/*;
ALGEBRA
AN
ELEMENTARY TEXT BOOK
Uniform with Part I
PART II
Completing the Work and containing an Index
to both Parts.
640 pp., post 8vo.
BY THE SAME AUTHOR
AN INTRODUCTION TO ALGEBRA
For the Use of Secondary Schools and Technical
Colleges.
Third Edition. Crown 8vo.
Or may he had in two separate Parts.
I have kept the fundamental principles of the
subject well to the front from the very beginning.
At the same time I have not forgotten, what
every mathematical (and other) teacher should
have perpetually in mind, that a general proposi
tion is a property of no value to one that has not
mastered the particulars. The utmost rigour of
accurate logical deduction has therefore been less
my aim than a gradual development of algebraic
ideas. In arranging the exercises I have acted
on a similar principle of keeping out as far as
possible questions that have no theoretical or
practical interest. — Preface.
AGENTS
America . . The Macmillan Company
60 Fifth Avenue, New York
Australasia . Oxford University Press
205 Flinders Lane, Melbourne
Canada . . The Macmillan Co. of Canada, Ltd.
St. Martin's House, 70 Bond Street, Toronto, 2
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276 Hornby Road, Bombay
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North Beach Road, Madras
ALGEBEA
AN ELEMENTAEY TEXTBOOK
FOR THE
HIGHER CLASSES OF SECONDARY SCHOOLS
AND FOR COLLEGES
BY
G. CHRYSTAL, M.A., LL.D.
HONORARY FELLOW OF CORPUS CHRISTI COLLEGE, CAMBRIDGE ;
PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF EDINBURGH
PART I.
FIFTH EDITION
A. & C. BLACK, Ltd.
4, 5 & 6 SOHO SQUARE, LONDON, W.l.
1926
Printed in Great Britain.
'■ I should rejoice to see . . . morphology introduced into tha
elements of Algebra. '— Sylvester
fa. i
9j
PtihlUhed July 1S86
Reprinted, with corrections and additions, 1889
New impressions 1893 and 1898
Reprinted, with corrections and additions, 1904 and 1910
Reprinted in 1920 and 1926
PKEFACE TO THE FIFTH EDITION.
In this Edition considerable alterations have been made
in chapter xii. In particular, the proof of the theorem
that every integral equation has a root has been amplified,
and also illustrated by graphical considerations.
An Appendix has been added dealing with the general
algebraic solution of Cubic and Biquadratic Equations ;
with the reducibility of equations generally ; and with the
possibility of solution by means of square roots. As the
theorems established have interesting applications in Ele
mentary Geometry, it is believed that they may find an
appropriate place in an Elementary work on Algebra.
G. CHEYSTAL.
29th June 1904.
PEEFACE TO THE SECOND EDITION.
The comparatively rapid sale of an edition of over two
thousand copies of this volume has shown that it has, to
some extent at least, filled a vacant place in our educational
system. The letters which I have received from many
parts of the United Kingdom, and from America, containing
words of encouragement and of useful criticism, have also
strengthened me in the hope that my labour has not been
VI PREFACE
in vain. It would be impossible to name here all the
friends who have thus favoured me ; and I take this oppor
tunity of offering them collectively my warmest thanks.
The present edition has been thoroughly revised and
corrected. The first chapter has been somewhat simplified ;
and, partly owing to experience with my own pupils, partly
in consequence of some acute criticism sent to me by Mr.
Levett of Manchester, the chapters on Indices have been
recast, and, I think, greatly improved. In the verification
and correction of the results of the exercises I have been
indebted in a special degree to the Rev. John Wilson,
Mathematical Tutor in Edinburgh.
The only addition of any consequence is a sketch of
Horner's Method, inserted in chapter xv. I had originally
intended to place this in Part II. ; but, acting on a sugges
tion of Mr. Hayward's, I have now added it to Part I.
To help beginners, I have given, after the table of
contents, an index of the principal technical terms used in
the volume. This index will enable the student to turn up
a passage where the " hard word " is either defined or other
wise made plain.
G. CHKYSTAL.
Edinburgh, 11th October 1889.
PREFACE TO THE FIRST EDITION.
The work on Algebra of which this volume forms the first
part, is so far elementary that it begins at the beginning of
the subject. It is not, however, intended for the use of
absolute beginners.
The teaching of Algebra in the earlier stages ought to
consist in a gradual generalisation of Arithmetic ; in other
words, Algebra ought, in the first instance, to be taught as
Aritlimetica Universalis in the strictest sense. I suppose
that the student has gone in this way the length of, say, the
solution of problems by means of simple or perhaps even
cpiadratic equations, and that he is more or less familiar
with the construction of literal formulae, such, for example,
as that for the amount of a sum of money during a given
term at simple interest.
Then it becomes necessary, if Algebra is to be any
thing more than a mere bundle of unconnected rules, to
lay down generally the three fundamental laws of the
subject, and to proceed deductively — in short, to introduce
the idea of Algebraic Form, which is the foundation of all
the modern developments of Algebra and the secret of analy
tical geometry, the most beautiful of all its applications.
Such is the course followed from the beginning in this
work.
Vlll PREFACE
As mathematical education stands at present in this
country, the first part might be used in the higher classes
of our secondary schools and in the lower courses of our
colleges and universities. It will he seen on looking through
the pages that the only knowledge required outside of
Algebra proper is familiarity with the definition of the
trigonometrical functions and a knowledge of their funda
mental additiontheorem.
The first object I have set before me is to develop
Algebra as a science, and thereby to "increase its usefulness
as an educational discipline. I have also endeavoured so
to lay the foundations that nothing shall have to be un
learned and as little as possible added when the student
comes to the higher parts of the subject. The neglect of
this consideration I have found to be one of the most
important of the many defects of the English text books
hitherto in vogue. Where immediate practical application
comes in question, I have striven to adapt the matter to
that end as far as the main general educational purpose
would allow. I have also endeavoured, so far as possible,
to give complete information on every subject taken up, or,
in default of that, to indicate the proper sources ; so that
the book should serve the student both as a manual and
as a book of reference. The introduction here and there of
historical notes is intended partly to serve the purpose just
mentioned, and partly to familiarise the student with the
great names of the science, and to open for him a vista
beyond the boards of an elementary textbook.
As examples of the special features of this book, I may
ask the attention of teachers to chapters iv. and v. With
respect to the opening chapter, which the beginner will
PREFACE IX
doubtless find the hardest in the hook, I should mention
that it was written as a suggestion to the teacher how
to connect the general laws of Algebra with the former
experience of the pupil. In writing this chapter I hud to
remember that I was engaged in writing, not a book on the
philosophical nature of the first principles of Algebra, but
the first chapter of a book on their consequences. Another
peculiarity of the work is the large amount of illustrative
matter, which I thought necessary to prevent the vagueness
which dims the learner's vision of pure theory ; this has
swollen the book to dimensions and corresponding price
that require some apology. The chapters on the theory of
the complex variable and on the equivalence of systems of
equations, the free use of graphical illustrations, and the
elementary discussion of problems on maxima and minima,
although new features in an English text book, stand so
little in need of apology with the scientific public that I
offer none.
The order of the matter, the character of the illustra
tions, and the method of exposition generally, are the result
of some ten years' experience as a university teacher. I
have adopted now this, now that deviation from accepted
English usages solely at the dictation of experience. It
was only after my own ideas had been to a considerable
extent thus fixed that I did what possibly I ought to have
done sooner, viz., consulted foreign elementary treatises.
I then found that wherever there had been free considera
tion of the subject the results had been much the same.
I thus derived moral support, and obtained numberless hints
on matters of detail, the exact sources of which it would be
difficult to indicate. I may mention, however, as specimens
PREFACE
of the class of treatises referred to, the elementary text
books of Baltzer in German and Collin in French. Anion"
the treatises to which I am indebted in the matter of theory
and logic, I should mention the works of De Morgan, Pea
cock, Lipschitz, and Serret. Many of the exercises have
been either taken from my own class examination papers
or constructed expressly to illustrate some theoretical point
discussed in the text. For the rest I am heavily indebted
to the examination papers of the various colleges in Cam
bridge. I had originally intended to indicate in all cases
the sources, but soon I found recurrences which rendered
this difficult, if not impossible.
The order in which the matter is arranged will doubt
less seem strange to many teachers, but a little reflection
will, I think, convince them that it could easily be justified.
There is, however, no necessity that, at a first reading, the
order of the chapters should be exactly adhered to. I think
that, in a final reading, the order I have given should be
followed, as it seems to me to be the natural order into
which the subjects fall after they have been fully com
prehended in their relation to the fundamental laws of
Algebra.
With respect to the very large number of Exercises,
I should mention that they have been given for the con
venience of the teacher, in order that he might have, year
by year, in using the book, a sufficient variety to prevent
mere rotework on the part of his pupils. I should much
deprecate the idea that any one pupil is to work all the
exercises at the first or at any reading. We do too much
of that kind of work in this country.
I have to acknowledge personal obligations to Professor
TREFACE XI
Tait, to Dr. Thomas Muir, and to my assistant, Mr. R E.
Allardice, for criticism and suggestions regarding the
theoretical part of the work ; to these gentlemen and to
Messrs. Mackay and A. Y. Fraser for proof reading, and
for much assistance in the tedious work of verifying the
answers to exercises. In this latter part of the work I
am also indebted to my pupil, Mr. J. Mackenzie, and to
my old friend and former tutor, Dr. David Rennet of
Aberdeen.
Notwithstanding the kind assistance of my friends and
the care I have taken myself, there must remain many
errors both in the text and in the answers to the exercises,
notification of which either to my publishers or to myself
will be gratefully received.
G. CHRYSTAL.
Edinburgh, 2&h June 1880.
CONTENTS.
CHAPTER I.
FUNDAMENTAL LAWS AND PROCESSES OF ALGEBRA.
PAGE
Laws of Association and Commutation for Addition and Subtraction 27
Essentially Negative Quantity in formal Algebra ... 8
Properties of . . . . . . . . 11
Laws of Commutation and Association for Multiplication . . 12
Law of Distribution . . . . . . .13
Laws of Association, Commutation, and Distribution for Division . 1419
Properties of 1 . . . . . . . . 17
Synoptic Table of the Laws of Algebra .... 20
Exercises I. ........ 22
Historical Note ....... 24
CHAPTER II.
MONOMIALS — LAWS OF INDICES — DEGREE.
Laws of Indices
Theory of Degree, Constants and Variables
Exercises II. .
2529
30
31
CHAPTER III.
THEORY OF QUOTIENTS — FIRST PRINCIPLES OF THEORY OF NUMBERS.
Fundamental Properties of Fractions and Fundamental Operations
therewith ......
Exercises III. ......
Prime and Composite Integers ....
Arithmetical G.C.M. .
Theorems on the Divisibility of Integers
Remainder and Residue, Periodicity for Given Modulus
Arithmetical Fractionality ....
The Resolution of a Composite Number into Prime Factors is unique
General Theorem regarding G.C.M., and Corollaries .
The Number of Primes is infinite
Exercises IV. .
3336
36
38
39
41
42
43
44
44
47
48
XIV
CONTENTS
CHAPTER IV.
DISTRIBUTION OF PRODUCTS ELEMENTS OF THE THEORY OF RATIONAL
INTEGRAL FUNCTIONS.
Generalised Law of Distribution .....
Expansion by enumeration of Products ; Classification of the Products
of a given set of letters into Types ; S and II Notations
Principle of Substitution
Theorem regarding Sum of Coefficients
Exercises V.
General Theorems regarding the Multiplication of Integral Functions
Integral Functions of One Variable
Product of Binomials
Binomial Theorem ....
Detached Coefficients
Addition rule for calculating Binomial Coefficients, with a General
isation of the same
x n±y7i as a Product ....
Exercises VI. .
Exercises \Il. ....
Homogeneity ....
General forms of Homogeneous Integral Functions
Fundamental Property of a Homogeneous Function
Law of Homogeneity
Most general form for an Integral Function
Symmetry .....
Properties of Symmetric and Asymmetric Functions
Rule of Symmetiy ....
Most general forms of Symmetric Functions
Principle of Indeterminate Coefficients
Table of Identities ....
Exercises VIII.
PAGE
49
5154
54
55
56
57
5969
60
61
64
66
68
69
70
7175
72
73
74
75
7579
76
77
78
79
81
83
CHAPTER V.
TRANSFORMATION OF THE QUOTIENT OF TWO INTEGRAL FUNCTIONS.
Algebraic Integrity and Fractionality ....
Fundamental Theorem regarding Divisibility .
Ordinary DivisionTransformation, Integral Quotient, Remainder
Binomial Divisor, Quotient, and Remainder .
Remainder Theorem ......
Factorisation by means of Remainder Theorem
Maximum number of Linear Factors of an Integral Function of a;
New basis for the Principle of Indeterminate Coefficients
Continued Division ......
85
86
8693
93
96
97
98
99
102
CONTENTS
XV
One Integral Function expressed in powers of another .
Expansion in the form Ao+Ai(aj — ai) + A%(x — «x)(a; — a2) + A$(xai)
(x(t2){xa 3 )+ . . .
Exercises IX. .
CHAPTER VI.
GREATEST COMMON MEASURE AND LEAST COMMON MULTIPLE.
G.C.M. by Inspection ....
Ordinary process for Two Functions .
Alternate destruction of highest and lowest terms
G.C.M. of any number of Integral Functions .
General Proposition regarding the Algebraical G.C.M.,
regarding Algebraic Primeness
L.C.M.
Exercises X. .
CHAPTER VII.
PAOE
105
107
108
with Corollaries
112
113
117
119
119
122
124
FACTORISATION OF INTEGRAL FUNCTIONS.
Tentative Methods . . . . . . .126
General Solution for a Quadratic Function of a; . . 128137
Introduction of Surd and Imaginary Quantity . . 130133
Progression of Real Algebraic Quantity .... 130
Square Root, Rational and Irrational Quantity . . . 132
Imaginary Unit ....... 132
Progression of Purely Imaginary Quantity .... 133
Complex Quantity ....... 133
Discrimination of the different cases in the Factorisation of
ax' 2 + bx + c ....... 134
Homogeneous Functions of Two Variables . . . .136
Use of the Principle of Substitution . . . . .136
Use of Remainder Theorem . . . . . .138
Factorisation in general impossible . . . . .139
Exceptional case of ax 2 + 2hxy + by 2 + 2gx + 2fy + c . . .140
Exercises XI. . . . . . . . .142
CHAPTER VIII.
RATIONAL FRACTIONS.
General Propositions regarding Proper and Improper Fractions 144147
Examples of Direct Operations with Rational Fractions . 147150
Inverse Method of Partial Fractions .... 151159
General Theorem regarding decomposition into Partial Fractions . 151
Classification of the various species of Partial Fractions, with
Methods for determining Coefficients . . . .153
Integral Function expansible in the form 2{a u + ai(x a)+ . . . +
a r . 1 (xa) 1 — 1 }(xpY(xy) t . . . ..... 155
Exercises XII. ....... 159
XVI
CONTENTS
CHAPTER IX.
FURTHER APPLICATIONS TO THE THEORY OF NUMBERS.
TAOE
Expression of an Integer as an Integral Factorial Series . . 163
Expression of a Fraction as a Fractional Factorial Series . . 165
Scales of Arithmetical Notation .... 167175
Expression of an Integer in a Scale of given Radix . . . 167
Arithmetical Calculation in various Scales . . . .169
Expression of any Fraction as a Radix Fraction . . .170
Divisibility of a Number and of the Sum of its Digits by r  1 ; the
"Nine Test" ....... 174
Lambert's Theorem . . . . . .176
Exercises XIII. ....... 177
CHAPTER X.
IRRATIONAL FUNCTIONS.
Interpretation of scW" ....... 180
Consistency of the Interpretation with the Laws of Indices Examined 182
Interpretation of a; . . . . . . .185
Interpretation of x~ m ....... 186
Examples of Operation with Irrational Forms . . 187189
Rationalising Factors ...... 189198
Every Integral Function of \Jp, \Jq, \Jr, &c, can be expressed in
the linear form A + B\Jp + C\/q + 'D\Jr + . . . + ~E\/pq + . . .
+Y"sfpqr+. . . . . . . . .193
Rationalisation of any Integral Function of \Jp, \Jq, \Jr, &c. . 195
Every Rational Function of \/p, \Jq, \/r, &c, can be expressed in
linear form . . . . . . .196
General Theory of Rationalisation .... 197198
Exercises XIV. ....... 199
Historical Note' ....... 201
CHAPTER XL
ARITHMETICAL THEORY OF SURDS.
Algebraical and Arithmetical Irrationality
Classification of Surds .....
Independence of Surd Numbers
Expression of \/{a + \/b) in linear form
Rational Approximations to the Value of a Surd Number
Extraction of the Square Root
Square Root of an Integral Function of x
Extraction of Roots by means of Indeterminate Coefficients
Exercises XV. ......
203
204
205207
207
210215
211
215
217
218
CONTENTS
XV11
CHAPTER XII.
COMPLEX NUMBERS.
PACK
Independence of Heal and Imaginary Quantity . . . 221
Twofolduess of a Complex Number, Argand's Diagram . . 222
If x + yi = x' + y'i, then x = x', y = y' ..... 224
Every Rational Function of Complex Numbers is a Complex Number 224227
If <(>(x + yi) = X + Yi, then $(xyi)=XYi ; if <p(x + yi) — 0, then
<f>(xyi) = ....... 226
Conjugate Complex Numbers ...... 228
Moduli. ........ 229
If x + yi = 0, then  x + yi  = ; and conversely . . . 229
 <p(x + yi)  = \/{<P(x + yi)<P(xyi)}; Particular Cases . . . 2l;0
The Product of Two Integers each the Sum of Two Squares is the
Sum of Two Squares ...... 230
Discussion by means of Argand's Diagram . . . 232236
Every Complex Number expressible in the form r(cos 6 + i sin d) ;
Definition of Amplitude ...... 232
Addition of Complex Numbers, Addition of Vectors . . 233
\z, + z 2 + . . . + z n \^\z 1 \ + \z 2 \ + . . . +\z H \ . . .234
The Amplitude of a Product is the Sum of the Amplitudes of the
Factors ; Demoivre's Theorem ..... 235
Root Extraction leads to nothing more general than Complex
Quantity 236244
Expression of ij{x + yi) as a Complex Number . . . 237
Expression of Mx + yi) as a Complex Number . . . 238
Every Complex Number has n nth roots and no more . . 240
Properties of the nth roots of ±1 ..... 240
Resolution of x"± A into Factors ..... 243
Every Integral Equation has at least one root ; Every Integral
Equation of the ?ith degree has n roots and no more ; Every
Integral Function of the nth degree can be uniquely resolved
into n Linear Factors ..... 244250
Upper and Lower Limits for the Roots of an Equation . . 247
Continuity of an Integral Function of z . . . 248
Equimodular and Gradient Curves of/(r) .... 248
Argand's Progression towards a Root ..... 249
Exercises XVI. 251
Historical Note ....... 253
CHAPTER XIII.
RATIO AND PROPORTION.
Definition of Ratio and Proportion in the abstract
Propositions regarding Proportion
Examples .....
Exercises XVII. ....
VOL. I
255
257264
264266
267
b
XV111
CONTENTS
Ratio and Proportion of Concrete Quantities
Definition of Concrete Ratio
Difficulty in the case of Incommensurables
Euclidian Theory of Proportion
Variation ....
Independent and Dependent Variables
Simplest Cases of Functional Dependence
Other Simple Cases .
Propositions regarding "Variation"
Exercises XVIII.
CHAPTER XIV.
PAOE
26827.3
269
270
272
273279
273
274
275
276
279
ON CONDITIONAL EQUATIONS IN GENERAL.
General Notion of an Analytic Function
Conditional Equation contrasted with an Identical Equation
Known and Unknown, Constant and Variable Quantities
Algebraical and Transcendental Equations ; Classification of Integral
Equations .....
Meaning of a Solution of a System of Equations
Propositions regarding Determinateness of Solution
Multiplicity of Determinate Solutions
Definition of Equivalent Systems ; Reversible and Irreversible
Derivations .....
Transformation by Addition and Transposition of Terms
Multiplication by a Factor ....
Division by a Factor not a Legitimate Derivation
Every Rational Equation can be Integralised .
Derivation by raising both sides to the same Power .
Every Algebraical Equation can be Integralised ; Equivalence of the
Systems, Pj = 0, P 2 = 0, . . . P„ = 0, and L, P, + L 2 P 2 + . . . + L n P„ = 0,
P 2 = 0, . . . P„ = ....
Examination of the Systems P = Q, R=S; PR=QS, R = S
On Elimination ....
Examples of Integralisation and Rationalisation
Examples of Transformation
Examples of Elimination
Exercises XIX. ....
Exercises XX. .....
Exercises XXI. ....
CHAPTER XV.
VARIATION OP A FUNCTION.
Graph of a Function of one Variable .
Solution of an Equation by means of a Graph
Discontinuity in a Function and in its Graph
281
282
283
283
284
286288
289
289
291
292
293
294
295
296
297
298
299301
302
304
305
306
308
310
313
315
CONTENTS
XIX
326
PACK
Limiting Cases of Algebraic Operation . 318322
Definition of the Increment of a Function .... 322
Continuity of the Sum and of the Product of Continuous Functions . 323
Continuity of any Integral Function ..... 324
Continuity of the Quotient of Two Continuous Functions; Exception 324
General Proposition regarding Continuous Functions . . 325
Number of Roots of an Equation between given limits . . 326
An Integral Function can change sign only by passing through the
value ; Corresponding Theorem for any Rational Function
Sign of the Value of an Integral Function for very small and for very
large values of its Variable ; Conclusions regarding the Number
of Roots .......
Propositions regarding Maxima and Minima .
Continuity and Graphical Representation of f{x, y) ; Graphic Surface
Contour Lines ......
f(x, y) = Q represents a Plane Curve ....
Graphical Representation of a Function of a Single Complex Variabl
Horner's Method for approximating to the Real Roots of ar
Equation ....
Multiplication of Roots by a Constant
Increase of Roots by a Constant
Approximate Value of Small Root
Horner's Process
Example
Extraction of square, cube, fourth, . .
Exercises XXII.
roots by Horner's Method
328
330
331
334
335
338346
338
339
340
341
342345
346
347
CHAPTER XVI.
EQUATIONS AND FUNCTIONS OF FIRST DEGREE.
Linear Equations in One Variable ....
Exercises XXIII. .......
Linear Equations in Two Variables — Single Equation, Onefold Infin
ity of Solutions ; System of Two, Various Methods of Solution
System of Three, Condition of Consistency
Exercises XXIV. .......
Linear Equations in Three Variables — Single Equation, Twofold In
finity of Solutions ; System of Two, Onefold Infinity of Solu
tions, Homogeneous System ; System of Three, in general Deter
minate, Homogeneous System, Various Methods of Solution
S3'stems of more than Three ....
General Theory of a Linear System .
General Solution by means of Determinants .
Exercises XXV. .......
Examples of Equations solved by means of Linear Equations .
Exercises XXVI. .......
349,
350
351
352
364
364
365372
373
374376
376
379383
383
XX
CONTENTS
PAGE
Graph of ax + b ....... 385
Graphical Discussion of the cases 6 = ; a = ; a = 0, b = . ». 388
Contour Lines of ax + by + c ...... 389
Illustration of the Solution of a System of Two Linear Equations . 390
Cases where the Solution is Infinite or Indeterminate discussed
graphically ....... 391
Exceptional Systems of Three Equations in Two Variables . . 393
Exercises XXVII. ...... 394
CHAPTER XVII.
EQUATIONS OF THE SECOND DEGREE
ax 2 + bx + c = has in general just two roots
Particular Cases ....
General Case, various Methods of Solution
Discrimination of the Roots
Exercises XXVIII. ....
Equations reducible to Quadratics, by Factorisation, by Integralisa
tion, by Rationalisation
Exercises XXIX., XXX.
Exercises XXXI.
Reduction by change of Variable ; Reciprocal Equations
Rationalisation by introducing Auxiliary Variables
Exercises XXXII. ....
Systems with more than One Variable which can be solved by means
of Quadratics ....
General System of Order 1x2
General System of Order 2x2; Exceptional Cases
Homogeneous Systems
Symmetrical Systems
Miscellaneous Examples
Exercises XXXIII.
Exercises XXXIV.
Exercises XXXV.
396
397
398
400
401
402406
406
407
408413
413
413
414427
415
416
418
420
425
427
429
430
CHAPTER XVIII.
GENERAL THEORY OP INTEGRAL FUNCTIONS.
Relations between Coefficients and Roots .... 431
Symmetric Functions of the Roots of a Quadratic . . .432
Newton's Theorem regarding Sums of Powers of the Roots of any
Equation ........ 436
Symmetric Functions of the Roots of any Equation . . 438
Any Symmetric Function expressible in terms of certain elementary
Symmetric Functions ...... 440
CONTENTS
XXI
Exercises XXXYI.
Special Properties of Quadratic Functions
Discrimination of Roots, Table of Results
Generalisation of some of the Results
Condition that Two Quadratics have Two Roots in common
Lagrange's Interpolation Formula .
Condition that Two Quadratics have One Root in common
Exercises XXXVII. .
Variation of a Quadratic Function for real values of its Variable ;
Analytical and Graphical Discussion of Three Fundamental Cases,
Maxima and Minima
PAGE
415
447453
447
449
450
451
452
453
458
461
462
463
Examples of Maxima and Minima Problems .
General Method of finding Turning Values by means of Equal Roots .
Example, y = x 3  9x ,2 + 24a; + 3 . .....
Example, y= (x 2  Sx + 15)/* ......
General Discussion of y = (ax 2 + bx + c)j(a'x 2 + b'x + c'), with Graphs of
certain Particular Cases ..... 464467
Finding of Turning Values by Examination of the Increment . 468
Exercises XXXVIII .469
CHAPTER XIX.
SOLUTION OF PROBLEMS BY MEANS OP EQUATIONS.
Choice of Variables ; Interpretation of the Solution .
Examples ....•••
Exercises XXXIX. ....••
CHAPTER XX.
ARITHMETIC, GEOMETRIC, AND ALLIED SERIES.
Definition of a Series ; Meaning of Summation ; General Term
Integral Series . . . . ■
Arithmetic Progression
Sums of the Powers of the Natural Numbers
Sum of any Integral Series .
ArithmeticGeometric Series, including the Simple Geometric Series
as a Particular Case
Convergency and Divergency of Geometric Series
Properties of Quantities in A. P., in G.P., or in H.P.
Expression of Arithmetic Series by Two Variables
Insertion of Arithmetic Means
Arithmetic Mean of n given quantities
Expression of Geometric Series by Two Variables
Insertion of Geometric Means
Geometric Mean of n given quantities
471
472476
476
480
482488
482
484487
487
489494
495
496502
496
497
497
499
499
500
XX11
CONTENTS
Definition of Harmonic Series .
Expression in terms of Two Variables
Insertion of Harmonic Means
Harmonic mean of n given quantities
Propositions regarding A.M., G.M., and H. M.
Exercises XL. .....
Exercises XLI. .....
Exercises XLII. ....
CHAPTEE XXL
LOGARITHMS.
Discussion of a x as a Continuous Function of as
Definition of Logarithmic Function
Fundamental Properties of Logarithms
Computation and Tabulation of Logarithms
Mantissa and Characteristic .
Advantages of Base 10
Direct Solution of an Exponential Equation
Calculation of Logarithms by inserting Geometric Means
Alteration of Base ....
Use of Logarithms in Arithmetical Calculation
Interpolation by First Differences
Exercises XLIII. ....
Historical Note ....
CHAPTER XXII.
PAGE
500
501
501
501
501
502
505
507
509
511
512
513519
514
515
516
517
519
519523
524
527
529
THEORY OP INTEREST AND ANNUITIES.
Simple Interest, Amount, Present Value, Discount . . 531532
Compound Interest, Conversion Period, Amount, Present Value,
Discount, Nominal and Effective Rate . . . 533535
Annuities Certain, Accumulation of Forborne Annuity, Purchase Price
of Annuity, Terminable or Perpetual, Deferred or Undeferred,
Number of Years' Purchase .... 536540
Exercises XLIV. ....... 540
APPENDIX.
Commensurable Roots, Reducibility of Equations
Equations Soluble by Square Roots
Cubic ......
Biquadratic, Resolvents of Lagrange and Descartes
Possibility of Elementary Geometric Construction
Exercises XLV. ....
RESULTS OF EXERCISES.
543546
546548
549
550
551
553
555
INDEX OF PRINCIPAL TECHNICAL TEEMS
USED IN PAET I.
Addition Eule for binomial coefficients,
66
Afline of a complex number, 223
Algebraic sum, 10
Algebraical function, ordinary, 281
Alternating function, 77
Amount, 532
Amplitude of a complex number, 236
Annuity, certain, contingent, termin
able, perpetual, immediate, deterred,
forborne, number of years' purchase,
536 et seq.
Antecedent of a ratio, 255
Antilogaritlim, 518
Arganddiagram, 222
Argand's progression, 249
Argument, 524
Arithmetic means and arithmetic mean,
497
Arithmetic progression, 482
Arithmeticogeometric series, 491
Association, 3, 12
Auxiliary variables, 380
Base of an exponential or logarithm, 511
Binomial theorem, 62
Characteristic, 514
Coefficient, 30
Commensurable, 203
Common Measure and Greatest Common
Measure (arithmetical sense), 38, 39 ;
algebraical sense, 111
Commutation, 4, 12
Complex number or quantity, 133, 221
Conjugate complex numbers, 228
Consequent of a ratio, 2f>5
Consistent system of equations, 288
Constant, 30
Continued division, 102
Continued proportion, 256
Continuity of a function, 317, 323, 324,
336
Contour lines of a function, 333
Couvergency of a series, 493
Conversionperiod (for interest), 533
Degree, 30, 58
Degree of an equation, 284
Derivation of equations, 290
Detached coefficients. 63 91
Determiuateness of a system of equa
tions, 286
Differences (first), 521
Discontinuity of a function, 317
Discount, 532
Discriminant, 134, 141
Distribution, 13, 49
Divergency of a series, 493
Divisibility (algebraical sense), 85
Divisibility (arithmetical sense), 38
Efjminant (or resultant), 415. 430
Elimination, 298
Equation, conditional, 282
Equation and equality (identical), 22
Equimodular curves, 248
Equiradical surds, 204
Equivalence of systems of equations,
289
Exponent, 25
Exponential function, 509
Exponential notation (exp a ), 511
Extraneous solutions, 294
Extremes and means of a proportion,
256
Factor (arithmetical sense), 38
Fractional (algebraical sense), 30, 85
Fractional (arithmetical sense), 43
Freehold, value of, 539
Function, analytical, 281
Function, rational, integral, algebraical,
58
Geometric means and geometric mean,
499
Geometric series, 489
Gradient curves, 248
Graph of a function, 312, 333
Graphical solution of equations, 313
Greatest Common Measure (algebraical
sense), 111
Harmonic means and harmonic mean,
501
Harmonic series, 500
Homogeneity, 71
Homogeneous system of equations, 418
Identity, identical, 22
Imaginary unit and imaginary quantity,
132
xxm
XXIV
INDEX
Increment of a function, 322
Indeterminate coefficients, 79, 100
Indeterminate forms, 318, 319, 320
Indeterminateness of a system of equa
tions, 286
Index, 25
Infinite value of a function, 315
Infinitely great, 318
Infinitely small, 318
Integral (algebraical sense), 25, 85
Integral (arithmetical sense), 37
Integral function, 58
Integral quotient (algebraical sense), 86
Integral series, 484
Integralisation of equations, 296
Integrogeometric series, 492
Interest, simple and compound, 531, 533
Interpolation, 524
Inverse, 5, 14
Irrationality (algebraic), 203, 240
Irrationality (arithmetical), 203
Irreducible case of cubic, 549
Irreducible equation, 545
Irreversible derivation, 290
Laws of Algebra, fundamental, 20
Least Common Multiple (algebraical
sense), 122
Limiting cases, 318
Linear (algebraic sense), 138
Linear irrational form, 193
Logarithmic function, 511
Manifoldness, 496
Mantissa, 514
Maxima values of a function, 330
Mean proportionals, 256
Minima values of a function, 330
Modulus (arithmetical sense), 43
Modulus of a complex number, 229
Modulus of system of logarithms, 519
Monomial function, 30
Negative quantity, 9
Ninetest, 175
Operand and operator, 4
Order of a symmetric function, 439
Partial fractions, 151
Pascal's triangle, 67
Periodicity of integers, 43
IInotation, 53
l'rime (arithmetical sense), 38
Primeness (algebraical), 120
Principal, 532
Principal value of a root, 182
Proper fraction (algebraical sense), 86,
144
Proportion, 256, 269
Proportional parts, 526
Quantity, ordinary algebraic, 130
Rate of interest, nominal and effective,
535
Ratio, 255, 268
Rational (algebraic sense), 144
Rational fraction, 144
Rationalisation of equations, 296
Rationalising factor, 190, 197
Reciprocal equations, 410
Reducibility of an equation, 545
Remainder (algebraical sense), 86
Remainder (arithmetical sense), 42
Remainder theorem, 93
Residue (arithmetical sense), 42
Resolvent of a biquadratic, 550
Resultant equation, 415
Reversible derivation, 290
Root of an equation, 284
Roots of a function, 313
Scales of notation, 168
Series, 480
Similar surds, 204
Snotation, 53
Solution of an equation, formal and
approximate numerical, 284
Substitution, principle of, 18
Sum (finite) of a series, 481
Sum (to infinity) of a series, 493
Surd number, 132
Surd number, monomial, binomial, &c,
203, 204
Symmetrical system of equations, 420
Symmetry, 75
Term, 30, 58
Transcendental function, 282
Turning values of a function, 330
Type (of a product), 52
Unity (algebraical sense of), 17
Variable, 30
Variable, independent and dependent,
273
Variation of a function, 311
" Variation " (old sense of), 273, 275
Weight of a symmetric function, 434
Zero (algebraical sense of), 11, 14
CHAPTEK I.
The Fundamental Laws and Processes of Algebra
as exhibited in ordinary Arithmetic.
§1.1 The student is already familiar with the distinction
between abstract and concrete arithmetic. The former is con
cerned with those laws of, and operations with, numbers that are
independent of the things numbered ; the latter is taken up
with applications of the former to the numeration of various
classes of things.
Confining ourselves for the present to abstract arithmetic,
let us consider the following series of equalities : —
2623 1023 2 623x3 + 1023x61
ITT*  3~~~~ 61x3
70272 00i
= = 384.
183
The first step is merely the assertion of the equivalence of
two different sets of operations with the same numbers. The
second and third steps, though doubtless based on certain simple
laws from which also the first is a consequence, nevertheless
require for their direct execution the application of certain rules,
of a kind to which the name arithmetical is appropriated.
We have thus shadowed forth two great branches of the higher
mathematics: — one, algebra, strictly so called, that is, the theory of
operation with numbers, or, more generally speaking, with quanti
ties ; the other, the higher arithmetic, or theory of numbers. These
two science? are identical as to their fundamental laws, but differ
widely in their derived processes. As is usual in elementary
textbooks, the elements of both will be treated in this work.
VOL. I B
2 REPRESENTATIVE GROUPS chap.
§ 2.] Ordinary algebra is simply the general theory of those
operations with quantity of which the operations of ordinary
abstract arithmetic are a particular case.
The fundamental laws of this algebra are therefore to be
sought for in ordinary arithmetic.
However various and complex the operations of arithmetic
may seem, it appears on consideration that they are merely the
result of the application of a very small number of fundamental
principles. To make this plain we return for a little to the very
elements of arithmetic.
ADDITION,
AND THE GENERAL LAWS CONNECTED THEREWITH.
