ thrice a multiplied by e s()uare, and
^ cube, setting each of them one place more to the right than
the former, and call the sum the subtrahend ; which must
hot exceed the resolvend ^ bdt if it does, then lilake the last
£gure e lessj and repeat the operation for finding the subtra
hend> till it be less than the resolvend.
4. From the resolvend take the subtrahend, and to there
tnainder join the next period of the given number tdr s hew
resolvend ; to which form a new divisor from the wlfole rOot
'now found ; and froith ihence another figure of the root, as
directed in Article 2, and so oii.
^
♦ The reason, for pointing the givcaa number into periods of
three figures eacb> is because the cube of one' figure never amounts
io more than three pl^es. And, for a similar, reason, a given
number is pointed into periods of four figures for the 4th root> tif
£ve figures for the 5th root, and so on.
And the reason for theother parts of the rule depends on the
algebraic formatioad of a ci^be : for, if the root consist of the two
parts a { h, then its cube is as follows ; (a + 6)' = a' + 3a*6f
3aA* + b^ j where ei is the foot of the first part a' y the resolvend
is 3a*6 + 3«i* f J?, which is aUo.the same as the three parts of
{he subtrahend ; also the divisor Is 3a* + 3tf, by which dividing
the first two terms ©f the re^lvend 3o*^ + ao*, gives b for the
le^kond part of the root 3 and so on.
ixAMPLK.
tft
AWTHMET^,
J^XAMPLC
To extsaet tke cube root of 48328544.
S X S* sr 27
S X ? 5B. .09
Divis^ 279
48228*544 ( 3$ '4 rpot.
27
■WP*<PV«^
21228 resoWend.
^ly^'fnr f " ) ' '^'
S X S* X 6 =c 162
3x3 X 6* = 324 Vadd
6» = 216
3 X 36*. == 3888
3 X 36 sc 108
38988
19656 subtrahend
1572544 resolvend.
5' X 36^ X 4 =
3 X 36 X 'V =
4^ =
15552
nSSSadd
64
1572544 subtrahcn*.
0000000 remainder.
Ex. 2/ Extraa the cube root erf" 57 1482' 19»
Ex. 3. Extnct the cube root of 1628' 1582.
£x« 4. Extract the cube root of 1332.
* II. T* 49ctract tie Cube Root by a short ITay*.
t. Bj trials, c»r by the table of roots at p. 90, &c, t^ke.
the nearest rational cube to the given nunaber, whether it be
greater or less ; and call it the assumed cube.
2. Then
f •.
* The method usually giveb for extracting the cube root, is so
exeedingly t^edioos> and difficult to be remembered, that various
pth^r appro'xiiuiiting rules have heeh invented, viz. by Newton,
llaphson, H^lley^ De Lagny, Simpson, Emerson, and several other
mathematioiaDs ; but; ho one that I have yet seen, is so simple in
^ts form, or seems to wiell adapted for general use, as ^hat above
given. This rule is the same in effect as Dr. Bailey's, rational
formula.
hmtbm nai doaUe the assumsdrcube^ is to iht stm of the
assumed cube and ctouble the given number, so isthe lOOt of
the assumed cube, to the root required^ nearly^ Otf As the
first sum Is to the difierence of the given and assumed cube^
so is the assumed root to the difference of the root» nearly.
3. Again, by using, in like manner, the cube of the root
last found as a new asstiiTietrcube, another root wiQ be ob
tained still nearer.  And $o on as £u: as we please ; uring d
ways the cube^ of the last found root, for the assumed cube.
EXAMPLE.
To £nd<the ctdse root of 31035*8.
Here we soon find that the root lies between 20 and 30,
and then between 27 and 98. Taking th^erefore 27, its cube
is 19683, which is the assumed cube. Then
. 2 2
39366 4207 16'
21035*8 19683
»■ < ■■
As 604018 : .617S46 :: 27 : 27 '60471
27
f . k
4322822
1235092
604018) 16673742 (276047 th« root neadf.
4is;9338
284
42
formoki) but m or e co m med t o usl y eapiessed ' , tnd^ie firstintesti^
gation of it was given in my Traets, p. 49. ^ Tl^e algsbraic fom
of it is this :
As P H^ 2a : A «f 29 ; : r ; a. Or^ . . \
As B + aA : p fio; a : : r ; a^ v> r j ]
^bere r is the given nittqber, .i^ Ite as^otattd^ieafrilt cobc r ^Ire
^be root of a^ and a the root of ? sought.
Again,
88 ARTTHMEnC. .
Agam, for a secoad opdratlony the cube of tfaif rooit le
91035*318645155623^ and the process bj the ktter method
will be thus :
21035318(545, &c. '
• . 42070637!29a 210358
210358 . 21085318645, &c.
As 6310643729 : diff. 481355 ;: 27*6047 :
thediff. 0002 10560
codseq. the r<>ot req. is 27604910560.
Ex. 2. To extract the cube root of •67.
Ex. 3. To extract the cube root of *01.
TO EXTRACT ANY ROOT WHATEVER*.
Let p be the given power or number, n the index of t&e'
power, A the assumed power, r its root, r the required root
of p. Then say.
As the sum of if + 1 times a and /i — 1 times p,
is to the sum of » + 1 times ? and » — 1 times a^
so is the assumed root r, to the required root R.
Or, as half the said sum of ii f 1 times A, and »* 1 times
r, is to the difference between the given and assumed powers,
so is the assutned root r, to the difference bet'^een the true
and assumed roots ; which difference, added or subtraaed^ as
the case requires, gives the true root nearly,
' ■ ■ ■ ■■ ' ■ ■ ■■ ■ ■ ■ ii
Thatis, «+l. A + «— 1. P :/f+l'. P. +«— 1. a :: r :R.
Or, «+1.4a+«— I. y* : PcQ A :: r : Rco r.
And the operation may be repeated as often as we please,
by using always the last found root for the assumed root, and
its nth power for the assumed power A.
mmmimim'mmm^aKt^ttm
* This is a very general approximating rule« of which that fbc
the cube root is a particuk.r case^ and is the best adapted for
practice^ andfor meinorjr, of any that I have yet seen. It was first
disooveied in this form by myself^ and the investigation and use of
it.wefe^ven at large ia my Tacts, p. 45, Dec.
EXAMPLE.
d£N£RAL tOOrtS.
i&
fXAMPLE.
"to extract the 5th root of 21035*8.
(
Here it appears that the 5th root is between 7*3 and 7'4.
^Taking 7*3, its 5th power is 20130'7 1593. Hence we have
p = 210358, « = 5, r =7*3 and A = 2073071593; then
«+l. iA + «— U 4^ • ^ CO A :: r : rco r, that is,
3x3073011593+11x210358 : 305 084 :: 73 :
3 2 73
6219214719
420716
1P426574779
420716 915252
2135588
2227 11 32 (02 13605 = 11 cor
7'3 = r, add.
1. What
2; What
3. Wliat
4. Wifiat
5. What
6. What
7. What
8. What
9* What
lOi What
11. What
12. What
13. 'What
OTHfiH iXAMPLES.
s the 3d root of 2 ?
sthe 3d root of 321 4?
s the 4th root of 2?
s the 4th root of 91 '4ft ?
s the 5th root of 2 ?
s the 6th root of 210358 ?
s the 6 th root of 2?
s the 7th root of 210358 ?
s the 7th root of 2 ^
s the 8th root of 210358 ?
s the 8th root of 2 ?
s the 9th root of 2 1 0S5'*S ?
sth^' 9th root of 2?
7321360 = R, true
to the last figure^
Ans.
1259921.
Ans.
1475758.
Axis.
1189207.
Ans« 31415999.
A^s.
I148699.
Ans.
5254037.
Ans.
1122462.
Ans.
4145392
Ans.
1104089.
Ans.
3470323.
Ans.
1090508.
Ans.
3022239.
Ans.
l080059.f
•riT
llie following is a Table of squares and cubes, as also the
Square, roots aild cube roots, of ah numbers from 1 to 1000,
JKrhich will be found very useful on many occasions, in nu
meral calculations, when xoots or powers are concerned.
A TAtiS
fO A TABLE OF SQUAWJSb CUMS^ and ROOTS.
^.
Number.
Square.
Cube.
Square Roo
t. « Cube Root.
1
I
1
1*0000000
1000000
3
4
8
14142136
1259921
3
9
. 27
17320508
1442250
4
l6
64
2'C
iJi • 1 ?i
1*587401
5
25
125
^^ ^^ ^^ ^y ^r ^^ ^^ ^^
2*2360680
' 1*709976
6
36
%\6
24494897
1^17121
7
49
343
26457513
1912933 
s
64
. 512
2*82842^1
2K)octoob
9. .
81
729
. 3ooooonf]
2*Qfi0nB4
10
100
1000
3*1622777
2*i8.4?l35
11
121
1331
3*3166248
2*223986
ji
144
1728
3*444 r 016
2*289428
13
169
2i97
36055513
235i335
14
196
2744
37416574
2*410142
g^5
2f5
3375
.38729833
J , _ ^_. .^ _
2*466212
250
4096
4913
4C
i:i:i:i:t:i:«
2 510842
17
289
^ ^^ ■mm' ^^ ■^^ m^ ■m^ '^^
4*1231056
 '2*571282
18
324
5833
42426407
2*620741
19
361
6859
4*3588989
2668402
20
4CX)
8000
4'4721360
27144I8
21
441
9261
45825757
2758923
22
484
10648
46904158
2*802039
23
529
12167
4*7958815
2843867
24
;?76 .
13624
48989795
2884499
25'
625
156!15
50000000
2924OJ 8
26 .
^7^
1757I5
50990195
2962496
27
7'^9
19683
51961524
3000000*
28
784
21952
' 5*2915026
3*036589
29
841
24389
53851648
3072317
30
900
27000
5477^^6
I 3107232
31
961
29791
5*5077644
3^141381
32
1024 .
32768
5*6568542
> 3*174802
33 '
IO89
35937
6*7445626
3*207534 '
34
1156
39304
58909519
" 3239612
35
1225
42875
59160798
3*271066
36
1296
, 46656
6'OOOOOOCi
1 . 3*3019i27
^37
1369
50653
6082/625
3^332222
^8
1444
54872
61644140
► 3361975
^
1521
59319
^*:M49980
3*391211
40
1600
64000
63245553
3*419952
41
I68I
68921
64031242
3*448217
42
1764
74O88
64807407
3*476027
w 4a
1849
79507
6*5574a85
3*503398
44
19^6
85184
66332496
3*530348
45
2^
.911^
6*7082039
3556893
46
2116
97336
6*7823300
3*563048
47
220^
103823
6*8556546
3*608826
48
2304
110592
6*9282032
3*634241
49
2401
117649
70000000
\ 3*659306
50
2500
125000
7*0710678
3*684031
•
r.
M
ri ■
SQlTARtS, CUBES, AND ROOTS.
91
Tf umber.
Square.
51
52
53
*54
*55
56
57
58
,m
6k
Gl
63
64
65
66
07
68
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
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90
91
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94
95
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97
98
99
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9801
10000
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132651
140608
148877
157464
166375
175616
195193
195112
205379
216000
22^81
238328
250047
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300763
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531441
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Square Root. * Cube Bool. .
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7'1414284
7'21 11026
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73484692
7'4l6l985
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7\8iQ2497
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80000000
80622577
8*1240384
81853529
82462113
83066239
83666003
8*4261498
8*4852814
85440037
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88317609
8888 1944
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90000000
90553851
9*1104336
91651514
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9*2736185
93273791
9*3808315
943398 11
9*4868330
9*5398920
9*5916630
9*6436508
916953597
97467943
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98488578
98994949
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10*0000000
I
3708430
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3*7562^96
3779763
38O2953
3825862
3*843501
3*870877
3*892996
3*914867
3936497
3957892
3*979057
4000000
4020726
4*041240
4061548
4081656
4101566
4 12 1285
4*140818
4*l60l68
4179339
4198336
4*217163
4*235824
4254321
4272659
4*290841
4*308870
4326749
4*344481
4362071
4'379519
4*396830
4*414005
4431047
4447060
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4497942
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4578857
4*594701
4610436
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4641589
^
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s.
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ARITHMETIC.
N amber.
Square.
10201
101
102
10404
103
10609
104
10816
105
11025
106
11236
107
11449
108
11664
109
11881
110
12100
111
12321
112
12544
113
12769
114
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115
13225
116
13456
117
13699
118
13924
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120
14400
121
14641
122
14884
123
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. 124
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125
15625
126
15876
127
16129
128
16384
129
16641
130
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131
17161
132
17424
133
176S9
134
17956
135
18225
136
18496
137
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1
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139
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140
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141
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142
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143
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144
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145
21025
146
21316
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148
21904
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22201 .
150
22500
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1030301
.IO612O8
1092727
1 1 24864
1157625
1191016
1225043
1259712
1295029
1331000
1367631
1404928
1442897
1481544
1520875
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1601613
1643032
1685159
1 728000
1771561
1815848
I86O867
1 906624
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2000376
2048383
2097152
2146689
2197000
2248091
2299968
5K3 52637
2406104
2460375
2515456
2571353
2628072
2685619
2744000
2803221
2863288
2924207
2985984
3048625
3112136
3176523
3241792
3307949
3375000
Square Root.
00498756
0p99i049
01488916
O198O39O
02469508
02959301
03440S04
03923048
04403065
0*4880885
05356538
05830052
0*6301458
06770783
0*7238050
07703296
0*8166538
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09087121
0*9544512
rooooooo
10453610
1*0905365
1'1355287
1*1803399
1 224972a
1 2694277
13137085
1*3578167
14017543
14455231
1*4891253
1*5325626
157583^
16189500
16619038
1*7046999
1 7473444
1789^261
18321596
18743421
19163753
195 82607
20000000
2*0415946
2*0830460
21243557
21655251
22065556
22474487
Cube Root.
4657010
4672330
4687548
4702669
4717694
4*732624
4747459
4762203
4'77^56
4'791420
4*805896
4*820284 i
4834588
4*848808
4862944
4876999
4*890973
4*904668
4*918685
4*932424
4*946088
4959675
4973190
4*986631
5*000000
5013298
5026526
5039684
5052774
5*065797
5*073753
5091643
5104469
5*117230
5129928
5*142563
5155137
5^167649
5*160101
5*192494
5204828
5*217103
5229321
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5253588
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5277632
5289572
5301459
5313293
SQUARES, CUBES, akd ROOTS.
9$
1
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Square Root.
CubeKoot.
151
22801
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5325074
152
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153
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154
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156
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157
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158
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159
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165
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168
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182
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183
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5677411
184
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ARITHMETIC.
Numb.
Square.
Ciibfe.
Sqaare Root.
Cube Root.
201
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8120601
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202
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5962731
213
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6000000
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SQUARES, CUtJES, and ROOTS: 95
'^imlb.
Square.
Cube.
Square Koot.
Cube Root
251
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15813251
158429795
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253
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270
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27000000
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Square Root.
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301
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27270901
173493516
6*701758
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302
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27543606
17378 1472
6*709172
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9I8O9
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318
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lb 13835/1
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SQUARES," CUBES, and ROOTS.
97
Cu^elt
OQt.
Numb. \ Square.  Culie.
351
352
353
354
355
356
357
358
359
360
361
362
303
364
365
366
367
368
370
371
372
373
374
375
376
377
378
379
380
381
382/
383'
384
335
386
387
38S
389
390
391
393
393
394
395
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398
399
400
■ ■ ■ m tA
123201
123i,04
124609
125316
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128881
129600
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131044
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132496
133225
133P56
134689
135424
136161
136900
137641
138384
139I29
135876
140625
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142884
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144400
145 161
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147456
148225
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149769
150544
151321
152100
152881
153664
154449
15523*6
156025
156816
157609
158404
1.59201
160000
Square Root.
43243551
43614208
43986977
44361 864
44738875
45118016
45499293
45832712'
46268279
46656000
47045881
47437928
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ARITHMETIC.
* Numb, i b<\\\sire.
50I
502
503
504
505
5Q7
506
srp
510
511
512
513
514
515
Si\6
517
5m
519
520
521
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523
524
525:
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527
528
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291<XX)
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298II6
299209
30030f
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11 ittii •
25751501
26506008
27263527
26024064
28787625
29554216
30323843
31(^512
31b72229
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225610283
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SQUARES, CUBES, a^j> ROOTS. 101
Numbr.
.551
552
553
554
555
556
557
558
5 9
500
5b i
5(52
503
564
565
566
567
508
56g
57P
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573
574
575
576
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585
580*
587
688
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303601
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311721
315844
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321489
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3200.1 1
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328329
329476
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332929
331084
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33(>400
337561
338724
339889
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342225
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316921
348*00
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35(MC>4
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10*91 123/7
1704X^1464
170,51875
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172S08693
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183250432
184220009
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193103552
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1951i20JO
190*122941
197J37368
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199 70701
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102
ARITHMETIC.
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i\umb
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.
(101
603
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605
606
607
60S
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610
611
612
613
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36/236
368449
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370831
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374544
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105
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7^1
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602
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851
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SQUARES, CUBES, and ROOTS. 109

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9
110 ARITHMETIC.
Of ratios, proportions, and PROGRESSIOKS
Numbers arc compared to each other in two difFerent
>yay$: the one comparison considers the difference of the two
numbers, and is named Arithmetical Relation ; and the dif*
ference sometimes the Arithmetical Ratio 2 the other consi
ders their quotient, which is called Geometrical Relation ^
and the quotient is the Geometricat Ratio. So, of these two
numbers 6 and 3, the difference, or arithmetical ratio> is
6 — S or 3, but the geometrical ratio is y 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 which it is compared} the Conse^
quenl. So, in the two numbers above, 6 is the antecedentf
and 3 the consequent.
If two or more couplets of numbers have equal ratios, or
equal differences, 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, 8, are geometri
cal proportionals, because 4 ^ 4 ^ ^> ^^^ same ratio.
To denote numbers as being geometrically proportional a
colon is set between the terms of och couplet, to denote' their
ratio; and a double colon, or else a'mark of equality, betweeri
the couplets or ratios. So, the four proportionals, 4, 2, 6, 3
&re set thus, 4 : 2 : : 6 : 3, uhicli, means, that 4 is to 2 as 6
is to 3 ; or thns 4 : 2 3= 6 : 3, or thus, 4 = yy both
which mean, that the ratio, of 4 to 2, is equal to the ratio
of 6 to 3. .
Proportion is tUstinguished into Continued and Disconti*
nued. When ' the difference or ratio of the consequent of
one couplet, and the antecedent of (he ndxt couplet, is not the
' same as the common difference or ratio of the coufplets, the
proportion is discontinued. 5o, 4, 2, ?, 6 are in discontinued
arithmetical proportion^ because. 4 — 2=fc8 — 6 =2, where
as 8—2 = €; and'4, 2, 6, 3 are in. discontinued geometrical
proportion, because 4 « . a 2,«btit f =' 3, which is not
the same. ' *
But when tlie difference or ratio ,qf every two succeeding
terms is the same quantity, the proportioh i§ &ai4 to be Conti
nued, and the num^jers thenwelyes make a seriets p/Cpntinued
. '• \ . ;, /i?jc6pQrtionals,
ARITHMETICAL PROieORTION. ill
Proportionalsr oc a progressiox^t So 2, 4» 6, 3 form an arith*
metical progression, because 4 — 2 = 6—4 = 8—6 = 2, all
the same common difierencc ; and 2, 4, 8, i6 a geometrical
progression, because $ = J = y^ r= 2, all the. same ratio.
When the following terms of a pro^rression increase, oy
exceed each other, it is called an Ascending Progression, ot
Series ; but when the terms decrease, it is a descfjhding one.
Ko, 0, 1, 2, 3, 4, &c. is ^ja ascending arithmetical progisession^
but 9, 7, 5, 3, 1, &c. is a descending arithmetical progression.
Also 1 , 2, 4, 8, 1 6, &c. is an ascending geometrical propession,
and 16, 8, 4, 2j^ 1, $cc* is a de^^ding geometrical profession.
,. ■.. i^ I i gStt
ARITHMETICAL PROPORTJONj^iPROGRESSlON.
In Arithmetical P/ogressionj^ the niimbers or terms hate all
the same cbmmon difference. . Also, the first and last terms
of a Progression, are called tlieTIxtremes j ai;id the? otter
terms, lying between them, the Means. The mose useful
part of arithmetical proportions, is tonttiined in the follow*
ing theorems : , » • . ,. . '
TilUoREM 1. When four quahtities are in arithmetical
proportion, the suaa^of the two ^tremes is equal to the dum
of the two means. Thus, of the four % 4, d, 8, here 2 ^
8=4 + 6 = iOi
* f
Thkokesvt 2. In any continued arithmetical progression,
the som of the two exti%mes is equal to the sum of any two /
means that are equally distant from them, or tqual to double
the middle term when there is an uneven number of terms*
Thps, in the terms I, 3, 5, it is 1 + 5 = S + 3 ;= 6.,
And in the series 2, 4, 6, 8, 10, 12, 14, it is 2 + 14 = 4
+ 12 = 6+10 = 8 + 8=16.
TtoEoKiSM 3. Th^ difference between the extreme^ terms
p£ an arithmetical progression, is equal to the common dif
ference of the series multiplied by one less than the nuipber
of the terms. So, of the ten terms, 2, 4, 6, 8, 10, 12, 14,
J 6, IS, 20, tbet <fom»dn differtoce is 2, and one less thaa
tho number of terms 9 j then.tli^ifferenceof th^ extremes
is 202 = 18, and 2 X 9 = IS also. , '. i.
Consequently,
112 AftltHiMETlC
Consequently tW greatest ttrm is <^at t6 the least term
^dded to the product of the eommon 4»ereiiee multiplied hf
'1 less than the humber of terms.
Theorem 4. The sum of all the terms, of any arithiue'
♦ical progression, is equal to tht sum of the two eiCtreines mul
tiplied by the* number of terms, and divided by 2; or the sum
of the two iBxtrcmes multiplied by the number of the tcrms^
gives double the sum of all the termt in the series. '
This is made evident by setting the termi of the series in
tin inverted order, under the same series in a direct order, and
lidding the corresponding termi together in tliat order. Thus^
in the series I, 3, 5, 7, 9, n, l^t 15 1
ditto inverted 15, 13, !!» 9> % 5, ^8, 1 ;
the sums are 16+ 16 + 16 + 16+ 16 + 16 + J6 + 16,
which must be double the sum of the single series, and is
equal to the sum of the extremes repeated as often as are the
number of the terms. * ■
From these theoreftis may reacfilf be fcund any one of
these Qye parts ^ the two exlremos, the number of terms, th^
common difference, and the ium of all the ternis^ when any
three of them are given ; is in the following problems :
PROBLEM I.
Xiivmihe ExtfrmeSf antlthe Number ofTemi$t to Jmdthe Suth
of^dl the Terms ^
Add the extremes together, midtipty the sum by the num^
ber of terms, and divide by 2. . '
EXAJ^WS.
1. The extremes being .3 and Wt and the nuic^wr of
teriQs 9 \ required the sum of the terms ?
19 ' ^
22 .
Or, *— ; — X 9 = ~ X 9 = 1 1 )c 9 = 99,
2 ) 198 ** 1 *
' , ^ the same answti\
Ans. 99
 2. It Is reqiwred t© find the number of all the strokes a
common clock strikes in ^ifie whole irtvohition of the ind^x,
or in W hours? Ans. 78v
1)
ARITHIMETICAL PROGRESSION. US
(
lEx. S, How m^ny strokes do the clock* of Venice strike
in the compass of the dzj, which go continually on from I
to 24 o'clock ? . Ans. 300*
4. What debt can be discharged in a^ ye^, by weekly
payments in soithmetical progression, the first payment
being 1/^ and the last or 52d payment 51 Si/ AnSj 135/ 4/.
'^
PROBLEM It.
Given the Extremes y and the Number <kf Terms ; to jind the
Common Difference,
SuBTjiACT the less extreme from the greater, and divide
the remainder by I less than the number of terms; for th<
i^ommoh difference.
£XAMPL£2(.
• (
1. The extremes bein^ 3 and 19, and the number of terms
; required the tomimon difference ?
19
3 ^ 19 3 16 ,
^ 91 8
8) 16
Ans. 2
• /
2. If the extremes be 10 and 10, and the number of terftii
t\ \ what is the. common difierence, and the sum of the
series ? Ans. the com. diff. is 3, and the sum is 840.
3, A* certain debt can be discharged in one year, by weekly
payments in arithmetical progression, the first payment being
1 J, and the last 5/ 3/ i what is the cdmmcin difference of the
terms ? Ans« 2i
PROBLEM III. '
Xjiven one o£ the Extremes^ the Common Difference^ and the
Number of Terms : to find the other Extreme .^ and the Sum of
the Series.
Multiply the common difference by 1 less thati the num
ber of teriiis, and the product will be the difference of the
fextremes i Therefore add the produtt to the less extreme, to
give the greater ; ot subtract it from the greater, to give the
less extreme*
Vol. I. I EXAMPLES.
I
114' AfttTHMETlC
EXAMPLES.
1. Giv^n the fcasf term 3, the connnoit difference ^, of an
arithmetical series of 9 terms ^ to finxl the greatest term, and
the sam of the series:
8
16
19 the greatest tertii
3 the least
22 sum
9 number of terms.
2 ) 198
99 the sum of the series.
.2. If the greatest term be 70, the common difference if
and the number of terms 21, what is the least term, ai\d the
sum of ^he series ?
Ans. The least term is 10, and the sum is 840.
3. A debt can be discharged in a year, by paying I 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/ 3/, and the debt is 135/ 45.
PROBLEM IV.
*
To find an Arithmetical Mean Proportional between Two Givetr
Terms,
Add the two given extremes or term* together, and take
half their sum for the arithmetical mean required.
EXAMPLE.
To find an arithmetical mean between the two numbers 4
and 14. TT
Here
14
4
IT) 1»
Ans. 9 the' mean required.
;>v^ P160BLE5*
ARITHMETICAL PROGRESSION. 115
PROBLik V.
to find Two Arithmetical Means between two Given Extremes.
Subtract the less extreme from the greater, and divide
the dijflference by % so will the quotient be the common dif
ference ; which being continually added to the less extreme^
or taken from the greater, gives the means.
EXAMPLE.
To find two arithmetical means between 2 and 8.
Here 8 .
2 .
3)6 Then 242 = 4 the one mean;
and 4 + 2 = 6 th« other mean;
com. dif. 2
PROBLEM Vl.
t$ find any Number of Arithmetical Means between Tw Given,
Terms or Extremes.
Subtract the less extreme from the greater, and divide
the difference by 1 more than the ntlmber of means required
to be found, wblch will give the common difference ; then
this being added continually to the least term, or subtracted
from the greatest, will give the mean terms required.
EXAMPLE.
To find five arithmeticarmeans between 2 and 14.
Here 14 .
2
6)12 Then by adding this com. dif. continually,
the means, are found 4, 6, 8, 10, 12.
^dn. dif. 2
See more of Arithmetical progfesslbn in. the Algebra.
I 2 GFOMETRICAL
116 ARITHMETIC.
I
/
GEOMETRICAL PROPORTION *^ PROGRESSION,
In Geometrical Progression the numbers or terms have
all the same multiplier or divisor. The mpst useful part of
G^metrical Proportions is contained m the following^
theorems*
Theorem 1. ,When four quantities are in geometrical
proportion, the product of the two extremes is equal to the
product of the two means.
. Thus, in the four 2, 4, 3, 6, it is 2 x 6 = 3,x 4 = 12.
And hence, if the product of the two means be divided by
one of the extremes, the quotient will give the other extreme.
So, of the above numbers, the prodirct of the. means 12 f 3
= 6 the one extreme, and 12 r 6 =2 the other extreme \
and this is the foundation and reason of the practice m the
Rule of Three.
Theorem 2. In any continued geometrical progression,
the product of the two extremes is equal to the product of
any two means that are equally distant from them, or equal
to the square of the mididle term when there is an uneven
number of terms.
■
Thus, in the terms 2, 4, 8, it is 2 x 8 =c 4 x 4 = 16.
And in the series 2, 4, 8, 16, 32, €4, 128,
it is 2 X 128 = 4 X 64 = 8 X 32 = 16 X 16 = 256*
Theorem 3. The quotient of the extreme terms of si
geometrical progression, is eqtial to the common ratio of the
series raised to the power denoted by 1 less than thi number
of the terms. Consequently the greatesit term is equal tof
the least terra multiplied by the said quotient.
So, of the ten terms 2, 4, 8, \^^ 32, 64, 128, 256, 512j
1024, the common ratio is 2, and one less than the numbef
of ternis is 9 ; then the quotient of the extremes is 10^4 4
2 = 512, and 2^ = 512 also.
Theorem
• x: V
GEOMETRICAL PROGRESSION. 117
TThborem 4. The suih of all the terms, of any geome
trical progression, is found by adding the greatest term to th^
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,
1024 — 2
(whoseratiois2),isl024+— — — = 1024 + 1022 = 2046,
jL — 1
The foregoing, and several other properties of geometrical
proportion, are demonstrated more at large in the Algebraic
part of this work. A few examples may here be added of
the theorems, just delivered^ with some problems concerning
mean proportionala.
EXAMPLES.
!• 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 simi 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
V so 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.
t
V  '.
To find One Geemetrical Mean Proportional between any Tnv$
Numbers,
9
MuLTiPXT the two numbers together, and extract the
square root of the product, which will give the mean propor
tional sought. ^
EXAMPLE*
To find a geometrical mean between the two numbers
S and 12.'
12
3
36 (6 the mean.
36 PROBLEM
lia ARITHMETIC.
PROBLEM II.
To findr Two Geometrical ]Hean Proportionals iet^uneen any Ttu§
Numbers,
Divide the greater number b7 the less, and extract tho.
cube root of the quotient, which will give the common ratio
of the terms. Then multiply the least given term by th<^
ratio for the first mean, and this mean again by the ratio foi'
the second mean : or, divide the^ greater of th^ two given
terms by the ratio for the greater mean, and divide this agaiA
by the ratiq for the less mean.
E3CAMPLE.
' To find two geometrical means between 3 and 24f.
Here 3 ) 24« (8; its cub^ root 2 is the ratio.
Then 3 x 2 =: 6, and 6 x 2 =s 12, the two means.
Or 2* 7 2 = 12, and 12 — 2 = 6, the same.
That is, the two means between 3 apd 24, are 6 and 12.
PROBLEM III.
s
To find any Number of Geornetrical Mea^s between Two Numbers^
Divide the greater number by the less, and exta'act 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 3d
iTpot for two means, the 4th 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^
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,&6x2 = 12, &12x2 = 24, &24x2=48.
pr96^2=48,&48r2=24,&24T2=:12, &a272=6.
' That is, 6,12, 24, 4?8, are the four means between 3 and 96.
Of
MUSICAL PROPORTION. l»
Of MlTSICiVL PROPORTION.
There is also a third kind of proportion, called Musical^
which being but of little or no common use^ a very short ac
count of it may here suffice.
Musical Proportion is when, of three numbers, the first
has the same proportion to the third, as the difference between
the first and second, has to the diderence between the second
and third.
As in these three, 6, 8, 12 ;
where 6 : 12 :: 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 be
tween the first and second has to the difference between the
third and fourth*
As in these, 6, 8, 12, 18;
where 6 : 18 :: H ^^ i 18  12>
that is 6 : 18 :: 2 : 6.
 When numbers are in musical progression, their recipro
cals are in/ arithmetical progression ; and the converse, that
is, when numbers are in arithmetical progression, their reci^
f rocals are in musical progression.
So in these musicals 6, 8, 12, their reciprocals 4i h A»
are in arithmetical progressioo \ for 4 + A = A = T >
^i T + T^T'Ta ^^^^ i^> ^^^ sum of the extremes is
equal to double the mean, which is the property of arithme«
ticals.
The method of finding out numbers in musical propor
tion is best expressed by letters iji Algebra.
FELLOWSHIP, OR PAJITNERSHIP.
Fellowship is a 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 rul^ ar^ adjusted the gains or loss or charges of
partnei;5
t
12« ARrrHMETia
psurtners in company ; 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 de
tachments; or the division of waste lands among* a number
of proprietors.
Fellowship is either Single or Double. It is Single, when
the shares or portions are to be proportional each to one sin
gle given number only 5 as when the stocks of partners are
all employed for the same time : And Double; when each
portion is to be propoirtional to two or more numbers ; as
when the stocks of partners are employed for different times.
SINGLE FELLOWSHIP.
GENERAL RULE.
Add together the numbers that denote the proportion of
the shares. Then say,
As the sum of the said proportional numbers.
Is to the whole sum to be parted or divided,
So is each several proportional number.
To the corresponding share or part.
Or, as the whole stock, is to the whole gain or loss.
So is each man's particular stock,
To his particular share of the gain or loss.
To PROVE THE Work. Add all the shares or parts to^
gether, and the sum 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, a$
shall be in proportion to each other as the three numbers l^
2 and 3.
Here 1 + 2 + 3 = 6, the sum of the numbers.
Then, as 6 : 240 : : 1 : 40 the 1st part,
and as 6 : 240 : : 2 : 80 the 2d part,
' also as 6 : 240 :: 3 : 120 the 3d part.
Sum of all 240, the proof.
Ex. 2«
\.
SINGLE FELLOWSHfP. 121
Ex.'2. Three persons, a,b, c, freighted a ship with 540 tuns
of wine ; of which, a loaded 1 10 tuns, B 97, and c the rest :
in a storm the seamen were obliged to throw overboard 85
tuns I how much must each person sustain of the loss i
Here 110+ 97 = 207 tuns, loaded by A and b j^
theref." 340 — 207 = 133 tuns, loaded by c.
110
110 : 27i tuns = a's loss j
97 : 24^ tuns =: b's loss ; .
133 : 33^ tuns =: ds loss ;
Hence, as 340 : 85
or as 4 : 1
and as 4 : 1
ajiso as 4 : 1
Sum 85 tuns, th^ 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 merchants, e, f, g, make a stock of 700/, of
which E contributed 123/, F 358/, and G the rest : by trading
they gain lg5/ 10/ ; what must each have of it ?
Ans. E must have 22/ Is Od 2^^q.
F    64 S 8 Off.
G    39 5 3 l/y.
5. A General imposing a contribution * of 700/ on four
villages, to be paid in proportion to the number of inhabitants
contained in each j the 1st containing 250, the 2d 350, the
3d 400, and the 4th 500 persons ; what part must each vil
lage pay ? Ans. the 1st to pay 116/13/ 4rf.
the 2d^   163 6 8
the 3d   186 13 4
the 4th   233 6 8
6. A piece of ground, consisting of 37 ac 2ro i4<ps, is
to be divided among three persons, L, M, and n, in propor
tion to their estates : now if l's estate be worth 500/ a year,
lid's 320/, and n's 75/ ; what quantity of land must each one
have ? ^Ans. l must have 20 ac 3 ro 39 {4^ ps..
M   • IS 1 30,*^.
,N .   3 2344I.
7. A person is indebted to o 57/15/, to p 108/ 3/ 8i,
to Q 22/ 10 J, and to r 73/; but at his decease, his effects
. * Contribution is a tax paid by provinces, towDs^ villages,
^c. to excuse tbem from being plundered. It is pffid an provir
sio)^$ or in money^ and sometimes in both.
arc
123 ARITHMETIC
are found to. be^ worth no more that! 170/ 14s ; how must it
be divided among his creditors ?
Ans, o must have 37/ 15j Bd2^^^^,
P ... 70 15 2 ^T^^.
Q^  .  14 8 4. OAVtV
R    47 U 11 2t^V
Ex. 8. A ship, worth 900/, being entirely lost, ofwhich gber
longed to s, \ to T, and the rest to y ; what loss will each
sustain, supposing 540/ of her were Insured ?
Ans. s will lose 457, r 90/, and v 225/,
9. Four persons, w, x, y, and z, spent among them 25/,
and agree" that w shall pay i of it, x 4> y , and z 4 ; that
is, their shares are to be in proportion as j, ^, \y and ^ :
what are their shares ? Ans. w must pay 9/ 8</ S^f^,
• X  '.6 5 3ff.
T   4 10 If^.
Z   3 10 SyV
10. A detachment, consisting of 5 companies, being sent
into a 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. 5 4 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*.
BOUBLE FELLOWSHIP..
Double Fellowship, as has been said, is concerned in
cases in which th r stocks of partners are employed or conti
nued for different times.
* Questions of this nature frequentiy occurring in military
>9ervice« 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 ^
]^avilaod. ^
RVLE,
DOUBLE FELLOWSHIP. 123
• I
Rule*.— Multiply each person s stock by the time of its
continuance ; then divide the quantity, as in Single Fellow
ship, into shares, in proportion to these products^ by sayings
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,
T^p the correspondent share of the gain or loss.
EXAMPLES.
I. A had in company 50/ for 4 months, and b had 60/ for
5 months ; at the end of which time they find 24/ gained ;
bow must it be divided between them ?
Here 50 60
4 5
200 + 300 =: 500
Then, as 5C0 : 24 : : 200 : 91 zz 9/ 12j = 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 ; hpw much ought each man to pay of
the rent f Ans. c must pay 23/ 5s 9rf.
p must pay 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/ lOi 6d lr^g^.
F . . 11 16*10 Oj^.
G   12 12 7 2t%.
4. A ship's company take a prize of 1 000/, which they
agree to divide among them according to their pay and the
time they have been on board : now the officers and midship
men have been on board 6 months, and the sailors 3 months ;
s
* The proof of this rule is as follows : When the times arc
equal/the shares of the gain or loss ar^ evidently as the stocky, as
JD Single Fellowship ; and when the stocks are equal, the shares
are as the times; therefore, when neither are equal, the shares must
be as their products.
the
124 ARITHMETIC.
the officers have 40j a month, the midshipmen SOSf and the
sailors 22/ a month ; moreover there are 4 officers, 12 mid*
thipmen, and 110 sailors : what will each man's share be ?
Ans. each officer must have 2S/ 2s 5d O^^^.
each midshipman  17 6 9 3i7^«
each sieaman   6 7 2 Oj^,
Ek. 5. H, with a capital of 1000/, began trade the first of
January, andj meeting with success in business, took in i as a
partner, with a capital of 1500/, on the fii^t of March fol
lowing. I'hree months after that they admit K as a third
partner, who brought into stock 2800/. After trading toge
ther till the end o^^ the year, they find there has been gained
1776/ 10/ ; how must this be divided among the partners ?
Ans. H must have 457/ 9s A^d
J    571 16 sy
K    747 3 14*
6. X, Y, and z made a jointstock for 12 months; x at
first put in 20/, and 4 months after 20/ more ; t put in at
first SO/, at the end of 3 months he put in 20/ more, and 2
months afiier 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 S^q.
Y   22 8 1 0^.
z   16 13 4 0.
SIMPLE INTEREST.
Interest is the premium or sum allowed for the loan, or
forbearance of money. The money lept, or forkorn, is called
the Principal. And the sum of the principal and its interest,^
added together, is called the Amount. Interest is allowed
at so much per cent, per annum ; which premium per cent,
per annum, or interest of 100/ for a year, is called the rate of
interest ; — So,
Whe»
SIMPLE INTEREST.
IfS
When interest is at 3 per cent, the rate is 3;
   4 per cent.   4;
   5 per cent.   5;
   6 per cent.   6;
But, by lawj 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 forborn only, for the whole time of forbearance.
As the interest of any sum, for any time, is directly propor
tional to' the principal sum, and also to the time of continu
ance; hence arises the following general rule of calcula
tion.
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.
NoUf 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, to
divide by 100, is done by only pointing off two figures for
decimals.
EXAMPLES.
1. To find the interest of 230/ 10/, for 1 year, at the rate
«f 4 per cent, per annum.
Here, As 100 : 4 ; : 230/ lOi : 9/ 4/ 4rf.
4
100) 9,22
20
Ans. 9/ 4/ 4 J.
320
Ex.2,
126 ARITHMETIC.
Ex. 2. To find the interest of 54^/ 1 5sy for 3 years, at 5 per
cent, per annum.
As 100 : 5 :: 54775 :
Or 20 : 1 :: 54775 : 273875 interest for 1 year.
3
/ 321625 ditto for 3 years.
20
J 32500
12
\
d 3 00 Ans. 82/ Ss Zd.
3. To find the interest of 200 guineas, for 4 years 7 months
»nd 25 days, at 4Y*pcr cent, per annum.
ds / ds
210/ As 365 : 945 :: 25 : /
44 or 73 : 9*45 :: 5 : 6472
5
840
105 73 ) 4725 ( •64721
345
945 inteiest for 1 yr. 530
4 19
3780 ditto 4v years.
6 mo =5 4 4725 ditto 6 month,
1 mo = 1^ 7875 ditto 1 month.
6472 ditto 25 days*
/ 439597
20
s 191940
12
d 23280
4* Ans. 45/ ids 2^;
q 13120
4. To find the interest of 450/, for a year, at 5 per cent,
per annum. Ans. 22/ lOj.
5. To find the interest of 715/ i2s.6df for a year, at 4j.
per cent, per annum. Ans» 32/ 4/ O^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/ 15j for 4 years, at 4 per
cent, per annum. Ans. 561183 ^d.
,  Ex. 8.
. ' COMPOUND INTEREST. 121
Ex. 8. To find the interest of 32/ 5s Sd, for 7 years, at 4.
per cent, per annum. Ans. 94 \2s \d.
9. To find the interest of 170/, for I4 year, at 5 per cent.
per annum. Ans. 12/1 5j.
^ 10. To find the insurance on 205/ 15j, for J of a year, af
4 per cent, per annum. Ans. 2/ 1/ \\d,
^ 1 1 . To find the interest of 319/ 6//, for 5^ years, at 3 per
cent, per annuin. . Ans. 68/ I5s 9\d.
12* To find the insurance on 1 07/, fof 117 days, at 4 per
cent, per annum. Ans. 1/ 1 2s Id.
— IS* To find the interest of 17/ 5/, for 117 days, at ^ per
cent, per annum. , Ans. bs Sd.'
' r4. To find the insurance on 712/ 61, for 8 months, at
74 per cent, per annum. Ans. 35/ t2s S^d.
Ni^^m The Rules for Simple Interest, serve also to calcu
late Insurances, or the Purchase of Stocks, or any thing else
that is rated at so much per cent.
See abo more on the subject of Interest, with the algebraical
expression and investigation of the rules, at the ^nd of the
Algebra, next following. 1
COMPOUND INTEREST.
Compound I^nterest, called also Interest uppn Interest,
Is that which arises from the principal and interest, taken
together, tis it becomes due, at the end of each, stated time
of payment. Though it be not lawful to lend mOhey at
Compound Interest, yet in purchasing annuities, pensions,
or leases in reversion, it is usual to allow Compound Interest
to the purchaser for his ready money.
Rules. — 1. Find the amount of the given principal, for the
time of the first payment, by Simple Interest. Then con
sider this amount as a new principal for the second payment,
whose amount calculate as before. And so on through all
the payments to the last, always accounting the last amount
as a new principal for the next payment. The reason of
which is evident from the definition of Compound Interest*
Or elsCi
2. Find the amount of 1 ^ound for the time of the first
payment, and raise or involve it to the power whose index
is denoted by the number of paynients. Then that power
multiplied by the gi^cen principal, will produce: the whole
amount.
\
/
128 ARltHMETIC. ,
amount. From which the said principal' being ^ubti^actdil/
leaves the Compound Interest of the same. As is evident
from the first Rule.
EXAMPLES.
I. To find the amount of 720/, for 4 years, at 5 per cent,
per annum.
Here 5 is the 20th part bf^.lOO, aiid the inter^t of 1/fora
year is ^V or '05, and its amount 1*05. Therefore^
1 . By the ist Rule. 2. By the 2d Rule.
I s d r05 amount of l/«
20)720 1st yr*s princip. 105
36 1st yr's interest.
1 • 1025 2d pow^rof it*'
2Q) 736 2d yr's pr^ncip. 1'1025
37 16 2d yr's interest.
1 2 1 5506 25 4th pow. of it,
20) 793 16 3dyr*sprincip. 720
39 13 94: 3d yt's interest.
/ 8751645
20) 833 9 9^ 4th yr's princip. 20
41 13 5 4th yr's interest ,
^ ^ / 3 2900
£ 875 3 3. the whole amo^ 12
■ or ans. required. ■
^34800
2. To find the amount of 50/, in 5 years, at 5 per centj
per annum, compound interest. Ans. 63/ 16/ 3^rf.
3. To find the amount of 50/ in 5 years, or 10 half,
years, at 5 per cent, per annum, compound interest, the in
terest payable halfyearly. Ans. 64/ Os Id*
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/ 2/ 0^.
5. To find the compound interest of 370/ forbom for 6
years, at 4 per cent, per annum. Ans. 98/ 3/ 4^.
6. To find the compound interest of 410/ forborn for 24
years, at 4^ pei* cent, per annum, the interest payable half
yearly. Ans. 48/ 4j 1 \\d*
7. To find the amount, at compound interest, of 217/,
forbom for 2^ years, at ^5 per oent. per annum, the interest
payable quarterly. Ans. 242/ i 3j 44rf.
Nate. See the Rules for Compound Interest algebraically
investigated, at the end of the Algebra.
ALLIGATION*
ALLIGATION* ■ , 129
ALLIGATION.
1 \
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 ia com
monly distinguished into two cases, Alligation Medial, and
Alligation Alternate*.
ALLIGATION MEDIAL.
Alligation Medial is the method of find ing the rate
or quality of the composition, from havii^g. the quantities
and rates of qualities of the several simples given* And it
is thus performed : '
* Multiply the quantity of each ingredient by 'its rate or
quality; then add. all the produclts together, and add also all
r • ■('...
• ■ '   — ■■  . f • j ■•■ ' I I ■ ■ ■ , ■  ■ , J  ; jq 1 ^_iJi«lM^M
i ' '  * " '
* Demonstration* The Rhle is thus riroyed by Algebra.
Let flj bg c be the quantities of the ingredients,
arid wi, «, p their rates, or qualities> or prices 3
then amy bn, cp iare their severar values,
atid am + bn f cp the sum of their values,
also a f 6 f c is the sum of the quantities,
and if r denote the rate of the whole Compositioh^ .
then a + 6 + c X r will be the value of the whole,
conseq.a {■ b + c X r zz am ^ bn ^ cp,
and fit am + 4« + cp^a \' i Hc, which is the Rule.
• > ■
Note, If an ounce or any other quantity of pure gpld be reduced
into I?4 equal, parts, these parts are called Caracts; but gold is often
mixed with sonie base metal^ which is called the Alloy , and the
mixture. is said to be of so many caracts fine, according to the.
proportion of pure gold contained in it ; thus, if 22^ caracts of
pare golfd, and 2 of alloy be mixed together, it is said to be 22
caracts fine.
If any,one of the simples be of litde or no value with respect to
the rest, its rat6 is supposed to be nothing ; as water mixed with
wine, and alloy with gold and silver.
VolL K ^ the
ru
the quantities together into another sum ; then divide the
former sum by the latter, that is, the sum of the product*
by the sum of the qutotities, and: the quotient will be the
rate or quality of the composition required*^
EXAMPLES*
I. If three sorts of gunpowder be imxed togetbeis ^rz*
50lb 9t I2d z pound, 441b at 9d, and 261b at Sd a pound ^
how much a pound is the composition worth i
Here 50, 44, 26 are the quantities,
and 12, 9, 8 the rates or qualities ;
then 50 x 12 = 600
44,X 9 = S96
26 X 8=208
120 ) 1204 ( lOrJv = 10^
Ans. The rate or price is 10^^ the poimd.
^•2, A comporitioa bein^ made of 5lb of tea at 7s per Ib^
9lb.at 8r 6^ per Ib^ and. 14ilbat 5/ lOi/per lb; whacis»^
♦lb of it worth ? Ans. 6s lO^rfr
5» Mixe44gsfioa»ofwfiieat4j' ICtfpcr gatf, with"7 gal
k)ns at 5s Sd per, gall, and 9 gallons at 5s Si/pjer^gall^
what is a gallon of this composition worth ? * Ans. 5s 4^.
4. A meahnan would mix 3 bushels of flour Jat ^i 5d^
per bushel, 4 bushels at 5$ 6d per bushel, and 5 bushels at
4j8^per bu⪙ what is the: worth of. a bushel of this
mixture ? • > Ans. 4r iid^
5. A farmer nvixes 10 bushels of wkeat. at .5s the bushd,
Vith 18 bushels of rye at 3x the bushel, and 20 bushels of
barley at 2s per l>ushel:' how much is a bushtel of the mixture
worth ? Ans; di^
6. Having melted tog^b^r. 7 *oz of gold of 22caracts fine,
1240Z of 21 caracts fine, aild 17 oz of 19 caracts^fine : I
Would know the fineness of tbfe composition ?
Ansl 20ff caraets fine.
1, Of what fineness Is that con^osit ion, which is made by
tiiixing,3lb of silyer of 9 oz fine, ,with 5lh 8 oz pf lOoaSr.
fine, fUid> lib lOoz of aUay.. Ansi 7j;0Z» fine.'
ALUGATIOK
[ 131 ]
ALLIGATION ALTERNATE..
: AttiOATiaN Alternate is the method of finding what
quantity of any number of simples, whose rates are given,
will compose a mixture of a given rate. So that it is the re
verse of Alligation Medial, and may be proved by it.
RULE I*.
1. Set the rates of the simples in ^ column under each
Other.^2. Connect, or link with a continued liae, 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 com'
ppund ;* and. each greater rate with one or any number of the
le»rftr^* Write the difference between the mixture rate^and^
that of each of the simples^ opposite the rate with which they
ale Unked.*4. Then if only one difference atand ;^ainst
any rate, it will be the quantity belonging to that rate ; but
if there be seV.eral, their suni will be the. quantity. ,
 TUfe' examples may be, proved by the rule for AlHgation
MediaL
f Demonst, By connecting the less rate to the gteikter, atid
placing the difference between them and the rate alternately, tho
quantities resulting are such, that there is precisely as much
gtified by 6ne 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 Rale^
In like manner, whatever the number of siibples may be, and
with how ttiany soever every one is linked, since it is always a
less with a greater than the ^roean price, there will be an equal
balance of loss and gain between every two, and consequently an
equal balance on the whole.  a. e* o. «
.It is obvious, from this Rale» that questions of this sort admit of
a gneat variety of answers \ for, having found one answer, we may
£nd as many more as we please,, by only multiplying or dividing
each of the quaptities found, by 2, or 3, or 4, ^c: the reason of
which is evident : for^ if two quantities, of two simples, toake a
balance of loss and gain, with respect to the mean price, so must
also the double or treble, the 4 or  part, or any other ratio of these
quantities, and so on, 0(2 in^mYt/m.
These kinds of questions are called by algebraists indeterminate
or unlimited problems 3 and by an analytical process, theorems may
be raised that will give all the possible answers.
' K2 EXAMPLES.
X
133 ARITHMETIC
EXAMPLES.
1. A merchant would mix wines at 16/, at 18j, and at
' 22s per gallon, so as that the mixture may be worth 20/ the
gallon : what quantity of each must be taken i
/^16"^ 2 at 16/
Here20< 18x j2 at 18/ ^
i^22j/ 4 + 2 =: 6 at 22/.
\Ans. 2 gallons at 16/, 2 gallons at 18/, and 5 at 22x.
2. How much wine at 6s per gallon, and at 4s per gallon,^
must be mixed together, that the composition may be worth
Ss per gallon ? Ans. 1 qt, or 1 gall, &c.
3. .How much sugar at 4rf, at 6rf, and zt \ld per lb, mUst
be mixed together, so that the composition formed by them"
may be worth Id per lb ?
Ans. 1 lb, or 1 stone, or 1 cwt, or any other equal quantity
of each sort.
4. How much corn at 2x 6rf, 3/ 8rf, 4j, and 4/ 8 J per
bushel, must be mixed together, that the compound may be
worth ifif lOd per bushel ?
Ans. 2 at 2s 6d. 2 at 3/ 8 J, 3 at 4/, and 3 at 4/ Sd.
'I
5. A goldsmith has gold of 16, of 18, of 23, and of 24
caracts fine : how much must he take of each, to make it ,21
caracts fine ? Ans. 3 of 1 6, 2 of 1 8, 3 of 23, and 5 of 24.'
6. It is required to mix brandy at 12/, wine at 10/, cyder
at 1/, and water at per gallon together, so that the mixture
may be worth 8/ per gallon ?
Ans. 8 gals of brandy, 7 of wine, 2 of cyder, and 4 of water.
' ' * . . . «
RULE II. ^
* •)•
When the whole composition is limited to a certain
iquantity : Find an answer as before by linking ; th^en say, as
the sum of the quantities, or differences thus determined, is
to the given quantity ; so is each ingre4ient, found by link
ing, to the required quantity of each*
examples.
^ . How much gold of 1 5, 17, 18, and 22 caracts fine, must
be mixed together, to form a composition of 40 oz of 20 ca^
racts fiuie ? .
Here
ALLIGATION ALTERNATE. 133
 2
  2
 2
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 caracts fine, and 25 oz of
22 caracts fine*.
Ex. 2. A vintner has wine at 4/, at 5/, at 5s 6i, and at 6s
a gallon; and he would make a mixture of 18 gallons, so
that it might be afforded at 5s 4rf per gallon ; how much of
each sort must he take ?
Ans. 3 gal. at 4 J, 3 at 5j, 6 at 5s 6t/, and 6 at 6/.
* A great number of qtiestions might be here given relating to
the specific gravities of metals, &c. but one of the most curious
may here suffice.
Hiero, king of Syracuse, gave orders for a crown lobe made
entirely of pure gold ; but suspecting the workman had debased
it by mixing it with silver or copper, he recommended the dis
povery of the fraud to the famous 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 j and by putting each
Separately into a vessel full of water, the quantity of water expelled
by them determined their specific gravities 5 from which, andtheit
given weights, the exact quantities of gold and alloy in the crowa
may be determined.
Suppose the weight of each crown to be lOlb, and that the
water expelled by the copper or silver was 92lb, by the gold 52lb,
and by the compound crown 64lb ; what will be the quantities of
gold and alloy in th€ crown r
The rates of the simples are 92 and 5*2, and of the compound
04 5 therefore •
fii I 9'^"^ ^^ of copper
"■* I 52^ 28 of gold
And the sum of these is 12 + 28 =: 40, which should have been
but 105 therefore by the Rule, ^
40 : 10 :: 12 : 3 lb of copper 1 ,
: 7lbofgold /t^e answer.
40 : 10 : ; 28
RULE
IS* ARITHMETia
HOLE III*,
When one of the ingredients is limited to a certain quaii«
tity ; Take the difference 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 se^
verally, so is the quantity given, to the several quantities
required.
EXAMPLES*
1. How much wine at 5/, at 5s 6df and 6s the gallon,
must be mixed with 3 gallons at 4s per gallon, so tbat thp
multure may be worth Ss 4d per gallon ?
+ 2 = 10
Here 64
+ ^ = 10
+ .4 = 20
+ 4 = 20
3
3 : 6
3 : 6
Ans. 3 gallons at 5sy 6 at bs 6^, and 6 at 6x«
ft. A grocer would mix teas at 12j, 10/, and 6x per lb,
with $!01b at 4/ per lb • how much of each sort must he tako
to make the composition worth %s per lb ?
Ans. 201b at 4/, lOlb at ^x, lOlb at lOi, and 201b at 12/.
3. How much gold of 15, of l7, and of 22 caracts fine,
must be mixed with 5 oz of 18 caracts fine, so that the com
position may be 20 caracts fine ?
Ans. 5 oz. of 15 caracts fine, 5 oz of 17, and 25 of 22«
•♦*"
« In the very same manner questions may be wrought when sc*
veral of the ingredients are limited to certain quant ities, by finding
first for one limits and then for another. The two last Rules can
peed no demonstration^ as they evideody result from the firstj the
1^904 of wl^icli h^ been already explained*
fOSITl[0H«
SINGLE POSITION. 1«
POSITION
Position is a method of performing certain questibnSf
which cannot be resolved by the common direct rules. It h
sometimes called False Position, or False Supposition} because
it makes a supposition of false numbers^ to work with the
same as if they were the true ones, and by. tl^ir means dis
covers the true numbers sought. It is sometimes ako called
Trialand«£rror, because it proceeds by trials of false num
bers, and thence finds out the true ones by a comparisoa of
the /rrvr/.'— Position is either Single or Double.
SINGLE POSITION.
Single Position is that by which a questmi k resotved
by means of one supposition only. Questions which have
their result proportional to their suppositionSj^ belong to
jingle Position : such as those which require the multiplica
tion or division of the number sought by any proposed num*
ber ; or when it is to be increased or dimiotthed bj ifiself^
«r any parts of itself, a cevtain proposed nutnbe» ot times*
The rule i& as follows :
Takb or assume any nwnbeF fiur that whick is required,
sad perform, the some operatttons with it, as aM described or
performed in the question* Thensay, As the resiite cS the
said operation, is to the position, or number assumed^ so is
the result in the question, to a fourth term, which will be
the number sought*.
^1^
* The reason of this Rule is evident^ becanse it is auppoaad tbajt
the results are proportional to the sapposUions.
Thus, na I a i I nz I 2g
a z
or — : a : : — : z.
n n
a Q z z
or — ± — &c : fl ; : — ± — &c s z,
n tn u m
and so on.
EXAMP LES«
1S«
ARITHMETIC.
EXAMPLES.
1. A person after spending y and ^ of his money, has yet
remaining 60/; what had he at first ?
Suppose he had at first 120/. Proof.
Now ^ of 120 is 40
i of it is 30
•§.ofl4«4is 4S
i of 144 is 36
their sum is 70
rhich taken from 1 20
their sum 84
taken from 144
leaves 50
Then, 50 : . J20 : ; 60
leaves 60 as
144, the Answer. per question.
2. What number is that, which being multipied by 7, and
the product divide4 by 6^ the quotient may be 21 ? Ans. 18,
S. What number is that, which being increased by i^ f,
^nd ^ of itself, the sum shall be 15 i Ans. 36.
4. A general, after sending X)ut a foraging i and J of his
men, had yej; remaining 1000 : what number had he in com
mand ? Ans. 6000.
5. A gentleman distributed 52 p€n9e among a number of
ppor people, consisting of men, women, and children ; to
each man he gave 6^, to each woman 4^, • and to each child
Sd : moreover there were twice as many women as men, and
thiice as ipany children as women. How many were there
of each i . Ans. 2 men, 4 women,' and 12 children.
6. One being asked his 'age, said, if ^ of the years 1 hav^
lived, be multiplied by 7, and f of them be added to the
product, the sum will be 219. What was his age ?
Ais. 45 years.
>OUBl.E
I 137 ]
DOUBLE POSITION,
Double Position is the method of resolving certain
questions by means of two suppositions of false numbers.
To the Double R ule of Position belong such questions as
have'their results not proportional to' their positions : such are
those, in which the numbers souglit, or their parts, or* their
multiples, are increased or diminished by some given absolute
number, which is no known part of the number sought*
RULE I*.
Take or assume any two convenient numbers, and proceed
with each of them separately, according to the conditions of
the question, as in Single Position ; and find how much earJi
result is different from* the result mentioned in the question,
calling thes« differences the errors ^ noting also whether the
results are too great or too little.
* Demonstr,^ The RuKs is founded on tliis supposition, namely,
thai the first errqr is to the second, as 'the difference between the
true and first supposed number^ is to the diffbr^ce between the true
and second supposed number 5 when that is no( the case, the exact
answer to the question cannot be found by this Rule.— That thot
Rule is true, according to that supposition^ may be thus proved.
I^t a and b be the two suppositions, and a and b their results,
produced by similar operation \ also r and s their errors, or the
differences between the results a and b from the true result n ;
and let x depote the number sought, answering to the true result
If of the question.
Then is n — a = r, and n — b = *. And, according to the
supposition on which the Rule is founded, r : s :: x— a: x—bs%
hence, by multiplying extremes and means, rx — r6 = ^«— *a j
4hen, by transposition, rx — «x = rb — sa j and, by div«iioi^
flf g(i • *
X — r iz the number sought, which is the rule when the^
r — « .
results ^re both too little;.
If the results be both too great, so that a and b are both greater
than N 5 then n — • a z: — r, and n — bzz — *, orr and s are both
negative ; hence — r ; — $ : nx — a : x ^b, but — r : —4 : : + r
: f *, therefore c ; s : : ^p— a ; x ^"b; and the rest will be ex
actly as in the former case. ^
But if one result a only be too little, and the other b too great,
or one error r positive, and the other s negative, then the theoreiu
becomes x = — ^ , wbioh la the Rule in this case, or when
the errors are unl&e,
Theo
13S ARITHMETia
Then multiply each of the said errors by the contrary sup
position, namely, tlie first positiQn by the second .error, jmd
the second position by the first error. Then,
If the errors are aJike, divide the difference pf th^ products
by the difierence of the errors, and the quotient will be tl^e
answer.
But if the errors are unlike, divide the siun of tl^e produces
by jhe sum of th^ errors, for the answer,
Noicy The errors are said to be alike, when tb^y are either
both too great or both too little^ ^md unlike, ^\^^ we is too
great and the other too Uttle.
EXAMPLES.
1. What number is th^t, which being multiplied by 6^
%!t^ product increased by 18, and the sum divided by 9, the
quotient shall ^e 20 ?
Suppose the two numbers 18 and 30* Then,
First Position.
18 Suppose
6 mult*
iSecopd Position.
30.
6
Rroof.
27
108
1^
« 1
add
180
18
162
18
9) 126
div.
9) 198
9} 1^
14
20
results
tri^e res.
22
20
20, •
OA pos. SO
errors unlike ^ 2
mujt. ' 18
1st
pos.
Igr ( 2 180.
rors ( 6 36
«
surn. 8 ) 216
sum of products
27
<
Answer
* . #
sought.
RULE 11.
Find, by trial, two numbers, as near the true number as
convenient, and work with them as in the question ; mark
ing the errors which arise from each of them.
Multiply the difierence of the two numbers assumed,, or
found by trial, by one of the errors, and divide th^. product by
the difference of the errors, wlien they are alik;e, but by their
sum when they are unlike.
AAi the quotient, last found, to the number belonging to
the said error, when that number is too little, but; subtracjt
It
»•
DOUBLE PpSmON. . 18?
It when too greats and the result will give the true quantitf
jBOUght *#
examples;
4
1. So, the foregoing example, worked by this 2d rulet
will be as follows :
t: I.
30 positions 18; their dif. 12
^ 2 errors + 6 ; least error 2
« . 9
sum of errors 8 ) 24 ( 3 subtr.
from the position 30
leaves the answer ^7
Ex. 2. A son asking iis father how old he was, received
this answer: Your age is no^ onethird of mine; but 4
years ago, your age was only onefourth of mine.. What then
are their two ages ? , Aas. 15 and 4^,.
3, A workman was hii^ed for 20 days, at 3/ per day, for
every day he worked ; but with this condition, that for
every day he played, he should forfeit* 1^. Now it so hap
pened, that upon t^e whole he had 2/ 4x to receive. How
piany of the days did he work f 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 4 times as much as
A. What sum did each begin with ? Ans. lOO guineas.
5. Two persons^ A and B^ have both the same income^
A saves 4^ of his ; but b, by spending 50/ per annum more
than A, at the end of 4 years finds himself 100/ in debt*
What does each receive and spend per annum ?
Ans. They recei^ 125/ per annmn; also A spends 1004
and B spends 150/ per annum.
■••^
1
* for since, by the supposition, rtnix — a : x^b^ there*
fore by divisioni r— ^ : t ; ; *— « ; «— ft, which is the ad Rule,
"" PRACTICAI,
iLd
liO » * AfHTHMETIC.
PRACTICAL QUESTIONS in ARITHMETIC.
Quest. 1. The swiftest velocity of a cannonball, is
about 2000 feet in a second of time. Then in what time,
at that rate, would such a ball be in moving from the earth
to the sun, admitting the distance to be 100 millions of
miles, and tlie year to contain S65 days 6 hours ?
Ans. Sy^^s^ years.
Quest. 2. What is the. ratio of the velocity of light to
that of a cannonball, which issues from the gun with a ve
locity of 150O feet per second; light passing from the sun to
the earth in 14 minutes ? Ans. the ratio of 782222 to I •
Quest. 3. The slow or paradestep being 70 paces per
minute, at 28 inches each pace, it is required to determine
at what rate per hour that movement is ? Ans. Ixfl miles.
Quest. 4. The quicktime or stcp^ in marching, being
55 paces per second, or 120 Y^r minute, at 28 inches each ;
then at what rate per hour does a troop march on a route,
and how Jong will they be in arriving at a garrison 20 miles
distant, allowing a halt of one hour by the way to refresh?
A r the rate is 3rV miles an hour.
^^^' 1 and the time 7 hr, or 7 h 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 completed only 220 yards of
the wall. It is required then to determine how many men
must be added to the former, that the whole number of them
may just finish the wall in the time proposed, at the same
rate of working. Ans. i men to be added.
Quest. €. To determine how far 500 millions of gui
neas will reach, when laid down in. a straight line touching
one another ; supposing each guinea to bean 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, they begin to go
round it, both the same way, at the same instant of time ; a
goes at the rate of 1 1 yards per minute, and B 34 yards in
S minutes j the question is, how many times will the wood
be gone round before the quicker overtake the slower ? J^
Ans. 1 7 times.
Quest,
. PRACTICAL QUESXIONS. Ut
Quest. 8. a can do a piece of work alone in 12 days^
and B alone in 14"^ 'in what time wiil they btrth together perf
form a like quantity of work ? Ans. 6^ days* ,
. QupsT. 9. A person who was possessed of a ^.share of a
copper mhie, sold J of his interest, in it for \ 800/ v what was
the reputed value of the whole at the same rate? Ans. 4000/*
QifEsT. 10. A person after spending 20/ more than \ or
his yearly income, had then remaining' 30/ more than the .
half of it ; what was his jncome ? Ans. 200/.
Quest. 11. The hour and mjnute hand of a clock are
exactly together at 12 o'clock; when^are they next together?
Ans. at I7V hr> or 1 hr, 5^^ rainr
Quest. 1 2., If a gentleman whose annual income is 1300/y.
spend 20 guineas a week ; whether will he save or run in
debt, and how much in the year ? i 'I Ans. save 403/*
' Quest. 13, A person bought •! SlQ.or^nge^ at 2 a penny,
and 180 more at 3 a penny; after ,>viiich, ^selling them out
again at 5 for 2 pence, whether did he gain or lose by the
bargain? '' Ans. he lost 6 pence. .
Qubst. 14. If a qujintity of provislopis s^ves ,1500 men
12 weeks, at thje^.*ate of ?0' ounces a;4ay for each man ;> how
many men will the .same provisions maintain for 20 weeks, at
the rate of 8 ounces a day for each'm^n ? Ans. 225Q ttitxu
Quest. 15. In the latitude*of London, the distance round
the earth,'measured on. the parallel q£ latitude^ is about 15550
miles; now as the earth turns round in.23iours 56 nijfij^t;es,
at what rate per hour is the city of J^p;i<J(xp carried by.jthis
motion from west to east J A,ns< ^4pr' tpiles anl?i>u^»
Quest. 16. A father left his son a fdrtune, J of which he
ran through in 8 months': f of the remamder lasted him 12
months longer ; after which he had bare 820/ left. What
sundr did the father bequeath his son?' > Ans/ 1913/6/ 8//.
Quest. 17. If 1000 men, besieged in a town, with pro
visions for 5 weeks, allowing each man 16 ounces a day, be
reinforced with 500 men more ; and supposing that they can
not be relieved till th^ erd of 8 weeks, how many ounces a
day must each man have, that the provision may last that
time ? \ Ans. 6. ounces. •
Quest. 18. A younger brother received 8400/, which
was just  of his elder brother's fortune : What was the
father worth at his death ? , Ans. 19200/.
'  Quest.
U2 ARlTHMfetiC
. Qvi,sr. 19. . A person, looking on his j^stchj v^s. a^k^d
Vlis^t >yas the time of^th^ day, who ansv^ere^i.It. U.bp(Wjeea
5 and 6 ;. but a more particular answer being required, he
uad that the hoitr and minute hands were then exaqtly toge
ther : What was the time ? Ans. 271^ >^i°' P^^ ^*
.^ Quest. 20. If 20 men tin perform a piece of work in
lb days, how many men will accomplish another ti^rice as 
large in onefifth of the time? Ans. 300*
... . .....
Quest* 21, A father devised y^^ of his estate tp 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
5i4/ 6s 8d different: Then what money did he leave the
widow the use of? ' Ans. 1270/ 1/ 94^.
Quest., 22* A person, making his will, gave tp^ ouj? child .
i% of his estate, and the rest to another. When these legacies
caoie to be paid the one turned out 1 200/ more than the
other : What did the testator die worth ? Ans. ^OOO/*
.Quest. 23» Two persons, a and B, travel between
£ondon and Lincoln, distant 100 miles, a from London,
and B from Lincoln, at the same instant. ' After 7 hoUrs they
meet on the road, when it appeared that a had rode \i miles
an hour more than b. At what rate per hour, then did
each of the travellers ride ? Ans. a 74V^nd b 6^4 miles.
Quest. 24. Two. petrsons, A and b, travel betwGeti Lon
don and Exeter, a leaves Exeter at 8 o'clock in the morrj*
ing'^ and walks at t}ie r^te of S miles an hour,wtthout inter*
mission; and b sets out from London at 4 o'clock the same
evening, and walks for Exeter at the ritte of 4 miles an hopr
cpnsfantly. , Now,.. supposing the distance betwejen the two
cities to be 130 miles, whereaSouts on the road will they
meet ? ; Ans. 691 miles from Exeter.
• QuesTi,25. One hiindred eggs being placed on the
ground, in ^ straight line, at the distance of a yard from each
other : How far will a .perspn travel who shall bring them
ope by one to a basket, which, is placed at one, yard from the.
first egg ? Ans. lOlOO yards, or 5 miles and 1300 yds.
Quest. 26. The ' clocks, of • Italy gd on to 24 hours:
Then how many strokes do they strike in one complete re
volution of the index? Ans. 300.
Quest. 27. One Sessa, an Indiaiijij having ipyented the
game of chess, shewed it to his prince, who was so delighted
with
wftK iti tfekf he promised^ litnf aby iWstr'd Ire shbuld' isle ; on'
wHith Sessa requested that he might be allofwed one griiri'of^
wheat for the first square on the cjiess boards 2 for the secoxid^
4 for the third, and so on, d9ubiing contihualiy, to 64; the
whold number of squares. Now, supposing a pint to contain
7(180 of theje grains, and one quarter or 8 busheU to be worth
^Is 6dy it is required to compute the value of* all the corji ?
Ans. 6450468216285/ lis 3d 3j4J4^*
. Quest. 26* A' person increased his estate annually by
100/ more than the j part of it ; and zt the end of 4 year^
found'that'his estate amounted to 10342/ 3/ 9rf. What had
h6' at' first ? Ans: 46oO/.
Quest. 29.' Paid 1012/ lOx for a principal of 750/,'takeii'
irf^7 years before : at what rate per ceiit. p^r aftnum did I'
piy interest ? Ansf. 5 pet cent*
■ • '> .  . '•
Quest. 30. Divide 1 000/ among A, b,' c j so as to give
A 120 more, and b 95 less than o
Ans. A 445^ b 2S0, c ^25
QuEST. 3U A person l)eing asked the hour of the day,
said, the time past noon is equal tooths of the time till
midnight. What was the time r Ans. 20 min. past ^
QuEsT. 82. Suppose that I have^ /y of a ship worth
1^00/; what part of her hiave I left after selling  of ^ of
ftiy share, and what is it worth? Ans. ^^j worth 185A
Quest. 33. Fart ItOO acres of land among A, b, q ; so
fliat B may have 100 morfc thkn A, and c 64 more than b.
Ans. A 312, B 412, c 476.
Q0EST. 34i What number is that, from which if there
be takien j of , ■ and 'to the remainder be 'added ^j of j^,
thfc^^utn will be 1 ? Ans. 9
QCEsT; S3; There is a number which, if multiplied by ^
of 4 of li, will produce ,1 ; what is the square of that
numbejr? ' ' • , ^ Ans. l,2^.
Quest. 36. What leugth must be cut off a board, S^
inclies broad, to contain a square fobt, cr as much as 12
incBkes in length and 12 in breadth ? Ans. 1614 i^ches.
J Quest. 37. What sum of money will amount to 1 3S/ 2s
IBd, in 15 months, at 5 per cent, per annum simple interest i
Ans. 130/.
Quest. 3'8. A father divided his fortune among his three
sons, A, B, c, giving a 4 as often as b 3, and g 5 as often as
B 6 J
144 ARITHMETIC.
B 6 ; what neas the whole legacy, supposing ^*s share wsfc*
4000/. Ans. 9500/.
Quest. 39. A young hare starts 40 yards before a grey
houndy 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 1 8 : hovr
long will the' course hold, and^what ground wiU bcTrun ov^,
tounting from the outsetting of the dog?
Ans. 60^5y sec. and 530 yards run.
UE^T. 40. Two young gentlemen, without private forr
tiffle, obtain commissions at the same time, and at the age of
ISy One thoughtlessly spends 10/ a year more than his pay ;
butjfi 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 »^sures his life for the amount; each bond costs him. 30
shillings, besides the lawful interest of 5 per cent, and to in
sure liis life costs hhn 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, perse
veres, 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 ansfwerthe effect of com
pound mterest.
Suppose these two officers to meet at the age 'of 50, when
each recieives from Government 400/ per annum ; that the
one, Seeding his past errors, is resolved in future to spend no
more thto. he actually has, after paying the interest for what
he owes,«^d the insurance oh his life.
The oth^lk having now something before hand, means in
future, to spen4 his* full income, without increasiog his stock.
It is desirable to know how much each has to spend per
annum, and wliat money the latter has by him to assist the
distressed, or leaVM^ those who deserve it ?
Ans. The reformed officer has to spend 66l 19/ lf*5389i
per annum.
The prudent officer has to spend 437/ 1 2^ 1 l^4379i
per annum.
And the latter has saved, to dispose of,7 52/ 1 9/ 9' 1 896^.
END OF THE ARITHMETIC.
[ 14i' 1
OF LOGARITHMS^
L
rOG ARITHMS are ihade to facilitate troublesonie caIcH^
lations in numbers. This they do^ because they perform
multiplication by only addition, and division by only subtrac
tidn, and raising of powers by multiplying the logarithm by
the Index of the pbwer j and extracting of roots by dividing
the logarithm of the number by the index of the root. For,
lagarithms are numbers sd contri?ed and adapted to othei^
numbers, that the sums and difierences of the former shall
correspond to, and show, tlie products and <uoti^nts of th^
fatter, &c.
Or, niore generally, logarithms are the numerical ^tpp^
nents of ratios ; Or they are a series of numbers in arith^
metical
Mhata
* The invention of Logdritbchs is due to Lord Napier, Baron of
Merckistton, in Scotland, and is properly considered as one of the
inost useful inventions of modern times. A tabl^ of tbe<ie numben
was first published by the inventor at Edinburgh, iii the year
16 1 4, in a treatise entitled Canon Mirificum liOgnrtthmorum ; which
was eagerly received by all the learned thfotighoat Europe. Mr.
Henry Briggs, then professor of geometry at Qresham College^
soon after the discovery, went to visit the noble nventor ; aft»r
which, they jointly undertook the arduous task of computing new,
tables on this subject, and redacing them to a more cdnvenieht
form than that which was at first thought of. But Lord Napier
dying soon after, the whole burden fell upon Mr. Briggs, who,
with prodigious labour and great skill, made an entire Canon, 9.0
cording to the new form, for all numbers from i to 'iOOncs and
from 9 XX,0\o 10 100, to 14 places of figures, and published it at
London in the year 1(J24, in a treatise euiitled Arithmetica Loga<>
jtlthmica, with directions for supplying the intermediate parts.
Vol.1. L ' This
U6 LOGARITHMS.
metical progression, answering to another series of numbers
in geometrical progression.
Th /^* ^» ^* ^* ** ^* ^' Indices, or logarithms*
tl> 2, 4, 8, 16, 32, 64, Geometric progression*
^ rO, 1, 2, 3, 4, 5, 6, Indices, or logarithms*
ll, 3, 9, 27, 81, 243, 729, Geometric progression.
{0, 1, 2, 3, 4, 5, Indices, or logs.
1, 10, 100, 1000, 10000, 100000, Geom. progres.
Or
Where it is evident, that the same indices serve equally
for any geometric series ; and consequently there may be an
This Canon was again published in Holland by Adrian Vlacq,.
in the year l6U8, together with the Logarithms of all the numbers
which Mr. £riggs had omitted , but he contracted them down to
10 places of decimals. Mr. Briggs also computed the Logarithms
of the sines^ tangents, and secants, to every degree^ and centesm,
or 100th part of a degree^ of the whole quadrant ; and annexed
them to the natural sines, tangents, and secants, which he had
before computed, to fifteen places of figures. These Tables,
with their construction and use^ were first published in the year
1633, after Mr. Briggs's death, by Mr. Henry Gellibrand, under
the title of Trigonometria Britannica.
Benjamin Ursinus also gave a Table of Napier's Logs* and of
tines, to every 10 seconds. And Chr. Wolf, in his Mathematical
Lexicon, sajs that one Van Loser had computed them to every
single second, but his untimely death prevented their publica
tion. Many other authors have treated on this subject ;' but as
their numbers are frequently inaccurate and incommodiously dis«
posed, they are now generally neglected. The Tables in most
repute at present, are those of Gardiner in 4to, first published in
the year 1742 3 and my own Tables in 8vo, first printed in the
year i78^> where the Logarithms cf all numbers may be easily
found from 1 to lOCXXXXX) , and those of the sines^ tangents, and
secants, to any degree of accuracy required;
Also, Mr. Michael Taylor's Tables iu large 4to, containing the
common logarithms, and the logarithmic sines and tangents to
cVery second of the quadrant. And, in France, the new book of
logarithms by Calletj the 2d edition of which, in 1795, has the
tables still farther extended, and are printed with what are called
stereotypes, the types in each page being soldered together into a
solid ^ass or block.
Dodson's Antilogaritbroic Canon is likewise a very elaborate
work, and used for finding the numbers answer'uig to any givea
logarithm. '
endless
tOGARITHMS. 147
endless T^rkty of systems of Ic^arithms^ to the samecosi
mon numbersy by onlyt chaoaging the second term^ 2, 3/or
IG, &c.^ of the geometrical series of whole numbers; and
by interpolation the whoie system of numbers may be made
to enter: the geometric series, and receive^thieir proportkmal
logarithms, whether integers or decimals.
It is alao apparent, from the nature of these ^ries,, that if
any two indices be added together, their sum. will be the
index of that number which is equal to the product of the
two terms, in the geometric progression, to which those in
dices bdpng^ Thus, the indices 2 and S> beifig* zsMfii. to
gether, make 5*; and the numbers 4t and .8, on the termts
corresponding to those indiees) being multiplied togetherj
tnake 32, which is the number answering to. i^e itidex 5.
" ■ ' • *
In like manner, if any one index. b6 subtracted from
another, the difierence willbe the index of that number
which is equal to the quotient of the twa tenns to which
those indices belong. Thus, the index 6> minus the index
4, is = 2' ; and the terms corresponding to those indices are
64 and 16, whose quotient is = 4, which is the number
answering to the index 2*
For the same reason, if the logarithm of any number
be multiplied by the index of its power, the product will
be equal to t^^e logarithm of that powef. Thus, the
index or logarithm of 4, in the above series, i» 2; and
. if this number be nmltiplied by 3j the product will be
=s 6 ; which is the logarithm of 64^ or'Ae' third pDwer
of 4. •' . / ■ ' . .' 
.>'
And, if the Icgaritbm of any number bexliyided by the
index of its root,.' the qtiOtient. yrill b&. eqiial to' the logarithm
of that root. Thus, the iskdex . or logarithm of:.64 .is..6;
and if this number be divided by 2, the quotient will b<e
= 3; whiihris the logaritHm of 8, or the sqi^arerroot
of 64.
'  * « . <
The logarithms most convenient for jpriactice, are such as
are adapted" to a geometric series iricreasmg in a tenfold pro
portion, as in the last of the above forms *; and are those
which are to be found, at present, in most of the common
tables on this subject. The distinguishing mark of this
system of logarithms is, that the index or logarithm of 10
is 1 5 that of 100 is 2 5 that of 1000 is 3 5 &c* And, in
L t, decimals.
\ .
148 LOGARITHMB.
^•cimils, the logarithm of *1 b — 1 ; that of '01 i$^9i that
of OOl is— 3} Slc* The log. of 1 being in every lystem.
Whence it fellows, that the logarithm of any numbtr be
tween 1 and 10, must be and some fractioittl partem s^
that of a number between 10 and lOQ, wiU be 1 and sopie
fractional parts \ and so on, fer any other number whatever.
^d since the integral part of a logarithm, usually called the
Index, or Characteristic, b always thus readily feund, it is
commonly omitted in the tables ; being left to be suppiied
by the operator himself, as occasion requires.
Another Definition of Logaritiuns is, that the logarithm of
any number is the index of that power of some other nmn^
her, which is eaual to the nven number. So, if there be
N zs #*, then i» is the log. of M ; where n may be either pcK
sitive or negative, or nothing, and the root r any number
whatever, according to the different systems of k^sirithms.
When nis TzOf then N is ss 1, whatever the value of r is^
which shows, that the k)^. of 1 b always 0, in every systepi
of logarithms. When jiis ::= 1. thenN b =? r/ so thatthe
radix r b always that number whose log. b 1, in every
system* When the raduL ris s 2*7l8281828<b59 &<:, the
indices n are the hyperbolic or Napier's log* of the i^umtMurs 
N; so that n is always the hyp. log. of the number N or
(2718 &c.)^
« But when die ra£x r is s 10, thaa the index n becomes
the common or Briggs's log. of the number N: so that Ihe
common log. of any number 10" or N, b n the index of
thsit power of lO which is equal to the said number. Thus
100, being the second power of 10, will have 2 for its loga^
rithm i and 1000, being the third power of 10, will have 9
for its logarithm: hence also, if 50 be s: iO'^^'^^ then
b 1*69897 Ae cofdmon log. of 50. And, in^genentl, the
following docuple series of terms,
viz. 10*, lO', 10*, 10% 10*, 10^, 10*, 10"», lOr*,
or 10000, 1000, 100, 10, 1, 1, 01, OOl, OOOl,
have 4, 8, 2, 1, 0, —1, —2, — S, —4,
for their logarithms, respectively. And from this ^alf.of
numbers and logarithms, the same properties itosily follow; ^
as above mentioned.
I'ROBLSM*
LOGARITHMS. H9
PROBLEM.
To compute the Logarithm to any of the Natural Numbers
1, 2, 3, 4, 5, isfc.
RULE 1*.
Take the geometric series, 1, 10, IGO, 1000, 10000, &c.
And apiply to it the arithmetic series, 0^ 1, 2, 5, 4, &c. as
logarithms.— Find a geometric mean between 1 and 10, or
between 10 and 100, or any other two adjacent terms of the
series, between which the number proposed lies.^Ii^ like
manner^ between the mean, thus found, and the nearest ex
treme, find another geometrical mean ; and so on, tilli you
arrive within the proposed limit of the number whose loga^
rithm is sought. — ¥ind also as many arithmetical means, in
the same order as you found the geometrical pnes, and these
will be the logarithms answering to the said geometrical
means.
EXAMPLE.
Let it be required to find the logarithm of 9.
Here the proposed number lies between 1 and 10.
First, then, the log , of 10 il 1, and the log. of 1 is ;
theref. ^_+0 f 2 = J ae 5 is the arithmetical mean,
and v/ 10 X 1 = V 10 — 3*1622777 the geom. mean j
hence the log. of 31622777 is '5.
Secondly, tl ie log, o f 10 is 1, and the log. of 81622777 is 'B\
theref. 1 f '5 ^ 2 = '75 is the arithmetical mean,
and v/10 X 31622777 = 5*6234132 is the geom. mean J
hence the log. of 5*6234132 is '75.
Thirdly, the log, of 10 is 1, and the log. of 562341 32 is 75 ,
theref. 1 + '75 r 2 = 875 is the arithmetical mean^
and v^ 1 X 56235 1 32 =» 7 4989422 the geom. mean ;
hence the log. of 74989422 is 875.
Fourthly, tfa elog. of lO ils 1 , and the log. of 749*9422 is '875 i
theref. 1 f 875 4 2 =s 9375 is the arithmetical mean,
and V iO X 7*4989422 » 8*6596431 the geonr. mean
hence the log. of 8*6596431 is *9S75.
» *^M*4M4
* The reader who wishes to inform himself more particularly
coBceming the history, Bature, and constraction of Logarithms,
may oonsoll the introdnction to my Math^malical Tables, lately
published, where he will §nd his curiosity amply gratiled.
Fifthly,
150 LOGARITHMS.
Fifthly, the l og.oflOi sl, and the log. of S*659643 1 is 9375;
theref. 1 +'9375 ^g = '96875 is the arithmetical mean,
and v'lOx 865964'31 = 9*3057204 the geom. mean 5
hence the log. of 9*3057204 is 96875.
Sixthly, the log. of 86596431 is 9375, and the, log. of
9 30 57204 is '96875 ;
theref. '9375 + '96875 ^ 2 = '95 3125 isthearith.mean,
and /8'6596431 x 93057204 = 89768713 the geo
metric mean ;
hence the log. of 89768713 is '953125.
And proceeding in this manner, after 25 extractions, it
tri?l Ije fbimd^hat the logarithm of 89999998 is 9542425;
which may be taBen for the logarithm of 9, as it differs so
little from it, that it is sufficiently exact for all practical pur
poses. And in this manner were the logarithms of almost all
the prime numbers at first computed.
RULE II*. '
Let h be the number whose logarithm is required to be
found ; and a the number next less than ^, so that h'^azz.Xy
the logarithm of a being known \ and let s denote the sum
of the two numbers a ^ h. Then
, 1. Divide the constant decimal 8685889638 &c, by /,
and reserve the quotient : divide the reserved quotient by
the square of /, and reserve this quotient : divide this last
quotient also by the square of /, and again reserve the
quotient : and thus proceed, continually dividing the last
quotient by the square of /, as long as division can be
made.
2. Then write these quotients orderly under one another,
the first uppermost, and divide them repectively by the odd
numbers, 1, 3, 5, 7, 9, 8cc, as long as division can be made;
that is, divide the first reserved quotient by 1, the second by
S, the third by 5, the fourth by 7, and so on.
3. Add all these last quotients ^together, and the sum will
be the logarithm of h =. a; thereJFore to this logarithm add
also the given logarithm of the said next less number a, so
will the last sum be the logarithm of the number h proposed*
f For the demonstration of thi^ rule, s^ npy Mathematics
Tables, p. 109, &c.
/ ' ' That
LOGARtTHMS,
J51
. That IS,
/; 111
Log. oi b IS log. a'\jx {1+ "57+ ■57 + '77^ + ^^'^
where n denotes the constant given decimal '8685889638 &c.
EXAMPLES.
Ex. 1 . Let it be required to find the log. of the number 2,
Here the given number b is 2, and the next less number a
is 1, whose log. is 0; also the sum 2  1 = 3 zz /, and its
square ,s* = 9. Then the operation will be as follows :
3)
•868588964
1 )
•289529654 (
[ 28952965*
9)
•289529G54
3 )
32169962 (
; 10723321
9 )
32169962
5 )
3574440 (
[ 714888
9 )
3574440
7)
397160 (
: 56737
9)
397160
9)
44129 1
; 4903
9 )
44129
11 ).
4903 (
[ 446
9 )
4903
13 )
545 '(
; 42
9)
545
15 )
61 (
4
9)
61
1
»
"
log. of T
 301029995
add log. 1
log. of 2
 000000000
 301029995
Ex. 2. To compute the logarithm of the number 3.
Here 3 = 3, the next less number ^ = 2, and the sum
« + ^ sr 5 = J, whose square s^ is 25^ to divide by which,
always multiply by '04. Then the operation is as follows:
5
25
25
25
25
'^5
•868588964
173717793
6948712
277948
11118
445
18
1
3
.5
"7
9
11
173717793
6948712
271948
11118
445
. 18
log. of 4 
log. of 2 add
•173717793
2316237
55590
1588
50
2;
'176091^60
3Q1029995
log. of 3 sought 477121255
Then, because the sum of the logarithms of numbers,
gives the logarithm of their product ; and the difference of
^he logarithms, gives the logarithm of the quotient of the
numbers ;
15?
LOGARITHMS.
pumbers ; from the above two logarithms, and the logarithm
of 10, which is 1, we may raise a great many logarithms', as
in the following examples ;
EXAMPLE S.
Because 2x2 = 4, therefore
to log, 2  •30102!:'995
add log. 2 '301029995 J
t^mmrn^^ I. ■ ■ ■
sum is log. 4 6020599911
FXAMPLE 4.
Because 2 x 3 = 6, therefore
to log. 2  301029995
add log. 3. 477121255
sum is log. 6, '778151250
EXAMPt^ 5.
Because 2^ = 8, therefore
tnulu by 3
301029995^
EXAMPLE 6.
Becayse 3* = 9, therefore
log. 3  4771212541^
muk. by 2 2
gives log. 9 954242 09
EXAMPLE 7.
Because V* = 5, therefore
from log. 1 1 OOOOr^OOOO
take log. 2 3010299951
leaves log. 5 69897000411
EXAMPLE ^r
Because 3x4 = 12, thereferer
to log. 3  477121255
add log. 4 602055(991
gives log. 8 903089987 gives log. 12 10791 S124«
■Wi W
TT
And thus, computing, by this general mle, the logarithms
tb the other prim^ numbers, 7, U, 13, 17 19, 23, &c, and
then using composition and division, we may easily find as
many logarithms as we please, or may speedily examine any
logarithm in the taUe *•
f^^^i^^mr^
TT^
>^»»
* Th^te are, besides these, many other ingenious methods,
which later writers have discovered for ^ding the logarithms of
numbers, in a much tasiet way than by the ori^nal inventor ; bat>
as ^bey cpoodt ^ uiderstood wthpci a knowledge of some of the
higher branches of tie mathematics^ it is thought proper to omit
thefn» ana to r^fer the reader to those works which are written
expressly on the subject. It would likewise much exceed the
limits of this cbfnp^pdium, to point out all the peculiar artifices
that are made use of for constructing an entire table of these num
bers } but any information of this kind^ which the learner may
"ivtsh to obtain^ may be found in my Tables, before mentioned.
Descrtftiw,
LOOARlTHliMS. US
Vescripiim and Use rfthe Tablr j/^Logae^thms.
HavinC explained the manner of forming a table of the
logarithms of numbers, greater than unity; tne next thing to
be done is^ to show how the logarithms of fractional quan
tities may be found. In order to this^ it may be observed, '
that as in the former case a geometric series is supposed to
increase towards the lefty from unity, so in the latter case
it is supposed to decrease towards the right hand, still be
ginning with unit ; as exhibited in the general description,
page 148, where the indices being made negative, still show
the logarithms to which they belong. Whence it appear^^
that as h 1 is the log. of 10, so — I is the log. of Vrr or '1 >
and as + 2 is the log. of 100, so  2 is the log. of ^w or
^01 : and so on*
Hence it appears in general, th^t all numbers which con«
sist of the same figures, whether they be integral, or frac
tional, or mixed, will have the decimal parts of their loga*
rithms the same, but diSering only in the index^ which will
be more or Jess, and positive or negative, according to the
place of the first figure of the number.
Thus, the logarithm of 2651 being 3423410, the log. rf
tVj or i^y or twyj ^c, part of it 5 wil^ be as'fbllows :
Numbers.
Logarithms.
2 6 5 1
3 4 2 3 4 1
2 6 5i
2423410
2 651
1423410
2*6 5 1
0423410
•2651
1423410
•0 2 6 5 J
2423410
2 6$!
3 423410
Hence it abo appears, diat the ixKlex of any logarithm, i%
ftlways less by 1 than the number of integer figures which
the natural number consists of; or it is equal to ihe distance
of the first figure firom the f^e of iinit% ox first place of in
tegers, whether on the kft', cnr on the right, of it : and this
index is constantly to b^ placed on the lefthand side of the
decimal part of the logarithm^
When there are integers in the given number, the index
IS always afiirmative ; but when there are no integers, the
index is negative,' and b to be marked by a short line drawn
l^efore it, or else above it. Thus,
A number having 1, 2, 3, 4, 5, &c, integer places,
l;ie index of it& log. is 0, 1, % 3, 4 &c. or 1 less than those
* places.
^ And
15* LQGAmTHMS.
And a decimal fraction having its first figure in the
1st, 2d, Sd, '4th, &c, place of the decimals, has always
— 1, —2, —3, —4, &c, for the index of its logarithm,
, It may also be observed, that though the indices of frac
tional quantities are negative, yet the decimal parts of their
logarithms are always affirmative. And the negative mark( — )
may be set either before the index or over it.
1. TO nND, IN THE TABLE, THE LOGARITHM TO AKY
NUMBER*.
1. If the given Number be less than 100, or consist of
only two figures \ its log. is immediately found by inspection
in the first page of the table, which contains all numbers
from 1 to 100, with their logs, and the index immediately
annexed in the next column.
So the log. oi5 is 0*698970. The log. of 23 is V361728.
The log. of 50 is 1 698970, And so on.
2. y^ the Number be more than 100 but less than 10000;
that is, consisting of either three or four figures; the
decimal part, of the logarithm is found by inspection in the
other pages of the table, standing against the given number, in
this manner; viz. the first three figures of the given number in
the first column of the page, and the fourth figure one of those
along the top line of it ; then in the angle of meeting are the
last four figures of the logarithm, and the first two figures of the
same at the beginning of the same line in the second column
of the page : to which is to be prefixed the proper index,
which is always 1 less than the number of integer figures.
So the logarithm of 251 is 2*399674, that is, the decimal
'399674 found in the table, with the index 2 prefixed, be
cause the given number contains three integers. And the
log. of 34*09 is 1532627, that is^ the decimal 532627
found in the table, with the index 1 prefixed, because the
given number contains two integers. .
3. But if the given Number contain more than four figures^;
take out the logarithm of the first four figures by inspection
in the table, as before, as also the next greater logarithm^
subtracting the one logarithm from the other, as also their
corresponding numbers the one from the other. Then say»
As the difference between the two numbers.
Is to the difference of their logarithms,
So is the remaining part of the given number.
To the proportional part of the logarithm.
I I .1 1 ' I ■■■■■'■ JL I
* See ,tfce table pf Logarithms^ after the Geometry, at; th^ endi
of this volume.
* . Whigh
LOGARITHMS.
155
'Which part being addejd to th« less logarichm^ before taken
out, gives the whole logarithm sought very nearly.
EXAMPLE.
To find the logarithm of the number 34t*0926.
The log. of 340900, as before, is 532627.
And log. of 341000 *  is 532754.
Thediffs.^are 100 and . 127
Then, as 100 : 127 : : 26 : 33, the proportional part.
. This added to •   332627, the first log.
Gives, witii the index, 1 •532(360 for the log. of 34'0926.
4. If the number consist both of integers and fractions, or
is entirely fractional ; find the decimal part of the logarithm
the same as if all its figures were integral ; then this, having
prefixed to it the proper index, will give the logarithm re
quired.
5. And if the given number be a proper vulgar fraction :
subtract the logarithm of the denominator from the loga,
rithm of the numerator, and the remainder will be the loga
rithm sought ; which, being that of a decimal fraction, must
always have a negative index.
6. But if it be a mixed number ; reduce it to an improper
fraction, and find the difference of the logarithms of the
numerator and denominator, in the same manner as before.
EXAMPLES.
l.Tofindthe.log. off J,
Log. of 37  1568202
Log. of 94  1*973 123
Dif. logl of f.  1595074
Where the index 1 isnegative.
2. To find the log. of 17.
First, n4iJ=\V Then,
Log, of 405
Log. of 23
Dif. log. of 17i^
2607455
1361728
1245727
II, TO FIND THE NATURAL NUMBER TO ANT GIVEN
LOGARITHM.
This is to be found in the tables by the reverse method
to the former, namely, by searching for the proposed loga
riit]im among those in the table, and taking out the corre
sponding number by inspection, in which the proper number
of integers are to be pointed off^, viz. 1 more than the
index. For, in finding the number answering to any given
logarithm, the index always shows how far the first figure
must
156 logarithms/
icnst be removed from the place of units, viz. to the Ifcfi;
hand, or" integers, v^hcn tlxe index is affirmative; but to the'
right hand, or decimals, when it is negative.
EXAMPLES*
So, the number to the log. P532888 » 34*1K
And the number of the log. 1*5S2882 is *34i 1.
But if the logarithm cannot be exactly found in the table ^
take out the next greater and the next less, subtracting the
one of these logarithms froiii the dther, as also their natural
numbers the one from the other, and the less logarithm from
the logarithm proposed. Then say>
As the difference of the first or tabular logarithms^
Is to the difference of their natural numbers.
So is the differ, of the given log. and the least tabular log.
To their corresponding numeral difference.
Which being annexed to the least natural number above
taken, gives the natural number sought, corresponding to
the proposed logarithm^
EXAMPLfi*
So, to find the natural number answermg to the gives
logarithm 1532708.
Here the next greater and next less tabular logarithms,
with their corresponding numbers, are as below :
Next greater 532754 its num. 34iOOO; given log. 532108
Next less ,532627 its num. 340900; next less 532627
Differences 127 — 100 — 81
Then, as 127 : 100 :: 81 : 64 nearly, the numeral difiTer.
Therefore 34*0964 is the number sought, marking oflTtwo
integers, because the index of the given logarithm is 1.
Had the index been negative, thus 1 532708, its ccmtc
sponding number would hxvt been *S4Q064, whoUy d«
cimaL
MWLTIPLI
£ 157 }
MULTIPLICATION, bt LOGARITHMS*
Ta^e out the logarithms of tlie j&ctorfi from the table^
then add them together, and their sum will be the logarithm
of the product required. Then, bj^ means of the table,
take out the natural number, answering to the sum, for the
product sought.
Ob^i^ving to add what is to be carried from the ijecimal
part of the logarithm to the affirmative index or indices, or
else subtract it from the negative.
* Also* adding the indices together yrhen they are of the
samfs kind, bom affirmative or both negative ; but subtract^
ing the less, from the greater, when uie one is affirmative
gnd the other negative, >nd prefixing the sign of the greater
to the remainder.
EXAMPLES*
. 1. To multiply 2314 by
5062. .
Numbers. Logs.
23.14  1'364.363
5062  0704322
Product 1 17*1347 2*068685
2. To multiply ^531 92^
by 3457291.
Numbers. Logs*
2581926  041 1^44
3457291  0538736
Prod. 892648  09506W
X To mult. 3902 and 59716
^4 '0314728 all together.
Numbers. Logs.
3902  p591287
 59716  2776091
•03 14728 2497935
>ro4. 7S3333  1865318
9
Here the — 2 cancels the 2,
and the 1 to carry from the
di^cimals is set down.
4.To mult.3*586, and21046,
and 08372, and 029.4ali
together.
Numbers. Logs.
3586 . 0'55461Q
21046  0323170
08372 1922829
00294 — 2468347
Prod. 0'1857618l268956
Here the 2 to carry cancels
the— 2, and there remains thir
I — 1 to set dawn .
DITISIOK
r 158 ]
DIVISION BY LOGARrrHMS.
RULE.
• * «
From the logarithm of the dividend subtract the loga
rithm of the divisor, and the number answering to the re
mainder will be the quotient required.
Observing to change the sign of the index of the divisor,
from affirmative to negative^ or from negative to affirmative^
then take the sum of the indices if they be of the same name,
cr their difference when of different signs, with the sign of
the greater, for the index to the logarithm of the quotient.
And also, when 1 is borrowed, in the lefthand place of
the decimal part of the logarithm, add it to the index of
the divisor when that index is affirmative, but subtract it
when negative ; then let the sign of the index arising from
hence be changed, and worked with as before.
EXAMPLES.
I. Tt> divide 24163 by 4567.
Numbers. Logs.
Dividend 24163  4383151
Divisor 4567  3659631
Quot. 5*29075 0723520
3. Divide 06314 by 007241.
Numbers. . Logs.
Divid. 06314 2800.'^05
Divisor 007241 —3859799
2.Tpdivide37149by5237e.
Numbers. Logs.
Dividend 37 1 49  1569947
Divisor. 52376  2719132
Quot. 871979 0940506
Herie 1 carried from the
Quot. 0709275 28508 15
4.Todivide7438byl29476. '
Numbers. Logs.
Divid. 7438 1871456 '
Divisor 129476 1112189
Quot. 057447  2759267
Here the 1 taken from •the
decimals to th6 —3, makes it — 1, makes it become —2, to
becopie— 2,whichtakenfroi^ set down,
the other — 2, . leaves re '
maming.
,' Note. As to the KuleofThree, or Rule of Proportion,
it is performed by adding fhe logarithms of the 2d and Sd
tcims, and subtracting that of the first term from their sum.
INVOLUTION
( 159 ]
mVOLUTION BT LOGARITHMS.
RtTLE. ' • . • '"
Ta«e out the logarithm of the given number from the
table. Multiply the log. thus found, by the index of the
power proposed. Find the number answering to the pro
duct, and it will be the power required.
Ifdie* In multiplying a logarithm with a negative index,
by an affirmative number, the product will be negative.
But what is to be carried from the decimal part of the loga^
rithtn,.will always be affirmative. And therefore their dif
ference will be the index of the product, and is always to be
made of the same kind with the greater. ^
EXAMPLES.
1. To square the number
.25791.
Numb. I'Og.
Root 2'5791   0411468
The index   2
Power 665174 0822936
2. To find the cube of
307146.
Numb. Log.
Root 307146   0*487345
The index   S
Power 289758 1 462035
8. To raise 09163 to the 4th
power.
Numb. Log.
Root 09163 —2962038
The index   4
Pow. 000070494  5 8^8 152
Here 4 times the negative
index being — 8,and 3 to carry,
the difference — 5 is thp index
of the product.
4. To raise 1 0045 to the
365th power.
Numb. Log.
Root 10045   0001950
The index   365
9750
11700
5850
Power 514932 0*711750
EVOLUTIOK
[ 160 i
'ZVOLXmON IT LOCillUtllMSl.
Take the log: of the fiv^ pmmber out of the tabled.
Divide the log. thus round by the index of the root. Thent
dur number antveriag to the quotientj iTillbf the root.
Hoii, When the index of the logsurithmj to be divided, W
negative, and does not exactly contain the divisor, without
some remainder, increase the index by such a number as will
make it exactly divisible by the index, carrying the units bor*
rowed, as so many teni> to the lefthand place of the decimalf
and then divide as in whole numbers*
Ex. 1 .To find the square root
of 365.
Numb. Log.
Power 365 2)2*562293
Root 1910496 128114.64
"Ex. 3. To find the 10th root
of 2.
Numb. Log.
Power 2  10 ) 0*301030
Root 1071173 0030103
Ex. 5. To find V' 093.
Numb. Log.
Power 093 t )  29684'»3
Root 304^959 ~ 14842414
Here the divisor 2 is con
tained exactly once in the ne
gative index —2, and there
Fore the index of the quotient
is —I.
Exi 2. To find the 8d foot of
12345. .
Numb. Log*
Power 12345 3)4091491
Rpot 231116 1363830J.
Ex. 4. To find the 365th root
of 1045.
Numb. Log*
Powerl'045 365)0^019116
Root 1000121 00000524
Ex. 6. To find the ^00048,
Numb. Log.
Power '00048 3)^4'681241
RcfQt 0782973  2893747
Here thedirisor 3, not beipjBT exact
ly contained lA — ^, it is augmented
by 2, to malw up 6, in which the di*
visor it contained just 2 times; then
the 2, thus bon;owed, being ci^rriedt^
the decimal figure (>, makes 2^f which
divided by d, ^ives 8, &c.
Ex. 7. To find 31416 x 82 x fj.
Ex. 8. To find 02916 x 7513 X ^
Ex. 9. As 7241 : 3*58 :: 2046 : ?
Ex. 10. As v'724 : v^4 : : 6927 : ?
ALGEBRA.
[ i«i J
■•;• •
A L G E B R A;
DEFINITIONS AND NOTATION.
1. xjLLCxEBRA i^ the scieoce of computing by sy^nhplju
It Is sometimes also called Analysis ; and is a general kind
of arithlnetic, or universal way of computation.
2. In this science, quantiti^^ of all kindaar^ re^es^ented by
the letters of the alphabet. Ahd tbeo(}^rations' to be per*
^ibrmed with them, as addition or sjbibtractiDQ^ &C9 aiie ile*
noted by certain simple character^ instead; of being ^i^e^s^d
hy words at^ength. ._.....
3. In algebraical questions, s6mt quantities^ are kh^w^ oit'
given, viz. those whose values are known: and othersun
known, or are to be found otit, vils. thos^ whos^ values Ve
not known. The fonrter of these are represented by the
leading letters of the alphabet, a, b^ r, dy &c \ and the lattef »
or unknoi^ quantities, by the final lettei^, z> j^ at, u^ &<f.y
4>. The fiiaracters lis^sd to denote the opiorations^ av^
chiefly the JFoUowing : ' 1.
+ signifies addition, and is named //wi** "
^ signifies subtraction, and is named minus. ,
X or . signifies multiplication, and is named into^ . j
i signifies division, and is named by*
•/ signifies the square root ; ^ the cube root s ^ thi
4th root, &c i and ^ the /7th root.
: : : : signifies proportion. • .
ss signifies equality, and is named 9^iud iff.
And so on for other operations.
Thus a^b denotes that . the number tepresented by.^ is
to be added to that represented by tfk
a — b denotes, that the number represented by j is to; be
subtracted from that repiiresented by ii.
aKn b denotes the difference pf a ahd ^, when it, id not
known which is the greater. 
Vot. I. M n^, oli*
162 ALGEBRA.
aBf or a X if or a.i^ expresses the product^ by multipli
cation^ of the numbers represented by a and i.
# r i^, orT, denotes, that the number represented by #
is to be divided by that which is expressed by i.
a : b :: e : dy signifies that tf is in the same proportion to t,
as r is to d»
X =^ a ^ b i^ CIS 7kn equation, expressing that x is equal
to the diiSerence of a and by added to the quantity r.
^a, or M^, denotes the square root of « j ^a, or ^i^, the
cube root of ^ ; and ^^a^ or a^ the cube root of the square of « ;
zho ^y or sT^f is the Mth root oia\ and ^a* or tf« is the
iith power of the mth root of a, or it is a to the ~ power.
d'^ denote* the square of as n^ the cube of ^ ; o^ the fotuth
p0wer oi a.: and if^ the Mth power of a,
d + ^ >^ r, or (rt 4 i^) <•, denoted the product of the compound
^uc^tit]^ n^h multiplied^ by the ^i^le quantity r. U»ng
the bar , or the parenthesis ( ) as a vi^culum^ to codtieGt
•Several aimjple quantities into one compound.
/ «.+ ira — ^*or7 — y, expressed like a fraction, means
flftft quotient oi<jL\ it.divided by a^b.
i/a^^^cd^ rov {fib^ + 4ri)^9 is the square root of the com*
I. ,1. I
poixn4 quantity /l^4«r^. And^v^^ + ^^f or r («^ + c£ff
denotes the product'of c into the square r<>ot of the coxnpocUid
quantity ^3 + r^. .,. •
fl H ^ — r , or (^j + ^ ^ f)', denotes the cube, or third
power, of the compound quantity a f ^ — r.
3tf denotes that the quantity a IS' to be taken 3 times, and
4 (o 4 ^) i^ 4* timed ^i + ^ And these numbers, 3 or 4y
showing how often the quantities are td be taken, or multi
plied, are called Coefficients. • .
Also \x denotes that x is multiplcd by ^ ; thus f x * or
 5. Like Quantities, are those which consist of the same
letters, and powers. As a and 3a ; ch* 2db and ^ab ; or
3flf*Ar and r bs^bc^
6. Unlike Quantities, are those which consist of different
letters, or dffierent powers. As a and ^^ or 2tf and d^\ or
%§ff' and %abc. •
 1. Simple
DEFINITIONS' ANiJ NOTATlOiJ. iss
it*
Simple Quantities, are those which consist of On^ term
onjy. As Say or Saby or Saic^
8. CompQund Quantities, are those which consist of two qj*
more terms. A& « 4 ^, or 2<i — Sc, or a + 2b ^ Sf.
9. And when, the compound quantity consists of tw^
terms, it is called a Binomial, as a + ^} when of three term^ ^
k is a Trinomial, as tf + 2^ ~ 3^r ; when of foiir terms^ a
Quadrinomial, as 2a '^ $b ^ /^ ^ 4d \ and so  on. Also, a
Multinomial or Polynomial, consists of many tefmsi
10. A Residual Quantity, is a binomial having one of the
terms negative. As a ^ ^^.
1 1 . Positive or Affirmative Qu^titiesj are those which are
to be added, oi' have the sign +. As a or + ^j, or ^J : for
when a quantity is found without a sign, it is understood tq
be positive, or have the sign + prefixed.
12. Negative Quantities, are those which ^ are to be sub**
tracted. As — ii, or '2tf^, or 3ai%
13. Like Signs, are either all positive ( + ), (Jr all nega
tive ().
14. Unlike Signs, are when some are positive ( *f ), and
©thers negative ( — ).
1 5. The Coefficient of any quantity, as shown above, is
the number prefixed to it. As 3, in the quantity Sab.
16. The Power of a quantity (a]y is its square {a^)f or
cube («^), or biquadrate (^'*), &c; callfed also, the 2d power,
or 3d power, or ^th power, &c.
17. The Index or Exponent, is the number which denotes
the power or root of a quantity. So 2 is the exponent of
the square or second power /»* ; and 3 is the index of the
cube or 3d power ; and ^ is the index of the square root, a^
or v^^ 5 and \ is the index of the cube root, d^y or ^a,
18. A Rational Quantity, is that which Has no radical
sign (y/) or index annexed to it. Ag tf, or Sab.
19. An Irrational Quantity, or Surd, is that which has
not an exact root, of is expressed by means of the radical
sign v/. As v^ 2, or v^j, or^a^y or ai^.
20. The Reciprocal of any quantity, is that quantity in
verted, or unity divided by it. So, the reciprocal of tf, or
a , I f. a , i
— ,is — y and the reciprocal of "T" is — :•
M2 21. The
i64 ALGEBRA.
21 • The letters by which any simple quantity is expressed^
may be ranged according to any order at pleasure. So the
product. of a and b,, may be either expressed by ah^ or ta ;
and the product of a, i, and r» by either abcf or acb^ or hacy
or bcof or cab^ or cba ; as it matters not which quantities are
placed or multiplied first. But it will be sometimes found
convenient in long operations, to place the several letters
According to* their order in the alphabet, as abc^ which order
adso occurs most easily or naturally ta the mind.
22. Likewbe, the several members, or terms, of which a
compound quantity is composed, may be disposed in any
order at pleasure, without altering the value of the signifi
cation of the whole. Thus, Sa — 2ab + Aabc may also be
\\Titten 3fl f ^abc — 2aby or ^abc f 3 j — 2ab^ or — 2ab + 3«
+ ^akcy &C5 for all these represent the same thing, namely,
' the quantity which remains, when the quantity or term 2ab
is subtracted from the sum of the terms or quantities Stf and
4ahc. But it is most usual and natural, to begin with a po*
sitive term, and with the first letters of the alphabet.
a
SOME EXAMPLES FOR PRACTICE,
In finding the numeral values of various expressions^ or
combinations, of quantities.
Supposing /7 = 6, and 4=5, and c = 4,. and rf = 1, anct
/ = 0. Then
1. Will a' + Zabc^ =^ 56 + 90  16 = UQ.
2. And 2^3 3^*3 ^ c^ = 432 540 + 64 =—44.
3. And a^ X a\ b2obc = 3« x 11240 =156.
a^ 216
4. And r + ^ =z — r + 16 = 12 + 16 = 2S..
a i 3c 18"
5. Arid v" 2^7+7^ or 2ac + r]^ = y/64i = 8.
6. And x/c + ^ • \ =5? .+ — = 7.
^ 2ac ' + e 8 /
„ ., 1 ^'/^'^^ 361 _35_
8. And s^h" ac^ V2ac + ^ = 1 + 8 = ^.
9. Andx/b^ac + ^?^cT^ = ^2524^ + S = 3.
10. And fl*^ + r  ^/ = 1^5.
n. And 9/7^  10^* + r = 24.
\
12. And
ADDITION. fSS
fl'=
tt. And — X rf =3 45.
c
13. And^X 7= ISi
c a ^
. ,flr + ^ if— 5
14. And 7 = 11^
c a
15. And 1^=46.
16. And — X ^ = 9: 4^ ::r
17. And*^x J— ^ = i . ^
18. And^i+^— r— rf = 8,
19. Andtf+^ — ^ — </ = 6»'
20. Andfl*r x J' = 144.
21. Andacd d:=:2S*
22. And tfV + y'e + rf ^^^ ^^
23. And 1 X — ^ = 18J.
. 24. Andv^/i^ +^^"i/tf*  ^* = 44936249.
25. And 3^^ + ^a^  ^' == 292497942.
56. And 4fl* — 3^1 ^a'^ab = 72.
ADDITION.
Addition,^ in Algebra, is the connecting the qnaptities
together by their proper signs, and incorporating or uniting
into one term or sum, such as are similar, and can b^ united.
As Sa + 2b — 2a = a + 2b j the sum.
• The rule of addition in algebra, may be divided into three
cases : one, when the quantities are like, and their signs like
also ; a second, when the quantities are lite, but their signs
unlike; and the third, when the quantities are unlike.
Which are performed as follows*.
CASE
* The reasons on which these operations are fou tided, will rea
dily app^ar^ by a little reflection on the nature of the quantities to
16« ALOEBRA.
CASE I.
TVben tie Quantities are Liiey and have Like Signs. .
«
Add the co efficients together, and set down the sum j
after which set the common letter or letters of the like
quantities, and prefix the common sign + or — •
be added, or collected together. For, with regard to the first ex
ample> 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 tiroes any thiqg and 5 times the same thing, collected
together, must needs make 8 times that thing. As if a denote a
shilling \ then \\a is 3 shiirmgj5> and ba is 6 shillings, and their sum
8 shillings. In like manner, — 2a6 and — 7«^> 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 to
g€Jther by means of the arithmetical operations denoted by their
signs ( and — , or of addition and subtraction ; which being of
contrary or opposite natures, the one coefficient must be sub
tracted from the other, to obtain the incorporated or united mass.
As to the third case> where the quantities are unlike, it is plain
that such quantities cannot be united into one, or otherwise added,
than by means of their signs : thus, for example, if t^ be supposed
to represent a crown, and h z shilling ; then the sum of a and b
can be neither 2a nor 26, that is neither 2 crowns nor 2 shillings,
but only 1 crown plus 1 shilling, that is a f 6*
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 incorporate into one mass, or alge
braic 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 affirmative and some ne
gative j and the relation of these quantities can in the whole or in
part be discovered ; such incoi'por^^tion 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, and sometimes subtraction. But
the paradox wholly arises from the scantiness of the name giyen 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 striking a balance, or any name to which a more ex
tensive idea may be annexed, than that which is usuj^Hy implied
by the word addition ; and the paradox vanishes.
Thus,
f
Thus, 2a added to 5a, mak;ps 8a.
And —2ab added to — 7a^, makes — 9fl*.
And 5a + 7^ added to la + Si, makes I2a + 10*
M7
OTHER EXAMPLES IfOR PRACTICE.
3tf
Sa
I2a
a
2a
Sla
\5z
 Six
 lix
hxy
2hcf
5hxy
kxy
Shxy
&xy
11 bxy
2z
Sjt* + Sxy
2ax — 4;
22
x" + xy
4ax — J?
4z
2x' 4 4.Afy
«Af — Sy
z
5x^ + 2afj?