§ 3.] When a group of things, no matter how unlike, is con
sidered merely with reference to the number of individuals it
contains, it may be represented by another group, the individuals
of which are all alike, provided only there be as many individuals
in the representative as in the original group. The members of
our representative group may be merely marks (l's say) on a
piece of paper. The process of counting a group may therefore
be conceived as the successive placing of l's in our representa
tive group, until we have as many l's as there are individuals
in the group to be numbered. This process of adding a 1 is
represented by writing + 1. We may thus have
+ 1, +1 + 1, +1 + 1 + 1, +1 + 1 + 1 + 1, &c,
as representative groups or "numbers." As the student is of
course aware, these symbols in ordinary arithmetic are abbreviated
int0 1, 2, 3, 4,&c.
Hence using the symbol " = " to stand for " the same as," or
"replaceable by," or "equal to," we have, as definitions of 1, 2,
3, 4, &c,
1= +1,
2= +1 + 1,
3 +1 + 1 + 1,
4= + 1 + 1 + 1 + 1, &c.
I ASSOCIATION IN ADDITION 3
And there is a further arrangement for abridging the repre
sentation of large numbers, which the student is familiar with as
the decimal notation. With numerical notation we are not
further concerned at present, but there is a view of the above
equalities which is important. After the group +1 + 1 + 1 has
been finished it may be viewed as representing a single idea to
the mind, viz. the number " three." In other words, we may
look at +1 + 1 + 1 as a series of successive additions, or we
may think of it as a whole. When it is necessary for any
purpose to emphasise the latter view, we enclose +1 + 1 + 1 in
a bracket, thus ( + 1 + 1 + 1) ; and it will be observed that pre
cisely the same result is attained by writing the symbol 3 in
place of + 1 + 1 + 1, for in the symbol 3 all trace of the for
mation of the number by successive addition is lost. We might
therefore understand the equality or equation
3= +1 + 1 + 1
to mean ( + 1 + 1 + 1)= +1 + 1 + 1,
and then the equation is a case of the algebraical Law of
Association.
The full meaning of this law will be best understood by con
sidering the case of two groups of individuals, say one of three
and another of four. If we wish to find the number of a group
made up by combining the two, we may adopt the child's process
of counting through them in succession, thus,
+ 1 + 1 + 1  +1 + 1 + 1 + 1 = 7.
But by the law of association we may write for +1 + 1 + 1
(+1 + 1 + 1),
and for +1 + 1 + 1 + 1
( + 1 + 1 + 1 + 1),
and we have + (+ 1 + 1 + 1) + (+ 1 + 1 + 1 +1) = 7,
or +3 + 4 = 7.
It will be observed that we have added a + in each case be
fore the bracket, and it may be asked how this is justified. The
answer is simply that setting down a representative group of
three individuals is an operation of exactly the same nature as
4 COMMUTATION IN ADDITION chap.
setting down a group of one. The law of association for addition
worded in this way for the simple case before us would be this :
To set down a representative group of three individuals is the same
as to set down in succession three representative individuals.
The principle of association may be carried further. The
representative group +3 + 4 may itself enter either as a whole
or by its parts into some further enumeration : thus,
+ 6 + ( + 3 + 4)= +6 + 3 + 4
is an example of the law of association which the student will
have no difficulty in interpreting in the manner already indi
cated. The ultimate proof of the equality may be regarded as
resting on a decomposition of all the symbols into a succession
of units. There is, of course, no limit to the complication of
associations. Thus we have
+ [( + 9 + 8) + { + 6 + ( + 5 + 3)}] + { + 6 + ( + 3 + 5)}
" = + ( + 9 + 8) + { + 6 + ( + 5 + 3)} + 6 + ( + 3 + 5),
= +9 + 8 + 6 + ( + 5 + 3) + 6 + 3 + 5,
= +9 + 8 + 6 + 5 + 3 + 6 + 3 + 5,
each single removal of a bracket being an assertion of the law
of association. The student will remark the use of brackets of
different forms to indicate clearly the different associations.
§ 4.] It follows from the definitions
3= + 1 + 1 + 1, 2 = +1 + 1,
that +3 + 2= +2 + 3;
and "by a similar proof we might show that
+ 3 + 4 + 6= +3 + 6 + 4= +4 + 3 + 6, &c. ;
in other words, the order in which a series of additions is arranged
is indifferent.
This is the algebraical Law of Commutation, and it will
be observed that its application is unrestricted in arithmetical
operations where additions alone are concerned. The statement
of this law at once suggests a principle of great importance in
algebra, namely, the attachment of the " symbol of operation " or
" operator " to the number, or, more generally speaking, " subject "
or " operand," on which it acts. Tims in the above equations
I SUBTEACTION DEFINED
the + before the 3 is supposed to accompany the 3 when it
is transferred from one part of the chain of additions to another.
The operands in +3, +4, and + 6 are already complex ; and
it may be shown by a further application of the reasoning used
in the beginning of this article that the operand may be complex
to any degree without interfering with the validity of the com
mutative law ; for example,
+ {+3 + ( + 2 + 3)} + ( + 6 + 8)
= + ( + G + 8) + { + 3 + ( + 2 + 3)} ,
of which a proof might also be given by first dissociating, then
commutating the individual terms +6, +8, '+ 3, &c, and then
reassociating.
The Laiv of Commutation, thus suggested by arithmetical considera
tions, is noiv laid down as a general law of algebra ; and forms a
part of the definition of the algebraic symbol + .*
SUBTRACTION.
§ 5.] For algebraical purposes the most convenient course is
to define subtraction as the inverse of addition ; or, as is more
convenient for elementary exposition, we lay down that addition
and subtraction are inverse to each other, t By this we mean
that, whatever the interpretation of the operation + b may be,
the operation  b annuls the effect of + b ; and vice versa.
Thus,  is defined relatively to + by the equation
+ a b + b= + a (1),
or + a + bb +a (2).
These might also be written f f
* See the general remarks in § 27.
t Here we virtually assume that if x + a = y + a, then x = y. See Hankel,
Vorlcsungen il. d. Complcxen Zahlen (Leipzig, 1867), p. 19.
tt It may conduce to clearness in following some of the above discussions to
remember that the primary view of a chain of operations written in any order
is that the operations are to be carried out successively from left to right ;
for example, if we think merely of the last addition, +2 + 3 + 5 + 6 in more fully
expressive symbols means + (+2 + 3 + 5) + 6, that is, + 10 + 6; +a+b+c means
+ ( + a + b) + c; +a~b + c means +(+ab) + c; and so on. We may here re
mind the reader that, in ordinary practice, when + occurs before the first member
of a chain of additions and subtractions, it is usually omitted for brevity.
6 LAWS OF COMMUTATION AND ASSOCIATION chap.
+ ( + ab) + b= + a (T),
+ ( + a + b)b = + a (2').
From a quantitative point of view we might put the matter
thus : the question, What is the result of subtracting b from a 1
is regarded as the same as the question, What must be added to
+ b to produce + a 1 and the quantity which is the answer to
this question is symbolised by + ab. Starting with the defini
tion involved in (1) and (2), and putting no restriction upon the
operands a and b, or, what is the same thing from a quantitative
point of view, assuming that the quantity + ab always exists, we
may show that the laws of commutation and association hold for
chains of operations whose successive links are additions and sub
tractions. We, of course, assume the commutative law for addi
tion, having already laid it down as one of our fundamental laws.
§ 6.] Since + a c + c= + a by the definition of the mutual
relation between addition and subtraction, we have
a+bc=ac+c+bc;
= a  c + b + c  c,
by law of commutation for addition ;
= ac + b (1),
by definition of subtraction.
Also abc = ac + cbc,
by definition ;
= acb + cc,
by case (1) ;
= a~cb (2).
by definition
Equations (1) and (2) may be regarded as extending the law
of commutation to the sign  .* We can now state this law
fully as follows : —
±a±b= ±b±a;
* It might be objected here that it has not been shown that  c may come
into the first place in the chain of operations. The answer to this would be
that +acb may either be a complete chain in itself or merely the latter
part of a longer chain, say p + a  c  b. In the second case our proof would
show that p+ac1>=pc + a b J and the nature of algebraic generality
i FOR ADDITION AND SUBTRACTION 7
or, in words, In any chain of additions and subtractions the different
members may be written in any order, each with its proper sign
attached.
Here the full significance of the attachment of the operator
to the operand appears. Thus in the following instance the
quantities change places, carrying their signs of operation with
them in accordance with the commutative law : —
+ 32 + 11= +3 + 112,
= +31 + 12,
= 21 + 1+3.
§ 7.] By the definition of the mutual relation between addi
tion and subtraction, we have
a + ( + bc)= + a + ( + bc) + cc,
=a+bc (1).
Again, by the definition,
p + bc + cb =p + bb,
=p.
Hence a( + bc) = a( + bc) + bc + cb;
= a( + bc) + ( + bc) + cb,
by case (1) ;
= a + cb,
by the definition ;
= ab+c (2),
by the law of commutation already
established.
§ 8.] The results in last paragraph, taken along with those of
§ 3 above, may be looked upon as establishing the law of associa
tion for addition and subtraction. This law may be symbolised
as follows : —
±(±a±b±c± &c.) = ± ( ± a) ± ( ± b) ± ( ± c) ± &c,
with the following law of signs,
+ ( + «)= + a, ( + «)=  a,
+ (  a) =  a, («)=+ a.
requires that +acb should not have any property in composition which it
has not per se. As to all questions of this kind see § 27.
8 NONARITHMETICAL CASES chap.
The same may be stated in words as follows : — If any number of
quantities affected with the signs + or  occur in a bracket, the bracket
may be removed, all the signs remaining the same if + precede the
hracket, each + being changed into  and each  into + if precede
the bracket.
In the above symbolical statement double signs ( ± ) have
been used for compactness. The student will observe that with
three letters 2x2x2x2, that is, 16, cases are included. Thus
the law gives
+ ( + a + b + c)= + a + b + c,
+ (a + b + c)=  a + b + c,
(a + b + c)= + ab  c, &c.
§ 9.] It will not have escaped the student that, in the as
sumption that + a  b is a quantity that always exists, we have
already transcended the limits of ordinary arithmetic. He will
therefore be the less surprised to find that many of the cases
included under the laws of commutation and association exhibit
operations that are not intelligible in the ordinary arithmetical
sense.
If a = 3 and b = 2,
then by the law of association and by the definition of sub
traction
+ 32= +1 + 22,
= +1,
in accordance with ordinary arithmetical notions.
On the other hand, if
a = 2 and b = 3,
then by the laws of commutation and association and by the
definition of subtraction
+ 23= +2( + 2+l),
= +221,
=  1 + 2  2,
=  1.
Here we have a question asked to which there is no ordinary
arithmetical answer, and an answer arrived at which has no
meaning in ordinary arithmetic.
I ESSENTIALLY NEGATIVE QUANTITIES 9
Such an operation as + 2  3, or its algebraical equivalent,
 1, is to be expected as soon as we begin to reason about
operations according to general laws without regard to the appli
cation or interpretation of the results to be arrived at. It must
be remembered that the result of a series of operations may be
looked on either as an end in itself, say the number of in
dividuals in a group, or it may be looked upon merely as an
operand destined to take place in further operations. In the
latter case, if additions and subtractions be in question, it must
have either the + or the  sign, and either is as likely to occur
and is as reasonably to be expected as the other. Thus, as the
results of any partial operation, + 1 and  1 mean respectively
1 to be added and 1 to be subtracted.
The fact that the operations may end in results that have no
direct interpretation as ordinary arithmetical quantities need not
disturb the student. He must remember that algebra is the
general theory of those operations with quantity of which ordinary
arithmetical operations are particular cases. He may be assured
from the way in which the general laws of algebra are established
that, when algebraical results admit of arithmetical meanings,
these results will be arithmetically right, even when some of the
steps by Avhich they have been arrived at may not be arithmetic
ally interpretable. On the other hand, when the end results
are not arithmetically intelligible, it is merely in the first instance
a question of the consistency of algebra with itself. As to what
the application of such purely algebraical results may be, that
is simply a question of the various uses of algebra • some of these
will be indicated in the course of this treatise, and others will
be met with in abundance by the student in the course of his
mathematical studies. It will be sufficient at this stage to give
one example of the advantage that the introduction of algebraic
generality gives in arithmetical operations. + a  b asks the ques
tion what must be added to + b to give + a. If a = 3 and b — 2,
the answer is 1 ; if a = 2 and b = 3, then, arithmetically speaking,
there is no answer, because 3 is already greater than 2. But if
we regard + a  b as asking what must be added to or subtracted
10 REDUCTION OF AN ALGEBRAICAL SUM OHAP.
from + b to get + a, then the evaluation of + ab in any case
by the laws of algebra will give a result whose sign will indicate
whether addition or subtraction must be resorted to, and to what
extent ; for example, if a = 3 and b = 2, the result is + 1, which
means that 1 must be added ; if a = 2 and 5 = 3, the result is
 1, which means that 1 must be subtracted.
§ 10.] The application of the commutative and associative
laws for addition and subtraction leads us to a useful practical
rule for reducing to its simplest value an expression consisting
of a chain of additions and subtractions.
We have, for example,
+ab+c+def+g
= + a + c + d + gb e /,
= + (a + c + d + g)  (b + e + /),
= + { + ( a + c + d + g)(b + e+f)} (1),
=  { + (b + e+f)(a + c + d + g)} (2).
If a+c+d+g be numerically greater than b + e+f, (1) is
the most convenient form ; if a + c + d + g be numerically less
than b + e + f, (2) is the most convenient. The two taken to
gether lead to the following rule for evaluating a chain of
additions and subtractions : — *
Add all the quantities affected ivith the sign + , also all those
affected with the sign  ,• take the difference of the two sums and affix
t/ie sign of the greater.
Numerical example : —
+35+6+8910+2
= + (3 + 6 + 8 + 2)(5 + 9 + 10),
= +1924,
= (2419)= 5.
§ 11.] The special case +aa deserves close attention. A
special symbol, namely 0, is used to denote it. The operational
definition of is therefore given by the equations
+ a  a=  a + a = 0.
In accordance with this Ave have, of course, the results,
* Such a chain is usually spoken of as an "algebraical sum."
I PROPERTIES OF H
b + = b = b  0,
and + =  0,
as the student may prove by applying the laws of commutation
and association along with the definition of 0.
§ 12.] It will be observed that 0, as operationally defined, is
to this extent indefinite that the a used in the above definition
may have any value whatever.
It remains to justify the use of the of the ordinary
numerical notation in the new meaning. This is at once done
when we notice that in a purely quantitative sense stands for
the limit of the difference of two quantities that have been made
to differ by as little as we please.
Thus, if we consider a + x and a,
+ (a + x) — a = + aa + x = x.
If we now cause the x to become smaller than any assignable
quantity, the above equation becomes an assertion of the identity
of the two meanings of 0.
MULTIPLICATION.
§ 13.] The primary definition of multiplication is as an ab
breviation of addition. Thus + a + a, + a + a + a, + a + a + a + a,
&c, are abbreviated into + a x 2, + a x 3, + a x 4, &c. ; and, in
accordance with this notation, + a is also represented by + a x 1.
a x 2 is called the product of a by 2, or of a into 2 ; a is also
called the multiplicand and 2 the multiplier. Instead of the
sign x , a dot, or mere apposition, is often used where no am
biguity can arise. Thus a x 2, a. 2, and a2 all denote the same
thing.
.§ 14.] So long as a and b represent integral numbers, as is
supposed in the primary definition of multiplication, it is easy to
prove that
a x b = b x a ;
or, adopting the principle of attachment of operator and operand,
with full symbolism (see above, § 4),
x a x b = x b x a.
12 COMMUTATION AND ASSOCIATION chap.
The same may be established for any number of integers, for
example,
x a x b x c — x a x c x b = x b x c x a, &c.
In other words, The order of operations in a chain of multiplication
is indifferent.
This is the Commutative Law for multiplication.
§ 15.] We may introduce the use of brackets and the idea of
association in exactly the same way as we followed in the case
of addition. Thus in x a x ( x b x c) we are directed to multiply
a by the product of b by c. The Law of Association asserts
that this is the same as multiplying a by b, and then multiplying
this product by c. Thus
x a x ( x b x c) = x a x b x c.
The like holds for a bracket containing any number of factors.
In the case where a, b, c, &c, are integers, a proof of the truth
of this law might be given resting on the definition of multi
plication and on the laws of commutation and association for
addition.
§ 16.] Even in arithmetic the operation of multiplication is
extended to cases which cannot by any stretch of language be
brought under the original definition, and it becomes important
to inquire what is common to the different operations thus com
prehended under one symbol. The answer to this question,
which has at different times greatly perplexed inquirers into the
first principles of algebra, is simply that what is common is the
formal laws of operation which we are now establishing, namely,
the commutative and associative laws, and another presently to be
mentioned. These alone define the fundamental operations of
addition, multiplication, and division, and anything further that
appears in any particular case (for example, the statement that
 x  is \ of §) is merely a matter of some interpretation,
arithmetical or other, that is given to a symbolical result demon
strably in accordance with the laws of symbolical operation.
Acting on this principle we now lay down the laws of com
mutation and association as holding for the operation of multi
plication, and, indeed, as in part defining it.
i FOR MULTIPLICATION LAW OF DISTRIBUTION 13
§ 1 7.] The consideration of composite multipliers or com
posite multiplicands introduces the last of the three great laws
of algebra.
It is easy enough, if we confine ourselves to the primary
definition of multiplication, to prove that
+ ax( + b + c) = + a x b + a x c,
+ a*( + bc)= + a x b  a x c,
( + ab)x( + cd)=+axcaxdbxc + bxd.
These suggest the following, which is called the Distributive
Law : —
The product of two expressions, each of which consists of a chain
of additions and subtractions, is equal to the chain of additions and
subtractions obtained by multiplying each constituent of the first expres
sion by each constituent of the second, setting down all the partial
products thus obtained, and prefixing the + sign if the two constituents
previously had like signs, the  sign if the constituents previously had
unlike signs.
Symbolically, thus : —
(±a±b) x(±c±d)
= (± a) x (± c) + (± a) x (± d) + (± b) x (± c)
+ ( ± b) x ( ± d),
with the following law of signs : —
( + a) x ( + c) = + ac, ( + a) x (  c) =  ac,
(  a) x ( + c) =  ac, (  a) x (  c)  + ac.
There are sixteen different cases included in the above equation,
as will be seen by taking every combination of one or other of
the double signs before each letter.
Thus ( + ab)( + c + d)
= + ac + ad be bd ;
(ab)(c + d)
= + ac ad + bcbd ;
and so on.
There may, of course, be as many constituents in each
bracket as we please. If, for example, there be m in one
14 PROPERTY OF
CHAP.
bracket and n in the other, there will be mn partial products
and 2 m+n different arrangements of the signs.
Thus ( + ab + c)(d + e)
  ad + bd  cd + ae  be + ce ;
and so on.
The distributive law, suggested, as we have seen, by the
primary definition of multiplication, is now laid down as a law
of algebra. It forms the connecting link between addition and
multiplication, and, along with the commutative and associative
laws, completes the definition of both these operations.
§ 18.] By means of the distributive law we can prove another
property of 0. For, if b be any definite quantity, subject without
restriction to the laws of algebra, we have
+ baba= + bx( + aa) = ( + aa)x( + b),
 bx( + aa) = ( + aa)x(b),
whence = ( + i)x0 = 0x( + J) = (J) x = 0x(J) ;
or briefly b x = x b = 0.
DIVISION.
§ 19.] Division for the purposes of algebra is best defined as
the inverse operation to multiplication : that is to say, the
mutual relation of the symbols x and f is defined by
x a^b x b x a (1),
or
*
x a x b r b = x a (2).
From a quantitative point of view, this amounts to defining
the quotient of a by b, that is, a i b, as that quantity which,
when multiplied by b, gives a.
In a f b, a is called the dividend and b the divisor. Some
times a is called the antecedent and b the consequent of the
quotient.
Another notation for a quotient is very often used, namely, T or
' b
a/b. As this is the notation of fractions, and therefore has
a meaning already attached to it in the case where a and b
are integers, it is incumbent upon us to justify its use in another
* See second footnote, p. 5.
I QUOTIENT AND FRACTION 15
meaning. To do this we have simply to remark that b times =,
that is, b times a of the bih parts of unity, is evidently a times
unity, that is, a ; also, by the definition of a^b, b times arb is a.
Hence we conclude that 7 is operationally equivalent to a^b in
the case where a and b are integers. No further justification is
necessary, for when either a or b, or both, are not integers, 7
loses its meaning as primarily defined, and there is no obstacle
to resrardinc; it as an alternative notation for a^b.
In the above definition we have not written the signs + or 
before a and b, but they were omitted simply for brevity, and one
or other must be understood before each letter. We shall continue
to omit them until the question as to their manipulation arises.
§ 20.] Since division is fully defined as the inverse of multi
plication, we ought to be able to deduce all its laws from the
definition and the laws of multiplication.
We have*
X a X biC = xdvCXCxtrC,
by definition ;
= x a^c x b x cfr,
by law of commutation for
multiplication ;
— x a^c x b (1),
by definition.
Again, x a^b^c  xdvCxof Jic,
by definition ;
= xa^rC~bxc^c,
by case ( 1 ) ;
= XOrCfi (2),
by definition.
In this way we establish the law of commutation for division.
* Here again the remark made in the third note at the foot of p. 5 applies,
namely, axbic primarily means, if we think only of the last operation, the
same as (a x b) 4 c ; a^bxc the same as (aTb)x c ; and so on.
As in the case of + a, when x a comes first in a chain of operations, x i
in practice usually omitted for brevity.
is
16 COMMUTATION AND ASSOCIATION IN MULTIPLICATION chap.
Taking multiplication and division together and attaching
the symbol of operation to the operand, we may now give the
full statement of this law as follows : —
In any chain of multiplications and divisions the order of the
constituents is indifferent, provided the proper sign be attached to each
constituent and move with it.
Or, in symbols, for two constituents,
*a*b = * b*a,
there being 4 cases here included, for example,
H« xi= x b ~a,
iaib = Tb~a > and so on.
§ 21.] By the definition of the mutual relation between
multiplication and division, we have
xax(x&jc)= x a x (^ x b 7 c) x c 7 c,
= x a x brc (1).
Again, since
x p x brc x c— b = x p xbhb,
= xp,
therefore xai(xb7c)= x a f ( x b f c) x b 4 c x c — b ;
= xa7(xb7c)x(xb~c)xc7b,
by case (1) ;
= x a x cib,
by definition ;
= x a~b x c (2),
by the law of commutation
already established.
These are instances of the law of association for division and
multiplication combined, which we may now state as follows : —
When a bracket contains a chain of multiplications and divisions,
the bracket may be removed, every sign being unchanged if x precede
the bracket, and every sign being reversed if ■— precede the bracket.
Or, in symbols, for two constituents,
l(lalb) = l(la)l{lb),
I AND DIVISION PROPERTIES OF 1 17
with the following law of signs : —
x ( x a)  x a, x(Tfl)=Td,
4 ( x rt) = 4 a, 4 ( 4 a) = • ".
In the above equation eight cases are included, for example,
x(iox})= 4 a x 6,
4 ( r a x 6) = xflrJ,
■4 ( 4 a 4 6) = x a x b,
and so on.
§ 22.] Just as in subtraction we denote the special case
+ a  a by a separate symbol 0, so in division we denote x a 4 a
by a separate symbol 1 . From this point of view, 1 has a purely
operational meaning, and we can prove for it the following laws
analogous to those established for in § 11 ; —
x ara= r a x a = 1,
6x 1=6 = 641,
x 1 = 4 1.
Like 0, 1 has both a quantitative and a purely operational
meaning. Quantitatively we may look on it as the limit of the
quotient of two quantities that differ from each other by a
quantity which is as small a fraction as we please of either. For
example, consider a + x and a, then the equation
(a + x) 4 a = a 4 a +zra
= 1 + X 4 H
becomes, when x is made as small a fraction of a as we please,
an assertion of the compatibility of the two meanings of 1.
It should be noted that, owing to the onesidedness of the
law of distribution (that is, owing to the fact that in ordinary
algebra 6 + ( x a fc) = x (b + a) 4 (6 + c) is not a legitimate trans
formation), there is no analogue for 1 to the equation
b x = 0,
which is true in the case of 0.
§ 23.] If the student will now compare the laws of commuta
tion and association for addition and subtraction on the one hand
and for multiplication and division on the other, he will find them
to he formally identical. It follows, therefore, that so far as these
VOL. I C
18 PRINCIPLE OF SUBSTITUTION chap.
laws are concerned there is virtually no distinction between addi
tion and subtraction on the one hand and multiplication and
division on the other, except the accident that we use the signs
+ and — in the one case and x and r in the other, — a conclusion
at first sight a little startling. This duality ceases wherever the
law of distribution is concerned.
§ 24.] We have already been led to consider such expressions
as + ( + 2) and + (  2), and to see that + a may, according to
the value given to a, be made to stand for + ( + 2), that is, + 2, or
+ (  2), that is,  2. The mere fact that a particular sign, say
+ , stands before a certain letter, indicates nothing as to its
reduced or ultimate value ; the sign + merely indicates what
has to be done with the letter when it enters into operation.
In what precedes as to division, and in fact in all our general
formulae, we may therefore suppose the letters involved to stand
for positive or negative quantities at pleasure, without affecting
the truth of our statement in the least.
For example, by the law of distribution,
(a  b) (c + d) = ac + ad be  bd ;
here we may, if we like, suppose d to stand for — d'.
We thus have
(a  6) {c + (  d')} = ac + a( d') bcb{ d'),
which gives, when we reduce by means of the law of signs
proper to the case,
(a  b) (c  d') = ac  ad'  be + bd',
which is true, being in fact merely another case of the law of
distribution, which we have reproduced by a substitution from
the former case. This principle of substitution is one of the most
important elements in the science ; it is this that gives to
algebraic calculation its immense power and almost endless
capability of development,
§ 25.] We have now to consider the effect of explicit signs
attached to the constituents of a quotient. As this is closely
bound up with the operation of the distributive law for division,
it will be best to take the two together.
I DISTRIBUTION OF A QUOTIENT 19
The full symbolical statement of this law for a dividend
having two constituents is as follows : —
(± a ±b) + (±c) = (±a) + (±c) + (±b) + (±c),
with the following law of signs,
( + a) T ( + c)  + a f c, ( + a) f (  c) =  of c,
(  a) r ( + c) = are, (  a) 5 (  c) = + a f c.
Or briefly in words —
In division the dividend may be distributed, the signs of the partial
quotients following the same law as in multiplication.
The above equation includes of course eight cases. It will
be sufficient to give the formal proof of the correctness of the
law for one of them, say
( + a — 6)t( — c)=  a f e + 6 f e.
By the law of distribution for multiplication, we have
( — a — c + b T c) x (  c) = + (a f c) x t'  (b i c) x c ;
 + ab,
by the definition.
Hence ( + a— i)r( — c) = ( — arc + Jrc) * (  c) f (  c) ;
= — afe + 6Tc,
by the definition.
§ 26.] The law of distribution has only a limited application
to division, for although, as just proved, the dividend may be
distributed, the same is not true of the divisor. Thus it is not
true in general that
a ~ (b + c) = a v b + a ~ c,
or that a f (b — c) = a ■— b — a f c,
as the student may readily satisfy himself in a variety of ways.
§ 27.] As we have now completed our discussion of the
fundamental laws of ordinary algebra, it may be well to insist
once more upon the exact position which the}^ hold in the
science. To speak, as is sometimes done, of the proof of these
laws in all their generality is an abuse of terms. They are
simply laid down as the canons of the science. The best evi
dence that this is their real position is the fact that algebras ai'e
20 POSITION OF THE FUNDAMENTAL LAWS chap.
in use whose fundamental laws differ from those of ordinary
algebra. In the algebra of quaternions, for example, the law
of commutation for multiplication and division does not hold
generally.
What we have been mainly concerned with in the present
chapter is, 1st, to see that the laws of ordinary algebra shall be
selfconsistent, and, 2nd, to take care that the operations they lead
to shall contain those of ordinary arithmetic as particular cases.
In so far as the abstract science of ordinary algebra is con
cerned, the definitions of the letters and symbols used are simply
the general laws laid down for their use. When we come to the
application of the formulae of ordinary algebra to any particular
purpose, such as the calculation of areas, for example, Ave have
in the first instance to see that the meanings we attach to the
symbols are in accordance with the fundamental laws above
stated. When this is established, the formulae of algebra become
mere machines for the saving of mental labour.
§ 28.] We now collect, for the reader's convenience, the
general laws of ordinary algebra.
Definitions connecting the Direct and Inverse
Operations.
Addition and subtraction
+ a — b + b= + a,
+ a + b — b= + a.
Multiplication and division
x a f b x b = x a,
x. a * bib  x a.
For addition and subtrac
tion —
Law of Association.
For multiplication and divi
sion —
±(±a±b)= ± ( ± a) ± ( ± b),
with the following law of signs :—
The concurrence of like signs gives the direct sign ;
The concurrence of unlike signs the inverse sign.
SYNOPTIC TABLE OF LAWS
21
Thus—
+ ( + a) = + a, + (  a) =  a, \ x ( x a) = x a, x ( h a) = s a,
 (  a) = + a,  ( + a) =  a. j = ( = a) = x a, f ( x a) = 4 a,
Law of Commutation.
For addition and subtrac
tion
sion
For multiplication and divi
the operand always carrying its own sign of operation with it.
±a±b= ±b±a,
Properties of and 1.
±5 +
+
+ a — a,
±b0 = ±b,
0.
1 = x a~a,
*bx 1 = *£>= 1 = *&,
Xl= Tl.
Law of Distribution.
For multiplication —
(± a ± b) x (± c ± d) = + (± a) x (± C ) + (± a) x (± d)
+ (±b)x(±c) + (±b)x(± d),
with the following law of signs : —
If a partial product has constituents with like signs, it must
have the sign + ;
If the constituents have unlike signs, it must have the
sign  .
Thus—
+ ( + a) x ( + c ) = +ax c, + ( + a) x (  c) = axe,
+ (a)x(c)= + a x c, +(a)x( + c)= axe.
Property of 0.
0x6 = 5x00.
For division —
( ± a ± b) I ( ± c) = + ( ± a ) f ( ± e) + ( ± 6) * ( ± c),
with the following law of signs : —
22
EXERCISES I
CHAP.
If the dividend and divisor of a partial quotient have like
signs, the partial quotient must have the sign + ;
If they have unlike signs, the partial quotient must have tho
sign  .
Thus—
+ ( + a) — ( + c)= + a~c, +( + a)r(c) =
+ .(  a) 7 (  c) = + a + c, + (  a) 4 ( + c) =
N.B. — The divisor cannot be distributed.
 a f c,
— aTc
Property of 0.
r 6 = 0.
N.B. — Nothing is said regarding &r0. This case will be
discussed later on.
The reader should here mark the exact signification of the
sign = as hitherto used. It means "is transformable into by
applying the laws of algebra, without any assumption regarding
the operands involved."
Any " equation " which is true in this sense is called an
"Identical Equation," or an "Identity"; and must, in the first
instance at least, be carefully distinguished from an equation the
one side of which can be transformed into the other by means of
the laws of algebra only when the operands involved have particular
values or satisfy some particular condition.
Some writers constantly use the sign = for the former kind
of equation, and the sign = for the latter. There is much to
be said for this practice, and teachers will find it useful with
beginners. We have, however, for a variety of reasons, adhered,
in general, to the old usage ; and have only introduced the
sign = occasionally in order to emphasise the distinction in cases
where confusion might be feared.
Exercises I.
[In working this set of examples the student is expected to avoid quoting
derived formulae that he may happen to recollect, and to refer every step to
the fundamental principles discussed in the above chapter.]
i EXERCISES I 23
(1.) Point out in what sense the usual arrangement of the multiplication
of 365 by 492 is an instance of the law of distribution.
(2.) I have a multiplying machine, but the most it can do at one time is
to multiply a number of 10 digits by another number of 10 digits. Explain
how I can use my machine to multiply 13693456783231 by 46381239245932.
(3.) To divide 5004 by 12 is the same as to divide 5004 by 3, and then
divide the quotient thus obtained by 4. Of what law of algebra is this an
instance ?
(4.) If the remainder on dividing X by a be R, and the quotient P, and
if we divide P by 6 and find a remainder S, show that the remainder on
dividing N by ab will be aS + R.
Illustrate with 5015+12.
(5.) Show how to multiply two numbers of 10 digits each so as to obtain
merely the number of digits in the product, and the first three digits on
the left of the product.
Illustrate by finding the number of digits, and the first three lefthand
digits in the following : —
1st. 3659893456789325678 x 342973489379265 ;
2nd. 2 64 .
(6.) Express in the simplest form —
(((( • • • (D • • • )))),
1st. Where there are 2« brackets ;
2nd. "Where there are 2?i + 1 brackets ; n being any whole number whatever.
(7.) Simplify and condense as much as possible —
2a {3a[a(ba)]}.
(8.) Simplify—
1st.
3 [4 5[6 7(8 9.1011)]},
2nd.
(9.) Simplify—
*ttt»M*AA)]}
l(2(3
(4 . . . (9(10ll)) . . . ))).
(10.) Distribute the
following products : — 1st. {a + b) x (a + b) ;
2nd
(r*6)x(a + 6) ; 3rd. (3a 66) x (3a + 66) ; 4th. (Ja £6) x (Ja + £6).
(11.) Simplify, by expanding and condensing as much as possible —
{(m + l)a+(n + l)b}{{ml)a + (nl)b}
+ {{m + l)a(n + l)b} {{ml)a(nl)b}.
(12.) Simplify
K)K)+K)K>
(13.) Simplify—
H)H)H)H)B)H>
(14.) Expand and condense as much as possible —
24 HISTORICAL NOTE chap, i
Historical Xote. — The separation and classification of the fundamental laws
of algebra has been a slow process, extending over more than 2000 years. It is
most likely that the first ideas of algebraic identity were of geometrical origin.
In the second book of Euclid's Elements (about 300 B.C.), for example, we have a
series of propositions which may be read as algebraical identities, the operands
being lines and rectangles. In the extant works of the great Greek algebraist
Diophantos (350 ?) we find what has been called a syncopated algebra. He uses
contractions for the names of the powers of the variables ; has a symbol f to
denote subtraction ; and even enunciates the abstract law for the multiplication
of positive and negative numbers ; but has no idea of independent negative quan
tity. The Arabian mathematicians, as regards symbolism, stand on much the
same platform ; and the same is true of the great Italian mathematicians Ferro,
Tartaglia, Cardano, Ferrari, whose time falls in the first half of the sixteenth
century. In point of method the Indian mathematicians Aryabhatta (476),
Brahmagupta (598), Bhaskara (1114), stand somewhat higher, but their works
had no direct influence on Western science.
Algebra in the modern sense begins to take shape in the works of Regiomon
tanus (14361476), Rudolff (about 1520), Stifel (14871567), and more particularly
Viete (15401603) and Harriot (15601621). The introduction of the various
signs of operation now in use may be dated, with more or less certainty, as
follows :  and apposition to indicate multiplication, as old as the use of the
Arabic numerals in Europe; + and , Rudolff 1525, and Stifel 1544; =,
Recorde 1557 ; vinculum, Viete 1591 ; brackets, first by Girard 1629, but not
in familiar use till the eighteenth century ; < > , Harriot's Praxis, published
1631 ; x , Oughtred, and ^, Pell, about 1631.
It was not until the Geometry of Descartes appeared (in 1637) that the im
portant idea of using a single letter to denote a quantity which might be either
positive or negative became familiar to mathematicians.
The establishment of the three great laws of operation was left for the present
century. The chief contributors thereto were Peacock, De Morgan, D. F. Gregory,
Hankel, and others, working professedly at the philosophy of the first principles ;
and Hamilton, Grassmann, Peirce, and their followers, who threw a flood of light
on the subject by conceiving algebras whose laws differ from those of ordinary
algebra. To these should be added Argand, Cauchy, Gauss, and others, who
developed the theory of imaginaries in ordinary algebra.
CHAPTEE II.
Monomials — Laws of Indices—Degree.
THEORY OF INDICES.
§ 1.] The product of a number of letters, or it may be num
bers, each being supposed simple, so that multiplication merely
and neither addition nor subtraction nor division occurs, is called
an integral term, or more fully a rational integral monomial (that
is, onetermed) algebraical function, for example, ax3x6nxa
xxxxxyxbxb.
By the law of commutation we may arrange the constituents
of this monomial in any order we please. It is usual and con
venient to arrange and associate together all the factors that are
mere numbers and all the factors that consist of the same letter ;
thus the above monomial would be written
(3 x 6) x (a x a) x (b x b) x (x x x x x) x y.
3x6 can of course be replaced by 18, and a further contrac
tion is rendered possible by the introduction of indices or ex
ponents. Thus a x a is written a 2 , and is read " a square," or
"a to the second power." Similarly b x b is replaced by b*, and
x x x x xhy x 3 , which is read " x cube," or " x to the third power."
We are thus led to introduce the abbreviation x 11 for x x x x x x . . .
where there are n factors, n being called the index or exponent*
while x n is called the nth power of x, or x to the nth. power.
§ 2.] It will be observed that, in order that the above defini
tion may have any meaning, the exponent n must be a positive
* In accordance with this definition x 1 of course means simply x, and is
usually so written.
26 RATIONAL INTEGRAL MONOMIAL chap.
integral number. Confining ourselves for the present to this
case, we can deduce the following "laws of indices."
I. (a) a m x a n = a m+n ,
and generally a™ x a n x aP x . . . = a m+n+p+ • • •
(J3) — = a m ~ n if m>n,
1
a n
a nm
if m < n.
II. (a m ) n = a mn = (ft' 1 )'
III. (a) \ab) m = a m b m ,
and generally (ahc . . . ) m = a m b m c m . .
W
¥ 1 '
To prove I. (a), we have, by the definition of an index,
a m x a n = (a x a x a . . . m factors) x (a x a x a . . . n factors),
= a x a x a . . . m + n factors, by the law of association,
= a ,n+n , by the definition of an index.
Having proved the law for two factors, we can easily extend
it to the case of three or more,
for a m x a n x a p = (a m x a 11 ) x a p , by law of association,
_ a m+n x a p^ \)y case already proved,
_ a (m+n)+p^ by case already proved,
_ a m+n+p .
and so on for any number of factors.
In words this law runs thus : The product of any number of
powers of one and the same letter is equal to a power of that
letter whose exponent is the sum of the exponents of these
powers.
To prove I. (fi),
a m
— = (a x a x . . ,m factors) 7 (ft x ft x . . . n factors),
by definition of an index,
II
LAWS OF INDICES 27
I
= a x a x a . . . m factors ~ a f a j . . . w divisions,
by law of association.
Now if m > ?i we may arrange these as follows : —
a m
— = (a x a x . . . m — n factors) x (a 4 a) x (a J a) ... w factors,
l>y laws of commutation and association,
= a x a x . . . m  « factors, by the properties of division,
= a m ~ n .
If wi < n, the rearrangement of the factors may be effected
thus : —
— = 7 (a y. a x ... n m factors) x (a f a) x (a r a) . . . wi factors
= ha"™,
1
,71Ml
It is important to notice 'that I. (ft) can be deduced from
I. (a) without any further direct appeal to the definition of an
index. Thus, if m > n, so that m  n is positive,
gmn x a n _ a mn)+n ^ J. ( a ) }
Hence
Therefore, by the definition of x and J :
a m " n x a n 4 a n = a'" f a".
Again, if m < n, so that n  m is positive,
a m x ft nm _ (jm+to^ ty J ^
= a n , by the laws of + and 
Hence
a m x a n  m _j_ fl n  m _ a « _l_ (f n  j»
Therefore, by the definition of x and ~ ,
a m = a n ~a n  m .
Hence, by the laws of x and f ,
a m f a n = a n f a n ' m ~ a n ,
= (a n 7a n )ra n  m t
= 1 ra n  m .
28 LAWS OF INDICES
CHAP.
To prove II.,
(a m ) n = a m x a m x . . . n factors, by definition,
= (a x a x . . .7)i factors) x (a x a x . . . m factors)
x . . ., n sets, by definition,
ran factors, by law of association.
= a x a x
= a mn , by definition.
To prove III. (a),
(ab) m = (ab) x (ab) x . . . m factors, by definition,
= (a x a x , . . m factors) x (6 x b x . . . m factors),
by laws of commutation and association,
= a m b m , by definition.
Again, (abc) m = {(ab)c} m ,
= (ab) m c m , by last case,
= (a™b m )c">, by last case,
= a m b m c m , and so on.
Hence the with power of the product of any number of letters
is equal to the product of the wth powers of these letters.
To prove III. (ft),
(a\ m
A = (a ~ b) x(a~b)x. . . m factors, by definition,
= (a x a x . . . m factors) H (b x b x . . . m factors),
by commutation and association,
= a m ^ b m ,
_ a m
In words : The mth power of the quotient of two letters is
the quotient of the mt\\ powers of these letters.
The second branch of III. may be derived from the first
without further use of the definition of an index. Thus
(d\ " l /(l \ " l
v x bm = [b x h ) ' by IIL (a) >
= a''\ by definition of x and ~ .
Hence /a\ m
t 1 x b 1 " + b m  a 11 ' J b">
that is,
ii LAWS OF INDICES 29
§ 3.] In so far as positive integral indices are concerned, the
above laws are a deduction from the definition and from the
laws of algebra. The use of indices is not confined to this case,
however, and the above are laid down as the laws of indices
generally. The laws of indices regarded in this way become in
reality part of the general laws of algebra, and might have
been enumerated in the Synoptic Table already given. In this
respect, they are subject to the remarks in chap, i., § 27. The
question of the meaning of fractional and negative indices is
deferred till a later chapter, but the student will have no diffi
culty in working the exercises given below. All he has to do is
to use the above laws whenever it is necessary, without regard
to any restriction on the value of the indices.
§ 4.] The following examples are worked to familiarise the
student with the meaning and use of the laws of indices. At
first he should be careful to refer each step to the proper law,
and to see that he takes no step which is not sanctioned by
some one of the laws of indices, or by one of the fundamental
laws of algebra.
Example 1.
(a 3 b 2 c 5 ) x (a 5 b s c u )+(a 4 b 3 c 15 )
= a s a 5 bb*c 5 c u 4 « 4 4 b 3 4 c ls , by commutation and association,
= « ! ^'^" 4 « 4 4 b 3 f c 15 , by law of indices, I. (a),
= (a 3 + 5 4 a*) x (b+ 6 f b 3 ) x (c 5 +" = c 15 ), by commutation and
association.
= «W<x6W3 xc s+ni5 j by law of indices, I. (/3),
= a*bh.
Example 2.
(15cVV) 2 '< (t^ttkY
= 15V)V)V) a x nf£p> b )' laws of indices, III. (a) and III. (£),
~(3x4) 2 (^vr y (a) '
32 x 523,3,610
= ~3 2 x 4yy T ' by l  (a) and n >
= 3 4 3 2 x i> 4 4 x x s x if 4 1/ x z ]0 4 z™,
= 5~i 2 xx g 4y 2 ,
30 ALGEBRAICAL INTEGRALITY AND FRACTIONALITY chap.
THEORY OF DEGREE.
§ 5.] The result of multiplying or dividing any number of
letters or numbers one by another, addition and subtraction
being excluded, for example, 3 x a x x x b ~ c — y x d, is called a
(rational) monomial algebraical function of the numbers and letters
involved, or simply a term. If the monomial either does not
contain or can be so reduced as not to contain the operation of
division, it is said to be integral; if it cannot be reduced so as
to become entirely free of division, it is said to be fractional. In
drawing this distinction, division by mere numbers is usually
disregarded, and even division by certain specified letters may be
disregarded, as will be explained presently.
§ 6.] The number of times that any particular letter occurs
by way of multiplication in an integral monomial is called the
degree (or dimension) of the monomial in that particular letter ;
and the degree of the monomial in any specified letters is the
siim of its degrees in each of these letters. For example, the
degree of 6 x a x a x x x x x x x y x y, that is, of 6a 2 x 3 y 2 , in a is 2,
in x 3, in y 2, and the degree in x and y is 5, and in a, x, and y 7.
In other words, the degree is the sum of the indices of the
named letters. The choice of the letters which are to be taken
into account in reckoning the degree is quite arbitrary ; one
choice being made for one purpose, another for another. When
certain letters have been selected, however, for this purpose, it
is usual to call them the variables, and to call the other letters,
including mere numbers, constants. The monomial is usually
arranged so that all the constants come first and the variables
last; thus, x and y being the variables, we write 32a 2 bcx 3 y' ; and
the part 32a 2 bc is called the coefficient.
In considering whether a monomial is integral or not, division
by constants is not taken into account.
§ 7.] The notion of degree is an exceedingly important one,
and the student must at once make himself perfectly familiar
with it. lie will find as he goes on that it takes to a large
extent in algebra the same place as numerical magnitude in
arithmetic.
IT
NOTIONS AND LAWS OF DEGREE 31
The following theorems are particular cases of more general
ones to be proved by and by.
The degree of the product of two or more monomials is the sum
of their respective degrees.
If the quotient of two monomials be integral, its degree is the excess
of the degree of the dividend over that of the divisor.
For let A = cx[i/ l ":"ni> . . .
At tjl Ml' »' )>'
= cx y z u' ...
where c and c are the coefficients, x, ij, z, u . . .the variables, and
/, m, v,p . . ., /', m', ri,p' ■ ■ ■ are of course positive integral numbers.
Then the degree, d, of A is given by d = I + m + n +p + . . ., and
tbe degree, d', of A' by d' = 1' + m' + n + p + . . .
But A x A' = (cxhf n z n uP . . .) x (c'x y m z n u p . . .)
/ a l+V m+»i' n+n! p+v'
= (c x c)x y z ir . . .
the degree of which is (I + V) + (m + m') + (w + n) + (p + p') . . .,
that is, (l + m + n+p . . .) + (/' + m! + n' +_// + . . . ), that is,
d + d', which proves the first proposition for two factors. The
law of association enables us at once to extend it to any number
of factors.
Again, let Q = A = A', and let Q be integral and its degree 8.
Now we have, by the definition of division, Q x A' = A. Hence,
by last proposition, the degrees of A and A' being d and d', as
before, we have d = 8 + d', and thence 8 = d  d'.
As an example, let A = G///, A' = T.r 7 // 3 , then A x A' =
42x*y, and A 4 A' = fc 2 /. The degree of A x A' is 24, that is,
14 + 10 ; that of Aj A' is 4, that is, 14  10.
The student will probably convince himself most easily of
the truth of the two propositions by considering particular cases
such as these ; but he should study the general proof as an
exercise in abstract reasoning for on such reasoning he will have
to rely more and more as he goes on.
Exercises II.
[Wherever it is possible in working the following examples, the student
should verify the laws of degree, §§ 57.]
11.) Simplify— 5 7 x 12 4 x 32' x (3 2 x 4 2 x 5) 2
(3 x 15 x 2 3 ) 10
32
EXERCISES II
CHAP II.
(2.
(3.
(4.
(5.:
(6.:
(7.:
(8.;
(9.
(io.;
(ii.
(12.
(13.
(14.
(15.
(16.;
(17.
"Which is greater, (2 2 ) 2 or 2' ? Find the difference between them,
Simplify —
2
Simplify
/ \a 4 b 3 x 3 y 3 J '
2(2 2 ) 2
36W6W 3
81a 4 b 3 <? *
Express in its simplest form —
c y \ 2 / a*Vh?
<a 3 bWtf) X \aWc 3 x 3 y'
Simplify— /45« 3 &V\ 2 /2i3a 4 b 4 c 4 x\ 2
Simplify —
Simplify —
Simplify —
Simplify —
Simplify —
27a 2 6 2 c ) *\ 180a 2 bc J
,„ * x*y 9 /3xV\ 2
{.rhfzj 7 X (;//l~Y) 7 X (A 5 /)
S) 7 *
gas/ Va;?//
X"'J
aj^J x (y
J («p?)« x (a^ r yp \ J
I (a r +») r « /
(z« 6 x ^ c ) a x C^)
/r
(a! a Xic ) o T(jc o + c ) c
Simplify— /x^\p+*^ (xp^\v"!<1
Simplify —
 f £*Y x /^' Y" I _i_ {(^ x ( X m)m} x f^m)* x (jgljw^
Prove that
{yz)^(zxYP(xy)P"
_ {xyz)P+^^
(yt~ l z r  1 )P{z r ' i xP 1 Y{xP l yi l Y xPifz r
Distribute the product —
i V i
aP + r a» + r
Distribute —
If ?n. = ft*, n = aM, a z ={mfn x Y ; show that a:?/z = lc
CHAPTEE III.
Fundamental Formula relating to Quotients or
Fractions, with Applications to Arithmetical
Fractions and to the Theory of Numbers.
OPERATIONS WITH FRACTIONS.
§ 1.] Before proceeding to cases where the fundamental
laws are masked by the complexity of the operations involved,
we shall consider in the light of our newlyacquired principles
a few cases with most of which the student is already partly
familiar. He is not in this chapter to look so much for new
results as to exercise his reasoning faculty in tracing the opera
tion of the fundamental laws of algebra. It will be well, how
ever, that he should bear in mind that the letters used in the
following formula? may denote any operands subject to the laws
of algebra ; for example, mere numbers integral or fractional, single
letters, or any functions of such, however complex.
§ 2.] Bearing in mind the equivalence of the notations ,
ajb, and a —■ b, the laws of association and commutation for
multiplication and division, and finally the definition of a
quotient, we have
y b = O) * (P b ) ^xariirJ,
= a 7 b rp x p,
— arb;
that is, ^7 = 7
pb b
VOL. I D
34 ALGEBRAICAL ADDITION OF FRACTIONS chap.
Kead forwards and backwards this equation gives us the
important proposition that ice may divide or multiply the numerator
and denominator of a fraction by the same quantity without altering
its value.
§ 3.] Using the principle just established, and the law of
distribution for quotients, we have
a
±
6
1
± qa
qb
pb
qb'
± qa
±pb
qb
that is, To add or subtract two fractions, transform each by multiplying
numerator and denominator so that both shall have the same denomi
nator, add or subtract the numerators, and write underneath the
common denominator.
The rule obviously admits of extension to the addition in
the algebraic sense (that is, either addition or subtraction) of any
number of fractions whatever.
Take, for example, the case of three : —
b d f bdf bdf bdf Yb '
± adf ± cbf ± ebd , , , , . ^ . .
= J ,,; , by law ol distribution.
bdf
The following case shows a modification of the process, which
often leads to a simpler final result. Suppose b  Ic, q lr; then,
taking a particular case out of the four possible arrangements
of sign,
a p a p
b q Ic lr'
_ ar pc
Icr Ire
_ ar  pc
Icr
Here the common denominator Icr is simpler than bq, which is
Ihr.
The same result would of course be arrived at by following
in MULTIPLICATION AND DIVISION OF FRACTIONS 35
the process given above, and simplifying the resulting fraction
at the end of the operation, thus : —
a p air  pic .
(ar  pc)l
= far '
by using the law of distribution in the numerator, and the laws
of association and commutation in the denominator ;
ar cp
§ 4.] The following are merely particular cases of the laws
of association and commutation for multiplication and division : —
©«(a)>+«>"<HA
= a J b x c f d,
= a x c7b —el,
= (ac) t (bd),
ac
= bd'
or, in words, To multiply two fractions, multiply their numerators
together for the numerator, and the denominators together for the
denominator of the product.
Again,
© + ©<«■»> +<•+<>.
= a x d 7 b f c,
 (ad) 7 (be),
ad
= Tc'
also
\C
■©
by last case. In words : To divide one fraction by another, invert
the latter and then midtiply.
§ 5.] In last paragraph, and in § 2 above, we have for
36 EXERCISES III CHAr.
simplicity omitted all explicit reference to sign. In reality we
have not thereby restricted the generality of our conclusions, for
by the principle of substitution (which is merely another name
for the generality of algebraic formulae) we may suppose the p,
for example, of § 2 to stand for  w, say, and we then have
(  o>)a a
C^jb = b '
that is, taking account of the law of signs,
— wa a
~ub = b'
and so on.
Exercises III.
(1.) Express in its simplest form —
K» V 2
x — y yx
(2.) Express in its
simplest form —
a h
ab ba
(3.) Simplify
P + Q PQ
PQ P + Q'
where
T> = x + y, Q=x~y,
(4.) Simplify—
1 afltf
x + y
1+
x + y
(5.) Simplify—
1 1 1
ab ac be
a?(bc?
(6.) Simplify
( h " \ I h * \
\ a + bj \ ab)
(7.) Simplify— _1_ _J_ 1x
, 8 . )Siml , %  fr^ + iy^ hl y
(9.) Simplify— fa b\/f0 &\
\b a) • \b* a 2 )'
Ill ARITHMETICAL INTEGRALITY AND DIVISIBILITY 37
(10.) Simplify— a{a b) b{n + b)
a+b a b
(11.) Simplify— 1x 1+a:
l+x + x 2 1x + x"'
(12.) Simplify x
1+a . {x + lfx *
1 x 2 + x+l '
1 + x
(13.) Simplify— 2_/l 1\
a + b 2 ab\a b)
w+ (k '
(14.) Show that &_ (a 2 a 2 ) 2 (a 2 ^) 2
a 2 J a+ a 2 (a a ft 2 ) Wft 2 ")
is independent of x.
(15.) Simplify— «■
6°
' J
(16.) Simplify— 1
1
a 2b
a 2b
a 2b
(17.) Simplify a + b
1
a + b +
1
a ~b +
a + b
APPLICATIONS TO THE THEORY OF NUMBERS.
§ 6.] In the applications that follow, the student should look
somewhat closely at the meanings of some of the terms employed.
This is necessary because, unfortunately, some of these terms,
such as integral, factor, divisible, &c, are used in algebra generally
in a sense very different from that which they bear in ordinary
arithmetic and in the theory of numbers.
An integer, unless otherwise stated, means for the present a
positive (or negative) integral number. The ordinary notion of
greater and less in connection with such numbers, irrespective of
their sign, is assumed as too simple to need definition.* When
* This is a very different thing from the algebraical notion of greater
and less. See chap. xiii. , § 1. It may not he superfluous to explain
38 PRIME AND COMPOSITE INTEGERS chap.
an integer a can be produced by multiplying together two others,
b and c, b and c are called factors of a, and a is said to be exactly
divisible by b and by c, and to be a multiple of b or of c. Since
the product of two integers, neither of which is unity, is an
integer greater than either of the two, it is clear that no integer
is exactly divisible by another greater than itself.
It is also obvious that every integer (other than unity) has
at least two divisors, namely, unity and itself; if it has more, it
is called a composite integer, if it has no more, a prime integer.
For example, 1, 2, 3, 5, 7, 11, 13, . . . are all prime integers,
whereas 4, 6, 8, 9, 10, 12, 14 are composite.
If an integer divide each of two others it is said to be a
common factor or common measure of the two. If two integers
have no common measure except unity they are said to be prime
to each other. It is of course obvious that two integers, such as
6 and 35, which are prime to each other need not be themselves
prime integers. We may also speak of a common measure of more
than two integers, and of a group of more than two integers
that are prime to each other, meaning, in the latter case, a set
of integers no two of which have any common measure.
§ 7.] If we consider any composite integer N, and take in
order all the primes that are less than it, any one of these either
will or will not divide N. Let the first that divides N be a,
then N = «N], where N, is an integer ; if N, be also divisible by a
we have Ni = «N 2 , and N = fl(</N 2 ) = a 2 N 2 ; and clearly, finally,
say N = a*N a , where N a is either 1 or no longer divisible by a.
N a (if not =1) is now either prime or is divisible by some
prime >a and <N a , and, a fortiori, <N, say b ; we should on
the last supposition have N« = Z^N^, where N /3 <N a , and so on.
The process clearly must end with unity, so that we get
N = aW
where a, b, . . . are primes, and a, /3, . . . positive integers. It
here the use of the inequality symbols 4=, >, <, >, <f ; they mean
respectively "is not equal to," "is greater than," "is less than," "is not
greater than," "is not less than." Iustead of > , <t we may use <,> which
may b; read "is equal to or less than," " is equal to or greater than."
in ARITHMETICAL G.C.M. 39
is to be observed that a", b , . . . are powers of primes, and
therefore, as we shall prove presently, prime to each other. It
is therefore always possible to resolve every composite integer into factors
that are powers of primes ; and we shall presently show that this
resolution can be effected in one way only.
§ 8.] If a be divisible by c, then any integral multiple of a, say ma,
is divisible by c; and, if a and b be each divisible by c, then the algebraic
sum of any integral multiples of a and b, say ma + nb, is divisible by c.
For by hypothesis a  ac and b = fir, where a and ft are in
tegers, hence ma = viae  (ma)c, where ma is an integer, that is, ma
is divisible by c. And ma + nb = mac + nfic = {ma + nfi)c, where
ma + n(3 is an integer, that is, ma + nb is divisible by c. The
student should observe that, by virtue of the extension of the
notion of divisibility by the introduction of negative integers,
any of the numbers in the above proposition may be negative.
§ 9.] From the last article we can deduce a proposition which
at once gives us the means of finding the greatest common measure
of two integers, or of proving that they are prime to each other.
If a =pb + c, where a, b, c, p are all integers, then the G.C.M. of
a and b is the G.C.M. of b and c.
To prove this it is necessary and it is sufficient to show —
1st, that every divisor of b and c divides a and b, and, 2nd, that
every divisor of a and b divides b and r.
Since a = pb + c, it follows from § 8 that every divisor of b
and c divides a, that is, every divisor of b and c divides a and b.
Again, since a = pb + c, it follows that c — a —pb • hence, again
by § 8, every divisor of a and b divides c, that is, every divisor
of a and b divides b and c. Thus the two parts of the proof are
furnished.
Let now a and // be two numbers whose G.C.M. is required ;
they will not be equal, for then the G.C.M. would be either of
them. Let b denote the less, and divide a by b, the quotient
beings and the remainder c, where of course c<b* Next divide
b by c, the quotient being q, the remainder d ; then divide c by
d, the quotient being r, the remainder e, and so on.
* For a formal definition of the remainder see § 11.
40 ARITHMETICAL G.C.M.
CHAP.
Since a > b, b>c, c> d, d>e, &c, it is clear that the re
mainders must diminish down to zero. We thus have the
following series of equations : —
a =pb + c
b = qc + d
c = rd + e
I = vm + n
m = ten.
Hence the G.C.M. of a and b is the same as that of b and c, which
is the same as that of c and d, that is, the same as that of d and e,
and finally the same as that of m and n. But, since m = wn, the
G.C.M. of m and n is n, for n is the greatest divisor of n itself.
Hence the G.C.M. of a and b is the divisor corresponding to the
remainder in the chain of divisions above indicated.
If n he different from unity, then a and b have a G.C.M. in
the ordinary sense.
If n he equal to unity, then they have no common divisor
except unity, that is, they are prime to each other.
§ 10.] It should be noticed that the essence of the foregoing
process for finding the G.C.M. of two integers is the substitution
for the original pair, of successive pairs of continually decreasing
integers, each pair having the same G.C.M. All that is necessary
is that j), q, r, . . . be integers, and that a, b, c, d, e, ... be in
decreasing order of magnitude.
The process might therefore be varied in several ways.
Taking advantage of the use of negative integers, we may some
times abbreviate it by taking a negative instead of a positive
remainder, when the former happens to be numerically less than
the latter.
For example, take « = 4323, & = 1595,
we might take 4323 = 2 x 1595 + 1133
or 4323 = 3x1595462;
the latter is to be preferred, because 462 is less than 1133. In practice the
negative sign of 462 may be neglected in the rest of the operation, which may
be arranged as follows, for the sake of comparison with the ordinary process
already familiar to the student : —
Ill
riUME DIVISORS
41
1595)4323(3
4785
462)1595(3
1386
209)462(2
418
44)209(5
220
11)44(4
44 •
G.C.M.=11.
By means of the process for finding the (x.C.M. we may prove
the following proposition, of whose truth the student is in all
probability already convinced by experience : —
If a and b be prime to each other, and h any integer, then any
common factor of ah and b must divide h exactly.
For, since a and b are prime, we have by § 9,
a=pb + c
b = qc + d
c = rd + e
I  vm + 1
>(1). Hence <
r ah pbh + ch
bh = qch + dh
eh = rdh + eh
Ih = vmh + h
•(2).
Now, since any common factor of ah and b is a common
factor of ah and bh, it follows from the first of equations (2) that
such a common factor divides ch exactly, and by the second that
it also divides dh exactly, and so on ; and, finally, by the last of
equations (2), that any common factor of ah and b divides h
exactly.
In particular, since b is a factor of itself, we have
Cor 1. If b divide ah exactly and be prime to a, it must divide h
exactly.
Cor. 2. If a' be prime to a and to b and to c, &c, then it is
prime to their product abc . . .
For, if a' had any factor in common with abc . . ., that is,
with a(bc . . .), then, since a' is prime to a, that factor, by the
proposition above, must divide be . . . exactly ; hence, since a'
42
REMAINDER AND RESIDUE
CHAP.
is prime to b, the supposed factor must divide c . . . exactly, and
so on. But in this way we exhaust all the factors of the pro
duct, since all are prime to a'. Hence no such factor can exist,
that is, a' is prime to abc . . .
An easy extension of this is the following : —
Cor. 3. If all the integers a', b', c', . . . be prime to all the integers
a,b,c,. . ., then the product a'b'c' . . . is prime to the product abc . . .
A particular case of which is
Cor. 4. If a' be prime to a (and in particular if both be primes),
then any integral power of a' is prime to any integral power of a.
§ 11. J It is obvious that, if a and b be two integers, we can
in an infinite number of ways put a into the form of qb + r, where
q and r are integers, for, if we take q any integer whatever, and
find r so that a  qb = r, then a = qb + r.
There are two important special cases, those, namely, where
we restrict r to be numerically less than b, and either (1) positive
or (2) negative. In each of these cases the resolution of a is
always possible in one way only. For, in case 1, if qb be the
greatest multiple of b which does not exceed a, then a  qb — r,
where r <b ; hence a = qb + r ; and in case 2, if qb be the least
multiple of b which is not less than a, then a  q'b  r', where
r' < b. Also the resolution is unique ; for suppose, in case 1, that
there were two resolutions, another being a = \b + p, say; then
qb + r = xb + p, therefore r — p  (x  q)b ; hence r  p is divisible
by b ; but, r and p being each positive, and each numerically <b,
r  p is numerically less than b, and therefore cannot be divisible
by b. Hence there cannot be more than one resolution of the
form 1. Similar reasoning applies to case 2.
r and r' are often spoken of as the least positive and negative
remainders of a with respect to b. When the remainder is spoken
of without qualification the least positive remainder is meant. If
a more general term is required, corresponding to the removal
of the restriction r numerically < b, the word residue is used.
It is obvious, from the definitions laid down in § 6, that a is
or is not exactly divisible by b according as the least remainder of a
with respect to b does or does not vanish.
ill ARITHMETICAL FRACTIONALITV 43
The student will also prove without difficulty that if the re
mainders of a and of a' with respect to b be the same, then a  a' is
divisible by b ; and conversely.
Cor. If q be a fixed integer (sometimes spoken of as a modulus),
then every other integer can be expressed in one or other of the forms
bq, bq+ 1, bq+ 2, . . ., bq + (q I),
where b is an integer.
For, as we have seen, we can put any given integer a into
the form bq + r, where r~^>q, and here r must have one of the
values 0, 1, 2, . . ., (q \).
Example. Take q = 5, then
= 0.5, 1 = 0.5 + 1, 2 = 0.5 + 2, 3 = 0.5 + 3, 4 = 0.5+4;
5 = 1.5, 6 = 1.5 + 1, 7 = 1.5 + 2, 8 = 1.5 + 3, 9 = 1.5 + 4;
10 = 2.5, 11 = 2.5 + 1, 12 = 2.5 + 2, 13 = 2.5 + 3, 14 = 2.5 + 4;
and so on.
It should be noticed that, since bq + (q  1) = (b + \)q 1,
bq + (q  2) = (b + l)q  2, &c, we might put every integer into
one or other of the forms
bq, bq± 1, bq± 2, . . . , &c.
For example,
8 = 2.52, 9 = 2.51, 10 = 2.5, 11 = 2.5 + 1, 12 = 2.5 + 2.
The above principle, which may be called the periodicity of
the integral numbers with respect to a given modulus, is of great
importance in the theory of numbers.
§ 12.] When the quotient a/b cannot be expressed as an
integer, it is said to be fractional or essentially fractional ; if a> b,
a/b is called in this case an improper fraction; if a<b, a proper
fraction.
Hence no true fraction, proper or improper, can be equal to an
integer.
Every improper fraction a/b can be expressed in the form q + rjb,
where q is an integer and rjb a proper fraction. For, if r be the
least positive remainder when a is divided by b, a = qb + r, and
ajb = (qb + r)jb = q + rjb, where q and r are integers and r < b.
If two improper fractions ajb and a'jb' be equal, their integral
parts and their proper fractional parts must be equal separately. For,
44 THEOREM REGARDING G.C.M. CHAP.
if this were not so, we should have, say ajb = q + rjb, a'jb'
= q 1 + r'/b', and q + rjb = q' + r'jb' ; whence q  q = r'/b'  r/b =
(r'6  rb')/bb'. Now r'b < b'b and rb' < bb', hence r'b  rb' is numeric
ally < 66'. In other words, the integer q  q' is equal to a proper
fraction, which is impossible.
§ 13.] We can now prove that an integer can be resolved info
factors which are powers of primes in one way only.
For, since the factors in question are powers of primes, the}''
are prime to each other. Let, if possible, there be two such
resolutions, namely, a'b'c . . . and a"b"c" ... of the same integer N.
Since a'b'c! . . . = a"h"c" . . . , therefore a'b'c' ... is exactly divisible
by a". Now, since a" is a power of a prime, it will be prime to
all the factors a', b', c, . . . save one, say a', which is a power of
the same prime. Moreover, such a factor as a' (that is, a power
of the prime of which a" is a power) must occur, for, if it did
not, then all the factors of a'b'c' . . . would be prime to a", and
a" could not be a factor of N. It follows, then, that a' must be
divisible by a".
Again, since a"b"c" . . .= a'b'c' . . ., therefore a"b"c" ... is divis
ible by a', and it follows as before that a" is divisible by a'.
But, if two integers be such that each is divisible by the
other, they must be equal (§ 6) ; hence a" = a'.
Proceeding in this way we can show that each factor in the
one resolution occurs in the other.
§ 14.] Every remainder in the ordinary jwocess for finding the
G.C.M. of two positive integers a and b can be expressed in the form
± (Art  Bo), where A and B are positive integral numbers. The
upper sign being used for the 1st, 3rd, 5th, &c, and the lower for the
2nd, 4th, &c, remainders.
For, by the equations in § 9, we have successively —
(i);
(2);
(3);
c =
+
{apb}
d =
h
 qc = b 
q(a  pb),


[qa  (1
+ pq)b]
e =
c 
 rd,
=
{a
 pb] + ?
•{qa(\ +
pq)b],
=
+
{(l+gr)
a  (j) + r
+pqr)b]
in THEOREMS REGARDING G.C.M. 45
and so on. It is evident in fact that, if the theorem holds for
any two successive remainders, it must hold for the next. Now
equations (1), (2), and (3) prove it for the first three remainders;
hence it holds for the fourth ; hence for the fifth ; and so on.
In the chapter on Continued Fractions, a convenient process
will be given for calculating the successive values of A and B
for each remainder. In the meantime it is sufficient to have
established the existence of these numbers, and to have seen a
straightforward way of finding them.
Cor. 1. Since g, the G.C.M. of a and b, is the last remainder, we
can always express g in the farm —
g = ± (Aa  Bb) (4),
where A and B are positive integers.
Cor. 2. If a be prime to b, g = 1 ; hence, If a and b be two
integers prime to each otlier, two positive integers, A and B, can
always be found such that —
Aa  Bb = ±1 (5).
X.B. — It is clear that A must be prime to B. For, since ajg
and b/g are integers, I and m say, we have, from (4),
1 = ±(Al Bm) ;
hence, if A and B had any common factor it would divide 1 (by
§ 8 above).
Cor. 3. From Cor. 1 and § 8 we see that every common factor
of a and b must be a factor in their G.C.M.
A result which may be proved otherwise, and will probably
be considered obvious.
Cor. 4. Hence, To find the G.C.M. of more tlian two integers a, b,
c, d, . . ., we must first find g the G.C.M. of a and 6, then g the
G.C.M. of g and c, then g" the G.C.M. of g and d, and so on, the last
G.C.M. found being the G.C.M. of all the given integers.
For every common factor of a, b, c must be a factor in a and
b, that is, must be a factor in g ; hence, to find the greatest com
mon factor iii a, b, c, we must find the greatest common factor
in g and c ; and so on.
From Cor. 2 we can also obtain an elegant proof of the
conclusions in the latter part of § 10.
46 EXAMPLES chap.
Example 1. To express the G.C.M. of 565 and 60 in the form A565  B60.
We have 565 = 9x60 + 25, 60 = 2x25 + 10, 25 = 2x10 + 5, 10 = 2x5.
Hence the G.C.M. is 5, and we have successively
25 = 5659x60;
10=602(5659x60}
=  {2x56519x60} ;
5 = 252x10
= 565  9 x 60 + 2 { 2 x 565  19 x 60 }
= 5x56547x60.
Example 2. Show that two integers A and B can be found so that
5A7B = 1.
We have 7 = 1x5 + 2, 5 = 2x2 + 1; whence 2 = 75, 1 = 52(75)
= 3x52x7.
Hence A = 3, B = 2 are integers satisfying the requirements of the
question.
Example 3. If a, b, c, d, . . . be a series of integers whose G.C.M. is g,
show that integers (positive or negative) A, B, C, D, . . . can be found
such that
g = Aa + Bb + Cc + T)d + . . .
(Gauss's Disquisitioncs Arithmetics, Th. 40).
Find A, B, C, D, when a=36, 6 = 24, c=18, ^=30.
This result may be easily arrived at by repeated application of corollaries
1 and 4 of this article.
Example 4. The proper fraction p/ab, where a is prime to b, can be de
composed, and that in one way only, into the form
a' V ,
 + T1;
a b
where a' and b' are both positive, a' <a, V <b, and k is the integral jiart of
a'/a + b'/b ; that is to say, or 1, according to circumstances.
Illustrate with 6/35.
Since a is prime to b, by Cor. 2 above,
Aa B6=±l ;
multiplying this equation by ^j/ab, we have
±p h~ =JL h »)■
o ' a ab '
If the upper sign has to be taken, resolve pk and^B as follows (§ 11) : —
pA = lb + b' (b' positive <b),
pB—ma~a' (a' positive <a).
Then (1) becomes
V 7 «' b' '
£=:lm++ T (2.
ab a b
Now, since p/ab is a proper fraction, the integral part on the righthand side
of (2) must vanish ; hence, since the integral part of a'/a + b'/b cannot exceed
1 , we must have I  m = 0, or I  m —  1 .
in NUMBER OF PRIMES INFINITE 47
If the lower sign has to be taken in (1), we have merely to take the
resolutions
pA = lbb' (b' positive <b),
pB = 7)ia + a' (a' positive <a),
and then proceed as before. We leave the proof that the resolution is unique
to the ingenuity of the reader.
Illustration. 35 = 5 x 7.
Now 3x52x7 = 1 (see Example 2 above) ;
whence — = —(3 x 5  2 x 7),
.55 .35
_18 12
~7 5'
2x7+4 3x53
~ 7 5 *
4 3
2 +r 8 + ,
■Wj.