Sax — 5^
5^
4;v* ^ Sxy
7a* — 2y
15if* + 15*y l^ax — 15;
Sxy
X^xy
22xy
11 xy
lixy
ixy
12/
7/
2/
4/
/
3/
4/1 •^ 4#
Ai  5*
^a — *
3a  2*
2tf  7*
30  ISacf  Sxy
23 — lOxY  44ry
14  14*1  7xy
10  16;rt  5xy
16  20Ar^  xy
Sxy — 3jp + 4ai
Sxy * 4* + 3a*
3iA[y •«" 5ac + 5a*
xy — 2x + ab
ifxy ^ ;if + lot
i m» % > " > " ■
^U^
iii . i m
CASS
\$$, ALGEBRA.
CASl^ II.
r
tf^ifu tie Qjuantitiei ari Lite, but have Unlike Signs :
Add all the affirmative coefficients into one sum, and all
the negative ones into another, when there ^e several of i
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.
So + 5a and — Sa, united, make + 2a.
And — 5a and + Sa, united, make ^ 2a.
m
OTHEH EXAMPLES FOR PRACTICE.
r 5a + Sax^
\ 4a + irax*
+ 6fl — Sax^^
— 3a — eax":
fa + 5ax^
+
8*3 + 3^
—
5x^ + 4y
—
16*' + 5y
+
3x^  1y
+
2x^ — 2y
+ 3a — 2fl**  8*3 f i(5y
—
3i»*
+
sby
+ 4tf* + f
—
5a^
+
9by
 4ab + 12
' /
lO/i*
—
loby
+ lab ~ 14.
+
lOfl*
—
19by
+ ab + $
+
Ua^
—
2by
 Sab  10
■ <
I
— Zax'^ + 10^ ax + 3;^ + ^^x^
+ ^M^ — 3x/ax ^ y — ,5tf*^
I r
+ 5ax^ + 4x/ax + 4)^ + 2ax'^
 6fl*^ — 12 V ax — 2y + 6^*^
CASE
— /
ADDITION. 1G9
CASE III.
When the Quantities are Unlike.
Having collected together all the like quantitlesj,^ as ia
the two foregoing cases, set down those that are unlike> one
aft^r another, with their proper signs.
%xy
2ax
Sxy
6ax
SXAMPLES.
— 40;'*+ 3xy
+ 4;r*— 2xy
3^7+ 4x*
4.r^— 8a:'
4^x— 130+30:^
5x* +3/ir+9jr'
i
Ixy 4jr^ + 90
'2xy + Sax
lax + 8 j:* + 7^^
9r*/
7x*)»
+ Saxy
4jr*j?
•
14tfjr— 2.r*
5ax^3xy
8/ 4aj'
3jr* + 26
9+lQV'/jjr5;
2jr+ 7v/jrj> + 5j^
5y+ 3^i7X*4;
10— 4v/Ar+4;
•
4x*y
— 6;r/
'+3/jr
Ix'y
4v/x — , Sy
2v^a:y+14r
Sx + 2j^
9 + 3^jrjr
3/1* + 9 + a^— 4
2/j 8+2/1*— 3j:
4a:f2iJ*+18 —7
12+ tf3.r*2y
Add /f + i and 3/i — 5A together.
Add Stf— 8a: and 3/i— 4a: together.
Add 6a:5* + /i + 8to — 5/i 4a:+4i3.
Add /I +: 2*3r 10 to 3i— 4/i + 5c + lOand 5*r.
Add fl + i and a— h together.
Add 3/1 + i 10 to c^d^a and —4^ + 2^3^—7.
Add 3/1* + b'^c to^*3tf' + fc4.
Add ^' + ^V*' tP aA* /i^r + A*.
Add 9tf— 8* + 10Ar6//7r + 50 to 2ar3/i— 5r + 4A +
W10.
SVBIE ACTION/
170 ALGEBRA.
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 comi>osing the subtrahend : ranging the like
quantities under each other, as in AdditicMi.
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*.
\
EXAMPLES*
From 7tf*3* 9x*  4y + 8 SJcyS + 6jr— y
Take 3a* 8* 6x^ + 5;^  4 My1 ^ 6jr4y
I r m. ^ i ' i..i. .Ill
Rem. 4a* + 5^ Sjt*— 19^+12 44:j^ + 4+ 12^ + Sy
From 5xy—\ 6 4y*— Sj;— 4 — 20— 6j:;5xjr
Take 2ji'j^4 6 2/ + 2j?+4 3xy9j; + S^'2ay
Rem. 7jry— 12 2/— 5;^8  28 + St "Sa^y^^ay
*~«—
From 8jr*3ff6 5^xy'{2a:^xy lx^+2^x—l8 + Si
Take 2x*j;+2 7 4/ xy { 3  2xy 9^12 +5^+j:
Rem.
I
* This rule is founded on the consideration, that addition and
subtraction are opposite to each other in their nature and operation^
^s are thje signs h and — , by which they are expressed and repre
sented. So that, since to unite a negative quantity with a positive
one of the same iiind^ has the effect of diminishing it> or subduct
ing an equal positive one from it, therefore to subtract a positive
(which is the opposite of uniting or adding) is to add the equal
negative qu'antity. In like manner, to subtract a negative quan
tity, is the same in effect as to add or unite an equal positive one.
So that, by changing the sign of a quantity from f to — , of
from — to +, changes its nature from a subduottve quantity to an
additive one j and any quantity is ip effect subtracted^ by barely
changing its sign,
5xy
MULTIPLICATION. 171
Sxy  30 7^3 _2 {a + h) Sjt/ + QOax/{xy +^ 10)
Ixy  5Q 2;r=4 {a + h) 4xy j V2a*y\xy + 10)
From /J 4 ^> take «— ^.
From 4ii + 4^, take 3 + <»• '
From 4tf — 4^, take 3a + 55.
From 8^1—1 2jr, take 4/ar — Sjt.
From 2jr— 4a~2^ + 5, take 85* + a + 6t.
From Sa^b + cd 10, take c ^^ad, '
From 3a+h+ cd 10, takei 10 43zi.
From 2^l5 + Ir^—^c + be by take ^a^'c + b\
From tf3 + 35V + ab^ahc, take b" + (^b^abc.
From l2x + 6a'U 4<40, take 45  3<i + 4ar + 6i 10.
From 2j:— 3fl + 45 46^50, take 9tf+^ +656r4a
From 6«45 12t: + 12jr, take 2x8^ + 455^,
MULTIPLICATION.
This consists of several xases, according as the fattors ar?
simple or compound quantities.
CASE i^ When both the Factors are Simple Quantities :
First multiply the coeiEcients of the two terms together,
then to the product annex all the letters in those termsj
which will give the whole product required.
Ngte ^. Like signs, in the factors, produce + > suid unlike
jsigns — , in th^, products,
I
{:XAMPLES,
* That this rule for the signs is true, may be thus shown.
1. When + a is to be multiplied by + c; the meaning is, that
{ <2 is to be taken as many times as ther<^.are units in c ; and since
the sum of any number of positive terms is positive^ it follows that
j a X } c ip^kes f ac,
% When
2
lOa
2b
ALGEBRA.
•
EXAMPLES.
 Sa la
+ 2^ ^c
ex
i 1 1
20ab
Cab
2Sac
i'2iax
4ac
Sab
Od'x
4x
2x^y
— ifXy
 xy
\2abc
36tf*jr*
6xy
+4xy
9
Sax
4x
1
— ax
6c
+Sxy
4
— Sxyz
— Sax
CASE II.
' f
♦
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.
2. When two quantities are to be multiplied together, the re
sult will be exactly the same, in whatever order they are placed $
for a times c is the same as c times a, and therefore, wjiep — a if
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 foljows that
— a X  c, or + '< X — c make or produce — ac.
3. When — « is to be multiplied by — c; here — a is to b«
subtracted as often as there are units in c: but subtracting nega
tives is the same thing ^s adding affirmatives, by the demonstration
of the rule for subtraction 5 consequently the product is c times a,
•r + ac
Otherwise. S^nce a — a zr 0, therefore {a — a) X — c is also
=: 0, because multiplied by any quantity, is still but 5 and
since the first term of the product, or a X — c is 3: — ac, by the
second case ; therefore the last term of the product, or — a x — c,
niust be { ac, to make the sum iz 0, or — ac ^ ac zzQ', that is,
— a X — c =: + ac
EXAMPLES.
MULTIPLICATION. ng
EXAMPLES*
SaZc Sac 4b 2a^Sc+5
2a 3a be
10a*
•6ac
•
12j7
4a
2ac
i
4xy
X
9fl V  1 2ab 2a^bc  Uc" + 5bc
25c lb 4xb + 3ab
— 2a iab
CASE III.
When both the Factors are Compound Quantities;
Multiply every term of the multiplier by eVery term of
the mulriplicand, separately ; setting down the products one
after or under another, with their proper signs ; and add the
several lines of products all together for the whole product
required.
« + * SxiQy 2x^'\xy2f
« + ^ 4x — by Sjt— 3y
a^kab I2x^+Sxy 6x^ h Sx^y 6x/
+fl4 + ** IBxy—lOy"  6^*y  3x/ + 6/
m^\2ab\b^ ISj;'*— 7xy10f 6x^3x^y'9xy''^ey^
m^b x^j^y a^kab^b
— b ^^{y ' a ^b
^ :_
abb"^ +yx'' +/  a^b  ab""  b^
^ ♦  ^» ^*'f2;!^*+/ a^ * * b^
Notf>
~i
174 ALGEBRA.
Note. In the multiplication of compound quantities, it is
the best way to set them down in order, according to the
powers and the letters of the alphabet. And in multiplying
them^ begin at the lefthand side» and multiply from the left
hand towards the ri^ht, in the manner that we write, which
is contrary to the way of multiplying numbers. But in
setting down the several products, as they arisei in the second
and following lines, range them under the like terms in the
lines above^ when there are such like quantkiee ; which is
the easiest way for adding them up together.
In many cases, the multiplication of compound quantities
is only to be performed by setting them down one after
another, each within or under a vinculum, with a sign of
multiplication between them. As {a ^ b) x {a — b) x Sabf
era {■ b . a — i . Zab.
exa;mples for practice.
1. Muhiply lOac by 2a. Am. 20d'c.
2. Multiply Sa^2b by Zb. . Ans. 9a^b~6b\
3'. Muhiply Sa + 2i.by 3a 2b. Ans. 9a^4l^.
4. Multiply jr*  ary + / by jr 4 ;>. Ans. a;^ + f.
5. Mukiply a^ ^ i^b + ab" ^ P by a^b. Ans. a*K
6. Multiply a^ + ab + b''hYa''ab + b\
7. Muhiply 3j:^'2^y + 5 byx^ + 2a:y'^6.
8. Multiply Sa^2ax + 5jr* by Sd'Uxl:^. '
9. Multiply Sx^ + 2jry + 3/ by 2:c^Sxy + 3/.
10. Multiply d' + ab +b^bYa2b.
DIVISION.
4
\
Division in Algebra, like that in numbers, is the con^rse
of multiplication j and it is performed like that of numberp
also, by beginning at the Tefthand side, and dividing all the
p^rts of the dividend by the divisor, when they can be so
divided ; or else by setting them down like a fraction, the
dividend over the divisor, arid then abbreviating the fraction
as much as can be done. This will naturally divide into
the following particular cases.
CISE
DIVISION.
175
CASE I.
When th^ Diwsor and Dividmd are both Simple QuantitUs;
Set the terms both down as in division of numbers, either
the 4ivisor before the dividend, or below it, li're the deno
minator 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 fraction, by dividing them by their common
measure. ^ .
Note.' Like signs in the two 'factors make + in the quo
tient ; and unlike signs make — ; the same as in multipli
cation *^
EXAMPLES.
1. To divide 6^^ by 3 J.
Here Sab f 3tf, or Za ) Gab ( ox
6ab
Sa "
2b.
t
2. Also c^c ^ — "^ 1 ; and abj:
c
T b.vy :
atx a
^ bxy  ji •
3. Divide I6x*by 8^.
Ans. 2jr.
4. Divide 12j*j:* by  'id'x.
Ans. — 4*r.
5. Divide — \5ay^ by ^ay.
Ans. — 5j.
6. Divide ■ ISax^y by — Sdfjrz. .
Ans.  •
•
* Because the divisor multiplied by the quotient, must produce
the dividend. Therefore,
1 . When both the terms are + , the quotient must be + ; be
cause + in the divisor X f in the quotient, produces + in the
dividend.
2. When the terms are both — , the quotient is also f ; be«
cause ^ in the divisor X + in the quotient, produces — in the
dividend.
3. When one term is + and the other — , the quotient mast be
—J because + in the divisor x — in the quotient produces —
ki the dividend, or — in the divisor X + in the q^uotient gives *
Ih the dividend.
So that the rule is general; viz. that like signs give {, and
unlike signs give — > in the quotient.
C4SE
/»
176 ALGEBRA,
CASE II.
When the Dividend is a Compound Quantiify and the Divisor
Simple one :
Divide every term of the dividend by the divisor^ a$ in
the former case.
EXAMPLES.
ab + b'' a + b
1. {ab + *^) ^2*,or ^ =  ^ia + ib.
lOab + I5aa: ^, . „
2. (lOab + I5ar) r 5a, or j^ = 2* + 3.r.
30flfz48z • . _
3. (30^2 48«) r z, or * = 30^48.
4. Divide 6^?^ Stfo: +ahj 2a.
5. Divide 3^* 15 + 6.r + Qa by 3r.
6. Divide 6fl^r + 12fl^J79fl'3 by 3tf*.
7. Divide. 10^*x— 15r*— 25jr by 5^.
8. Divide lot^bc iBacr" + $ad^ by  Sac.
9. Divide 15^ + 3ay 18/ by 2U.
10. Divide ZOdb" + 60^^' by6a*.^
CASE III. ,
When the Divisor and Dividend are both Compound Quantities:
1 . Set them down as in common division of hurabers^
the divisor before the dividend, ttrkh a small curvecJ line
between them, and ranging the terms according to the
powers of some one of the letters in both, the higher
powers before th6 lower. ^
2. Divide the first term of the dividend by the first term
of the divisor, as in the first case, and set the result in the
cjuotient.
3. Multiply the whole divisor by the term thus found, and
' subtract the result from the dividend.
4. To this remainder bring, down as tnany terms of the
dividend as are requisite ' for the next operation, dividing as .
before ; and so »n to the end, as ih common arithmetic.
Not~e4
F
DIVISION. m
Noie. If the divisor be not exactly contained in the divi
'dend^ the quantity which remains after the operation is
finishddy may be placed over the divisor, like a vulgar frac
tion, and set down at the end of the quotient, as in common
arithmetic*
EXAMPLES.
tf* ab
a^4?
^—2) a'— 6fl* + 12a8 (fl*4tf + 4
r
+
8
■8
az
+
+

4<i
4«
il
«) a^ + «» (,
+ »'

N
«3
P

• '.•^'
Tot. I. N «+*)
lis ALGEfeRA.
2^*
a + x) a^'ix* ( tf'a'x + ff4P'jr»
*+*
»*
+v*
•
 3«*
— «**
«'«»
/!»«».
1
*»>
«.'
8**
—
■2**
EXAMPLES FOR PRACTltE.
!. Divide u* + 4j;r + 4;ir* by a + 2;r. Ans. a + 2x.
2. Divide tf^3fl*« + 3iia*z' hy a— z.
Ans. II*— 2/jz + »*•
5. Divide 1 by 1 + ii. Ans. 1 tf + tf*a' + &c,
4. Divide 12;j«*ld2 by 3*6.
Ans. 4w» + 84r* + 16jr + 32*
A. Divide «55fl** + I'Oa^*^  .lOay + 5/j** ** by n*
24f* + F: Ans.. «'  3^** + 3^**  ^.
6. Divide 482396tf«*— 6*0^2 + ISOtf' by 2a;— 3^.
7. Divide ^*3**P»* + SS^x^^x^ by ^33*'x + 3**'^«^
8« Divide fl^— o;^ by ^ — x
9. Divide a' + 5a^x + Sax" + *' by a + x.
10. Divide a* + 4fl*J* 32** by « + 2*.
11. Divide '24fl*** by 3a 2b.
tSS!
i^» i^i i M S <
ALGEBRAIC FRACTIONS.
Algebraic Fractions have the same names and rules
of operation, as numeral fractions in common aritjj^netic } as
..appears in the following Rules and Cases.
CASE
JfRACIIONS. 17SI
CfSB Z.
To Reduce a Mixed Quantity to ah Improper Fraction.
Multiply the integer by the denominator of the fraction^
and to the product add the huinerator> dr cozmect it wit^
its proper sign, .+ of — ) tli^n the denominator beipg set
under this sum> will give the improper friction required*
1* Reduce S4>.atid d —  t6 improper fractions.
3x5+4 15 +4 : 19 , ^
First, 5f ^ j^ = — ^ = J, the Answeir^
. ■ * ax x^b a^—i . .
And. tf — = — — r^— 2= the Answer.
^ X X X
2* Reduce a + — and a ^ t6 improper fractions;
. o a
fl* axb+d* at^d"
First, 4+7 = — *7 ==
t
the Amwer.
And, a — ^ — ^— = '^^ = the Answer.
a . a a
. 3. Reduce 5^ Jo an improper fraction^ Ans. y *
4. Redu(^e 1 to an improper fi^ction^ Ans. * '
fiC X
6i Reduce 2i» — — to an improper frractioh^
4^ " ■ *
6. ite4uce 12 4 ^t to an impfopiir fraction*
7. Reduce * H — ' ^ to sin improper fraction*
c
8« Reduce 4 ^2x ^ to an improper fraction' '
tASE It.
To Redwte an Improper Praction to a Jf%Ie wr Mixed Quantity!
DitiiJi; the numerator! by thcf denpniinator, for the in*
te^al part ; and set the remainder, if any, oter the deno^^
xmnator, for the fractional part j the two joined together will
be the mi^ed gji^antity required. '
1$0 ALGEBRA.
EXAMPLES.
. 1. To reduce — and — r^ to mixed quantities.
3 p
First, y 5= 16 r 3 = 54, the Answer required.
And, f*±^= ^M^ 5 4 = « + J. Answer.
"2. To reduce !flzif! and 5fl±if! to mixed quantities.
First, ^^^^^ = 2ar  3^ ^ ^ = 2«  — . Answer.
c c
And, ^^"^^"^ = i^HK*^ Tfl+or = 3a: + — . Ans.
3. Reduce — and r — — to mixed quantities.
3jr*
Ans. 64> and 2* .
4. Reduce —  and I— to whole or mixed quan
2a ab
tities.
3jr*^ 3/ • , 2j^^  2y3  , . .
5. Reduce ^, and to whole or mixed
quantities.
6. Reduce ■ — to a mixed quantity.
7. Reduce ^ to a mixed quantity.
3a3 + 2a* 2a— 4?
CASE III.
To Reduce Fractions to a Common Denominator*
Multiply every numerator^ separately, by all the deno^
minators except its own, for the new numerators ; and all the
^ denominators together, for the common denominator.
When the denominators have a common divisor, it will be
^better, instead of multiplying by the whole denominators, to
multiply only by those parts which arise from dividing by the
common divisor. And observing also the several rules and
directions as in Fractions in the Arithmetic*
EXAMPLES.
/
FRACTIONS. 181
EXAMPLES.
a  i
1. Reduce — and — to a common denominator.
X z
Here — and — = £ and — , by multiplying the terms of
X % xz x%
the first fraction by z, and the terms of the 2d by x*
/I V b
2. Reduce , — , and — to a common denominator.
X b c
Here — , , and — = ^ — , — > and , by multiplying the
X h c hex hex hex
terms of the 1st fraction by hc^ of the 2d by cx^ and of the
3d by hx.
I*
3. Reduce — and — to a common denominator*
X 2c
Ans. and — •
2ex 2eK
4. Reduce — anJ^^^^^I — to a common denominator.
h 2c
ifOe  , SaH2i*
*Ans. rr—, and — r .
2bc' 2bc.
' 5. Reduce — and — , and 4«f. to a common denominator*
3* 2?
Ans, and and
%cx Qex Qex
6. Reduce — and — and 2b + 7. to fractions having a
6 4 b *
common denominator. Ans. — r— and ^ and ,
. 243 243 243
7. Reduce — and and — III to ^ commori deno
3 4 fl+3
minator.,
8. Reduce — and — and —  to a common denominator.
4a* 3a 2a
CA8B
n2 ALGEBRA,
CASE IV.
To find the Qriofist Common Meastfti rf the Tkrms jf #
practian*
Divide the greater term by the less, and thje last divisor
by the last remainder, and so on till nothing remains ; then
the divisor last used will be the common measure required 9
just the same as in common numbers.
But note, that it is proper to range the quantities according
to the dimensions of some letters, as is shown in division.
And note also, that all'the Itoers or figures which are com
mon to each term of the divisors, must be thrown out of
them, or must divide theip, before they are used in the
operation.
£3(AMPLBS.
J. To find the greatest common measure of ^ rr.
«* + **) ac" + i^
6r m+b) ac" + 6^{c^
Therefore the greater Cbmmon measure isa + i.
2. To^nd the greatest common mea^r^e of ^
or a+ i )d' + 2aL+f(a+L
ai + ^
Therefore <f + ^ is the greatest common divisor.
9« To find the greatest cozjunon devisor of; — rrr,
ao+29
Ans* «— 2«
4. Ta
FRACTIONS. 183
4. To find the greatest common divisor of — ^ ^ ,^ «
Ans. fl*  f.
5, Fiadtne greatest com. measure of ^ .t\ ^ •;r^ — t ^ % z '
CASE T. i^
I
To Reduce a FractUn to its Lowest Terms*
Find the greatest common measure^ as in the last pro
blem. Then divide both the terms of the fraction by tjie
common nxeasure thus founds and it will reduce it to its lowest
terms at once^ as was required. OriKvide the terms by any
quantity which it may appear will divide them both, as in
arithmetical fractions.
\
EXAMPLES.
ai + b* .
1. Reduce ■ . , ^ *o i^s lowest terms.
or + or
ab + i^) ac" + he*
dra + b ) 0c* + be'{c'
0^ + ^V*
Here ab + i^U divided by the common factor b»
Therefore a + i^s the greatest common measure, and
ab + b^ b
hence a + b ) , , . = t*** the. fraction required
2. To reduce ; .■ ^, — rrr to its least terms.
c^+2bc + b
tf* + 2fc + *" ) r'  b^c ( c
 2*^V2*V) c' + 2bc + b^
orc + b) i^ + 2bc+b^(c + i
ji'+ be
be + b'
bc + b''
Thtjrefore
lU ALGEBRA.
Therefore ^ 4 ^ is the greatest common measure) and
hence r + *) , ^ = , , is the fraction required*
S. Reduce r — rr^ to Its lowest terms. Ans. ^ , , ^ ■
fl* — J* 1
4. Reduce — — rr to its lowest terms. Ans. ^ . ^ ^
fl* — **
5. Reduce r — ^ ,, , ^ ^ — n to its lowest terms. ^
7. Reduce , . ^ . . ... to its lowest terms.
a* + 2ai + b^
CASE vi.
T^ tf^i/ Fractional Quantities together.
If the fractions have a common denominator^ add 73\ 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> and then add them as before.
EXAMPLES.
T ^ J ^
1. Let ~ and — be given, to find their sum.
3 *
T a A 4tf 3tf la
Here 5 + t"^ i^ ^ T^ ^ 75 '* ^'^^ ^^°^ required
3 4 12 12 12
.> 2i.nrl — 4'r\ 4\r\A 4ViA«« ».««««
a b , c
% Given 7, — ^,anQ~7, to find their sum.
PC a
the sum required*
3. Lit
,*\i\
FRACTIONS. IBS
Sx* 2ax
* 3 Let a rand i H — ^ be added together.
b c , °
^x^ , Sax * 3rjr* , , 2a5j: ^
Here^— +* + — =^^^+*+>^j
2abx~^3cx*  *
= a + i H r , the sum required.
^. *., 4!X ^ 2x , . 20*j:+6flx
4. Add r and rr together. Ans. — ■ ^ , ^.
5. Add J — and — together. Ans. ^a.
^ Ajj 2^—3  5tf , . . 9a — 6
6. Add and e together, Ans. "^
4 8^ , 8
7. Add 2a J •— to 4a H ^ — ^. Ans, 6a +
,**na
5 '4 20
So" a+b
8. Add 6a, and — r and —7 together.
5a 6a 3a + 2
9. Add — , and  and —  — together.
3a a
10. Add 2a, and — and 5 + — together.
, , . , , ^ , Sa  5a
11. Add 8a + — and 2a— — together.
4 o
CASE VH.
* To Subtract one Fractional Quantity from another .
Reduce the fractions to a common denominator, as in
addition, if they have not, a* common denohiinator.
Subtract the numerators from each othler, and under their
difference set the common denominator, and the work is
done.
* In the addition of mixed quantities^ it is best to bring the
fractional parts only^o a common denominator, and to annex theiv
sum to the sum of the integers, with the proper sign. And the
same lule n^ay be observed for mixed quantities in subtraction
also.
(XAMPUS.
^
1S6 ALGEBRA.
EXAMPLES.
1 . To find the difference of *f and ~. 
4 7
Hcrc^l?=ii?^l^=,^is the dMTerence remiircd .
4 7 28 28 28
£• To find the difference of and
Here
•4r 3^
4c Sb I2bc \2bc
6tf?— 3ii— 124awr+ lefc . , ■»./*• • »
— i ' IS the difference required.
\2bc ^
3. Required the difference, of — f and ^.
9 7
4. Required the difference of 6« and — .
4
5. Required the difference of — a^d — •
4 3
6. Subtract?* from ?l±f.
c b
7.Take!l±.^fromll±i.
9 5
U. Take 2a^t^l^ from 4^i + £f.
CASE Till. .
Jb Multiply Fractional Qjianiities together*
Multiply the numerators together for a new numerator,
and the denominators for a new denominator*.
* 1. When the numerator of one fraction, and the denominator
of thf»other^ can be divided byisome qqaptttyj which is common
to both, the quotients may be used inAtfsad of them.
a. When a fraction is to be nulti(^ie4 by an integer^ the pro
duct is found either by i^ultiplying the uum^i^tort or dividing th&
denominator by it; and if the integer be the s^e with the deno
minator^ the numerator may be taken for the product.
EXAMPLES.
FRACnONB;
i*r
EXAMPLES.
I. Required to find the product of ^ and — .
Here .— — r = r
8x5 40
— the product required.
2d'
2. Required the'product of — , ^, and — .
f^iliiiiif = i!f! = !i the i*oduct reqal^ ■
3x4x1 «4 , J4r *^
3. Required the product o£ — and •
^ ^ b 2a ^c
Here i —  — i = ! the product requum.
4. Required the product of — and — .
5. Required the product of— and — 
4 3flf
6. To multiply ^, and — ^i and  together.
3r
7. Required the product of 2fl+ — and — ,
2c b
8. Required the product of — « — » and ^ — ,
Sbc a {rb
9. Required the product of 8^, and ^ T , and ^ *" *
10. Multiply «+4 ^byariL + _?l.
*^' 'Za 4<i* 24? 4j?*
CASE IX.
2it Divide one Fractional Quantity by another.
Divide the numerators by each other, and the denomina
tors by each other, if they will exactly divide. But^ if not^
then invert the terms of the divispr, and multiply by it
exaetly is in muhi{£catiQ«i^.
*'^^ — — ^^ — ^^^ — ^^^^ — ,^ — .._ ^ _ _, ^ ^^^^^ — ^_.,.^^^,^ ^^^.^ — . — ^__ — ^^.^ — .^
^ 1. if tin £raedoot'to be divided have a csramon deomninator^
take Ibe mmefto r t£ tbe dividend for a new nuineratDO and
the numeratQjT of the Avisos ibr the new denomMiAtor.
2. Wheii
Ito ALGEBRA.
EXAMPLES.
1 Required to divide — by^.
Sa 5c
2. Required to divide rr by tv
„ 3# 5r Sa 4d \2ad Sad ,
• 3a* 2a ,
4. To divide ^ . .» by  — r^.
3tf^ tf+i _ 8a*x(a+ ^) _ 3tf
is the quotient r.equiredl
5. To divide r by t*.
4 ^ 12
6. To divide — by 3x.
7. To divide —  — by — .
9 ^ o
' 8. To divide * —  — : by ^.
2^:— 1' ' 3 ,
4x Sa
9. To divide r by t
5 ^ 5h
2a "b Sac
10. To divide —J by .TT.
11 n ^ 5a^^5b^ , 6a*+5a*
11. Divide ^ . ^ I 77 by r.
2a*  4ai h 24* ^ 4a  4*
i2. When a fraction is to i>e divided by any quantity, it is tho
Wne thing whether the numerator be divided by it, or the deno
minator multiplied by it.
3. When the two numeratprs, or the two denominators, can be
divided by some common quantity, Jet that be done, and thequo
tients used instead of the fractions first proposed.
INVOLUTION.
[ 18i> ]
INVOLUTION.
Involution is the raising of powers from any proposed
root ; such as finding the square, cube, biquadrate, &c, of
any given quantity. The method is as follows :
* Multiply the root or given quantity by itself, as many
tipes as there are units in the index less one, and the last pro
duct will be the power required.— Or, in literals, multiply
the index of the root by the index of the power, and the
result will be the power, the same as before.
Note. When the sign of the root is +> all the powersof
it will be 4 ; but when the sign is — , all the even powers
will be .+f and all the odd powers — ; as is evident from
multiplication.
exampCes.
«, the root
tf* = square
tf' z= cube
tf * =r 4th power
41* = 5th power
&c.
— 2a, the root
f 4^* = square
'— 8/j^ = cube
+ 16^1* == 4th power
J 32a* = 5th power
2ax^
, the root
+
+
3A
4a^j?*
=: cube
27*^
1 6a*x*
■ I I «.
81**
= 4th power.
a*, the root
a^ = square
a^ = cube
if = 4th power
**°= 5th power
&c.
 Zab\ the root
+ 9a*i* = square
 27a'** = cube
+ Sia'^b* = 4th power.
 243a**'° = 5th power.
, the root
2i
^ = square
«
W
=: cube
16*
 = biquadrate
* Any powerof the product of two or more quantities, is equal
to the same power of each of the factors, multiplied together.
And an^ power of a fraction, is equal to the same power of the^p
numerator, divided by the like [ibwer of the denominator.
Also, powers or roots of the same quantity, are multiplied by
one another, by adding their exponents ; or divided, by sobtract
ing their exponents.
Hius, «^ X a* =««+*=; ff^. And a^^a« or =: a = a.
a*
L
190 AIX;^£BRA.
j:'a = root iji+tfnrool
;r* + tfjr
X 4 tf
+ aj^ +2fl*x + fl'
«' + 3tfT* + 3i^x + a'
jf*— 2ar + a* square
s ^ a
jc^— 2aj;* +<!**
jr53Ar*+ Mxa^
Ae cubes^ or third powers, olx^a and x + ^.
EXAMPLES. FOR PRACTICE.
1. Required. the cube or 3d power of So'.'
2. Reqjuired the 4th powi^.of 2i7'j.
3. Re<uired the 3d powej of — 4iiV.
4. To find the biquadrat^ of — ^.
5. Retfuired the 5th power of a — 2x.
6. To find.the 6th power of 2d^.
Sir Isaac ICiTewtom^s Rdle for raising a Binomial to any
Power whatever *.
1. To find the Terms without the Co^efficiertts. The index of
the first, or leading quantity, begins with the index of the
given power, and in the succeeding terms decreases conti
nually by I, in every term to the last; and in the 2d or
following quantity, the indices of the terms are O, 1, 2, 3, 4,
&c, increasing always by 1. That is, the first teno will con»
tain only the 1st part of the root with the same index, or of
— 1 — — ' ' — ■■ I , ji ■ — ■
* Tbb ruW expressed in general tennsi is as follows t
M 2 3
^ ' 2 2 3
Nate* The sum of the coe£ident»^ in every power, is equd to
the number 2, wl^n raised to that power. Thus 14 1 =: 2 in
the first power ; I + 2 + 1 = 4 == 2^ in the square ; i 4. 3 4 3
j+ 1 =: 8 =: 2^ in the cube, or third power ; and so on* '
the
ItJVOLtfTIiDN. , l$l
I
fllel^ame height as the inteticled 'power : and the kst teHn of
the^ series will contain only the 2d part of the given teot,
when raised aho to the same height'of the intended p6wer:
but an the other or intermediate terms will contain the ]pro^
ducts of some powers of both the members of the root, in
such sort, that the powers or indices of the 1st or leading
•lembet will always decrease by 1» while those of the 2d
member always increase by 1.
2; To find the Co^iffictents. The first coeifficient is always
1, and the second' is the same as the index of the intended
' jpew^er ; to find the 3d coefficient, multiply that tA die ^d
term by the index of the leadkig letter in the sameterm^ and
divide the product by 2 ; and so on, that is, multiply the co
efficient of the term last found by the index of the leading
quantity in that terhi, and divide ^he product by the number
of terms to that place, and it will give the coefficient of the
term next following; which rule will find all the coefficients^
one after another.
, Nate. The whole number of terms Will be 1 more than the
index of the given power i and when both tel^ms of the root
are +, all the terms of the power will be + ; but if the '^er
cond term be — , all the odd terms wilt /be' +, afed all the
even terms — , which causes the terms to be + and — alter*
nately. Also the sum of the two indices, in each term, is
always the same number, viz. the index of the required
power: and, counting from the middle of the series, both
ways, or towards the right and left, the indices of the two
terms are the same figures at equal distances, but mutually
changed places. Moreover, the coefiicients are the saihe
numbers at equal distances from the middle of the series,
towards the right and left ; so by whatever numbers the
increase to the middle, by the same in the reverse order they
decrease to the end.
' tiAMPLES.
\\ Let d^^x be involved to the 5th power. '
The terms without the coefficients» by the ist rule,
#iU be
a\ it%, c^x^y c^x^y #jr*^ jt^
and the coefficients, by the 2d irule, will be
. , ^5x4 10X3...1P„x2 ^>^\
l.^> 2 ' "^^^ 4^5^
or, 1,5, 10, 10, 5', \s
Therefore the Ath power altogether is
#5 + Ba\v + 10#»^* + lO^^jr^ + Sax^ + x\
But
y
192 ALGEBRA.
But it is best to «et down both the cOhefficiehts and the
powers of the letters at once, in ope line, without the inter
mediate lines in the above example^ as in the example here
below. ^
2, Let fl — X be involved to the 6th power.
The terms with the coefl5cients will be
a^^ea^x + 15tf*x* 20^x3 + ISd'j^Gax^ + jt*. •
S. Required the 4th power of ^ — jr.
Ans. fl*  ^x + e(^x^  ^ap^ + «♦.
' And thus any other powers may be; set down ^t once^ in
the same manner ; which is the be$t way.
EVOLUTION.
Evolution is the reverse of Involution^ being the method"
of finding the square rootj cube root> &c>. of any given
quantity, whether simple or compound.
CASE I. To find the Roots of Simple QuMntities.
Extract the root of the coefficient, for the numeral
part \ and divide the index of the letter or letters, by the
mdex of the power, and it will give the root of the literal
part ; then annex this to the former, for the whole root
sought*.
■■iiiUi
* Any even root of an affirmative quantity^ may be either 4
or — : thus the square root of + c^ is either ^ a,ov — a ; be«
cause + a X + a = + a*, and —a X — a r= + a^ also.
But an odd root of any quantity will have the same sign as the
quarittty itself: thus the cube root of f a* is f a, and the cube
root of— a^is — ajfbr + ax +ax+a=:+«^, and —a X
— fl X ^ azu  a^.
Any even root of a negative quantity is impossible; for neither
+ a x^a, nor —a X — a can produce —a*.
Any root of a product, is equal to the like root of each of the
fBictors multiplied together. And for the root of a fraction, take
^he root of the numerator, and the root of the denominator.
EXAMPLES.
EVOLUTION. 19.S
EXAMPLES.
1. The square root of 4fl*, is 2/i,
3*
2. The cube root of 8«^ is 2a^ or 2a.
3. Ihe square root of "^^> ^'' V^"5^j js ^\/5.
4. ' The cube root of ' j , xs — r — ^2^,
5. To find the square root of 2^***. Ans. fli*v^
6. To find thfe cube root pf — 64«^^ Ans. 4a**.
7. To find^the square root of^r, Ans. 9,abA/^*
8. To find the 4th root of^U^b^. Ans. 3aby/b.
9. To find the 5th root of  32a*^. Ans. —^ab^b.
CASE II.
To find the Square Root of a Compound Quantity*
This is performed like as in numbersi, thus :
i . Range the quantities according to the dimensions of
one pf the letters^ and set the root of the first term in the
quotient. ^ ,
2. Subtract the square of the root, thus founds from the
first term, and bring down the next two terms to tike re
mainder for a dividend) and take double the root for a
divisor.
3. Divide the dividend by the divisor, and annex the re
sult both to the quotient and to the divisor.
4. Multiply the divisor, thus increased, by the term last
set in the quotient, and subtract the pnx^uct from the
dividend.
And so on, always tlie same, as in common arithmetic.
EXAMPLES.
1 • Extract the square root of j* ^Q}h + e^j*i*  4o3* + *♦.
a'^^^a^ + ^a^b^^d^ + ** ( fl*2fli + i* the root.
2<l*  fitf* )  4tf ^^ + 6<i**** \
Vot. I. O 2. Find
1
194 ALGEBRA.
< ^
2. Find the root of a^ + 4^a^b + lOaH"^ + l2aP + *♦.
3. To find the square root of fl*'+ 4^' + 6«' + 4* + 1.
Ans. fl* + 2« + l.
4. Extract the square root of a^ — 2a^ + 2£^—a + ^. "
.Ans. a:*— jr +4.
5. It is required to find the square root of a* at.
CASE III.
To find the Roots of any Powers in Genera/,
This is also done like the same roots in numbers, thus t
Find the root of the first term, and set it in the quotient.
— Subtract its power from that term, and bring down the
second term for a dividend. — Involve the root, last found, to
the next lower power, and multiply it by the index of the
given power, for a divisor. — Divide the dividend by the di
visor, and set the quotient as the next term of the root. —
Involve now the whole root to the power to be extracted ;
then subtract the power thus arising £rom the given power,
and divide the first term of the remainder by the divisor first
found; and so on till the whole is finished^.
EXAMPLES.
* As this method, in high powers^ ipay be thought too labo
rious, it will not be improper to observe, that the roots of com
pound quantities may sometimes be easily discovered, thus':
Extract the roots of some of the most simple terms, and connect
them together by the sign + or — , as may be judged most suit
able for the purpose. — Involve the componnd root, thus found, la
the proper power , then, if this be the same with the given quan
tity, it is the root required. — But if it be found to differ oiily in
some of the signs, change them fcom + to — , or from — to +,
till its power agrees with the given one throughout
Thus,
EVOLUTION. 195
EXAMPLES.
i. To find the square root of a*2a^t+Sa'k'^2aP + i\
— — ■ I . I. ■■ • !■ I
2. Find the cube root of a^^a^ + 21a^  44a' + 63^*
 54fl + 27.
f«6^i + 21a*44tf'+ 63fl*— 54fl + 27 ( d'2a + $.
3tf^)6a5
ii*  6^5> i2fl*  8fl' = (^~2j)3
a**6^* + 21fl*44fl5+63a*54a +27 = {a^^2a3y.
3. To find the square root of «*  2ab + 2ax + ^* 
^x + j^. . Ans. a—b + j:.
4. Find the cube root of a^  Sn* + 9fl^13/i' + ISa*
12a +,8. Ans. fl*tf + 2.
5. Find the 4th root of 81a*  216a'* + 216a***  96a*»
+ 16*^ Ans. 3a 2*.
6. Find the 5th root of a^  10a* + 40a' 80a* + 80/i
— 32. . Ans. a— 2.
!• Required the square root of 1 — :r*.
8. Required the cube root of l—x'.
i«i«i
Thusy in the 5th example^ the root 3a— 2*, Is the differenpe of
the roots of the first and last terms ; aad in the 3d example, the
root a—* + x> is the sum of the roots of the ist^ 4th^ aod 6th
terms. The same may also be observed of the 6th example, where
the root is found from the first and last terms.
O 2 SURDS.
196 ALGEBRA.
SURDS.
SuKDs are such qutntkics m ha^« no exact rbot; 4nd are
usually expre^ed by fractional indices, or by means of the
radical sign a/. Thus, 3^, or ^/S, denotes the square root
of S ; and 2^ or i/2S or V*, the cube root of the s^are of
2 ; where the numerator shows the power to trhich the
qutotity is tb be raised, and the denoniinator its root,.
pHoblem I.
Ta Reduce a Rational Quantity to the Form of a Surd.
Raise the given quartity to the power denoted by the.
index of the surd ; then over or before this nevr qiljiatity set
the radical sign, and it will be of the form required.
EXAMPLES.
J • To reduce 4 to the form of the sqyare root.
First, 4* = 4 x 4:=16; then Vl6 is the answer.
2. To reduce Sa^ to the form of the cube root*
First, 3^' X^d" X Za" = {^a^f == 27a%
then ^'Itf or (27/1^)"^ is the answer:
3. Reduce 6 to the form of the cube root.
Ans. (216)7or^l0.
4. Reduce \ab to the form of the square root.
Ans. /fiV.
5. Reduce 2 to the form of the 4th root. Aiw. (U)^.
I
6. Reduce a'^ to the form of the 5th root.
7. Reduce a^x to the forhi of the square root.
S. Reduce a—x to the form of the cube root,
PROBLEM II.
' * I. "
To Reduce^ Quantities to a Common Index.
1. Reduce the indices of the gi^eh quantities to a comi
mon denominator, and involve each of mem to the power
denoted by its numerator ; then 1 set ov^ the common ^^
npminator will form the common index. Or,
2. If
I
t
SURDS. 197
2. If tht 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 $sud quantities, with tbeir new indices, set the given
ind^x, and they will make the equivalent quantities SQughiji
EXAMPLES.
I 1 '
1. Reduce 3^ and 5'^ to a common ind^x.
Here 4 and I = ^^ and 1%.
Therefore 3^ and 5tI=(3s)tV ^nd {5^)^=^'^S^ and '^S*
= '°/243 and ^^25.
1
2. Reduce (^ and AT to the same common index J*
Here, 4.44 3=x^=3f the 1st indes^
and 1^4 = 1 X 4^ = f the 2d index.
Therefore (tf®)^ and (H^)^, or x^cf and ^S^ are the quan
tities.
3. Reduce 4^ and 5^ to the common index ^.
Ansl 2567)Tand2S^.
II
4. Reduce d^ and x^ to the common index 4
Ans. (fl*)^and(.r^)"«'.
 5. ' Reduce a^ and .r^ to the same radical sign.
Ans. iy/fl* and a/x^*
6. Reduce (a + x)'^ and (a— j?*)^ to a common index.
I I
7, Reduce {a + ^)^ and (a— ip to a commoti mdex.
PROBLEM III. ,
T& Reduce Surds to more Simpte Terms.
Find put the gre^te^t power contained in, or to divide the
given surd 5 take its root, and set it before the quotient pr
the remaining quantities, with the proper radical sign b^
tween them.
EXAMPLES*
1. To reduce v/82 to simj^er terms.
Here v^32=: ^16 x2=^\6 x V2 =^4xV2=:4^2.
2. To reduce ^320 to simpler terms.
^320 = 3/64 X 5 =4/^4 X 4/5 = 4 X 4/5 = 4^/5.
, S. Reduc
19S ALGEBRA.
3. Reduce v/ 75 to its simplest terms. ^ Ans. 5a/S.
4. Reduce %/f to simpler terms. Ans. rrV^^^*
5/ Reduce ^189 to its simplest terms. Ans. 3'/7.
6. Reduce V^^V to its simplest terms. Ans. :(/10,
7. Reduce ^ISe^b to its simplest terms. Ans. Sax^Sb.
Nate, There are other cases of reducing algebraic siu>ds
to simpler forms, that are practised on several occasions ; one
instance of which, on account of its simplicity and usefulness,
may be here noticed, viz. in fractional forms having com*
pound surds in the denominator! multiply both numerator
and denominator by the same terms of the denominator, but
having one sign changed, fr9m + to — or from — to +>
which will reduce the fraction to a rational denominator.
xi r^ A ^2Q + ^/12. • ^/5rf^3
Ex. To reduce ^^^^3 , multiply it by z^s^%' *°^
. ^ . 16 + 2v^l5 , ' ^, .^3v/154^5
It becomes — ^ — ; =8+^^15. Also, if — ,, ^ . — 7;
, . , . , s/lSx/S ... 65  Yv'^S
multiply It by ,, ^ ' — TT, and it becomes — =
'^ ^ ^ v/15 — ^5 15 — 5
6535x/3 _ IS — 7^/3
10 ■* 2 '
PROBLEM IV.
To add Surd Quantities together.
»
i. 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.—
S. Then, if the surd part be the same in them all, annex it
to the sum of the rational parts, with the sign of multiplica
tion, and it will give the total sum required.
But if the surd part be not the same in all the quantities,
they can only be added by the signs + and — .
EXAMPLES.
1. Required toadd v'l^ and v'32 together.
First, 1/^8= /9x 2=3v/2; and '•32=v^I6x 2=4v/2:
Then,.3v/2 + 4v/2 = (3 + 4) ^2 = 1^.2 = sum required.
2. It is required to add ^375, and VI 92 together.
First, V375=V125X3=:5V3; and ^102= V64 x 3 =4^3 :
Then, 5^3 + 4V3 = (5 + 4) V3 = 9?/3 = sum required.
3. Kequired
'
.; SURDS. isy
S. Required the sum of V27 and V48. Ans". 7 V^3.
4. Required the sum. of V50 and V72. Ans. 11 \^ 2.
5. Required the sum of ./I and /^
Ans. 4 ViV o** A^l^
6. Required the sum of ^56 and ^189. Ans. 5^/7,
7. Required the sum of V and^^ ' Ans* ^l^.
8. Required the sum of 3 Va^b and 5Vl 6a^i»
PROBLEM V.
To find the Difference of &urd Quantities,
Prepare the quantities the same way as in the last rule;
then subtract 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, they can only
be subtracted by means of the sign — . ,
EXAMPLES.
I
1. To find the difference between V320 and V 80.
First, v^320= v'64 x 5—8^/5', and/80=: ViGx 5=:4^^5,
• Then S^ 5 — 4 • 5 = 4 ^Z 5 the difference sought.
2. To find the difierente between ^128 and ^54.
First, 3/128=3/64 x 2=4^2 ; and 4/54= ^27 x 2=3J/2.
Then 4^ — 3^ =v^2, the difference required.
3. Required the difference of ^15 and v^48. Ans. ^3.
4. Required the difference of 3/256 and ^32. Ans. 2?/4.
5. Required the difference of \/^ and ^^. Ans. ^V6.
6. Required the difference of ^/^ and 3/y . An*. f^l$.
7. Find the difference of v'24tf^^* and ^/54ab\
' * • Ans. {a2by^{Sb''2ab)^ 6a.
' PROBLEM VI.
To Multiply Surd Quantities together •
Reduce the surds to the same index, if necessary ; next
multiply the rational quantities together, and the surds toge
ther ; tnen annex the one product to the other for the whole
product required ; which may be reduced to more simple
terms if pecessary.
EXAMPLES.
200 ALGEBk A.
EXAMPLES,
!• Rei^ulred to find the product of 4^^/12 and Sv^2.
Here, 4x3xv'12xV2 = A2v/12x2=12v'24=12^4x6
= 12 X 2 X \/6 = 24 v' 6. the product required,
2. Required to mukiply ^^i by ^.
Here ^x^>?/^x/ = ^x^^=^\xl/ii=^^xix Vl8 =
^^18, the product required.
3. Required the product of 3^2 s^d 2V'8. Ans. 24.
4. Required the product of ^ and V12. Ans. ^^6.
5. To find the product of j^s/i ^^^ Av^t« Ans. ^^ 15.
6. Required the product of 2^14 and S^4. Ans. 12^7.
ft 4
7. Required the product of 2a^ »nd a^. Ans* 2a\
8. Required the product of {a + 3)^ and {« + *) ♦.
9. Required the product of 01.^ + \/* and 2x— v^*.
10. Required t^ product of (a + 2v/i)^, and («  2v/*)'
11 J, Required the product of 2x' and Sx"*.
i J.
12. Required the product of 4;? and 25^**
/
PROBLEM mi.
t
y
To Divide one Surd Qmntity 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 give the whole quotient
required 5 which may be reduced to more simple terms if
requisite;.
EXAMPLES.
1. Requh^d to divide 6^/96 by 3 v' 8.
Here 6 r 3 .V(96 i 8)ifc2v^l2 =: 2v/(4xa) :?? 2x2v/3
= 4v/3, the quotient required.
a Required to divide 12^280 by 3^5.
Here 12 f 3 = 4, and 280 r 5 = 56 = 8 x 7 = 2^7 ;
Therefore 4 x 2 x4/7 = 8^7, is the quotient i^equired. •
3. Let
SURD8. rSW
S. Let 4i/50)be divided by 2V'5. Ans. .2^ 10
4. Let 6V100 be divided by 3^5. Ans. 2^20.
5. Let Iv^tW be divided by ^Vf. Ans. ^^y/B.
6. Let ^;^j% be divided by ^. Ans. ^l^SO.
7. Let 4 v'/i, or Jo^, be divided by j^^. Ans. ^^.
8. Let ^^ be divided hja^.
9. To divide Sa^ by 4/i *. "*
PROBLEM VIIX.
To Involve or Raise Surd Quantities to any Power*
■ >
Raise both the rationd part and the surd patrt. Or mul»
Uply 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. Required to £nd the square of ii\
Fin^t, ()*=ix 1=^2^, and {a^Y=za^^^^a^:^a.
Therefore (^j^)* = iV^i ^s the square required.
2. Required to find the square of \a'^.
Fkst, i X 1 = ^, and {a^f = a^'a\/a\
z
Therefore {ia^Y ^ ^fi^l/a ia the square required.
I
* 3. Required to find the cube of .v/6 or f x e''.
First, {\y =z f X I X f z= ,V» and (fi^f =r 6* = 6^6 j
Theref. (v/6)3 = ^s^ x 6^/6 = ^^^s/^ytbQ cube required.
4. Required the square of ,2 V2. An«k 4^4.
I •
5.. Required the cube of 3^, or /S. Ans. Sy'S.
6« Required the 3d power of y V 3* . Ans,  v^ 3.
7* Required to find the 4th power of iy/2. Ans. 4»
8. Required
2M ALGEBRA.
S. Required to find the mth power of tf".
9. Required to find the square of 2 + VS.
PROBLEM IX.
To Evolve or Extract the Roots of Surd Quantities*.
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 IGv'S.
First, v'lS = 4, and (6^^= 6^"^^ = 6^ ; '
II ' ^
theref. (16 V^6)^ =: ^.6\zz 4jJ/6, is the sq. root required.
2. Required to find the cube root of rr ^^*
Fust, i/,V = h and (^S)^ = 3^"="^ r: 3^ j
theref. (^ V3)'''=j. . 3^ = ^5^3, is the cube root required.
3. Required the square root of 6^ Ans. 6\/6^
4. Required the cube root of ^tf^i. Ans. ial/6,
5. Required the 4th root of I6a\ Ans. 2Va.
X.
6. Required to find the mth root of x" •
7. Required the square root of a* ~ 6a Vi + 9^.
* The square root of a binom ial or r esidual surd, a ^ b,
m^h, may be foand thus : Take \/a« — ^ =z cj
or
. 7 .flf + C fl—C
then ^/a +6 = •— r— + v^tt;
— ^ — r a fc a — c
and v^o 6 =z •/— ^T"'
Thus, the square root of 4 + 2v^3 = 1  ^3 ^
and the square root of 6— 2\/5 := ^5 — 1 . .
But for the cube, or any higher root, qo general rule is
known.
inhntte
SURDS. 20S
INFINITE SERIES.
An Infinite Series is formed either from division, dividing
by a compound divisor, or by extracting^ the root of a coih
pound surd quantity ; and is such as, being continued, would
run on infinitely, in the manner of a continued decimal
fraction.
But, by obtaining a few of the first terms, the law of the
progression will be manifest; so that the series may thence be
continued, without actually performing the whole operation.
PROBLEM I^
To Reduce Fr^actional Quantities into Infinite Series by Division.
Divide the numerator by the denominator, as in commcm
division; then the operation, continued as far as maybe ,
thought necessary, will give the infinite series required.
EXAMPLES.
2ah
1 . To change r into an infinite series.
a ^ a
. . 2i* 2*^ 2**
M+b)2iA..(2b hT r + &<^
2ab + 2b^
2i*
 2**—
a
2*»
a ■
2i} 2b*
2i*
23'
*
,3 .&c.
2. Let
a04 ALGEBRA.
t
2. Let be changed into an infinite series.
1 — a
I  tf ) I . (l+a + tf*+^ + a^ + &c.
1  a
a
a*
«*a»
a'
S. Expand ^^ into an infinite series*
Ans. — X (1  — hi r + &c.)
4. Expand — ^. — 7 into an infinite series.
a <r Or
1  ^ .
5. Expand r—r — into an infinite series.
Ans. 1 — 2.r + 2j:*— 2;ir' + 2x*, &c j
«* . . .
€. Expand  — \ — rrr into an infinite series.
.2* 34* 44'
Ans. 1 ' 1 — r r> &c.
fl ' a* a'*
1
7. Expand •; — ; —  = 4> into an infinite series.
1 4 1
PROBLEM II*
To Reduce a Compound Surd into an Infinite Series.
Extract the root as in common arithmetic ; then the
Operation, continued as far as may be thought necessary, will
give the series required. But this method is chiefly of use
in extracting the square root, the operation being too tedious
for the higher powers.
EXAMPLES.
INFINITE SERIES. fiOi
EXAMPLES.
i. Extract theitoot of rf^— ;r* in an infinite $erte.
2a Sa^ I6a^ 128a7
a"
'Ta^^
or* a:* . ;i"*
2^:^)
4a* 8a* 64a*
a 4a3 ' 8«* 64a'
6
*• *»
r*+i
8iJ* • 16a'
&c.
640^
2. Expand \/ 1 + 1 = \/^, into an ininite series,
Abs. 1 + i  i + rt  xIt &c
3. Expand V'l — i i^^^o ^^ infinite series.
Ans. liliV— i4t*^
4. Expand ^/a^ + x into an infinite series.
 '  *»
5. Expand Va^2bx — or* to an infinite series.
PROBLEM III.
3i Epitract any Root of a Binmud : or to Riduce a ^ BimmicJ
Surd into an Infinite Series*
Thi jj trill bp done by substituting the particular letters of
the binomial, with their proper signs, in the following
general theorem or formula, viz.
(p + PQ^) n =p n +  ^^^—— B(^+ —^ C(^+ &C.
and
206 ALGEBRA.
and it will give the root required : observing that p dehotes
m
the first term, (^the second term divided by the first, T the
index of the power or root ; and A> b, c» d, 8cc, denote the
several foregoing terms with theii: proper sighs.
EXAMPLES.
1.. To extract the sq. root of a^+ ^> in an infinite series.
Here p = tf*> Q = «f and — = — : therefore
m m r
F n = (a*) Q = (a^)^ zz a ss At the 1st term of the series.
— AQ = i X a X — = — 5= B, the 2d term.
m^n 12 4* 4» 4* , ,
—  — B(i X — r X :r" X "1 = — rr, = ^j the 3d term
2« ^ 4 2fl a* 2.4a^
m2« 1—4 4* 4* 34*  ,'
3^c^ ; — X  — 3 X ^= 5;^:^ = D the.4th.
> 4* 3.4*
Hencea+g^53j^3 + 5;j;^  ^^^^^' '
4* 4* 4^ 54»
1
2. To find the value of ;^ r, oritsequal (a— a;)"*kin
(a— x)*
an infinite series*.
* Note. To facilitate the application 'of the rule to fiactional
examples, it is prgper to observe, that any surd may be taken fi*om
tb^ denominator of a fraction and placed in the numerator^ and
rice versa, by only changing the sign of its index. Thus,
^ =1 X jr^ or only jr« 5 and—— = 1 X (a +4)"^ of
(« + 4)« y and • ^ .^ = tt^a + jp)« j and f! == x^ X «^ ; alse
— 1
(^ _ ' ^.y i = («Hx*)* X (o«^) 'i &c.
H«rt
Here p
INFINITE SERIES. »)7
r
=fl, o== = — ^~ •*"> and — = — = — 2 ; there*. ♦
«», 1
p » =: fa)"* = a"* r:— :; = a, the 1st term of the series*
* ^ .a*
— AQ=:— 2 X — X = , = 2a"'^ =x B, the 2d term.
2« ^ ^ a^ ^ fl*
w — 2« ^ Sot* X 4ir' . t ,
Hence at* + 2a"'a: + 3a"*jr* + 4flr*jr'. + &c, or
1 . 2x Sx^ AiX^ . 5x^ «.,.'• J
~7 H i + —7 + — r H s ^c, IS the series required,
;«* ' ^' ' a* ^ ^5 ^ ^^ . ^
3. To find the value of^ — » in an infinite series.
a — x
x^ x^ s^
Ans; a + x { +5+r&c
4. To expand V ^^^^^ or ^^,^^^4 in a series.
. 1 X* 3x* 5^** ,
Ans. ^rT + "—7 — TTij aCC.
a*
5. To expand tt: in an infinite series*
*^ (a — Of
2J 3i* 4J^ 5J* ,
Ans. 1 +— + ^ + ^ + "^ *^
a a* a* a*
e. To expand Va^x^ ox (a* jt*)"^ in a series
* x^ x?" 5x^
X ^ w %^— «
■*"*•''" 2a ~ 8a' T6a»~"l28a^
7. Find the value ofMa'i') or (a'A')^ in a series.
Ans.a — ^g^&c.
S. To find the value of V(a^ + ^0 or («*+^0^^"^ series.
x^ 2x^° 6jr**
Ans.a+^^^,+ j5^&c.
9. To
90%
ALGEBRA.
9. To find the square root of ' n in an infinite serief*
I
. b M x^ ^^
AnS. 1 ;;; +;;£ '"^ ^^'
10. Find the cuW root of
a'
e^ + b^
2a'
in a series.
63 3J6
Ans. I — — r + rx
&c.
ARITHMETICAL PROPORTION.
Arithmetical Proportion is the relaticxi between
two numbers with respect to their difference.
Four quantities are in Arithmetical Proporti(Mi, when the
difference between the first and second is equal to the dif
ference between the third and fourth. Thus, 4, 6, 7, 9,
and Uy a { di b^ b { d, are in arithmetical proportion.