5 7
N.B. — If negative numerators are allowed, it is obvious that p/ab can
always be decomposed (sometimes in more ways than one) into an algebraic
sum of two fractions a'/a aud b'/b, where a' and V are numerically less than
a and b respectively. For example, we have 6/35 = 3/5  3/7 = 4/7  2/5.
Example 5. If the n integers a, b, c, d. . . . be prime to each other, the
proper fraction pjabed . . . may be resolved in one way only into the form
a 8 y 8 ,
abed
where a, 8, y, 5, . . . are all positive, a<a, 8<b, y<c, $<d, . . and £
has, according to circumstances, one or other of the integral values
0, 1, 2, . . „»l.
(Gauss's Disquisitiones Arithmetical, Th. 310).
This may be established by means of Example 3.
Example 6. Work out the resolution of Example 5 for the fraction
10729/17017.
§ 15.] We conclude this chapter with a proposition which is
as old as Euclid (ix. 20),* namely —
The number of prime integers is infinite.
For let a, /3, y, . . ., k he any series of prime integers what
soever, then we can show that an infinity of primes can be
derived from these.
In fact the integer a/3y . . . k + 1 is obviously not exactly
* Most of the foregoing propositions regarding integral numbers were
known to the old Greek geometers.
48 EXERCISES IV
CHAP. Ill
divisible by any one of the primes a, (3, y, . . ., k. It must
therefore either be itself a prime different from any one of the
series a, (3, y, . . ., k, or it must be a power of a prime or a
composite integer divisible by some prime not occurring among
a, (3, y, . . ., k. We thus derive from a, /?, y, , . ., k at least
one more prime, say A. Then from a, (3, y, . . ., k, A we can in
like manner derive at least one more prime, \x ; and so on ad
infinitum.'"
Exercises IV.
(1.) If the two fractions A/B, a/b be equal, and the latter be at its lowest
terms, prove that A = /xa, B = /Jib, where /x is an integer.
(2.) Prove that the sum or difference of two odd numbers is always even ;
the sum or difference of an odd and an even number always odd ; the product
of any number of odd numbers always odd ; the quotient of one odd number
by another always odd, if it be integral.
(3. ) If a be prime to b, then —
1st. (a + b) n and (ab) m have at most the G.C.M. 2 m ;
2nd. a m + b m and a m b m have at most the G.C.M. 2 ;
3rd. a + b and a 2 + b"  ab have at most the G. C. M. 3.
(4.) The difference of the squares of any two odd numbers is exactly
divisible by 8.
(5.) The snm of the squares of three consecutive odd numbers increased
by 1 is a multiple of 12.
(6. ) If each of two fractions be at its lowest terms, neither their sum nor
their difference can be an integer unless the denominators be equal.
(7.) Resolve 45738 and 297675 into their prime factors.
(8.) Find the G.C.M. of 54643 and 91319, using negative remainders
whenever it is of advantage to do so.
(9.) Trove that the L. CM. of two integers is the quotient of their product
by the G.C.M.
(10.) If pi, gi, g 3 be the G.C.M. 's, h, h, h the L.C.M.'s, of b and c, c and
a, a and b respectively, G the G.C.M., and L the L.C.M., of the three a, b,
c, show that
, , T abcG
1st. L = ;
2nd. I;= jm ,
I* V \9tfsgi
a).
(11.) When x is divided by y, the quotient is u and the remainder u ;
show that, when x and uy are divided by v, the remainders are the same, and
the quotients differ by unity.
* On this subject see Sylvester, Nature, vol. xxxviii. (1888), p. 261.
CHAPTER IV.
Distribution of Products— Multiplication of Rational
Integral Functions— Resulting General Principles.
GENERALISED LAW OF DISTRIBUTION.
§ 1.] We proceed now to develop some of the more important
consequences of the law of distribution. This law has already been
stated in the most general manner for the case of two factors,
eacb of which is the sum of a series of terms : namely, we multiply
every term of the one factor by every term of the other, and set
down all the partial products thus obtained each with the sign
before it which results from a certain law of signs.
Let us now consider the case of three factors, say
(a + b + c + . . .)( a ' + b' + c' + . . .)( a " + b" + c" + . . .).
First of all, we may replace the first two factors by the process
just described, namely, we may write
(aa' + ah' + ac' + . . . + ba' + bb' + bc' + . . .) (a" + b" + c" + . . .).
Then we may repeat the process, and write
aa'a" + aa'b" + aa'c" + . . .
+ ab'a" + ab'b" + ab'c" + . . .
+ ac'a" + ac'b" + ac'c" + . . .
+ baa" + ba'b" + ba'c" + . . . &c,
where the original product is finally replaced by a sum of
partial products, each of three letters. AYe have simplified the
matter by writing + before every term in the original factors, but
the proper application of the law of signs at each step will pre
sent no difficulty to the student.
VOL. I E
50 GENERALISED LAW OF DISTRIBUTION chap.
The important thing to remark is that Ave might evidently
have arrived at the final result by the following process, which
is really an extension of the original rule for two factors : —
Form all possible partial products by taking a term from each
factor (never more than one from each) ; determine the sign by the law
of signs (that is, if there be an odd number of negative terms in the
partial product, take the sign  ; if an even number of such vr none,
take the sign + ). Set down all the partial products thus obtained.
Cor. The number of terms resulting from the distribution of a
product of brackets which contain I, m, n, . . . terms respectively is
I x m x n x . . . For, taking the first two brackets alone, since
each term of the first goes with each term of the second, the
whole number of terms arising from the distribution of these is
I x m. Next, multiplying by the third bracket, each of the I x m
terms already obtained must be taken with each of the n terms
of the third. We thus get (I x m) x n, that is, I x m x n terms.
By proceeding in this way Ave establish the general result.
It should be noted, hoAvever, that all the terms are supposed
to be unlike, and that no condensation or reduction, owing to like
terms occurring more than once, or to terms destroying each
other, is supposed to be made. Cases occur in § 2 below in
which the number of terms is reduced in this Avay.
If the student have the least difficulty in folloAving the aboA T e,
he "will quickly get over it by working out for himself the results
stated below, first by successive distribution, and then by apply
ing the law just given.
(a + b)(c + d)(e+f)
— ace + acf+ ade + adf+ bee + bcf+ bde + bdf
(2x2x2 = 8 terms) ,
(ab)(cd)(ef)
— ace  acf ade + adf bee + br/+ bde  bdf ;
(ab)(cd)(e+f+g)
— ace + acf + acg  ade  adf  adg  bee  bef  beg + bde + bdf+ bdg
(2x2x3 = 12 terms).
§ 2.] It Avas proved above that in the most general case of
distribution the number of resulting terms is the product of the
numbers of terms in the different factors of the product. An
examination of the particular cases where reductions may be
iv ENUMERATION OF PRODUCTS 51
afterwards effected will lead us to some important practical
results, and will also bring to notice certain important principles.
Consider the product (a + b) (a + b). By the general rule the
distribution will give 2x2 = 4 terms. We observe, however,
that only two letters, a and 6, occur in the product, and that
only three really distinct products of two factors, namely, a x a,
a x b, b x b, that is, a 3 , ab, b 3 , can be formed with these ; hence
among the four terms one at least must occur, more than once.
In fact, the term a x b (or b x a) occurs twice, and the result of
the distribution is, after collection,
(a + b)(a + b) = a 2 + 2ab + b\
This may of course be written
(a + b) 2 = a 2 + 2ab + b 2 (1).
Similarly (a  bf = a 2  2ab + b 2 (2).
In the case (a + b) (a — b) = a  b 2 (3),
the term ab occurs twice, once with the + and again with the 
sign, so that these two terms destroy each other when the final
result is reduced.
Before proceeding to another example, let us write down all
the possible products of three factors that can be made with two
letters, a and b. These are a 3 , a 2 b, ab 2 , b 3 , four in all.
Hence in the distribution of (a + b) 3 , that is, of (a + b) (a + b)
(a + b), which by the general rule would give 2x2x2 = 8 terms,
only four really distinct terms can occur. Let us see what terms
recur, and how often they do so. a 3 and b 3 evidently occur each
only once, because to get three a's, or three b's, one must be
taken from each bracket, and this can be done in one way only.
a"b may be got by taking b from the first bracket and a from
each of the others, or by taking the b from the second, or from
the third, in all three ways ; and the same holds for ab 3 . Thus
the result is
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 (4).
In a similar way the student may establish for himself that
(a  b) 3 = a 3  Za 2 b + 3ab 2  b 3 (5),
(a ± by = a* ± ia 3 b + 6a 3 b 2 ± lab 3 + b* (6),
52 COUNTING OF RECURRENCES chap.
and, remembering that the possible binary products of three
letters, a, b, c, are a 2 , b 2 , c 2 , be, ca, ab, six in number, that —
(a + b + c)*= a 2 + b 2 + c 2 + 2bc + 2ca + 2ab (7),
(a + b c) 2 = a 2 + b 2 + c 2 2bc2ca + 2ab (8),
&c.
The ternary products of three letters, a, b, c, are a 3 , a 2 b, a 2 c,
ab 3 , ac 2 , abc, b 3 , b 2 c, be 2 , c 3 . The enumeration is made more certain
and systematic by first taking those in which a occurs thrice,
then those in which it occurs twice, then those in which it occurs
once, and, lastly, those in which it does not occur at all.*
Bearing this in mind, the student, by following the method
we are illustrating, Avill easily show that
(a + b + c) 3 = (a + b + c) (a + b + c) (a + b + e),
= a 3 +b 3 + c 3 + 3b 2 c + Uc 2 + 3c 2 a + 3ca 2
+ 3a% + 3ab 2 + 6abc (9),
from which again he may derive, by substituting (see chap, i.,
§ 24)  c for c on both sides, the expansion of (a + b  c) 3 , and so
on. He should not neglect to verify these results by successive
distributions, thus : —
(a + b + c) 3 = (a + b + cf(a + b + c)
= (a 2 + b 2 + c 2 + 2bc + 2ca + 2ab) (a + b + c\
= a 3 + ab 2 + ac 2 + 2abc + 2ca 2 + 2a 2 b
+ a% + b 3 + be 2 + 2b 2 c + 2abc + 2ab 2
+ ca 2 + b 2 c + c 3 + 2bc 2 + 2c 2 a + 2abc
= &c.
It is by such means that he must convince himself of the
coherency of algebraical processes, and gain for himself taste and
skill in the choice of his methods.
* There is another way of classifying the products of a given degree which
is even more important and which the student should notice, namely, according
to type. All the terms that can be derived from one another by interchanges
among the variables are said to be of the same type. For example, consider
the ternary products of a, b, c. From a 3 we derive, by interchange of b and a, b 3 ;
from this again, by interchange of b and c, <? : no more can be got in this way,
so that a 3 , b 3 , c 3 form one ternary type ; b"c, be", ca, car, a"b, ab", form another
ternary type ; and abc a third. Thus the ternary products of three variables
fall into three types.
IV S AND li NOTATIONS 53
Let us consider one more case, namely, (b + c) (c + a) (a + b).
Here even all the ten permissible ternary products of a, b, c cannot
occur, for a 3 , b a , c 3 are excluded by the nature of the case, since
a occurs in only two of the brackets, and the same is true of b
and c. In fact, by the process of enumeration and counting of
recurrences, we get
(b + c)(c + a) (a + b) = be + b 2 c + ca 2 + c 2 a + ab 2 + a 2 b + 2abc (10).
In the product (b  c) (c  a) (a  b) the term abc occurs twice
with opposite signs, and there is a further reduction, namely,
(b  c) (c  a) (a  b) = be 2  b 2 c + ca 2  c 2 a + ab 2  a'b (11).
2 Notation. — Instead of writing out at length the sum of all the terms of
the same type, say bc + ca + ab, the abbreviation Hbc is often used ; that is to
say, we write only one of the terms in question, and prefix the Greek letter
2, which stands for "sum," or, more fully, "sum of all terms of the same
type as." The exact meaning of 2 depends on the number of variables that
are in question. For example, if there be only two variables, a and b, then
"Lab means simply ab ; if there be four variables, a, b, c, d, then Hab means
ab + ac + ad + bc + bd + cd. Again, if there be two variables, a, b, 2« 2 6 means
a?b + ab~; if there be three, a, b, e, 1a?b means a 2 b + ab 2 + a*c + ac 2 + b 2 c + bc*.
Usually the context shows how many variables are understood ; but, if this
is not so, it may be indicated either by writing the variables under the 2,
thus 2a6, or otherwise.
abed
This notation is much used in the higher mathematics, and will be found
very useful in saving labour even in elementary work. For example, the
results (4), (9), and (10) above may be written—
{a + b) 3 = Za s + 3?,ab;
(a + b + c) 3 = 2<t 3 + 32a 2 6 + 6abc ;
(b + c) (c + a) (a + b) = ~2,ab + 2abc.
By means of the ideas explained in the present article the reader should
find no difficulty in establishing the following, which are generalisations of
(1) and (9) :—
(a + b + c + d+ . . . ) 2 = 2a 2 + 22«& (12),
(a + b + c + d+ . . . ) 3 = 2« 3 + 32« 2 & + 62aZ>c (13),
the number of variables being any whatever.
IT Notation. — There is another abbreviative notation, closely allied to the
one we have just been explaining, which is sometimes useful, and which often
appears in Continental works. If we have a product of terms or functions of
a given set of variables, which are all different, but of the same type (that is,
derivable from each other by interchanges, see p. 52), this is contracted by
writing only one of the terms or functions, and prefixing the Greek letter II,
which stands for "product of all of the same type as." Thus, in the case of
three variables, a, b, c,
54
PRINCIPLE OF SUBSTITUTION
CHAP.
llab means a"b x air x arc x ac 2 x b'c x be" ;
H(b + c) means (b + c) {c + a) (a + b) ;
ri fb + c\ ( b + c \ ( c + a \ ( a + b
u {W+r?) means [FT?) [* + *) {*+&
and so on.
"We might, for example, write (10) above —
nib + c) = Zb 2 e + 2abc.
§ 3.] Hitherto we have considered merely factors made up of
letters preceded by the signs + and  . The case where they are
affected by numerical coefficients is of course at once provided
for by the principle of association. Or, what comes to the same
thing, cases in which numerical coefficients occur can be derived
by substitution from such as we have already considered. For
example —
(3a + 2b) 3 = {(3a) + (2b)} 3
= (3a) 3 + 3(3a) 2 (2b) + 3(3a)(2b) 2 + (2b) 3 ,
whence, by rules already established for monomials,
= 27 a 3 + 54a 2 b + 36ab 2 + 8b 3 .
(a2b + 5c) 2 = {(a) + (2b) + (5c)} 2
= (af + (  2b) 2 + (5c) 2 + 2(  2b) (5c) + 2(5c)(a) + 2(a) (  2b)
= a 2 + ib 2 + 25c 2  205c + lOca  iab.
The student will observe that in the final result the general
form by means of which this result was obtained has been lost,
so far at least as the numerical coefficients are concerned.
§ 4.] It is very important to notice that the principle of
substitution may also be used to deduce results for trinomials
from results already obtained for binomials. Thus from (a + b) 3 =
a 3 + 3a 2 b + 3ab 2 + b 3 , replacing b throughout by b + c, we have
{a + (b + c)} 3 = a 3 + 3a\b + c) + 3a(b + cf + (b + c) 3
= a 3 + 3a 2 b + 3a 2 c
+ 3a(b 2 + 2bc + c 2 )
+ b 3 + 3b 2 c + 3bc 2 + c 3 ;
whence (a + b + c) 3 = a 3 + b 3 + c 3 + 3b 2 c + 3bc 2 + 3c 2 a + 3ca 2
+ 3a 2 b + 3ab 2 + Qabc.
By association of parts of the factors, and by partial distri
bution in the earlier parts of a reduction, labour may often be
saved and elegance attained.
iv SUMS OF COEFFICIENTS DO
For example —
{a + b + c  d) (a  b + c + d)
= {(a+e) + (6  d)} {(a + c)  (b  d)) ;
= (a + c) 2 (bdy 2 ,
by formula (3) above ;
= (a 2 + 2ac + c 2 )  (b n   2bd + cP) ;
=a?P+(*cP+2ac+2bd.
Again,
(a + b + c) (b + c  a) (c + a  b) (a + b  c)
= {(b + c) + a}{(b + c)a}{a{bc)}{a + (bc)};
= {(b+c)*a?}{a?(bc)*\,
by a double application of formula (3) ;
= Ji 2 + 26c + c"  a} {a  b 2 + 2bc  c 2 } ;
= {2bc + {b" + c"  a)} {2bc  (b 2 + c 2  a")} ;
= {2bcf{b 2 + ca 2 )\
by formula (3);
= 46V  (6 4 + c ' + a 4 + 26 2 c 2  2c V  2a 2 6 2 ) ;
= 26 2 c 2 + 2c 2 a 2 + 2a 2 6 2  a*  ¥  c\
a result which the student will meet with again.
§ 5.] There is an important general theorem which follows
so readily from the results established in §§ 1 and 2 that we may
give it here. If all the terms in all the factors of a product be
simple letters unaccompanied by numerical coefficients and all affected
with the positive sign, then the sum of the coefficients in the distributed
value of the product will be I x m x n x . . ., where I, m, n, . . . are
the numbers of the terms in the respective factors.
This follows at once from the consideration that no terms
can be lost since all are positive, and that the numerical co
efficient of any term in the distribution is simply the number of
times that that term occurs.
Thus in formula? (4), (G), and (10) in § 2 above we have
1+3 + 3 + 1 =2x2x2,
1+4 + 6 + 4 + 1 =2x2x2x2,
1 + 1 + 1 + 1 + 1 + 1 + 2 = 2x2x2,
&c.
In formulae (8) and (11) of § 2, and in the formula? of § 3,
the theorem does not hold on account of the appearance of
negative signs and numerical coefficients.
The following more general theorem, which includes the one
just stated as a particular case, will, however, always apply : —
The algebraic sum of the coefficients in the expansion of any
56
EXERCISES V
CHAP.
product may be obtained from the product itself by replacing each of
the variables by 1 throughout all the factors.
Thus, in the case of
(a + b  c) 2 = a 2 + b 2 + c 2  2bc  2ca + 2ab,
we have (1 + 1  l) 2 = 1 = 1 + 1 + 1  2  2 + 2.
The general proof of the theorem consists merely in this —
that any algebraical identity is established for all values of its
variables : so that we may give each of the variables the value 1.
When this is done, the expanded side reduces simply to the
algebraic sum of its coefficients.
Exercises V.
(1.) How many terms are there in the distributed product (a^+a^)
(h + b 2 + b 3 ) ( Cl + 0., + % + c 4 ) (ih + d. 2 + d 3 + d 4 + d 5 ) ?
Distribute, condense, and arrange the following : —
(2.) (x+y){xy)( a ?y*)(a?+y*)*.
(3.) ( x ^ + ,f){x i f)(x i + y i ).
(4.) { x + y)\xyf.
(5.) (x + 2y)\x2y)\
(6.) (b + c){c + a){a + b)(be)(ca)(ab).
(7.) (x + x + l) 3 .
(8.) (3a + 26l) 3 .
(9.) ( x *+x + l +  + K
\ X x
(10.) (a + b + cy, and {abc)\
(11.) Write down all the quaternary products of the three letters x, y, z ;
point out how many diiferent types they fall into, and how many products
there are of each type.
(12.) Do the same thing for the ternary products of the four letters a, b, c, d.
(13.) Find the sum of the coefficients in the expansion of (2a + 3b + Ac) 3 .
Distribute and condense the following, arranging terms of the same type
together : —
x . y z \ f x y
(H.) ~ + ^
o  c c a a b
b+c c+a
a + bj
(15.) [x + y + z)x{y + zx)y(z + xy)z(x + yz).
(16. ) {bc) {b + c a) + {c a) (c + a b) + {a b)(a + bc).
(17.) (b + c)(y + z) + (c + a)(z + x) + (a + b)(x + y)(a + b + c)(x + y + z).
(18.) 2u(b + ca)m(b+ca).*
* "Wherever in this set of exercises the abbreviative symbols S and IT are
used, it is understood that three letters only are involved. The student who
finds difficulty with the latter part of this set of exercises, should postpone
them until he has read the rest of this chapter.
IV
EXERCISES V 57
Show that
(19.) (x + yy = 2(.r + y") (X + y) 2  (.'."  y ).
(20.)
a i( x _ l)i _ 4a 3 b(x  a)(x  b) 3 + 6a"b"{x  a)(x  b)  ial?{x  af{x b) + b*{xa)*
= (a*  ia 3 b + 6a"b  iab 3 + 6*)a^.
(21.) (x  ay") (.,■>"  ay') = {xx'±ayy'f  a{xy , ±yxf ;
(a 2  ayf = (x 3 + Zaxtff  a[Zxhj + ay 3 )' ;
(x~  By 2  C; 2 + BCu) (x' 2  By' 2  Cz' 2 + BGu')
= {xx' + Byy'±C(zz' + Bmi')\ 2 B{a'y' + x'y±C(uz' + u'z)} 2
Q{xz' Byu'±{zx' Buy')\ 2 + BC{yz' xu'±{%ix' zy')) 2 .
Lagranrjc.
The theorems (21.) are of great importance in the theory of numbers ; they
show that the products and powers of numbers having a certain form are
numbers of the same form. They are generalisations of the formula; numbered
V. in the table at the end of this chapter.
Distribute, condense, and arrange —
(22.) 2a26cII(6 + c).
(23.) 2a(2a 2 + 26c) + 2r/2a 2  2(6 + cf.
(24. ) (6  e) (6 + cf + (c  a) (c + a) 3 + {ab) (a + bf.
(25.) Distribute
{{a + b)x*abxy + {ab)y 2 ){{ab)x 2 + abxy + (a + b)y 2 };
and arrange the result in the form
Ax* + Bx?y + Cx 2 y 2 + Dxy 3 + Ey*.
Show that
(26. ) { 3? y 3 + Bxy(2x + y)} 3 + { y 3 x 3 + 3xy(2y + x)} 3
=27xy{x + y) (x 2 + xy + y 1 ) 3 .
(27.) Z{2(x 2 + xy + y 2 )(x 2 + xz + z 2 )(y 2 + yz + z 2 ) 2 }=Z{Zyz} 2 .
(28.) i2a 2 (6 + c) 2 + 2a6c2a} = {26c} 2 .
(29.) 2(a6)(ac)={2a 2 26c}.
. (3«6c  26 3  a 2 d) 2 + 4(ac  b 2 ) 3 _ (Mtb  2c 3  d 2 a) 2 + i( db  c 2 ) 3
(30.) ^ _ d . 2
general theory of integral functions.
§ 6.] As we have now made a beginning of the investigation
of the properties of rational integral algebraical functions, it will
be well to define precisely what is meant by this term.
We have already (chap, ii., § 5) defined a rational integral
algebraical term as the product of a number of positive integral
powers of various letters, x, y, z, . . ., called the variables, multi
plied by a coefficient, which may be a positive or negative number,
or a mere letter or function of a letter or letters, but must not
contain or depend upon the variables.
58 GENERAL NOTIONS REGARDING chap.
A rational integral algebraical function is the algebraical sum of
a series of rational integral algebraical terms. Thus, if x, y, z, ... be
the variables, /, m, n, . . ., /', m', n', . . ., I", m", n", . . . positive in
tegral numbers, and C, C, C", . . . coefficients as above defined,
then the type of such a function as we have defined is
Cx l y m z n • ■• + Casty*'* 1 ' . . . + C"x l "y m "z n " ...+ &c.
For shortness, we shall, when no ambiguity is to be feared, speak
of it merely as an " integral function."
To fix the notion, we give a few special examples. Thus
(a) dx 3 + dxy + 2y 2 is an integral function of x and y ;
(/3) ax 2 + bxy + cy, a, b, c being independent of x and y, is an integral
function of x and y ;
(7) 3X 3  2.t 2 + 3x + 1 is an integral function of x alone ;
OS If  %
{5)  + T +  1 is an integral function, if x, y, z be regarded as the
a b c °
variables ; but is not an integral function if the variables be
taken to be x, y, z, a, b, c, or a, b, c alone.
Each term has a "degree," according to the definition of
chap, ii., § 6, which is in fact the sum of the indices of the vari
ables. The decrees of the various terms will not in general be
alike ; but the degree of an integral function is defined to be the
degree of the term of highest degree that occurs in it.
For example, the degree of (a) above in x and y is the 3rd, of (/3) the
2nd in x and y and the 1st in a, b, c, of (7) in x the 3rd, of (5) in x, y, z the 1st.
§ 7.] From what has already been shown in this chapter it
appears that, in the result of the distribution of a product of any
number of integral functions, each term arises as the product of
a number of integral terms, and is therefore itself integral.
Moreover, by chap, ii., § 7, the degree of each such term is the
sum of the degrees of the terms from which it arises. Hence
the following general propositions : —
The product of any number of integral functions is an integral
function.
The highest * term in the distributed product is the product of the
* By " highest term " is meant term of highest degree, by " lowest term "
term of lowest degree. If there be a term which does not contain the vari
ables at all, its degree is said to be zero, and it of course would be the lowest
term in an integral function, for example, +1 in (7) above.
IV
INTEGRAL FUNCTIONS 59
highest terms of the several factors, and the lowest term is the product
of their lowest terms.
The degree of the product of a number of integral functions is the
sum of tlie degrees of the several factors.
Every identity already given in this chapter, and all those
that follow, will afford the student the means of verifying these
propositions in particular cases. It is therefore needless to do,
more than call his attention to their importance. They form, it
may he said, the cornerstones of the theory of algebraic forms.
INTEGRAL FUNCTIONS OF ONE VARIABLE.
§ 8.] The simplest case of an integral function is that where
there is only one variable x, As this case is of great importance,
we shall consider it at some length. The general type is
2W n +PniX n ~ 1 + . . . +p v r.+p ,
where p , p lf . . ., p n are the various coefficients and n is a posi
tive integral number, which, being the index of the highest term,
is the degree of the function. The function has in general n + 1
terms, but of course some of these may be wanting, or, which
amounts to the same thing, one or more of the letters p ,p u . . .,p n
may have zero value.
§ 9.] When products of integral functions of one variable
have to be distributed, it is usually required at the same time to
arrange the result according to powers of as, as in the tjpical
form above indicated. We proceed to give various instances of
this process, using in the first place the method described in
the earlier part of this chapter. The student should exercise
himself by obtaining the same results by successive distribution
or otherwise.
In the case of two factors (x + a) (x + b), we see at once that
the highest term is x 2 , and the lowest ab. A term in x will be
obtained in two ways, namely, ax and bx ; hence
(x + a) (x + b) = x 2 + (a + b)x + ab (1).
This virtually includes all possible cases ; for example, putting  a for a
we get
(x + (  a)) {x + b) =x + ((  a) + b)x + (  a)b,
= x 2 + {a + b)x  ab.
60 DISTRIBUTION OF (x  ff x ) (x  C( 2 ) . . . (x  a n ) CHAP.
Similarly (a:  a) (x  b) = or + (  a  b)x + ab,
= o? (a + b)x + ab.
(x  a) (x  a) = o;" + (  a t a)x+ a 2 ,
= or  lax + a", &c.
Cases in which numbers occur in place of a and b, or in which x is affected
with coefficients in the two factors, may be deduced by specialisation or other
modification of formula (1), for example,
U'2)(.r+3)=.r 2 + (2 + 3)z + (2)( + 3),
= or + x6.
(px + q) (rx + s) =p ( x +
«\'*+i
=prx~+pri  + )x+pr — ,
=prx 2 + (rq +ps)x + qs,
which might of course be obtained more quickly by directly distributing the
product and collecting the pow y ers of x.
In the case of three factors of the first degree, say (x + a x )
(x + a 2 ) (x + a 3 ), the highest term is .r 3 ; terms in x 2 are obtained
by taking for the partial products x from two of the three brackets
only, then an a must be taken from the remaining bracket ; we
thus get a x x 2 , a 2 x 2 , a^z 2 ; that is, {a x + a B + a 3 )x 2 is the term in x 2 .
To get the term in x, x must be taken from one bracket, and a's
from the two remaining in every possible way ; this gives
(a^a 2 + a,{a z + a 2 a 3 )x. The last or absolute term is of course a^a^
Thus (x + rtj) (x + a.,) (x + a 3 )
= x 3 + (a, + a 2 + a 3 )x 2 + (rt,a 2 + a,a 3 + a 2 a 3 )x + a l a i a 3 (2).
By substitution all other cases may be derived from (2), for
example,
(x  a,) (x  a 2 ) (x  a 3 )
= x 3  (a, + a 2 + a 3 )x 2 + (a a a a + a,a 3 + a 2 a 3 )x  a x a 2 a 3 (3) ;
(x + 1) (x + 2) (x  3) = x  7x  G, and so on.
After what has been said it is easy to find the form of the dis
tribution of a product of n factors of the first degree. The result is
(x + «,) (x + a 2 ) . . . (x + a n )
= x n + P.a:' 1  1 + P^' 2 + . . . + P„_,a; + P n (4),
IV
BINOMIAL THEOREM 61
where P x signifies the algebraic sum of all the a's, P 2 the alge
braic sura of all the products that can be formed by taking two
of them at a time, P 3 the sum of all the products three at a time,
and so on, P n being the product of them all.
§ 10.] The formula (4) of § 9 of course includes (1) and (2)
already given, and there is no difficulty in adapting it to special
cases where negative signs, &c, occur. The following is par
ticularly important : —
= X n  P,':" 1 + P^ 2 ... + ( l)" 1 ^!* + (  l)»P n (I)
Here T t P 2 , &c, have a slightly different meaning from that
attached to them in § 9 (4) : P 3 , for example, is not the sum
of all the products of a 1}  a 2 , . . .,  a n , taken three at a
time, but the sum of the products of + a 1} + a 2 , . . ., + a n , taken
three at a time; and the coefficient of a; n ". 3 is therefore  P 3 ,
since the concurrence of three negative signs gives a negative
sign. As a special case of (1) let us take
(x  a) (x  2a)(x  Sa)(x  4a) = x*  P r x 3 + F 2 x 2  P^ + P 4 .
Here P, = a + 2a + 3a + 4a = 10a,
P 2 = 1 x 2a 8 + 1 x 3a" + 1 x 4a 2 + 2 x 3« 2 + 2 x 4a 2 + 3 x 4a 2
= 35 a 2 ,
P 3  2 x 3 x 4a 3 + 1 x 3 x 4a + 1 x 2 x 4a 3 + 1 x 2 x 3a 3
 50a 3 ,
P 4 = 1 x 2 x 3 x 4a' = 24a 4 .
So that (x  a) (x  2a) (x  3a) (x  4a)
 x  Wax 3 + 35a 2 /  50a 3 * + 24«\
§ 11.] Another important case of § 9 (4) is obtained by
making a l = a 2 = a 3 = . . . = a n , each = a say. The lefthand side
then becomes (x + a) n . Let us see what the values of P n P 2 , . . .,
P n become. Pj obviously becomes na, and P n becomes a n . Con
sider any other, say P r ; the number of terms in it is the number
of different sets of r things that Ave can choose out of n things.
This number is, of course, independent of the nature of the
things chosen ; and, although we have no means as yet of calcu
lating it, we may give it a name. The symbol generally in use
62 BINOMIAL COEFFICIENTS chap.
for it is n C r , the first suffix denoting the number of things chosen
from, the second the number of things to be chosen. Again,
each term of P r consists of the product of r letters, and, since in
the present case each of these is a, each term will be a r . All
the terms being equal, and there being n C r of them, we have in
the present case P r = n C/t r . Hence
(x + a) n = x n + naz n  1 + n C 2 a 2 x n ~ 2 + n Q 3 a z x n ~ z + . . . + a n ;
or, if we choose, since n C\ — n > rfin = 1, we may write
(x + a) n = x n + nC^x 71 ' 1 + n C 2 a 2 x n ~ 2 + ... + n O n . l cfl 1 x + n C n a n (1).
This is the " binomial theorem " for positive integral exponents, and
the numbers n C M n C 2 , n C 3 , . . . are called the binomial coefficients of
the nth order. They play an important part in algebra j in fact,
the student has already seen that, besides their function in the
binomial expansion, they answer a series of questions in the
theory of combinations. When we come to treat that subject
more particularly we shall investigate a direct expression for n C r
in terms of n and r. Later in this chapter we shall give a pro
cess for calculating the coefficients of the different orders by
successive additions.
By substituting successively a, +1, and  1 for a in (1)
we get
(x  a) n = x n  n G x ax n ~ 1 + n C 2 a 2 x n  2  n G 3 a 3 x n  3 + . . .
+ (  1)W W (2) ;
(x+l) n = x n + n C 1 x n  l + n C.p: n  2 + . . .+„C n (3);
(xl) n = x n  n C 1 x n  1 + n CfP*. . . + (l)\C n (4);
and an infinity of other results can of course be obtained by
substituting various values for x and a.
§ 12.] In expanding and arranging products of two integral
functions of one variable, the process which is sometimes called
the long rule for multiplication is often convenient. It consists
simply in taking one of the functions arranged according to
descending powers of the variable and multiplying it successively
by each of the terms of the other, beginning with the highest
and proceeding to the lowest, arranging the like terms under
one another. Thus we arrange the distribution of
iv LONG MULTIPLICATION 63
(x a +2x 2 +2x+l)(x 2 x+l)
as follows : — a? + 2x 2 + 2x + 1
x 2  x + 1
x + 2x* + 2x a + of
 x 4  2/  2x 2  x
+ z* + 2x 2 + 2x + 1
x* +
x* + x 3 +
x 2
+
x +
1
or again
(px*
+ qx + r)(rx 2
+
qx+p)
px 2
+ qx
+ r
rx*
+ qx
+ p
prx*
+ qrx 3
+ pqx 3
2 2
+ r x
2 2
+ qx
+ qrx
+ px
+ pqx
+ pr
prx K + (pq + qr)x 3 + (jf + q 2 + r 2 )x 2 + (pq + qr)x + pr.
The advantage of tins scheme consists merely in the fact that
like powers of x are placed in the same vertical column, and that
there is an orderly exhaustion of the partial products, so that
none are likely to be missed. It possesses none of the funda
mental importance which might be suggested by its prominent
position in English elementary textbooks.
§ 13.] Method of Detached Coefficients. — When all the powers
are present a good deal of labour may be saved by merely
writing the coefficients in the scheme of § 12, which are to be
multiplied together in the ordinary way. The powers of x can
be inserted at the end of the operation, for we know that the
highest power in the product is the product of the highest powers
in the two factors, and the rest follow in order. Thus we may
arrange the two multiplications given above as follows : —
1+2+2+1
11 + 1
1+2+2+1
1221
+1+2+2+1
1+1+1+1+1+1;
61
DETACHED COEFFICIENTS
CHA1".
•whence
(x 3 + 2.c 2 + 2x + l)(x 2  x + 1) = X s + x* + x 3 + %• + x + 1.
Again,
p
+ 2
+ r
r
+ 1
+ 1>
pr
+ qr
+ r~
+ M
+ qr
+ pq + pr
pr + (qr + pq) + (p 2 + q 2 + r 2 ) + (pq + qr) + pr ;
whence
(px 2 + qx + r)(rx 2 + qx +p)
 prx* + (pq + qr)x 3 + (p 2 + q 2 + r 2 )x~ + (pq + qr)x + pr.
The student should observe that the use of brackets in the
last line of the scheme in the second example is necessary to
preserve the identity of the several coefficients.
It has been said that this method is applicable directly only
when all the powers are present in both factors, but it can be
made applicable to cases where any powers of x are wanting by
introducing these powers multiplied by zero coefficients. For
example —
(x i 2x"+l)(x i + 2x 2 +l)
= (%* + Ox 3  2x 2 + Ox + l)(x* + Ox 3 + 2x 2 + Ox + 1) ;
1+02+0+1
1+0+2+0+1
whence
1+02+0+1
+0+0+0+0+0*
+2+04+0+2
+0+0+0+0+0*
+1+02+0+1
1 +0+0+02+0+0+0+1
x* + O.c 7 + 0x° + 0/  2x* + 0x a + 0/ + Ox + 1 ;
(x*  2x 2 + 1)(^ 4 + 2x 2 + 1) = x  2x* + 1.
iv DETACHED COEFFICIENTS 65
The process might, of course, be abbreviated by omitting the
lines marked *, which contain only zeros, care being taken to
place the commencement of the following lines in the proper
columns ; and, in writing out the result, the terms with zero
coefficients might be omitted at once. With all these simplifica
tions, the process in the present case is still inferior in brevity
to the following, which depends on the use of the identities
(A + B) (A  B) = A 2  B 2 , and (A + B) 2 = A 2 + 2AB + B 2 .
(x*  2x* + 1) (x* + 2x* + 1) = {(x* + 1)  2x*}{(x< + 1) + 2x 2 }
= (x* + I) 2  (2xJ
= x + 2x* + 1  4a; 4
= x*2x*+l.
The method of detached coefficients can be applied with ad
vantage in the case of integral functions of two letters which are
homogeneous (see below, § 17), as will be seen by the following
example : —
(J  xy + f) (x 3  2x*y + 2xf  f),
12+21
11 + 1
12+21
1422 + 1
+12+21
13 + 55 + 31,
= x"  Sx*y + Ssfy*  bx*y* + 2>x>f  y.
If the student Avill work out the above distribution, arrange
his work after the pattern of the long rule, and then compare,
he will at once see that the above scheme represents all the
essential detail required for calculating the coefficients.
The reason of the applicability of the process is simply that
the powers of x diminish by unity from left to right, and the
powers of y in like manner from right to left.
We shall give some further examples of the method of
detached coefficients, by using it to establish several important
results.
VOL. ] F
66 PASCAL'S ARITHMETICAL TRIANGLE chap.
§ 14.] Addition Rule for calculating the Binomial Coefficients.
We have to expand (x + l) 2 , (x + l) 3 , . . ., (x + l) n . Let us
proceed by successive distribution, using detached coefficients.
1 + 1 (The coefficients of x + 1 ),
1 + 1
1 + 1
+ 1 + 1
1 + 2 + 1 (The coefficients of (x + l) 2 ),
1 + 1
1 + 2 + 1
+1+2+1
1 + 3 + 3+1 (The coefficients of (x + l) 3 ).
The rule which here becomes apparent is as follows : —
To obtain the binomial coefficients of any order from those of the
previous order — 1st, Write down the first coefficient of the previous order;
2nd, Add the second of the previous order to the first of the same ;
3rd, Add the third of the previous order to the second of the same ;
and so on, taking zeros when the coefficients of the previous order run
out. We thus get in succession the first, second, third, &c, coefficients
of the new order. For example, those of the fourth order are
1 + (1 + 3) + (3 + 3) + (3 + 1) + (1 + 0),
that is, 1+4 +6 +4 +1,
which agrees with the result obtained by a different method
above, § 2 (6).
We have only to show that this process is general. Suppose
we had obtained the expansion of (x + l) n , namely, using the nota
tion of § 11,
(., + l)» = x n + n C lX nl + n C 2 .^ 2 + n C 3 .^ 3 + . . . + n C»_ r T + n C n .
Hence
(z+ l) n + 1 = (a;+ l)' l x(.c+ 1)
= (^ + n C l .r' l  1 + n C li ^ 2 + . • .+*C n _ 1 a> + n C n )(a;+l)
IV PASCAL'S TRIANGLE GENERALISED 67
using detached coefficients, we have the scheme
1 + n *~>i + nv'g + n^3 + • • • + tv^ni "i~ iv^n
1 + 1
I + »^i + n\ 2 + 7V~ n
1 + (1 + n C,) + ( U C, + M C 2 ) +....+ ( H C n _, + n C„) + ( n G n + 0).
Hence (s+l)»+ 1
= af+» + (1 + W C>" + (.0, + „cgaf»» + („C, + »C,)*» a + . . .,
in which the coefficients are formed from the coefficients of the
nth. order, precisely after the law stated above, namely,
This law is therefore general, and enables us whenever we
know the binomial coefficients of any rank to calculate those of
the next, from these again those of the next, and so on. A
table of these numbers (often called Pascal's Triangle) carried to
a considerable extent is given at the end of this chapter, among
the results and formulae collected for reference there.
§ 15.] "We may calculate the powers of x 3 + x z + x + 1 by
means of the following scheme, in which the lines of coefficients
of the constantlyrecurring multiplier, namely, 1 + 1 + 1 + 1, are
for brevity omitted.
Tower.
1st.
1
+ 1
+
1
+
1
+ 1
+
1
+
1
+
1
+
1
+
1
+
1
+
1
1
+ 2
+
1
+
1
+
1 +
1
1
2nd.
+
3
+
4
+
3
+
2 +
+ 1
+
2
+
3
+
4
+
3 +
2 +
1
+
1
+
2
+
3
+
4 +
3 +
2+ 1
1
+
1
+
2
+
3 +
4 +
3+ 2+ 1
3rd.
+ 3
+
6
+
10
+
12
+
12 +
10 +
6+3+1
+ 1
+
3
+
+
10
+
12 +
12 +
10+ 6+ 3 +
1
+
1
+
3
+
6
+
10 +
12 +
12 + 10+ 6 +
3 + 1
+
1
+
o
+
6 +
10 +
12 + 12 + 10 +
6 + 3 + 1
4th. 1 + 4 + 10 + 20 + 31 + 40 + 44 + 40 + 31 + 20+10 + 4+1
and so on.
68 X n ± if AS A PRODUCT chap.
The rule clearly is— Jo get from the coefficients of any order
the rth of the succeeding, add to the rth of that order the three preced
ing coefficients, taking zeros when the coefficients required by the ride do
not exist.
The rule for calculating the coefficients of the powers of
x n + x 71 ' 1 + x 1l ~ 2 + . . .+x+l is obtained from the above by
putting n in place of 3.
These results may be regarded as a generalisation of the pro
cess of tabulating the binomial coefficients. They are useful in
the Theory of Probability.
§ 16.] As the student will easily verify, we have
(xy)(x 2 + xy + f) = x 3 y 3 (1),
(as + y) (x 2  xy + f) = x 3 + y 3 (2).
The following is a generalisation of the first of these : —
If n be any integer,
(a;  y) (x n  1 + x n ~ % y + x n ~ 3 y 2 + . . . + xy n ~ 2 + y n ~ a ),
1 + 1 + 1 +. . .+ 1 + 1
11
1 + 1 + 1 +. . . + 1 + 1
11. . .111
1 + + + . . . + + 01
 x n  y n (3).
Again, n being an odd number,
(x + y) (x n ~ 1  x n ~hj + x n  3 y 2  • • ~ Xl f " 2 + V n ' *)>
(  sign going with odd powers of y)
1  1 + 1  . . .  1 + 1
1 + 1
11+1. . .1+1
+ 11+. . . + 11 + 1
1 + + + . . . + + 0+1
■ x tl + y n (4).
iv EXERCISES VI 69
And, similarly, n being an even number,
(x + y) (x n ~ l  x n ~ 2 y + x n ~ hf  . . . 4 xf ' 2  if ~ l )
= x n if (5).
The last two may be considered as generalisations of (2) and of
(x + if) (x  y) = x 2  if respectively.
Exercises VI.
(1. ) The variables being x, y, z, point out the integral functions among the
following, and state their degree : —
(a) Zx" + 2xy+Sy~ ;
, m 3 2 3
x xy y
(7) a?yz + yzx + z"xy + x 3 + y 3 + z 3 ;
.„ xyz* a?}/ 2 * oW
X'yz xyz xyz
Distribute the following, and arrange according to powers of x : —
(2.) 6{ajJ(a ! l)}{a!S(a!l)}+20{a!(a!l)}{ a ;(a;l)}.
x(x + l)(x + 3) a;(a;+l)(2a:+l)
(3.) g .
(4.) {(aj2)(a:3)+<a;3)(a;l) + (a:l)(a!2)}
x {(a;+2)(a;+3) + (a;+3J(a;+l) + (a:+l)(ar+2)}.
(5.) {x + a} {x + (b + c)x + bc} {x 3  (a + b + c)x" + (be + ca + ab)x  abc] .
(6.) {(x+p)(xq)(x + l)}{(xp)(x + q)(xl)}.
(7. ) (z 2  y") (x*  2f) (a?  3/) (a?  4tf) (a?  5y ).
(8.) {az + (6c)y} {te + (ca)?/} {<..>■ + («%} ;* and show that the
sum of the coefficients of xy and y 3 is zero.
(9.) Show that
(x + 4«) 4  1 Oa(x + £«) 3 + 35a 2 (.r + £a) 2  50a 3 (a; + a) + 24a 4
= (x 2 i« 2 )(ara 2 ).
(10.) Show that
( + ^)(+^)( + fl/)( + ^)( + ?2/)( + ^
_ a;y(a;2/)(gr)(rjp)(p g)
Distribute and arrange according to powers of x, the following : —
(11.) {(i+c)a! s +(c+a)aj+(a+J)}{(&c)ar J +(ca)a;+(ffl&)}.
(12.) (x n 'x + l){x 2 + x + l)(x' 1 2x + l){x i + 2x + l).
* In working some of these exercises the student will find it convenient to
refer to the table of identities given at the end of this chapter.
70 EXERCISES VI, VII chap.
(13.)! 5x 2  ix(x y) + {x y) 2 } (2s + By).
(14. ) (2x 2  Zxy + 2y 2 ) (2x 2 + Bxy + 2if).
(15.) {(x+x+l)(xx + l)(xl)}\
(16.) (x 3 x~ + xl){x* + x~ + x+l) 2 .
(17.) {lx*fr? + lx+%)(ix*+fr?hx+i)
(18.) (x*  ax*y + abxy" + bxy 3 + y*) (ax 2  abxy + by").
(19.) (x 2 + ax + b 2 ) 3 + (x 2 + ax b 2 ) 3 + (x 2  ax + b 2 ) 3 + (x 2 ax b 2 ) 3 .
(20.) (x i 2a 2 x 2 + a*)\
(21.) (.r 4 a 5 ) 3 .
(22.) (3*+)'.
(23.) (a + bx 2 )*.
(24.) {(^ + 2/ 3 )(^2/ 3 )} 9 .
(25.) (l+z + a: 2 + a; 3 + ai 4 ) 3 .
(26.) Calculate the coefficient of a; 4 in the expansion of (1+x + x 2 ) 8 .
(27. ) Calculate the coefficient of X s in (1  2x + 3x 2 + 4x? x A )\
(28.) Show that
{a + b) 3 (a 5 + b 5 ) + 5ab(a + b) 2 (a 4 + 6 4 ) + I5a 2 b 2 (a + b) {a 3 + b 3 )
+ B5a 3 b 3 (a 2 + b 2 ) + 70a i b i =(a + b)\
(29.) Show that
«Ci + n C 2 + „C 3 + . . . +„C„ = 2"1;
1 + «Co + n Ci + . . . = »Cj + n C3 + n Cs + . . . ;
(30. ) There are five boxes each containing five counters marked with the
numbers 0, 1, 2, 3, 4 ; a counter is drawn from each of the boxes and the
numbers drawn are added together. In how many different ways can the
drawing be made so that the sum of the numbers shall be 8 ?
(31.) Show that
(xy) 2 (x n  2 + x n  3 y+ . . . +xy n  3 + y n  2 ) = x n x n  l yxy n  1 + ij n .
Exercises VII.
Distribute the following, and arrange according to descending powers
of x: —
(1.) (3.i' + 4)(4;c + 5)(5;e + 6)(6a; + 7).
(2. ) (px + qr) (qx + rp) (rx +p  q).
(3. ) [x  a) (x  2a) [x  3a) {x  4a) (x + a) (x + 2a) (x + 3a) (x + 4a).
(4. ) (x 3 + 2,x 2 + 2x + 1) (or 3  3.x 2 + Bx  1 ).
(5. ) (ix 3 + \x 2 + \x + f ) (Ix 3 + \x 2 + kx + 1 ).
(6.) (x#{x*ix+l)(x+lHx*+fr+i).
7. lx 2 + x+ 7 ) (x 2 + 1 x +  JX + X+).
(8.) (2aj3)»
(9.) {{x+y)(x*xy+y*)}*.
(10.) (a; 2 l) 4 (u; + l) 10 .
iv HOMOGENEOUS FUNCTIONS 71
(11.) In the product (x + a)(x+b)(z+e), r disappears, and in the product
(x a)(x+b)(x + c), x disappears; also the coefficient of X in the former is
equal to the coefficient of a; 2 in the latter. Show that a is either or 1.
Prove the following identities : —
(12.) (bc)(xa) 2 +(ca){xb) 2 +(ab){xc) 2 + (bc)(ca){ab) = 0.
(13. ) 2(2a  b  c) (h  6) (h c) = 2(6  c){h  a).
(14.) (saf+(sb) 3 + (sc)* + 3abc = s 3 ,
where 2s = a + b + c.
(15.) (sa)*+(8b) 4 +(8c)*
= 2(s  b)%s  c) 2 + 2(s  c) 2 (s  a) 2 + 2(s  a) 2 (s  b) 2 ,
where 3s = a + b + c.
(16.) (as+bc)(bs+ ca) (cs + ab) = (b + c) 2 (c + a) 2 (a + 6) 2 , where s  a + b + c.
(17 '. ) s(s  a  d)(s  db)(s  c  d) = (s  a)(s b)(s  c) {s  d)  abed,
where 2s = a + b + c + d.
(18. ) 16(8  a) (s b)(s c) (s  d) = 4(6c + ad) 2  (6 2 + c 2  a 2  d 2 ) 2 ,
where 2s=a+b + c+d.
(19. ) 2(6  c) 6 =3n(i  cf + 2(2« 2  26c) 3 .
(20.) If U„ = (6 c) n + (c «)" + (« 6)», then
U n+3  (a 2 + b 2 + c 2 bcca ab)V n+1  (b  e) (e  a) (a  5)U„= 0.
(21.) If pi = a + b + c, p 2 = bc + ca + ab, p 3 = abc, s n = a n + b n + c n , show that
Si =lh , s 2 piSi  2p 2 , s s =piS 2  paSi + 3p 3 ,
S» =PlS„i pfin2 +PsS n 3
(22. ) If j)2= (b c)(ea) + (e a) (a  6) + (a  6) (6  c),
2i 3 =(bc){ca){ab),
s H = (b c) n + (c a) n + {a 6)»
show that
*2 =  S^pa , «3 = 3ps , Si = 2ps?, s s =  5p»ps,
s fi =  2pa 8 f 8pg 8 , s 7 = 7p 2 ps 5 20S7S3 = 21s 5 2 .
Homogeneity.
§ 17.] An integral function of any number of variables is said
to be " Homogeneous " when Vie degree of every term in it is the same.
In such a function the degree of the function (§ G) is of course
the same as the degree of every terra, and the number of terras
which (in the most general case) it can have is the number of
different products of the given degree that can he formed with
the given number of variables. If there be only two variables,
and the degree be n, we have seen that the number of possible
terms is n + 1.
72 HOMOGENEOUS FUNCTIONS chap.
For example, the most general homogeneous integral functions of x and y
of the 1st, 2nd, and 3rd degrees are *
Az+By (1),
Ax 2 + Bxy + Cy 2 (2),
Ax 3 + Bx 2 y + Cxy 2 + Vy 3 ( 3),
A, B, C, &c. , representing the coefficients as usual.
For three variables the corresponding functions are
Ax + By + Cz (4),
A* 2 + By 1 + Cz 2 + Vyz + Esse +¥xy (5),
Ax 3 + Bif + Cz 3 + Vyz 2 + P'tfz + Qzx 2 + Q,'z 2 x + Bxy 2 + Wx 2 y + Sxyz ( 6 ),
&c,
As the case of three variables is of considerable importance, we shall in
vestigate an expression for the number of terms when the degree is n,
We may classify them into — 1st, those that do not contain x ; 2nd, those
that contain x ; 3rd, those that contain x 2 ; . . . ; n + lth, those that contain x n .
The first set will simply be the terms of the ?ith degree made up with
y and z, n + 1 in number ; the second set will be the terms of the (?i — l)th
degree made up with y and z, n in number, each with x thrown in ; the third
set the terms in y and 2 of («2)th degree, nl in number, each with x 2
thrown in ; and so on. Hence, if N denote the whole number of terms,
N = (7!+l)+?l + (?ll)+ . . .+2 +1.
Reversing the righthand side, we may write
N= 1 +2+ 3 + . . . +n + (n + l).
Now, adding the two lefthand and the two righthand sides of these equali
ties, we get
2N=(»+2) + (»+2) + (» + 2)+. . . +(h + 2) + (« + 2);
= (n + l){n + 2),
since there are n + 1 terms each =n + 2.
Whence N=4(»+l)(»+2).
For example, let n = 3 ; N=£(3+ 1) (3 + 2)= 10, which is in fact the number
of terms in (6), above.
In the above investigation we have been led incidentally to sum an
arithmetical series (see chap, xx.) ; if we attempted the same problem for 4,
5, . . ., 7n variables, we should have to deal with more and more complicated
series. A complete solution for a function of the ?i.th degree in m variables
will be given in the second part of this work.
* Homogeneous integral functions are called binary, ternary, &c, accord
ing as the number of variables is 2, 3, &c. ; and quadric, cubic, &c, according
as the degree is 2, 3, &c. Thus (3) would be called a binary cubic ; (5) a
ternary quadric ; and so on.
IV HOMOGENEOUS FUNCTIONS 73
The following is a fundamental property of homogeneous
functions : — If each of the variables in a homogeneous function of the
nth degree be multiplied by the same quantity p, the result is the same
as if the function itself were multiplied by p n .
Let us consider, for simplicity, the case of three variables ;
and let
F = kxPtftf + A'«py tf' + . . .,
where p + q + r =p' + q + r' = &c, each = n.
If we multiply x, y, z each by p, we have
F = A(px)*(py)i(pzy + A'(pr)P'(pyy( P :y + . . . }
= ApP+<i+ r xPy ( }z r + A'pP'+<i'+ r 'xP'f'z r '+ . . .,
by the laws of indices. Hence, since p + q + r = p + q + r' = &c.
= n, we have
F' = p n { AxPy?z r + A'xP'y<i'z r ' +...},
= p n F,
which establishes the proposition in the present case. The
reasoning is clearly general.*
* This property might be made the definition of a homogeneous function.
Thus we might define a homogeneous function to be such that, when each
of its variables is multiplied by p, its value is multiplied by p n ; and define n
to be its degree. If we proceed thus, we naturally arrive at the idea of homo
geneous functions which are not integral or even rational ; and we extend the
notion of degree in a corresponding way. For example, (.c 3  y 3 )/(x + y) is
a homogeneous function of the 2nd degree, for ( (px) 3  {py) 3 )j{ (px) + (py) )
= p i (x 3  y 3 )/(x + y). Similarly \Z( t ' 3 + V 3 )' l/(* 2 + V") are homogeneous functions,
whose degrees are f and 2 respectively (see chap, x.) Although these ex
tensions of the notions of homogeneity and degree have not the importance of
the simpler cases discussed in the text, they are occasionally useful. The
distinction of homogeneous functions as a separate class is made by Euler in
his Introductio in Analysin fnfinitorum (1748), (t. i. chap, v.), in the course
of an elementary classification of the various kinds of analytical functions.
He there speaks, not only of homogeneous integral functions, but also of
homogeneous fractional functions, and of homogeneous functions of fractional
or negative degrees.
74 LAW OF HOMOGENEITY chap.
Example.
Consider the homogeneous integral function 3.>j 2  2xy + y 2 , of the 2nd
degree. We have
3(/w) 2  2( P x) ( P y) + (p2/) 2 = 3pV  2p"xy + p*y*,
= P *(dx i 2vy+y 2 )>
in accordance with the theorem above stated.
The following property is characteristic of homogeneous integral
functions of the first degree.
If for the variables x, y, z, . . . toe substitute Xx, + /j..r 2 , ky x
+ ay at Xz x + fxz 2 , . . . respectively, the result is the same as that
obtained by adding the results of substituting x„y Xi z lt . . . and x 2 ,
y,, z 2 , . . . respectively for x, y, z, . . . in the function, after multi
plying these results by X and fi respectively.
Example.
Consider the function Ax + By + Cz.
We have
A(X.ri + fix.,) + B(\2/! + ^ 2 ) + C(Xzi + ftna)
= AXa'x + BXyi + C\Sj + Afix? + Bp.y 2 + Cfiz*
= \( Aa?i + Byx + Czi) + fi{Ax. 2 + By 2 + Cz»).
This property is of great importance in Analytical Geometry.
§ 18.] Law of Homogeneity. — Since every term in the product
of two homogeneous functions of the mth and ?ith degrees re
spectively is the product of a term (of the mth degree) taken
from one function and a term (of the «th degree) taken from
the other, we have the following important law : —
The product of two homogeneous integral functions, of the mth and
nth degrees respectively, is a homogeneous integral function of the
(m + n)th degree,
The student should never fail to use this rule to test the
distribution of a product of homogeneous functions. If he finds
any term in his result of a higher or lower degree than that
indicated by the rule, he has certainly made some mistake. He
should also see whether all possible terms of the right degree are
present, and satisfy himself that, if any are wanting, it is owing
to some peculiarity in the particular case in hand that this is so,
and not to an accidental omission.
The rule has many other uses, some of which will be illus
trated immediately.
IV
SYMMETRICAL FUNCTIONS 75
§ 19.] If the student has fully grasped the idea of a homo
geneous integral function, the most general of its kind, he will
have no difficulty in rising to a somewhat wider generality,
namely, the most general integral function of the nth. degree in
in variables, unrestricted by the condition of homogeneity or
otherwise.
Since any integral term whose degree does not exceed the
nth may occur in such a function, if we group the terms into such
as are of the Oth, 1st, 2nd, 3rd, . . . , nth degrees respectively,
we see at once that we obtain the most general type of such a
function by simply writing down the sum of all the homogeneous
integral functions of the m variables of the Oth, 1st, 2nd, 3rd, . . .,
nth degrees, each the most general of its kind.
For example, the most general integral function of x and y of the third
degree is
A + Bx + Cy + Dx + Exy + Fy 2 + Gx 3 + Hxy + Ixrf + hf.
The student will have no difficulty, after what has been done
in § 17 above, in seeing that the number of terms in the general
integral function of the ?tth degree in two variables is
J(n+l)(n + 2).
Symmetry.
§ 20.] There is a peculiarity in certain of the functions we
have been dealing with in this chapter that calls for special notice
here. This peculiarity is denoted by the word " Symmetry "; and
doubtless it has already caught the student's eye. What we
have to do here is to show how a mathematically accurate
definition of symmetry may be given, and how it may be used
in algebraical investigations.
1st Definition. — An integral function* is said to be symmetrical
with respect to any two of its variables when the interchange of these
two throughout the function leaves its value unaltered.
* As a matter of fact these definitions and much of what follows are
applicable to functions of any kind, as the student will afterwards learn.
According to Baltzer, Lacroix (1797) was the first to use the term Symmetric
Function, the older name having been Invariable Function.
76 VARIOUS KINDS OF SYMMETRY chap.
For example, 2a + Bb + 3c
becomes, by the interchange of b and c,
2a + 3c +36,
which is equal to 2a + Sb + Be by the commutative law. Hence 2a + 3b + 3c is
symmetrical with respect to b and c. The same is not true with respect to
a and b, or a and c ; for the interchange of a and b, for example, would
produce 2b + Sa + 3c, that is, 3a + 26 + 3c, which is not in general equal to*
2a + 36 + 3c.
2nd Definition. — An integral function is said to be symmetrical
(that is, symmetrical with respect to all its variables) when the interchange
of any pair whatever of its variables would leave its value unaltered.
For example, Sx + Sij + Bz is a symmetrical function of x, y, z. So are
yz + zx + xy and 2(x 2 + y + z 2 ) + Zxyz. Taking the last, for instance, if we
interchange y and z, we get
2(x° + z + y) + 3xzy,
that is, 2(x 2 + y 2 + z 2 ) + Sxyz,
and so for any other of the three possible interchanges.
On the other hand, x' 2 y + yz + zx is not a symmetrical function of x, y, z,
for the three interchanges x with y, x with z, y with z give respectively
yx + xz + z~y,
o o o
zy + yx + xz,
xz + zy + ifx,
and, although these are all equal to each other, no one of them is equal to the
original function. It will be observed from this instance that asymmetrical
functions have a property — which symmetrical functions have not — of assuming
different values when the variables are interchanged: thus x 2 y + yz + z~x is
susceptible of two different values under this treatment, and is therefore a
twovalued function. The study of functions from this point of view has
developed into a great branch of modern algebra, called the theory of substitu
tions, which is intimately related with many other branches of mathematics,
and, in particular, forms the basis of the theory of the algebraical solution of
equations. (See Jordan, Traiti des Substihitions, and Serret, Cours d'Alg&bre
Superieure. )
All that we require here is the definition and its most elementary con
sequences.
3rd Definition. — A function is said to be collaterally symmetrical
\ X X X )
iii ttco sets of variables  ' " 2 ' ' r, each of the same number,
I a,,a. 2 ,. . ., a n )
* It may not be amiss to remind the student that for the present "equal
to" means "transformable by the fundamental laws of algebra into."
IV
RULE OF SYMMETRY 77
ivhen the simultaneous interchanges of two of the first set and of the
corresponding two of the second set leave its value unaltered.
For example, a?x + bhj + <?z
and (b + c)x + (c + a)y + (a + b)z
are evidently symmetrical in this sense.
Other varieties of symmetry might be defined, but it is
needless to perplex the student with further definitions. If he
fully master the 1st and 2nd, he will have no difficulty with the
3rd or any other case. At first he should adhere somewhat
strictly to the formal use of, say, the 2nd definition ; but, after
a very little practice, he will find that in most cases his eye will
enable him to judge without conscious effort as to the symmetry
or asymmetry of any function.*
§ 21.] From the above definitions, and from the meaning of
the word " ecpial " in the calculation of algebraical identities, we
have at once the following
Rule of Symmetry. — The algebraic sum, product, or quotient of
two symmetrical functions is a symmetrical function.
Observe, however, that the product, for example, of two
asymmetrical functions is not necessarily asymmetrical.
Thus, a + b + c and bc + ca + ab being both symmetrical, their product,
(a + b + c) ( be + ca + ab) = b"c + be 2 + c"a + ca + orb + ab 2 + 3abc,
is symmetrical.
Again, abc and abc 2 are both asymmetrical functions of a, b, c, yet their
product,
(<rbc) x (ab' 2 c") — a 3 b 3 c 3 ,
is a symmetrical function.
§ 22.] It will be interesting to see what alterations the
restriction of symmetry will make on some of the general forms
of integral functions written above.
Since the question of symmetry has nothing to do with
degree, it can only affect the coefficients. Looking then at the
* There is a class of functions of great importance closely allied to sym
metrical functions, which the student should note at this stage, namely, those
that change their sign merely when any pair of the variables are interchanged.
Such functions are called "alternating." An example is [y — z)(zx) (x — y).
Obviously the product or quotient of two alternating functions of the same
set of variables is a symmetric function. The term Alternating Function is
due to Cauchy (1812).
78 APPLICATION OF THE RULE chap.
homogeneous integral functions of two variables on page 72, we
see that, in order that the interchange of x and y may produce
no change of value, we must have A = Bin§17(l); A = C in
(2) ; A = D and B = C in (3).
Hence the symmetrical homogeneous integral functions of x and y of 1st,
2nd, 3rd, &c, degrees are
Ax + Ay (1),
Ax 2 + Bxy + Ay 2 (2),
Ax* + Bx 2 y + Bxy 2 + Ay 3 (3),
&c
The corresponding functions of x, y, z are
Ax + Ay + Az (4),
Ax 2 + Ay 2 + Az 2 + Byz + Bzx + Bxy (5),
Aa? + Ay 3 + Az* + Pyz 2 + ?y 2 z + Bzx" + Bz 2 x + Fxy 2 + T?x 2 y + Sxyz (6),
&c,
The most general symmetrical integral function of x, y of the 3rd degree
will be the algebraic sum of three functions, such as (1), (2), and (3), together
with a constant term, namely,
F + Ax + Ay + Bx 2 + Cxy + By + Bx 3 + Ex' 2 y + Exy 2 + By 3 .
And so on.
If the student find any difficulty in detecting what terms
ought to have the same coefficient, let him remark that they are
all derivable from each other by interchanges of the variables.
Thus, to get all the terms that have the same coefficient as a; 3 in
(6), putting y for x, we get y* ; putting z for x, we get z 3 ; and we
cannot by operating in the same way upon any of these produce
any more terms of the same type. Hence .r 3 , if, z 3 form one
group, having the same coefficient. Next take yz~ ; the inter
changes x and y, x and z, y and z produce xz 2 , yx 2 , yz 2 ; applying
these interchanges to the new terms, we get only two more new
terms — zx 2 , xy 2 ; hence the six terms yz 2 , y 2 z, zx 2 , £x, xy 2 , x 2 y form
another group ; xyz is evidently unique, being itself symmetrical.
§ 23.] The rule of symmetry is exceedingly useful in abbre
viating algebraical work.
Let it be required, for example, to distribute the product (a + b + c)
(a? + b 2 + c 2 bc caab), each of whose factors is symmetrical in a, b, c. The
distributed product will be symmetrical in a, b, c. Now we see at once that
the term a? occurs with the coefficient unity, hence the same must be true of
b 3 and c 3 . Again the term b'c has the coefficient 0, so also by the principles
of symmetry must each of the five other terms, be 2 , c"«, cu 1 , ab 2 , a 2 b, belonging
to the same type. Lastly, the term abc is obtained by taking a from the
first bracket, hence it must occur by taking b, and by taking c, that is, the
iv INDETERMINATE COEFFICIENTS 79
a&cterni must have the coefficient  3. We have therefore shown that
(ct + b + c) (a? + b 2 + c bcca ab) = a? + b* + c 3  Zabc ; and the principles of
symmetry have enahled us to abbreviate the work by about twothirds.
PRINCIPLE OF INDETERMINATE COEFFICIENTS.
§24.] A still more striking use of the general principles of
homogeneity and symmetry can be best illustrated in conjunction
with the application of another principle, which is an immediate
consequence of the theory of integral functions.
We have laid down that the coefficients of ah integral function
are independent of the variables, and therefore are not altered by
giving any special values to the variables. If, therefore, on cither
side of any algebraic identity involving integral functions we determine
the coefficients, either by general considerations regarding the forms of
the functions involved, or by considering particular cases of the identity,
then these coefficients are determined once for all. This has (not very
happily, it must be confessed) been called the principle of inde
terminate coefficients. As applied to integral functions it results
from the most elementary principles, as we have seen ; when
infinite series are concerned, its use requires further examination
(see the chapter on Series in the second part of this work).
The following are examples : —
(x + y) 2 = (x + y)(x + y), being the product of two homogeneous
symmetrical functions of x and y of the 1st degree, will be a
homogeneous symmetrical integral function of the 2nd degree ;
therefore (x + yf = Ax 2 + Bxy + Ay 2 (1).
"We have to determine the coefficients A and B.
Since the identity holds for all values of x and y, it must
hold when x = 1 and y = 0, therefore
(1 +0) 2 = A1 2 + B1 xO + AO 2 ,
1=A.
We now have (x + y) 2 = x 2 + Bxy + y 2 ;
this must hold when x = 1 and y   1,
therefore (1  1) 8 = 1 +B.1.( 1) + 1,
that is, = 2  B,
whence B = 2.
Thus finally (x + y) 2 = x 2 + 2xy + y 2 .
80 INDETERMINATE COEFFICIENTS CHAr.
This method of working may seem at first sight somewhat startling, but
a little reflection will convince the learner of its soundness. We know, by
the principles of homogeneity and symmetry, that a general identity of the
form (1) exists, and we determine the coefficients l>y the consideration that
the identity must hold in any particular case. The student will naturally ask
how he is to be guided in selecting the particular cases in question, and
whether it is material what cases he selects. The answer to the latter part of
this question is that, except as to the labour involved in the calculation, the
choice of cases is immaterial, provided enough are taken to determine all the
coefficients. This determination will in general depend upon the solution of
a system of simultaneous equations of the 1st degree, whose number is the
number of the coefficients to be determined. (See below, chap, xvi.) So fat
as possible, the particular cases should be chosen so as to give equations each
of which contains only one of the coefficients, so that we can determine them
one at a time as was done above.
The student who is already familiar with the solution of simultaneous
equations of the 1st degree may work out the values of the coefficients by
means of particular cases taken at random. Thus, for example, putting x=2,
y = 3, and x = l, y = i successively in (1) above, we get the equations
25 = 13A + 6B,
25 = 17A + 4B,
which, when solved in the usual way, give A = 1 and B = 2, as before.
We give one more example of this important process : —
By the principles of homogeneity and symmetry we must have
(x + y + z) (x 2 + y 2 + z  yz  zx  xy)
= A(.r> + y 3 + z 3 ) + B(yz 2 + y 2 z + zx 2 + z 2 x + xy 2 + x*y) + Cyxz.
Putting sc=l, y = 0, z — 0, we get 1=A.
Using this value of A, and putting x=\, y = l, z = 0, we get
2xl = 2 + Bx2,
that is, 2 = 2 + Bx2,
therefore 2B = 0,
and therefore B = 0.
Using these values of A and B, and putting x=1, y=l, 2 = 1, we get
3xO=3 + C,
that is, = 3 + C,
therefore C =  3 ;
and we get finally
(x + y + z) (x 2 + y 2 + z 2 yzzxxij) = x 3 + if + z z 3xyz (2),
as in § 23.
§ 25.] Reference Table of Identities. — Most of the results given
below will be found useful by the student in his occasional calcu
lations of algebraical identities. Some examples of their use
TABLE OF IDENTITIES
81
have already been given, and others will be found among the
Exercises in this chapter. Such of the results as have not
already been demonstrated above may be established by the
student himself as an exercise.
(x + a) (x + b) = x" + (a + b)x + ab ;
(x + a) (x + b) (x + c) = x 3 + (a + b + c)x*
+ (be + ca + ab)x + abc ;
and generally
(x + «,) {x + a 2 ) ...(/• + a n ) = x n + P,.^ 1 " 1 + P^" 2
+ . . . + P n _^ + P w (see §9).
(x ± yf  x 2 ± 2xy + if ;
(x ± yf = x 3 ± 3x 2 y + Zxif ± if j
&c. ;
the numerical coefficients being taken from the following
table of binomial coefficients : —
>
Hi)
Tower.
Coefficients.
1
1
1
o
1
2 1
3
1
3 3
1
4
1
4 6
4
1
5
1
5 10
10
5
1
6
1
6 15
20
15
6 1
1
7 21
35
35
21 7 1
8
1
8 28
56
70
56 28 8 1
9
1
9 36
84 126 126 84 36 9
1
10
1
10 45 120
210
252 210 120 15
10 1
11
1
11 55
165
330 462 462 330 165
55 11 1
12
1 12 G6
220 495
792 924 792 495
220 66 12 1
&c.
Mil.)
* This table first occurs in the Arithmctica Integra of Stifel (1544), in
connection with the extraction of roots. It does not appear tli.it he was
aware of the application to the expansion of a binomial The table was dis
cussed and much used by Pascal, and now goes by the name of Pascal's
Arithmetical Triangle. The factorial formula' for the binomial coefficients (see
the second part of this work) were discovered by Newton.