Arithmetical Progression is when a series of quantities
have all the same common difference, or when they either
increase or decrease by the same common difference. Thus>
2, 4, 6, 8, 10, 12, &c, are in arithmetical progression, hav
ing the common difference 2; and a, a ^ d, a + 2dy fl + 3</,
a + 4^/5 a + 5dy &c, are series in arithmetical progression^
the eommon difference being d.
The mpst useful part of arithmetical proportion is con
tained in the following theorems :
«
1. When four quantities are in Arithmetical Proportion,
the sum of the two extreme^ is equal to the sum of the two
means. Thus, in the arithmetical 4, 6, 7, 9, the sum 4 +
9 = 6 + 7=:l3: and in the arithmetical fl, a+rf, A, pkd^
the sum a^b^dzia + b+d.
2. In any continued . arithmetical progression, the sum of
the two extremes is equal to the sum of any two terms at an
equal distance from them.
tAus,
ARITHMETICAL PllOPORTION. *p^
Thus, ifAe ^^ries bfe 1, 3,5, 7, 9, 1 r, &c.
Th^a 1 + 11 = 3 + 9 = 5 + 7 = 12. . ■
S. Tlie fast t^rm of an^ increasing arithmetical series, is
equal to th« first term increased by the product of the
common 'difference multiplied by the ntimber of tenn^ lessr
one ; but in.a.4^rea^g>;Serie9r t,^^ is^t tQrmhiii<e^ual to. the
first term lessened by i^inesaid ^pduct^
/ Thus, the 20th tewti of the^ s«fies 1 , »V 5is7, 9j &c, is sss'
1 + 2 (20 1 J «= r 4 2 i^ 19 = 1 ^^W^zz'SQ. •
And the «th term of tf, a^d, a^^Qd, ar^Sd, «'4i/,'&c,
4. The sum of all th^ fvrffisimHty series in arithmetical
progression, is equal to half the sum of the two extreines
multiplied by the number ojTterm^. ^  ^ ,.. t.
Thus, the sum.of 1, 3, 5,%4>i &ci coi^iinved to the lOfli
. (1 + 19) X 10 '20x10 : •
term, is =5^ ^ =: — r — a: lo x lO =s 100,
Andthisvmo£ntermso{a,a+d^a + 2d, a+^d^toa^md, "
her ...:
\
1^ The first^tern)^,^ jaAHnKreasing arithmetti:al seriei^is 1,
the comman, difference 2^ and thts namber of teritos'421 } re«
<)uired the si^^pf ^^ series ?. , . ~ r i ^^ :: I
First, 1 +2 x: 20^ 1 +40'== 41, isltHe lifet^term.
14:41' ' '^^" '' ■ I ^ ^ > ^
Thenf ^  ^ fe'«0 ==21 x 20 =i: 420, {heWm required.
•. .' *' ' i  « L w: j: 4 . .if/j* .'■t ! ^'^ff /^
2k The first term of a decreosing^arithnietical so^es i# 1 99^ *
like common dffiTcnrence'l^j and tht dumbe'r of tettxik 37 ; re
quired th* ^m 6f the serieaf ? • :i '^j ' ^
; JFirst, 1 9d  3 . e^ = 1 dp  ISiS h J, iS'lhe last term,' . ,;, ,
^ Then ^ ^ X 67 = lOQ ?^ ^7 ;=,,6700, .th,e sum, ric^r
quired.'' ^^ ' ■ ■'■"'' ' ^ ••'^ • '• '' ' •^^;
^3. Tp find the^sjim pf JOO^terins ©fjtjbe. ns^fllil^nUi^i^bm
Y>VL 1. , P . ^« He<]vured
4. * Required the ^um pf 9^ terma of Ike ^d omilbers
1, 3, 5, 7, 9, &c. ._ . . . . • ; An* 9811.
5. T2^ ^st term of a decres^ng.oH^ui^tic^i^rit^ls lO,
the oamojQn. diff$i!9nce f, and the Dtin^r of te^m^Sl )
i^eqaked^.supapf, the serves? . r. Ana. 140.
6. <)biejhitiidl«d' stoned beiR^ 'plai!e4«ii the ^6uhd, ins
straight line, at the dktanc^ of ^' yitrdft frtittt each bther ;
how far . viU zi^spri ^ tra?el» . whp \Ah9li bring tben\ ooe by
one to a iyA9kst%_y^ck is pi^ed 2 y^ds ftom^ihe first
stone f ^ . Ans. 11 miles ^d 840 yards*
*v .'!..' /I *•• ! o? 'r.
APPXICATJON OF ARITHMETICAjL PROGRESSION
' ' TO MBLfTA^T 'AfTAIl^S. ^' ;
■ .1 X "i • ■• ■■  • • ■
OCTESTION I.
A Triangular Battalion f, consisting of thirty ranks,
in which tiie first rank is forifled of vone man biify, the
second
* The sum of aoy miAiber in) df IMUH: 4f the arithmetical
aeries of odd number \,Z,5f 7, 9, &C is equal to the square (n^)
pf that number. That is,
If 1 1^ '3, ^^ ^i 9» he, be (he nnoiiMn/than ^!U ^
. ^^i %\t^A\ 6\ Uc, be the suiiB 0£ Bpa, d^ dse, tormi;
Thus, + 1 =: 1 or I2,the*tfm^i tfertn, •
I .lb ^3 f?i 4or2^^the8u*o68t^ms . 
4 + 52= o or 3^, the sum of 3 term9,
, 9 + vj s: igor4>thefium.of4 t6rm4>.icfr * .
For, by the 3d theorem, 1+2 («— 1) = l'+ 2«— 2 = 2ii— 1
i8thQli(5tterm^wfaenitfa9e''Otmberof'jlbrms ig.n; ti» tBia hist t^ui
2»— \/Mi t^ flrat tpfro. h give^2n ^ sum of thexearfarciBes, or
n half the sum of the extremes ; then, < by the 4tb thi^oretpi nXfi
=r 71/^ is the sum of , all. the term?. Hence it appears in general>
that half the suioa df the extremes;, is dlways iht satne as the Hub*.
ber of the tei^s n ^ ,^Dd that the sum of all the tcmk,'k the ^ame
as ihi^squai^ of the same ndiriber, n^.
See more on Arithmetical Proportion in the ArithoqtetiCj^.
p. 111.
'y.^'iti^ti^lM bittiiron, is tp be understood, a body of Iropps.
Wg^ in the form of 9 triangle^ in which the Jl^ks exdedd ^ch
ARITHMETICAL t^ROGRESSION. 21 1
rtcond of S, the third of 5 5 and so on : What is the
strei^[di of such a triangular battalion ?
'^ Ans#eri 900 iiten.
* J
<^J£ST20K II,
A detachment having 12 snccesshedays til march, with
orders to advance the first docf only i ieagae$j the second S^,
and so on, increasing I4 league each day's inarch: What is
the length of the whole marth^ and what is the last day's
inarch?
Ansjver, the last day's march is I8j l^^igues, and 12S
leagues is the length of the whole march*
qjJESTION III.
A brigade of sappers*^, having carried on 1$ yards of
3ap the first night, the second only 13 yards, and so on,
decreasing 2 yards every night, till at last Uiey carried on in '
one night only 3 yards : What is the number of nights thef
Were employed ; and what is the whole length of tne sap ?
Answer, they were employed 7 nights, and thq length of
the whole sap was 63 yards.
n . ' » i
other by an equal number of men : if the first ranl^ consist of one
man only^ and the difference between the ranks be also 1, tbea
its form is that of an^equilateral triangle , and when the dljfbrenc^
between tlw ranks is more than i, its form may then be an
isosceles or scalene triangle. The practice of forming troops fa
this order, which is now laid aside^ was formerly held in greatet
esteem than forming them in a tfolid square^ a^ admitting of a
greater fronts especially when the troops were to make simplj a
stand on ail sides.
^ 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 wnen the first party is tired, t^e
second takes its place, and so on, till each man in turn has heed
at the head of the sap. A sap is a snudl ditch, between 3 and ^
feet in breadth and depth , and is distinguished froin the trench
by its breadth only, the trench having between 10 and 15 feet,
breadth. As an encouragement to sappers, the pay for all the
work carried on by the whole brigade,, is given to the survivors*
F 2 QJJESTfQN
212 ALGEBRA.
QUESTION IV,
A number •£ gabions ^ being given to he placed in she
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 f
Answer, the difference between the ranks will be 1» and
the number of gabions in the six ranks will be 39.
' * « •
QUESTION V. *
Two detachments, distant from each other 37 leagues, and
both 'designing to eccupy an advantageous post equidistant
from each other^s camp^ set out at different times ; the first
detachment increasing e^^ery 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* diiys'
faiarch ; What is ^he number of leagues marched by each
detachment each day ?
The progression ^, 2^xy, 3,^, 5^, 6^y answers the con*
ditions of the first detachment : and the progression 14> 3<»
^Ti ^h smswers the conditions of the second detachment.
QUESTION VI.
A^eseitery in>his flight, travelling at the rate of 8 leagues
a day ; and a detachment of dragoons being sent after him^
with orders tamiarch the first day only 2 leagues, the second
5 leagues, the third 8 leagues, and so on : What is the
number of days necessary for the detachment to overtake the
deserter^ and what will be the number of leagues marched
before He is overtaken ?
Answer, 5 days are necessary to overtake him \ and coij
^quently 40 leagues will be the extent of the march.
*•■
* Gabions are baskets^ open at both ends, made of ozicr twigs,
smd of a cylindrical form : those made use of at the trenches are
2 feet wide, and about 3 feet high $ whicb^ being filled with earthy
serve as a shelter from the enemy's fire: and those made use of
to construct batteries, are generally higher and broader. There
is another sort of gabipn, made use of to raise a low parapet : its
height is from 1 to 2 ftpt, and 1 foot wide at top, but somewhat
les^ at bottom, to give room for placing the muzzle of a firelock
between them : these gabions sen^e instead of sand bags. A sand
bag is generally made to contain about a cubical foot of earth.
• QUESTION
PILING OP BALLS.
(UTESnOH TII.
A convoy • tSstant 95 leagues, having orders to join its
camp, and to march at the rate of 5 leagues per day; its
escort departing at the same timci with orders to march the
first day only hidf a league, and the last day 94 leagues;
and both the escort and convoy arriving at the same time : ,
At what distance is the escort m>m the convoy at the end of
«ach march ?
OF COMPUTING SHOT OR SHELLS IN A FIHISHED PILE.
'Stiot and Shells are generally piled in three different
forms, called triangular, square, or oblong piles, according
as their base is either a triangle, a square, or a rectangle.
Fig. 1. C G Fig. 2,
ABCD, fig. 1 , is a triangular pile,
EFGH, £g. 2, is a square pile.
E A Fig.
ABCDEP, iig. 3, is an oblong pile.
* fi^ convoy is generally meant a snp^y of ammunition or
provisions, conveyed to a town or army. The body of men that
{Hard this supply^ is cslled Mcort.
A triangular
314 ALGEBRA.
A triangulir pile is formed by the continual Ikying of
triangular horizontal co.urse^ of shot one above another, i%
such a manner, as that the sides of these courses, called rows,
decrease by unity from the bottom row to the top row^*
which ends always in 1 shot.
A square pile is £Drm^d by the continual laying of square
horizontal courses of shot one above another, in such a man
ner, as that f he sides of these courses decrease by unity from
the bottom to the top row, which ends also in 1 shot.
In the triangular and the square piles, th& sides or faces
being equilateral triangles, the shot contained in those faces
form an arithmetical progression^ having for first tem^unity,
and for lait term and number of terms, the shot contained
in the bottom row ; for the number of horizontal rows, or
the number counted on one of the angles from the bottom to
the top, is always equal to those counted on one side in the
^ttom : the sides or faces in either the triangular or square
piles, are called arithmetical triangles ; and the nuoaibers
contained in these^ are called triangular numbers: abc, fig. 1,
£FG, fig. 2, are arithmetical triangles.
The oblong pile may be conceived as formed from the
square pile abcd ; to one side or face of which, as ai), a
number of arithmetical^triangles equal to the face have been
added : and the number of arithmetical triangles added to the
square pile, by means of which the oblong pile is formed,
is always one less than the shot in the top row ; or, which
is the same, equal to the difference between the bottoooTrow
of the greater ^ide and that of the lesser.
QIJESTIOH Till.
To find the shot in the triangular pile abcd, fig. ], the
b6ttdm row ab consisting of 8 shot.
SOLUTION.
The proposed pile consisting of S horizontal courses, eack
of which forms an equilateral triangle ^ that is, the shot
contained in these being in an arithmetical progression, of
which the first and last term, as also the number of terms,
•are known ; it follows, that the sum or these particular
^ourses, or of the 8 progressions, will be the shpt contained
in the proposed pile i then .
The
PILING OS BALLS.
tlS
Hie sW of the fitst or lowet >
^ trianj^oiaf coarse vttlbe 5
8+ 1 X 4 ss'^e
the second 
' ( 1
* <
7+a X3^=gf
the itivd \ ' m .' ^ * «.
6 + 1x3 &: 2t
§
the fourth 
5 ^ I X 2i s= 15
the fiita • •* • 
♦ + I xS = 19
the sixth   
3 + 1 X li =s $
the seventh * 
2 + 1x1 = 3
the eighth  
1 + ix i* I
X
Total  I20sl
J
in the pile propo$e(4
qt^ESTioN IX.
To find the shot of the square pile xfgHj fig. 2, the bot^
torn row £F consisting of 8 shot.
SOLUTION.
The bottom row containing 8 shot> and the second only 7 ;
that isj the rows forming the progression, S, 1, 6, 5, 4, 5,2, 1 p
in which each of the terms being the square root of the shot
contained in each separate square course employed ill formin^r
the square pile j it follows, that the sum of the squares of
these roots will be the shot required : and the sum of the
squares, divided by 9» 7, €, 6, 4> S^ 2, 1 , being 204, expresses
the shot ia the proposed pile.
QJTESTION X.
To find the shot of the oblong pile abc^def, fig. 3 ; in
which >F «s 16, sind xc =b 7.
SOLUTION.
The ^oblong pile proposed, consisting of the square pile
ABCD, whose bottom, row is 7 shot ; besides 9 arithmetical
triangles or progressions, in which the first and last term^ ?s
also the number of term$, are known ^ it follows^ that,
if to the contents of the square pile  14Q
we add the sum of the dth progression  252 ,
their total gives the contents required  S92 shot*
REMARK I.
The shot in the triangulin: and the square piles, as also
the dbiot in each, horizontal course, may at once be ascer<^
tained
v^
ALGEBRA.
tauned by the followmg table : the Optical c<diimn a, coiu*
taint the shot in the bottom row, from 1 tx> 30. izicliisiTe ;
the column b contains the triangular numbers, cr number
of eslch course ; the column c contains the sum of the
triangular numbers, that is, the shot eontaihed in a trian^
gular pile, commonly called pyramidal numbers ; the column
B contains the square of the numbers of the column a, that
is, the shot contained in each square horizontal course ; and
the column £ contains' the sum of these squares or shot in a
square pile.*
c
B
A
D
E
Pyramidal
Triangular
Natural
Square of
thj* vtAtuml
Sumofthese
square
numbers.
numbers.
numbers.
numbers.
LUC llV«*iilflil
Qumbers.
1
I
1
. 1
I
4
3
2
4
5
10
6
S
9
14
20
^ 10
4
16
30
35
15
5
25
55
56
21
6
36
91
34
28
7
49
140
120
36
8
64
204
165
45
9
81
285
220
55
10
100
385
28(J
66
11
121
506
♦ 364
78
12
144
650
455
91
13
169
8I9
5d0
105
14
19^
1015
680
120
15
225
1240
816
136
16
^56
1496
969
153
17
289
I78d
1140
171
18
324
2109
1330
190
19
,361
2470
1540
i 210
20
400 2870 t
^Thus, the bottom row in a triangular pile, consisting of
9 shot, the contents will be 165 ; and when of 9 in the square
pile, 285.— ^In the same manner, the contents either of a
square or triangular pile being given, the shot in the bottom
roynfc may be easily ascertained.
The contents of any oblong pile by the preceding table
may be also with little trouble ascertained, the less side not
exceeding 20 shot, nor the difference between the less and
the greater side 20. Thus^ to find the shot ia an oblong pile,
... the
PILING OF BALLS. <11
the less side bein^ 15, and the ^eaten95, we are first to
imd the contents of the square pile, by means of which the
oblong pile may be conceived to be formed $ that is, we tare
to find the contents of a square pile, whose bottom ro# is
15 shot ; which being 1240, we are, secondly, to add these
1240 to the product 2400 of the triangular number 120»
answering to 15, the number expressing the bottom row of
the arithmetical triangle, multiplied by 20, the number of
those triangles; and their sum, being 3640, expresses the
number of shot in the proposed oblong pile.
REMARK II.
\
The following algebraical expressions, deduced from the
investigations ot the sufns of the powers of numbers in
arithmetical progression, which are seen upon many gunners*
callipers *, serve to compute with ease and expedition the shot
or sheUs in any pile. •
That serving to compute any triangular ) ^^^ + 2x /i f 1 x^
pile, is represented by I 6
\
That serving to compute any square ) ^ f 1 x 2fg 4 1 x ft
pile, is represented by ) 6
In each of these, the letter n represents the number in the
bottom row : hence, in a triangular pile, the number in the
bottom row being 30 ; then this pile will be 30 + 2 x SO + I
X V ^ 49$0 shot or shells. In a square pile, the number
in the bottom row being also 30; then this pile will be
30 + 1 X 60 + I X ?/ = 9455 shot. or shells.
That serving to compute any obiong pile, is represented by
2/1+1 +3«ix« + 1 X « . ,11. 1 J
■ ' ■ • — i in which the letter n denotes
* Callipers are large compasses, with bowed shanks, serving to
take the diameters of convex and concave bodies. The gunners'
callipers concisx of two thin rules or plates, which are moveable
quite round ^ joint, by the plates folding one over the other : the
length of each rule or plate is t> inches, the breadth about 1
inch It is usual to represent, on the plates, a variety of scales,
tables, proportionH, 6kC, such as are esteemed useful to be kuovq
by persons employed about aitillery ; but; except the nfMsuring
of the caliber of shot and cannon^ and the measuring of saliant and
reentering at^gles, none of the articles, with which the callipers
are usually filled^ are essentia^ to that instrument.
the
fit ALGEBRA.
the number of coimcSy aBcl the letter m the number of shat^
less ODe> in the top row : hence, in an oblong pile the num
ber of courses being 30» and the top row 31 ; thb pile wiU
be 60 + 1+90 X 30+1 x V" = 23405 shot or shells.
GEOMETRICAL PROPORTION,
Geometrical Proportion contemplates the relation of
quantities con^dered as to what part or what multiple one
is of another, or how often one contains, or is contained
in, another.— Of two quantities compared together, the first
is called the Antecedent, and the second the Consequent.
Their ratio is the quotient which arises from dividing the
one by the other.
Four Quantities are proporticmal, when the two couplets
have equal ratios, or when the first is the same part or mul
tiple of the second, as the third is of the fourth. Thus,
3, 6, 4, 8, and a, ar^ b, br, are geometrical proportionals.
or bt
For ^ = . = 2, and — = — = r. And they are stated
thus, 3 : 6 :: 4 : 8, &c.
Direct Proportion is when the same relation subsists be
tween the first term and the second, as between the third and
the fourth : As in the terms above. But Reciprocal, or
Inverse Proportion, is when one quantity increases in the
same proportion as another diminishes : As in these, S, 6, 8,
4 ; and these, a, ar^ br^ b^
The Quantities are in geometrical progression, or con
tinuous proportion, when every two terms have always the
same ratio, or when the first has the same ratio to the
second as the second to the third, and the third to the
fourth, §cc. Thus, 2, 4, 8, 16, 32, 64, &c, and a, ar^ at^,
ar^, ar^y ar^, &c, are series in geometrical progression.
, The most useful part of geometrical proportion is con
tained rtr the following theorems ; which are similar to those
in Arithmetical Proportion, using multiplication for addi
tion, &c.
. 1, When
r
GEOMETRICAL PROPORTION. 819
1. W^jen four qujuitities are in geometrical j>roportion,
the product of the two extremes i$ equal to the product of
the two means. As in these, 3, 6, 4, 8, where 3x8=6
X 4f=i24:i aad in tht^se, a» ar% b^ br^ where ax kr^ar x
2. When four quantities are in geometrical proportion,
the product of the means divided by either of the extremes
gives the other extreme. Thus, if 3 : 6 : : 4 : 8^ then
6x4 6x4
■ ■ =: 8, and ■ = 3} also if tf : ar :i b : br, then
= ir, or — r— = a* And this is the foundation of the
a or
Rule of Three.
9* In any continued geometrical progression, the product
of the two extremes, sekI that of any other two terms,
^quaUy ^stanl: fr(»n them, are equal to each other, or equal
to the square of the middle term when there is an odd
number of them. So, in the series 1, 2, 4, 8, 1(5, 32, 64, &c,
it is 1 x 64 = 2 X 32 = 4" X 16 = 8 x 8 = 64.
4. In any continued geometrical series, fhe last term is
f qual to the first multiplied by such a power bf the ratio as
is denoted by 1 less than the number of terms. 'Thus, in the
series, 3, 6, 12, 24, 48, 96, 8cc, it is 3 x 2^ = 96.
5« The sum of any series in geometrical progression, is
{bund by multiplying the last term by the ratio, and dividinr
the difference of this product and the first term by the difc
ference between I and the ratio. Thus, the sum of 3, 6,
192 X 2 — 3
12, 24, 48, 96, 192, is — ^ = 384 3 == 381. And
2— i
the suro of n terms of the series j, or, ar^, ar^^ ar^, &c, to
, . ar^"^ X r— tf af—a r" — 1
I jg rz ~ = a.
' r—l r 1 r — 1
6. When four quantities, «, tfr, ft, Ar, or 2, 6, 4, 12, are
proportional ; then any of the following forms of those quan
tities are also proportional, viz.
1. Directly a : ar :: b : Ar ; or 2 : 6 : : 4:12.
2. Inversely, ar :a : : ir : A j ar 6 : 2 : : 12 : 4.
3. Alternately, j : 6 : : «r : ir; or 2 : 4 : : 6:12.
4. Coxn^
9ib AtCEBIlA.
4. Comp6\md€dlyfO:a+ar::b:b+kp\ or2:S::4:I6.
5. Dividediy, a i ur^a iihi br~b\ or 2 : 4 :: 4 : 8.
6. Mixed, ar^g znr^a :: br^bibr^b\ or 8 : 4 :: 16 : S.
7. Multiplication, aciarc: : fc : brc ; or 2.3 : 6.3 : : 4 : 12*
8. Division, — :—:;>: ir; or 1 : 3 :: 4 : 12.
c c
9. The numbers a, b, c, d^ are in harmonical proportion,
when A I d II a^nh I czo d\ or when their reciprocals
1111
•^% "T* ""■> ^> are in arithmetical proportion*
EXAMPLES.
1. Given the first term of a geometrical series 1, the ratio
2> and the number of terms 12 ; to find the sum of the series?
First, 1 X 2" = 1 X 2048, is the last ternu
i„ 2048x2 — 1 4096 — 1 , , .
Then — ss ■ == 4095> the sum required.
2. Given the first term of a geometric series f, the rati©
I, and the number of terms 8 ; to find the sum of the series?
Krst, I X {\y = I X TTT = TTTf »s the last term.
Then (f ^3^ X 4) 4 {\i) = (iyW) ^ 1 = fH X x
=: f 44'> ^^ ^^^ required.
3. Requited the sum of 12 terms of the series 1, 3, 9, 27,
SI,&c. Ans. 265720.
4. Required the sum of 12 terms of the series 1, f, ^t
Tjy Tr> &€• Ans. ttHtt*
5. Required the sum of 100 terms of tha series I, 2, 4, 8,
16, 32, &c. Ans. 1267650600236229401496703205375.
3ee more of Geometrical Proportion in the Arithmetic*
I
I
SIMPLE EQUATIONS.
An Equaticm is the expression of two equal quantities,
with the sign of equality (=•) placed between them. Thus,
10^4 =x 6 is an equation, denoting the equality of the quan
tities J.0 — 4 and 6.
Equations
SIMPLE EQUATIONS. 221
Equations are either simple or compound. A Simple
Equation, is that which contains only one power of the un
known quantity, without including different powers. Thus,
x^a Si b + c, or ax^ = 4, is a simple equation, containing
only one power of the unknown quantity iZ*. But jt*— Sat
:;^&^ is a compound one.
Reduction of Equations, is the finding liie value of the
unknown quantity. And this consists in disengaging that
quantity, from the known ones ; or in ordering the eqtia^
tion so, that the unknown letter or quantity may stan<i
alone on one side of the equation, or of the mark of equality,
without a co^efficient ; and all the rest, or the known quan
tities, on the other si€le.«4n general, the unknown quantity
i$ disengaged from the known ones,> by performing always
the reverse operations. So, if the known quantities are con
nected with it by + or addition, they must be subtracted ; if
by minus (—), or subtraction, they must be added'; if by
multiplication, we must divide by them ; if by division, we
^ust multiply ; wheQ it is in any power, we must extract
the root ; and when in any radical, we must raise it to the
power. As in th^ foUdwiiiff particular rules ) which are
tbundedon the general principle of performing equal operas
tiQits on equal quantities; in which case i is evident that
the results must still be equal, whether by equal additions,
or ^ubtraoliops, or multiplicatipns^ or divisions, or roots, or
powers,
FARTXCULAIt RULE I,
«
When known quantities are connected with the unknown
by + or — } transpose them to the other side of the ecjua
tion, and change their signs. Which is only adding or sqb'
tracting the same quantities on both sides, in order to get
all the unknown terms on one side of the equation, and ail
the icnowi pnes on (he other side *«
  Thusy
' » Wii I " ■ ■ . ■! . ■'■■ * '■' ^ " ^1 ■■■»■>!
♦ Here it is earnestly recommended that the pupil be aqf
customed^ at every line or stejj in the reduction of the equation $».
to name tbo particular operation to be performed on the equation
in the last line^ in order to produce the next form or state of the
eqoation> in applying each of these rules, according as the particular
^rm of th^ equation may require 5 applying them according to the.
order
2^2  ALGEBRA. •
Thus, i(x 4 5=S ; then transposing 5 givte jr=r8 — 5==$.
And, if a; — 3 {7=^9 then transposing the 3 and 7, gives
j:=:9 + 3— 7 = 5.
Alsb, if X ^ s f b=s cd: then by transposing dr ^nd 5,
itisx =: a — b + cd.
In like manner, if 5jc — 6 = 4x + 1 0, then by transposing
6 and 4jr, it is 5r.4?a: = 10 + 6, or .r = 16,
RUtE II.
"When the unknown tferm is multiplifed by aiiy quantity ;
idiride all the terms of the equation by it.
Thus, if /7jr=^5 — 4a; then dividing by a, gives x ^i^if.
And, if 3x + 5 xs 20 ; then first tron^osing 5<fi^^ SJt:
sc 15 } and then by dividing by 3, it is x = 5.
In like manner, i{ax+3ab=4€^^f then by dividing by <j, it
4r* 4:^^
is x+Sb = ; and then transposing 8^, gives x = — 3A
RULE III.
When the unknown term is divided by any quantity ; we
tnust then multiply all the terms of the equation by that di
visor ', which takes tt away.
Thus, if— =i= 3 +2 : then mult, by 4, gives a: = 12 4 8 = 20*.
4
And, if — = 3i + 2^  rf;
a • ' ■
then by mult, a, it gives x = Sab + 2ac — ad.
5
Then by trani^posing 3, it is ^x = 10.
^d multiplying by 5, it is 3x = 50.
Lastly dividing by 3 gives j; = 16y.
order in which tbey are here placed , and beginning every line
with the words Then by, as in the following specimen^ (^ Bt
amjples > which two words will always bring to his recoUectioo^
that He is to pronounce what particular operation he is to perform
dn the last \\{ie, in order to give the next , allotting atways a
single line for each operation^ and ranging the equations neatly
just under each otlier^ in the several lines^ as they are successively
produced.
Rule
SIMPLE l£QUATIONS. S&S
RULE lY.
Whek the unkn(^#n; ^tlamtiiy is incloded in any root or
surd : transposes tiie rest; of the terms, if thtsre be any, bjr
Rule 1 ; then raise e^ch side to such a power as is deao^
by the index of the surdf viz* square each side when it is
the square root ; cube each side when it is the cube ro9t; &c.
which clears that radical.
Thus, if V.r— 3 = 4f ; then transposing 3j gives Vt=s7 j
And squaring both sides giires r = 49.
And, if V2J+To K « : .
Then by squaring, it becomes 2x + 10 = 64 j
And by transposing 10, it is 2x =: 54/^
Lastly, dividing by 2, gives x zz 27.
Also, if3/3.r+4 + 3 =6: . v
Then by transposing 3, it is ^3^ + 4 ^ 5;.'"
And by cubing, it is 3;i' + 4 = ^7 ; * ^
Also, by transposing 4, it is Sjt ^=: 23 ;
Lastly, dividing by 3, gives x ss 7.J,
R«L£ V.
I
I
%
r
Whe j? that side of the equation which contains the un
known quantity is a complete power, or can easily be reduced
to one, by rule 1, 2, or 3 : then extract the rbbt of the said
power on both sides of the equation ; that is, extract the
square root when it is a square power, or th^ cube root
when it is a cube, &c.
Thus, if JT* + Sj? + 16 = 36, or {x + 4)' = 36 :
Then by extracting the roots, it is r + 4 = 6 ;
And by transposing 4, it is x ;= 6 — 4 = 2.
Andif 3^*19 = 2l +3^.
Then, by transposing 19, it is 3j:^,r= 75 ;
And dividing by 3, gives .r* =s 25 ;
And extracting the root, gives jr = 5.
' Also, if fr*— 6=s 24.
Then transposing 6, gives f x* = 30 ;
And multiplying by 4, gives 3:r* = 120;
Then dividiag by 3, gives x* = 40 $
Lastly, extracting the root, gives x = V40 = 6*324555«
RULE
.f2iy ALGEBRA.
EULE VU
:. When there is any analogy or proportion, it. is to be
changed into an equation, by multiplying the two extreaae
tdrois together ) and the two means together, and xmaking
the one. product equal to the other.
Thus, if 2x : 9 : : 3 : 5.
Then, mult, the extremes and means, gives lOx =: 27 v
And (dividing by 10, gives ^r =: 2^,
And i{ ^x : a :: 5i :2c.
Then mult, extremes and means gives icx = S^t ;
And multiplying by 2, gives Sex si \Oab ;
\Oab
Lastly, dividing by 5r, gives a: = — •
Also, if 10— X : fjr : : 3 : 1.
Then mult, extremes and means, gives 10<r r= 2r ;
And transposing ir," gives 10 = 30*^
Lastly, dividing by 3, gives 3 j. = ;c,
nvm VII.
When the same quantity is found on both sides of ait
equ?.tion, with the same sign, either plus or minus, it may be
left out of both : and when every term in an equation is
either multiplied or divided by the same quantity, it may b^
strucl? ou^ of them all.
Thus, i[3x + 2az^2a + b: \ ,
Then, by taking away 2^, it is 3x =: k^
And, dividing by 3, it is :r = ^6.
Also if there be 4ax + 6ah = lac.
Then striking out or dividing by fl, gives 4j: + 6i =; 7r,
Then, by transposing 6^, it becomes ^x r: 7^—6^;
And then dividing by 4? giv^s ;r = ^r— J^.
Again.ifxJ=V>.J. . ; , .
Then, taking away the ^^ it becomes .r =: */ 5
And taking away the 3's, it is 2jr =: 10 ;
Lastly, dividing by 2 gives ^ n 5,
MISCELLANEOUS KXAMpLESt
i\
1. Given 7;r 18 = 4«r + 6 ; to find the value of \r.^
First, transposing 1 8 and 5j; gives 3;v 2: 24 )
Then dividing by 3, gives jr =r 8.
1. . ^' Qlvesx
SIMPLE EQUATIONS. . 2t&
I
% GivenSO— 4a:12 s=s^— Ipjr; to find i*.
Firtt transposing 20 and 12 and IOjT, gives 6x » 84 ;
Then dividing by 6, gives j: = 14.
^. Let 4tfjr 5^ = 3<& + 2<? be given ; to find x.
First, by trans. 5* and 3dXi.it is ^ax^Sdx = 5* + 2r;
5ft42r
Then dividing by 4fl— 3</, pves a: = ■ ^> •
4* Let 5;r*— 12x c= 9 J? + 2jr* be given; to find *.
. Fii^, by dividing by x, it is 5j: — 12 =: 9 + 2jr ;
Then transposing 12 and 2x, gives ^jt rt 21 ;
Xastly, dividing by 3, gives j; =» 7.
5. Giveii 9fljr'— ISabx^ ^ 6ax^ + I2ax^ \ to find x.
First, dividing by SoJi^, gives Sx— 5^ = 2jr + 4;
Then transposing Bb zad 2Xf gives x :=:, 5b h 4.
*
6. Let r +j = 2be given, to find X.
I 3 4 5
First, multiplying by S, gives x^^x + ^x =s 6 ;
Then multiplying by 4, gives x { Y^ == 24.
Also njultiplying by 5, gves J7r = 120 ;
Lastly, dividing Dy 17, gites x = 7^.
^ ^. ^ .r— 5 jr ; i*— 10 ^ "
7. Given — ; — + — = 12 — : to find x.
3 ^ ^ .. S '■ ^ . . .^^., .^ .
First, mult, by S, gives x—S f 4x =,*36 — jr +'lO ;
Thep transposing 5 and x, gives 2jr f* l^ = 51 ;
And multiplying by 2, gives 7jr =: 1 02 ;
Lastly, dividing by 7, gives jrt=:14, ,
Sx ' '
^ S^ Let V'^T" + '^ ~ ^^> ^ givai ; to find x.
First, transposing 7frgiyes ^^x = 3 ;
Then squaring the equation, glves j^ tsz>9\
'Then dividing by 3, gives ^ as= 3 ;
1/astlyi multipiying by 4, gives a: = 1 2.
. 9.'' Let* 2^ + 2 Vi?47r* =^J^=L, be^gitreti; tb firid x.
Fiht, mult, by ^tf*+ jr*, gives 2xVa* + a:* + 2^* + 2^"
Then transp. 2a* and2jr%gives2;rv'a* + jr*=3«*— 2x'5
xoVoLI. Q Th'^
\ i*r
226 ALGEBRA.
i • I. ii'V .H'ft.
Then by squiring, it is 4a:*x^+ x*^3tf*  ar* 5
That is, ^a^x^ + 4x* = 9tf *  1 2/i*a;* + 4jr* j
By taking 4x* from both sides, it is 4tfV=9ii*— 12tfV j
Laftly extracting the root, gives v ^^
CXIMPLBS FCfR Pfticmd^ '
1. Given ^»— 5 + 16 = 21 ; to find n. Ato. at = 5*
2. Given 9* — 15 £rT + 6j to find x. Ans. x = 4..
3. Given 83i'+ 12=30— 5x+4; to find ;r. Ansjr=7;
4. Given x + pr— i* =185 to find x. Ans, or r: 12.
5. Given S*'+4;» +2=5*  4 j to find at. Ans. * = 4;
6. Givgn 4fljr + ^a— 2 cr ^jr— *x j to find x.
Ans. X =:
9a+3^
7. Giv€4i \x^\x + T^ = T> ^o fi»d r. Ans. ^ ^/if*J 4ji
8. Given v^4+x = 4 — v^jr; to find jr. Ans. xzz2\.
X*
9. Given 4ta + x ^ r — — ; to deter, x* Ans. ^= — 2a.
4a+x
V
10. Given V4ji* + jr*x=;J/4** + jr*j to find x.
Ans. X tSL'V
2^
^MMM
4a
11. Given V;c + V^2a + x zr ^ ; to find x.
Ans. 4? irfiPir
12. Given r~*" +t — ^ =2 2* ; to find x.
l+2x 1 — ; 3*
Ans. X fSr^V— r—
13, Given j + x = Va* + x VW + x* to find k.
Ans. j: ts a^
a
w
SIMPLE'EQUATIONS. «27
OV REDUCING DOUfiLB, TRIPLE, .ftc. £<^A'fiIONs/ CON
TAINING TWO, THREE, OR MORS UNX3K0WN jqUAN
TITIES.
i»ROBL$M I. ;
7i Exterminate Two Unknown Sluantities; Or^ to Reduce
the Two Simple ^puLtiom tontetining fkentfto a Single
4m. ,
RULE L
Find the T?lue pf one of the pn^nown letters, in terms
W ^e other qiantjti^^ in eskfix 4af the (equations, by the
lðods already explained. Then put those two values
equal to each other tor a new equation, with only oi&e yn
known quantity in it, iiriiose value is to be found .as .before.
N&te. It is ev ident that we must first begin to find the
values of that letter which are easiest to be found in the two
proposed equati^is*
EXAMPLES*
1. Given [ll t%Zu]^ '^ ^^ * ^^^
17 — Sv
In the lstequat.traf»p.SyanddAv.by2,gives;rs%im^;
14 4 22/
Jbi the 2d transp. 2v and div. by 5, gives * =s r^J
Putting these two values equal, gives "^^ ^»
Then mult, by 5 and 2, gives 28 + 4y =z SS 15yj
Transposing 28 and 15y, gives 19y = 57;
And dividing by 19, gives ^ =ai S.
And hence, ffp = 4.
Or, to do the same by finding two valuc^ of ^, thus:
In the 1st equat. tr. 2k and div. by ^, gives^ =
s
5:^— 14
In the 2d tr. 2yand 14, anddiv. by 2, ja^vesj^ ;=■
art
«,,.,, 1 1 . 5«— 14 172*
Jrutting these two values equal, gives t— rr= . ■' ■
Mult, by 2^ and by 3, gives lBx^.^2. =s. ^^ t'4^
" Q2 Transp.
2U ALGEBRA^ ^
Tramp. 42 and 4r, gives^ 19x cr J6 ;
Dividing by 19, gives x ;= 4.
Hence j^ = 3, as before.
' d. Given J^j;j;=;^> to find* and;.
Ans. * = « + *, and j^ = ^—44
St Given 3* + ^ = 22, and Sy+«f = 18 ; to find x and >.
Ans. X s 6} and j^ :^ 4.
4. Given {jj + ?{= ^, { ; to findirand,.
Ans. jr = 6, and v = 3.
'^. 2* 3v 22 3*p 2ir 67  .
,5. Given — l'f = — , and— +^«rr; to find /?
and y. Ans. x = 3, and ; = 4.
6. Given * + 2j^ sz /, and i^* 4/ = i/* ; to find x and jr.
Ans. X = — ^, and jr =— jj—
7. Given ;» — 2v n rf, and xiyiiaib; to find x and y.
ad , W
Ans. X = r, and f = r
RULE U.
Find the value of one of the unknown letters, in ooly one
of the equations, as in the former rule ; and substitute this
value instead of that unknown quantity in the other equation,
and there will arise a new equation*^ with only one unknown
quantity, whose value is to be found as before.
" . ... *
Nate. It is evident that it i5,l?est to begin first with that
letter whose value is easiest found in the given equations.
EXAMPLES* ' ,
1. Given ^l^ 1 1 Zu]> to find ' and j.
This will admit, of fom* ways of solution; thus : First,
17— 3v
in the 1st eq. trans. 3y and div. by 2, gives x =s — ^~^*
. 85/^ I5y
This val. subs, for x in the 2d, gives — ^ — 2y = 1 4 j
Mttlt. by 2, this becomes 8S  liy 4y =: 28 j
Transpv
SIMPLl EQUATIONS. 22§
Transp. 15y and 4y and 28, gives 57 c= 19y ; ,
And dividing by 19> gives 3 szy.
Then^=: — s~ = 4.
2dly, in the 2d trans! 2y and div. by S^ gives ^ = — ^A
5
This subst. for :r in the 1st, gives 2! + 3j^;s 17 j
Mult, by 5, gives 28 + 4y + 15y = 85 j
Transpos. 28, gives 19y =: 57 ;
And dividing by 19, gives y =: 3.
Then x = ' ' = 4, as before.
3dly, in the 1st trans. 2r and div. by 3, gives j>= — ^^^^5
3
This subst* for ^ in the 2d, gives 5:c — = 14 .
Multiplying by 3 gives 15*1;' ^ 34 + 4jr s= 42 $
Transposing 34, gives . 1 9x = 76 ;
And dividing by 19, gives x = 4.
„ 172;r ^
Men^e^ 1=: —  — <*ss 3, as before.
4thly,in the 2d tr. 2^ and 14 and div. by 2, gives y=: ^"^ \
This substituted in the 1st, fives 2x + = IT;
Multiplying by 2, gives i 9x — 42 =; 34 j
Transposing' 4<2, gives 19x == 76 ;
And dividing by 1 9, gives x r: 4,
5x— 14
Hence y = — r — =: 3, as beforfc.
2. Given 2;r + 3y =: 29, and 3x  2jy — 11 ; to find *
andy. . Ans. jt = 7, and j^ == 5.
, . 3. Given {J 1 J f ^2} ' '" ^'^ ^ '"'^>
Aris. a? = 8, and v = 6.
4. Given
299 AIGBBRA*
4. GiTCi! {^'J^jl I '^1} ; to fftid T^iy.
' Am^ jr =: 6, and y = 4.
5. Given r + 3^ = 21, and ~ + S>r = 29 ; to find ;r
and jf^ Ans. *r ss 9, and j^.= 6.
«. Given 10  ^ i + 4, and ^ + ~  2 =
— I 1 ; to find X and y. An$. x zz 8, and v s: 6,
7. Given x : y : : 4 : 5^ and ap'— / = 37 j to find x and jr.
Ans. tT = 4j and y = 3.
I.ULE III.
I/ET the given equations be so multiplied, or divided^ ScCt^
, and by such numbers or quantities, as will make the terms
which contain one of the unknown quantities the same ia
both et[uations ; if they afe not the sarte when £rst pro
posed.
Then by adding or siibtracting the equations, according
as the signs may require, there will remain a new equa
tion, with only one unknown quantity, as before. That isj^
add the two equations when the signs are unlike, but sub
tract them when the signs are alike, to cancel that common
term.
Note. To make two unequal terms become equal, as abovcn
multiply each term by the coefficient of the other.
EXAMPLES.
^^^^" {S + S^ = le} ' ^^ ^^ * ^^^ J'
Here we msly either make the two first terms, containing n^
equal, or the two £d terms, containing j^, equal. To make
the two first terms equal, we must multiply tji€ 1st equation
by 2, and the 2d by 5; but to mlike the two 2d terms equal,
we must multiply the i st equation by 5, ancl the 2d by 3 5
as follows*
1. B7
SIMPLE EQUATIONS. • 231.
i
1. Bj making the two first terms equal :
Mult, the 1st equ. by 2, gives lOor— Gy ss 18 ^
And mult, the 2d by 5, gives 10.r + 25j> = 80 y
Subtn the upper froi^ the under^ gives Sly a 62
' And dividing by 51 gives ys£ 2*
d + 3f
Hfince> from the Ist given equ* .r s: ■ / ■ ■ f ' ==. 9i»
• £• By making the twQ 24 terms equal:
Mult, the 1st equat. by 5, gives 2Sx — 1 5jf =5 45 ;j
And milt. the 2d by 3, gives 6ur f 15; =;; 4S ;
Adding these two, gives Six s 93 ;
And dividing by 31,^ gives x =s S.
^ 5x — 9
^ Hence^ from the 1st equt^y = — ''^— s: 2*
MISCELLANEOUS fiXAMPLSS*
;rhS vf6
1. Given —  + 6t =^ 21, and ^~ + 5x ;a 2S 5 to
» s
find X and jiv Ans. x ^ 4, and ;^ =9 S*
2. Given — ^ + 10 = 13, an4 ^ J  + 5 = 12 ;
to find X and y. Ans*> str 5, and jr s: 3.
3. Given— j^+— =slO,and ^ + ^^^9
to find X and y. Ans. ^ = 8, and y =: 4*
4. Given Sx + 4y = 38,and4x— Sy = 9; tofindxandy.
Ans. x :;;: 6, and y sb 5«
FROBLSM I.
To ExterffunaU Three or More Unknown Quantities i Or, to
Reduce tie Single Equations, containing them, to a Singly
one.
nxsiA.
This may be done by any of the thret methods i^ the Iat
problem: viz.
1. After the manner of the first rule in the la$t problemi
find the value oip one of the unknoven letters in each of th«
given equations : next put Vrro of these values equal to each
other, and then one of these and a third value equal, and so
on for all the values of it ; which gives a ntwsct of equations,
with
i3« * ALGEBRA.
i
I
t
with which the same process is to be repeated, and so on
till there is only one equation, to be reduced bj the rules for
a single equation.
2. Or, as in the 2d rule of the same problem, find the
value of one of the unknown quantifies in one of the equa»
tions only; then substitute this value instead of it in the
other equations ; which gives a new set of equations to be
resolved ^ before, by repeating the operation.
3. Ots as in the 3d rule, reduce the equations, by multi
plying or dividing themi so as to make som^ of the terms to
agree: then,. by adding or subtracting them, as the signs
may require, one of the letters may be exterminated, &c, as
before.
EXAMPLES.
1. Given I ;ir + 2j/ + 32 = 16 >• ; to find 4f,^, and a»
(Ar+3j/ + 4z = 2l)
1. By the 1st method :
Transp. the terms containing^ and z in each ^ua.. gives
•  » =s 9 — J/ — SB,
4?= 21 — 3y — . 42}
Then putting the 1st and 2d values equal, and the 2d and 3d
values equal, give
9 — y  as = 16 — 2j/  3«,
16 _ 2y  32S = 21  8j^  4z ;
In the 1 St trans. 9, 25, and 2yj gives j^ = 7 — 2z ;
In the 2d trans. 16, 3z, and 3y, gives j/'= 5 —  a;;
Putting these two equal, gives 5—2= 7 — 2zj
Trans. 5 and 3z, gives 2 = 2..
Hence 3^ = 52 = 3, and x = 9— j^— 2 =; 4.
Jdly. By the 2d method :
From the 1st equa. x = 9—^ — 2;
This value of «: substit. in the 2d and 3d, gives
9 f y + 22 == 16,
.U + 2y+.32 = 21;
In the 1st trans. 9 and 2z, gives 3/ =; 7 — 23^;
This substit. ii the last, gjves 23 — 2 == 21 j.
Trans, z and 21, gives 2 = 2.
Pence again J/ =; 7 — 2z == 3, and x = 9— 3^2 = 4.
3dly. By
SIMPLE EQUATIONS. 3SS
Sdly. By the 3d method : subtracting the 1st tqu. from
the 2d, and the 2d from the 3d, gives
J/ + 25f = 7,
y + z=: 5i
Subtr. the latter from the former, gives z = 2,
Hence y = 5 — z = 3, and x =^ 9—7/—Z =i 4.
C x+ 1/+ z^lSl
\. Given < x + 3j/ + 25r = 38 >.;
L ^+j;y + i2 = ioy
to find X, j^, and s,
«
Ans. a? = 4, ^ = 6, 5r = 8.
= 271
^. Given ^ x + yj/ + z = 20 >• ; to find jr,3^, and s.
z = 163
Ans. X = 1, j^ = 20, at = 60.
4, Given jt — j/ = 2, jr — ;a =: 3, and y — a = 1 j to
find a;*, ^, and z, Ans. r = 7 ; j/= 5 ; 2; = 4.
2jr + 3y + 4z = 34T
5. Given ^Sx + iy + 5z = 46^ i to find x, y, ajid z.
4^ + 5y+ 6z =583
A COLLECTION OF QUESTIONS PRODUCING SIMPLE
^ EQUATIONS.
Quest. 1. To find two numbers, such, that their sum
shall be 10, and thair difference 6.
Let JT denote the greater number, and 1/ the less *.
Then, by the 1st condition x + j/ = 10,
And by the 2d   a; — 3/ = 6,
Transp. y in each, gives jr = 10 — ^,
and X = 6 + y J
Put these two values equal, gives 6+^ = 10*j/;
Transpjos. 6 and — t/^ gives  2^ == 4 j
Dividing by 2, gives   ^ = 2.
And hence     x =:^ 6 +j/ = 8.
* In all these solutions^ as many unknown letters are always
used as there are unknown numbers to be founds purposely the
better to exercise the modes of reducing the equations : avoiding
the short ways of notation^ which> though g'^ving a shorter solu
tion, are for that reason less usefiil to the pupil, as affording less
exercise io practising the several rules in reducing equations.
Quest. 2.
SS4 ALGJEBRA.
Quest. 2. Divide lOQ/. simong A» u, c, se tlut a may
have 20/. more than b, and b 10/. more thaa c.
Let A' = a's share, j^ = b's, and z = c's.
Then :t' + j^ + » = 100,
X =cj/ +20,
j^ =5 z 4 10.
In the 1st substit. y + 20 for x^ gives 2y + 2 + 20 = 100;
In this substituting 2 + 10 for j<, ^ves Ss; + 40 :s 100 ;
By transposing 40, gives • Sf; ^ 60 ;
And dividing by S, gives   z = 20.
Htocej^ = 2 + 10 = 30, and x = j/ + 20 = 50.
«
Quest. 3. A prize of 500/. is to be divided between two
persons, so^^as their shares may be in proportion as 7 to 8;
required the share of each.
Put X and y for the two shares ; then by the question,
7 : 8 : : jr :y, or muk. the extremes
and the means, ly = Sx,
and a:Hj/ = 500;
Transposing J/, gives x = 500 — j^ ;
This substituted in the 1st, gives 7j/ = 4000 — St/^
By transposing 8y, it is 15j^ = 4000 ;
By dividing by 15, it gives y = 266^1
And hence x = 500— j/ =' 233.
Quest. 4. Whatnumber is that whose 4th part^exceeds
its 5th part by 10 ?
Let X denote the numba* sought.
Then by the question ^x — jr = 10 ;
 By mult, by./4, it becomes x — ^:r = 40 j
By mult, by 5, it gives x = 200, the number sought.
Quest. 5. What fraction is that, to the numerator of
which if 1 be added, the value will be 4 } but if 1 be add^
to the denominator, its value will be j ?
X
Let  denote the fraction.
^ 3/
Then by the quest. = », and —^ — = '.
The 1st mult, by 2 andy, gives 2^ + 2 = v ;
The 2d mult, by 3 and J^ + 1, is ,3^ =:^ + J ;
The upper taken from the under leaves .r— 2 = 1 j
By transpos. 2, it gives x z=z s. ,
And hence J/ = 2^ + 2 = 8 ; and the fraction is .
Quest. «.
\
SIMPI^'EQUATIONS. 285
Quest. 6. A hbourer engaged to serve for SO days on
these conditions : that for every day he worked^ he was to
receive 20d, but for wu^ry day he played, or was absent, he
was to forfeit \()d* 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 lie was
idle ?
Let j: be the days worked, and^ the days idled«^
Then 20x is the pence earned^ and iOy th^ forfeits ;
Hence, by the question  x +j/ ^ 80,
and 20x — IC^ = 240;
The 1st. mult, by 10, gives lOx + lOj^ = 300 j
• These two added give  30jr = 540 ; •
This div. by 30, gives  ;r = 18, the days worked j
Hence .  j^=30— a:=sl2, the days idled.
Quest. 7. Out of a cask of wine, which had leaked :^way ,
80 gallons were drawn ; and then, being gaged, it jappeared
to be half full; how much did it hold?
Let it be supposed to have held x gallons.
Then it would have leaked ^x gallons,
Conseq. there had been taken away ^x + 30 gallons.
Hence ix=zix + 30 by the question.
Then mult, by 4, gives 2x = a: + 120; 
And transposing x, gives x 2= 120 the contents.
Quest. 8. To divide 20 into twd such parts, that 3 times
the one part added to 5 times the other may make 76.
Let X and^ denote the two part^.
Then by the question   :r + 3^ = 20,
and Sx + 5t/ = 76.
Mult, the 1st by 3, gives  3r + 8j^ =s 60;
Subtr. the latter from the former, gives 2y = 16 ;
And dividing by 2, gives   ^ = 3.
Hence^ from the 1st,  x s= 20 — j/ = 12. ,
QUBST. 9. A market woman bought in a certain number
pf 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 twopence, and
by so doing, contrary to expectation, found she lost 3^.;
what number of eggs had she ?
Let X ss number of eggs of each sort.
Then will ^x = cost of the first sorty
And ^ = cost of the s^coi^sott ^
But
285 ALGEBRAr
Bat 5 r 2 : : 2r (the whole number of eggs) : ^ ;
Hence ^x = price of both sorts, at 5 for 2 pence ;•
Then by the question ^r + jx— 4ir = 3 ;
Mult, by 2, gives  *• + f^— 1^ = 6 ;
And mult, by 3, gives 5x — V^ = 18;
Also malt, by 5, gives x = 90, the number of eggs ol
each sort.
Quest. 10. . Two persons, a and b, engage at play.
Before they begin, a has SO guineas, and b has GO. After
a certain number of games won and lost between them, a
rises with three times as. many guineas as B. Query, how
many gttineas did A win of B ?
Let X denote the number of guineas A won. '
Then a rises with 80 + x.
And B rises with 60— x ; •
Theref. by the quest. 80 + x = 1 80  Zx\
Transp. 80 and 3 a;, gives 4x = 100 ;
Anc^ dividing by 4, gives x =: 25, the guineas won.
QUESTIONS FOR PRACTICE.
1. To determine two numbers such, that their difference
may be 4, and the difference of their squares 64.
Ans. 6 and 10.
2. To find two numbers with these conditions, viz. that
half the first with a 3d part of the second may make 9,
and that a 4th part of the first with a 5th part of the se .
cond may make 5. Ans. 8 and 15«
. 3. To divide the number 20 into two such parts, that a
3d of the one part added to a fifth of the other, may
make 6. Ans. 15 and 5.
4. To' find three numbers such, that the sum of the 1st
and 2d shall be 7, the sum of the 1st and 3d 8, and the
sum of the 2d and 3d 9. Ans, ^^ 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 ta
have a shilling. What then were their two shares ?
Ans. The son 2000/. s^nd the daughter 800/.
6. Hiree persons. A, B, c, make a joint contribution,
which in the whole amounte to 400A : of which sum b con
tributes
SIMPLE EQUATIONS. ?37
tribiiites twice as much as a and 20/. more \ and c as much
as A and b together. What sum did each contjribute ?
Ans. A 60A B UO/. and c 200/1
7. A person paid a bill of lOOA with half guineas and
crowns, vsing 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 lO of your guineas, and then I shall
liave 5 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 pocke't, where he spends 2 shillings; he then
borrows as much money as he had left, and going to another
tavern9 he there spends 2. shillings also; then borrowing
< again as much money 'as was left, he went to a third tayerQ^
whi^e likewise h^ speixt 2 shillings; and thus repeating. the.
« same at a &urth tavern, he then had nothing remaining.
What sum had he at first ? Ans. 3/. 9 J.
10. A man with his wife and child dine together at an
inn. The landlord charged 1 shilling for the child; and
for the woman he charged as much as for the child and i^%
much as for the man ; and for the man he charged as much
as for the woman and child together.. How much was that
for each ? Ans. The woman 20d. and the man 32rf.
4
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 \ of the brandy. How
much was there of each ?
'^S.' Ans. Brnndy 15, cyder 21, wine 24
12. A general, disposing his army into a square form,
finds that he has 284 nien more than a perfect square ; but
increasing the side by 1 man, he then wants 25 men to be
a complete square. Then how many m^n had he under his
command ? Ans. 24000.
13. What number is that, to which if 3, 5, and 8, be
severally added, tl^e three sums shall be in geometrical pro
gression? Ans. 1.
13. The stock of three traders amounted to 860/. the
shares of the first and second exceeded that of the third
SS8 ALGEBRA.
hf 240 ; and the sum of tbe 24 and Sd extdeded the fint
by 260, What was the share of each ? ^
Ans. The 1st 200, the 2d 300, the 3d 260.
15*. What two ntimbers are those, whicht being in* the
ratio of S to 4, their product is equal to 12 times their swni
Ans. 2 i and 29.
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 shilling apiece less than
they did ; but that if there had beea 3 fewer in company.
tliey most have paid a shilling apiece more than they did.
What then Was the number of persons in company, what
each paid, and what was the whole reckoning ?
Ans. 24 persons, each paid 7i. and the whole
reckoning 6 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 wor#e on theM, it
makes the first horse worth double tbe 2d : but when he
places the better saddle on the 2d horse, and the worse on
the first, it makes the 2d horse worth three times the 1st,.
What then were the values of the two horses t
Ans. The Ist (i/., and the 2d 9/.
IS. 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»
19fc 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 number*^ are those , whose difference, sum,
and product, are to each other, as the three numbers 2,
3, 5 ? Ans. 2 and lO.
21. To 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. 1 ^, 21, 27.
22. It is required to divide the number 24 into two such
parts, that the quotient 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,
answer<»d, 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
^o
QUADRATIC EQUATIONS. Wl^
ttffeen fitfm the difference of ttueir age«, the remainder will
be edual to the age of the yotmger. What then were their
nfges? ^ Ans. SCand 12.
24. To find four numbers such, that the sum of the 1 st,
2d, and Sd shall be 13 ; the sum of the 1st, 2d, and 4th,
15 ; the sum. of the 1st, 3d, and 4th, 18 ; and lastly th^
sum of the 2d, 3d, and 4th, 20. Ans. 2, 4, 7, 9.
25. To divide 48 into 4 such parts, that the first increased
liy 3, the second diminished by 3, the third multiplied by 3,
and the 4th divided by 3, may be all equal to each other.
Ans. 6, 12, 3, 27.
QUADRATIC EQUATIONS.
Quadratic Equations are either simple oh compound.
A simple quadraticequatton, is that which involves' the
jliqtare of the unknovOvi iquantityonly.. Asax^ = h. And
the ^ktion of such quadratics has been aheady given 'in
simple equattoms.
A compound qtladr^atic equation, is that^which contains
the square of the unknown quantity in one term, and the ^
ifilfst power in' another term. As ax^ ^^ hx ^c.