VOL. I G
82
TABLE OF IDENTITIES
 1i n
(x ± yf t % = (a t y) 2 
(x + y) (x y) = x 2 y 2 ;
(x ± y) (x 2 T sqf + y 2 ) = x 3 ± / ;
and generally
(x _ y) ( x « i + x n ~ h) + . . . + «y n ~ 2 + f ~ l ) = a*
(z + y) (sc n  1  » n  2 y + . . . t ™/ n " 2 ± y n  1 ) = aJ« ± y»
upper or lower sign according as w is odd or even.
(x* + f)<f + y' 2 ) = {xx' T yy') 3 + (^ ± FT;
(.<"  *■)(*"  2/' 2 ) = (a* ± ^7 ~ (*y ± 3*Ti
(/ + y 2 + «■)(*/■ + y" + *' 2 ) = ixx' + yy' + zz') 2 + (yz'  y'z) 2
+ (zx'z'x) 2 + (xy'x'y) 2 ;
(x 2 + y 2 + z 2 + u 2 )(x' 2 + y' 2 + z" + u' 2 ) = (xx' + yy' + zz' + uu') 2
+ (xy  yx' + zu'  uz') 2
+ (xz'  yu'  zx' + uy') 2
+ (xu' + yz'  zy'  ux') 2 .
(x 2 + xy + f) (x 2  xy + y 2 ) = x* + x 2 y 2 + y\
( a + l ) + c + d) 2 = a 2 + b 2 + c 2 + d 2 + 2ab + 2ac + 2ad
+ 2bc + 2bd + 2cd ;
and generally
(a l + a 2 + . . . + a n ) 2 = sixm of squares of a ti a 2 , . . ., a,
+ twice sum of all partial products two and two.
(a + b + c) 3 = a 3 + b 3 + c 3 + U 2 c + 3bc 2 + Zc'a + Sea* + 3a b
+ Sab 2 + Gabc
= a 3 + b 3 + c 3 + 36c (6 + c) + 3ca(c + a)
+ Sab (a + b) + Saba
( a + b + c) (a 2 + b 2 + c 2 bc ca  ab)  a 3 + b 3 + c 3  3abc
(b  c) (c  a) (ab)=  a\b  c)  b\c  a)  c\a  b),
= a(b 2  c 2 ) + b(c 2  a 2 ) + c(a 2  b%
  bc(b c) ca(c  a)  ab(a  b),
= + be 2  b 2 c + ca 2  c 2 a + ab 2  a 2 b.
CHAP.
(III.)
MIV.)
Uy.y
(VI.)
I (VII.)
(VIII.)
(IX.)
* These identities furnish, inter alia, proofs of a series of propositions in
the theory of numbers, of which the following is typical :— If each of two
integers he the ram of two squares, their product can he exhibited in two ways
as the sum of two integral squares.
iv EXERCISES VIII 83
(b + c)(c + a)(a + b) = a s (b + c) + b 2 (c + a) + c 2 (a + b) + 2abc, \
= bc(b + c) + ca(c + a) + ab(a + b) + 2abc, I (XI.)
= be 2 + b 2 c + ca 2 + c 2 a + ab 2 + ab + 2dbc. J
(a + b + c) (a 2 + b 2 + c 2 ) = bc(b + c) + ca(c + a) + ab(a + b) \ ,
+ a 3 + b 3 + c\ j (XIL)
(a + b + c) (Lc + ca + ab) = a 2 (b + c) + b 2 (c + a) + c 2 (a + b) \ /VTTT
+ 3abc, J (X111)
(b + c a) (c + a b) (a + b  c) = a 2 (b + c) + b 2 (c +•«) (
+ c\a + b)a 3 b 3 c 3  2abc, I ( '
(a + 6 + c)(  a + 6 + c)(a  b + c)(a + b  c) = 2b 2 c 2 + 2c 2 a 2 ) „
+ 2a*b*a t b i c*. / (XV.)
(6  c) + (c  a) + (a  b) = ; \
a(b c)+ b(c a) + c(a  b) = ; V (XVI.)
(6 + c)(fi  c) + (c + a)(c a) + (a + b) (a b) = 0. )
Exercises VIII.
(1.) Write down the most general rational integral symmetrical function
of x, y, z, u of the 3rd degree.
(2.) Distribute the product {x"y + y"z + z"x) (xy' + yz^ + zx 2 ). Show that
it is symmetrical ; count the number of types into which its terms fall ; and
state how many of the types corresponding to its degree are missing.
(3.) Construct a homogeneous integral function of x and y of the 1st
degree which shall vanish when x = y, and become 1 when x =5 and ?/ = 2.
(4.) Construct an integral function of x and y of the 1st degree which
shall vanish when x = x', y = y', and also when x—x", y = y".
(5.) Construct a homogeneous integral function of x and y of the 2nd
degree which shall vanish when x = x', y = y', and also when x — x", y = y", and
fchall become 1 when aj=l, y = l.
(6.) If A(a:3)(iB5) + B(aJ5)(aj7) + C(a!7)(a!3)=8iB120 for all
values of x, determine the coefficients A, B, C.
(7.) Show that 5.r 2 + 19a;+18 can be put into the form
l{x  2) {x  3) + m(x  3) (x  1) + n{x  1 ) (x  2) ;
and find I, m, n.
(8.) Assuming that {x  1) (x  2) (a; 3) can be put into the form
l(xl) {x +2){x + 3) + vi{x2)(x + B)(x + l) + n(x  3) (*+ 1) (»+ 2),
determine the numbers 7, m, n.
* Important in connection with Hero's formula for the area of a plane]
triangle.
84 EXERCISES VIII chap, iv
(9.) Find a rational integral function of x of the 3rd degree which shall
have the values P, Q, R, S when x = a, x=b, x = c, x = d respectively.
(10.) Find the coefficients of yz and yz in the expansion of
(ax + by + cz) (ax 4 by 4 c 2 z) (a 3 x 4 b 3 y 4 <?z).
(11.) Expand and simplify 2(</ 2 + z 2  x 2 ) (y + z  x).
Trove the following identities : —
(12.) (ad + bc) 2 + (a + b + cd)(a + bc + d)(b + d)(bd) = (b 2 d 2 + ab + cd) 2 .
*(13.) 2(b 2 + c 2 a 2 + bc + ca + ab) 2 (c 2 b 2 ) = 4(b 2 c 2 )(c 2 a 2 )(a 2 b 2 ).
(14. ) 2(ca  b 2 ) (ab  c 2 ) = (26c) (26c  2a 2 ).
(15.) 2(&c'  b'c) (be"  b"c) = 2a 2 2rt'«"  2aa'2W.
(16.) 31% 4 z)  62?/~ = 2a(2.c 1) (2jj 2)  Saj(a1) (a!2).
(17.) Z(b 2 + c 2 a")/2bc=(ip 1 2>2lh 3 fy3)/ty3, where Pl =  2a, p, = Zbc,
p 3  abc.
(18. ) 1% 4 ;) 2 4 2x 2 y 2 z 2  2x*(y 4 z) 2 = 2(2 r ) 3 .
(19.) S(a: +yz){(y z) 2 (zx)(x y)\ = 2a* Zxyz.
(20. ) n(a± b±c±d) = 2a 8  42a 6 6 2 4 62« 4 i 4 4 42aW  40a 2 6 2 c 2 rf2.
(21.) Show that
(x 3 4 y 3 4 s 3  3ays) a = X 3 4 Y 3 4 Z 3  3X YZ, where X = x 2 4 2y;, &c. ;
also that
CZx 3  3xyz) (2a;' 3  2>x'y'z) = S(a»' 4 ysf* 4 ?/~~) 3  3II(»/ 4 yz' 4 y'z).
(These identities have an important meaning in the theory of numbers. )
(22.) Show that, if n be a positive integer, then
lh+h~. ■ .!(»even) = 2('^L + L + . . .+!)■
1i+J. . . + l(»odd)=2/l + _l_ + . . .+1
n \n + l n + 3 2n
(Blissard).
* In this example, and in others of a similar kind, 2 is not used in its
strict sense, but refers only to cyclical interchanges of a, b, c; that is, to
interchanges in which a, b, c pass into b, c, a respectively, or into c, a, b
respectively. Thus, 2cr(&c) is, strictly speaking, =0; but, if 2 be used in
the present sense, it is a 2 (b c)+ b 2 (c a)+ c 2 (a  b).
CHAPTER V.
Division of Integral Functions— Transformation of
Quotients.
§ 1.] The operations of this chapter are for the most part
inverse to those of last. Thus, A and D being any integral
functions of one variable x* and Q a function such that
D x Q = A, then Q is called the quotient of A by D ; A is called
the dividend and D the divisor. We symbolise Q by the nota
tion A 7 D, A/D, or =, as explained in chap. i.
The operation of finding Q is called division, but we prefer
that the student should class the operations of this chapter under
the title of transformation of quotients.
A and D being both integral functions, Q will be a rational
function of x, but will not necessarily be an integral function.
When the quotient can he transformed so as to become integral, A
is said to be exactly divisible by D.
IFhen the quotient cannot be so transformed, the quotient is said
to be fractional or essentially fractional.
It is of course obvious that an essentially integral function cannot
be equal, in the identical sense, to an essentially fractional function.
§ 2.] When the quotient is integral, its degree is the excess of the
degree of the dividend over tlie degree of the divisor. For, denoting
* For reasons partly explained below, the student must be cautious in
applying many of the propositions of this chapter to functions of more vari
ables than one ; or at least in such cases he must select one of the variables
at a time, and think of it as the variable for the purposes of this chapter.
86 THEOREM REGARDING DIVISIBILITY chap.
the degrees of the functions represented bj 7 the various letters
by suffixes, we have
therefore, by chap, iv., § 7, m =p + n, that is, p = m  n.
§ 3.] If the degree of the dividend be less than that of the divisor,
the quotient is essentially fractional. For, m being <??, suppose, if
possible, that the cpiotient is integral, of degree p say, then
therefore m=p + n; but p cannot be less than by our hypo
thesis, and m is already less than n, hence the cpiotient cannot
be integral, that is, it must be fractional.
§ 4.] If A, D, Q, R be all integral functions, and if A =
QD + R, then R will be exactly divisible by D or not according as A
is exactly divisible by D or not.
For, since A = QD + R,
A _ QD + R R
therefore ~ = j  Q.
Now, if A be exactly divisible by D, A/D will be integral, and
A/D  Q will be integral, that is, R/D will be integral, that is, R
will be exactly divisible by D.
Again, if A be not exactly divisible by D, A/D will be
fractional. Hence R/D must be fractional, for, if it were
integral, Q + R/D would be integral, that is, A/D would be in
tegral, which is contrary to hypothesis.
INTEGRAL QUOTIENT AND REMAINDER.
§ 5.] The following is the fundamental theorem in the
transformation of quotients.
A^ and D n being integral functions of the degrees m and n respect
ively, we can always transform the quotient A 7n /D„ as follows . —
A ffl _ p ^i
XJ n XJ n
V INTEGRAL QUOTIENT AND REMAINDER 87
where P TO _ n is an integral function of degree m  n, and R (if it do
not vanish) an integral function whose degree is at most n 1.
This transformation is effected by a series of steps. "We shall
first work out a particular case, and then give the general proof.
Let Ah = 8a* + 8* 5  20a 4 + 40* 3  50* 2 + 30* 10,
D 4 = 2* 4 + 3* 3 4* 2 + 6*8,
multiply the divisor D 4 by the quotient of the highest term of the dividend
by the highest term of the divisor (that is, multiply D4 by 8* 6 /2* 4 = 4* 2 ), and
subtract the result from the dividend A 6 . "We have
A 6 = 8* fi + 8* 5  20k 4 + 40* 3  5 Ox + 30*  1
4* 2 D 4 = Sx e + 1 2* 5  1 6* 4 + 24.C 3  32a?
A 6 4*'D 4 =  4ar» 4* 4 + 16* 3  18a; 3 + 30* 10
= A 3 say ;
therefore A 6 = 4* 2 D 4 + A 5 (1).
Repeat the same process with the residue A 5 in place of Ae, and we have
As=  4a*  is* + 16* 3  18* 2 + 30*  10
 2*D 4 =  4* 5  6* 4 + 8*'  1 2.7r + 1 Qx
A 3 + 2*D 4 = 2x*+ 8X 3  Qx + 14* 10
= A 4 say;
therefore A 5 =  2*D 4 + A 4 (2).
And again with A 4 ,
A 4 = 2x* + 8x*  6* 2 + 14*  10
1 x D 4 = 2** + 3*"  4* 2 + 6*  8
A 4 D 4 = 5* 3 2* 2 + 8* 2
= A 3 say ;
therefore A 4 =D 4 + A 3 (3).
Here the process must stop, unless we agree to admit fractional multi
pliers of D 4 ; for the quotient of the highest term of A :! by the highest term of
D 4 is 5* 3 /2* 4 , that is, f/*, which is a fractional function of *. Such a con
tinuation of the process does not concern us now, but will be considered below.
Meantime, from (1) we have
A 6 =4*D 4 + A 5 (4) •
and, using (2) to replace A 3 ,
A 6 = 4*D 4  2*D 4 + A 4 (5 ) ;
and finally, using (3),
A 6 = 4* 2 D 4 2*D 4 + D 4 + A 3 ,
= (4* 2 2* + l)D 4 + A 3 (6).
Hence A 6= J4* 2 2a; + l)D 4 + A,
D 4 D 4
= 4* 2 2* + l + 1 4>' ;
•U4
88 INTEGRAL QUOTIENT AND REMAINDER CHAP.
or, replacing the capital letters by the functions they represent,
8 ,> ; 6 + 8a; 5  20.r 4 + 40.T 3  50.r 2 + 30.r  10
2x* + 3x :i  ix + 6x  8
. , 5,r 3 2x 2 + 8a;2 ...
= ^2*+l+ ^ + 8a ,_ 4a . + fla ,_ 3 (7).
Since 64 = 2, it will be seen that we hare established the above theorem
for this special case. It so happens that the degrees of the residues A 5 , A 4 ,
A 3 diminish at each operation by unity only ; but the student will easily see
that the diminution might happen to be more rapid ; and, in particular, that
the degree of the first residue whose degree falls Tinder that of the divisor
might happen to be less than the degree of the divisor by more than unity.
But none of these possibilities will affect the proof in any way.
We shall return to the present case immediately, but in the first place we
may give a general form to the proof of the important proposition which we
are illustrating.
§ 6.] Let A m = 2V m +ja l x m ~ 1 + p 2 x m ~ 2 + &c  1
D n —q v x n + q^ 1 ' 1 + q a x n ~ 2 + &c.
Multiplying D n by the quotient p$? n jq x n , that is, by (p /qo) xm ~ n >
and subtracting the result from A,„,, we get
= A m _, say,
whence, denoting p /q by r for shortness, we get
Am = tx D n + A m _ ! ( I ).
Treating A m _t in the same way, we get
A m _, = SX D n + A m _ 2 {')•
And so on, so long as the degree of the residue is not less
than n, the last such equation obtained being —
A„ = m>D„ + K (3),
where E is of degree n  1 at the utmost. Using all these
equations in succession we get
_ (rtfnn + stfmn1 + . . . + w )J) n + R ;
whence, dividing both sides by D n , and distributing on the
right,
A _ ? r m  » T)
v THE ORDINARY DIVISION TRANSFORMATION UNIQUE 89
A R
^ = rs m  " + sx m  n ~ 1 + . . .+«;+'=,
which, if we bear in mind the character of R, gives a general
proof of the proposition in cpiestion.
§ 7.] We have shown that the transformation of § 5 can
always be effected in a particular way, but this gives no assur
ance that the final result will always be the same. The proof
that this really is so is furnished by the following proposition: —
The quotient A/D of two integral functions can be put into the
form P + R/D, where P and R are integral functions and the degree
of R is less than that of D, in one way only.
If possible let
1 A TV R '
and D = P+ D'
where P, R and P', R' both satisfy the above requirements ;
then p + ? = P' + ^ ;
D D '
subtracting P' + =r from both sides, we have
R' R m
r r ~ 1 ) D '
whence —  = P  P'.
D
Now, since the degrees R and R are both less than the degree
of D, it follows that the degree of R'  R is less than that of D.
Therefore, by § 3, the lefthand side, (R'  R)/D, is essentially
fractional, and cannot be equal to the right, which is integral,
unless R'  R  0, in which case we must also have P  P' = 0,
that is, R = R', and P = P\
§ 8.] The two propositions of §§ 5, 7 give a peculiar import
ance to the functions P and R, of which the following definition
may now legitimately be given : —
If the quotient A/D be transformed into V + R/D, P and R being
90 CONDITIONS FOR EXACT DIVISIBILITY chap.
integral and R of degree less than D, P is called the integral quotient,
and R the remainder of A when divided by D.
§ 9.] We can now express the condition that one integral
function A may be exactly divisible by another D. For, if E be
the remainder, as above denned, we have, P being an integral
function, —
^P + ?
whence, subtracting P from both sides,
 Ax* 
 Ax* 
Ax* + 16a; 3  18a; 2 + 30a; 10
6a; 4 + 8a; 3  12a; 2 + 16a;
2x* + 8/ 6a; 2 + I Ax 10
2»* + 3a; 3  4a 2 + 63 8
2x 4 + Sx 3  4/ + 6.r  8
4a: 2  2x + 1
A B
D D
Now, if A be exactly divisible by D, A/D will be integral, and
therefore A/D  P will be integral. Hence R/D must be integral ;
but, since the degree of R is less than that of D, this cannot be
the case unless R vanish identically.
The necessary and sufficient condition for exact divisibility is there
fore that the remainder shall vanish.
When the divisor is of the nth degree, the remainder will in
general be of the (n  l)th degree, and will contain n coefficients,
every one of which must vanish if the remainder vanish. In
general, therefore, when the divisor is of the nth degree, n conditions
are necessary to secure exact divisibility.
§ 10.] Having examined the exact meaning and use of the
integral quotient and remainder, we proceed to explain a con
venient method for calculating them. The process is simply a
succinct arrangement of the calculation of §§ 5, G. It will be
sufficient to take the particular case of § 5.
The work may be arranged as follows : —
8x° + 8x*  20a; 4 + 40a; 3  5 Ox 2 + 3 Ox  10
8a; 6 + 1 2/ 16a: 4 + 24a; 3 32a' 2
5x 3  2x 2 + 8x 2
v DETACHED COEFFICIENTS 91
Or, observing that the term  10 is not waited till the last
operation, and therefore need not be taken down from the upper
line until that stage is reached, and observing further that the
method of detached coefficients is clearly applicable here just as
in multiplication, we may arrange the whole thus : —
8+ 8  20 + 40  50 + 30  10  2 + 3  4 + 6  6
42 + 1
8 + 1216 + 2432
I
 4 4 + 1618 + 30
 4 6 + 812 + 16
2+ 8 6 + 14
10
2+ 3 4+ 6
8
5 2+8 2
Therefore, Integral quotient = 4/  2x + 1 \
Remainder = 5x 3  2x 2 + 8x  2.
The process may be verbally described as follows : —
Arrange both dividend and divisor according to descending powers
of x, filling in missing powers with zero coefficients. Find the quotient
of the highest term of the dividend by the highest term of the divisor ;
the result is the highest term of the " integral quotient."
Multiply the divisor by the term thus obtained, and subtract the
result from the dividend, taking down only one term to the right beyond
those affected by the subtraction ; the result thus obtained will be less in
degree than the dividend by one at least. Divide the highest term of
this residt by the highest of the divisor ; the result is the second term
of the " integral quotient."
Multiply the divisor by the new term just obtained, and subtract,
&c, as before.
The process continues until the result after the last subtraction is
less in degree than the divisor; this last result is the remainder as
above defined.
§ 11.] The following are some examples of the use of the
" long rule " for division.
92
Example 1.
EXAMPLES
i. 19 „ 5 l_
I 36 36 r + 36
1 _ 5 i 19 _ 6 i 1
,x a; + 4
1i + l
5
3ff
"A
i
•B
1
V
— 141
J 1 "}
CHAP.
+
The remainder vanishes, therefore the division is exact, and the quotient is
Example 2.
l+p +q
I a
{x 3 +px 2 + qx + r)i[xa).
+r 1a
1 + (a +p) + {a 2 + ap + q)
{a+p) + q
(a+p){a 2 + ap)
{a? + ap + q) + r
(a 2 + ap + q) (a 3 + a 2 p + aq)
(a 3 + a 2 p + aq + r)
Hence the integral quotient is
x 2 + (a+p)x + (a 2 + ap + q) ;
and the remainder is
a 3 + a 2 p + aq + r.
The student should observe the use of brackets throughout to preserve the
identity of the coefficients.
Example 3.
(re 4  2>a 3 b + 6a 2 b  Zab 3 + 1*) 4 (a 2 ab + b 2 ).
1st. Let us consider a as the variable. Since the expressions are homo
geneous, we may omit the powers of b in the coefficients, and use the numbers
merely.
13 + 63 + 1 I 11+1
11 + 1
2+53
2+22
12 + 3
a 2 2ab + Sb 2
31 + 1
33 + 3
" 22
2ab 3  2b*
BINOMIAL DIVISOR
93
whence
i i 3a*b + 6crbSab* + b i
= a?2ab + M +
2a't?  2b*
a 2 ab + b 2 '
We must then arrange according
a 2 ab + b'
'2nd. Let us consider b as the variable,
to descending powers of b, thus—
(6 4  3ab 3 + 6ab°  3a?b + a 4 ) + {b ab + a).
Detach the coefficients, and proceed as before. It happens in this particular
case that the mere numerical part of the work is exactly the same as before ;
the only difference is in the insertion of the powers of a and b at the end.
Thus the integral quotient is b 2 2ba + 3a", and the remainder is 26a 3 2a 4 ,
whence
— , , ,., =Za2ab + b + — — r .
a 2 ~ab + b 1 a  ab + b 2
§ 12.] The process of long division may be still further
abbreviated (after expertness and accuracy have been acquired)
by combining the operations of multiplying the divisor and sub
tracting. Then only the successive residues need be written.
Thus contracted, the numerical part of the operations of Example
3 in last paragraph would run thus : —
13+63+1 1 11+1
2+53+i;i2+3
31 + 1
22
BINOMIAL DIVISOR — REMAINDER THEOREM.
§ 13.] The case of a binomial divisor of the 1st degree is of
special importance. Let the divisor be x  a, and the dividend
p^ 1 + p x x
n i
+ Pr
,,«2 +
+ Pn&+Pn
Then, if we employ the method of detached coefficients, the
calculation runs as follows : —
P0+P1
PoPo«
+ Pa +
+ Pni+Pr.
(P&+Pi)+F»
(P&+Pi)(Poa+Pi)a
!
+ (p<fl.+p 1 )
+ (Po° 2 +Pia+pJ
(j><P+Pi*+P*)+pa
(j><P+Pi* + Pi) ~ (Po* +p x o.+ p a )a
(p a + P,a + p,a + p 3 )
94 RULES FOR COEFFICIENTS OF chap.
The integral quotient is therefore
pjFl + {p o a + Pl )x n ~* + {p/ + Pl a +pX' 2 + ••'
The law of formation of the coefficients is evidently as follows: —
The first is the first coefficient of the dividend ;
The second is obtained by multiplying its predecessor by a and
adding the second coefficient of the dividend ;
The third by multiplying the second just obtained by a and adding
the third coefficient of the dividend ; and so on.
It is also obvious that the remainder, xchich in the present case is
of zero degree in x (that is, does not contain x), is obtained from the
last coefficient of the integral quotient by multiplying that coefficient by
a and adding the last coefficient of the dividend.
The operations in any numerical instance may be con
veniently arranged as follows : — *
Example 1.
(2x i Sx 2 + 6x4) + (x2).
2+03+ 6 4
+ 4 + 8 + 10 + 32
2 + 4 + 5 + 16 + 28
Integral quotient = 2a? + 4a; 2 + 5x + 1 6 ;
Remainder = 28.
The figures in the first line are the coefficients of the dividend.
The first coefficient in the second line is 0.
The first coefficient in the third line results from the addition of the two
ahove it.
The second figure in the second line is obtained by multiplying the first
coefficient in the third line by 2.
The second figure in the third line by adding the two over it.
And so on.
Example 2.
If the divisor be x + 2, we have only to observe that this is the same as
* The student should observe that this arrangement of the calculation of
the remainder is virtually a handy method for calculating the value of an
integral function of x for any particular value of x, for 28 is 2 x 2 4  3 x 2 3
+ 6x24, that is to say, the value of 2x*  Zx* + 6x  4 when x2 (see § 14).
This method is often used, and always saves arithmetic when some of the
coefficients are negative and others positive. It was employed by Newton ;
see Horsley's edition, vol. i. p. 270.
v DIVISOB, AND FOR REMAINDER 95
x{2); and we see that the proper result will be obtained by operating
throughout as before, using  2 for our multiplier instead of + 2.
(2a*  3a? + 6b 4) Ha; +2)
= (2a, 4  3ar + 6»  4) = (x  (  2)).
2+03+ 64
04+810+8
24 + 5 4 + 4.
Integral quotient = 1)?  Ax + 5x  4 ;
Remainder =4.
Example 3.
The following example will show the student how to bring the case of any
binomial divisor of the 1st degree under the case of a; a.
3f '  2X 3 + 3x 2  2x + 3 _ 3.x 4 2j? + 3 x 2  2.v + 3
3a: + 2 ~ 3(x + f)
( 3b 4 2a; 3 + 33? 2a; + 3 "1
Transforming now the quotient inside the bracket { } , we have
32+ 32+3
— 01 s _ 34 i 1 04
SiJ.n_58i.lS5
I ntegral quotient = Sx 3  4a? + ^x  Af .
Whence
Remainder — y 7 5 .
3a: 4 
2 S5 »+8a?2a!+3 f A.it,  , W
_te + 2 = *t 3a! 4x + ^ x ¥+ «s(l
185
— ar* ^x +5 a. #?+•> , .>•
Hence, for the division originally proposed, we have —
Integral quotient =a?a? + ^aj :]; ;
Remainder = V\ 5 .
The process employed in Examples 2 and 3 above is clearly
applicable in general, and the student should study it attentively
as an instance of the use of a little transformation in bringing
cases apparently distinct under a common treatment.
§ 14.] Reverting to the general result of last section, we see
that the remainder, when written out in full, is
p a n +_p 1 a" 1 + . . . + £>„_!<_ +p n .
96 REMAINDER THEOREM CHAP.
Comparing this with the dividend
IV'" + Pi*" " + • • • + Pn  i x + Pn ,
we have the following " remainder theorem " : —
When an integral function of x is divided by x  a, the remainder
is obtained by substituting a for x in the function in question.
In other words, the remainder is the same function of a as
the dividend is of x.
Partly on account of the great importance of this theorem,
partly as an exercise in general algebraical reasoning, we give
another proof of it.
Let us, for shortness, denote
jj x 11 + p t x n  l + . . ■+p n  1 x+ p n by f(x),
/'(a) will then, naturally, denote the result of substituting a for
x in f(x), that is,
p o n +p l a 11  1 + . . . +!>„_, a +p n .
Let x(') denote the integral quotient, and R the remainder,
when f(x) is divided by x  a. Then x( x ) i s an integral function
of x of degree nl, and R is a constant (that is, is independent
of ,»■), and we have
X a ^ X a
whence, multiplying by x  a, we get the identity
f( X ) = (X  a) X (x) + R.
Since this holds for all values of x, we get, putting x = a
throughout,
/(a) = (a  a) X (a) + R,
where R remains the same as before, since it does not depend on
x, and therefore is not altered by giving any particular value
to .''.
Since x(«) i s finite if a be finite, (a  a)x(a) = x x («) = ;
and we get finally
/(a) = R,
which, if we remember the meaning of /(a), proves the "re
mainder theorem."
v FACTORISATION BY REMAINDER THEOREM 97
Cor. 1. Since x + a  x  (  a), it follows that
The remainder, when an integral function f(x) is divided by
X + a, isf(a).
For example, the remainder, when v 4  2>x* + 2a?  5x + 6 is divided by
as+10, is (10) 4 3(10) 3 + 2(10)5(10) + 6 = 13256.
Cor. 2. The remainder, when an integral function of x, f(x), is
divided by ax + b, is /(  bja).
This is simply the generalisation of Example 3, § 13, above.
By substitution we may considerably extend the application
of the remainder theorem, as the following example will show: —
Consider p m (x") m +p m  1 (x» J" 1  1 + . . . +Pi(x")+p and x n  a". Writing
for a moment £ in place of x n , and a in place of a", we have to deal with
Pm$ m +p m iZ m ~ 1 + • • ■ +]h^+Po and £<x. Now the remainder, when the
former of these is divided by the latter, is p m a" l +p m ia m  1 + . . . +Pia + p .
Hence the remainder, when p m (x") m +p m i{x n ) m  1 + . . . +piaf*+po is divided
by x n a n , is 2} m (a") m + p m i(a») m  1 + . . . +p x a n +2)o
APPLICATION OF REMAINDER THEOREM TO THE DECOMPOSITION
OF AN INTEGRAL FUNCTION INTO LINEAR FACTORS.
§ 15.] If a„ a 2 , . . . , a r be r different values of x, for which the
integral function of the nth degree f(x) vanishes, where r < n, then
f{x)  (x  c^) (x  Og) . . . (x  a. r )cj> n _ t (.r), cf) n _ r (x) being an integral
function of x of the (n  r)th degree.
For, since the remainder, /(a,), when f(x) is divided by x  a„
vanishes, therefore f(x) is exactly divisible by x  a„ and we have
where <£ n _i(.>') is an integral function of x of the (n  l)th degree.
Since this equation subsists for all values of x, we have
/(a,) = (a a  a!)4>ni(<h),
that is, by hypothesis, = (a. 2  <*,)<£„_ ,(a 2 ).
Now, since a! and a 2 are different by hypothesis, a.,  a t 4= ;
therefore <f> n i(a 2 ) = 0.
Hence, <f> n \( 1 ') is divisible by (x  a,),
that is, </,„ _ v (x) = (x  a 2 )<f> n _ . 2 (x) ;
whence f(x) = (x  a t ) (x  a 2 )<j> n _ 2 (x).
VOL. I H
98 FACTOKISATION BY MEANS OF chap.
From this again,
=/( a 3) = ( a 3 ~ a ( a 3 ~ a z )<£n_ a (a3),
which gives, since a n a. 2 , a 3 are all unequal, <$> n  2 { a 3) = ; whence
<£»s(a) = (■''  a 3 )<£m 3 (a) J so that
f(x) = (x  a,) (x  o s ) (a;  a 3 )<£ w _ 3 (.r).
Proceeding in this way step by step, we finally establish the
theorem for any number of factors not exceeding n.
Cor. 1 . If an integral function be divisible by the factors x  a„
x  a 2 , . . . , x  a n all of the 1st degree, and all different, it is
divisible by their product ; and, conversely, if it is divisible by the
product of any number of such factors, all of the 1st degree and all
different, it is divisible by each of them separately. The proof of
this will form a good exercise in algebraical logic.
Cor. 2. The particular case of the above theorem where the
number of factors is equal to the degree of the function is of
special interest. We have then
f(x) = (x  a,) (X  a 3 ) . . . (X  a ?l )P.
Here P is of zero degree, that is, is a constant. To determine it
we have only to compare the coefficients of x n on the left and
right hand sides, which must be equal by chap, iv., § 24. Now
f(x) stands for p^ 1 + p^ 1 ' 1 + . . . +p n i x +Pw Hence Vp ,
and we have
In other words — If n different values of x can be found for which
the integral function fix) vanishes, then f(x) can be resolved into n
factors of the 1st degree, all different.
The student must observe the "if" here. We have not
shown that n such particular values of x can always be found,
or how they can be found, but only that if they can be found the
factorisation may be effected. The question as to the finding of
aj, a 2 , , . . , &c, belongs to the Theory of Equations, into which
we are not yet prepared to enter.
§ 16.] The student who has followed the above theory will
naturally put to himself the question, " Can more than n values
v THE REMAINDER THEOREM 99
of x be found for which an integral function of x of the nth degree
vanishes, and, if so, what then 1 " The following theorem will
answer this question, and complete the general theory of factorisa
tion so far as we can now follow it.
If an integral function of x of the nth degree vanish for more than
n different values of x, it must vanish identically, that is, each of its
coefficients must vanish.
Let a„ a 3 , . . . , a,! be n of the values for which f(x) vanishes,
then by § 15 above, if p be the coefficient of the highest power
of x in f(x), we have
f(x) = p (x  a,) (X  a 2 ) . . . (x  a n ) ( 1 ).
Now let ft be another value (since there are more than n) for
which /(.r) vanishes, then, since (1) is true for all values of x, we
have
=/(/?) =MP  «,) (fi " «.) ... 08  «n) (2).
Since, by hypothesis, a 1} a 2 , . . . , a n and /? are all different,
none of the differences /5  a„ /3  a 2 , . . . , (3  a n , can vanish,
and therefore their product cannot vanish. Hence (2) gives p = 0.
This being so, f(x) reduces to p^x 111 +) 2 a;' l ~ 2 + . . . +p n jX
+ p n . We have now, therefore, a function of the {n  l)th degree
which vanishes for more than n, therefore for more than (n  1),
values of its variable. We can, by a repetition of the above
reasoning, prove that the highest coefficient J9, of this function
vanishes. Proceeding in this way we can show, step by step,
that all the coefficients of /(./•) vanish.
As an example of this case the student may take the following : —
The integral function
Q3  7 ) (3  0) (3 y) + (ya) (x  y) (x  a) + (a  /3) (x  a) (x  /3)
+ (/3 7 )( 7  )(a/3)
vanishes when x — a, when x = p, and when x = y ; but it is only of the 2nd
degree in x. We therefore infer that the function vanishes for all values of x,
that is to say, that we have identically
{Py){3Bp){xy) + (ya.){xy){xa) + (afl(xa)(xp)
+ (/*7)(7a)(ft/9) =0.
That this is so the reader may readily verify by expanding and arranging the
lefthand side.
100 INDETERMINATE COEFFICIENTS CHAP;
Cor. If two integral functions of x, whose degrees are m and n
respectively, m being > n, be equal in value for more than m different
values of x, a fortiori, if they be equal for all values of x, that is to
say, identically equal, then the coefficients of like powers of x in the
two must be equal.
We may, without loss of generality, suppose the two functions
to be each of degree m, for, if they be not equal in degree, this
simply means that the coefficients of x n+1 , x 1l + 2 , . . . , x m in one
of them are zero. We have therefore, by hypothesis,
p ( & m +p l X m ^ 1 + . . . +p m = q<pi m + qiZ m ~ 1 + • • . +2m»
and therefore
(p ~q )x m +(p 1 q 1 )x m  1 + . . . +(p m qm) = 0,
for more than m values of x.
Hence we must have
Po?o = 0, Pifr^O, . . ., p m q m = 0;
that is,
Po = % lh = So • • • » Pm = 2W
This is of course merely another form of the principle of
indeterminate coefficients. The present view of it is, however,
important and instructive, for we can now say that, if any
function of x can be transformed into an integral function of x, then
this transformation is unique.
§ 17.] The danger with the theory we have just been ex
pounding is not so much that the student may refuse his assent
to the demonstrations given, as that he may fail to apprehend
fully the scope and generality of the conclusions. We proceed,
therefore, before leaving the subject, to illustrate very fully the
use of the remainder theorem in various particular cases.
To help the student, we shall distinguish, in the following
examples, between identical and conditional equations by using
the sign '* = " for the former and the sign " = " for the latter.
Example 1. To determine the value of the constant k in order that
may be exactly divisible by a; + 2.
The remainder, after dividing by x + 2, that is, by x (  2), is (2) s
EXAMPLES
101
+ 6(  2) 2 + 4(  2) + /, that is, 8 + k. The condition for exact divisibility is
therefore 8 + & = 0, that is, k= 8.
Example 2. To determine whether
80? 2a 3 7a 2 (1)
is divisible by (x + 1) (x 2).
If we put x—  1 in the function (1) we get
32 + 72 = 0,
hence it is divisible by x+1.
If we put x = 2 we get
248142 = 0,
hence it is divisible by X  2.
Hence by § 15 it is divisible by (x + l)(x2). The quotient in this case
is easily obtained, for, since the degree of (1) is the 3rd, we must have
Bx"2a?7x2 = (x+l){x2){ax + b) (2),
where a and b are numbers to be determined.
If we observe that the highest term 3a 3 on the left must be equal to the
product x x x x ax of the three highest terms of the factors on the right, we
see that SxP^ax 3 , hence a=3. And, since the product of the three lowest
terms of the factors on the right must be equal to  2, the lowest term on the
left, we get  2&=  2, that is, b = l. Hence finally
3a 3  2x  7x2=(x+l) (x  2) (Bx+1).
Example 3
If n be a positive integer,
when
is divided by
the rem. is
that is
x n  a' 1
X a
a"  a"
always.
x n  a"
x + a
{a)" a"
if n be even,  2a n if n be odd.
x" + «"
xa
a" + a' 1
2a" always.
x" + a"
x + a
(  a)" + a"
if n be odd, 2a" if n be even.
Hence x n a" is exactly divisible by xa for all integral values of n, and by
x + a if n be even. x n + a" is exactly divisible by x + a if n be odd, but is
never exactly divisible by z — a (so long as «=#0). These results agree with
those given above, chap, iv., § 16.
Example 4. To prove that
a"'{b  c) + b 3 {c a) + c^{ab)= (a + b + c) (b  c) (c  a) (a  b).
First of all regard F^a^b  c) + b 3 {ca)+c s (ab) as a function of a. P
is an integral function of a of the 3rd degree ; and, if we put a = b,
Y = b 3 (b c) + ¥(c b) + c s (b  b)
= 0;
102 CONTINUED DIVISION chap.
and similarly, if we put a = c, P = ; therefore P is exactly divisible both by
a  b and by a  c.
Again, regard P as a function of b alone. It is an integral function of b,
and vanishes when b — c, hence it is exactly divisible by bc. We have,
therefore,
T = Q{ab){ac)(bc).
Since P, regarded as a function of a, b, and c, is of the 4th degree, it
follows that Q must be an integral function of a, b, c of the 1st degree.
Hence, I, m, n being mere numbers which we have still to determine, we
have
w\b  c) + b 3 (c  a) + c 3 (a  b) = (la + mb + nc) {bc) (ac) (ab)
= —{la + mb + nc) (b c){c a) (a  b).
To determine I we have merely to compare the coefficients of a 3 b on both
sides. It thus results by inspection that 1 = 1; and similarly m = l, n = l;
the last two inferences being also obvious by the law of symmetry. We have
therefore finally
a s (b  c) + b\c a) + c 3 (a b)=  (a + b + c)(bc) (ca)(a b).
Example 5. To show that
Pee 2b 2 c 2 + 2c 2 a 2 + 2a 2 b 2  a*  b*  c 4
= (a + b + c) (  a + b + c) (a  b + c) (a + b  c).
First, regarding P as an integral function of a, and dividing it hy a + b + c,
that is, by a  (  b  c), we have for the remainder
2J 2 c 2 +2c 2 (  b  cf + 2b\  b cf (  4c) 4  6*c*
== 2b 2 c 2 + 2b 2 <? + 4.bc s + 2c 4 + 2¥ + ib 3 c + 2b 2 c 2
 6 4  ib 3 c  6b 2 c 2  Abe 3 c^b^c 4 
=0.
Hence P is exactly divisible by a + b + c.
Observing that the change of a into  a, or of b into  b, or of c into  c
does not alter P, all the powers of these letters therein occurring being even,
we see that P must also be divisible by a + b + c, ab + c, and a + bc. We
have thus found four factors of the 1st degree in a, b, c, and there can be no
more, since P itself is of the 4th degree in these letters. This being estab
lished, it is easy to prove, as in Example 4, that the constant multiplier is
+ 1 : and thus the result is established.
EXPANSION OF RATIONAL FRACTIONS IN SERIES BY MEANS OF
CONTINUED DIVISION.
§ 18.] If we refer back to §§ 5 and 6, and consider the analysis
there given, we shall see that every step in the process of long
division gives us an algebraical identity of the form
V DESCENDING CONTINUED DIVISION 103
where Q' is the part of the quotient already round, and R' the
residue at the point where we suppose the operation arrested.
For example, if we stop at the end of the second operation,
8/ + 8/  20/ + 40/  50/ + 30a  10
2/ + 3/  4/ + 6.c  8
*_ 2/ + 8/  6/ + 14s 10
~~ X+ 2/+3/4/+6*8 '
Again, instead of confining ourselves to integral terms in x, and
therefore arresting the process when the remainder, strictly so
called, is reached, we may continue the operation to any extent.
In this case, if we stop after any step we still get an identity of
the form
where
Q' = AxP + BxP 1 + . . . + Kx + L + M/:c + . . .+T/& (2),
and
R' = U^ 1  1 + Vx n ~ 2 + . . . + Z (3).
This process may be called Descending Continued Division.
For example, consider
^+2.r 2 + 3a;+4
x 2 + x + 1
and let us conduct the division in the contracted manner of § 12, but insert
the powers of x for greater clearness.
a 2 + a;+l
x 3 + 2x 2 + Sx + i
+ x 2 + 2x + 4
x + d
.2 3 1_ 2_
x i + x 5
X + 1+ + . + . + 
X iV" ft,
X
3 2
x ar
L I.
x 2 x 3
2 1_
x*
3 2
+ ar* a*
whence ^±g±£±i = 1 2 3 1 2 ^faa^
ar+sc+l a; a; 2 a, 5 ar 4 a^ 5 a^+a+l
an identity which the student should verify by multiplying both sides by
ar + a+l. '
104: ASCENDING CONTINUED DIVISION chap.
§ 19.] When we prolong the operation of division indefinitely
in this way we may of course arrange either dividend or divisor,
or both, according to ascending powers of x. Taking the latter
arrangement we get a new kind of result, which may be illus
trated with the fraction used above.
We now have
ixx + 3x 3 2x*x 5
i + 3x + 2x+ x 3
 x2x 2 + x 3
 x + 2z 3
+ Bx 3 + x i
 2x i  3ar>
 x r ' + 2x 6
+ 3x 6 + x 7 ,
wh ence 5 = 4  x  x + Bx 3  2x*  x 5 + , , ( o )•
x 2 + x+l 1+x + x 2
And, in general, proceeding in a similar way with any two
integral functions, A^ and D n , we get
where Q" = A + Bx + . . . + Kafl (7),
K" = L + M.c + . . . + T.:" 1 (8).
This process may be called Ascending Continued Division.
§ 20.] The results of §§ 18, 19 show us that we can, by the
ordinary process of continued long division, expand any rational fraction
as a "series" either of descending or ascending powers of x, continu
ing as many terms as ice please, plus a " residue," which is itself a
rationed fraction.
And there is no difficulty in showing that, when the integer q is
given, each of the expansions (2) and (6) o/§§ 18 and 19 is unique.
The proof depends on the theory of degree, and may be left
as an exercise for the reader.
These series (the Q' or Q" parts above) belong, as we shall
see hereafter, to the general class of " Recurring Series. "
The following are simple examples of the processes we have been describ
ing:—
■ l+x + x 2 +. . .+.<" f n —  (9).
1 x \x
111 1 l'.<"
— 53? • "x^^x (10) 
EXPANSION THEOREMS 105
ls+a?. . . +(l)»x» [ ?\ (11).
1 + a; ' 1 + a:
a; a' 2 a: 3 a!" 1 + x
•i" r .3 • • • ,,„ t , ..
l + ar + . . . +x n
= 1 x+ x"**  a;»+ 2 + a?»+ 2  ai 2 "+ 3 + . . .
a /r+])(n+l)
i r iir+r _ ^.nr+r+l . . (13).
1 + x + . . . + a;"
1 , „ „ , , ,, (/i + 2)a;»+ 1 (n + l)a.'»+ 2
(1 a) 2 ' v ' (1 a) 2
(14).
EXPRESSION OF ONE INTEGRAL FUNCTION IN POWERS OF
ANOTHER.
§ 21.] "We shall Lave occasion in a later chapter to use two
particular cases of the following theorem.
If P and Q be integral functions of the rnth and nth degrees
respectively (m > n), then P may always he put into the form
P = R + R,Q + R 2 Q 2 + . . . + iy^ (1),
where R , E,, . . . , R^ are integral functions, the degree of each of
which is re — 1 at most, and p is a positive integer, which cannot
exceed mjn.
Proof. — Divide P by Q, and let the quotient be Q and the
remainder R .
If the degree of Q be greater than that of Q, divide Q by
Q, and let the quotient be Q, and the remainder Rj.
Next divide Qi by Q, and let the quotient be Q., and the
remainder R 2 , and so on, until a quotient Qp_i is reached
whose degree is less than the degree of Q. Q^,, for con
venience, we call also R^. "We thus have
P = Q Q + R
Qo = Q,Q + R,
Hpi = Kp
106 EXPANSION THEOREMS CHAP.
Now, using in the first of these the value of Q given by the
second, we obtain
P = (Q,Q + R,)Q + R„,
= R + R,Q + Q,Q 2 .
Using the value of Q, given by the third, we obtain
P = R + R 1 Q + R,Q 2 + Q 2 Q 3 (3).
And so on.
We thus obtain finally the required result ; for, R , R,, . . .,
Rp being remainders after divisions by Q (whose degree is n),
none of these can be of higher degree than n1; moreover,
since the degrees of Q , Q„ Q., . . ., Qpi are m  n, m  2n,
m  3n, . . ., m np, p cannot exceed m/n.
The two most important particular cases are those in which Q = a:a and
Q = x 2 + (3x + y. We then have
T = a + a 1 (xa) + . . .+a„{xa)",
where a , «i, • • • , «« ai ' e constants ;
P = (oo + b&) + (rti + hx) (x° + (3x + y) + . . . +(a p + b p x)(x" + fix + y) p,
where a , a u . . ., a p ,bo,h, • • •, ^ are constants, a,ndj»mj2.
Example 1.
Let P=5« s ll£c 2 +l0aj2,
Q = x1.
The calculation of the successive remainders proceeds as follows (see
§ 13):—
5 11 +10 2
+ 5  6+4
5  6 + 4 + 2
0+51
5  l+ 3
0+5
6+ 4
1 + 5;
and we find
5... 3 1U" + 10a:2 = 2 + 3(o; 1) + 4(3 l) 2 +5{x l) 3 .
Example 2.
P = x« + 3x 7 + 4z 6 + 4a; 2 + Sx + 1,
Q = x 2 x + l.
The student will find
R = 11 x, • Ri =  22a! + 7, R; = 1 9a  22, R 3 = 7x + 15, R 4 = 1
v EXPANSION THEOREMS 107
so that
Y = n.r + {22x+7)(x 2 x+l) + {19x22)(x i x+l) ,i
+ (7x + 15) (.c 2  x + 1 ) 8 + (k 2  x + 1 ) 4 .
§ 22.] If a,, a 9 , . . ., a rt &e w constants, any two or more of
which may be equal, then any integral function of x of the nth degree
may be put into the form
A + A,(.c  a,) + A 2 (x  a l )(x  a, 2 ) + A 3 (x  a x )(x  a 2 )(x  a 3 ) + . . .
+ A H (x  a,) (a;  a 2 ) . . . (x  a n ) (1 ),
where A , A,, A 2 , . . . , A n are constants, any one of which except A n
may be zero.
Let P ?i , be the given integral function, then, dividing P n by
x  a, , we have
P» = P»i(s «,) + Ao (2),
where A is the constant remainder, which may of course in any
particular case be zero.
Next, dividing P n _! by (x — a.,), we have
Pn.^Pn^O + A, (3);
and so on.
Finally, P : = A n (.?;  a n ) + A n _ , (n + 1 ).
Using these equations, we get successively
P n = A + A,(x  a,) + (x  a,)(x  a.,)? n _ 2 ,
= A + A,(.?;  «,) + A 2 (x  a t ) (x  «,)
+ (xa 1 )(xa s )(xa a )'P n  a ,
P n = A + A,(.?;  «,) + A,,(r  <?,) (a  a,) + A 3 (x  a,) (x  a 2 ) (x  a 3 )
+ . . . + A n (x  «,) (./•  a, 2 ) . . . (x  a n ).
This kind of expression for an integral function is often
useful in practice.
Knowing a priori that the expansion is possible, we can, if we choose,
determine the coefficients by giving particular values to x. But the most
rapid process in general is simply to carry out the divisions indicated in the
proof, exactly as in Example 1 of last paragraph.
Thus, to express a 3 ! in the form A„ + A\(x 1)4 A..(.r 1) {x 2)
+ A 3 {x  1) (,r  2) (x  3), we calculate as follows : —
108
EXERCISES IX
CHAl'.
1 +0 +0 1
+1 +1 +1
1 +1 +l0
+2 +6
1 + 3 + 7
+3
1+6
Henceo. 3 l = + 7(al) + 6(a:l)(a:2) + (xl)(a.'2)(a;3).
Exercises IX.
Transform the following quotients, finding both integral quotient and
remainder where the quotient is fractional.
(1.) (sc 5 5a 8 +5ar ! l)/(ar ! +3a;+l).
(2. ) (6a: 6 + 2a; 5  19a 4 + 3lar 5  37a; 2 + 27a:  7)/(2a 2  3a + 1 ).
(3. ) (4a; 5  2x* + Zx 3 aj+1 )/(a; 2  2a; + 1).
(4. ) (as 5  8aj+ 15) (x" + Sx + 15)/(x  25).
(5. ) {(as 1) {x 2) (x  3) (as  4) (x  5)  760(.i  6) + 120(a:  7)} + (a!  C)
(x7).
(6.) (x 6 + 4a: 5  3a 4  16x 3 + 2x 2 + x + 3)/{x 3 +4x 2 + 2x+l).
(7.) (27a* + 10as s +l)/(3a?2a: + l).
(8. ) (x?  9a? + 23a;  15) (a:  7)/(as s  Sx + 7).
(9. ) (a; 3 + fix* + && + t& + i)/(x" ~ ix + 1 ).
(10.) (a; 4 + fce 3 + ia 2 + Aa: + i)/(x 2 + 2aj + 1 ).
(11.) (a/ + T ^)/(2a:+l).
(12. ) (a; 2  a; + 1 ) (x s  1 )/{x* + ar» + 1 ).
(13.) (ai l Vflty u )/(a:y).
(14. ) (9« 4 + 2o 2 fi s + &*)/(3a a + 2«fe + 6 2 ).
(15.) (a 7 + 6 7 )/(a + 6).
(16.) (a; 4 + y 4  7xy°)l{x> + Zxy + y").
(17.) (x 5  2afy + 4.r 3 2/ 2  8a*Y + 1 6a*/ 4  32y s )ftx*  8)/).
(18. ) (a; 4 + 5arfy + 7a?V + 15xif + 12y 4 )/(a + Ay).
(19.) (l+.r + x 2 + a? + a^ + x 6 + x 7 + a* + a; 9 + a; 15 )/(l
(20.) (a; 6 5a^ + 8)/(a^ + a; + 2).
(21 . ) {abx 3 + {ac  W )x 2  (a/+ cd)x + <//} /(aa:  d).
(22. ) {a 2 6 2 + uV + fe 2 c 2 + 2a?bc  2ab~c  2abc" } 4 { « 2 ■
(23. ) (1 + b+c  be  Ve ' bc n )/(l  be).
(24.) {(ax + by) 3 + (axby)*(aybxf+(ay + bx) 3 }/{(a + b)hr3ab(x>y)}.
(25. ) { (a 2 + b) 3 + b 3 a 3 }j { (a + b) n  ba}.
(26.) {(x 2 + xy + y y(x i xy + i/) i }l{x i + Zxhj + y i }.
(27. ) { (x + yY x> /}/(a? + *>/ + 2/ 2 ) 2 .
(28.) {(x+l) 6 .r 6 l}/{a? + x + l).
(29.) {o6(a? + i/)+arj/(o s + 6 2 )}/{o6(ar'y s )a^(a 2  5 2 )}.
(30.) (a !i + 2a 3 b + 2a 2 b 3 3b' i )/(ar2ab + b).
t' 5 + X 6 ).
(«S)(ae)}.
v EXERCISES IX 109
(31.) (x i Zx 2 2;r + 4)/(.r + 2).
(32.) (jc* 4a?Zia?+76x+V)5)f(x7).
(33. ) Find the remainder when' a?  6x 2 + 8a:  9 is divided by 2x + 3.
(34.) Find the remainder when pa? + qx 2 + qx+p is divided by x 1 ; and
find the condition that the function in question be exactly divisible by x 2  1.
(35.) Find the condition that Ax 2m + Bx">y n + Cy 2 " be exactly divisible by
]\v m + Qy".
(36.) Find the conditions that x 3 + ax" + bx + c be exactly divisible by
x +px + q.
(37. ) If x  a be a factor of x 2 + 2ax  36', then a . = ± b.
(38.) Determine X, /u, v, in order that x 4 + Sx 3 + \x 2 + fix + v be exactly
divisible by (x 2 l)(x + 2).
(39.) If x 4 + Ax 3 + Qpx 2 + 4 qx + r be exactly divisible by x 3 + Bx 2 + 9x + S,
find p, q, r.
(40.) Show that px 3 + (p 2 + q)x 2 + {2pq + r)x + q 2 + s and px 3 + (p 2  q)x 2
+ rxq 2 + s either both are, or both are not, exactly divisible by x 2 +p>x + q.
(41.) Find the condition that (x m + x m ~ l + . . . +l)/(a," + a: n  1 + . . . +1)
be integral.
(42.) Expand l/(3.c + l) in a series of ascending, and also in a series of
descending, powers of x ; and find in each case the residue after n + 1 terms.
(43.) Express l/iarax + x 2 ) in the form A + Bz + Cx 2 + Dx 3 + R, where A,
B, C, D are constants and R a certain rational function of x.
(44.) Divide 1+ x + ^ + 1^3 + • • ■ t>ylsc
(45.) Show that, if y<\, then approximately 1/(1 +y) = l y, 1/(1 y)
= 1 + ?/, the error in each case being 100;/ 2 per cent.
Find similar approximations for 1/(1+?/)" and 1/(1 y) n , where n is a
positive integer.
(46.) If a>l, show that «">1 +n{a 1), n being a positive integer.
Hence, show that when n is increased without limit a n becomes infinitely
great or infinitely small according as«> or <1.
(47.) Show that when an integral function f(x) is divided by (xai)
(x — ao) the remainder is {/(aj)(a;ai) /(ax)(a;ao)}/(a2 aj). Generalise
this theorem.
(48.) Show that f(x)/(a) is exactly divisible by xa; and that, if
f(x)=p x n +piX n  1 +p&c n  2 + . . . +p n , then the quotient is x( x ) pox n ~ l
+ (p a+pi)x n  2 + (p l) a 2 +p l a.+2h2)x n  3 +. . . +(p a n ~ l +p x a n ~ 2 + . . .+p n  l ).
Hence show that when f(x) is divided by (xa) 2 the remainder is
X(o) (za)+/(a),
where f(a)=p a n +pia n  1 +. . . +p„,
x(a)np a n  1 + (nl)2ha n  2 +. . . +p n  1 .
(49.) If x"+2hx"~ 1 +. • . +p n and x" 1 + q x x n  2 + . . . +q»i have the
same linear factors with the exception of a: a, which is a factor in the first
only, find the relations connecting the coefficients of the two functions.
(50.) If, when y + c is substituted for x in x n + aix n ~ 1 + . . . +a„, the
110 EXERCISES IX chap. V
result is y n + &i2/ n_1 + . . . +b n , show that b n , £>„_i, . . . , b x aie the remainders
when the original function is divided by x  c, and the successive quotients
by xc. Hence obtain the result of substituting y + 3 for x in a^l&B*
+ 20x 3 17x 2 x + B.
(51.) Express (x 2 + 3x + l)* in the form A + B(a: + 2) + C(a: + 2) 2 + &c., and
also in the form Ax + B + {Cx + B)(x 2 + x + l) + ('Ex + F)(x + x + lf + kc.
(52.) Express x A + x? + x 2 + x + l in the form A + Ai(x + 1) + A 2 (x+l)(x+3)
+ A 3 (a: + l)(a; + 3)(a; + 5)+A4(a: + l)(a; + 3)(a; + 5)(.r + 7).
(53.) If, when P and P' are divided by D, the remainders are R and R',
show that, when PP' and RR' are divided by D, the remainders are identical.
(54.) When P is divided by D the remainder is R ; and when the integral
quotient obtained in this division is divided by D' the remainder is S and the
integral cuotient Q. R', S', Q' are the corresponding functions obtained by
first dividing by D' and then by D. Show that Q = Q', and that each is the
integral quotient when P is divided by DD'; also that SD + R = S'D' + R',
and that each of these is the remainder when P is divided by DD'.
CHAPTEE VI.
Greatest Common Measure and Least Common
Multiple.
§ 1.] Having seen how to test whether one given integral
function is exactly divisible by another, and seen how in certain
cases to find the divisors of a given integral function, we are
naturally led to consider the problem — Given two integral
functions, to find whether they have any common divisor or not.
We are thus led to lay down the following definitions : —
Any integral function of x which divides exactly two or more given
integral functions of x is called a common measure of these functions.
The integral function of highest degree in x zchich divides exactly
each of two or more given integral functions of x is called the greatest
common measure (G.C.3I.) of these functions.
§ 2.] A more general definition might be given by suppos
ing that there are any number of variables, x, y, z, u, &c. ; in
that case the functions must all be integral in x, y, z, u, &c, and
the degree must be reckoned by taking all these variables into
account. This definition is, however, of comparatively little
importance, as it has been applied in practice only to the case of
monomial functions, and even there it is not indispensable. As
it has been mentioned, however, we may as well exemplify its
use before dismissing it altogether.
Let the monomials be 432aWy 4 2, 270aWyV, 90a7v; 3 yV,
the variables being x, y, z, then the G.C.M. is x 2 ifz, or C'V':,
where C is a constant coefficient (that is, does not depend on the
variables x, y, z).
The general rule, of which the above is a particular case, is
as follows : —
112 G.C.M. BY INSPECTION
CHAP.
The G.C.M. of any number of monomials is the product of the
variables, each raised to the lowest power * in which it occurs in any
one of the given functions.
This product may of course be multiplied by any constant
coefficient.
G.C.M. OBTAINED BY INSPECTION.
§ 3.] Returning to the practically important case of integral
functions of one variable x, let us consider the case of a number
of integral functions, P, P', P", &c, each of which has been re
solved into a product of positive integral powers of certain factors
of the 1st degree, say x — a, x  (3, x  y, &c. ; so that
P =p(xa)«(xP)\x y y...,
P' =p'(x  a) a '(x  (3) b '(x  yY' ...,
V"=p"{;xar'{xpr{xyr ...,
By § 15 of chap, v., we know that every measure of P
can contain only powers of those factors of the 1st degree that
occur in P, and can contain none of those factors in a higher
power than that in which it occurs in P, and the same is true for
P', P", &c. Hence every common measure of P, P', P", &c, can
contain only such factors as are common to P, P', P", &c. Hence
the greatest common measure of P, P', P", &c, contains simply all the
factors that are common to P, P', P", &c, each raised to the lowest
power in which it occurs in any one of these functions.
Since mere numbers or constant letters have nothing to do
with questions relating to the integrality or degree of algebraical
functions, the G.C.M. given by the above rule may of course be
multiplied by any numerical or constant coefficient.
Example 1.
P=2x 2 6a: + 4 = 2(al)(a2),
P' = 6a: 2  6z  12 = 6(x + 1 ) (a;  2).
Hence the G.C.M. of P and P' is a 2.
Example 2.
P = a*  5as* + 7a: 3 + x 2  8x + 4 = (x  1 )(x + l)[x 2) 2 ,
P' = x e  7x>+ 17a 4  Ux 3  10a: 2 + 20a; S = (x l)(o. + l) (a;2) 3 ,
P" = x 5 Zx* a? + 7a,"  4 = (x  1 ) (x + l) 2 (x  2) 2 .
The G. C. M. is (a;  1) (a>+ 1) (as  2) 2 , that is, x 4 4x 3 + 3a; 2 + 4x  4.
* If any variable does not occur at all in one or more of the given func
tions, it must of course be omitted altogether in the G.C.M.
vi CONTRAST BETWEEN ALGEBRAICAL & ARITHMETICAL G.C.M. 113
§ 4.] It will be well at this stage to caution the student
against being misled by the analogy between the algebraical and
the arithmetical G.C.M. He should notice that no mention is
made of arithmetical magnitude in the definition of the algebraical
G.C.M. The word " greatest " used in that definition refers
merely to degree. It is not even true that the arithmetical
G.C.M. of the two numerical values of two given functions of x,
obtained by giving x any particular value, is the arithmetical
value of the algebraical G.C.M. when that particular value of x
is substituted therein ; nor is it possible to frame any definition
of the algebraical G.C.M. so that this shall be true.' 1 "
The student will best satisfy himself of the truth of this remark by study
ing the following example : —
The algebraical G.C.M. of x~2,x + 2 and x 2 x2 is x2. Now put
sc=31. The numerical values of the two functions are 870 and 928 respectively ;
the numerical value of x 2 is 29 ; but the arithmetical G.C.M. of 870 and 928
is not 29 but 58.
LONG RULE FOR G.C.M.
§ 5.] In chap. v. we have seen that in certain cases in
tegral functions can be resolved into factors ; but no general
method for accomplishing this resolution exists apart from the
theory of equations. Accordingly the method given in § 3
for finding the G.C.M. of two integral functions is not one of
perfectly general application.
The problem admits, however, of an elementary solution by
a method which is fundamental in many branches of algebra.
This solution rests on the following proposition : —
If A = BQ + R, A, B, Q, R being all integral functions of x,
then the G.C.M. of A and B is the same as the G.C.M. of B and R.
To prove this we have to show — 1st, that every common
* To avoid this confusion some writers on algebra have used instead of the
words "greatest common measure" the term "highest common factor." We
have adhered to the timehonoured nomenclature because the innovation in
this case would only be a partial reform. The very word /actor itself is used
in totally dilferent senses in algebra and in arithmetic ; and the same is true
of the words fractional and integral, with regard to which confusion is no less
common. As no one seriously proposes to alter the whole of the terminology
of the four species in algebra, it seems scarcely worth the while to disturb an
old friend like the G.C.M.
VOL. I I
114 LONG KULE FOR G.C.M. chap.
divisor of B and R divides A and B, and, 2nd, that every common
divisor of A and B divides B and R.
Now, since A = BQ + R, it follows, by § 4 of chap, v., that
every common divisor of B and R divides A, hence every common
divisor of B and R divides A and B.
Again, R = A  BQ, hence every common divisor of A and B
divides R, hence every common divisor of A and B divides B and R.
Let now A and B be two integral functions whose G.C.M. is required;
and let B be the one whose degree is not greater than that of the other.
Divide A by B, and let the quotient be Q,, and the remainder R,.
Divide B by R,, and let the quotient be Q 2 , and the remainder R 2 .
Divide Ri by R 2 , and let the quotient be Q 3 , and the remainder R 3 ,
and so on.
Since the degree of each remainder is less by unity at least than the
degree of the corresponding divisor, R 15 R 2 , R 3 , &c, go on diminishing in
degree, and the process must come to an end in one or other of two ways.
I. Either the division at a certain stage becomes exact, and the
remainder vanishes ; t
II. Or a stage is reached at which the remainder is reduced to a
constant. i i
Now we have, by the process of derivation above described,
A = BQ X + R,
B  R^a + R 2
R, = R 2 Q 3 + R 3
M 1 )
Hence by the fundamental proposition the pairs of functions
A ( B  R, ( Ro 1 R w 2 \R>ii ( ..li v, flVP fV, P eomp ci HM
B IRJR, jR, / • * • R B _ 1 J  R„ /
In Case I. H n = and R ;i _ 2 = Q,iR«i. Hence the G.C.M. of
R, t _ 2 and Rnu that is, of Q^R^! and R n _i, is R»_i, for this
divides both, and no function of higher degree than itself can
divide R, t _i. Hence R 7l _! is the G.C.M. of A and B.
In Case II. R ;l = constant. In this case A and B have no
G.C.M., for their G.C.M. is the G.C.M. of R ;l _, and R n> that is,
their G.C.M. divides the constant R n . But no integral function
VI
MODIFICATIONS OF LONG RULE
115
(other than a constant) can divide a constant exactly. Hence
A and B have no G.C.M. (other than a constant).
If, therefore, the process ends with a zero remainder, the last divisor
is the G.C.M. ; if it ends with a constant, there is no G.C.M.
§ 6.] It is important to remark that it follows from the
nature of the above process for finding the G.C.M., which con
sists essentially in substituting for the original pair of functions
pair after pair of others which have the same G.C.M., that we
may, at any stage of the process, multiply either the divisor or the
remainder by an integral function, provided we are sure that this
function and the remainder or divisor, as the case may be, have no
common factor. We may similarly remove from either the divisor or
the remainder a factor which is not common to both. We may remove
a factor which is common to both, provided we introduce it into the G. C M.
as ultimately found. It follows of course, a fortiori, tluit a numerical
factor may be introduced into or removed from divisor or remainder at any
stage of the process. This last remark is of great use in enabling us
to avoid fractions and otherwise simplify the arithmetic of the pro
cess. In order to obtain the full advantage of it, the student
should notice that, in what has been said, "remainder" may mean,
not only the remainder properly so called at the end of each sepa
rate division, but also, if we please, the " remainder in the middle of
any such division'' or "residue," as we called it in § 18, chap. v.
Some of these remarks are illustrated in the
examples : —
following
Example 1.
To find the G.C.M. of x 5  2x*  2x? + 8x 2  7x + 2 audz 4 
x 4  ix + 3
ix + 3.
X*
X s
2x i 2x 3 + 8x 2 ~ 1x + 2
 Ax 2 + Zx
2)
 2x 4  2r? + 12x 2  10a; + 2
ar*+ X s  6x 2 + 5x 1
x*  4a: + 3
x 3  6x 2 + 9xi
x*  4x + 3
x i Qx 3 + Qx 2  4x
3) 6ar» 9a; 2 +3
23? 2.x 2 +1
2.r 3 12a: 2 l18a:8
9) 9a^18x + 9
X +1
a. 3 6x 2 + 9a4
x +2
c 2  2x+l
116
EXAMPLES
CHAP.
X 3 
 6a; 2 + 9a;  4
 '2a; 2 + x
4) 
4a; 2 +8a:4
a; 2 2a; + l
a 2 2a:+l
x 2 2a;+l
a; +1
Hence the G.C.M. is xr2x + l.
It must be observed that what we have written in the place
of quotients are not really quotients in the ordinary sense, owing
to the rejection of the numerical factors here and there. In
point of fact the quotients are of no importance in the process,
and need not be written down ; neglecting them, carrying out the
subtractions mentally, and using detached coefficients, we may
write the whole calculation in the following compact form : —
_ o
122+ 8 7 + 2
'22 + 1210 + 2
1 + 1 6+ 51
1 6+ 94
4 +
1
84
2 + 1
1+0+0 4+3
69+ 0+3
23+ 0+1
918 + 9
1 2 + 1
^3
j9
zx
G.C.M., x 2
Example 2.
Required the G.C.M. of 4x 4 + 26a, 3 + 41a; 2  2a; 24 and 3a!* + 20a? + 32a; 2
 8a;  32.
Bearing in mind the general principle on which the rule for finding the
G.C.M. is founded, we may proceed as follows, in order to avoid large num
bers as much as possible : —
4 + 26 + 41
x2 1+ 6+ 9 +
2
6 +
24
8
2 + 12 + 18+ 12+ 16
7 + 44+ 68+ 16
1+29 + 146 + 184
+23 23 + 138 + 184
1+ 6+ 8
3 + 20 + 32
2+ 5
26
32
56
53318424
1+ 6+ 8
53
The G.C.M. is a 2 + 6a: + 8.
Here the second line on the left is obtained from the first by subtracting
the first on the right. By the general principle referred to, the function
se 4 + 6ar 3 + 9a? + 6a; + 8 thus obtained and 3a; 4 + 203 3 + 32a; 2  Sx  32 have tlie
same G.C.M. as the original pair. Similarly the fifth line on the left is the
result of subtracting from the line above three times the second line on the
right.
VI
SECOND RULE FOR G.C.M. 117
If the student be careful to pay more attention to the prin
ciple underlying the rule than to the mere mechanical application
of it, he will have little difficulty in devising other modifications
of it to suit particular cases.
METHOD OF ALTERNATE DESTRUCTION OF HIGHEST AND
LOWEST TERMS.
§ 7.] If /, m. p, q be constant quantities (such that Iq  mp is not
zero), and if
P = ZA + mB (1),
Q = M + qB (2),
where A and B, and therefore P and Q, are integral functions, then
the G.C.M. ofP and Q is the G.C.M. of A and B.
For it is clear from the equations as they stand that every
divisor of A and B divides both P and Q. Again, we have
qP  wQ = q(IA + mS)  m (pA + qB) = (Iq  mp)k (3),
pP + ZQ = p(IA + nzB) + I(pA + qB) = (Iq  mp)B (4) ;
hence (provided Iq  mp does not vanish), since I, p, m, q, and
therefore Iq  mp, are all constant, it follows that every divisor of
P and Q divides A and B. Thus the proposition is proved.
In practice I and m and p and q are so chosen that the
highest term shall disappear in I A + m*B, and the lowest in
pA + qB. The process will be easily understood from the follow
ing example : —
Example 1.
Let A = 4ar* + 26a 3 + 41a; 2  2x  24,
B = 3a; 4 + 20a? + 32a; 2  8x  32 ;
then 3A + 4B = 2ar> + 5a; 2 26a;56,
4 A  3B = 7x* + 44a. 3 + 68a; 2 + 16a;.
Rejecting now the factor x, which clearly forms no part of the G.C.M., we
have to find the G.C.M. of
A' = 7a;' , + 44a: 2 + 68a;+16 I
B' = 23? + 5a; 2 26a: 56.
Repeating the above process —
2A' 7B' = 53a 2 + 318a; +424,
7A' + 2B'=53a 3 + 318.v 2 +424a\
118 TENTATIVE PROCESSES
CHAP.
the G.C.M. of which is 53a; 2 + 318a; + 424. Hence this, or, what is equivalent
so far as the present quest is concerned, x 2 + Qx + 8, is the G.C.M. of the two
given functions.
When the functions differ in degree, we may first destroy
the lowest term in the function of higher degree, divide the
result by x, and replace the function of higher degree by the
new function thus obtained. We can proceed in this way until
we arrive at two functions of the same degree, which can in
general be dealt with by destroying alternately the highest and
lowest terms.
Detached coefficients may be used as in the following
example :
Example 2.
To find the G.C.M. of % A  3ar> + 2a; 2 + x  1 and x 3  x 2  2x + 2, we have
the following calculation : —
A
B
13+2+11
112+2
A' = (2A + B)/*
B' = B
25+3+0
112+2
A" = A'/x
B" =(A' +2B')
25 + 3
37 + 4
A"' = (4A" + 3B")/x
B'"= 3A" + 2B"
11
11
Hence the G.C.M. is x1.
The failing case of the original process, where Jq mp = 0, may be treated
in a similar manner, the exact details of which we leave to be worked out as
an exercise by the learner.
§ 8.] The following example shows how, by a semitentative
process, the desired result may often be obtained very quickly : —
Example.
A = 2a; 4 3a, 3 3a; 2 + 4,
B = 2x i x s 9x i + ix + i.