All compound quadratic equations, after being properly
r^dticed, fall under the thr^e following forms, to which
they miliist ' always be reduced by preparing them for sola
tion.
1. x^ 'Y oxsz b
2. a^  ax 1=: b
3. jr*  oT =  A
The general method of solving quadratic equations, is by
what is caUed completing the square, which is ^s follows :
1. Reduce the proposed equation to a proper simple form,
as usual, such as the forms above ; namely, by transposing
all the terms which <C6ntain the unknown quantity to one
side of the elation, and the known terms to the other;
placing the square term first, and the single power second ;
dividing the equation by the co*efficient of the square or
first term, if it has one, and changing the signs of all the
terms, when that terhi happens to do negative, as that
term must always be made positive before the solution.
Then the proper solution is by completing the square as
folio W8 viz.
2. Complete
xYr
S40 ALGEBRA.
2. Complete the unknown side to a squarei in this man
ner»' viz. Take half the coefficient of the second term, and
square it } which square add to both sides of the equation,
then that side which contains the unknown quantity will be
a complete square.
3. Then extract the square root on both sides of the
equation *, and the value of the unknown quantity will b^
determined.
* ^l^l^he square root of any quantity may be either + or — ,
therefore all quadratic equations admit of two solutions. Thus,
the square root of + n« is either + nor.ii5 for+nx+it
and — n X — n are each equal to + n*. But the square root of
— n*, or \/— n% is imaginaiy or impossible, as neither + n nor
— n, when squared, gives — n\
So, in the first form, x^+ax:nb, w here x  f ia Is found =:
^b + i^*, the root may be either + ^FTT^, or — i/hTj^S
since either of them being multiplied by itself produces b f ^\
And this ambiguity is expressed by writing the uncertain or double
Vign ± before •/^ + ^' ; thus x =z ± •^ + ^a" — ^a. •
In this form, where x r: ± \^b + ^a^ — itf» the first value of
X, viz. X zr + \/b + Ja« —^a, is always affirmative; for since
^* 4 ft is greater than ^*, the greater square must neces
sarily have the greater root ; therefore \/ft + ^a* will always
be greater than •Jo*, or its equal ia, and consequendy +
^b + .^* — ^ will always be affinnative.
The second value, viz. x iz — ^b f i«* — i^ will always be
negative, because it; is composed of two negative terms. There
fore when x' + ax — b, we shall have x zz + ^/b + Ja' — ^a
for the affirmative value of x, and x r: — ^ft f Ja^ — ^a for
the negative value of x. % * .
In the second form, where x = ± v^6 4. ^^ f ^a the first
value, viz. x = + v^ft i a^ + fl is always affirmative, since it
is composed of two affirmative terms. But the second value, viz.
JP = ^ v/^ + Ja^ + io, will always be negative j for since
ft + Jos is greater than Ja^ therefore ^/b + ^a* wi ll be gr eater
than \/ia% or its equal ia 5 and consequently ^ v^ft + ia' + ia
is always a negative quantity. »
Therefore,
QUADRATIC EQtJA^tONS Sit
determined, making the' root of the known side either + or
— , which will giv6 two roots of the Equation, or two values
of the unknown quantity*
'Notey 1. The root of the first side of tKe equation, is
always equal to the rpdt of the first term, with half the
coefficient of the second tei^m joined to it, with its sign,
whether '+ or — • ' ' *
2. All equations, in, which there are two terms including
the unknown qijiantjt jr, and which have the index of the one
just double that^ of the other, are resolved like quadratics,
by completing the square, as above.
1"
Thus, x^ + ^^ =* K or ^^ + ^^ ™ *> o** ^ + ox^ :2:^
are the same as quadraticsi and the value ,of the unki^qwo
quantity may be determined accordingly* ./
t
■ ^ i < ■ M <*■ III I ' 1*1 I II I II ■ I III II ■ ' I I II I U ' 1 •
Therefore, when a^ — ax.zzb, we shall have x zf  f . y^^ 4 ^
+ la for the affirmative value of a:; and j? =* — y6 f ia^+^a
for the negative vake of a? j i^ that' in both the first and second
forms> the unknown quanti^ has alwi^sbtro values^ one of which
is positive^ and the otb^r negative.
But, in the third form, where ar =z ± v'ifl* — ^ + 1«* both^tho
values of X will be positiv e/ when Ja^ is greater tli^n 0, FoxthfS
first vjflue, viz. ar' =: f ^/^a^  b + Ja will then be affirmative,
being composed of two affirmaiiveteirmd; !«>
The secQnd value, viz* .^r 5= .;t;. Vi^^ — 6 4 . J^ y a^jna*
tive also 5 *for since ^o^ ,5 jrreatei; than Ja* — 6^, therefore, Vi** ot
^a is greater than \/Ja^ r 6^ aqd consequently r \/Ja*— 6.^ fa
will always be an affirmative quantity. So that, when afi ^ ax
= ^— 5, we shall have a: =: 4 \/o* — ^ + ia, and,alsp x = —
v^Jo* — 6 + f aj for the valuesDf X, botti positive^ * '
But in thifc third forpa, if b be greater than ^a?, the solution of
th6 proposed question wilt be impossible/ For ^ince the sqUdre of
any quantity (whether jthat quantity be affirmative or negative)
is always amrinative, the square 'robt^of a negdfcive quaotily ik im
possible, and cannot be assigned. But when b is greater than
Ja^ thenirt*'— b is a negative quantity j add therefore its root
V'Jo* — ^ is impossible, or imaginary; consequently, in that case,
j: n ^a ± ^^a^ — b, or thfe two toots or values of Jf, are\>otli
impossible, 01 imaginary qu^tities; .1
Voii. L R ' EXAMPLES.
^^
34f ALGEBRA.
EXAMPLES.
1. Given jt* + 4j: = 60 •, to find r.
First, by completing the square, or^ '\'4fX + 4 ^ i4\
Then, by extracting the root, jr + 2 =z ± 8 j
Then, transpose 2, gives x = 6 or — 10, the tw« roots.
2. Given x*— 6x + 10 = 65 ; to find 4%
First, trans. 10, gives j;^— 6r = 55 ;
Then by complet. the sq. it is x^—'Sx + 9 =: 64 j
And by extr. the root, gives x^S =: ±: S'f
Then trans. 3, gives j* = II or — 5.
5. Given 2r* + ar30 as 60; to find x. ^
 First by transpos. 20, it is 2jr* + Bar rr 90;'
Then div. by 2, gives .r* + ^'*'= 45*;
And by compL the sq. it is jt* + 4jr f 4 = 49;
"Then extr. th e roo t , it is ar +^ =^ ± 7}' " '
«. Ahd iransp. 2, gives Jr = 5 or — 9.
4.' Given So:'  Ix V '^* = ^4^ to find .r. \ . . ' .
. First diy. by 3, giv^s ,ar*— jt + 9f= 21;
. Tbeja transpos. ^3, giye« jt*— itr ni'— I:} i ' : • J . i
And compl. the sq. gives ar^^x + ^ =s ^; t
Then extr. the root gives or — 4^ = dt Jr*
^ And transp. 4^, gives x = J or ^.
J5. Given i^*— ^ +.30i =,52* ; tp find jt.
First by transpos. 30 j., it is Jjr*— ^jr as 22^;
Then mult, by 2 gives jr*— l^r = 44^ ;
And by compl. the sq! it is x*— !rH ^ = 44$;
.' Then extr. the root, gives \r—^ = t 6^;
And transp. J, gives o^ = 7 or — 6y.
6. Given ijr*— ^jr = r ; to find x.
he
First by div. by a, it is 1:* x » •^;
' Then compl. the sq. rives x* x + :~x = 1 r»
And extrac. the root, gives X =. ± ^^j^;
r,^, i . ,4ac + i* A
Then transp. —, gives x = ± ^_^_+ —
7.. Given x* — 2<wr* =: h\ to find r.
First by compl. the sq. gives x^ * 2a4r* + a* i= a* +*s ^
QUADRATIC EQUATIONS. 24^
And extract, the root, gives x* ^a'=i ±, ^/et + h\
Then transpos. j, gives a;*=±v'^* + * + *»
iri tm ■*■
And extract, the root, gives ^ = ± v'* i V^.^ + ^*
And thus, by always tising similar words at each lihe, the
pupil will resolve the following examples.
EXAMPLES. FOR PRACTICE.
) I..
1. Given x*— 6x— 7 == 33 j to find ^r, i^wJf^ 10.
2. .Given x*— 5jr— 10 = 44; to find x. Ans. ar = 8.
3. Given 5jr* + 4x — 90 = 114 ; to find x, Ans. x = 6
4. Given J^x*— ^x + 2 » 9; fo find x. . A6s. > = 4.
5. Given 3x*— 2x* = 40 ; to find x. Ans. x == 2.
6. Given fr— ^JV^*^ = ^il to find x,. Ans. x = 9.
7. Given ix* + x =  ; to find x. Ans. x = '727766.
S. Given 05^ + 4x^ =» 12 i to find .r.
Ans. .r =^ = 1259921.
9. Given x* + 4x = tf* + ^5 to find x. '
. Ans. jr = y a*^^ 2.
•<^lSTIpNS PRODUCING qUADRATiC EqUATIOMS.
1. To find two numbeps v^hose difference is 2, and
product 80.
Let X and y denote the two required numbers *. *
Then the first condition gives x—yzz 2, f
And the second gives xy = 80.
Then trsnsp. y in the 1st gives x » y + 2 ;
This value of x substitut. in the 2d, i8^*42y 5= 80;
Then comp. the square gives j^ + 2y + l=8i;
And extrac. the root gives ^ +1=9^
And transpos. 1 gives y = 8;
« And therefore x ssj/ 42 = 10.
* These questions^ like those in simple equations^ are also
solved by using a^ many unknown letters, as are the numbers
required, for the better exercise io reducing equations ; not aim«
ing at the shortest modes of solution, which would not' afford so
much useful practice.
R 2 2. To
244f ALGEBRA.
2. Tq' divide the number 14 into two such parls> that' their
product may be 48.
■ > •
Xet X and 7/ denote the two numbers.
. T^ien i^e Ist condition gives x + t/ « 14,
. j,Apd the 2d gives xy ==48.
Then transp. ^ m the 1st gives * = 14 3/ ;
This value subst. for x in the 2d, is 143/^ = 48 ;
Changirig all the signs, to make the square positivejj
' ^vesy14y=48;
 Then compl. the square gives /— 14j; + 49 = 1 >
And extrac. the root gives 5^— 7 = i 1;
Then tnnspos. 7> gives > .= aor «> the two parts.
» , . ■ ' '
S. Given the sum of two nunibers = 9, and the sum of
their' sqiiareV = 45 ; to find those niimbers.
Let jr and y denote the two numbers,
* Then by* the 1st condition x + y = 9.
. A"nd'bythe2d:r*+/^45. ' '
Then transpos. y ixK tho. 1 st.gives xz^S^y^
Tl^% value '.sjbst. in jJ>e^2d gives 81  18;; + 2/ = .45 ^
Then transpos. .8J, give? 2/ ISy = — 36 •,
And dwiding by 2 gives / — 9y ±= — 1 8 ,'
8 Z — 9 .
,: The^ con>pL the sq. gives y^9y ^ V
And extrac. the root gives ;;—  = ± i ;
Then transpos. ^ gives j? == 6 or 3, the two numjpers.,
4. What two numbers arc those, whose sum^ pro4uct,' and
difference of their squares, are all e^ual to each other ?
' Let .«• ^ndj/ denote the two numbers.
Then the ist^md 2d expression give x + y sz xy^
And the 1 st and 3d give at + j? = *"*—/•
Then the la^ eqoa. div. by x + y^ gives 1 ^ x^y ^
'Axi^ transpos. y^ gives y ^ \ i^ x\
This vaL substit. in the 1st gives. 2;; + 1.3±:y +31^
And transpos. 2^, gives 1 3= /— y 5
Then complet. the sq. gives i isc y* —)••{ ^f ;
And extracting the root giVea iV 5 = y — 4 j
And transposing ^ gives 4>v/5 + t == y >
And therefore \r = y f 1 = ^^ 5 + 4.
And if. thc^e. expressions be turned intp numbers, by ex
tracting the root of 5, .&c, they give ar = 2*6I8a + ,
andy= 16180 +•
. ' * ' 
5. Theye are four numbers in arithmetical progression, of
which
QUADRATIC "E^^UATIONS. fe4S
wMch the product of the two extremes is 22f,'and that of
the means 40; what are the numbers? . ^
Let X = the less extreme,
and J/ ^ the common diflferencc^'
 Then x, .r+y, jr+2y, jr+Sy, will be thefourntJmbcrs.
Hence by the Irt c<»dition i* + diry « Sa,
And by the 2d x^ { 3xj/ + 2y =? 40,<
Then subtracting the first from the 2d ^ives 2y* =5 18 j
And dividing by 2 giv^y* — 9j
And extracting the root gives y = 3. .
Then substit. 3 for y in* tie 1st, gives 4:* .+ 9t = 22;
And completing the square gives :r* + 9.r + y =s '»j
Then extracting the root gives > + 4 =*: V,>
And ilransposing f give^ x =» 2 the least number.
Hence the four numbers «re 2, 5, 8, 11.
6. To find S numbers in geometrical progression^ whose
fjMnn shall be 7, and the sum of their squares 21, .
Let XyT/j and z denote the three numbers sought.
Then by the 1st condition 0% ±=j/%
And by the 2d x + 1/ + ;ar =7,
And by the 3d jr* +y + «* = 21.
Transposing y in the 2d giy?es x + z =7 —3/$
Sq. this equa. gives x"^ + 2xz +;8^ = 49 — 14j^+j/^}
Substi. 2y* for 2jr2f, gives x'^+ 2y*+ ;2^= 49 — 14y y* ;
Siibtr. j/* from each side, leaves ^* + y* + ^= 4?9 — 1 4y ;
Putting the two values of x' + v* + 2* 7 ^ i ^ n ia..
.0 • 1 V •^ ' ' > 21=49' 14V;
equal to each other, gives 3 "^ '
Then transposing 21 and 14y, gives 14y = 28;
And dividing by 14, gives y = 2.
Then substit. 2 fory in the 1st equa. gives xz =; 4,
And in the 4th, it gives ^4^ = 5;
Transposing z in the last, gives x ^ S^z%'
This substit, in the next above, gives 5lf—z^ :^ 4; :
Changing all the signs,.gives «*— 5a == 4^ .
Then completing the square, gives «* — 5a; + V ^ ii
And extracting the root gives 3 — 4 = i'4»
Then transposing 4, gives z and x = 4 and 1} the twa
, . other numbers;
So that the three numbers are }, 2, 4» '
QUESTIONS FOR PRACTICE. 
1. Wha,t number is that which added to its square makes
♦2? Ans. 6.
a Ta
S46 ALGEBRA.
2, To find two numbers such» that the less may be to the
greater as the greater is to 12j and that the sum of their
squares may be 45. Ans. S and €•
S. What two numbers are those, whose difference i$ S^
and the difference of their cubes 98 ? Ans. S and 5*
4. What two numbers are those whose sum is 6, and the
sum 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.
6. To divide the number 1 1 into two suqh parts, that the
product of their squares may be 784. Ans. 4 and 7.
7. To divide the number 5 into two such parts, that the
sum of their alternate quotients may be 44, that is of the
two quotients of each part divided by the other.
Ans. 1 and 4.
8. To divide 12 into two such parts, that their product
may be equal to 8 times their difference. Ans. 4 and 8.
9. To divide the number 10 into two such parts, that the
square of 4 times the less partf may be 112 more than the
^uare of 2 times the greater. Ans. 4 and 6.
10. To find two numbers such, that the sum of their
squares may be 89, and their sum multiplied by the greater
may prbduce 104. Ans. 5 and 8.
11. What number is that, which being divided by the
product of its two digits, the quotient is 5^ ; biit when 9 is
subtracted from it, there remains a numberilNaving the same
digits inverted ? Ans. 32.
12. To divide 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. To find three numbers in arithmetical Sprogression,
such that the sum of their squares may be 56^ and the sum
arising by adding together 3 times the first and 2 times the
second and 3 times me third, may amount to 28. ^^
Ans.^'2, 4, 6.
14. To divide the number IS into three such parts, that
their squares may have equal differences, andj that the sum
of those squares may be 75. * Ans. 1, 5, 7,
15. To find three numbers having equal differences, so
that their sum may be 12, and the sum of their fourth powers
962. Ans. 3, 4, 5.
16. To
CUBIC, &c. EQUATIONS. 2«
16. To find three numbers having equal differences, and
such that the square of the least added to the product of the
two greater may make 28, but the square of the greatest
added to the product of the two less may 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
9^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. To 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.
10. To find fpur numbers such, that the first may be to the
second as the third to the fourth ; and that the first may be
to the fourth as 1 to 5 ; also the Second to the third as 5 to
9 ; and the sum of the second and fourth may be 20. ,
Ans. 3, 5, 9, 15.
20. To find two numbers such, that their product added
^o their sum may make 47, and their sum taken firom the
sum of their squares may leave 62. Ans. 5 and 7.
RESOLUTION OF CUBIC AND HIGHER
EQUATIONS.
A Cubic Equation, or Equation of the 5d degree or
power, is one that contains the third power of the unknown
quantity. As a?^— (ur* j^ bx =:c,
A Biquadratic J or Double Quadratic, is an equation that
contains the 4th power of the unknown quantity :
As x^ — ax^ f bx^—cx = d*
An Equation of the 5th Power or Degree, is one that
contains the 5th power of the unknown quantity :
As pe^^ax* + bx^^cxP' + dx =^ e.
And so on, for all other higher powers. Where it is to
be noted, however, that all the powers, or terms, in the
equation, are supposed to be freed from surds or fractional
exponents*
There are many particular and prolix rules usually given
(pr the solution of some of the abovementioned powers
or
(tiS ALGEBRA. .
:or equations* • But they may be all readily solved by the
following easy rule of Double Position, sometimes called
TrialandError.
RULE.
«
1. Find, by trial, two numbers, as near the trtie roofas
you can, aiid substitute them separately in the given equation,
instead of the imknown quantity 5 and find how much the
terms collected together, according to their signs + or — ,
differ from the absolute known term of the equation, mark
ing whether these errors are in excess or defect.
2. Multiply the difference of the two numbers, found or
taken by trial, by either of the errors, and divide the pro
duct by the difference of the errors,. when they are alike^
but by their sum when they are unlike. Or say. As the
difference or sum of the errors, is to the difference of the
two numbers, so is either error to the correction of its sup
posed number.
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 nearer \ and, by proceeding in
like manner as above, a root will be had still nearer than
before. And so on to any degree of exactness required.
Note ] . It is best to employ always two assumed numbers
that shall differ from each other only by unity in the last
jBgure on the right hand j because then the difference, or
multiplier, is only I. It is also best to, use always the least
error in the above operation.
Note 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 morQ
figure. Then substitute this corrected result in the equation^
for the uftknown letter, and if the result prove too little,
substitute also the number next greater for the second sup
position ; but contrarywise, 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
^us last corrected number, and the next greater or less, *s
th^
CUBIC, &c. EQUATIONS,
249
the case may require^ carrying the third corrected number
to eight figures; because each new operation commonly
doubles the number of true figures. And thus proceed to
any extent that may be wanted.
Examples.
Ex. 1. To find the root of the^ubic equation x^ +**+
9c = IDO, or the value of * in it.
Here it is soon found that
x lies between 4 and 5. As
sume therefore these two num
bers, and the operation will be
3IS fellows :
1st Sup. 2d Sup.
4  X  5
16  ^* . 25
64
X'
84  sums 
100 but should be
125
155
100
Again, suppose 4*2 and 4'3,
and repeat the work as foU
lows:
1st Sup. 2d Sup.
4'2  jr  4*3
1764 ^ js* ^' 1849
74088  x'  79507
95928
100
sums
102297
100
— 16  errors  +55
the sum of which is 71.
Then as 71 : 1,:: 16 : 2.
Hence »r = 4'2 nearly.
— 4072 errors + 2297
the sum of which is 6*369.
As 6369 fl :: 2297 : 0*036
This taken fiom  4* 300
leaves x nearly = 4*264
Again> suppose 4264, and 4265, and work as follows: '
4264 ^ X  4265
18*181696  ;^*  . 18*190225
77526753
99972448
100
X
sums
775S1310
100036535
100
0027552  errors  +0O36533
the sum of which is 064P87.
Then as 064087 : 001 : : 027552 : 00004299
To this adding  4264
gives X very nearly = 42644299
The
250
ALGEBRA.
The work of the example above might have been much
shortened, by the use of the Table of Powers in the" Arith
metici which would have given two or three figures by in
q>ection. But the example has been worked out so particu
larly as it is, the better to show the method.
Ex. 2. To find the root of the equation x^ — 15jr* + 63x
ss 50, or the value of x in it.
Here it soon appears that x is very little above 1.
Suppose therefore 1*0 and 11 >
and work as follows :
10 
11
63*0 
15
1 *
6^x 
15jr*
x^ 
sums 
errors 
sum •f the
:1 ::1:'03
100
693
1815
1331
49 
50
52'4S1
50
1 
3481
As 3*481 ;
+ 2451
errors.
correct.
Hence x'=
1*03
nearly.
Again, suppose the two num*
bers 103 and 102, &c, ad
follows :
103  AT  I '02
6489  63Jf 64*26
 159135 — 15;v* 156060
1092727 x^ 1061208
50069227 sums 49*715208
50 50
+ 069227 errors — •284793
•284792
As 35401 9 : 01 : : 069227 :
0019555
This taken from 1 '03
leaves 4: nearly = 1*02804
Note 3. Every equation has as many roots as it contains
dimensions, or as there are units in the index of its highest
power. That is, a simple equation has only one value of
the' root ; but a quadratic equation has two values or roots,
a cubic equation has three roots, a biquadratic equation ha$
four roots, and so on.
And when one of the roots of an equation has been found
by approximation, as above, the rest may be found as follows.
Take, for a dividend, the given equation, with the known
term transposed, with its sign changed, to the unknown side
of the equation ; and, for a divisor, take x minus the root
just found. Divide the said dividend by the divisor, and
the quotient will be the equation depressed a degree lower
th^n the given one.
Find
CUBIC, &c. EQUATIONS. 251
Find a root of this new equation by approximation, as
before, or otherwisei and it will be a second root of tlye
original equation. Then, hj means of this root, depress
the second equation one degree lower, and' from thence
£nd a third root, and so on, till the equation be reduced to
a quadratic ; then the two roots of tjiis being found, by the
method of completing the square, they will make up the
remainder of the roots. Thus, in the foregoing equation,
having found one root to be 1 "02804, connect it by minus
with r for a divisor, and the equation for a dividend, &c, as
follows :
X  J 02804) x^ • ISx" + eSx  50 ( x* — 1397l96r +
4863627 = 0.
Then the two roots of this quadratic equation, or '
X*  13'97196jr se  48*63627, by completing the square,
are 6'57653 and 739543, which are also the other two
roots of the given cubic equation. So that all the three
roots of that equation, viz. x^— 15x^ + 6Sx = 50,
d fi« 576*58 J^^d the sum of all the roots is found to b^
j»T«ofi'.iofl5, beine equal to the coefficient of the
and 739543 V ^ 1 ^r 1 • 1 • t_ 1 i
>2d term of tlje equation, which the sum of
1500000 V^^ roots always ought to be, when thejr
/ are right.
sum
Note 4. It is also a particular advantage of the foregoing
rule, that it is not necessary to prepare the equation, as for
other rules, by reducing it to the usual final form and slate
of equations. Because the rule may be applied at once to an
unreduced equation, though it be ever so much embarrassed
by surd and compound quantities. As in the following
example :
Ex. 3. Let it be required to find the root x of the equation
Vl*4**(*' + 20)' + v/i96A?"(*' + 24)* =z 114, or the
value of X in it.
By a few trials, it is soon found that the value of x is but
little above 7. Suppose therefore first that at is = 7, and
then X = B.
First,
t$2 ALG£BRA.
Firsts when .r s= 7j Second, when x ±z B.
I 47906  ^144x*— (a:* + 20)»  46'476
65384  v^ 196^* (a:* +24)2 . 69*283
■ > ■ ■ ■ ■ I 'I >' II
1 1 3290  the sums of these  1 1 5*759
114*000  the true number  IH'OOO
—0710  the two errors  +1*759
+ 1759
■ V
As 2469 : 1 :: 0*710 : 0*2 nearly
70
Therefore t = 7*2 nearly
Suppose again jt = 72, and then, because it turns out toa
great, suppose jc also = 7*1, &c, as follows :
Supp. jr' = 7*2 Supp. ;r =: 7'I
47*990  *Vl44:r*~(^* + 20)* ^. 47973
66*402  ^/196jr*~(a:* + 24)*  . 65*904
114*392  the sums of these  113*877
1 1 4*000  the true number  1 1 4*000
+0*392  the two errors  0*123
0*123
■ I '
As 515 : 1 :: '123 : 024 the correction,
7*100 add
Therefore X = 7*124 nearly the root required.
Note B, The same rule also, among other more difficult
forms of equations, succeeds very well in what are called
exponential ones, or those which have an unknown quan*
tity in the. exponent of the power ; as in the following
example :
• Ex. 4. To find the value of x in the exponential equation
x"" = 100.
For more easily resolving such kinds of equations, it is
convenient to take the logarithms of them, and then com,
pute the terms by means of a table of logarithms. Thus,
the logarithins of the two sides of the present equation are
ff X log,
CUBIC, &c. EQUATIONS.
235
X X iog. of iT = 3 the log. of 100. Then, by a few trials,
it is soon perceived that the value of x is somewhere be
tween the two numbers S and 4, and indeed nearly in the
middle between them, but rather nearer the latter than the
former. Taking therefore first x = S'5, and then n 2*6,
and working with the logarithms, the operation will be as
follows :
First Supp. X =« 3*5.
Log. of 35 =: 054.4068
then 3*5 x log. 2^'S^ 1*904238
the true number 2OQOObO
error, too little, — •095762
002689
Second Supp. x = 3'6.
Log. of 36 = 0556303
then 36xlog.36=200&689
the true number 2000000
error, too great, +.002689
•098451 sum of the errors. Then,
As '098451: '1 ; : '002689 : 000273 the correction
taken from 360000
leaves  359727 = x neatljr.
On trial, this is found to be a very small matter too little.
Take therefore again, x = 359727, and next = 3*59728,
and repeat the operation as follows :
* Firtt, Supp. i" 2= 359727.
Log. of 3^9727 is 0*555973
359727 X log.
 of 359727 = 19999854
the true number.20000000
error^ too little, 00000146
 00000047
Second, Supp. or = 359728.
Log. of 359728 is 0*555974
3*59738 X log.
of 359728 = 19999953
the true number 20000000
error, too little,  00000047
00000099 diff. of the errors. Then,
As 0000099 : 00001 : : 0TO0047 : 0*00000474747 the cor.
added to  359728000000
gives nearly the value oi.^ = .359728474747
Ex. 5. To find the value of x in the equation jt^ + lO^r*
+ 5x = 260. Ans. x = 41 179857.
Ex. 6. To find the value of x in the equation ;r' — 2;r = 50.
Ans. 38648«i. ^^^
Ex. 7.
H .
U4 ALGEBRA.
Ex. ?• To find the value of x in the equation jr' + 2x*
— 2Sx = TO. Ans. x = S'lS^S?.
Ex. 8. To find the value of x in the equation x^  IIa:^
+ 54* = 350. Ans. x = 14"95407.
Ex. 9. To find the value of x in the equation x^— 3x*
— 75x = 10000. Ans. x = 102609.
Ex. 10. To find the value of x in the equation 2x* 16x'
+ 40x*— SOx =  1. Ans. x = 1284724.
Ex. 1 1 . To find the value of x in the equation x^ + 2Af*
+ 3*3 + 4x* + 5;t = 54S21. Ans. x = 8414455.
Ex. 12. To find the value of x in the equation x* =
1234567^9. Ans. x = 8*6400268.
Ex. 13 Given 2x*7x5 f llx^Sx = 11, to find x.
Ex. 14. To find the value of x in the equation
{3x'' 2v/ Jr + 1 )^  (t*  4Xv^x + 3s/x)^ = 56.
Ans. X = 183i0877.
2a resolvf Cubic Equations by Cariarfs Rule.
•Though the foregoing general method, by the applica^
tion of Double Position, b^ the readiest way, in real practice^
of finding the roots in numbers of cubic equations, as well
as of all the higher equations universally, we may here add
fhe particular method commonly called Cardan's Rule, for
resolving cubic equations, in case any person should choose
occasionally to employ that method.
*
The form that a cubic equation must necessarily have, to
be resolved by this rule, is this, viz. z^ :e js =» 5, that is,
wanting the second term, or the term of the 2d power z\
There/ore, after any cubic equation has been reduced down
to its final usual form, x^ + px^ 4 ^x = r, freed from the
coefficient of its first term, it will then be necessary to take
away the 2d term px^ ; which is to be done in this manner :
Take \py or \ of the coefficient of the second term, and
annex it, with the contrary sign, to another tmknown letter
s, thus z—jpf then substitute this for x, the unknown
letter in the original equation x' +^* + ^x = r, and there
will result this reduced equation z^ ^ a» zz i, of the form
proper for applying the following, or Cardan's rule. Or
take r = far, and d = 4^, by which the reduced equation
takes this form, ;&' )c ^cz^ 2d.
Then
CUBIC, &c. EQUATIONS. 255
Then substitute the values of ^r and d in this
form, z =l^d+ V(^* + ^0 +{/^ V(^*+f^),
or * = V^+V(^T^^^7^^,
and the value of the root z, of the reduced equation s^ 4b
az = i, wjll be obtained. Lastly, take x = «— j^* which
will give the value of x, the required root of the original
equation x^ + px^ + qx czr^ first proposed.
One root of this equation being thus obtained, then de
pressing the original equation one degree lower, after the
manner described p. 250 and 251, the other two roots of
that equation will be obtained by means of the resulting
quadratic equation.
N^e, When the coefficient a, or c, is negative, and c* is
greater than d''^ this is called the irreducible case, because
then the solution cannot be generally obtained by this rule.
Ex. To find the roots of the equation x^^Sx^ + lOxsrS.
First, to take away the 2d term, its coeflicient being ^ 6y
its 3d part is — 2 ; put therefore ;r = « + 2 ; then
4r' = z3^6x*+ 12a f 8
 6x^ =  ez^  24.Z  24
J^lOx = + lOz + 20
theref. the sum z^ ♦ — 2z + 4=8
or z' * — 22 = 4
Here then «= — iS, ^ = 4, ^== — , rf = 2.
Tiieref. 4/£/4v/K+^) = e/^i5+ v/(4,V)= ^2 + / i^o=:
i/2 + V^V^ = 157735
and 3 /i^(^H^ ^) = >^^2^/(4:^V)=^2  ^/'^^ =
4/2 — V>/3 = 042265
then the sum of these two is the value of z = 2.
Hence :r zz z + 2 = 4, one root of x in the eq. r'— 6x* +
lOo: = 8.
To find the two other roots, perform the division, &c, as
in p.. 251, thus :
x4) J^— 6jr*+ 10jr8(jr"2;r + 2=B0
2x^+ IOt
2ar*+ Sjt
2ar— 8
2ar— 8
Hence
tS6 ALGEBRA.
*
Hence x^—2it = — 3, or a?f — 2ir +1 r: — 1, and r — 1
= ± \/'l I r c=: 1 + v' —1 or = .1  v'  *» the two
other sought.
Ex.2. Tofindtherootsof jt'— 94:*.+ 28Jr = 30.
Ans. X :i= 3, or = 3 +v/ — 1, or =3 — v/ — I.
Ex. 3. To find the roots of x'^rTf* + 14r =t 20.
Ans. x = 3, or = i + ^ — 3, or =* 1 — iy/' — 5.
OF SIMPLE: INTEREST.
As the interest of any sum*, ft)r anytime, is directly, pro
portional to the priocijxil sum^ and. to the time ; there^re
• the interest of i pound) for 1 yoar> beine multiplied by any
given principal sum', and by the time of its foH^earance^ in
y^ars and part$> will give it^ interest for that time. That isj
if there be put :
#
r = the rate of interest of 1 poimdper annum,
p = any principal sum lent,
/ = the time it is lent for, and
a = the amount or sum of principal and interest; then
IS prt = the interest of the sum p; for th^time /, and conseq.
p +prt or p X {I +rt) = a, the'amount for that time.
From this expression, other theorems can easily be de
duced, fo}r finding arty of the qtiaiui£jes abpVe mentioned: ^
which theorems, collected together,. will be as below i
1st, a "=: p + prfy the amount,
^^' ^ =" 1 4 rt *^^^ principal,
^  a—p ,
3d, r = — y the rate,
pt
4th, / = , the time.
pr
»
For Example. Let it be required to find, in what time
any principal sum will double itself, at* any rate of simple
interest.
In this case, we must use the first theorem, a = /) + prty
in which the amount a must be made = 2p, or double the
principal, that* i», /) + prt ;= 2/?, ovprt = /;, or r/ = 1 ;
and hence / «= — .
r
Here,
COlaPOUND INTEREST, $Sl
Here^ r being thfe iaterest of 1/^ for 1 yearj it follows,
that the doubling at simple interest, is equal to the quotient
t)f any sum divided by its interest for 1 yfear. So, if the
Ihate o^ interest be 5 per cent, then 100 ^ 5 =i^2d, is^the
time of doubling at that rate.
Or the 4th theorem give^ at 6nc6
a—p ^p — p 2—11
^ = — ^=± r~ = — 1—==— i the same as before*
pr pr TV
k!
Compound iNXEREST?
•■ . »
Besides the quiantities concerned io Simple ]^tfrest
hamely^
p ^,the principal sum, *
r =r the ratie or interest of 1/. for 1 year,
a = the whole aittoiim of the principal and interest,
i = the time,
th^re is another qUatltity employed in Compound^ Interest,
viz. the ratio of the x^te of interesti which is the amount
pf 1/. for 1 time of payment, and which here let be denoted
by R, viz.
R. ax i 4 ^j the amount of lU for 1 time.
Then the particular, amounts for the several tiities rpay
he thus coiiipflted, viz; As I/« is to its amount for any time,
isb is any proposed priiicipal sum> to its amount for die same
time \ diat is, as
1/. : R : : p : pR, the Ist yearns amount,
12. : R : : pR : j^R*, the 2d year's amount,
1/. t R : ; pR* t j^R', the 3d year's amount,
and so on. . , .
Therefore, in general, pBl^ =i a is the amount for the
t year,, or t time of piiyiiaent. Whence the following general
theorems are deduced : ,
Ifit, a = /RV the amount^
a
2d, ^ = n^t ^^ principal,
3d, R = .J/—, the ratio,
^ _ log. of <7 — log. of^ ,
: Vol. I. S From
1>5«
ALGEBRA.
,v.
From which, anyone 6( th^'qu&mmes imify be foElndr
Wh^n the rcsft are giteh^
A3 to the whole interest, it is found.by barely subtracting
the principal p from the amount a.
Example\ Suppose it be required to find, jn how many
years any principal, sum will double itself, at any proposed
rate of cote[^Qttn4 interest.
In this case the 4th theorem must be employed, making
41 = 2/ ; and then it is
log. fl— log./ log. 2^— log. p _ log. 2
"" log. Rk "" log. R. " log. R'
So, if the rate of interest be 5 per cent, per annum; then
lR BB 1 + *05 rr 1*05 j and hence . ^ .
log. 2 301030 , ^ ^^^^ , . '
/ = .:; — ^— : = ■ ^ ,^^ = 142067 nearly ;
log.J05 021189 ^ ■
that is, any sum doubles' itself in 14^ years nearly^ at the
rate of 5 per cent, per anmiixi compound interest* .
Hence, and from the like question iti Simple Interest^
' above given, are deduced the times in Which any sum
doubles itiself, at several rates of interest, both simple and
compound; Viz. ' • .
ii^mt^^i
At'
2
24
3
3i
4
5
6
7
8
9.
10
''AtSimp.Int
>
per cent, per annum
interest, 1/. 6r any
other sum, will
double itself in the
following years.
m
50
40
33
28f
25
22
20
16.
14
12^
IH
10
At Comprint.
•*»—«•
ill 350028 .
280701
234498
2014S3
176730 .
157473 fr
142067 §
118957*
!0'2448
'90065
804S^
72725
.1
The
f
COMPOUND INTEREST
2SB
The following Table wili very mttdi facilitate caloilitipiis
of compound interest on any sum, for any number of yeacSf
at various rat«s of imef est
The Amounts of 1/. in any Number of Tears.
Yrs.
3
34
.4
■■4,i
. 1 '^i
>
1*0300
1*0350
10400
1^0450
1*0500
106DO
2
10509 10712
IO8I6
10920
11025
11236
3
10927
11087
]124g
1J412
1*1576
11910
4
V1255
11475
11699
11925
12155
1*2625
5
1.1593
11877
12167
12462
l!?763
13392
6
11941
1*2293
12653
1*3023
1*3401
14165
7
1*2299
13723
1*3159
13609
1*4071
1*5036 ^
8
1*2668
13168
13686
1*4221
1*4775
15939
9
13048
1 3629
1*4233
1*4661
1*5513 1
i*669il
10
13439
1*4106
1*4802
15530
16289
17909
n
13842
14600
j5«9^
xm^
17103
18933.
12
14258
1*5111
I'60lO
16959
17959
20l2»
13
1*4685
J 5649,
16651
l'77'^2
1*«856
213'^
14
15126
l*6l87*
1*7317
18519
19799
22609
15
15580
1Q753
18O09
1*9355
20789
23966
16
16047
17340
1*8730
2*0224
2*1829
25404
17
16528
17947
19^79
21134
22920
26928
18
17024
18575
20258
2*2035
24066
28543
19
17535
19225
2 1068
2*30/0
25270 30256
20
18061
i'989B
21911
24U7
26533 32071
\
The use of this Tafeje, which contains all the poivers, :E^j
to the 20th power, or the anjounts of 1/, is cniefly tp c;^T
culate the interest, or the aoiount of any principal sum, for
any time, not more than 20 years.
For example, let it be required to find, to how much 92$I*
will ai^punt in 15 years, at the rate of 5 per cent per fttmuisi
^oropQund interest.
In the table, on the line 15, and in the column 5 per cent.
is the amount of 1/, viz.   2*0789
this multiplied by the principsil  523 ^
gives the amioimt
or
and therefoi*e the interest is
10S72647
1087/. 5/. '$IJ.
564/. 5s. 3^,
Noie 1. When the rate of interest i» to be determined tt^
any other time than a fear ; its suppose to 4 a ye^r, 4^^^a
year, &c ; the rules are still th« Mfiie \ '• but theii^ s^mHH
S 2 express
^260 ALGEBRA.
«
express, that time» and r must be taken the amount (or 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 deter
mined in the following manner :
. 1//, For any time which is some aliquot part of a year :—
Find the amount of 1/. 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 prin
cipal sum, will produce the amount of the given sum as
required.
2</, 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 before.— And in these calculations, the operation by loga
rithms will be very useful. '
OF 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
_ commenced ; and the latter such as will not liegin till some
particular, event has happened, or till after some certain
time has elapsed.
When an annuity is forborn for some years, or the pay*
moits not made for that time, the annuity is said to be in
Arrears.
An annuity may also be for a certain number of years ;.
or it may be without any limit, and then it is called a Per
petuity.
The Amount of an annuity, forborn for any number of
•years, is the sum arising from the addition of all the annul
.f i^s for that number of years j together with the interest due
Upon eAchjafter it becomes due. .
^i. : 4j The
ANNUITIES.' 261
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 forborn, or lent for ;
ft = the amount of 1/. for 1 year i
m = the amount of the annuity ;
V = its value, or its present worth.
Now, 1 being the present value of the sum R, by propor
tion the present value of any other sum a, is thus found :
a •
as R : 1 : : tf ; — • the present value of a due 1 year hence.
In like manner — is the present value of u due 2 years
a a A u o
hence ; for r : 1 : : — : ~y So also — ^ ^, — j, &c, will
be the present values of a, due at the end of 3, 4, 5, &:c,
years respectively. Consequently the sum of all diese, or
— +I+ A+r: + &c = {— + 5+i + ri&<^)x«»
R R* R' k* ^R R* R^ R* '
continued to n terms, will be the present value of all the n
years' annuities. And the value of the perpetuity, is the
sum of the series to infinity.
But this series, it is evident, is a geometrical progression,
.1 . ' .
liaving — both for its first term and common rsrtio, and the
R ~ .•»
number of its terms n ; therefore the sum v of all the terms^
or the present value of all the annual payments, will be
Jl \^ l^
R R R" R" — 1 tf
« = — y— x«,or=j^3^x
1
R ^ . ■■
When the annuity is a perpetuity; n being infinite, R**
. 1
is also infinite, and therefore the quantity — becomes = 0,
therefore " x ^ also = ; consequently the exprei
R — 1 R
sion becomes barely v = — jr ; that is; any annuity divided
by the interest of 1/. for } year, gives the value of .the per
petuity. So, if the rate of interest be 5 per cent.
Then lOO/i r 5 = 20a is the value of the perpetuity at
S per cent : Also IQOa r 4 = 25a is the value of the per
petuity
r
26^ ALGEBHA.
• • • •
petuifj' at 4 f)6r teni : Atid 100^ r 3 £= SS^ h tlie valut of
the perpetuity ^t 5 pcp cent t ^nd so on.
Again, because the amount of 1/. in n years, fe H", its
increase in that time will be R*^— I ; but its interest for one
single ytdr, or the annuity answering to that increase, is
R — 1 ; therefore as R — 1 is to r" — 1, so is a to iw ; that
r"— 1
is, m =: r X a. Hencc^ the several cases relating to
R — 1
Annuities in Arrear, will be resolved by the foUowipg
equations :
m = X a ■= vR" i
R — 1
. • R" i~ 1 a m
R  i . R  1 ^
t»  1 R"  1
WR — W f tf
■ log.
log. >yf — log. v _^ ^i
"" log. R ^ log. R *
log. ^ » log. V
. X.pg. R =2 — ^ '
■•r "
#
n
1 1 fl
^R.P V^ R  1
'.Ih this last theorem, r denotes the present value 6f an
annuity in reversion, after p years, or not commencing till
after the first j& years, being found by taking the difference
between the two values 7 x — r and x ;;:, for
R — 1 R" R — I RP
n jeTLi^ and p years.
But the amount and present value of any annuity for any
number of years, up to 21, will be most readily found by the
two following tables.
I 
TABLE
AumimES,
2$$
TABLE I.
The Amoiint of an Anhiiity of 1/. at Compound Interest.
im
atSperc.
M «^^
"^
1
2
r 3:
4
5
6
7
8
9
10
11
12
J3
14
15
16
■.17
a8
19
20
21
3f per cJ 4 per c.
.10000
^3'OaOO
1*0000
2T03i0
.^fOpQgrl 3iQQ2.
41836
5^091
d4684
7*6625
d8923
101591
1 1 4^9
128078
141920
156178
1 70863
185989
201569
>2l'5?6l6
334144
268704
286765
42149
53625
65502
7.7794
90517
103685
11*7314
131420
146020
161130
176770
192957
209710
227050
24^^7
^3572
282797
10000
20400
ai2i6
42^465
54163
66330
7*89.83
92142
105828
12*0061
13*4864
15:0258
166268
182919
203236
218245
S3^75
25*6454
«7*6712
297781
4^ per c.
3026951319692
10000
2*0,450
31370
4278:?
54707
67169
8*0192
9*3800
10*8021
122892
138412
154640
171599
1 89321
207841
227193
247417
26*8J51
290636
31'3714
'337831
5 per c. 6 per c.
10000
20500
31525
43101
55256
6*8019
81420
9*5491
110266
i«*5779
142068
15*9171
177130
195986
215786
236575
258404]
281324
305390
33*0660
135*7193
1*0000
20600
61836
43746
56371,
69753
83938 j
9*8975
114913
131808
U»97i6
166699
18*8821
210151
232760
256725
282129
309057
3376001
S6'7856l i
399927 1
tab;l^ 11. Th^ Pr^s^t ^Vs^lue of an Annuity of 1/.
Yrs.
1
2
3
4
5
6
7
8
9
10
11
12
\3
14
15
16
17
18
19
20
21
at9p^c<
09709
l9i35
28286
37171
45797
54172
62303
70197
7*7861
85302
92520
99540
106350
112961
11 9379
125611
131601
137535
143238
148775
15*4150
3t per^.[4 per c
09662
1*8997
28016
3*6731
45151
53286
61145
68740
7:0077
83166
9*01 16
96633
iG3027
:i 09205
115174
12*0941
126513
13*1897
137098
142124
14*6980
09615
18861
2*775 1
36299
44518
52421
60020
67327
74353
8110J9
87605
93851
9*9857
10563 1
111184
116523
12 1657
li6593
131339
135903
140292
4t per c.
0956&
1*8727
27490
35875
43900
51579
589i7
65959
7268S
79127
8*5289
9 1186
9*6829
102228
107396
1 1 2340
117072
121600
125933
130079
134047
5 perc.
ii< >i
0*9524
18594
27233
35460
43295
50757
5>864
64632
71078
7*7217
8*3054
8B633
9*3936
9*8986
10*3797
108378
112741
116896
12*0853
124622
•12*8212
li ■
per e.
09434
1*8334
2673^
34651.
42124<
4*9173
55824
62098
6*8017
7*3601
7*8869
83838
88527
929501
9*7^23
10*1059
104773
108276
11*1581
11*4699
117641
Ta
sH ALGSBRA.
n find thf Ammnt cf any annuity forbore a eertain numbir tf
years*
TiiKE out the amount of 1/. from the £r$t table^ for the
proposed rate and time ; then multiply it by the given
annuity; and the prodyct will be the amount , for the same
number of years, and rate of int^r$st^^And the converse to
find the rate or time,
£xaf^. To £nd how much an annuity of 50/. will amount
to in 20 years, at Si per cent, compound interest.
On the line of 20 years, and in the column of 3, f^r cent,
$tands 28*2797^ which is the amount of an annuity of l/« for
the 20 years. Then 282797 x 5Q, givw 1413985/. a:;
■ 1413/. 19/. 8d. for the answer required.
* To find the Present Value of anyannwtt^foranj number of
j^^tfrx.— Proceed here by the 2d table, in the same manner as
iabove for the Ist tabl^, and the present; vorth require will
be found.
Exam, \. To find the present value of an annuity of SOU
which is to continue 20 years, at 3^ per cent.f^y the table,
the present value of 1/. for the given >ate and time, is
142124 ; therefore 142124 x 50=710*62/. or 710/. 12/. 4rf.
IS the present value required^ * .
Exam. 2. To find the present value of an annuity of 20A
to commence 10 ye^rs hence, and then to continue for l\
jezef longer, or to terminate 2 1 years hence, at 4 per cent.
mterest. — In such cases as this, we hate to find the difierence
between* the present values of two equal annuities, for the
two given time? y which therefore will be done by $ubtracti
ing the tabular value of the one pieriod from thatdf the^
Other^ and then multiplying by the given annuity. Thus,
tabular value for 21 years 14*0292
ditto for  10 years 8 1105^
.■I . ■ 1
the difference 5'918i
multiplied by 20
H. ' I' "
■ I
gives  '1J8466/.
or   1 1 8/. 7/. 34*/. the zns9ftr\
ENP OP THE ALGEBRA.
•a
sess
i^MM*^
55SS55
!te«9MBavs
GEOMETRY.
•  DEFINITIONS.
1. jhL point is that which has position,
h%it no magnitude, nor dimensions ^ neither
length, breadths nor thickness.
2. A Line is length, without breadth or
thickness.
3. A Surface or Superficies, is an extension
or a figure of two dimensions, length and
bri^adth ', but without thickness.
4. A Body or 6oli4, is a figure of three di
mensions, namely, lepgth, breadth, and depth,
or thickness.
5. Lines are either Sight, or Curved, or
JVIixed of these two.
6. A Right Line, or Straight Line, lies all
in the same direction, between its extremities ^
^d is the shortest distance between two points.
When a Line is mentioned simply, it means
t Right Line*
7. A Curve continually changes its direction
between its extreme points.
8. Lines are either Parallel, Oblique, Per
pendicular, or Tangential.
9. Parallel Lines are alw&ys at the same per
pendicular distance; and they nevermeet, though
ever so far produced.
10. Oblique lines change their distance, and
would, meet, if produced on the side of the
least distance*
IL One line is Peipendicular to another,
when it inclines not more on the one side
than
t:v
— *
266 GEOMETRY.
•— / .^«.
than the ©ther, or when the angles on both
sides of it are equaL
12. A line or circfe is Tangehtial, or a
Tangent to a circle, or other curve, when it
touches it, without cutting, when both are pro
duced.
13. An Angle Js the inclination or opening
of two lines, having difierent directions, and
meeting in a point*
14. Angles are Right or Oblique, Acute, or
Obtuse,
15. A Right Angle is that which js made by
cuae iine. perpendicular to another. Or when
the angles on each side are equal to One an*
Qther^ they are right angles.
; 16. An Oblique Angle is that which is
made by two oblique lines ; and is either le$s
or greater than a nght angle.
l7. An Acute Angle is less than a right
angle.
1. IS. An Obtuse Angle is greater than a right
angle. /■ v ' '
i9. Superficies are either Plane or Curved.
20. A Plane Superficies, or a Plane, is that with which
a right line may, every way, coincide. Or, if the line touch
the plane in two points, it will touch it in every point. But,
if Bot, it is curved.
21. Plane Figures are bounded either by right lines or
curves.
22. 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 miny sides as angles ; the least
number being three.
23. A figure of three sides and angles is called a Triangle.
And it receives particular denominations from the relations
of its sides and angles.
24. An Equilateral Triangle is that whos^
three sides are all .equal.
25.. An Isosceles Xnangle is that which has?
two sides equal.
26. A
DEFINITIONS.
26t
^
zx
[
* 26. A Scalene Triangle is that whose three
sides are all unequal: ^ . '
27. A Rightangled Triangle is that which
has one rightangle. >
.28. Other triangles are Oblique angled, and
are either Obtuse or Acute. ^
29. An Obtuseangled Triangle has one ob
tuse an<rle.
SO. Alt AcuteAngled Triangle has all its
diree angles acute
8K A figure of Four sides and angles is
called a Quadrangle, or a Quadrilateral
'32. AParallelogram is a quadrilateral which
&as both its pairs cf opposite sides parallel.
And it takes the following particular name$»
viz. Rectangle, Square, Rhombus, Rhomboid.
33. A Rectangle is a parallelogram having
a right angle.
34*. A Square is an equilateral rectangle;
having its length and breadth equal.
35. A Rhomboid is an obliqueangled pa
rallelogram.
S6. A Rhombus is an equilateral rhomboid;
having all its sides equal, but its angles ob
lique.
37. A Trapezium is a quadrilateral which
hath not its opposite sides parallel.
'* . . '
. 38. A Trapezoid has only.one pair of oppcH
site sides paralleK
39, A Diagonal is a iitae joining any two
opposite angles of a quadriiateraL
40^ Plane figures that have more than four sides are, in
general, called Polygons : and they receive other particular
names, according to the number of their sid,es or angles.
Thus,
41. A Pentagon is a polygon of five sides; a Hexagon, of
six sidfcs;*a Heptagon, seven; an Octagon, eight; a No
fiagon, nine ^ a Decagon, ten ; an Undecagon, eleven ; and a
Dodecagon, twelve sides»
42. A
C7
i6i
GEOMETRY.
42. A Regular Poljgon has all its skies and all its angk§
equol. — If they are not both equal, the polygon is Irregular.
43. An Equilateral Triangle is also a Regular Figure of
three sides, and the Square is one of four j the former being
also called a Trigon, and the latter a Tetragon.
44. 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 a regular figure.
45. A Circle is a plane figure bounded by
a curve line, called the Circumference, which
55 every where equidistant from a certain point
within, called its Centre.
The circumference itself is often called a
circle, and also the Periphery.
46. The Radius of a circle it a line drawn
from the centre to the circumference.
47. The Diameter of a circle is a line
drawn through the centre, and terminating at
the circumference on both sides.
4S. An Arc of a circle is any part of the
circumference.
5 I
49. A Chord is a right line joining the
extremities of an arc.
50. A Segment is any part of a circle
bounded by an arc and its chord.
51. A Semicircle is half the circle^ or a
segmeilt cut off by a diameter.
The half circumference is sometimes called
the Semicircle.
52. A Sector is any part of a circle which
IS bounded by an arc, and two radii drawn to
i$s extremities.
53. A Quadrant, or Quarter of a circle, is
a sector having a quarter of the circumference
for its arp, and its two radii are perpendicular
to each other. A quarter of the circumference
» sometimes called a Quadrant.
•j^
5*. Th«
JJEPlMrtlONS.
26!l
/k
D K
V.
54. The Height or Altitude of a figure is
ft perpendicular let fall from an angle, or its
vertex, to the opposite side, called the base.
65, In a rightangled triangle, the side op
posite the right angle is* called tlie Hypothe
nuse 'y and the other two sides arecafted the
X^egs, and sometimes the Base and Perpends .
cukr.
56. When an angle is denoted by three
letters, of which one standi at the angular
point, and' the other two on the two sides,
that which stands at the angular point is read
in the middle.
57. The cirCunrference of every circle is
supposed to be divided into 360 equal parts,
called Degrees : and each degree into QO Mi
nutes, each minute into 60 Seconds, and so on.
Hence a semicircle contains 1 80 degrees, 2XidL
a quadrant 90 degrees.
58. The Measure pf an angle, is an arc of
any circle contained between the two lines
which form that angle , the angi,ilar point being
the centre; and it is estimated by the number
of degrees contained in that arc.
59. Lines, or chords, are said to be Equi
distant from the centre of a circle, when per
pendiculars drawn to them from the centre
are equal. .
60. And the right line on whichthe Greater ^
Perpendicular falls, is said to be farther from
the centre. .
6 k 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, >
62. An Angle On a segment, or an arc, is that which is
contained by two lines, drawn from any point in the opposite
or supplemental part of the circumference, to the extremi
ties of the arc, and containing the arc between them.
63. An angle at the circumference, is that
whose angular point is any where in the cir
cumference. And an angle at the centre, is
that whose angular point is at the centre,
1.^ \ . 6% A
270
GEOMETRY.
64. A right»Uned 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.
65. A rightlined figure Circumscribes a
circle, or the circle is Inscribed in it, when all
the sides of the figure touch the circumference
of the circle.
66. One rightlined figure is Inscribed in
another, or the latter Circumscribes the former,
when all the angular points of the former
are placed in the sjuies of the latter.
67. A Secant is a line that cuts a circle,
lying partly within, and partly without it.
68. Two triangles, or other rightlined figures, are said tp
be mutually equilateral, when all the sides of the one are
equal to the cotresponding sides of the other, each to each r
and they are said to be mutually equiangular, when the
angles of the one are respectively equal to those of the other.
69. Identical figures, are such as are both mutually equi
lateral and equiangular ; 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;
the two becoming as it were but one and the same figure.
• 70. Similar figures, are those that have all the angles of
the one equal to all the angles of the other, each to each, ind
the sides about the equal angles proportional;
74. The Perimeter of a figure, is the sum of all its sides
taken together.
72. A Proposition, is something which is either proposed
to be done, or to be demonstrated, and is either a problem or
a theorem.
4
73 A Problem is something proposed to be done.
74. A Theorem , is something proposed to be demonstrated.
75. A Lemma, is something wliich is premised, or demon
strated, in order to render what follows more easy.
76. A CoroUary, is a consequent truth, gained immedi
ately fiom some preceditig truth, or demonstration.
* 77. A Scholium, is a remark or observation* made upon
something going before it; ' * '^ *'
' # ^ AXIOMS.
[ 211 ]
, «
AXIOMS.
» • . ■ ' •
i. Thincs which are equal to the sam^ thing are equal
to each other.
5. When equals are added to equals^ the wholes are equal.
, 3. When equals are taken from equals, the remains arc
equal. " * , \
4. When equals are added to unequals, the wholes are
unequal.
5. When equals are taken from unequal% the remains are
: unequal. ' y
6. Things which are double of the same tbiag) or eqml
things, are equal to each other.
, 7. Things which are halves, of the samd thingi tsfe eqUaL
• ' . ,
^. Every whole is equal to all its parts taken together. '
. 9. Things which coincide, or fill the same space, are iden
tical, or inutually equal in all their parts.
10. All right angles are equal to one another..
1 1» Angles that have equal measures, or arcs, are equaL
THEOKEM 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 aec, 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 tq ^he angle f \
then will the two triangles be identi
cal, or equal in all wsp^ts, ■ ^ , ;
.. for xioiiceive the triatijjfe. £BC it^ be applied to, of ^e«d
On^ the t]4angle VB^yitk iMin a ttiAnii^ that tho^poiiit fc may
S72
GEOMETRY.
 \
coincide with the point f^ and the side ac with th< $ide Jiff
which is equal to it*
Then, since the angle F k equal to the angle c (by hyp.)^
the side bc will fall on the side £f. Also, because AC is
equal to df, and BC equal to £F (by hyp.)> the point A will
coincide with the point d, and the point B with the point £ ;
consequently the side ab will coincide witfi the side db.
Therefore the two triangles are identical^ and have all their
other corresponding parts equal (ax. 9), namely, the side ab
equal to the side D£, the angle A to the angle t>, and th^
angle B to the angle e. <^ £. d,
THEOREM tU
When Twb Triangles have Two Angles and the included
Side in the one, equ^ 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, t)EF
have the angle A equal to the angle
D, the angle B equal to the angle £,
and the side ab equal to the side D£;
then these two triangles will be idetl^
tical.
«
For, cenceive the triangle abc to be placed on the triangle
PEF, in such manner that the side ab may fell exactly on the
equal side de. Then, since the angle a is equal to the angle
D (by hyp,)> the side ac must fall on the side df ; and, in
like manner, because the angle B is equal to the. angle £, the
side BC must fall on the side ef. Thus the three sides of the
triangle abc will be exactly placed on the thr^e sides of the
triangle def: consequently the two triangles are identical
(ax. 9), having the other two sides ac, Bfc, equal to. the two
DF, EFy and the remaining angle c equal to the remaining
angle f. <^ e, d.
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 equal.
U the triangle abc have the side AC equal
to the side bc; then will the angle d be
equal to the angle a.
For, conceive the angle c to be bisected,
or divided into two «qua^>aru, by the line
CD, making the angle A(ii> equidi.to the
angfe*BCD. Then,
THEOREMS.
273
Then, the two triangles acd, bcd, have two sides and
the comained angle of the ,one, equal to two sides and the
contained angle of the other, viz. the side ac equal to bCj
the angle acd equal to bcd, and the side cd common;
therefore these two triangles are identical, or equal in all
respects (th, 1); and consequently the angle a equal to the
angle b. q.. e. d.
CotqL 1. Hence the line which bisects the verticle angte
of an isosceles triangle, bisects the base, and is also perpendi
cular to it.
Corol. 2. Hence too it appears, that every equilateral tri
angle, is also equiangular, or has all its angles equal.
THEOREM IV.
When a Triangle has Two of its Angles equal, the Sides
Opposite to them are also equaL
If the triangle abc, have the angle a
equal to the angle B, it will also have the side
AC equal to the sideBc.
For, conceive the side a B to be bisected
in the point d, making ad equal to db;
and join dc, dividing the whole triangle into
the two triangles acd, bcd. Also conceive
the triangle acd to be tiuned over upon the
triangle bcd, so that ad may fall on bd.
Then, because the line ad is equal to the line db (by k0.),
the point A coincides with the point B,.and the point d with
the point d. Also, because the angle a is equal to the angle
B (by hyp.), the line ac will fall on the line bc, and the ex
tremity c of the side ac will coincide with the extremity c
of the side bc, because dc is common to both ; consequently
the side ac is equal to BC. Q E. D.
CoroL Hence every equiangular triangle is also equila*
' teral*
THEOREM v. ^
When Two Triangles have all the Three Sides in the one,
equal to all the Three glides 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 j then shall
the two triangles be identical, or have
^ their angles equal, vijs. those afigles
Vol. I. T thuc
(ijrt>t4^l^t^
» *
i74
GEOMETkY.
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 bngest
equal sides, and draw the line cd.
Then, in the triangle acd, because the side AC is equal
to AD (by hyp.), the angle acd i^ 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 Be is equal
to BD. Hence then, tne angle acd being equal to the angle
ADC, and the angle bcd to the angle bdc, by equal addi«
tions 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, (by 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 ab& to the
angle abd. q. £• d.
THEOREM VI.
When one Line meets another, the Angles which it
makes on the Same Bide 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 to
gether, 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. 15).
But when the angles are unequal, suppose BE drawH per
pendicular to CD. Then, since the two angles ebc, ebd,
are right angles (def. 15), and the angle ebd is equal to the
two angles eba, abd, together (ax. 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, q. £. D*
CoroL 1. Hence also, conversely, if the two angles ABC,
abd, on both sides of the line ab, make up together twa
right angles^ then cb and bd form one contixraed right
Ijne CD.
Corol.
THEOREMS. ?75
CoroL 2. Henccy all* the injgles 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 twQ
right anglesi. \
Corol. S. And, as all the angles that can be made on th«
other side of the line CD are also equal to two right angles ;
therefore all the angles that c^m be made quite round a point
s, by any number of lines, are equal to four right angles.
CoroL 4. Hence also the whole circumfer
ence of a circle, being the sum of the mea
sures, 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 ene
right angle.
THEOREM VII.
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
ABD. equal to the angle ceb.
For, since the line CE meets the line
AB, the two angles aec, beg, taken
together, are equal to two right angles (th. ^).
In like manner, the line be, meeting the Jdne CD, majces
the two angles bec, bed, equal to two right angles. , \
Therefore the sua\ 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 VIII.
When One Side of a Triangle is produced, the Outward
Angle is Greater than either of the two Inward Opposite
. Angles.
T2 Let
276
GEOMETRY.
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
•r c.
For, conceive the side bc to be bi
sected in the point £, and draw the line
ACT, producing it till bf be equal to A£i
and join bf.
Then, since the two triangles A EC, bef, have the side
AE =s the side ef, and the side ce = the side be (by suppos.)
and the included or opposite angles at s also equal (th. 7),
therefore those two triangles are equal in all^ respects
(th. I), and have the angle c = the corresponding angle
EBF. But tlie angle cbd is greater than the angle ebfj
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 bi
sected, it may be shown that the outward angle ABG, or its
equal CBD, is greater than the other angle a.
J>B
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, havbig the side
AB greater than the side ac ; then will
the angle acb, opposite the greater side
AB, be greater than thp single 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. 8). But the angle acd
is equal to the said outward angle adc, because ad is equal
to AC (th. 3). Consequently the angle acd also is greater
than the angle B. And since the angle acd is only a part
of i\cB, much more must the whole angle ACB be greater
than the angle b. q. e. d.
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 i^. But it cannot be equal, for
then
THEOREMS. 277
•
thtn the angle c would be eqyal 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 supposi
tion. The side ab, then, being neither equal to AC, nor less
than it, must necessarily be greater, q^. fi. D.
« THEOREM X. v
The Sum of any Two Sides of a Triangle is Greater than
the Third Side.
Let ABC be a triangle ; t^en will the » p
^um of any two of its sides be greater than ...••* i
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 (by 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 op
posite to the greater angle (th. 9), the side ad (of the tri
angle abd) is greater than the side ab. But ad is equal
to AC and cd, or ac and cb, taken together (by constr.) ;
therefore ac + cb is also greater than ab. <^ e. d.
Cors/. The shortest distance between two points, is a singly
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
difference 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 (by constr.), the opposite
angles d and abd are equal (th. 3). But the angle cbd
is less than the angle abd, and consequently also less than
the equal angle d. And since the greater side of any triangle
is
rts
GEOMETRY.
IS opposite to the greater angle (th. 9), the side co (of ihoj
triangle bcd) is less than the side bc. q. £• D.
THEOHEM XII.
When a Line Intersects two Parallel Lines, it m^es the
Alternate Angles Equal to each other.
Let the line ef cut the two parallel
lines AB, CD $ then will the angle aef be
equal to the alternate angle efd.
For if they are not equal, one of them
must be greater than the oth^ ; kft it be
£FD for instance which is the greater, if
poissible ; and conceive the line fb to be
drawn ; cutting off the part or angle efb equal to the angle
JLEF ; 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 (th, 8)^^ and
tunce these two angles also are equal (by the 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. <^ £. d.
CoroL Right lines which are perpendicular to one, of two
parallel lines, are also perpendicular to the other.
T.HEOB.EM XIII.
When a Line, Cutting Two other Lines, makes the Al
ternate Angles Equal to each other, those two Lines are Pa^
rallel.
Let the line ef, cutting the two lines
as, CD, make the alternate angles aef,
PFE, equal to each other; then will ab
be parallel to cd..
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. J 2). But
the angle aef is equal to the angle efd (by hyp.) There
fore the angle efd is equal to the angle efg (ax. 1) ;.that
is, a part is equal to the whole, which is impossible. There^
fore no line but cd can be parallel to AB. q. e. d,
CoroL Those lines which are perpendicular to the same
line, are parallel to each other.
theorem
^■V i
THEOREMS.
m
THEOREM XIV.
When a Line cuts two Parallel Lines, the Outwar4
Angle is Equal to the Inward Opposite one, on the Same
Side ; and the two Inward Angles, on the Same Side, equal
to two Right Angles.
Let the line £F cut the two parallel
lines AB, CD; then will the outward angle
EGB be equal to the inward opposite
angle ghd, oh the same side of the line
£v; and the two inward angles bgh,
GHp, taken together, will be equal to
two right ^gles.
For, since the two lines ab, cd, are
parallel, the angle agh is e<^ual to the alternate angle ghd,
(th. 12). But the angle agh is equal to the opposite an?l^
£GB (th. 7). Therefore the angle ZGB is also equal to the
angle ghd (ax. l). (^ e. d.
Again, because the two adjacent angles EGB, bgh, are
together equal to two right angles (th. 6) ; of which the
single EGB has been shown to be equal to the angle ghd;
therefore the two angles bgh, ghd, taken together, are
^so e<val to two right angles. v
Coroi. 1. And, conyers^y, if one line meeting two other
lines, make the angles on the same side of it equal, those
two lines are parallels*
CoroL 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.
G
H
D
THEOREM XV,
Tho3L LInfs which are Parallel to the Ssune Lines tfC
Parallel to each oth^r*
Let the Lines ab^ cd, be each of
them parallel to the line ef ; then shall A. '*
the lines ab, cp, be parallel to each q^
other, _
For, let the line gi be perpendicular ^^
to EF. Then will this line be also per
pendicular to both the lines ab, cp (corol. th. 12), andcon
i^(uentl^ the two lines aB cd, are parallels (cor<^ th. 13),
Qi E. d;
THEOREM
—J
feftO GEOMETRY.
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 CBDbe equal to the sum
of the two inward opposite angles a
and c.
For, conceive be to be drawn p;v
rallel to the side ac of tlie triangle.
Then bc, meeting the two parallels AC, ^E, makes the
alternate angles c ^d cbe equal (th. 12). And ao,
cutting the same two parallels ac, bp, makes the inward
and outward angles on. the same side, A and ebd, equal to
each other (th. 14). Therefore, by ecfual additions, the
sum of the two angles a and c, is equal to the sum of the
two cbe and ebd, that i$ to the wl^ol^ angle CBD (by
ax 2). <^£. D,
THEOREM XVII,
In any Triangle, the sum of all the Three Angles is tqual
to Two Right Angles.
Let ABC be any plane triangle ; then
the sum of the three angles a + b + c
i^ equal to two right angles.
For, let the ride ab Be produced to d. / ^ ^^
Then the outward angle cbd is equal J> !>
to the sum of the two inward opposite
angles a + c (th. 16), To each of these equals add the
inward angle b, thep 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. 6). There
fore also the sum of the three angles ofthe triangle a+b + c
is equal to twp right angles (ax, 1). q. E. d. •
Corgi. 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 equiangular,
Corol. 2. If one angle in one triangle, biB equal to on^
angle in another, the sums of the remaining angles will also
be equal (ax. 3),
Cord.
THEOREMS. «si
V
\
Coroh 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.
CoroL 4. The two least angles of every triangle are acutet
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
tl>e quadrangle into two triangle, aec, 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 io four right angles (ax. 2).
q. E. D.
Coroh L Hence, if three of the angles be right ones, the
fourth will also be a right angle.
Cor^l. 2. 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 .ABCD^ be any figure ; then the
sum of all its inward angles, a + 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 angled (th. 17) ;
^erefbre the sum of the angles of all the triangles is equal
to twice as many right angles as the figure has sides. . But
the sum pf all the angles about the point P, which are so
^ ' ' many
282 GEOMETRY.
many of the angles of the triangles, but no part of the in
ward ai}«{k^s of the polygon, is equal to four right angles
(corol. 3, :h. ^), arid niUbt be deducted out of the former
Sum. II<'nc( ii follows that the sum of all the inward angles
c ^' T.or^rori alone, a + b+c + d + e, is equal to twice
as ..^y iight angles as the figiu*e has sides, wanting the
sai«i iour right angles, q. £. D*
THEOREM XX.
"When every Side of any Figure is prq^uced out, the
Sum of all the Outward Angles thereby made, is equal to
Four Right Angles.
Let A, B, c, &c, be the outward
singles of any polygon, made by pro
ducing all the sides ; then will the sum
A + B + c + D + Ej of all those outward
angles, be equal to four right angles.
For every one of thes^ 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* 6). And tbere
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 sum of all the inward angles, with four
right angles, is equal to twice as many right angles as the
figure has sides (th. 1 9). Therefore the sum of all the in<*
ward and all the outward angles, is equal to the sum of all
the inward angles and four right angles (by ax. 1). From
each of these take away all the inward angles, and there
remains 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 Indefinite 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^ &c, be line^ drawn from
the given point a, to the indefinite line de^
of which AB is perpendicular. Then shall
the perpendicular AB be less than ac, and AC
less than ad, &c. ji c Jft II
foFf ib^ ^gle B bemg a rigbt one^ the
angle
THEOREMS. sas
angle c is acute (by 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 adja
cent angle aCd will be obtuse (by th. 6) ; consequently the
angle d is acute (corol. 3, th, 17), and therefore is less than
the angle c. And «ince the less side is opposite to the lesi
angle, therefore the side ac is less than the side Ap.
t q. E. !>•
CoroL A perpendicular is the least distance of a givea
point from a line.
THEOREM XXII.
The Opposite Sides and Angles of any Parallelogram ai^
equal to each other 5 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 pa
jrallel, as also the sides ad and Bc (defin.
32), 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 (eor. 2, th. 17),
namely, the angle A equal to the angle c, which are twp of
the opposite angles of the parallelogram*
Abp, if to <tbe equal angles abd, cdb, be added the
equal angles cbd, adb, the wholes will be equal (ax. 2)^
hamdy, the whole angle abc to the whole adc, which
are the other two opponte angles of the parallelogram.
, (^ E. D,
Again, since the two triangles are mutually equiangular ,i
^nd have a side in each equal, viz. the common side bd 5
therefore the two triangles are identical (th. 2), or equal in
all respects, namely, the side ab equal to the opposite sidef
DC, and AD equal to the opposite side BC, and the wiiole
triangle abd equal to the whol^ triangle scd. iq. s. b.
Carets
284 GEOMETRY. ^
CoroL 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.
CoroL 2. Hence also, the sum of any two adjacent angles
of a parallelogram is equal to two right angles.
THEOREM XXIII.
Evert Quadrilateral, whose Opposite Sides are Equal, is a
Parallelogram, or has its Opposite Sides F^llel.
Let ABCD be a quadrangle, having the
opposite sides equal, namely, the side ab
equal to pc, 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 (by hyp.), they are
also mutually equiangular (th. 5), or have their correspond
ing angles equal ; consequently the opposite side$ are paralUl
(th. 13); viz. the side AB parallel to Dc, and AD parallel to
BC, and the figure is a parallelogram, q. £. d.
THEOREM XXIV. '
^HOSE 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, which join their extremes, be also equal
and parallel. [See the fig. above.]
For, draw the diagonal bd. Then^ because ab and dc are
parallel (by hyp.), 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 con
tained angle abd equal to the contained angle bdc, they
have the remaining sides and angles also respectively equal
(th. I); consequently AD is equal to Bc, and also parallel
to it (th. 12). (^ E. D.
THEOREM XXV.
Parallelograms, as also Triangles, standing on the
Same Base, and bettyeen the Same Parallels, are equal to
^ach other. ' .
Let
THEOREMS.
UBS
Let ABCD, ABEF, be two parallelo
jrams, and abc, abf, two' triangles,
staiiding on the same base ab, and be
tween the same parallels ab, de ; then
will the parallelogram 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 aid, and the angle d equal to the angle
BCE j(th. H) ; the two triangles, adf, bce, are therefore
equiangular (cor. 1, th. 17); and haying the two correspond
ing sides, AD, BC, equal (th. 22), being opposite sides of a
parallelogram, these two triangles are identical, or equal in
all respects (th. ^). If each of these equal triangles then be
taken from the whole space abed, there will remain the
parallelogram abef ip the one case, equal to the parallel*
ogram abcd in the other (by 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). (^ e. d.
Corol. 1. Parallelogramsv or triangles, having the same
base and altitude, ai*e 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.
Corol. 2., Parallelograms, or triangles, having equal bases
^nd altitudes, are equal. For, if the one figure be applied
with its base on the other, the bases will couicide or be the
same> because they are equal : and so the two figures, having^
the same base and altitude, are equal.
THEOREM XXVI.
If 2^ 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 Pa
rallelogram.
Let abcd be a parallelogram, and abe
a triangle, on the same base ab, and between
the same parallels ab, dej then will the pa
rallelogram ABCD be double the triangle
ABE, or the triangle half the parallelogram .
For, draw the diagonal ac of the paral
lelogram, dividing it into two equal parts
(th* 22). Then because the triangle?, abc,
ABE,
286
GEOMETRY.
ABE, on the same base, and between the same parallek, are
equal (th. 25) ; and because the one triangle abc is half the
parallelogram abcd (th. S2), the other equal triangle abb is
also equal to half the same paralielogram abcd. a. £. d.
Cor<^» 1 . A triangle is equal to half a parallelogram of the
same base and altitude^ because the altitude is the perpendi
cular distance, between the parallels, which is eriery where
equal, hj the definition of parallels.
CoroL 2. If the base of a parallelogram be half that of a
triangle, of the same altitude, or the basq of the triangle be
double that of the parallelogram, the two figures will be
equal to each other.
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 ep,
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 (th. 1), because they
have the two ^ides ab, bc, and the contained angle B,
equal to the two side's ef, fg, and the contained angle f
(by hyp). But these equal triangles are the, halves of the
respective rectangles. And because the halves, or the tri
angles, are equal, the wholes, or the rectangles db, if f» are
also equal (by ax. 6). <^ E. D. .
CoroL The squares on equal lines are also equal ; for
every square is a species of rectangle.
N ' THEOREM XXVJkl.
The Complements of the Parallelograms, which are
about the Diagonal of any Parallelogram, are equal to each
other. '
Let AC be a parallelogram, at} a dia^
gonal, EiF parallel to ab or dc, and
GiH parallel to ad or bc, making ai,
ic complements . to the parallelograms
EG, HF, which are about the diagonal
DB : then will the complement Al be
equal to the complement ic.
For,
I) Gr
C
eN '
/r
>r
A JI
15
THEOREMS.
S81
For, since the diago&al )>B bisects the thtee parallelograms
Ac; EG, HF (th. 22); thcffcfore, ^ the whdte triangle dab
being equal to the ^^hole triangle dcb, and the parts DEly
IHB, respectively equal to the parts DGi, ifb^ the remaining
parts Ai, ic, must akio be equal (by ax. S). q. E. d.
THEOKEM 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 Distance
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
AE 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 A E 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*
Now, since triangles, or parallelograms, of equal bases
and altitude, are equal (corol. 2, th. 25), the parallelogram
dg is equal to the parallelogram he, and the triangle CGB
equal to the triangle chb ; consequently the line BC bisects,
or equally divides, the parallelogram af, and abcd is the
half of it. Q. E. D.
(^ , H C
THEOREM XXX.
• •
The Sum of all the Rectangles contained undar one
Wh<^ Line, and the several Parts of another Line, any way
divided, is Equal to the Rectangle contained undei* the Two
Whole Lines. •
Let AD be the one line, and AB the
other, divided into the parts kE, ef,
fb ; then will the rectangle contained by
AD ahd AB, be equal to the sum of the
rectangles of ad and ae, and ad and ef,
and AD and fb : thus expressed, ad . ab
*s 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,
^ BH,
EjnB
S8S
GEOMETRY.
SELC
A — t i j
EH» Fcv But these rectangles afe contain
ed by ADand AB» eg and bf, fh and fb ;
which are equal to the rectangles of ad
and A^ AD and £Ff ad and FB9 because
AD is equal to each of tlie two, bg^'fh.
Therefore the rectangle ad • ab is equal to the sum of aU
the ether rectangles aj> . ae, ad . bf, ad . FS. q^ e. d.
CoroL 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.
I
"n
THEOREM XXXI.
The Square of the Sum of two Lines, is greater than th6
Sum of their Square s> 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 £ H p
lines AC, CB : then will the square of ab
be equal to the squares of AC, cB, together ^
with twice the rectangle ®f Ac . cb. That
is, ab* = 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 I.
From the lines en, gi, which are equal, being each
equal to the sides of the square ab or bd (th. 22), take the
parts CF, OF, 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 a parallelo
gram. Hence the £gure hi is equilateral : and it has all
its angles right ones (corol. 1, th. 22); it is therefore a
sqtiare 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 wjkole square ad is made up of the four
figures, viz. the two squares af, fd, and the two equal rect
angles EF, FB. That is, the square of ab is equal to the
squares of AC, cb, togeth^ with twice the rectangle of AC,
CB. q. E. D.
CoroL 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
1?HEOR£M&N
289
THEOREM XXni.
Thb Square of the Difference of tw^o lines, is less tLaii
tke Sum of their Squares^ by Twice the Rectangle of the.
said Lines.
 Let Ac» Bc> be any two lines, and ab
their diSerence : then will the square of ab
be less than the squares of ac, bc, by
twice the rectangle of ac and bc. Or,
AB* s= AC* + BC* — 2AC . BC.
For, let ABDE be the square on the di&
ference ab, and acfg the square on the
line AC. Produce £d to h ; also produce
Db and Hc, and draw ki, making bi the sqiislr^ of the
other line bc.
Now it is visible that the square ad is less than the two
squares af, bi, by the two f ectangles ef^ di; But gf is
equal to the one line ac, and gE or fh is equal to the other
line BC ; consequently the rectangle bf, 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 Ht will be equal to the whole
FCy or equal to ac ^ and consequently 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. Q^ E. D*
THEOREM XX^llt.
TttB Rectangle under the Siim and Difference of two
Lines, is equal to the Difference of the Squares, of those
Lines.
Let ABj AC, be any two unequal lines ; E ly If
then will tl>e difference of the squares of
AB, AC, be equal to a rectangle undet
their sum and difference. That is.
&
TJ
b i>
3
A
AB* — AC*?=AB + AC . AB T 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 ttliro squares AD, Af, is evi
VoL.L U dendy
I
290 CEOMETRY.
dently the t^o rectangles ef, rb. But the rectangles XiPy^
Biy are equal, bei!ng contained under eq[ual lines ; lor bk and
BH are each equal to AC, and ge is equal to cb, being each^
equal to the difference between ab and AC, or their equals
AE and AG. Therefore the two bf, kb, are equal to the two
KB, Bi, or to the whole kh ; and consequendy kh is equal
to the difference of the squares ad, af. But kh is a rect
angle contained by DH, or the sum of ab and AC, and by KS^
or the difference of ab and AC, Therefore the difference of
the squares of AB, AC, is equal to the rectangle under tfaeiff
sum smd difference* q,. £. D.
THfiORKM XXXIV.
In any Rightangled Triangle, the Square of the Hypo*
fhenuse, is equal to the Sum of the Squares of the other tw«
Sides.
Let ABC be a right<<uigled triangle,
having the right angle c ; then will th«
square of the hypothenuse ab, be equal
to the sum of the squares of the other
two sides AC, CB, Or ab* s= ac*
+ BC*.
For, on AB describe the square ae,
and on ac, cb, the squares A6, bh;
then draw CK parallel to ai> or be ;
and join ai, bf, cd, cb.
Now, because the line AC meets the two CG, cb, so as tq
make two right angles, these two form one straight line gb
(corol. 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 equals add the common angle bac, so will
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
(tt. 26); in like manner, the parallelogram *k is douNe
the triangle ACD, on the same base ad, and between the
same parallels ad, ck. And since the deubles of equal
things, are equal (by ax. 6); therefore the square AG is equal
t« the parallelogram *AK.
In
THEOREMa.
29i
In like manner, the other square Bk is proved equal to
the. other parallelogram bk. Consequently the two squared
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 th»
square on the greatest side. q. e. d.
CoroL 1 . Hence, the square of either of the two less sides.
Is equal to the diflFerence of the squares of the hypothenuse
and the other side. (ax. 3);. or, equal to the rectangle con
tained by the sum and diflFerence of the said hypothenuser
9nd other side (th. 33).
.CoroL 2. Hence also, if two rightaingled triangles have
two sides of the one equal to two corresponding sides of
the other; their third sidefs will alsd be equal, and th^
triangles identical;
THEOREM XXXV.
In any Triangle, the Difference of the Squares of the
two slides, is Equal to the Diderence of the Squares of thc^
Begihems of the Base, or of the two Lines, or Distances,
include between th^ Extremes of the Base and the Perpen^*^
dicuiar.
Let ABd b^ any triangle, having
CD perpendicular to Ai ; 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 Ac* is equal to ad* + Of \ >t ; „.^^
and Bc* is equal to bD* + cD* f
Theref. the difference between ac* and Bd*,
is equd to the difference between ad* + cd*
and BIT* + CD%'
* or equal to the difference Between ad* and bd*,
by taking away the common square cd* q. b. d«:
Corol, The rectangle of the sum and difference of the
two sides of any triangle, is equal to the rectangle of the
^um and difference of the distances between the perpendK»
culai: 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
' tirithout the triangle,
i . F 2  That
r
m GEOMETRY.
That IS, AC + BC • AC * BC =5 AD + BD . AD — BD
————*< ■ ■ ■
Or, AC 4* BC • AC — BC = AB. AD — BD in the 2d figure.
And AC { BC . AC — BC = AB. AD + BD iQ the Ist figure.
THBOREM XXICVI.
In any Obtuseangled Triangle, the Square of the Side
subtending the Obtuse Anglei 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 perpen
dicular tp 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* + BG* + 2ab . BD. See the 1st fig. above»
or below. *
For, since the square df the whole line ad is equal to the
squares of the parts ab, bd, with twice the rectangle of
the same parts ab, bd (th. SI); if to each of these equsds
there be added the square of cD, then the squares of ad, cd,
will be equal to the squares of ABy BD^ CD, with twice the
rectangle of ab, bd (by ax. 2). .
But the squares of ad, cd, are equal to the sqnare of ac;
and the squares of bd^ cd, equal to the square of BC (th. 34) ;
therefore the square of ac is equal to the squares of ab, bc,.
together with twice the reaangle of ab^ bd. (^ k. d.
THEOREM JCXXVII.
In any Triangle, the Square of the Side subtending an
Acute Angle, is Less than the Sqiuares 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 Q 'C
the angle A acute, and cd perpen y
dicttlartoAB; thenwill the square y^ j
©f BC, be less tjian the squares of jT I
AB, AC, by twice the rectangle a W
of AB, AD. That is, bc* = AB*
+ AC* — 2AB . ad.
J. *.
fi.V,
THEOREMS. «93
For, in fig. 1, AC* is = bc» + ab* + 2ab . bd (th. 36).
To each of these equals add the square of AB, ^
then is AB* + Ac* = bc* + 2ab* + 2ab . bd (ax. 2),
or = bc* + 2ab . AD (th. 30).
Q.£. D.
Again, in fig. 2, AC* is = ad* + DC* (th. 34).
And AB* = AD^ + db* + 2 ad . db (th. 31).
Theret ab*+ ac* = bd* + dc* + 2ad* + 2 ad , db (ax. 2),
^or = bc* 4 2ad* + 2ad . db (th. 34),
or = BC* + 2ab .ad (th. 30). Q^ E. D.
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 Equ<il to the
Sum of the Squares of the other Two Sides.
Let ABC be a triangle, and cd the line c
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 £ DlirB
$um of the squares of CD, bd •, or ac^ f ' ^
CB* = 2cD* j 2db*.
For, let CE be perpendicular to the' base ab. Then,
since (by th. 36) AC* exceeds the sum of the two squares
ad* and CD^ (or bd* and cd*) by the double rectangle
2aj> . D£ (or 2bd . D£) ; and since (by th. 37 )^ bc^ is less
than the same .sum by the said double rectangle :; it is mani
fest that both AC* and bc* together, must be equal to that
sjun twice taken ; the excess on the one part making up the
ilefiect en the other, q. e. d.
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 tQ 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 equ^ to the square of cb, together
with the rectangle of ad and db. That iS
Ac* = cao* + AD • db.
£§♦
GEOMETWt.
For, let CE bisect the vertical ai^e ; then ^ill ft alsf
bisect the base AB perpendicularly, making ae =r eb (cor. 1,
th. 3).
But, in the^triangle Acp, obtuse angled at d^ ^he squarp
ac* is ^ CD* + AD* + 2ad . DE (th. 36),
or 5= CD* + AD . AD 4 2de (th. 30),
or = CD* + AD . AE + DE,
or = CD' + AD . BE + D^i
or = CD* 4 AP . DB.
Q. E. Df
ITIEOllEM XL*
In ahy Parallelogram, the two Diagonals Bisect each
pther ; an^ the Sum of their Squares is equal to the Sum of
the Squares of all the Four Sides of the Parallelogram*
Let ABCD b^ a parallelogram, whos^
diagonals intersect each other in e : then
will A^ be equal to ^c, 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 = jEc, and Bp == ED,
and Aq* + BD* =;: AB* + BC* + cp* + DA*.
For, the triangles a^^b, dec, are equiangular, because
they have the opposite angles at e equal (th. T), and the two
lines AC, bd, meeting 'the parallels ab, dc, make the
angle bajb equal to the angle dce, and the angle ab]^ equal
to the angle CD^, and th^ side ab equal to the side Dp
(th. 22); therefore thes^ two triangles are identical, and
have their corresponding sides equal (th. 2), viz. ae = EC,
and BE = ED. :" ^
*
' Again, sinde ac is bisected in e, the sum of the squares
AD^ + DC* = 2a^^ + 2de* (th. 38). 
In like manner, ab* + bc* == 2ae* + 2be* or 2de*.
• Theref. fB'f bc*+cd*+pa* = 4?ap*+4de* (ax. 2).
But, because the square of a whole line is equal to ^
^mes the square of half the. line (cor. tl^. 31), that 13, ac* =5
jl^AE*, and bd* ±= 4de*. ' ' '
' Theref. ab* + bc? + CD* + da* s= ac* '+ Bf)« (ax. 1 ).
' ^ ■■'■ ■ '"' ^'" ' . ■"'■■■ ^ • (J. E.D.
THEOREM
THEOREMS. 391
TpiBOUBM XU.
If? a Line^ drawn through or from the Oetitre 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 Chor4.
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 bisected in
the point d, cd will be perpendicular to
For, 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 (by hyp.); they have ail the three sides
of the oqe, equal to all the three sides of the other, and so*
have their apglea also equal (th. 5). 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. 1 1).
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 A£b bisected in the point ^, or have ab equal eb.
For, having drawn CA, cb, as before,. Then, in the tri
angle 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 9f and the angles at d are equal (def. 11); therefore
their third angles are also equal (corol. I, 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 bch, 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.
CoroL Hence a line bisecting any chord at right angles,
Kisses through the centre of the circle.
THEOREM XLIU
If More than Twp Equal Lines can be drawn from any
Point within a Ciycle to the 'Circumference, that Point wiH
he the Centre.
ms
GEOMETRY.
Let ABC be a circlei and d a point
within it : then if any three lines^ ^9
JOB, DC, drawn from the point d to the
circumferenqe, be equal to e^h other^
the point D will be the centre.
For, draw the chords ab, BCj which
let be bbected in the points £, c, and
join DE, DF.
I Then, the two triangles, dais, dbi,
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 identical, an4
have the angles at e equal to each other (th. 5) ; conser
quentiy D£ is perpendicular tp the middle of the chord ab
(def. 11), and therefore passes through the centrie of the
circle (corol. th. 41).
In like manner, it may be showi) that df passes thrqugh
the centre. Consequently the point d is the centre of the
circle, and the three equal lines da, db, dc, are radii.
ft; ?• ^?
theorem xliii.
>
If two Circles touch one another Internally, the Centres of
the Circles and the Point of Contact will be all in the
Same Right Line.
Let the two circles abc, ade, toucl^
one another internally in the point a ; then
will the point A and the centres of those
circles be all in the s^me ri&;ht line.
For, let F be the centre of the circle
ABC, through w^iich 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 surti 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 p being suppo^
the centre of the inner circle, its two radii, ga, gd, are
equal to each other 5 consequently gd will also he greater
than GB. But ade being the inner circle, cu is necessarily
' less
V »
THEOREMS
3sn
less tlian cb. So that cd is both greater and less than CB;
whkh is absurd. Con^^uently the centre g cannot be ont
of the line afc. a. £. d.
THEOREM XLIV.
J
If two Circles Touch one another Externally, the Centres
!of the Circles and the Point of Contact will be ^ in
the Same Right Linew
Let the two circles abc, ade, touch one
another externally at the point a ; then will
thje point of contact a and the centres of the
two circles be all in the same right line.
' For, let F be the centre of the circle abc,
jhrough which draw the diameter afc, and
produce it to the other circle at E. Then, if
the ccntreof 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, f and c
being the centres of the two circles, the two radii GA, CD^
are equal, as are also the two radii af, fb« Hence the s;um
of ga, af, is equal to the sum of gd, bf ; and therefore
this latter sum also, gd. bf, is greater than gf, which is
absurd. Consequently me centre G cannot be out of the
)ine ef. q. e. p.
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 this centra g; then will
these two chords ab, cd^ be equal to each
other.
For, draw the two radii ga, go, and
the two perpendiculars ge, gf, which are
the equal distances Aom the centre G.
Then, the two rightaneled triangles, GAE, Gcf, having
the side ga equal the side gc^ ana the side gs equal the
side
49S
GEOMETUT;
sde GWf said the angle at x equal to ib0
abgle at f, therefore the two trtanfi^es
GAB, GCF, are identical (cor. 2, th. S4)«
and have the line ae equal the line cf.
But AB is the douUe of ae, and cd is the
double of CF (th. 41)^ therefore a8 b
equal to cd (by ax. 6). q. e. d.
Again> if the chord ab ^ equal to the chord co ; then
will their distances from the centre, G^t gf, also be equal to
each other.
ToVf since ab is equal cd by soppoeition, the half abi^
equal the half of. Also the radii ga, gc, being equal, as
well as the right angles £ and f,. therefore the third sides
are equal (cor. 2, tlu 34), or the distance GE equal the diis
tance gf. q. e. d» 
2UEJI
THEOREM XL VI.
A Une Feq>endicular to the Extremity of a Radius, is a
Tangent to the Circle.
L'et the line adb be perpendicular to the
radius CD of a circle ; then shall ab touch
the circle in the point d only.
For, from any other point £ 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 (th. 17, cor. 3),
and consequently less than the angle d. But the greater
side is always opposite to the greater angle (th. 9); there
fore the side ce is greater than the side cd, or greater than
its equal cf. Hence the point e is without the circle ; aUd
the same for every other point in the line ab. Consequently
the whole line is without the circle, and 'meets it in the'
point D only.
.w *'., . ~
THEORBiif
THEOREMS. «d»
THEOREM XLTn.  
I I
When a Line is a Tangent to ia Circle, a Radius drawn to
the Ppint of Contact is Perpendicular to the Taagent.
Let the line ab touph the circumference of a circle at th^
point d; then will the radius cd be perpendicular to thi*
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 froni
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 conse
quently (th. 21) it is perpendicular to that line.
CoroL Hence, conversely, a line drawn perpendicular to
a tangent, at the point of eontact> passes through die centra
of the circle.
THEOREM XLVITI.
The Angle formed by a Tangent and Chord is Measured bjr
Half the Arc of that. Chord.
Let ab be a tangent to a circle, and cd
a chord drawn 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.
For, draw the radius ec to the point of
contact, and the radius £F perpendicular to
the chbrd at h. . '
Then, the raius ef, being perpendicular to the chord
CD, bisects the arc cfd (th. 41)« Therefore cf is half the
arc CFD.
in the triangle ceh, the angle h being a right one, the
sum of the two remainiiig angles £ and c is equal to a right
angle (cofoL 3, th. 17), which is equal to the angle bce,
because. the radius c£ is perpendicular to the tangent. Froia
each of these equals take awaf the common part or angle c^
and there remains the angle £ equal' to the angle bcd.
But the angle e is measured by the arc of (def. 57;, which
Is the half of^CFD \ therefore the equal angle bcd must
also have the same measure^ namely, half the arc cfd of
^$he chord CD.
Again,
soo
GEOMITRY.
Again> the line gef, being perpendicular
to tnc chord CD, bisects the «rc cg0
(th» 41). Therefore co is half Ae arc
CGD. Now, since the line ce^ meeting
SG, makes the sum of the two angles at E^
equal to two right angles (th. 6)» and the
fine XJX makes with ab the sum of the two
angles at c equal to two right angles } if from these^ twa
equal sums there be taken away the parts or angles ceH and
mcH, which have been proved equal, there remains the angle
CEG equal to the angle ach. But the former of these,
C£G» being an angle at the centre, is measured by the arc
CG (def. 57) ; consequently the equal angle acd must also
bave the same measure CG> which is half the arc qgd of the
chord CD. Of e« d.
CorcL 1. The sum of two right angles U n^asured bjr
lialf the circumference. For the two angles qcd, Acb»
which make up two right andes, are measured by the arcs
CF> cG, which make up halt the circumference^ fg being
a diameter*
CdroL 2. Hence also one right angle must have for it^
measure a quarter of the circunuerencej or 90 degrees.
THEOREM XLIX.
An Apgle at the Circumference of a Qrcle, 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 »e passing
tlirough the point of contact A. Then, the
angle dag 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 difference^
or angle bag, must be measured by half the arc bc, which
it stands upon. q. £• p. x.
THEOREM
t.
THEOREMS, 901
THEOREM t
AU Angles in the Same Segment of a Circle^ or Standing on
' the Same Arc, are Equal to each othen
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
win the angle c be equal to the angle d.
For each of these angles is measured by
lialf the arc aeb ; and thus, having equal
measures^ they are equal to each other (ax. 11).
*
THEOREM U.
An Angle at the Centre of a Orcle is Double the Angle at
the Circumference, when both stand on the Same An:.
Let c be an angle at the centre c, and
D an angle at the circumference, both stand
ing 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. 57), and the angle atthecircum*
ference 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 o.
THEOREM LU.
f An Angle in a Semicircle, i;^ a Right Angle.
If ABC or adc be a Semicircle ; then
any angle D in that semicircle, is a right
angle.
For, the angle n, at the circumference,
is measured by half the arc abc (th. 49),
that is, by a quadrant of the circumference.
But a quadrant is the measure of a right
angle (corol. 4, th. 6 ; or corol. 2, th. 48). Therefinre the
angle d is a right angle*
THEOREM
aos GEOMETRT.
THEO&SM LIII.
The Angle formed bjr a Tangent to a Circle, and a Chord
drawn fsom the Point of Contact, is Equal to the Anglt ,
in the Alternate Segment*
If AB be a tangent, and AC a chord, dnd
]> any angle in the alternate segment adc ;
then will the angle D be equal to the angle
BAG made by the tangent and chord of the
lore AEC.
For the angle d, at the circumference,
is measured by half the arc aec (th. 49) ;
and the angle bag, made by the tangent and chord, is also
measured by the same half arc Asc (th. 48) ; therefore these
two angles are equal (ax. 11)*
THEOREM tlV.
The Sum of any Two Opposite Angles of a Quadrangle
Inscribed in a Circle, is Equal to .Two Right Angles.
iiET ABCD be any quadtilaterat inscribed
in a circle ; then shall the sum of the two
opposite angles a and c, or b and i>, be
equal to two right angles.
For the angle A is meascrf ed by half the
;rrc dcb, which it stands on, and the angle
t by half the arc dab (th^ 49) ; therefore
Ae sum of the two angles a and c is measured by half the
turn of these two arcs, that is, by half the circumference.
But half the circumference is the measure of two right
angles (corol. 4, th. 6) ; therefore the sum of the two oppo
site angles a and c is equal to two right angles. In like
inanner it is shown, that the sum of the other two opposite
angles, d tnd b, is equal to two right anglesw <^ £• b.
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
AECD, inscribed in a circle, be produced
to £ ; the outward angle dae will be equal .
to the inward opposite angle c*
For,
\ /
•THEOREMS. .463
r For, the sum of the two adjacent angles DAfi ancToAB is
equal to two right angles (th, 6) ; and the sum of the two
opposite angles c and "bkh 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. l). From each of these equals taking away the com
mon angle dab, there remains the angle dae equal the
angle c. q. £. p.
TttEOREM LVI*
Any Two Parallel Chords Intercept Equal ArcsSi
LfiT the two chords ab, c&, be parallel :
then will the arcs ac, bi>, be equal; or
AC = bo.
■ ^
For, draw the line bC« Then, becaiise
the lines ab, cd, are paraUel, the alternate
angles b and c are equal (th. 12). But the
angle at the circumference B, is measured by half the ar«
AC (th. 49) ; and the other equal angle at the circumference
c is measured by half tiie arc bd : therefore the halves of thef
arcs AC, bd, and consequently the arcs themselves^ are alsa
equal. Q. £. D«
*rME0REM Ltir.
"When % Tangent and Chord are PaJr^llel to each odieri thejr
Intercept Equal Arcs.
Let the tangent abc be parallel to the
chord DF* ; then are the arcs bd, bf, equal y
that is, bd = BF.
For, draw the chprd bd. Then, be
cause the lines ab, df, are parallel, the al
ternate angles D and b are equal (th. 12).
But the angle b, formed by a tangent and chord, is measured
■b^ half the arc BD (th. 48) ; and the other angle at the cir
cumference D is measured by half the arc bf (tn. 49); there
for^ the arcs bd, bf, are equali . (^ B. d. '
THEOHEM
5M GEOMETRY.
THEOREM LVIII.
The Ancle 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 two arcs ac.
For,, draw the chord af parallel to cd.
Then, because the lines af, cd, are parallel,
and AB cuts them> the angles on the same
aide a andi\D£B 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 £ is also
measured by half the sum of fd and db*
Ag2un, because the chords af, cd, are parallel, the arcs ac^
FD, are equal (th. 56) i 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 ^nm, is also
measured by half the former, q. £. d.
THEOREM LIS.
The Angle formed. Without a Circle, by two Secants, h
Measured by Half the Difference of the Intercepted
Arcs*
Let the angle X be formed by two se>
cants EAB and ecd; this angle is measured
by half the difference of the two ztcs
AC, DB, intercepted by the two secants.
Draw the chord af parallel to CD. Then,
beca^ise the lines af, cd, are parallel, and
AB cuts them, the angles on the same side a
and BED are equal (th. 14). But the ^igle A, at the circum*
ference, is measured by half the arc bf (th* 49), or of the
difference of Df and DB : thenefore the equal angle £ if
also meas.ured by half the difference of df, DB.
Again, because the chords af, cd, are parallel, thf arcs
AC> FD, are equal (th* 56) ; therefore the difib^nce of the
two
TttEOREMS.
soi
ttro arcs At, DB, IS equal to the. difference of the twopp,
DB. Consequently the angle E, which is measured by half
the latter difference;, is also measured by half the former.
q. £. 01
THEOUBM LX.
Th^ Angle formed by Two Tangents, is Measured by Hilf
the Difference of the two Intercepted Arcs.
Let £B^ £b^ be two tangents to a circle
at the points a, c; then the angle £ is
measured by half the difference of the two
arcs CFA, CQA*
r
For, draw the chord af parallel to ed;
Then, because the linies af, ed, are pa
rallel, iand eb meets them, the angles on
the same side a and £ are equal (th. 14').
But the angle A, formed by the chord af ahd tangent AB,
is measured by half the arc AF (th. 48) j therefore the* equal
angle E is also measured by half the same arc Af, or half the
difference of the ards cfa and cf, or cga (th. 57).
CoroL In like manner it is proved, that
the angle E, formed by a tangent ec»,
and a secant eab, is measured by half
the difference of the two intercepted arcis
t A and CFB, '
J) F
Theorem lxu
When two Lines, meeting a Circle each in two Points, Cut
one another, either Within it or Without it ; the Rect
angle of the Parts of the one, is Equal to the Rectangle of
the Parts of the other ; th^ Parts of each being measured
from the point of meeting to the two intersections with ^
the circumference.
Vol. L
X
Let
X)^
GEOMfeTRt.
Let* the two lines ab, ^d, meet each
other in E^ then the rectangle of ae, eb,
will be equal to the rectangle of CE, EP.
Qr^ A£ • EB = C£ . £0.
FoTf through the point £ draw the dia
meter FG ; also, from the centre h draw
ihe radius dh« and drai;^ hi perpendi
cuhr to CO.
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. Also, because h is the centre of the
circle,' and the radii dh, fh, gh, are all equal, the line Bc;
is equal to the sum of the sides dh, he ; and ef is equal to
their difference.
But the rectangle of the sum and difference of the two
tildes 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,
XG, is equal to the rectangle of ae, eb. Consequently the
rectangle of ae, eb, is also equal to the rectangle of CE £>
(ax. 1). (^ E. D.
Corol. 1. When one of the lines in the
second case, as de, by revolving about the
point E, comes into the position of the tan
gent EC or ED, the two points c and D
nmning into one^ then the. rectangle of cE,
ED, becomes the square of ce, because cb
and DE are then equal. Consequently the
rectangle of the parts of the secant, ae . eb,
is equal to the square of the tangent, €e\
Carol. 2. Hence both the tangents ec, ef, drawn from
the saine point £, are equal ; since the square of each is equal
to the same rectangle or quantity ae • eb*
THEOREM LXU.
ta Equiangular Triangles, the Rectangles of the Correspoiid
ing or Like Sides^ taken aljiernately, are Equal.
Lfit
THEOREMS: S07
, L^T ABCf DEF, be two equiangular
triangles, having the angle a = the
angle d, the angle b •= the angle e,
afld the angle c =^ the anglp 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 thrdugi the
three points b, c, g, conceive a circle bcgh to be described^
ftieeting CA produced at H, and join Gif .
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 (th. 7) :
therefore the trlanglfe agiI is equiangular to the triangle
acb, and consequently to the triangle dfe alsoi But the
' two like sides AG, df, are also equal by supposition; conse
quently the two triangles agh, dfE, are identical (th. 2),
having the two sides AG, AH, equal to the two df, db, each
to each. •
But the rectangle ga . AB is equal to the rectangle
HA . AC (th. 61): consequently the rectangle d^ * ab is
equal the rectangle de . AC q. e. d. .
THEOREM LXIII.
iThe Rectangle of the^two Sides of any Triangle, is Ecjual to
the ReCtangl^ of the Perpendicular on the third Side and
the Diameter of the Circumscribing^ Circle. .
Let CD be the perpendicular, and cU
the diameter of the circle about the triangle
ABC 5 then the, rectangle CA . cb is ±s the
rectangle cd . cB.
For, join BE : theii in the two triangles
_ACD, ECB, the angled A and E are equal,
standing on the same arclBC (th. 50) j 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).
X 2 THEOREM
N
io# GEOMETRY.
tHEOKEM LXIV.
The Scpare of a line bisecting any Angle of a Tnangfe^
together with the Rectangle of the two Segments of the
opposite Side, is Equal to the Rectangle of dbe two other
Sides including the bisected Angle.
LsT CD bisect the angle, c of the triangle
IBc J then the square cd* + the* rectangle
AD . DB is 2=: the rectangle Ac . cB.
For» let CD be produced to meet the cir
cumscribing circle at e, and join ae.
Then the two triangles ace, bcd, are
equiangular : for the angles at c are equal
by suppositioDf and the angles B and E are equal, standing
on the same arc AC (th. 50) v consequently the third angles
at A and d are eqvtal (coro). I, th. 17): also ac, cd, and
GE, CB, are like or corresponding sides,, being opposite to
equal angles : therefore the rectangle ac . Cb is = the
rectangle cd • ge (th. 62). But the latter rectangle cef. ce
is = cr^ + the rectangle cd . de (th. 30) ; therefore also
the former rectangle AC • cb is also = cd*  cd. • db, or
equal to CD*^ + AD . db,. since C2> . de is = ad . db (th. 61)«
q. E« D.
THBOREM LXV.
The Rectangle of the two Diagonals of gny Quadrangle
Inscribed in a Circle, is equal to the sum of the two Rect
angles of the Opposite Sicbss*
Let abcp be any quadrilateral inscribed
in a circle, and ag, bd, its two diagonals :
then the rec^tangle AC . bd is = the rect
angle ab . DC + Jthe rectangle ad . bc
For, let CE be drawn, making the angle
BCE equal to the angle dca. Then thetwa
triangles acd, bce, are equiangular ; for the angles A and
b are equal, standing on thesame.arc dc; and theangles*
dca, bcb, are equal by supposition ; consequently the third
angles adc, B£c,,are also equal t also, Ac, bc, and ad^ be>
are like or corresponding sides, being opposite to the equal
angles : therefore the rectangle AC . BE is =s^ the rectangle I
A.D . bc (th.. 62)»
Again,
THEOREMS. 3<»
Again, the two triangles abc, D£C are eqniangidar : for
tlie angles bag, bdc, are equal, standingon the same arc bc;
and the angle dce is equal to the angle fiCA, by adding the
common tingle ace to the two «qua:l angles dca, bc£ ; there
fore the third angles £ and abc are also equal : but AC, Dc>
and AB, DE, are the like sides : therefore the cectangle AC •
"DE is = the rectar^le ab . dc (th. 62),
Hence, by equal additions, the sum of the rectangles
AC • BE + AC . DE is = AD . SO  AB . DC. Bllt the
•ibrmer «um of tb^ rectangles AC . be + ac . 0E is = the
rectangle AC . bb (tb« 30): therefore the same rectangle
AC . bd is equal tp the latter suro^ the rect. ad . Bc + the
rect. AB . DC (ax. I). <^ e. n.
OF RATIOS AND PROPORTIONS,
DETlNiTIONS.
' Def. 76. Ratio is the proportion or relation which one
magnitude bears to another magnitude of the same kind,
with respect to quantity.
JVi^. The measure> or quantity, of a ratio, is conceived,
by considering what part or parts the leading quantity, called
the Antecedent, is of the other, called the Consequent ; or
what part or parts the number expressing the quantity of the
•former, is of the number denoting in like manner the latter.
So, the ratio of a quantity expressed by the number 2, to' a
like quantity expressed by the number 6, is denoted by 6
divided by 2, or  qr 3 : the number 2 being 3 times con
tained in ^, or tlie third part of it. In like manner, the ratio
of the quantity 3 to 6, is measured by ^ or 2 ; the ratio of
.4 to 6 is ^ or 1.; that of 6 to 4 is f or ; &c.
77. Proportion is an equality of ratios. Thus,
78. Three quantities are said to be Proportional, when the
ratio of the first to the second is equal to the ratio of the
second to the third. As of the three quantities A (2), b (4),
c (8), where 4 = J. = 2, both the same ratio.
19. Four quantities are said to be Proportional, when the
ratio of the first to the sec6nd,*is the same as the ratio of the
third to the fourth. As of the four, a (2), b (4), c (5), D (10),
•where i = V* = ^> ^^^ ^^® ^^"^^ ratio.
«ie GEOMETRY.
■
Noie. To denote that four quantities, a, b> c, d^ are pro*
portional, thej are usually stated or placed thus, a : b : : c : Q;
^nd read thus, a is to B as c is to d. But vhen three
€uantities are proportional, the middle one is repeated, and
they are written thus, a : B : : fi : c.
80. Of three proportional quantities, this middle one is
said to he a Mean Proportional between the other two j ^d
.the last, a Third Proportional to the first Jtnd second.
81. Of four proportional quantities, the last is said to be
a Fourth Proportional to the other three, taken in orfier.
' ' ' ' . • . ' .
82. Quantities are said to be Continually Proportional, or
in Continued Proportion, when the ratio is the same bietween
every two adjacent terms, viz. when the first is to the second,
a^ the second to the third, as the thijd to the fourth, as the
fourth to the fifth, and so on, all in the same common ratio*
As in the quantities 1, 2, 4, 8, 16, &c; where the com«
mon ratio is equal to 2.  ' ' 
83. Of any number of quantities, A, b, c, d, the ratio of
the first A, to the last D, is said to be Compounded of the
ratios of the firs^ to the second, of the second to the third,
and so on to the last. . *
, 8'k Inverse ratio is, when the antecedent is made the
consequent, and the consequent the antecedent. — ^Thus, if
1 : 2 : : 3 : 6 5 then inversely, 2 : 1 : : (5:3.
> ■  * •
SS, Alterrjate proportion is, when antecedent is compared
with antecedent, and '^consequent with consequent. — Asy if
1 : 2 : : 3 : 65 then>.by alternation, or permutation, it will be
1 :3 ;;2 ;6. ' *.
»
S(), Compounded ratio is, when the sura of the antecedent
and consequent is conipared, either with the consequent, or
with the antecedent.— Thus, if 1 : 2 ! : 3 : 6, then by compo
sition, 1 + 2 : 1 : : 3 + 6 : 3, and J + 2 : 2 ; : 3 + 6 : 6.
87. Divided ratio, is when the'difference of the antecedent
and consequent is compared, eikher with the antecedent or
with the consequent. — Thus, if 1 : 2 : : 3 : 6, then, by division,
2^1 : 1 :: 63 : 3, and 2— 1 : 2:: 63 : 6.
^ ■ I « I
Note. . The term Divided, or Division, Here means sub
tracting, or parting; being used in the sense opposed to^com
pounding, or adding, in def. 86.
TH£ORSM
THEORIJMS, a>I
THEOREM LXVI.
Equimultiples of any two Quantities have the same Ratio as
the Quantities themselves,.
Let a and b be any two quantities, and f»A, wb, any
equimultiples of them*, m being any number whatever : then
will mA and mB h^ve the same ratio as a and B;, or
A : B : : mA : mB.
ror — ^ = •— , the sa:me ratio.
Corol. Hence, like parts of quantities have the same ratio
as the wholes ; because the wholes are equimultiples of th«
like p9rt5i or A and i ax^e like parts of i»a and tn^*
THEORBM LXVJI.
If Four Quantities, of the Same Kind; be Proportionals^
thf y will be in Proportion by Alternation or Permutation,
or th^e Antecedents will have the Same Ratio as the Con^
' sequents.
Let a : b : : f»A : mB ; then will a '" mWi^ x mB,
For 5;: », an^ — • ^ niy both the same ratio,
THEOREM LXVnx.
If Four Quantities be Proportional ; they ^ill be in Pro*
portion by Inversion, or Inversely,
Let a : b : : mA : ms ^ then will B : a : : /»B : mA^
mtmm A A
For — =  — , both the same ratio.
;;;b B
theorem lxix.
Jf Four Quantities be Proportional ; they will be , in Pro*
portion by Composition and Division*
Let a : b :: mA : mB'f
Then will B ± A : A : : /wB ± iwA : mA,
and B ± A : b : : /wb ± mA : i»b.
_ mA A mB B
ffor, _, , — = 2"; — 9 and
m^±mA, » ± a' mB±mA » ;t A
318 GEOMETRY.
Coroi. It appears from hence, that the Sum of the Greatest
and Least of four proportional quantities, of the same kind,
exceeds the Sum of the Two Means. For, since —  
A : A + B : : ffifA : niA + nm% where A is tiie lea^t, and
mK + w» the greatest ; then f« + 1 • A + mB, the sum of
the greatest and least, exceeds « + i . A + P the sum o^
the two means.
THEOREM IXX.
If, of Four Proportional Quantities, thiire be taken mj
Equimultiples whatever of the two Antecedents, and any
Eqoimnltiples. whatever of the two Consequents ; the
'quantities resulting will still be proportional.
Let A : b : : ffiA : /«b ; also, let px antl pmi^ be any
equimultiples of the two antecedents, and ^B and qm^ any
equimultiples of the two  consequent* \ then will  r   
^A : ^B : : pmA. : qm'B, '
for  — = ^^, both the jame ratio.
pmk pA
THEOREM LXXI.
If there be Four Proportional Quantities, and the iwfl
Consequents be ,eifher Augmented, or Diminished by
Quantities that have the Same Ratio as the respective
Antecedents ; the Results and the Antecedents will still
be Proportionals.
Let A : b : : mA : /tie, and tiA and nmA any two quan:
pities having the sai; e ratio as the two antecedents j then will
A : B ± «A ; : fwA : mB ± nmA.
wB ± nmk B ± «A ^ , ,
ror = , both the same ratio.
mA K ^
THEOREM LXXIJ.
If any Number of Quantities be. Proportional, then any
one of the Antecedents will be to its Consequent, as the
Sum of all the Antecedents is to the Sum of all the Con
sequents.
Let a \ b :i tnA ', ntfi \\ nA '. wB, &c ; then will *   ^
A : B : : A + wA 4" ^A : : B + ^B + //B, &c.
_ 5 + WiB + «B B _
J or — =c •— , the same ratio,
A + IWA + /7A 4
theorem
THEOREMS. 4 It
THEOKEU IXXIir.
f a Whole Magnitude be to ?i Whole, as a Part taken from
tbe first, is to a Part taken from the other ; then th^ Re
mainder wijl be to the Remainder, as th^ whpl^ to the
whole.
JLet a : 3 : : — A : — B ;
n . n
tjpten will a:b::a a:b B^
n n
B^B B
For •■ ; ■  ^ ■ ' ' = — , both the same ratia^
• • Arr^ A A '
» 
THEOJIEM J.iXIV.
If any Quantities be Proportional ; ^heir Squares, or Cubes^
or any Like Powers, or Roots, of them, will also be Pro
portional.
Let a : b : : »a : ms ; then will a* : B° : : iw^a" : jw^b**.
For —; = — , both the same ratio.
«j"A° a' ■
THEOREM LXXV. *
^ there be tvo Sets of Proportionals 5 then the Products 01?
Rectangles of the Corresponding Terms will also be Pro
portional.
JjET a 2 b : : ota : fwBj
and p : D : : «c : «D J
^hen will AC : bd : : mnKC : mftBD^
rfmtBT> bd ,  , '
or ' = — , both the same ratio.
mtJAC AC
THEaHEM LXXVI.
Jtf Refer Quamtities be Proportional; the Rectangle or Product
of the two Extremes, will be Equal to the Rectangle or'
Product of the two Means. And the converse.
X»^r A : 8 : : mA : mB ;
then & A X nmszu X mAss foxB. as is evident.
theor;^
«M
GEOMETRY.
THEOREM .LXXVIT.
If Three Quantities be Continued Proportionals ; the Rect
angle or Product of the two Extremes, will be Equal to
the Square of the Mean. And the conveise.
Let a, mA, w*A be three proportionals,
or A : mk : : wa : m^K ;
' then b A X /»*A == «i'A% as is evident.
THROREM LXXVIII.
If any Number of Quantities be Continued Proportionals \
the Ratio of the First to the Third, will be duphcate or the
Square of the Ratio of the First and Second; and the Ratia
of the First and Fourth will be triplicate or the cube of
that of the First and Second ; and so on.
. Let a, mA, f»^A, im'a, &c, be proportionals $
. OTA
then IS = iM ,
nP'A
m'
= iM ; but =; «*; and =5
s
&c
TBEOREU LXXIX.
Triangles, and also Parallelograms, having equal Altitude^
are to each other as their Bases.
Let the two triangles aoc, def, have
&e same altitude, or be between the same
parallels ae, cf ; then is the surface of
the triangle adc, to the surface of the
triangle def, as the base ad is to the
base DS. Or, ad : de : ; the triangle
ADC, : the triangle def.
For, let the base ad be to thp base de, as any one nupa
ber m (2), to any other number » (3) ; and divide the respec
tive bases into those parts, ab, « bd, dg, gh, he, all
equal to one another ; and from the pcrints of division draw
the lines bc, fg, f», to the vertices c and f. Then will
these lines divide the triangles adc, def, into the. same
number of parts as their bases, each equal to the triangle
ABC, because those triangular parts have equal bases and
altitude (corol. 2, th. 25) ; namely, the triangle abc, equal
to each of th^e triangles bdc, dfg, gfh, hfb« Sq that
the triangle adC is 10 the triangliei qfE aithe nuimber of
THEOREMS.
S15
parts « (2) of the former, to the number n (3) of the latter,
chat is, as the base ad to the bas^ djk (cl6f. 79).
^ In like manner, the parallelogram adki is to the parallel
pgram defk, as the base ad is to the base de; each of
these having the same r;itio as the number of their partSj
m to «. Qi ;e. p.
THEOREM LXXX,
Triangles, and also Parallelograms, having Equal Bases,
to each other as their Altitudes,
>
Let ABC, BEF, be two triangles
having the equal bases ab, be, and
whose altitudes are the perpendiculars
CG, FPl \ then will the triangle ABC :
the triangle bef : : cg : fk.
' For, lei BK be perpendicular to AB,
and equal to €G; in which let there
be taken Bx. s? fh ; drawing Ak and AL«
»
Then, triangles of equal bases and heights being equal
(corol. 2, th. 25), the triangle abk is = ABC, ind the tri*
atigle ABL == B^F. But, considering now abk, abl, as two
triangles on the bases bk, bl, and having the same altitude
AB, these will be as their bases (th, 19), namely, the triangle
ABK: the triangle abl : : bk : bl.
But the triangle abk = abc, and the trialigle abl = bef«
also BK = CG, and bl = fh.
Theref. the triangle abc : triangle bef :: cg : fh.
«
And since parallelograms are the doubles of these triangles,
having the same bases and altitudes, they will likewise have
to each other the same ratio as their altitudes, q. £. D.
Coral. Since, by this theorem, triangles and parallelo^ams,
^hen their bases are equal, are to each other as thew alti
tudes 5 and by the foregoing one, when their altitudes are
equal, they are to each other as their bases ; therefore uni
versally, when neither are equal, they are to each other in
the compound ratio, or as the rectangle or product of their
}>ases and altitudes.
THX0S.EU
31$ GEOMETRY.
THEOEEU LXXXI.
If Four Lines be Proportional ; the Rectangle of the Ex
tremes will be Equal to the Rectangle of the Means.
And) conversely, if the Rectangle of the Extremes, of four
Lines, be equal to the Rectangle of the Means, the Four
Lines, taken alternately^ will be FfoportionaL
Let the four hues a, b, c, d, be
^oportionals, or A : b : : c : d; a,—
then will the rectangle of a and d be i^
fequal to the rectangle of b and c j j^'
or the rectangle A ^ d = 9 • c
A
1 f U
*^ i
For, let the four lines be placed
with their four extremities meeting
in a common point, forming at that
point four right angles ; and draw lines parallel tb them to
complete the rectangles p, q, r, where p is the rectangle
of A and D, Q^the rectangle of b and .c> and % the rect*
angle of b and d.
Then the rectangles p and R being betFe^n the same
p3trallels> are to each other as their bases a and B (th. 7:9/;
and the rectangles q and r, being between the same pa
rallels, are to each other as their bases c and D* 3ut the
ratio of a to b^ is the same as the ratia of c to p, by hypo*
thesis 'j therefore the ratio of p to r, is the same as the ratio
«of <^ to R 5 and consequently the rectangles P and q are
tqual. Q. B. D.
Again, if the rectangle of a and d, be equal to the
rectangle of b and c; these Imes will be proportional^ qr
A : b : : c : o.
For, the rectangles being placed the sjime as before : then,
because parallelograms between the same parallels, are to one
another as their bases, the rectangle P : g : : a : Bi anil
Q^ : R :: c : D. But as p and q^are equal, by supposition,
they have the same ratio to r, that is, the ratio of a to MP
equal to the ratio of c to d, or a : b :: c : d, q^ e. d.
Corol, I. When the two means, namely, the second and
third terms, are equal, their rectangle becomes a square of
the second term, which supplies the place of both the second
and third. And hence it follows, that when three lines arQ
proportionals, the rectangle of the two extremes is equal ta
a '
THEOREMS. z\1
the square of the mean ; and, conversely, if the rectangle of
the extremes be equal to the sq^uare of the meaa, the three
lines are proporti(»ials.
CoroL 2. Since it appears, by the rules of proportion lit
Arithmetic and Algebra, that when four quantities are pr«i
portionai, the prockict of the extremes is equal to the product
of the two means ; and, by this theorem^the rectangle of the
extremes is ^qual to the rectangle of the two mean^ » it fol
lows, that the area or ^pace of a rectangle is represented 03f
expressed by the product of its length and breadth multiplied
together. And, in general, a rectangle in geortietry is simi
lar to the product of the measures of its two dimensions of
length and breadth, or base and height. Also, a square, it
similar to, or represented by, the measure of its side multi
plied by itself. So that, what is shown of such products, is
to be understood of the sqtiares and rectangles.
CoroL 3. Since the same reasoning, as in this theorem,
holds for any parallelograms whatever, as well as for the
rectangles, the same property belongs to all kinds of paral
lelograms, having equal angles, and also to triangles, which
are the halves of parallelograms ; namely, that if the sides
about the equal, angles of parallelograms, or triangles, be
reciprocally proportional, the parallelograms or triangles
will be equal ; and, conversely, if the parallelograins or
triangles be equal, their sides about the equal angles will be
reciprocally proportional '
Corol. 4. Parallelograms, or triangles, having an angle in
each equal, are in proportion to each other as the rectangles
of the sides which are about these equal angles.
THEOREM LXXXII.
If a Line be drawn in a Triangle Parallel to one of its
sides, it will cut the two other 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 tri
angles dbe, dce, are equal to each other,
because they have the same base de, and
are between the same parallels j^e, bc
(th. 25). But the two triangles ade, bde,
on the bases adi, i>h, ha?e the saotie alti
tudei
)
J
SIS
GEOMETkt
tude; and the two triangtes 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>
therefore
the triangle Ade : bdk : : ad : db,
and triangle adr : cde : : ae : £c.
But BDE is ts CDE } and equals must ha^e to equals the
same ratio; therefore ad : db ;: ae : £c* q. e. d*
CS^ol. Hence, also, the whole lines ab, Xc, are propor
tional to their corresponding proportional segments (corol<
VIZ* AB : AC
and AB : ac
• *
ad
bd
AE,
C£«
THEOREM hXXXllU
A Line which Bisects any Angle of a Triangle, divides the
opposite Side into Two Segments, which are Propottional
to the two other Adjacent Sides.
LtT the angle acb, of the triangle abc, ^
be bisected by the line cd, making the
angle r equal to the angle / : then will the
segment ad be to the segment db, as the
,iide AC is to the side cb^ Or,    
AD : db : : AC : CB.
For, let BE be parallel to cd, meeting
AC produced at £• Then, because the line bc cuts the two
mrallels cd, be, it makes the angle cbe equal to the alter*^
Hate angle / (th. 12), and thereifore also equal to the angle
r, which is equal to / by the supposition. Again, because
the line ae cuts the two parallels dc, be, it makes the
angle E equal to the angle r on the same side of it (th. 14).
Hence, in the triangle bce, the angles b and e, being each
equal to the angle r, are equal to each other, and conse
quently their opposite sides cb, ce, are also equal (th, 3).
But now, in the triangle abe, the line cd, being drawn
par^lel to the side be, cuts the two other sides AB, A £, pro
portionally (th. 82), making ad to db, as is ac to cb or to
Its equal CB. Q. £. D*
THEORBM
THEOREMS.
519
Theore:^ Lxxxir. .
Equiangular Triangles are Similar^ or have their Like Sidct
Proportional,
Let ABC, DEF, be two equiangular tri
angles, having the angle a equal to che^
angle D, the angle B to the angle £, and
consequently the angle c to the angle v,
then will ab : ac : : D£ : d^.
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 con
tained angles a and d also equal, are iden
tical, or equal in all respects (th. 1 ), namely,
the angles b and c are equal to the angled G and if. Bat the
angles b and c are equal to the angles £ and f by the hypo*
thesis ; therefore also the angles g and h are equal to the'
angles £ and f (ax. 1), and consequently the line gh is pa^
rallei to the side ef (cor. 1, th. 14).
Hence then, in the triangle def, the line gh, being pa
rallel to the side ef, divides the two other sides propor
tionally, making dg : dh : : de : df (cor. th. 82). fiut
DO and DH are equal to ab and ac ; therefore also    "^
JB : AC : : DC : df. q. e. p.
THEOREM LXXXV.
Triangles which have their Sides Proportional, are Equi*
angular.
In the two friangles abc, def, if
AB : DE : : AC : df : : bc : ef ; the two
triangles" will have their corresponding
angles equal.
For, if the triangle abc be not equian
gular 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 equi*
angular, 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 tliat OG and
SG, being fourth proportionals to the same three quantitiest
Cr F
880
Gfi(»fETRY*
as well as the two pr» ef, the former D<f, £G, irdoici hi
equal to the latter, df, ef. Thus then, the tiKro tmngk^
DBF, d£g, having their three sides equal, lifroilld be identkal
(th« 5); which is absurd^ since their angles ate^ un^qual^
THEOREM LXXXVt
Triangles, which have an Angle in the on^ Equal to zh Angte
in the other, and the Sides about these angles Proportionals
Equiangular.
Let ABC, DEF, be two triangles, having
the angle a = the angle d, and the sides
AB, AC, proportional to the sides D^, Dt:
then will the triangle abc be equiangular
with the triangle def»
For, make dg = ab, and dh = AG
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. l), having the angles G
and H equal to the angles B and c« But, since the side^^
DG, DH, are proportional to the .ides 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
m and c* q« s. d.
THEOREM LXXXVII.
In a Right Angled Triangle, a Perpendicular from the Right
Angle, is a Mean Proportional betweex^ 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 rightangled triangle, and .
CD a perpendicular from the right angle
c 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 : »c : : ac : bd  attd
AB : AC : : AC : AD.
For J
THEOREMS*
i^i
For, the two triangles abc, adc, having the right angles
at c and D equal, and the angle a common, have their third
wangles equal, and are equiangular (cor. 1, th. 17). In like
manner, the two triangle^ 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, apc, bdc,
being equiangular, will have their like sides proportional
(th. 84);
VIZ. Ab : ci> * : CD
: db;
and A£ t AC :: AC
: AD}
and AB : BC : : BC
: BD.
^ <^E. D*
CoroL Because the angle in a semicircle is a right angle
(th. 52) ; it follows, Ihat if, fiom khjr point c in the peri
phery of the semicircle, a perpendicular be drawn to the
diaxtieter ab ; and the two chords ca, dB^ be drawn td
the extremities of the diameter : then are ac, bc, cd, the
mean proportionals as in this theorem, or (by th. 17)^ *• * ^
tD* ac AD . DB; AC* ss AB • AD; and BC^ = AB . dDk
THEOREM LXXXVIII*
jEquiangular or Similar Triangles, are to each other as the
'Squareis of their Like Sides.
Let abc, def, be two equi
angular triangles, ab and de
being two lik6 sides : then will
the triangle' abc Ije to the tri
angle DBF, as the square of AB
is to the $quaxe of D£, or as
AB* to de\
For, ht At and dn be the
isquares on ab and dk; also draw their diaj^onals bk, eM, and
the perpendicuhrs cg, fh, of the two triangles.
Then, since equiangular triangles have their like sides
proportional (th. 84), in the two equiangular triangles abc,
def, the side ac : df : : ab : de ; and in the two acg,
DFH, the side Ac : df : : cg : fh ; therefore, by equality
CG :, FH : : ab : de, or cg : ab :: fh : DE.
But because triangles on equal bases are to each . other as
their altitudes, the triangles ABC, Abk, on the same base
ab, are to eath other, as their altitudes cg, ak, or ab:
Vol. L Y and
Z22
GEOMETRY.
and the triangles def^ dem, on the same base PE are as their
altitudes fh> dm^ or d£ ;
that isy triangle abc : triangle abk ; : CG : AB9
and triangle def : triangle dem : : f h : ds.
But it has been shown that cG : ab : : FH : i>£ }
theref. of equality A abc : aabk : : abef : Adem,.
or alternately, as a abc : adef : ; >^ abc t A dem.
But the squares al, dn, being the double of the triangles
ABKj D£M« have the same ratio with them \
therefore the A abc : adef : : square al : square dn*
. , (i. E. i^
THEOREM LXXXIX.
All Similar Figures are to each other, as the Squares of their
Like Sides.
Let abcd£ 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
ABCDEbe tothefigurcFGHiK,
as the square of.AB to the
square of eg, or as ab* to fg*.
For, draw £e, bd, gk, g^x dividing the ^^es into a»
equal number of triangles, by Knes from two equal angles
b and G^
The tsffo figures being similar (by stippos.)i they are equi
angular, and have their like sides proportional (def. 67).
Then, since the angle A is ^^ the angle f, and the side^
ab, A£, 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^
, imd the sides «c, CD, proportional to the sides gh. Hi, are
also ecfuiangdar. 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 cde, hig, there will remain
the equals bde, gik 5 so that the two triangles bde, gik,
having two angles equal, are also equiangular. Hence each
triangle of* the one figure, is equiangular with each corre*
sponding triangle of the other.
But equiangular triangles are similar, and are propottfofial
to the squaj'es of their tike sides (tit. 88).*
TWeforr
THEO&tMS.
t2%
Therefore th^ A ab» : a fgk : t ab* :sg%
and A BCD :' A ghi : : bc* : gh*
and A bdb : A gik : : de^ : IK
.>
But ds the two polygons are similar^ their like sides are pro
portional, iand consequently their squares also proportional i
so that all the ratios ab* to pg% ^d bc* to gh*, and de* to
1K% are equal among themselves, and consequently the cor
responding triangles also, abe to fgK) and bcd to ghi, and
jBDE to GIK, have all the same ratio, viz. that of ab'^ to fg* :
and hence all the antecedents, or the figure abode, have to
mil the con^quents, or the figure fghik, still the «ame ratio^
Viz. that of AB* to FG* (th. 72). q. e. d. ^
THEOREM XC«
Similar Figures Inscribed in Circles, have their Like Side«,
and also their Whole Perimeters, in the Same Ratio as this
Diameters of the Circles in which they are Inscribed.
iiET ABCDE, fghik,
be two similar figures,
inscribed in the circles
whose diametof's are al *
and FM ', then will each
side A6, BC, &c, of the
t)ne figure be to the like
dde gf, gh^ &c> of the
"other figure, or the whole perimeter ab + bc + &c, of tlie
one figure, to the whole perimeter fg + GH 4" &c, of the
Other figure, as the dis^meter al to the diameter f^.
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*5ides have the same ratio
(def. 67) ; therefore the two triangles abc, fgh, have the
angle b =st the angle G, and the sides ab, bc, proportional
to the two sides fg, Ch, 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. l). And sjince the
wangle abl. = fgm, being both right angles, because in a
.semicircle ; therefore the two triangles abl, fgm, having
two anglCj* ecjoal, are equiangular ; and cpnsequentiy their
.if 4 GEOMETRr.
like sides afe profjortional (th. 8*) ; hence AB : ><J : ; the
diameter al : the diameter fm.
In like manner, each side bc, cd, &c, has to each side
CH,m, &c, the same ratio of al to fm; and consequently
the sums of them are still in the same ratio j viz, AB + bc f
c©, &c : FG + CH + HI, &c : i the diam. AV : the diami.
FM (th. 72). £. B. D. .
t
THEOREM XCi;
Similar Figures Inscribed in Circles, are to each other as
the Squares of the Diameters of those Circles.
Let abcdb, fghik,
be two similar figures, in'
scribed 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*.
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, this 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 abcde, fghik, are also to each other as the
squares of the diameters al* to fm* (ax. l). <^ £• d.
theorem xcii.
The Circumferences of all Circles are to each other as their
Diameters.
Let d, di denote the diameters of two circles, and c, <:,
their circumferences ;
then will d :.i/ : : c : r, or D : c : : rf : r. '
For (by theor. 90), similar polygons inscribed in circles
have their perimeters in the same ratio as the diameters of
those circles.
Now, as this property belongs to all polygons, whatever
the number of the sides may be ; conceive the numbe^r of the
sides to be indefinitely great, and the length of each inde
finitely smalU till they coincide with the chrcumference of
the
THEOREMS. 325
t
the circle, and be equjd to it, indefinitely near. . Then the
perimeter of the j)olygon of an infinite number of sides, is
the same thing as the circumference of the circle. Hence h:
appears that the circumferences of the circles, being the same
as the perimeters of such polygons, are to each other in the
i^ame ratio as the diameters of the circles, q^ s. d.
THEOREM XCllI.
t
The Areas or Spaces of Circles, are to each other as the
Squares of their Diameters, or of their Radii.
Let a, ay denote the areas or spaces of two circles, aad
JD, dy their diameters; then a :^ : : d^ : i/^.
Fot (by theorem 91) similar polygons inscribed in circles
are to each other as the ^squares of the diameters of th^
circles.
Hence^ conceiving the numher of the sides of the poly
gons to be increased more and more, or the length of the
aides to become less and less, the polygon apprpaches nearer
and nearer to the circle, till at length, by an infinite ap<>
proach, they coincide, and hecome in effect equal ; and then
it follows, that the spaces of the circles, which are the same
as of the polygons, will be to each other as the squares of the
diametersof the circles. (^ £. J>.
• CorQh The spaces of circles are also to each other as the
squares of the circumferences ; since the circumlFerences are
in the same ratio as the diameters (by theorem 92).
THEOREM XCIV.
The Area of any Circle, is Equal to the Rectangle of Half
its Circumference and Half its Diameter.
Conceive a regular polygon to he ^^f^
'inscribed in the'ciTcle^; and radii drawn to //\
all the angular points, dividing it into as j/ .
xnaAy equal triangles as the polygon has K .
. ^idesji one of which is abc, of which the \x /
jiltituid^ is the perpendicular CD from the S^F^li
centre to the base ab.'
Then the triangle abc, being equal to
a rectan^e bif half the base and equal altitude (th. 26, cor. 2),
is equal to the^ectangle of the half base ad aod the a^titudeoD ;
£onse«>
I
S^6
GEOMETRY,
consequently the whole polygon, or all
the triangles added together which com
pose it, is equal to the rectangle of the
conunop altitude cd, and the halves of all
the sides, or the half perimeter of the po
lygon.
Now, conceive the number of sides of the polygon to ht
indefinitely increased ; then will its perimeter coincide with
the circumference of the circle, and consequently the alt^
tude CD will become equal to the radius, and the whole
polygon equal to the circle. Consequently the space of the
circle, or of the polygon in that state, is equal to the rect^
angle of the radius and half the circumference/ ^ £• n.
9e
OF PLANES AND SOLffiS,
\ •
DEFINITIONS.
DjtF. S8. The Common Section of two Pl^es, is the
line in which they meet, to cut each other*
89. A Line is Perpendicular to a Plane, when it is per*
. pendicular to every line in that plane which meets it.
90. One Plane. is Perpendicular to Another, when every
Kne of the one, which is perpendicuJar to the line of their
common section, is perpendicular to the other*
91. The Inclination of one Plane to another, or the ^gle
they form between them, is the angle contained by two
lines, drawn from any point in the common section, and at
right angles to th« same, one of these lines in each plane^
92. Parallel Planes, are such as being produced ever so
far both ways, will never meet, or which are evcf]^ where at
an equal perpendicular distance*
9&. A. Solid Angle, is that which is made by three or
more pltoe irfglcs, meeting tach pthcr in the sam^ point.
94f. Similar
J
DEFINITIONS
327
.'94. SiirvilafS^Uds^ contained by plane figures, 3re such as
liave all their so)id angles equal, each to each, and are bound
ed by the same number of similar planes, alike placed*
^ 95. A Prisrn, is a solid whose ends are parallel, equal, and
like plane figures^ and its sides, connecting those ends, are
parallelograms.  •
> •
96. A Prism takes particular names according to the figure
of its base or ends, whether triangular, square, rectangular,
pentagonal, hexagonal, &c*
?7. A Right or Upright Prism, is that which has the
planei, of the jides perpendicular to the plane* of the endU
or bi^e.
t^
hi
9a» A Porallelopiped, or Parallel opipedon, is
.^ . '^m bounded by six parallelograms, every
o. , lie two of which are equal, alike, and pa
■•'9. A Rectangular Parallelopipedon, is that whose bound
ing i lanes are all \rectangles, which are perpendicular to each
©her.
100. A Cube, is a square prism, being bounded
by six equal square sidei or faces, and are perpen*
dicalar to each other.
I (;1 . A Cylinder is a round prism, havij^g cir
cles for its ends ; and is conceived to be formed
by I he rotation of a right line about the circum
ferences of two equal ^nd parallel circles, always
parallel to the axis.
1 02. The Axis of a Cylinder, is the right line
joinirrg the centres of the two parallel circles, about vhich
the ligure is described.
10^. A Pyramid, is a solid, whose base is any
rightlined jiane figure, and jts sides triangles,
having all their vertices meeting together in a
point above the base, called the Vertex of the
pyramid.
104. A pyramid, like the prism, takes particular names
'.from the figure of the base. ' ^
105. A Cone, is a round pyramid, having a cir
.cula^ base, and is qonceived to be generated by
the rotation of a right line about the circum
.ference of a circle, one ^nd of which is fixed at
. a point above the piw^e pf tjiat circle.
' ' • ^ 106. The'
528 GEOMETRY.
J Of, The Axis of a cone, is the right line, joining th«
vertex, or fixed point, and the centre of the circle about
H^hich the figure is described.
107. Similar Cones and Cylinders, are such as have their
altitudes and the diameters of their bases proportional.
108. A Sphere, is a solid bounded by one curve surface^
which is every where equally distant from a certain point
within, called the Centre. It is conceived to be generated
by tlie rotation of a semicircle about its diameter, which re
mains fixed.
109. The Axis of a Sphete, is the right line about which
th^ semicircle revolves; and the centre is the same as that of
the revolving semicircle.
110. The Diameter of a Sphere, is any right line passing
through the centre, and terminated both ways by the surface.
1 H. The Altitude of a Solid, is the perpendicul^ drawn
from the vertex to the opposite side or base.
THEOREM XCV,
A Perpendicular is the Shortest Line which can be drawfi
from any Point to a Plane.
Let ab b^ perpendicular to the plane ju
DE ; then any other line, as AC, drawn \
from the same point a to the plane, will \
be longer than the line ab. ^ ^"ul \
In the plancf draw the line Bc, j^oining <?^
the points B, o. ' '
 'Then, because the line ab is perpendi
cular to the plane de, the apgle B is a right angle {def. 90),
and coi^sequently greater than the angle c ; therefore the
line AB> opposite to the less angle, is less than any other lint
AC, opposifjB the greater angle (th. 21), q. E. D.
THEOREM XCVI.
A Ferpendic^lar Measures the Distance of any Point from 4
PJane,
.The distance of one point from another is measured by a
right line joining them^ because this is the shortest line which
can be drawn from one point to another. So, also^ the
^stance from a point to a line, is measured by a perpendi
ipular^ l^ecause this line is the shortest which can be drawn
fronat
THEOREMS.
S29
from tlie point to the line. In like manner, tbc distance
from a point to a plane, must be measured by a perpendicular
jdrawn from that point to the plane, because this is the
shortest line which can be drawn from the point to the
j>tane»
THEOREM XCVII.
. The Common Section of Two Planes, is a Right Line,
Let acbda, akbfa, be two planes
cutting each other, and A, b, two points £,
in which the two planes meet ; drawing . \ F*
the line ab, this line will be the common
intersection of the two planes.
For, because the right line ab touches
the two planes jn 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, *
^he cpmmon intersection of the two planes is a right line.
^
^
u
$
THEOREM XCVllh
I
\i a Line be Perpendicular to two other Lines, at their
Common Point o!F Meeting ; it will be Perpendicular to
the Plane of those Lines.
Let the line ab make right angles with
the lines AC, ad 5 then will it be per
pendicular to the plane cde which passes
jthrough these lines.
If the line ab were not perpendicular to
the plane cde, another plane might pass
through the point a, to which the line ab ^
would be perpendicular. But this is im
possible , for, since the angles bag, bad, are right angles^
this other plane must pass through the points g, d.. Hence,
this plane passing through the two points a, c, of the line
AC, and through the two points a, d, of the line ad, it will
"pass through both these two lines, and therefpre he the same
jplane with the former, q. e. d.
TH]£0R(M
sso
GEOMETRY.
THJEbUBM XCIX.
If Two Lines be Perpendicular to the Same Plane, thejr wM
be Parallel to each other.
S
>
Let the two lines ab, cd, be both per
pendicuVar to the same plane ebdf ; then
will Afi ha paralLed to cD.
«
For, join B, i>, by the line bd in the '
plane. Then, because the^ lines ab, cd, .
are j^^erpendicular to the plane £F, they are
both perpendicular to the line bd (def. 90) in that plane^
and cor sequent ly they are parallel to eacl^ other (corol.
th. I'i). Q. E. D.
CoroL It two lines be parallel, and if one of them be
perpendicular to any plane, the other will also be perpendi
cular to the same plane.
THEOREM C
If Two Planes Cut each other at Right Angles, and a Line
be drawn in one of the Planes Perpendicular to their
. Common Intersection, it will be Perpendicular to the
. other Plane.
Let the two pknes acbd, abbf, cut
each other at right angles; and the line^
CG be perpendicular to their common sec
tion ab ; then will cg be also perpendicular
to the other plane aebf.
For, draw eg perpiendicular to ab.
Then, because the two lines gc, gEj are
perpendicular to the common intersection
^ ABy the angle cge is the angle of inclination of the two
planes (def. 92). But since the two planes cut each other
. perpendicularly, the angle of mclination cge is a right
smgle. And since the line cq is perpendicular to the. two
lines GA, ge> in the plane aebf, it is therefore perp^ndi*
cular to that plane (th. 9^)* q. e. v.
TKBORSM
THEOREMS
SSI
THEOI^EM CI.
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 asbf; these planes
xnake with each other two angles whose sum is equal to two
right angles.
For, through any point g, in the common section ab^
draw CD, ef, perpendicular to ab. Then, the line co
^akeswjth EFtwo angles together equal to two right angles.
But thesie two angles are (by def. 92) the angles of inclina
tioii of the two planes. Therefore the two planes make
jingles with each other, which are together equal to two
right angles*
CoroL In like manner, it may be demonstrated, that planes
icfhich intersect, have their vertical or opposite angles equal;
also, that parallel planes have their alternate angles equal ;
(ind so on, as in parallel lines.
THfcOUEM CII.
y
If Two Planes be Parallel to each other ; a Line which is
Perpendicular to one of the Planes, will also be Perpendi»
Cular to the other.
I
tiET the' two planes CD, ef, be parallel,
jind let the line ab be perpendicular to the
plane cD ; then shall it also be perpendi
cular to the other plane ef.
For, from arty point G, in the plane £F,
draw GH perpendicufar to the plane cd, and
4raw^AH, BG.
Then, because ba, gh, are both perpendicular to the
plane cd, the angles k and h are both right an'gles. Attd
because the planes cd, j^f, are parallel, the perpendiculars
3A, gh, are equal (def, 93). Hence it follovrs that the
lines ;3G, AH, are parallel (drf. 9). And the line ab being
perpendicular tp the line ah, is also perpendicular to the
parallel line bg (cor. th. 12).
In'likemanner it is ptoved, that the line ab is p«rpen
4i£Dlar to aU otl^r 4ines whic^^a be drawn from^ the point b
in
532
GEOMETRY,
in the plane ef* Therefore the line ab is perpendicular tm
the whole plane £f (def. 90). q. b. d.
THEOREM cm.
If Two Lines be Parallel to a Third Line, though not in the
same Plane with it ; they will 6e Parallel to each other*
Let the lines ab, cd, be each of them ^ .
parallel to the third line £F, though not in
the same plane with it ; then will ab be pa*
lallel to cj>.
For, frem any point G in the line ef, let
GH, Gi, be each perpendicular to ef, in the
planes eb, ed, of the proposed parallels.
Then, since the line ef is perpendicular
to the two lines GH, gi, it is perpendicular
to the plane ghi of those lines (th. 9S). And because SF
is perpendicular to the plane <yHi, its parallel ab is also per
pendicular to that plane (cor» th. 99), 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 sam(^
plane, these two lines are paiydlel (th. 99). q. e. d*
theorem cit.
If Two Lines, that meet each other, be Parallel to Twq
other Lines that meet each other, though not in the sanje
^Plane with them j the Angles contained by those Lin^s
will be equal.
Let the two lines ab, bc, be parallel to
the two lines de, ef ; then will the ajigle
ABC be equal to the angle def.
For, make the lines ab, bc, de, ef, all
lequal 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. Therefore adj cf, are equal and parallel
(th. 15); and consequently also ac, df (th. 24). Hence,
the two triangles abc, d35f, having aU^heir sides equ^l,
THEOREMS
S3*
^ach to <^acli> have their angles also equal, and consequently
the angle ABC = the angle def. (^ £. d.
THEOREM cr.
The Sections made by a Plane cutting two other Parallel
Planes> are also Parallel to each other.
3LBT the two parallel planes ab, cd, be
cut by the third plane efhg, in the lines fXSL=::^K
£F, GH : these two sections ef, GH,.will
be parallel.
Suppose EG, FH, be drawn parallel to
each other in the plane efhg ; also let
EI, FK, be perpendicular to the plane cd ;
and let IG,KH, be joined.
Then eg, fh, being parallels, and ei, fk, being both
perpendicular to the plane CD, are also parallel to each other
(th. 99) ; conisequently the angle hfk is equal to the angle
CEi (th. 104). But the angle fkh is also equal to the angle
fcio, being both right angles; therefore the two triangles are
equiangular (cor. 1 th. 17) ; and the sides fk, ei, being
the equal distances between the parallel planes (def. 93), it
follows that the sides fh, eg, are also equal (th. 2). But
these two lines are parallel (by suppos.}, as well as equal;
consequently the two lines e^, gh, joining those equal pa
Tallek, are also parallel (th. 24). <^ e. d.
theorem cvr.
If any Prism be cut by a Plane Parallel to its Base, the Sectioa
will be Equal and Like to the Base.
LsT AG be any prism, and il a plane '
parallel to the base AC ; then will the plane
IL be equal and like 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 paral
lel, by hypothesis ; and two parallel planes,
cut by a third plane, having parallel sections
(th. 105); therefore IK is parallel to ab, and kl to BC, and
I,M to CD, and im to ad. But ai and bk are parallels
(by def. 95) ; consequently ak is a parallelogram ; and the
opposite sides ab, ik, are equal (th. ^2). In like manner,
it
S34
GEOMETRY.
It is shown that kl is =s bc» attd IM = ci>y
uid IM ^ AD» 'or the two planes Ac, il, are
mutually equilateral. But these two planes,
havingtheir corresponding sides parallel, have
tlie angles contained by them also equal
(th. 104), namely, the angle A = the angle t,
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, tL, have all their corresponding
sides and angles equal, or they arc equal and like. Qi £. d»
s
THEOREM CVII.
If a Cylinder be cut by a Plane Parallel to its Base> the
Section will be a Circle, Equal fo the Base.
Let af be a cylinder, and ghj any
section parallel to the base abc; then will
CHI 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 il, I, 4 ,
and join the goints as in the figure.
Then, since kl, ci, are parallel (by
def» l0'2) ; and the plane Ki, meeting the
two parallel planes ABC, GHi, makes the two sections KC, Lr,
parallel (th. 105) j the figure klic is therefore a paraI»
lelogram, and consequently has the opposite sides li, kg,
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 lines, dr;iwn from the point L to
the circumference of the section ghi, are all equal to radii
of the base 5 consequently Giii is a circle, and equal to ABC.
(^ £. D.
THEORElvr CVIII.
All prisms and Cylinders, of Equal Bases and Altitudes^ ave
Equal to each other. .
Let AC, DF, be two
prisms, and a cylinder,
en equal bases ab, de,
and having equal alti»
tudes £c, FF ; then will
^the solids AC, DP, be
cquaL
FoT) let FQ, Rs, be
any
C
J^^
Q,
IB
i^C3s
THEOREMS. ■ 355
any two sections parallel to the bases, and e^idlstant from
them. Then, by the last two theoi^ms, the section pq^is
equal to the base ab, and the section ks equal to the base ■
DK. But the bases ab. DE, are equal, by the hypothesis;
therefore the sections pa, rs, are equal also. In like manDer,
it m»y be shown, that any other corresponding sections are
equal to one another.
Since then every section in the prism ac, is equal to its
_ corresponding secfion in the prism or cylinder DF, the prisms
and cylinder themselves, which are composed of an equal
sumheror all these equal sections, must also be eiual H..E.D.
Carol. Every prism, or cylinder, is equal to a rectangular
parallelopipedon, of an equal base and altitude.
TUSORBM cix.
Rectangular Parallelopipedons, of Equal Altitudes, are to
each other as their Bases.
\XT AC, EG, be two rectan
rularparallelopipedons, having
the equal altitudes ad, eh ;
thea will the solid ac be to the
solid EG, as the base ab is to
the base ef.
For, let the proportion of the
base AB to the base ef, be that
of anyone number m (3) to any
Other number n (2). And conceive AB to be didded into m
equal parts, or rectangles, ai, i.k, mb (by dividing an into
that number of equal parts, and drawing II, K^f, parallel
to Bn). And let ef be divided, in like manner, into n equal
parts, or rectangles, eo, pf : all of these pans of both bases
being mutually equal among themselves.. And through the
lines of division let the plane sections lr, ms_, pv, pass
parallel to aq, et.
"nien, the parallelopipeJons ar, ls, mc, ev, pg, are all
equatj having equal base; and altitudes. . Therefore the solid
AC is to the solid eg, as the number of parts in the former,
to the number of equal parts in the latter ; or as the number
of parts in ab to the number of equal parts in ef, that is, as
the base ab tOjbe base Et. q. e. d.
CsrU. From this theorem, ajid .the corollary to the last, it
appears^ that all prisms and cylinders of equal alfitudest »»
336
GEOMETRY.
to each other as their bases ; every prism at^d cylinder beiagf
equal to a rectangular parallelopipedon of an equal base and
altitude.
THEOREM ex.
Rectangular Parallelopipedons, of Equal Bases^ are to each
other as their Altitudes.
L
B
^'
Ct
^. 4
a
C
Let ab, cd, be two rectan
gular paraUelopipi^ons, stand
ing on the equal bases ae,cf;
then will the solid ab be to the
solid CD, as the altitude £ B is to
the altitude fd.
For> let AG be a rectangular
parallelopipedon on the base ae^
and its altitude eg equal to the altitude fd of the solid CD.
Then ag and en are equal, being prisms of equal ba^S
and altitudes. But if hb, hg, be considered as bases, the
solids ab, AGy of equal altitude ah, will be to each other
as those bases hb, hg. But these bases hb, hg^ being
parallelograms of equal altitude he, are to each other a^
their bases eb, eg ; therefore the two prisms ab, ag, are
to each other as the lines eb, eg. But ag is equal to
CD, and EG equal to fd; consequently the prisms aCCD,
are to e^ch other as their altitudes eb, fD^ that is^   **
'ab : CD :: eb : fd, q^ e. d.
CoroU 1. From this theorem, and the corollary to theorem
108, it appears, that all prisins and cylinders, of equal bases^
are to one another as their altitudes.
CoroL 2. Because, by corollary 1, prisms and cylinders are
as their altitudes, when their bases are equal. And, by the
corollary to the last theorem, they are as their bases, 'when
their altitudes are equal. Therefore, universally, when nei
ther tire equal, they are to one another as the product of their
bases and altitudes. And hence also these products are the
proper numeral meaisures of their quantities or magnitudes^
THEORBM CXI.
Simijar Prisms and Cylinders are to each other, as the
Cubes of their Altitudes^ or of any other Like Linear Di*
mensions.
Let abcd, efgh, be two similar prisms ; then will the
prism CD be to the prism gh, as, ab^ to sif' or ad' to £h'.
 ^ ' For
o
p
Pot the solids M'e to e^ch other tis
the product of their bases and alti
tudes (th. 110, cor. 2), that Ts, as
AC . Alb to £G . £H. But the bases,'
i>6ing similar planes, tire to each
other as the squares* of their like ^^ [y'*' j;
sides, that is, ac to eg as ab^ to
£F^; therefore the solid cd is to
the solid Gii, a^ ABf . ad to ^f' . EH. 
But KD and fh, being . similar planes, have their like side^
proportional, that is, ab : ef : : ad : bh,      ^
qr AB* : £f* : : ad*: eh**: thereCore ab*. ad : ef*. sh : : ab^ :,ef%
or : : ad' : eh' ; conseq. the solid cd : solid gu : ; ab' :
15f' ;: AD* t bh'* <^ e. d.
6i
THEOREM exit
In any Pyramid, a Section Parallel to the Base is similar to
the Base; and these two planes are to each other as the
Squares of their Distances from th<» Ydrtexi
Let abcd be a pyramid, and e^o a sec'^
tion parallel to the base BCD, also Aiii a
line perpendicular to the two planes at H and
I : then will bd, ]sg, be two siniilar planes,
and the plane BD will be to the plane bg, as
AH* to AI*.
For, join ch, ti. TheUj because a plane
cutting two parallel planes, makes parallel
sections (th. 105), therefore the plane ABC,
meeting the two parallel planes bd, eg, mak^ the secticrns
Bc, ef, parallel : In like manner^ the plane acd makes
the sections CD, fG, paraUel^ Again, because two pair of
parallel lines make equal angles (th. 104), the two ef, fg,
which are parallel to BC, CD, make the angle bfg equal
rthe angle bcd. And in like mannef 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 equian^
gular.
Again, the three lines ab, Ac, ADj making with the
parallels bc, eF, and cb, fg, tqual angles (th. 14)j and
the angles at a being common, the two triangles ABC, aef,
&re equiangular, as also the two triangles ACD, afg, and
have thcrerore their like sides proportional, namely,   
VOL.L Z AC
d58
GEOMETRY.
AC : AF
BC
BP : : CD : VG. And in
like manner it may be shown, that aU the
lines in the plane fg, are proportional to all
the corresponding lines in the base bd.
Hence these two planes, having their angles
equal, and their sides proportional, are
similar, by def. 68. Ip — ^
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 there
fore their like sides proportional, namely, ac : af : : ah : ai,
or AC* : AF* :: ah* : ai*. Consequently the two planes
BD, EG, which are as the former squares ac*, af*, will
be also as the latter squares ah*, ai'^ that is,     
BD : EG :: ah* : Ai*. Q. E. D.
THEOREM CXIII.
X
t
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
p^vallel 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 paralld planes; and let the planes
ACE, ADE, pass through the axis of the
cone AK.E, 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 DB (th. 105). And because the triangles formed by
these lines are equiangular, kh : EC : : ak : ae : : Ki : ed.
But EC is equal to ed, being radii of the same circle ; there
fore 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 Gj^i, which is therefore a circle (def. 4?4).
Again, by similar triangles, al : af : : ak :
:: KI : ED, hence al* : af* :: Ki* : ed*; but Ki*
circle ghi : circle bcd (th. 03); therefore A L* :
circle ghi : circle bcd. q. e. d.
AE or
ED* : :
af"^ ;:
THEOHEM
THEOR^MS^
S30
TkEOR^M ckiv; . ^
All Pjramidsy and Cones, of Equal Bases and Altitudes, art*
Equal io one another^*
Let abc^ def,
ie any pyramids and
cone, of ^qiial ba^es
BC, £F, and equal ,
altitudes AG, dh:
then will the pyra,
mids and cone abc
and DEF, be equal*
For, parallel to the
bases and at equal distances AK, po, from the vertices,
suppose the planes tK, lm, to be di^wn. ' .......
Then, by the two. preceding theorems, *
DO* : DH* i: LM : sf, and
AN* 2 AG* : : IK : BC.
But since an*, ag*, are equal to d6% dh*,
therefore iK : BC : : lm :.ef. But bc is equal to fcp,
by hypothesis j therefore ik is also equal to lm. «
In lik^ thaiinef it is shown, that any other sections, at
equal distance from the vertex, are equal to each other.
Since theii, every section in the cone, is 6qual to the cor
responding section iii the pyramids, and the heights are equal,'
the solids abc^ d£f, /:omposed of all those sections, must be
equal also; q. IS. jb.
TilEOR^M CXV.'
Every Pyramid is tlie l*hird Part of a Prism o^ the Same'
Base and Altitude.
Let abcd£f bef a prism, and ubsp a ^
pyramid, on the same triangular base d£^:
then will the pyramid BDef be a third part
tf the prism abcdef*
For, in the plancfs of the thi^e sidrfs df thcf
?rism, draw the diagonals bf, bd, cd.
i'hen the two planes bdf, bcd, divide the
whole prism into the three pyramids bdef, ^abc, 'iJBCF^^
^hich are proved to be all equal to one another, as follows.
Since the opposite end^ of the prism are equal to each other,
ihe pyramid nfhbse base is abc and vertex D, is equal to the
a 2 pyraxjrwd
f
Sit)
CEOMETRT.
pyramid whose base is def and vertex B
(th. 114), being pyntmids of equal base .
and altitude.
But the latter pyramid, whose base is
B£F and vertex b> is the same solid as the
pyramid whose base is bef and vertex t>i
and this is equal te the third pyramid
whose base is bcf and vertex d, bei^ig py
.ramids of the same altitude and equal ba^es
BEF, seF.
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 pyra
mid, q. E. D.
Hence also, merj pyramid, whatever its figure may be, is
the third part of a prism of the same base and altitude ;
since the base of the prism^, whatever be its figure, may^
divided into triangles, and the whole solid into trian^lar
prisms and pyramids.
CoroL Any cone is the third part oJF 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 pyramid, 6f equal base and altitude.
ScholU/m, Whatever has been demonstrated of the propor
tionality of prisms, or cylinders, holds equally true of pyra
mids, or cones ; the former being always triple the latter j*
viz. that similar pyramids or cones are as the cubes of their
like linear sides, or diameters, or altitudes, &c. And the
same for all similar solids whatever, viz. that they are in pro
portion to each other, as the cubes of their like linear dimen
sions, since they are composed of pyramids every way similar.
THEOREM CXVI,
If a Sphere T^e cut by a Plane, the Section will be a Grcle^
Let the sphere aerf be cut by the
plane abb ; then will the Sisction adb
be a circle.
Draw the chord ab, or diameter of
the seotion ; perpendicular to which, or
to the section adjb, draw the axis of the
sphere ecg.f, thrpugh the centre c,
which will bisect the chord ab in the
point G (th. 41). Also, join cA, CBj
an^
THEOREWS.
341
9nd draw cd^ cd^ to any point d in tb^ pemmetm? of the
section AOB.
Thenj because CG is perpendicular to the plane adb, it
is perpendicular both to ca and gd (def. 90). So that cga,
COD are two rightangled triangles/ having the perpendicular
CG common y and the two hypothenuses^cAy cd^ equal, being
both radii of the sphere ; therefore the .third sides ga^ gd,
are also equal (cor. 2» th. 34). p like manner it is shown,
that any other line, drawn from the centre G to the circum
ference of the secdcm adb, is equal to ga or gb ; conse
quently that section is a circle.
CfTol* The section through the centre, is a circle having
the same centre and diameter as the sphere, and is called a
great circle of the sphere ; the oiher plane sections being
Uttle Circles.
THEOREM CXVII.
Every Sphere is TwoThirds of its Circumscribing Cylinder.
. Let abcd be a cylinder, circum
scribing the sphere efghj then will
the sphere efgh be twothirds of the
cylinder abcd.
For, let the plane ac be a section of
the sphere and cylinder through the
centre i. Join ai, bi. Also, let fih
be parallel to ad or bc, and eig and
KL parallel to ab or DC, the base of
the cylinder ; the latter line kl meeting Bi in M, and the
circular section of the sphere in n.
Then, if the whole plane hfbc be conceived to revolve
about the line hp as an axis, the square fg will describe
a cylinder ag, and the quadrant ifg will descrihe a hemi
sphere efg, and the triangle ifb will describe a cone iab.
Also, in the rotation, the tliree lines or parts kl, kn, km, as
radii, will describe corresponding circular sections of those
solids, namely, kl a section of the cylinder, kn a section of
the sphere, and km a section of the cone.
Now, fb being equal to Fi or IG, and kl parallel to
fb, then by similar triangles IK is equal to km (th. 82). Arid
since, in the rightangled triangle ikn, in* is equal to ik*
+ KN* (th» 34; J and because kl is equal to the radius ig
or
343
GEOMETRY.
or IN, and KM rr IK, therefore kl* is
jequal to km* + kn*, or the square of
the longest, radiusi of the sud circular
sections, is equal to the sum of the
squares of the two others. And ber
cause circles are to each other as the .
squares of their diameters, or of their
raidii, therefore the circle described by
KL is equal to both the circles de
scribed by KM and kn ; or the section of the cylinder, U
equal to both the corresponding sections of the sphere and
cone. And as this is always the case in every parallel posir
tion of KL, it follows, that the cylinder eb, which is com
posed of all the former sections, is equal to the hemisphere
SFG and cone iab, .which are composed of all the latter
sections.
But the cone iab is a third part of the cylinder eb
(cor. 2, th. 115) 5 consequently the hemisphere efg is equ4
to the remaining twothirds ; or the whole sphere bfgh
equal to twothirds of the whole cylinder abcd. q^ e. d.
CoroL 1. A cone) hemisphere, and cylinder of the same
base and altitude, are to ieach other as the numbers 1, 2, 3.
CoroL 2. All spheres are to each other as the cubes of their
diameters ; all these being like parts of their circumscribing
cylinders.
Corol. 3. From the foregoing demonstration it also api^
pears, that the spherical zone or frustrum egnf, is equal
to the difference between the cylinder eglo and the cone
IMQ, all of the 3ame common height ik. And that the
spherical segment pfn, is equal to tl^e difference between
the cylinder ablo and the conic frustrum. A(^I9> all of tbo
iwne common altitude FK,
V * »
♦ ■ *
fROBLEMS.
■■:.x fy^V^'i—^'
t s" ]
PROBLEMS.
PROBLEM !•
Ar
To Bisect a Line ab ; that Is, to divide it into two Equal
Parts.
From the two centres a and B, with
any «qual radii, describe arcs of circles, in
tersecting each other in c and d; and
draw the Une cd, which will bisect the
given line ab in the point £.
For, draw the radif ac,,. bc, ad, bd.
Then, because all these four radii are equal,
and the side cd common, the two triangles
ACO, BCD, are mutually equilateral : consequently they are
also mutually equiangular (th. 5), and have the angle ag^
equal to the angle bce.
Hence, the two triangles ace, bce, having the two sides
AC, cE, equal to the two sides Be, ce, and their contained
angles equal, are identical (th. 1), and therefore have the
^de A£ equal to eb« q^ b. d.
w
problem II.
To Bisect an Angle bac.
From the centre a, with any radius, de
scribe an arc, cutting off the equal lines
AD, AB ; and from the two ceni^res d, e,
with the same radius, describe arcs intersect
ing in F ; then draw af, which will biisect
the angle a as required.
For, join dp, ef. Then the two tri
angles ADF, AEF, having the two sides
AD, DF, equal to the two AE, EF (being equal radii), an4
the side af common, they are mutually equilateral ; conse*
quently they are also mutually equiangular (th. 5), and have
the angle baf equal to the angle caf. .
Sciolium. In the same manner is an ^c of a circle b)^
PROBLEM
3H
GEOmTBLY.
PROBLEM III>
t
At a Given P<Mnt c, in a Line ab, to Erect a Perpendicular^
From the given point c, with any radius^
cut off any equal parts cp, C3j of the given
line; and, from the ty^ centres d and e,
with anyone radius, describe arcs intersecting
is F ; then join cf, which will be perpendi
cular as required.
For, draw the two equs^ radii df, £F. Then die tw^
triangles cdf, cef, having the two sides cd, of, equal tp
die two CB, EF> and cf common, are mutually equilajteral;
consequently they are also mutually eqmangular (th. 5), and
have the two adjacent angles at c^qual to each other; there^
fore the line cf is perpendicular to ab (def. 1 1).
Otherwise.
Wh^n the Given Point c is near the End of the lin^.
From any point d, assumed above the
line, as a centre, through the given point
c describe a circle, cutting the given line
at E ; and through e and the centre D,
draw the diameter edf ; then join cF,
which will be the pci^ndicular required.
For the angle at c, being an angle in a semicircle, is a
right angle, and therefore the line cf is a perpendicular
(bydef. 15),
PROBLEM ly.
From a Given Point a, to let fall a Perpendicular on a
' given Line Be. *
From the given point A as a centre, with ^ ^
any convenient radius, describe an arc, cut
ting the giving line at the two points d and
E ; and from the two centres n, E, with
any radius^ describe two arcs, intersecting
at F 3 then draw agf, which will be per
pendicylar to ec as required.
For, draw the equal radii An, AE, and
DP, £.F. Then the two triangles adf, aef, having; the two
sides AD> df, equal to the two ae, ef, and af common^ are
mutually
%:
PHOBIXMS. SU
inBtually equilateral; consequently they are also mutuallf
equiangular (th. 5), and have the angle dag eqnal the angle
BAG. Hence then, the two triangles adg, aeg, haying
the two sides AD, ACry equal to the two ae, AGt and their
included angles equals are therefore equiangular (th. 1), and
have the angles at G equal; consequently a g is perpendicular
to BC (def. 11). "
Otherwise.
When the Given Point is nearly Opposite the end of the
Line.
From any point d, in the given line
B€, as a centre, describe the arc of a
circle through the given point A, cutting 3—. /''' J JgC
90 in B ; ai3 from the centre b, with the *• .. ^\
radius ea, describe another arc, cutting
the former in f ; then draw agf, which
will be perpendicular to bc as required.
For, draw the equafl radii da, df, and ea, ef. Then the
two triangles dae, dfe, will be mutually equilateral ; conse
quently they are also mutually equiangular (th. 5), and,have
the angles at d equal. Hence, the two triangles dag, dfg,
having the two sides da, dg, equal to the two df, dg, and
the included angles at d equal, have also the angles at G
equal (th. l); consequently those angles at G are right
angles, and the line AG is perpendicubr to dg*
PROBLEM v«
At a Given Point a, in a Line ab, to make an Angle Eqpial
to a Given Angle c.
From the centres a and c, with any one
radius, describe the arcs de, fg. Then,
with radius de, and centre f, describe an
arc, cutting fg in o. Through G draw
the line AG, and it will form the angle re
quired.
For, conceive the equal lines or i:adii,
DE, FG, to be drawn. Then the two triangles cde, afg,
being mutually equilateral) are mutually equiangular (th. S),
and have the angle at a equal to the angle c«
FROBLBM
S46 GEOMETRT.
PROBLEM VI.
Through a Given Point a, to draw a Line ParaQel ta a
Given Line Bc.
From the given point a draw a line ad
to any point in the given line bc. Then
draw the line eaf making the angle at A
equal to the angle at d (by prob. 5); so
shall £P be parallel to bc as required.
For, the angle d being equal to the alternate angle A) the
lines BC> bf, are parallel, by th. 1 9.
PROBLEM VII.
To Divide a Line ab into any proposed Number of Equal
Parts.
Draw any otlier line ac, forming any
angle with the given line ab ; on which
set off as many of any equal parts, ad, de,
£F, PC, a^ the line ab is to be divided into. •
Join BC ; parallel to which draw the other
lines FG, £H, Di : then these will divide
AB in the manner as required. — ^For those parallel lines di»
vide both the sides ab, ac, proportionally^ by th. 82.
PROBLEM VIII.
To find a Third Proportional to Two given Lines ab/ac.
Place the two given lines ab, ac,
forming any angle at a j and in ab take a a
also AD equal to ac. Join bc, and A C
draw DE parallel to it ; so will AE be r
the third proportional sought. ^^^^C\
For, because of the parallels bc, dk, ^'^^ — Tiri
the two lines ab, ac, are cut propor
tionally (th. 82) ; so that ab : ac : : ad or Ac : AE ; there
fore A£ is the third proportional to ab, ac.
m
PROBLEM ly. .
' To find a Fourth Proportional to three Lines ab, Ac, ad.
Place two of the given lines ab, ac, making any
angle at A; also place ad on ab. Join bc; and paralld
to
PROBLEMS.
3«
to It draw ds : $o shall ae be the fourth
proportional as required.
For, because of the parallels Bc, de,
the two sides ab, ac, are cut propor .
tionally (th. 82) ; so that   »  
AB : AC : : AD : AE.
PROBLEM X.
To find a Mean Proportional between Two Lines ab, bc
Place ab, ^c, joined in one straight a rB
line AC : on which, as a diameter, describe 15 — c
the semicircle adc ; to meet which erect
the perpendicular £p \ and it will be the
mean proportional sought, between AB
and bc (by cor. th. 87). ^ A^ oTlfe
, .1)
PROBLEM XI.
To find the Ceptre of a Circle.
Draw any chord ab ; and bisect it per
pendicularly with the line ep, which will be
a diameter {th. 41, cor.). Therefore cd
bisected in o^ will give the centre, as re
quired..
PROBLEM XII.
To describe the Circumference of a Circle through Three
Given Points a, b, c.
From the middle point b draw chords
BA, Bc, to the two other points, and bi
sect these chords perpendicularly by lines
meeting in o, which will be the centre.
Then from the centre o, at the distance
of any one of the point3, as OA, describe
^ circle, and it will pass through the two
Other points b, c, as required.
For, the two rightangled triangles oad, obd, having the
sides ad, pB, equal (by constr.), and od common with
the included right angles a^ d equal, have their third sido«
OA, OB^ also equal (th. I ). And in like manner it is shown,
that oc is equal to ob or OA. So that all the three oA, be,
pc, being equil^ will be radii of the same circle.
' PROBLEM
MS
GEOMETRY.
1
PROBLEM XIII.
To draw a Tangent to a Circle, through a Given Point a»
When the given point a is in the cir
cnmference of the ciicle : Join a and the
centre o ; perpendicular to which draw
BAc» and it will be the tangent, by th, 46.
But when the given pdint a is out of
the circle: Draw ao to the centre oj
on which as a diameter describe a semi
circlei cutting the given circumference in
D ; through which draw badc, which
will be the tangent as required.
For, join do> Then the angle ado»
in a semicircle, is a right angle^ and con
sequently AD is perpendicular to the ra
dius ix>> or is a tangent to the circle (th. 46).
PROBLEM XIV.
On a Given Line b to describe a Segment of a Circle> to^
Contain a Given Angle a
At the ends of the given line make
angles dab, dba, each equal to the
given angle c. Then draw ae, B£>
perpendicular to ad, bd ; and with the
centre £, and radius ea or eb, describe
a circle ; so shall afb be the segment
required, as any angle f made in it will
be equal to the given angle c.
For, the two lines ad, bd, being
perpendicular to the radii ea, eb (by c<Mistr.), are tangents
to the circle (th. 46) ; and the angle A or b, which is equal
to the given angle c by construction, is equal to the angle f
in the alternate segment apb (th. 53).
problem XV.
To Cut off a Segment 'from a Circle, that sihall Contain a
Given Angle G.
Draw any tangent ab to the given
circle ; and a chord ad to make the
angle Cab equal to the given angle c ;
then dea will be the segment required,
any angle E made in it being equal to
the given angle c.
For
PROBLEMS.
S4f
For 4he angle A, made by the tangent and chord, which
is equal to the given angle c by construction, is also equal
to any angle E in the alternate segment (th. 53). '
PLOBL£M 2;VI«
To make an Equilateral Triangle on a Given Line A3.
« •
From the centres a and b, with the
distance iLB describe arcsy intersecting in c.
JDraw AC, bc, and abc will be the equi
lateral triangle.
For the equal radii ac, bc, are, each of
them, equal to ab.
PROBLEM XVII.'
To make a Triarigle with Three Given Lines ab, Ac, bc
With the centre a, anddistance ao,
describe an arc. With the centre B» and
distance B€, describe another su'c, cutting
the former in c. Draw ac, bc,. and
ABC will be the triangle required.
For the radii, or sides of the triangle,
Ac, BC, are equal to the given lines AC
^c, by construction.
PROBLEM XV m.
To make a Square on a Given Line ab«
Raise ad, bc, each perpendicular and
^qual to AB ; and join dc ; so shall abcd
ibe the square sought.
For all the three sides ab, ad, bc, are
«qual) by the construction, and dc is equal ^
and parallel to ab (by th. 24); so that all the
four sides are equa]^ and the opposite ones are parallel.
Again, the angle A or B, of the parallelogram, being a right
angle, the angles are all right ones (cor. 1, th. ^2), Hence,
then, the ii^re, having all its sides equals and all its angles
xightf is a square (def. 34).
4
J \
PROBLEM
S50
GEOMETRY.
PROBLBM XIX. •
To make a Rectangle, or a Parallelogram} of a Given Lengtfi
and Breadth, ab, bc.
Erect ad, bc» perpendicular to ab, and
each equal to bc ; then join Dc, and it is
done»
The demonstration is the same as the last
problem.
And in the same manner is described any oblique paral
lelogram, only drawing ad and BC to make the given oh"
lique angle with ab, instead of perpendicular to it*
PROBLEM XX.
To Inscribe a Circle in a Given Triangle ABC.
Bisect any two angles a and b, with
the two linesvA d, bd. From the inter
section D, which will be the centre of
die circle, draw the perpendiculars de,
DF, DG» and they will be the radii of the
circle required.
For, since the angle DAE is equal to
the angle dag, and the angles at £, g,
right, angles (by constr.), the two triangles ade, adG, are
equiangular ; and, having also the side ad common^ they are
identical, and have the sides de, dg, equal (th. 2). In like
manner it is shown> that dp is equal to de or dg.
Therefore, if with the centre D, and distance DE, a
drcle be described, it will pass through all the three point*
E, F, G, in which points also it will touch the three sides of
the triangle (th. 46), because the radii de^ df, dg, are per
pendicular to them. ^
PROBLEM XXI.
To Describe a Circle about a Given Tjiangle abc^
Bisect any two sides ieith two of the
perpendiculars db, df, dg, and d will ho^
the c^itre.
For, join da, db, dc. Theft the two
rightangled triangles DA£,DBE,have the
two sides DE, ea, equal to the two de, ^. ^ . ^
EB, and the included angles at e equal :
those two triangles are therefore identical
PROBLEMS.
351
(th. 1), ^nd have the side da equal to DB. In like manner
it !s shown, that dc is also equal to da or db. So that all
the thre^ da, db, dc, being equal, they are radii of a circle
passing through A, B, and c. * . ,
PROBLEM XXII.
To Inscribe an Equilateral Triangle in a Given Circle. •
Through the^ centre c draw any dia
meter AB. From the point b as a centre,
with the radius bc of the given circle,
cLescribe an arc dce. Join ad, ae, de,
and ad£ is the equilateral triangle sought.
For, join db, do, eb, ec. Then dcb
is an equilateral triangle, having each
side equal to the radius of the given cir
cle. In like manner, bc£ is an equilateral triangle. But
the angle ade is equal to the angle ABB pr cbe, standing
on the same arc ae % also the angle aed is equal to the
angle cbd, on the same arc ad ; hence the triangle dae has
two of its angles, ade, aed, equal to the angles of an
equilateral triangle, and therefore the third angle at a is
also equal to the same ; so that triangle is equiangular, and
therefore equilateral.
problem XXIII.
To Inscribe a Square in a Given Circle.
Draw two diameters Ac, bd, crossing
at right angles in the centre e. Then
join the four extremities a, B, c, d, Vith
right lines, and these will form the in
scribed square abcd*
For the four rightangled triangles
aeb, bec, ced, dea, are identical, be
cause they have the sides ea, eb, ec, ed,
all equal, being radii of the circle, and the
four included angles at e all equal, be
ing right angles, by the construction. Therefore all their
third sides ab, bc, cd, da, are equal to one another, and the
figure ABCD is equilateral. Also, all its four angles, a, b, c, d,
are right ones, being angles in a semicircle. Consequently,
the figure is a square.
problem
nsi GEOMETRY.
prob;.£M XXIV.
Tp OescrU)e a Square about a GIvoi Circle;
Draw ^Dro cliameters ac, BDCro$smg
at right angles in the centre e. Then
through their four extremities draw.FG,
IH, parallel to AC^ and ft, gh, parallel
to BDj and they will form the square
FGHI.
Forj the opposite sides of parallelo
grams being equals fg and IH are each
equal to the diameter Ac^ and fi and gh each equal to the
diameter bd ; so that the figure is equilateral. Again, be
cause the opposite angles of parallelograms are equal, all the
four angles f^ g> h, i, are right angles, being equal to the
opposite angles at £. So that the figure fghi, having its
sides equal, and its angles right ones, is a squarej and its sides
touch the circle at the four points a, b, c, d, being perpen
dicular to the radii drawn to those points.
PROBLEM XXV.
To Inscribe a Circle in a Given Square*
Bisect the two sides ^o, fi, in the points a and s
(last fig.)» Then through these two points draw ac parallel
to FG or iH, and bd parallel to fi or CH. Then the point
of intersection e will be the centrcj and the four lines^EA,
£B, EC, £D, radii of the inscribed circle.
For, because the four parallelograms ef, eg, eh, ei, have
their opposite sides and angles equal, therefore all the four
lines EA, £B, EC, ED, are equal, being each equal to half a
side of the square. So that a circle described from the centfe
X., with the distance £A, will pass through all the points
A, 9, c, D, and will be inscribed in the square, or will touch
its four sides in those points, because the angles there are
right ones.
PROBLEM XXVI.
To Describe a Circle about a Given Square,
(see fig. Prob. xxiii).
Draw the diagonals /c, bd, and their intersection t
will be the centre.
For the diagonals of a square bisect each other (th. 40),
making E4, eb, eg, ed, all equal, and consequently thes^
are radii of a circle passing through the four points a, b, c, d.
PROBLKM
*
■i
/ .. .»
1^1
1»RDBLEM XXYU;
.!■ '}
Td Cut a CiVen L»n« ja Ejttreme and Me^an Ratb»
XjsT. AB be the eiven line to be divK
in e^ictreine and mean ratio, that is, so
divided
so as
tKat the whole line maybe to the greater .
p^t,»a$ the greater part is to the less part.
I>ra1?:BC perpendicular to. AB, and equal
to .half AB* Join AC Vrapd with tentre c .
and disftance Qit, describct the circle. sd
then with centre A arid distance ad, 4e*
scribe the arc DB ; so shall ab be divided in
£ in extreme and mean ratioj or so that
AB : A£ :: AE : eb'. .
For, produce Ac to thd circumferencef at ^. Then, ABfF
being a secant, and ab a tangent, because b U a right angle :
^ therefore the rectangie^AF.AD is^qual to ab* (cor. 1 th. Gl);
consequently the means and extremes of these are proportional
(th. 17), viz. AB : af or ad f rip : : ad : Ab. But a«
is equal to ad by construction^ and ab = 2bc r: d^j
therefore, ab : Ate' + ab :: a^ : ab^
mid by division^ ab : ▲£ : : ab ; sb; «
t
PROBLEM XXYllU
To Inscribe an Isosceles Triangle in a Givefn Circle, that
shall have each of the Angles at the Base Double the'
Angle at the Vertex. ^
iDKAW any diameter ab of th^ givea 
circle ; and divide thji radius CB, in thef
point d, in extreme and mean ratio, by the
last problem. From, the pomt b apply the
chords bB, bf, each equal to the greatef
part CD. Then join Afe, af, ef j and Aef
will be the triangle requif ed*
For, the chords Bg,' b1*, befing equals
their arcs are equal j therefore the supplemental arcs and
chords AE, AF^ are also equal 5 consequently the triangle ajsp
is isosceles, arid haa the angle B equ^l to the angle F$ aiscX
the angles at G are right angles. "V
Draw cf and Dt. Then, »c : «d : : cd : Bd, or
Be : Bf : : bp : fiiD'by constr. And ba : bf : : bf : bg
^by th. S7)i But bC = ^^ba j therefore bg = ^bd = gd j
therefore the two triangles cbf, cpFi are identical (th. 1),!
Vox.. L A a and
SI4
GEOMETRY.
and each equiangular to ibf and act (th. 87). ^ Tberefori
their doubles, BFb, af£ are ^isosceles and eqtiiangfiitar> as
well as the triangle bcf ; having the two sides bc» cf, equals
and the angle b common w^h the triangle bfiX Buf cet
is ^ DF or BF ; th^efore the angle c == the angle dfc
(th. 4) ; consequently the angle BDi, which is equal to the
sum of these two equal angles (th. 16), is double of one of
them c; or the equal angle b or cfb double the angle C.
So that cbf is an isosceles triangle, having each of its two
equal angles double of the third angle c. Consequently the
triangle abf (which k has been shown is equiangular to the
triangle cbf) has sdso each of its mg\e» at the base double
the angle a at the vertex^
PROBLEM XXIX.
To Inscribe a Regular Pentagon in a Given Crcle.
Inscribe the isosceles triangle abc
having each of the angles abc, acb,
double the angle ]6ac (prob. 28). Then
bisect the two arcs adb, aec, in the
points D, E ; and draw the chords ad, d^,
a£, £€, so ^ shall ADBCE be the inscribed
equilateral pentagon required.
For, because equal angles stand on equal arcs, and, double
angles on double arcs, aho the angles arc, acb, being each
double the angle bag, therefore the arcs adb, aec, subtending
the two former angles, one each double the arcs Be subtending
the lat;ter. And since the two former arcs are bisected in D
and £, it follows that all the five arcs ad, db, bc, ce, ea,
are equal to each other, and consequently the chords also
which subtend them, or the five sides of the pentagon, are
all equal.
Note* In the construction, the points d and e are most
easily found, by applying bd and C£ each equsri to bc^
problem' XXX.
To Inscribe a Regular Hexagon in a Circle.
Apply the radius Ao of the given circle
3s a chord, ab, bo, cd, &c, quite round the
circumference, and it will complete the re
gular hexagon abcdep.
For, draw the r.uiii ao, bo, co, i5o, Eo,
Fo, completing six equal triangles } of*
which any one, as Abo, being equilateral
(bf
PROBLEMS^ iU
(jby constr.) its i£ree angles ^ all equal (cor. 2, th. 3), and
any one <»f them, as^bs, is onethird of the whole, or of tw^
right iandes (th. IT), or bnesixtii of four right angles. But
the whole circumfeirence is the measure of fchir right angles
(ipor. 4, th. 6). Therefore the arc ab is onesixth of the
tircumference of the circKj and consequently 4ts chord Ap
one side of an isquilateral hexagon inscribed in the circle*
And the sanie bf the other thor<k*
Cpr'of. The side pf a regular hexagoh is equal to th6 radius
bf the circumscribing circle, or to ihi chord of oii^si^h
{tet df thie circumference:
ipROBtfiM xtxi: 
To descnbe a Regular Pentagon or Hexs^n about a Circle.
In the given circle inscAbe i^ regular
polygoh of the szm6 name br hiuhber
of sid^s, ' as abcde, by one of the
foregoing problems. Then to all its
angular points draw tangents (by'
prob. 13), and these will form the cir*
cumscribing polygon required.
For, all the chords, or sides 6t / .
the insc^ribing figure, ab. Be, &c, beiiig equal, and all thl*
radii Oa, ob, &c, being equal, all the vertical angles about the
point 6 are equal. But the angles OBF, oaf^ oXd, obc,
made by the tangents and radii, are right singles; therefore
b£F 4 OAF = tWd Hght iUigles, and oAg + 6qg = two
right angles j consequently, also, Ab& + Af5 =£ two right
angles, and Aob + agb :£: twd right angles (cor. 2, th. 18):
Hence, then, the angles Aofi + Afs being =at aob + acb»
of which aob is = aoe } consequently the remaining angles
F and G are also equal; In the same manhclr it is shown^
that all the angles f, G Hy I, k, are equal.
Again, the tangents from the same point #e, f a^ ari equal>
as alsb thie tstogents AG, gb (con 2) th. 61,) ; and the angles
F and G of the isosceli^ triangles afe, ag^, are 6qual^ as
weir as their opposite sides ae, ab j coiisequently those two
triangles are identical (th. l)i and have thWr oth^r sidei
£F, fa, Ag, gb, dll equal, and fg equal to the double of
iny one of them: In like manner it is shown, that all the
bther sides gh, hi^ ik, kF, are equal tb FcT, or double of th^
tangents gb, bh, &c.
Hence, then, the circumscribed figure is bcith equilaterat
and equiangular, which was tb be showa*
A a 2 C,ro/.
ceotfisfrsuH
'Cor$l Tl\ei
icirdc touches^theiBftddlnQfdie:^j€t
To Iivscribe a Circle In a Regular Polygon*
BisjEOT any two sides of the polygon
by the perpendiculars go^ fp» and tmir
intersection o. will be the centre of the
inscribed circle^ and OG or of will b<r
the radius.
For the perpendicul^s tq the t^ngei^ts
AF, AG, pass through the centre (cor.
th« 47); and the insciibed circle touches .
the middle points f, G by the last corollary, .AlsOf the twm
sides AG, AO, of the rightangled. triangle aoO« being equa)
to the two sides AFf Ao,. of the rightangled triangle aof, the
third sides of, og, will also be equal (cor. th. 45). Therefore
the circle described witH the centre o and radius oa, will
pa^ through f, and will touch the sides in the points g ami
F. And the same for all the other sides of the figure^
PROBLEM XXXIII.
To Describe a Circle about a Regular Polygon.
BisBCT any two pf the angles, c and Dj
\vith the lines co, do 5 then their inter
section o will be the centre of the cir
cumscribing circle } and QC, or OD, will
be the radius.
For, draw ob, oa, oe, &c, to the
angular points of the giten polygon.
Then the triangle ocb is isosceles, having the angles at c
and D equal, being the halves of the equal angles of the
polygon BCD, ODE ; therefore their opposite sides Co, do,
are equal (th. 4)« But the two triangles ocd, ocb, having
the two sides oc, cd, equal to the two oc, cb, and the in
cluded angles ocdj ocb, also ^qual, will be identical (th. l),
and have their third sides bo, od, equal. In like manner it
is shown, that all, the lines. o a, ob, oc, od, oe, ai'c equal.
Consequently a circle described with the centre o and radius
oa, will pass through all the other artgular points, B, c, D,
&c, and will circumscribe the polygon.
pkoblsm
■FROSLEMS.
857
5 IJ »
PROBLEM ZXXIV.
To make a Square Equdi to the Sum ol two ^w I^ore Given
^Squares^
Let ab and ac be the sides of two
given squares. Draw two iadefihite
lines AP, AQ9 at right singles 'to each
other I in which place the sides ab^ ac,
of the given square^ ; jom bc ; then a
square described on bc, will be equal to
the sum of the two squares described oh
AB and AC (th. S4).
in the ^amelnanaer, a sqiiare may beiisnadie jequai , to tbi^
sutli of the tfaoree ($rfiii>re giVen sqi^ares. Foiyif 4^9 AQgAHh
be taken as the sides of the given square^^i then,^ makio^
' A£ = Bc^ AD =: AD, and drawing de^ it is evident that the
square on de will be equal to the sum of the three squares
on AB, AC5 AD. Andjjo on for more squares.
PkOBLlE^ XXXV. •
To make a Square Equal to the Difference x^yf two Given
Squares.
Let ab and A'c, tikett in the sa'me
straight line> be equ^l to the sides of th^
two given squares.^Fh5iil th^ cfehtte A,
with the distance ab, describe a circle ; and
make cd perpendicular to ab, meeting the
circumference in d : so shall a square described on cD be
equal to ad^— ac% or ab*— ac% as requiri^ ^cpr, th* 34);
pRbBLfini xxxf I.
To make a Triangle Eqtod to ^ Given Qtiadn^de ABcb.
Draw the diagonal ac, ana parallel
to it i^, meting ba produced at £, and.
join C£ ; then will the triangle C£B be
equal to th^ given quadrilateral abcd.
For, the two triangles ace, acd, her
ing on the same base ac, and between
the same parallels ac, D£, are equal (th^ 25) ^ therefore,
if ABC be added to each^ it wjU znake bce equal to abc^
(ax. 2).
PiU)BL£M
S5»
GEOMETRt.
PROBLEM ZXXVir.
To make a Triangle Eqoal l;o a Given Pentagon abcde*
DiCAw DA and db, and also ef.
eC) parallel to them, meeting ab pro
duced at F and g ; then drai^ df and
T>G ; so shall the triangle dfg be equal
to the given pentagon abcd'e. •
For th^ triangle d^a = deAi and
the triangle dgb ss dcb (th. 25)}
therefore! by adding DAB'to the equals^
the tinms are equd (ax. 3), that is, dab «f DAF + DBG
SB' DAB 4 p^E + DBC> or the triangle dfg as to the
{>entagon ABODE*
^
t^
PHOBUEM XXXVIlt.
f t
To make a Rectangle Equal to a Given Triangle Ape.
' Bisect the base ab In d; then raisf
D£ aiid BF perpendicular to ab, and
meeting cf parallel to ^b^ at B and f :
so shall DF be the rectangle equal to the
given trismgle abc (by cor. 2, th. 26).
PROBLEM XXXIX.
To make a Square Equal to a Given Rectangle abjqO*
FB.0DUCE one side ab, till ^e be
equal to the other side bc. On as as
a diametei* describe a circki meeting
BC produced at f : then will bf be the
side bf the square bfgh, equal to the
given rectangle bd, as required; as ap^
pears by cor. th. 87, and th. 77.
vT ^.«'* .*
ATTT
*.; »
APPLICATION
i 350 ]
•t *
APPLICATION o» ALGEBRA
TO
 ' GEOMETRY.
W H E N it is proposed to resolve a gtometrical problem
algebraically, or by algebra, it is proper, in the first place^
to draw a figure that shatltepresent the several parts or con
'ditions of the problem^ and to suppose that figure to be the
true one* Then, hairing considered attentively the nature of
the problem, the figure is next to be prepared for a solution,
if necessary, by producing or drawing such lines in it as ap«
j)ear mcfitx^onducive to tbit end. This done, the usual sym«
bols or letters, for known and.unknown quantities, are em*
ployed to denote the several parts of the figure, both the
knbwn and unknown parts, or as many of them as necessary;
as also such unknown line or lines as may be easiest found,
whether required ot not. Then proceed to the operation,
by observing the relations that the several parts of the £gure
have to each other \ from which^ and the proper theorems
in the foregcMug elements of geometry, make out as many
equations independent of each other, as thel« are unknown
quantities employed in them : the resolution of which equa
tions, in the same manner as in arithmetical problems, will,
^determine the unknown quantities, and resolve the problem
proposed. » .
As no general rule can be given for drawing the lines, and
selecting the fittest . quantities to, substitute for, so as always
to bring out the most simple conclusions, because different
problems requure different mode^ of solution^ the best way to
gain experience, is to try the solution of the same problem
in different ways, and then apply that which succeeds best,
to other cases of the sanaeicind, when they afterwards occur.
The following particular directions, however, may be of
some use.
\4t% In preparing the figure, by drawing lines, let them bo
either parallel or perpendicular to other lines in the Bgurey
Or so as to form similar triangles. And if an angle be given;
it will be proper to let the perpendicular be opposite to that
anglei ftnd to fall from one end of a given line^ if possible.
2rf,
SCO APPLICATION or AtGEBRA
2di In selecting the qu^mities proper to substitute fopj^
those are to be phosen, whether required or not, which li^
nearest the known or given parts of the figure, and by means
of which this 9j^jt gdj^ceqt p^t$Aiy3^Iexp/jpss!Bd by addi
tion and subtraction only, without using surds.
3^9 When two lines or quantities are alike related to other
parts of the figure or problem, the best way i§, not to make
Dse of either* pf them separately, but to substitute for their
sum, or diflfer^nce, or rectangle, or the sum of their alternate
quotients, or for some line or lines, in the figure, to ^hidi
tKey have botji thp ^aipe t^hx\on*^. • \ t ,
. . 4/A, When thei ayea^ or the p^metpr, of a. figure, b p^^n^
pat such parts of it as have <^y .^ temotfi .rdation /to the
parts required: it is sometime! .of pse to assume asuathies
figure similar to the proponed om^f having. one side iiqualito
linity, or some other known qnastity. Fpr, hepce th» qthtc
parts of the figure may be &und, by tbe/kifow2\ proportions
pj'the like sides, Or parts, ind h> an equ;iti6a hei«btainfid?
For {examples, takf tJb^ following .pcobleins, : j  '«
Jn a RighUangkd Triangle^ having f^ven thf JB^sf (&)^ and iik
.^ Sum 7f the Hyfoihenusc and Berpfndict^Iar (9) ; to, find ba^
these two Sides. ' . > , ,
Li^T. ABC represent the proposed triangle,
rightangled at b. Put the base ab = 3 =:^ ^
and the sum AC + pc of tlie hypothehus.^ ' ^
ihd perpendicular = 9 == j; also, let 'jr iae
pote the hypothenuse AC, and jj; the perpen
dicular BC.
Then by the question r'^f+j'—'Cj
and by theorem 34,   "   or* =i / 1 ^J
By transpos. y in the 1st jcqu. gives ^ = / — j;,
"JThis valup of X substi. in the 2d,
gives      ^*  '2/y J* / =: f + (%
Takingaway/onbothsideslcayes r — 2sy = ^%
Bv tr^spo?, ^sy and i% gives  /* — i^ == 2/v,
■ ' ^ J*  ^*
And dividing hy2s, gives r •. ■ •.'. ■ =^ jp ?s 4?=?:]9C»
Hfnce x=^j — y==5;= Ap.
' N. B. In thi§ solQt.}pns and the^fQlIowipg onfs^ ^^ iiota^;
^opi§ miidc? b$ jising ^ m^l tmb^9V© l^Umi ^ ?ftd jr^ a%
ther^
there are tHikpearo ;»cleff^f the triangle^ a^par^e letter far
each; ippceft^igiKfi ,o ais^ng .only pije ^\nknoMrn letter iicir
ox\f «de, Jttid i^pfcee^ng ^e othfff mknown side in ticvms
of that letter ai\d the^iyen sum or difference of tli^e sides;
though this'l^at4:er, way Y'^ould render the sqlation shorter and
SQDlier J because the fbrn^er way, gives occasion for more and
better practice in reducing equations, which is the very end
aQd reason for which these problems are given at all«
t * JP.ROBLEM II
/« a Rightangled Triangle J having given the Hypothenuse (5) ;
and the Sum of the Base and P&rpendicular (7) ; tofiftd both
these two Sides.
^ Let A3C represebtt the prp.posed triartgle, rightangled at
p. Put the given hypQthen.usie Ac = 6 zr ^?, and the sum
AB + BC of the base and perpendicular = 7 = / ; also let r
denote the base ab, and j> the perpendicular fiC.
. Then by the question < • •  ;r rf y^ s
< ^A {yy tl^^rem 34 .  .^ t *• ^* +/ == «*
,JJy Irajnspos. yin.tb^ 1st, gives • :c v?: s r y
Byaabstitu. th^v^lu. for r, gives, x* r 2jy + 2j?* = tf*
By transposing /% gives  ^ 2/ — ^J)? = a* — j*
By dividipg.t)y"2, gives    f — sy zn ^a^ — ^
JBy .cpmpl^^iiig tjbte s<jLiare; gives / — j;y + ^^^ = 4^* — i^,
By eEtracjfeij3g the root, gives  y ^ is ^s/^d^ — ^s* *
J^^^ transposing JsT, gives   y = Jj" ± V'i^^ — :J/ =
4 and 3, the values of x and j».
PROjBLEM III.
In a Rectangle i having giveft the Diagonal {10\ and the Penmen
terj or Sum of all the pour Sides (28) 5 toJin4 (^'^h of the Sides
severally.
Let abcd be the proposed rectangle;
and put the diagonal AC = 10 = dy and
half the perimeter AB + BC or Ap +
PC = i 4 = a ; also put one side ab = .r,
and the other side BC =? y. Hence, by ^
rightangled triangles,     .r* + / = //*
And by the question    ^+^ = a
Thk^n by transposing)} in the 2d, givers x :ss a — y
Thisvaluesubstitutedinthe lst,£iye* a^ r 2^ + 2/ =» ^ .
Transposing
362
APPUCATION OP ALGEBRA
Transposing tf*, gives    2)>* — ^ay nt' ^ if
And dividing by 2, gives   j^ — jrjr =: 4rf* — 4** .
By completing the square, it is / — «^ + ^ =s ^ * * ^a*
And extracting the root, gives y — ■J^srv'i^— J^*
And transposing ia, «vcs  y = i^ ± /i^ "• i** = •^
or 65 the values oix and y.
PROBLEM IV.
Having given the Base and Perpen£eular of any Triangles ta
find the Side of a Square Inscribed in the same*
Let ABC represent the given ,triangle»
and EFGH its^ inscribed square. Put the 
base AB = ^/the perpendicular CD = ^
and the side of the square gf or gh =*
Di == X i then will ci = CD — di =x
« — or. '
Then, because the like lines in the
Similar triangles abc, gfc, are propor*
tional (by theor. 84, Geom.)> ab : cD : : GB : ci, that
hf h I a ', \ X \ a ^ X* Hence» by multiplying extremes
and means, ai^ix sz ax, and transposing ix, gives ai ^ax
+ hx^ then dividing by a + i, gives x = — jj = or
or GH the side of the inscribed square : which therefore is
of the same magnitude, whatever the species or the angles of
the triangles may be.
PROBLEM V.
In an Equilateral Triangle, hatting given the lengths of the three
Perpendiculars, drawn from a certain Point within, on the
three Sides ; to determine the Sides.
Let ABC represent the equilateral tri
angle, and D£, DF, DG, the given per
pendiculars from the point d. Draw the
lines DA, DB, DC, to the three angular
points; and let fall the perpendicular cu
on the base ab. Put thje ttu*ee given per
pendiculars DE = tf, DF = h, DG = c,
and put jr =z AH or bh, half !th« side of
the equilateral triangle. Then is ac or bc = 2 x, and by
right angled t riangles the perpendicular CH = v^AC*— ah*
= V4.r* — X* SB v'S^r* = x^ a.
Now,
% TO GEOMETRY. M
Now, since the area or space of a rectangle, is expressed
by the product of the base and height (cor. 2, th. ^1 GeQlD.)!^
and that a triangle is equal to halfa rectangle of equal base
and height (con 1, th. 26), Jt follows that,
the whote triangle abc is =tAb x ch =ir x x/S=n4r*^St
the tfiangle^BD 5= ^ab x t)G = j: x r = «r,
the triangle bcd =^ ^bc x de = o: x « = «x,
the triangle acd = ^ac x dp = a:* x * = ^x.
But the three last triangles make up, or are equal tm^ llie
whole former, or great triangle;
that is, ^v^3 :=: ax ^hx ^ cx'^ hence, dividing byx^i^
X ^3 :=: a +i +^> and dividing by /f, "
X = jjr — ^, half the side of the triangle sought.
.. Also, 3ince the whole perpendicular ch is = x^% it it
therefore =: ^ + J + ^. That is, the whole perpencBcofar
CH, is just equal to the sum of all the three smaller perpen
diculars DE + DF + DO taken together, wherever the point
D is situated.
PROBLEM VI.
In a Rightangled Triangle, having given the Base (S)»
and the Difference between th^ Hypothenuse and Perpendi
cular (1) 3 to £nd both these two Sides.
PROBLEM VII,
In a Rightangled Triangle, having given the Hypotbenme
(5), and &e Difference between the Base and Perpendknhr
(J } $ to determine both these two Sides.
PROBLEM VXII.
Having given the Area, or Measure of the Space, <if a
Rectangle, inscribed in a given Triangle \ to iletermiiie the
Sides of the Reetangle.
... . . ^   , .
^ PROBLEM IX.
In a Triangle, having ^ven the Ratio of the two Sidfls^
together with both the Segments of the Base, made by a
t^erpendicular from the Vertical Angle ; tp determine the
Sides of the Triangle.
PILOBLEM X.
In a Triangle, having given the Base, the Sum of the
other two SidtS and the Length of a Lint drawn from the
Vertical
• I
SM APPLICATIOH 69 ALGEBRA
Vertical Angle to the Middle tkftheiBate; to find the odes
of the Triangle*
PR09JUEM XI.
In tt Triangle, having given the two Sides abotit At
Vertical Angle, with the Line bisecting that Angle^ and tei^
ininating in iht Base i to find the Base*
PROBIJSM XIZ.
To determine a Rightangled Triangle; Imving given the
Lengths of two Lines drawn from the acute angles^ to the
Middle of the opposite Sides.
PROBLEM Xllt.
To determine a Rightangled Triangle; hiiving giretl the
Perimeter^ »ad the Radius of its Inscribed Circle.
PROBLEM XIV.
T^o determine a Triangle ; having given the Base> the
Perpendicular, and the Ratio of the two Sides.
PROBLEM XV.
To determine a Rightangled Triangle ; having given the
H jpothenuse> and the Side of the biscribed Square.
* •
PROBLEM XVI.
To determine the Radii of three Equal Circles, described
in a giv^n Circle, to touch each other and also the Circum
ference of the given Circle.
^ PROBtEM XVlI.
In a Rightangled Triangle, having given the Perimeter, or
Sum of all the Sides, and the Perpendicular let fall from the
Right Angle on the Hypothenuse ; to deterihine the Tri
angle, that is, its Sides.
PROBLEM XVin..
To. determine a Right^angled Triangle; having given the
Hypothenuse, and the Difference of two Lines drawn from
tlie two acute* angles to the Centre of the Inscribed Ciicle.
PROBLEli
/"
' /
TO GEOMETRY. 365
PROBLEM XIX.
To determine a Triangle; having given the Base, the Per
pendicular, aikd the Difi^nce of tHe tgvvt) other Sides.
PROBLEM XX. •
To determine a Triangle; having given the Base, the Per
pendicular, and the Rectangle or Product of the two Sides.
N PROBLEM :^xi.
To determine a Triangle ; having given the Lengths of
diree Lines drawn from the three Angles, to the Midcfie o£
the opposite Sides.
PROBLEM XXII.
In. a Triangle, having given all the.thsee Sides; to find
the Radius or the Inscribed Circle.
PROBLEM XXIII.
To determine a Rightangled Triangle; having given the
Side of theInscribed Square, and the Radius of the Inscribed
Circle.
PROBLEM XXIV.
To determine a Triangle, and the Radius of the Inscribed
Circle ; having given the Lengths of three Lines drawn
from the three Angles, to tlie Centre of that Circle.
PROBLEM XXV.
To detefirSnc a Rightangled Triangle ; having given the
Hypothenuse, and the Radius of the Inscribed Circle.
PROBLEM XXVI
To determine a Triangle; having given the Base, the
line bisecting the Vertical Angle, and the Diameter of the
7 Circumscribing Circle.
LOGARITHMS
LOGARITHMS
OF THE
NUMBERS
tROtt
1 tQ 1000.
aBBSB
N.
Log.
N.
Log.
N.
Log.
N.
76
Log.
1
26
1414973
51
1707570
1880814
riT. ♦ »:rt:fi
«
2
03O103O
27
1431364
52
1716003
77
1886491
S
0*477121
28
1447158
53
172*276
78
18^2095
4
0*602060
29
146239S
54
1732394
79
1897627
5
0698970
'so
1
1477121
56
1740363
80
1903090
€
0778151
Isi
1491362
56
1748188
81
1908485
7
0845098
32
1505150
57
1755875
82
1913814
8
0903090
33
1518514
58
1763428.
83
1919078
9
0^54243
34
1531479
59
1770852
84
1924279
10
1000000
35
15440P8
60
1778151
85
1929419
11
1041393
36
1556303
«1
1785330
86
1934498
12
1'079181
37
1568202
62
1792392
87
1939519
13
Ml 3943
i38
1579784
63^
1799341
68
1944483
14
1146128
139
1591065
64
1 806 1 80
89
1949390
15
1176091
40
1602060
65
1812913
90
1954243
16
1204120
41
1612784
66
1819544
9iL
1959041
17
1 ^30449
42
1623249
67
1826075
92
1963788
IS
1255273
43
1633468
68
1832509
93
1968483
19
1 278754
44
1643453
69
1*838849
94
1973128
20
1301030
45
1653213
70
1845098
95
1977724
21
1822219
46
1662758
71
1851258
96
1982271
22
1342423
47
1 672098
72
1857833
97
1986772^
25
1361728
48
1681241
73
1863323
98
1991226
24
1380211
49
1690196
74
1869232
99
1995635
25
1*397940
50
1 698970
75
1875061
'100
2000000
N« B. In the following table, in the last nine colunlns of eech page, where
the first or leading figures change from 9's to O's, large dots are now in
troduced instead 9f the 0*s through the rest of the line, to catch the eye^
and to indicate that from thence the corresponding natural number ia
the^rst column stand's in the next 'lower line, and its annexed first tiiv'C*
figures of the Logarithm ia the second ^column;
LOftARrrHMS.
>67
N.'
1
2
3
4
5
6
7
8
9
100
DOOOOO
0434
0868
1301
1734
2166
2598
3029
3461
31891
101
4^4321
4751
5181
5609
6038
6466
6894
7321
7748
8174
102
8600
9026
9451
9876
•300
•724
1147
1570
1993
2415
103
012831^
3259
3680
4100
4521
4940
5360
5779
6197
6616
104
7033
7451
7868
8284
8700
9116
9532
9947
•361
•775
105
021189
1603
2016
2428
2841
3252
3664
4075
4486
4896
106
5306
5715
6125
6533
6942
7350
7757
8164
8571
8d78
107
9384
9789
• 195
•600
1004
1408
1812
2216
2619
3021
lOS
033424
3826
4227
4628
5029
5430
5830
6230
6;£29
7028
109
7426
7825
8223
8620
9017
9414
9811
•207
•602
•998
no
041393
1787
2182
2576
2969
3362
3755
4148
4540
4932
111
5323
5714
6105
6495
6885
7275
7664
8053
8442
8830
112
9218
9606
9993
•380
•766
1153
1538
1924
2309
2694
113
053078
3463
3846
4230
4613
4996
5378
5760
6142
6524
114
6905
7286
7666
8046
8426
8?<05
9185
9563
9942
•320
115
060698
1075
1452
1829
2206
2582
2958
3333
3709
4083
116
4458
4832
5206
5580
5953
6326
6699
7071
7443
7815
117
8186
8557
8928
9298
9668
••38
•407
•776
1145
1514
118
071882
2250
2617
2985
3352
3718
4085
4451
4816
5182
119
.5547
5912
6276
6640
7004
7368
7731
8094
8457
8819
120
9181
9543
9904
•266
•626
•987*
1347
1707
2p67
2426
121
082785
3144
3503
3861
4219
4576
4934
5291
5647
6004
122
6360
6716
7071
7426
7781
8136
8490
8845
9198
9552
123
9905
•258
•611
•963
1315
1667
2018
2370
2721
3071
124
093422
3772
4122
4471
4820
5169
5518
5866
6215
6562
,125
6910
7257
7604
,7951
8298
8644
8990
9335
9681
•026
126
100371
0715
1059
1403
1747
2091
2434
2777
3119
3462
127
3804
4146
4487
4828
5169
5510
5851
6191
6531
6871
128
7210
7549
7888
8227
8565
8903
9241
9579
9916
•253
129
110590
0926
1263
1599
1934
2270
2605
2940
3275
3609
130
3943
4277
4611
4944
5^78
5611
5943
6276
6608
6940
131
7271
7603
7934
8265
8595
8926
9256
9586
9915
•245
132
120574
0903
1231
1560
1888
2216
2544
2871
3198
3525
133
3852
4178
4504
4830
5156
5481
5806
6131
6456
6781
134
7M)5
7429
7753
8076
8399
8722
9045
9368
9690
••12
135
130334
0655
0977
1298
1619
1939
2260
2580
2900
3219
136
3539
3858
4177
4496
4814
5133
5451
5769
6086
6403
137
6721
7037
7354
7671
7987
8303
8618
8934
9249
9564 .
138
9879
• 194
•508
•822
1136
1450
1763
2076
2389
270^
139
143015
3327
3639
3951
4263
4574
4885
5196
5507
5818
140'
^ 6128
6438
6748
7058
7367
7676
7985
8294
8603
8911
141
9219
9527
9835
• 142
•449
•7^6
1063
1370
1676
1982
142
152288
2594
2900
3205
3510
3815
4120
4424
4728
5032
143
5336
5640
5943
6246
6549
6852
7154
7457
7759
8061
144
8362
8664
8^65
9266
9567
9868
• 168
%^^^
•769
1068
145
1^1368
1667
1967
2266
2564
2863
3161
3460
3758
4055
14^
,*4353
4650
4947
5244
5541
5838
6134
6430
6726
7022
147
7317
7613
7908
8203
8497
8792
9086
9380
9674
9968
148
170262
0555
0848
1141
1434
1726
2019
2311
2603
2895
149
3186
3478
3769
4060
4351
4641 4932
5222
5512
580i>
>
9
36S
LOGARITHBte.
N.
1
2
8
6959
4 I^J^l €
. 7 '
8
f ^
150
176091
6381
6670
7248
7536
7S25
8113
8401
86S9
151
8977
9264
9552
9839
• 126
•413
•699
•986
1272
1558
152
181844
2129
2415
2700
2985
3270
3555
3839
4123
4407
.153
4691
4975
5259
5542
5825
6108
6391
6674
6956
7239
154
7521
7803
80H4
8366
8647
8928'
9209
9490
9771
••51
155
190332
0612
0892
1171
1451
1730
2010
2289
2567
2846
156
3125
3403
3681
3959
4237
4514
4792
5069
5346
5623
151
5899
6l7d
6453
6729
7005
:728fl
7556
7832
8107
8382
158
8657
8932
9206
9481
9755
••29
•303
•577
•850
1124
159
201397
1670
1943
2216
2488
2761
3033
3305
3577
3848
160
4120
4391
4663
4934
5204
5475
5746
6016
6286 6556 1
161
6826
7096
7365
7634
7904
817'3
8441
8710
8979
9247
162
9515
9783
««51
•319
•586
•853
1121
1388
1654
1921
163
212188
2454
2720
2986
3252
3518
37»3
4049
4314
4579
.164
4844
5109
5373
5638
5902
6166
6430
6694
6957
7221
165
7484
7747
8010
8273
8536
8798
9060
9323
9585
9846
166
220108
0370
0631
0892
1153
1414
1675
1936
2196
2456
1G7
2716
2976
3236
3496
3755
4015
4274
4533
4792
5051
168
5309
5568
5826
6084
6342
6600
6858
7115
7372
7630
169
7887
8144
8400
8657
8913
9170
9426
9682
9938
• 1*93
no
230449
0704
0960
1215
1470
1724
1979
2234
2488
2743
171
2996
3250
3504
3757
4011
4264
4517
4770
5023
5276
172
5528
5781
6033
6285
6537
6789
7041
7292
7544
7795
173
8046
8297
8548
8799
9049
9299
9550
9800
••50
• 300
174
240549
0799
1048
1297
1546
1795
2044
2293
2541
2790
175
3038
3286
3534
3782
4030
4277
4525
4772
5019
'5266
176
5513
5759
6006
6252
6499
6745
6991
7237
7482
7728
177
7973
8219
8464
8709
8954
9198
.9443
9687
9932
• 116
178
250420
0664
0908
1151
1395
1638
1881
2125
2368
2610
179
2653
3096
3338
3580
3822
4064
4306
4548
4790
5031
180
5273
5514
5755
5996
6237
647'7
6718
6958
7198
7439
181
7679
7918
8158
8398
8637
8877
9116
9355
9594
9333
182
260071
0310
0548
0787
1025
1263
1501
1739
1976
2214
183
2451
2.688,
2925
3162
3399
3636
3873
4109
4346
4582
184
4818
5054
5290
5525
5761
5996
6232
6467
6702
6931
185
7172
7406
7641
7875
8110
8344
8580
8812
9046
9279
186
9513
9746
9980
•213
•446
•679
•912
1144
1377
1609
187
271842
2074
2306
2538
2770
3001
3233
3464
3696
3927
188
4158
4389
4620
4S50
5031
5311
5542
5772
6002
6232
189
6462
6692
6921
7151
7380
7609
7838
8067
8296
8525
190
8754
8982
9211
9439
9667
9895
• 123
•351
•578
•806
191
281033
1261
1488
1715
1942
2169
2396
2622
2849
3075
192
3301
3527
3753
3979
4205
4431
4656
4882
5107
5332
193
5557
5782
6007
6232
6456
6681
6905
7130
7354
7578
194
7802
8026
8249
8473
8696
8920
9143
9366
19589
9812
195
290035
0257
0480
0702
0925
1147
1369
1591
1813
2034
196
2256
2478
2699
2920
3141
3363
3584
3804
4025
4246
197
4466
4687
4907
5127
5347
5567
5787
6007
6226
6446
198
66()5
6884
7104
7323
7542
7761
7979
8198
841.6
8635
199
8863
9071
9289
9507
9725
9943
• 161
• 878
•595^*813
0^ NUMBERS
S09
N.
1
2
3
4
5
6
7
8
9
200
301030
1247
1464
TesT
1898
2114
2831
2547
2764
2980
201
S196
3412
3628
3844
4059
4275
4491
4706
4921
5136
202
5351
5566
5781
5996
6211
6425
6639
6854
7068
7282
203
7496
7710
7924
8137
8351
8564
8778
8991
9204
9417
204
9630
9843
..56
.268
.481
.693
.906
1118
1330
1542
205^
311754
1966
2177
2389
2600
2812
3023
3234
3445
3656
206
3867
4078
4289
4499
4710
4920
5130
5340
5551
5760
207
5970
6180
6390
6599
6809
7018
7227
7436
7646
7854
208
8063
8272
8481
8689
8898
9106
1
9314
9522
9730
9938
209
320146
0354
0562
0769
0977
1184
1391
1598
1805
2012
210
2219
2426
2633
2839
3046
3252
3458
3665
3871
4077
211
4282
4488
4694
4899
5105
5310
5516
5721
5926
6131
212
6336
6541
6745
6950
7155
7359
^563*
7767
7972
8176
213
8380
8583
8787
8991
9194
9398
9601
9805'
• • • o
.211
214
330414
0617
0819
1022
1225
1427
1630
1832
2034
2236
215
2438
2640
2842
3044
3246
3447
3649
3850
4b51
4253
216
4454
4655
4856
5057
5257
5458
5658
5859
6059
6g60
217
6460
6660
6860
7060
7260
7459
7659
7858
8058
8257
218
8456
3656
8855
9054
9253
9451
9650
9849
..47
.246
219
3404.44
0642
0841
1039
1237
1435
1632
1830
2028
2225
220
2423
2620
^817
3014
3212
3409
3606
3802
3999
4196
221
4392
4589
4785
4981
5178
5374
5570
5766
5964^
6157
922
6353
6549
6744
6939
7135
7330
7525
7720
7915
8110
223
8305
8500
86<i4
8889
9088
9278
9472
9666
9860
..54
224
350248
0442
0636
0829
1023
1216
1410
1603
179t5
1989
225
2183
.2375
2568
276J
2954
3147
3339
3532
3724
3916
226
4108
4301
4493
4685
4876
5068
,5260
545^
5643
5834
227
6026
6217
6408
6599
6790
6981
7172
7363
7554
7744
228
7935
8125
8316
8506
8696
8886
9076
9266
9456
9646
229
99S5
..25
.215
• 404
.593
.783
.972
1161
1350
1539
230
361728
1917
2105
2294
2482
2671
2859
3048
3236
3424
2*1
3612
3800
3988
4176
4363
4551
4789
4926
5113
5301
232
5488
5675
5862
6049
6236
6423
6610
6796
6988
7169
23S
7356
754i2
7729
7915
8101
8287
8473
8659
8845
9030
234
9216
9401
9587
9772
9958
. 143
.328
.513
.698
.883
235
371068
1253
1437
1622
1806
1991
2175
2360
2544
2728
23$
2912
3096
3280
3464
S647
3831
4015
4198
4382
4565
237
4748
4932
$115
5298
5481
5664
•5846
^m^
6212
6394
23%
} 6577
6759
6942
7124
7306
7488
7670
7852
8034
8216
?39^
8398
9580
876118943
9124
9306
9487
9668
9849
..30
240
3802 U
0392
0573
0754
0934
1115
1296
1476
1656
1837
241
2017
2197
2377
2557
2737
2917
3097
3277
3456
3636
248
8815
3995
♦174
4353
4533
4712
4891
5070
5249
5423
243
5606
5785
b^^^
6142
6321
6499
6677
6856
7034
7212
244
•390
7568.
7746
7923
8101
S279
8456
8634
8811
8989
245
.9166
9343
9520
9698
9875
..b\
.228
.405
.5&2
.759
246
S90935
U12
1288
1464
1641
1817
1993
2169
234^
2521
847
2697
2873
3048
3224
3400
3575
3751
39.2(5
41Q1
4277
248
^U52
4627
4802
4977
5152
5326
5501
5676
5850
6025
249
6199
6374
6548
6722
6896
7071
7245
7419
7592
7766
. VJ
QUU.Ii
\
JBl
}
I
»'
:rtor
L0GAR1THM&
K
1
2
3
4 ^
5
6
7
8
9
t50
397940
8114
8287
8461
8634
8808
8961
9154
9328
9501
^51
9674
9847
..20
.192
.365
.538
.711
.883
1056
1228
252
401401
1573
1745
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1375
438
, 1474
1573
1672
1771
1871
1970
2069
2168
2267
2366
439
2465
2563
2662
2761
2860
2959
3058
3166
3255
3354
440
3453
3551
3650
3749
3847
3946
4044
4143
4242
4340.
441
4439
4537
4636
4734
4832
4931
'5029
5127
5226
5324
442
5422
5521
5619
5717
5815
5913
6011
6110
6208
6306
4'43
6404
6502
6600
6698
6796
6894
6992
7089
7187
^2&\
444
7383
7481
7579
7676
7774
7872
7969
8067
8165
S262
445
8360
8458
8555
8653
8750
8848
894S
9043
9140
'9237
446
9335
9432
9530
9627
9724
98^1
991,9
..1=6
. 113
.21,0.
447
650308
04O5
0502
0599
0696
0793
0890
0987
1084
ilfi
448
1278
1375
1472
1569
1666
1762
1^59
1956
3053
2150
,449
2246
2343
2440^25361
2633
2730
2826
2923
3019
3116
374
LOGARITHMS
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496.
497
149^
499
653213
4177
5138
6098
7056
8011
8965
9916
660865
1813
2758
3701
4642
5581
6518
7453
8386
9317
670246
1173
2098
3021
3942
4861
5778
6694
7607
8518
9428
680336
1241
2145
3047
3947
4845
5742
6636
7529
«420
9309
690196
1081
1965
2847
3727
4605
5482
6356
7229
8101
3309
4273
5235
6194
7152
8107
9060
..U
0960
1907
2852
3795
4736
5675
6612
7546
8479
9410
0339
1265
2190
3113
4034
4953
5870
6785
7698
8609
9519
0426
1332
2235
3137
4037
4935
5831
6726
7618
8509
9398
0285
1170
2053
2935
3815
4693
5569
6444
7317
8188
2
3405
4369
5331
6290
7247
8202
9155
. 106
1055
2002
2947
3889
4830
5769
6705
7640
8572
9503
0431
1358
2283
3205
4126
5045
5962
6876
7789
8700
9610
0517
1422
2326
3227
4127
5025
5921
6815
7707
8598
9486
0373
1258
2142
3023
3903
4781
5657
6531
7404
8275
3502
4465
5427
6386
7343
8298
9250
• 201
1150
2096
3041
3983
4924
5862
6799
7733
8665
9596
0524
1451
2375
3297
4218
5137
6053
6968
7881
8791
9700
0607
1513
2416
3317
4217
5114
6010
6904
7796
8687
9575
0462
1347
2230
3111
3991
4868
5744
6618
7491
8362
3598
4562
5526
6482
7438
8393
9346
.296
1245
2191
3135
4078
5018
5956
6S92\
7826
9759
9689
0617
1543
2467
3390
4310
5228
6145
7059
7972
8882
9791
0698
1603
2506
3407
4307
5204
6100,
6994
7886
8776
9664
0550
1435
2318
3199
4078
4956
5832
6706
7578
8449
3695
4658
5619
6577
7534
8488
9441
.391
1339
2286
3230
4172
5112
6050
6986
7920
8852
9782
0710
1636
2560
3482
4402
5320
6236
7151
8063
8973
9882
0789
1693
2596
3497
4396
5294
6189
7083
7975
8865
9753
0639
1524
2406
3287
4166
5044
5919
6793
7665
8535
3791
4754
5715
6673
7629
8584
9536
.486
1434
2380
3324
4266
5206
6143
7079
8013
8945
9875
0802
1728
2652
tS574
4494
5412
6328
7242
8154
9064
9973
0879
1784
2686
3587
4486
5383
6279
7172
8064
8953
9841
0728
1612
2494
3375
4254
5131
6007
6880
7752
8622
38^8
4850
5810
6769
7725
8679
9631
.531
1529
2475
3418
4360
5299
6237
7173
8106
9038
9967
0895
1821
2744
3666
4586
5503
6419
7333
8245
9155
..63
0970
1874
2777
3677
4576
5473
6368
7261
8153
9042
9930
0816
1700
2583
3463
4342
5219
6094
696S
7839
8709
Tl
3984
4946
5906
6864
7820
8774
9726
.676
1623
2569
3512
4454
5393
6331
7266
8199
9131
.60
0988
1913
2836
3758
4677
5595
6511
7424
8336
9246
. 154
1060
1964
2867
3767
4666
5563
6458
7351
8242
9131
. .19
0905
1789
2671
3551
4430
5307
6182
7055
7926
4080
5042
6002
6960
7916
8870
9821
.771
1718
2663
3607
4548
5487
6424
7360
8293
9224
. 153
1080
2005
2929
3850
4769
5687
6602
7516
8427
9337
.245
1151
2055
2957
3857
4756
5652
6547
7440
8331
9220
. 107
0993
1877
2759
3639
4517
5394
6269
7142
8014
879618883
OP NUMBERS.
915
N.
500
698970
501
9838
502
700704
503
1568
504
2431
505
3291
506
4151
507
5008
508
5864
509
6718
510
7570
511
8421
5152
9270
51S
710117
514
0963
515
1807
516
2650
517
. 3491
518
4330
519
5167
520
6003
521
6838
5^2
7671
525
8502
524
9331
525
720159
526
0986
527
1811
528
2634
529
3456
530
4276
531
5095
532
5912
533
6727
534
7541
5^5
8354
536
9165
537
9974
538
730782
539
1589
540
2394
541
3197
542
3999
543
4800
544
5599
545
6397
546
7193
547
7987
548
8781
549
9572
9037
9924
0790
1654
2517
3377
4236
5094
5949
6803
7655
8506
9355
0202
1048
1892
2734
3575
4414
5251
6087
6921
7734
8585
9414
0242
1068
1893
2716
3538
4358
5176
5993
68Q9
7623
8435
9246
..55
0863
1669
2474
3278
4079
4880
5679
6476
7272
8067
8860
9651
9144
..11
0877
1741
2603
3463
4322
5179
6035
6888
7740
8591
9440
0287
1132
1976
2818
3659
4497
5335
6170
7004
7837
sees
9497
0325
1151
1975
2798
3620
4440
5258
6075
6390
7704
8516
9327
. 136
0944
1750
2555
3358
4160
4960
5759
6556
7352
8146
8939
9731
9231 9317
98 . 184
0963 1050
1827 1913
2689 2775
3549 3635
4408
5265
6120
6974
7826
8676
9524
0371
1217
2060
2902
3742
4581
5418
6254
7088
7920
87511^834
9580
0407
1233
2058
2881
3702
4522
5340
6156
6972
7785
8597
9408
.217
1024
1830
2635
3433
4240
5040
5838*
6635
7431
8225
9018
9810
4 I 5
9404
.271
1136
1999
2861
3721
4494 1 4579
5330
6206
7059
7911
8761
9609
0456
1301
2144
2986
3826
4665
5502
6337
7171
8003
9663
0490
1316
2140
2963
3784
4604
5422
6238
7053
7866
8678
9489
.298
1105
1911
2715
3518
4320
5120
5918
6715
7511
8305
9097
9889
6
5436
6291
7144
7996
8846
9694
0540
1385
2229
3070
3910
4749
5586
6421
7254
8086
8917
9745
0573
1398
2222
3045
3866
4685
5503
6320
7134
7948
8759
9570
.378
1186
1991
12796
3598
4400
5200
5998
6795
7590
8384
9177
9968
9491
.338
1222
2086
2947
3807
4665
3522
6376
7229
8081
8931
9779
0625
1470
2313
3154
3994
4833
5669
6504
7338
8169
9000
9828
0655
1481
2305
3127
3948
4767
5585
6401
7216
8029
8841
9651
.459
1266
2072
2876
3679
4480
5279
6078
6874
7670
8463
9256
..47
9578
.444
1309
2172
3033
3393
4751
5607
6462
7313
8166
9015
9863
0710
1554
2397
3238
4078
4916
5753
6588
^421
8253
9083
9911
0738
1563
2387
3209
4030
4849
5667
6483
7j397
8110
8922
^732
•540
1347
2152
2956
3759
4560
5359
6157
6954
7749
8543
9335
. 126
9664
.•531
1395
2258
3119
3979
4837
5693
6547
7400
8251
9^00
9948
0794
1639
2481
3326
4162
5000
5836
6671
7504
8336
9165
9994
0821
1646
2469
3291
4112
4931
5748
6564
7379
8191
9003
9813
• 621
1428
2233
3037
3839
4640
5439
6237
7034
7829
8622
9414
.205
19751
.617
1482
2344
3205
4065
4922
^778
6632
7485
8336
9185
..33
0879
1723
2566
3407
4246
5084
5920
6754
7587
8419
9248
..77
0903
1728
2552
3374
4194
5013
5830
6646
7460
8273
9084
9893
.702
1508
2313
3117
3919
4720
5519
6317
7113
7908
8701^
9493
.284
37«
LOGARITHMS
JN.
550
1
2
3
4
5
6
7
8
9
740363
0442
0521
0560
0678
0757
0836
0915
0994
1073
551
1152
lli30
1309
1388
1467
1546
1624
1703
1782
1860
552
1939
2018
2096
2175
2254
2332
2411
2489
2568
2646
553
2725
2304
2S82
2961
3039
31*18
3196
3275
3353
3431
554
3510
35S8
3667
3745
3823
3902
3980
4058
4136
4215
655
4293
4371
4449
4528
4606
4684
4762
4840
4919
4997
556
5075
5153
5231
5309
5387
5465
5543
5621
5699
5777
557
^ 5855
5933
6011
6089
6167
6245
6323
6401
6.479
6556
558
6634
6712
6790
6868
6945
7023
7101
7179
7256
7334
559
7412
7489
7567
7645
7722
7800
7878
7955
8033
8110
560
8188
8266
8343
8421
8498
8576
8653
8731
8808
8885
561
8.963
9040
9118
9195
9272
9350
9427
9504
9582
9659
562
9736
9814
9891
9968
..45
. 123
.200
.277
.354
.431
56'3
750508
0586
0663
0740
0817
0894
0971
1048
1125
1202
564
1279
1356
1433
1510
1587
1664
1741
1818
1895
1972
565
2048
2125
2202
2279
2356
2433
2509
2586
2663
2740
566
28(6
2893
2970
3047
3123
S200
3277
3353
3430
3506
567
3583
3660
3736
3813
3889
3966
4042
4119
4195
4272
568
4348
4425
4501
4578
4654
4730
4807
4883
4960
5036
569
5112
5189
5265
5341
5417
5494
5570
5646
5722
5799
570
5875
5951
6027
6103
6180
6256
6332
6408
6484
6560
571
6636
67J2
6788
6864
6940
7016
7092
716a
7244
7320
572
7396
7472
7548
7624
7700
7775
7851
7J927
8003
8079
573
8155
8230
8306
8382
845«
8533
8609
8685
8761
8836
574
8912
898^
9063
9139
9214
9290
9366
9441
9517
9592
575
9668
9743
9819
9894
9970
..45
. 121
. 196
.272
.347
576
760422
0498
0573
0649
0724
0799
0875
0950
1025
1101
577
1176
1251
1326
14p2
1477
1552
1627
1702
1778
1853
578
1928
2003
2078
2153
2228
2303
2378
2453
2529
2604
579
2679
2754
2829
2904
2978
3053
3128
3203
3278
3353
580
3428
3503
3578
3653
S727
3802
3877
3952
4027
4101
581
4176
4251
4326
4400
4475
45 5Q
4624
4699
4774
4848
582
4923
4998
5072
5147
5221
5296
5370
5445
5520
5594
583
5669
5743
5818
5892
5966
6041
6115
6190
6264
6338
584
6413
6487
6562
6&36
6710
6785
6859
6933
7007
7082
585
7156
723Q
7304
7379
7453
7527
7601
7675
7749
7823
586
7898
7972
8046
8120
8194
8268
8342
8416
8490
8564.
587
8668
8712
8786
8860
8934
9003
9082
9156
9230
9303
588
9377
9451
9525
9599
9673
9746
9820
9894
9968
..42
589
770115
0189
0263
0336
0410
0484
0557
0631
0705
0778
590
0852
0926
0999
1073
1146
1220
1293
1367
1440
1514
591
1587
1661
1734
1808
1881
1955
2028
2102
2175
2248
992
2322
2395
^468
2542
2615
2688
2762
^835
2908
2981
593
3a55
3128
3201
3274
3348
3421
3494
3567
3640
3713
594
3786
3860
3933
4006
4079
4152
4225
4298
4371
595
4517
4590
4663
47S6
4809
4882
4955.
5028
5100
5173
596
5246
5319
5392
5465
5533
561^1
M»S
5756
5829
5902
597
5974
6047
6120
6193
6265
6398
64ii
7l»T
6483
6556
6629
598
6701
6774
6846
6919
6991^
7064
7209
7282
7354
599
7427 7499
7572
7644
7717
7789
7862
7934
8006
8079
'▼,
OF NUMBftftS.
3l^
600
O
1 2
3
4
5
6 7
8
9
778151
8224
8296
8368
8441
8613
85851
8658
8lf30
8802
601
8874
8947
9019
9091
9163
9236
9308
9380
9452
9524
602
^596
9669
9741
9813
9885
9957
..29
.101
.173
. 245
603
7S0317
0389
0461
0533
0605
0677
0749
0821
0893
0965 '
^4
1037
1109
1181
I25;i
1324
1396
1468
1540
f6l2
16^4
605
17S5
1827
1899
1971
2042
2114
2186
3258
2329
2401
.606
2473
2544
2616
2688
2759
2831
2902
2974
3046
3117
607
3(89
3260
3332
3403
3475
3546
3618
3689
3761
3832
60d
3904
3975
4046
4118
4189
4261
4332
4403
4475
4546
609
4617
4689
47G0
4831
4902
4974
5045
^116
518*7
5259
610
5330
.5401
5472
5543
5615
sese
5757
5828
5899
5970 '
611
6041
6112
6183
6^54
6325
6396
6467
6538
6609
6680
612
6751
6S22
6893
6964
7035
7106
nil
7248
7319
7390
613
7460
7531
7602
7673
7744
7815
7885
7956
8027
9098
614
8168
8269
8310
8381
8451
8522
8593
8663
8734
8804
615
8875
8946
9016
9087
9157
9228
9299
9369
9440
9510
616
9581
9651
9722
9792
9863
9933
...4
..74
.144
.215
617
790285
0356
0426
0496
0567
0637
0707
0778
0848
0918
618
0988
1059
1129
1199
1269
1340
1410
1480
1550
1620
619
1691
1761
1831
1901
1971
2041
2111
2181
2i25«
2322
620
2892
2462
2532
2602
2672
2742
2812.
2862
2952
3022
621
3092
3162
3231
3301
3371
3441
3511
3581
3651
3721
622
3790
3860
3930
4000
4070
4139
4209
4279
4349
4418
623
4488
4558
4627
4697
4767
4836
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