Every common divisor of A and B divides A  B, that is,
 2x a + 6.c 2  ix, that is, rejecting the numerical factor  2,
x(x 2 — 3x + 2), that is, x (x  1) (.c  2). We have therefore merely
to select those factors of x(x\)(x2) which divide both A
VI PROPOSITIONS REGARDING G.C.M. 119
and B. x clearly is not a common divisor, but we see at once,
by the remainder theorem (§ 13, chap, v.), that both x—\ and
x  2 are common divisors. Hence the G.C.M. is (x  1) (a; — 2),
or x 2  3a + 2.
§ 9.] The student should observe that the process for finding
the G.C.M. has the valuable peculiarity not only of furnishing
the G.C.M., but also of indicating when there is none.
Example.
k=xSx + \,
B = o; 2 4a;+6.
Arranging the calculation in the abridged form, we have
13+1 I 14+6
2+1 1+5
11 I
The last remainder being 11, it follows that there is no G.C.M.
G.C.M. OF ANY NUMBER OF INTEGRAL FUNCTIONS.
§ 10.] It follows at once, by the method of proof given in
§ 5, that every common divisor of two integral functions A and B is
a divisor of their G. CM.
This principle enables us at once to find the G.C.M. of any
number of integral functions by successive application of the
process for two. Consider, for example, four functions, A, B, C, D.
Let Gi be the G.C.M. of A and B, then G v is divisible by every
common divisor of A and B. Find now the G.C.M. of Gj and
C, G, say. Then Gr 3 is the divisor of highest degree that will
divide A, B, and C. Finally, find the G.C.M. of G 2 and D,
G 3 say. Then G 3 is the G.C.M. of A, B, C, and D.
GENERAL PROPOSITIONS REGARDING ALGEBRAICAL PRIMENESS.
§ 11.] We now proceed to establish a number of propositions
for integral functions analogous to those given for integral
numbers in chap, iii., again warning the student that he must
not confound the algebraical with the arithmetical results ;
120 PROPOSITIONS REGARDING G.C.M. chav.
although he should allow the analogy to lead him in seeking for
the analogous propositions, and in devising methods for proving
them.
Definition. — Two integral functions are said to be prime to each
other when they have no common divisor.
Proposition. — A and B being any two integral functions, there
exist always two integral functions, L and M, 'prime to each other, such
that, if A and B have a G.C.M. , G, then
LA + MB = G ;
and, if A and B be prime to each other,
LA + MB=1.
To prove this, we show that any one of the remainders in
the process for finding the G.C.M. of A and B may be put into
the form PA + QB, where P and Q are integral functions of x.
We have, from the equalities of § 5,
E^AQ.B (1),
E, = B  Q 2 E, (2),
E 3 = E x  Q 3 E 3 (3),
Equation (1) at once establishes the result for Ei (only here
P = l, Q= Q,).
From (2), using the value of E t given by (1),
E 2 = B  Q,(A  Q t B) = (  Q 3 )A + ( + 1 + Q.Q^B,
which establishes the result for E 2 .
From (3), using the results already obtained, we get
E 3 = A  Q,B  Q, 3 {(  Q g )A + ( + 1 + Q l Q B )B}
= (1 + QA)A + (<& Q 3  Q I Q,Q 3 )B,
which establishes the result for E 3 , since Q,, Q 2 , Q 3 are all in
tegral functions. Similarly we establish the result for E 4 , E 6 ,
&c.
Now, if A and B have a G.C.M., this is the last remainder
which does not vanish, and therefore we must have
G = LA + MB (L),
vi ALGEBRAIC PRIMENESS 121
where L and M are integral functions ; and these must be prime
to each other, for, since G divides both A and B, A/G ( = a say)
and B/'G ( = b say) are integral functions ; we have therefore,
dividing both sides of (I.) by G,
1 = La + M5 ;
so that any common divisor of L and M would divide unity.
If A and B have no G.C.M., the last remainder, R rt , is a
constant ; and we have, say, R n = L' A + M'B, where L' and M'
are integral functions. Dividing both sides by the constant R n ,
and putting L = L'/R w , M = M'/R n , so that L and M are still
integral functions, we have
1 = LA + MB (II.).
Here again it is obvious that L and M have no common divisor,
for such divisor, if it existed, would divide unity.
The proposition just proved is of considerable importance in
algebraical analysis. We proceed to deduce from it several con
clusions, the independent proof of which, by methods more
analogous to those of chap, iii., § 10, we leave as an exercise
to the learner. Unless the contrary is stated, all the letters
used denote integral functions of x.
§ 12.] If X be prime to B, then any common divisor of AH and
B must divide H.
For, since A is prime to B, Ave have
LA + MB=1,
whence
LAH + MBH = H,
which shows that any common divisor of AH and B divides H.
If A and B have a G.C.M. a somewhat different proposition
may be established by the help of equation (I.) of § 11. The
discovery and proof of this may be left to the reader.
Cor. 1. IfB divide AH and be prime to A, it must divide H.
Cor. 2. If A' be prime to each of the functions A, B, 0, <&c, it
is prime to their product ABC . . .
122 ALGEBRAIC PPJMENESS chav.
Cor. 3. If each of the functions A, B, C, . . . be prime to each
of the functions A', B', C, . . . , then the product ABC . . . is prime
to the product A'B'C . . .
Cor. 4. If A be prime to A', then A a is prime to A' a ', a and a'
being any positive integers.
Cor. 5. If a given set of integral functions be each resolved into a
product of powers of the integral factors A, B, C, . . ., which are
prime to each other, then the G.C.M. of the set is found by uniting
down the product of cdl the factors that are common to all the given
functions, each raised to the lowest power in which it occurs in any of
these functions.
This is a generalisation of § 3 above.
After what has been done it seems unnecessary to add de
tailed proofs of these corollaries.
LEAST COMMON MULTIPLE.
§ 13.] Closely allied to the problem of finding the G.C.M. of
a set of integral functions is the problem of finding the integral
function of least degree which is divisible by each of them. This
function is called their least common multiple (L.C.M.).
§ 14.] As in the case of the G.C.M., the degree may, if we
please, be reckoned in terms of more variables than one ; thus
the L.C.M. of the monomials Sx 3 yz 2 , 6.c 2 ?/V, 8xyzu, the variables
being x, y, z, u, is x s y 3 z*u, or any constant multiple thereof.
The general rule clearly is to write down all the variables, each
raised to the highest power in which it occurs in any of the mono
mials.
§ 15.] Confining ourselves to the case of integral functions
of a single variable x, let us investigate what are the essential
factors of every common multiple of two given integral functions
A and B. Let G be the G.C.M. of A and B (if they be prime
to each other we may put G = 1) ; then
A = aG, B = bG,
where a and b are two integral functions which are prime to each
vi LEAST COMMON MULTIPLE 123
other. Let M be any common multiple of A and B. Since M
is divisible by A, we must have
M = PA,
where P is an integral function of x.
Therefore M = PaG.
Again, since M is divisible by B, that is, by bG, therefore
M/bG, that is, TaG/bG, that is, ~Pa/b must be an integral function.
Now b is prime to a; hence, by § 12, b must divide P, that is,
P = GJ), where Q is integral. Hence finally
M = QabG.
This is the general form of all common multiples of A and B.
Now a, b, G are given, and the part which is arbitrary is the
integral function Q. Hence we get the least common multiple
by making the degree of Q as small as possible, that is, by making
Q any constant, unity say. The L.C.M. of A and B is therefore
abG, or (aG)(bG)/G, that is, AB/G. In other words, the L.C.M.
of two integral functions is their product divided by their G.C.M.
§ 16.] The above reasoning also shows that every common
multiple of two integral functions is a multiple of their least common
multiple.
The converse proposition, that every multiple of the L.C.M.
is a common multiple of the two functions, is of course obvious.
These principles enable us to find the L.C.M. of a set of any
number of integral functions A, B, C, D, &c. For, if we find
the L.C.M., L! say, of A and B; then the L.C.M., L 2 say, of L,
and C ; then the L.C.M., L 3 say, of L 2 and D, and so on, until all
the functions are exhausted, it follows that the last L.C.M. thus
obtained is the L.C.M. of the set.
§ 17.] The process of finding the L.C.M. has neither the
theoretical nor the practical importance of that for finding the
G.C.M. In the few cases where the student has to solve the
problem he will probably be able to use the following more direct
process, the foundation of which will be obvious after what has
been already said.
If a set of integral functions can all be exhibited as powers of a
124 EXERCISES X CHAP.
set of integral factors A, B, C, &e., which are either all of the 1st
degree and all different, or else are all prime to each other, then the
L.C.M. of the set is the product of all these factors, each being raised
to the highest power in which it occurs in any of the given functions.
For example, let the functions be
(,rl) 2 (a; 2 + 2) 3 (a: 2 + a + l),
(xl) 5 {x2) 3 (xBY(x 2 + x + l) 3 ,
then, by the above rule, the L.C.M. is
(xl) 5 (x2) 5 (xZ) i (x 2 +2f(x 2 + x + lf(x"x + l)\
Exercises X.
Find the G.C.M. of the following, or else show that they have no CM.
(1.) (x 2 lf, x 6 l.
(2.) a: 6 l, x i 2x 3 + Zx 2 2x + \.
(3.) z*x 2 + l, a 4 + ar + l.
(4.) jc 9 + 1, x ll + h
(5.) a?x 2 8x + \2, sP+ia? 3x 18.
(6.) a: 4 7a: 3 22a: 2 + 139x + 105, x A 8x* Ux 2 + \l6x + 70.
(7.) a 4 286a: 2 + 225, x* + 140? 480a: 2  690a:  225.
(8.) x 6 x*8x 2 + 12, a; 6 + 4a; 4 3a; 2 18.
(9.) x>  2a: 4  2a: 3 + 4ar + x 2, x 5 + 2x i 2x"'  8x 2 7x2.
(10.) x s + 6x 6  8.C 4 + 1 , a: 12 + 7a: 10  3a?  3a: 2  2.
(11.) 12x 3 + lBx 2 + 6x + l, 16a: 3 + 16a: 2 + 7a: + l.
(12.) 5ar 3 + 38a: 2 195a; 600, 4a; 3  15a: 2  38a; + 65.
(13.) 16a 4  56a, 3  88a; 2 + 278a; +105, 16a; 4  64a; 3  44a; 2 + 232a; + 70.
(14.) 7a 4 + 6a: 3 8a; 2  6a; + 1, lla: 4 + 15a; 3  2a: 2  5;<;+ 1.
(15.) a 4 + 64a 4 , (a; + 2a) 4  16« 4 .
(16.) 9a; 4 +4a; 2 + l, Zsj2x 3 + x 2 +l.
(17.) a; 3 + 3i?a; 2  (1 + 3p), px 3  3( 1 + Zp)x + (3 + 8p).
(18. ) x 3  3(a  b)x 2 + (4a 2  Zab)x  2a°{2a  3b),
x i {3a + b)x 3 +{5a 2 + 2ab)x 2 a 2 (5a + 3b)x + 2a 3 {a + b).
(19. ) 7iaf+ 1  (it + l)x n + 1, a"'  nx+{n  1 ).
(20.) Show that ar 3 +^a: 2 + g'a; + l, x 3 + qx 2 + px + 1 cannot have a common
measure, unless either p = q or p + q + 2 = 0.
(21. ) Show that, if ax 2 + bx + c, ex 2 + bx + a have a common measure of the
1st degree, then a±b + c — 0.
(22.) Find the value of a for which {x 3 aa; 2 + 19a'« 4] /{a: 3  (a+l)a 2
+ 23a;a7} admits of being expressed as the quotient of two integral
functions of lower degree.
(23.) If ax 3 + 3bx 2 + d, bx 3 + Bdx + e have a common measure, then
(ae 4bd) 3 = 27 {ad 2 + b 2 e) 2 .
VI EXERCISES X 125
(24.) Ax 2 + Bxy + Cy 2 , Ba?  2(A  C)xy  By 8 cannot have a common
measure unless the first be a square.
(25.) ax 3 + bx 2 + cx + d, dx 3 + cx 2 + bx + a will have a common measure of
the 2nd degree if
abc  a"b  b 2 d + acd ac 2  bed  a 3 + ad 2 d(ac  b<l)
acbd abcd a 2 d?
show that these conditions are equivalent to only one, namely, ac~bd =
or  dr.
(26.) Find two integral functions P and Q, such that
P(^ 2  Zx + 2) + Q(.x 2 + x + 1) = 1.
(27.) Find two integral functions P and Q, such that
P(2^7a; 2 + 7^2) + Q(2^ + a; 2 + xl) = 2a;l.
Find the L.C. M. of the following : —
(28.) a 5 ab\ a?+a*b, a 6 + b (i + a 2 b"(a 2 + b 2 ).
(29.) x 3 x 2 Ux + 24, x 3 2x 2 5x +6, x 2 4a + 3.
(00.) 3x 3 + x 2 8x + 4, 3x 3 + 7x 2 4, x 3 + 2a: 2  x  2, dx 3 + 2,r 2  3r  2.
(31. ) x 3  V2x + 16, x A  i.r 3  x 2 + 20,r  20, x* + 3x 3  llx 2  3x + 10.
(32.) x 6 + 2ax 5 + a 2 x 4 + 5a 5 x + a 6 ) x 3 + crx  ax 2  a 3 .
(33.) If x 2 + ax + b, x 2 + a'x + b' have a common measure of the 1st
degree, then their L.C.M. is
, aba'V . / , [bV\*\ , ,.,««'
(34.) Show that the L.C.M. of two integral functions A and B can always
be expressed in the form PA + QB, where P and Q are integral functions.
CHAPTEE VII.
On the Resolution of Integral Functions into
Factors.
§ 1.] Having seen how to determine whether any given
integral function is a factor in another or not, and how to deter
mine the factor of highest degree which is common to two in
tegral functions, it is natural that we should put to ourselves
the question, How can any given integral function be resolved
into integral factors ?
TENTATIVE METHODS.
§ 2.] Confining ourselves at present to the case where
factors of the 1st degree, whose coefficients are rational integral
functions of the coefficients of the given function, are suspected
or known to exist, we may arrive at these factors in various ways.
For example, every known identity resulting from the distri
bution of a product of such factors, when read backwards, gives
a factorisation.
Thus (x + y) (x  y) — x 2  y" tells us that of  y 2 may be re
solved into the product of two factors, x + y and x — y. In a
similar way we learn that x + y + z is a factor in x 3 + y 3 + z 3 — 3xyz.
The student should again refer to the tables of identities given
on pp. 8183, and study it from this point of view.
When factors of the 1st degree with rational integral
coefficients are known to exist, it is usually not difficult to find
them by a tentative process, because the number of possible
factors is limited by the nature of the case.
chap, vii TENTATIVE FACTORISATION EXAMPLES 127
Example 1.
Consider a ,2 12x + 32, and let us assume that it is resolvable into (xa)
{xb).
Then we have
a 2  12a! + 32 = x  (a + b)x + ab,
and we have to find a and b so that
«6=+32, a + b +12.
"We remark, first, that a and b must have the same sign, since their pro
duct is positive ; and that that sign must be +, since their sum is positive.
Further, the different ways of resolving 32 into a product of integers are
1x32, 2x16, 4x8; and of these we must choose the one which gives
a + b= +12, namely, the last, that is, a = 4, b — 8.
So that
a 2 12a + 32 = (a4)(a8).
Example 2.
x 3  2a 2  23a: + 60 = (x a)(x b) (x  c) say.
Here  abc — + 60.
Now the divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ; and we
have therefore to try a±l, a 1 ±2, x±3, &c. The theorem of remainders
(chap. v. , § 14) at once shows that x + 1, x  1, x + 2, x  2, are all inadmissible.
On the other hand, for a 3 we have (see chap, v., § 13)
1223 + GO
+ 3+ 360
1 + 120+
that is, a 3 is a factor ; and the other factor is a; 2 + a;20, which we resolve
by inspection, or as in Example 1, into {x 4) {x + 5).
Hence x*  2a 2  23a + 60 = {x  3) (a  4) (x + 5).
Example 3.
&x  1 9x + 1 5 = (ax + b)(cx + d).
Here ac= +6, bd=+ 15 ; and we have more cases to consider. "We might
have anyone of the 32 factors, a;±l, a; ±3, a' ±5, a' ±15, 2a; ±1, 2a ±3,
2x ± 5, 2xzk 15, &c. A glance at the middle coefficient,  19, at once excludes
a large number of these, and we find, after a few trials,
6x°  19a + 15 = (2a  3) (3a  5).
§ 3.] In cases like those of last section, we can often detect
a factor by suitably grouping the terms of the given function.
For it follows from the general theory of integral functions
already established that, if P can be arranged as the sum of a
series of groups in each of which Q is a factor, then Q is a factor
in P ; and, if P can be arranged as the sum of a series of groups
in each of which Q is a factor, plus a group in which Q is not a
factor, then Q is not a factor in P.
128
FACTORISATION OF
CHAP.
Example 1.
x 3  2x"  23a; + 60
= .>•■(./■ 2) 23(a: 2) + 14,
that is, x  2 is not a factor.
a; 3 2« 2 23a + 60
= a"(a;3)+a: 2 23a: + 60
= x 2 (x3) + x(xZ)  20a;+ 60
= x 2 {x  3) + x(x  3)  20(a;  3),
that is, a; 3 is a factor.
Example 2.
])x 2 + xy+pqxy + qy 2
= x{px + y) + qy(px + y) ,
that is, px + y is a factor, the other being x + qij.
Example 3.
x 3 + (m + n + 1 )a,"a + (m + ?i + mn)xa 2 + mna?
= x 3 + a; 2 a + (w + ?i) (ara + xa 2 ) + mn(xa 2 + « 3 )
= ar^ai + a) + (m + n)xa(x + a) + mna 2 {x + a)
= {a; 2 + (m + yi)a'a + mrea 2 } (x + a)
= {x(x + via) + ?ta(a; + ma)} (x + a)
= (x + ma) (x + ?ia) (x + a).
GENERAL SOLUTION FOR A QUADRATIC FUNCTION.
§ 4.] For tentative processes, such as Ave have been illustrat
ing, no general rule can be given; and skill in this matter is one
of those algebraical accomplishments which the student must
cultivate by practice. There is, however, one case of great im
portance, namely, that of the integral function of the 2nd degree
in one variable, for which a systematic solution can be given.
We remark, first of all, that every function of the form
x'+px + q can De made a complete square, so far as x is con
cerned, by the addition of a constant. Let the constant in
question be a, so that we have
x 2 + px + q + a = (./• + /5) 2 = x" + 2 fix + f3 2 ,
(3 being by hypothesis another constant. Then we must have
p = 2f3, q + a = ft 2 .
The first of these equations gives (32>j2, the second a = /3 2 q
 (jpffi ~ '1 Thus our problem is solved by adding to x 2 +px + q
the constant (jj/2) 2  q.
vii A QUADRATIC FUNCTION 129
The same result is obtained for the more general form,
ax* + bx + c, as follows : —
2 , / 2 6 c
ax + bx + c = a[x +  x + 
\ a a
Now, from the case just treated, we see that x 2 + (b/a)x + c/a is
made a complete square in x by the addition of (b/2a) 2 — c/a,
that is, (b 2  4:0c)/ia a . Hence ax 2 + bx + c will be made a com
plete square in x by the addition of a(b 2 — lac)/ la 2 , that is,
(b 2  lac)/ la. We have, in fact,
, , b 2 iac ( b\ 2
ax + bx + c + — ; = a[x + —) ■
la \ 2aJ
§ 5.] The j)rocess of last article at once suggests that
aaf + bx + c can always be put into the form a{(x + I) 2  m 2 },
where I and m are constant.
In point of fact we have
2i f « b c )
ax + bx + c = a< x~ +  x +  >
I a a )
..I ,t+ 9 > .+ (*)'.(>)'+:]
{ 2a \2aJ \2aJ a )
In other words, our problem is solved if we make I = b/2a and
find m, so that m 2 = (b 2  iac)/ia 2 .
This being done, the identity X 2  A 2 = (X  A) (X + A) at
once gives us the factorisation of ax 2 + bx + c ; for we have
ax 2 + bx + c = a {(x + I) 2  m 2 }
= a {(x + 1) + m } {(x + /)  m }.
Example 1.
Consider 6x 2  19a; + 15 ; we have
{'S*(S)'S*¥}
19 / 1 \ 2 1
Here 1=  — , and m= ( — ) ; so that our problem is solved if we take w = — •
VOL. I K
130 REAL ALGEBRAICAL QUANTITY chap.
We get, therefore,
"— + M(S)4}{(S)£}
=<¥)(^)
= (2as3)(3«5);
the same result as we obtained above (in § 2, Example 3), by a tentative
process.
Example 2.
Consider x 6  5x 3 + 6. We may regard this as (x 3 ) 2  5(,r 3 ) + 6, that is to
say, as an integral function of x 3 of the 2nd degree. We thus see that
x 6 5x 3 + 6 = (x 3 ) 2 ~5(x s ) + 6,
= (.k 3 3)(.b 3 2).
INTRODUCTION INTO ALGEBRA OF SURD AND
IMAGINARY NUMBERS.
§ 6.] The necessities of algebraic generality have already led
us to introduce essentially negative quantity. So far, algebraic
quantity consists of all conceivable multiples positive or negative
of 1. To give this scale of quantity order and coherence, we
introduce an extended definition of the words greater than and
less than, as follows : — a is said to be greater or less than b, according
as a b is positive or negative.
Example.
( +3)  ( + 2)= + 1 therefore + 3 > + 2 ; (3) (5) =+2 therefore  3 >  5 ;
( + 3)  (  5) = + 8 therefore +3> 5; (7) (3)=  4 therefore 7< 3.
Hence it appears that, according to the above definition, any
negative quantity, however great numerically, is less than any
positive quantity, however small numerically ; and that, in the
case of negative quantities, descending order of numerical magni
tude is ascending order of algebraical magnitude.
We may therefore represent the whole ascending series of
algebraical quantity, so far as we have yet had occasion to con
sider it, as follows : —
 CO
1 * ..£... ...+£... +1 ...+ oo
*
t
* The symbol oo is here used as an abbreviation for a real quantity as
great as we please.
vii RATIONAL AND SURD QUANTITY 131
The most important part of the operations in the last para
graph is the finding of the quantity m, whose square shall be equal
to a given algebraical quantity. We say algebraical, for we must
contemplate the possibility of (b 2  4ac)/4a s , say k for shortness,
assuming any value between  <x> and + qo . When m is such that
in 2 = k, then in is called the square root of k, and we write in = \/k.
We are thus brought face to face with the problem of finding
the square root of any algebraical quantity ; and it behoves us
to look at this question somewhat closely, as it leads us to a new
extension of the field of algebraical operations, similar to that
which took place when we generalised addition and subtraction
and thus introduced negative quantity.
1st. Let us suppose that k is a positive number, and either
a square integer = + k, say, or the square of a rational number
= + (*/A) 2 , say, where k and A are both integers, or, which is
the same thing [since (k/X) 2 = k'/'A 2 ], the quotient of two square
integers. Then our problem is solved if we take
in the one case, or
m — + k, or m =  k
m = + k/X, or m   k/X
in the other ;
for m*  ( ± «) 2 = k = k,
in
(^■©■*
which is the sole condition required.
It is interesting to notice that we thus obtain two solutions
of our problem ; and it will be afterwards shown that there are
no more. Either of these will do, so far as the problem of
factorisation in § 5 is concerned, for all that is there required is
any one value of the square root.
More to the present purpose is it to remark that this is the
only case in which m can be rational; for if m be rational, that
is, = ± k/X where k and A are integers, then m 2 = (k/A) 3 , that is,
k = (k/X) 2 , that is, k must be the square of a rational number.
2nd. Let k be positive, but not the square of a rational
132 DEFINITION OF THE IMAGINARY UNIT i CHAP.
number ; then everything is as before, except that no exact
arithmetical expression can be found for m. We can, by the
arithmetical process for finding the square root, find a rational
value of m, say v, such that m 2 = ( ± v) 2 shall differ from k by
less than any assigned quantity, however small ; but no such
rational expression can be absolutely exact. In this case m
is called a surd number. When k is positive, and not a square
number, as in the present case, it is usual to use \/k to denote
the mere (signless) arithmetical value of the square root, which
has an actual existence, although it is not capable of exact arith
metical expression ; and to denote the two algebraical values of
m by ± \/k. Thus, if k = + 2, we write m = ± v/2. In any
practical application we use some rational approximation of
sufficient accuracy ; for example, if k = + 2, and it is necessary
to be exact to the l/10,000th, we may use m ± 14142.
A special chapter will be devoted to the discussion of surd
numbers ; all that it is necessary in the meantime to say further
concerning them is, that they, or the symbols representing them,
are of course to be subject to all the laws of ordinary algebra.!
3rd. Let k be negative =  k', say, where k' is a mere arith
metical number. A new difficulty here arises ; for, since the
square of every algebraical quantity between  co and + oo (ex
cept 0, which, of course, is not in question unless k' = 0) is
positive, there exists no quantity m in the range of algebraical
quantity, as at present constituted, which is such that m 2 =  k'.
If we are as hitherto to maintain the generality of all algebraical
operations, the only resource is to widen the field of algebraical quantity
still farther. This is done by introducing an ideal, socalled imaginary,
unit commonly denoted by the letter i* whose definition is, that it is
such that
i 2 = 1.
It is, of course, at once obvious that i has no arithmetical
existence whatsoever, and does not admit of any arithmetical
expression, approximate or other. We form multiples and sub
multiples of this unit, positive or negative, by combining it with
* Occasionally also by t. t See vol. ii. chap. xxv. § 2841.
VII
COMPLEX NUMBERS 133
quantities of the ordinary algebraical, now for distinction called
real, series, namely,
oo... 1......0. ..+!•.. +1...+CO.
We thus obtain a new series of purely imaginary quantity : —
— oo *...— *.. .  \i. . . 0* . . . + \i . . . + i , . . + co i.
These new imaginary quantities must of course, like every other
quantity in the science, be subject to all the ordinary laws of
algebra when combined either with real quantities or with one
another. All that the student requires to know, so far at least as
operations with them are concerned, beyond the laws already laid
down, is the defining property of the new unit i, namely, i 2 — — 1.
When purely real and purely imaginary numbers are com
bined by way of algebraical addition, forms arise like!? + qi, where
P and q are real numbers positive or negative. Such forms are
called complex numbers ; and it will appear later that every alge
braical function of a complex number can itself be reduced to
a complex number. In other words, it comes out in the end
that the field of ordinary algebraical quantity is rendered com
plete by this last extension.
The further consequences of the introduction of complex
numbers will be developed in a subsequent chapter. In the
meantime we have to show that these ideal numbers suffice for
our present purpose. That this is so is at once evident ; for, if
we denote by »Jk' the square root of the arithmetical number
k\ so that ijk' may be either rational or surd as heretofore, but
certainly real, then m  ±i Jk' gives two solutions of the problem
in hand, since we have
m 2 = ( ± i Jk') 2
= ( ± i Jk') x ( ± i Jk'),
upper signs going together or lower together,
= (i 2 ) x ( </V Y
= (l)x(*')
= k\
§ 7.] We have now to examine the bearing of the discus
sions of last paragraph on the problem of the factorisation of
ax 2 + bx + c.
134 FACTORISATION OF QUADRATIC chap.
It will prevent some confusion in the mind of the student if
we confine ourselves in the first place to the supposition that a, b, c
denote positive or negative rational numbers. Then I = bj2a is in all
cases a real rational number, and we have the following cases : —
1st. If b 2  4ac is the positive square of a rational number,
then m has a real rational value, and
ox 2 + bx + c = a(z + I + m)(x + I  m)
is the product of two linear factors whose coefficients are real rational
numbers. Example 1, § 5, will serve as an illustration of this case.
2nd. If b 2 — iac is positive, but not the square of a rational
number, then m is real, but not rational ; and the coefficients in
the factors are irrational.
Example 1.
x 2 + 2xl=x" + 2x + l2,
= (a; + l) 2 (V2) 2 ,
= (x + l + yj2)(x + l^2).
3rd. If b 2  iac is negative, then m is imaginary, and the
coefficients in the factors are complex numbers.
Example 2.
Example 3.
a: 2 + 2a: + 5 = a: 2 + 2a' + l + 4,
= (a + l) 2 (2i) 2 ,
= {x + l + 2i){x + l2i).
x + 2x + 3 = x 2 + 2x + l + 2,
= (x + lf(i^2T~,
= (x + 1 + i\j2) (x + 1  i V2).
4th. There is another case, which forms the transition
between the cases where the coefficients in the factors are real
and the case where they are imaginary.
If b 2  iac = 0, then m = 0,
and we have ax 2 + bx + c = a(x + I) 2 ;
in other words, ax 2 + bx + c is a complete square, so far as x is
concerned. The two factors are now x + I and x + I, that is,
both real, but identical.
We have, therefore, incidentally the important result that
ax 2 + bx + c is a complete square in x ij ' tf  4ac = 0.
Example 4.
3a; 2  3a; + 1 = 3(a; 2  2. \x + J), = 8(»  h) 2 
* I'^iac is called the Discriminant of the quadratic function ax + bx+c.
vil FUNCTIONS RESUMED 135
§ 8.] There is another point of view which, although usually
of less importance than that of last section, is sometimes taken.
Paying no attention to the values of a, b, c, but regarding
them merely as functions of certain other letters which they may
happen to contain, we may inquire under what circumstances the
coefficients of the factors will be algebraically rational functions of
those letters.
In order that tins may be the case it is clearly necessary and
sufficient that b 2  iac be a complete square in the letters in
question, = P 2 say.
Then
/ b P\ / b P
= a ( x + — + — ) (x + —  —
\ 2a la J \ 2a lo
which is rational, since P is so.
If b 2  iac   P 2 , where P is rational in the present sense,
then
/ b P.\/ b P .
= a(x + —  + — n (x + —   —i
V 2a 2a / \ 2a 2a
where the coefficients are rational, but not real.
Example 5.
jtx 2 + {p + q)x J r q
«j»(«+i)(«+J).
= (x + l)(j>X + q);
a result which would, of course, be more easily obtained by the
tentative processes of §§ 2, 3.
136 HOMOGENEOUS QUADRATIC FUNCTIONS chap.
§ 9.] It should be observed that the factorisation for
ax 2 + bx + c leads at once to the factorisation of the homogeneous
function ax 2 + bxy + cif of the 2nd degree in two variables ; for
ax 2 + bxy + cf
,f z b ///"  lac ) f x b lb 1  iuc > .
( / b lb 2  \ac\ ") f / b lb 1  4ac\ )
= " { x+ (a; + V ~iH" \{ r+ (s " v tt) "I •
By operating in a similar way any homogeneous function of
two variables may be factorised, provided a certain nonhomo
geneous function of one variable, having the same coefficients,
can be factorised.
Example 1. From
a: 2 + 2x + 3={x+ 1 + is/2) {x + 1  z'V2),
we deduce
x 2 + 2xy + 3y n  = {x + (1 + i\/2)y} {x+(l  i\/2)y}.
Example 2. From
x f _ 2x  23a + 60 = (x  3) {x  4) (as + 5),
we deduce
a?  2a' 2 ;/  23ay 2 + 60?/ 3 = (.r  Zy) [x  iy) (x + 5y).
§ 10.] By using the principle of substitution a great many
apparently complicated cases may be brought under the case of
the quadratic function, or under other equally simple forms.
The following are some examples : —
Example 1.
xt + xhf + y^^ + y^ixyT,
= {x*+y*+xy)(x*+y*xy),
{(•^'(^^{(W'W}
{. + G + ^>}{. + G^>H + (^>}
vil EXAMPLES 137
Here the student should observe that, if resolution into quadratic factors only
is required, it can be effected with real coefficients ; but, if the resolution be
carried to linear factors, complex coefficients have to be introduced.
Example 2.
x 3 + y 3 ={x + y)(x 2 .ry + ij 2 )
*"»»{•+(£+#>}{
«#>)
Example 3.
.< i + y i = (x 2 +y 2 ) 2 2x 2 y 2
= (■<• + j/ 2 ) 2 W2xy?
= (x 2 + V&ey + y 2 ) [a?  ^2xy + y 2 ).
Again
a? + s/2xy + y=f x + ^fy Y + ~y 2
V2 y /V2.
X+ 2 1 ' \2 iy
x + ^(l + i)y] {*+^(l%}.
The similar resolution for x 2  \/2xy + y 2 will be obtained by changing the
sign of \J2. Hence, finally,
= {^f(l + i)y}{^^li) y }{,^lU)y}{x^ { li )y ).
Example 4.
x^i/^^yiy 6 ) 2
= (x e y , ')(x G + y R )
= {(* 2 ) 3 (2/ 2 ) 3 } {(* 2 ) 3 + (ir) 3 }
= (x 2  y 2 ) (x 4 + x 2 y 2 + y*) (x 2 + y 2 ) (x*  x 2 y 2 + y 4 )
= {x + y){xy) (x + iy) {x  iy) (x 4 + x 2 y 2 + y*) (x*  x 2 y 2 + y*),
where the last two factors may be treated as in Example 1.
Example 5.
2i 2 c 2 + 2cV + 2a 2 6 2  « 4  ¥  c 4
= 4i 2 c 2  (a b 2  c 2 ) 2
= (2bc + a 2  b 2  c 2 ) (26c  d 2 + b 2 + c 2 )
= {a 2 (bc) 2 } {(b + c) 2 a 2 }
= (a + b  c) (a  b + c) (b + c + a) (b + c  a).
* The student should observe that the decomposition x 2 + y 2 + xy =
(x + y+ \Jxy) (x + y \/xy), which is often given by beginners when they are
asked to factorise x 2 + y 2 + xy, although it is a true algebraical identity, is no
solution of the problem of factorisation in the ordinary sense, inasmuch as
the two factors contain \Jxy, and are therefore not rational integral functions
of a and y.
138 USE OF REMAINDER THEOREM chap.
RESULTS OF THE APPLICATION OF THE REMAINDER THEOREM.
§ 11.] It may be well to call the student's attention once
more to the use of the theorem of remainders in factorisation.
For every value a of x that we can find which causes the integral
function f(x) to vanish we have a factor x  a off(x).
It is needless, after what has been shown in chap, v., §§ 1316,
to illustrate this point further.
It may, however, be useful, although at this stage we cannot
prove all that we are to assert, to state what the ultimate result
of the rule just given is as regards the factorisation of integral
functions of one variable. If f(x) be of the wth degree, its coeffi
cients being any given numbers, real or imaginary, rational or
irrational, it is shown in the chapter on Complex Numbers
that there exist n values of x (called the roots of the equation
f(x) = 0) for which f(x) vanishes. These values will in general
be all different, but two or more of them may be equal, and one
or all of them may be complex numbers.
If, however, the coefficients of f(x) be all real, then there
will be an even number of complex roots, and it will be possible
to arrange them in pairs of the form X ± jxi.
It is not said that algebraical expressions for these roots in
terms of the coefficients of /(.<:) can always be found ; but, if
these coefficients be numerically given, the values of the roots
can always be approximately calculated.
From this it follows that f(x) can in all cases be resolved into n
linear * factors, the coefficients of which may or may not be all real.
If the coefficients of f(x) be all real, then it can be resolved into a
product of p linear and q quadratic factors, the coefficients in all of
which are real numbers which may in all cases be calculated approxi
mately. We have, of course, p + 2q = n, and either p or q may be
zero.
The student will find, in §§ 110 above, illustrations of these
statements in particular cases ; but he must observe that the
* " Linear" is used here, as it often is, to mean "of the 1st degree."
VII QUADRATIC FUNCTION WITH TWO VARIABLES 139
general problem of factorising an integral function of the nth.
degree is coextensive with that of completely solving an equation
of the same degree. When either problem is solved the solution
of the other follows.
FACTORISATION OF FUNCTIONS OF MURE THAN ONE VARIABLE.
§ 12.] Jflien the number of variables exceeds unity, the problem of
factorisation of an integral function {excepting special cases, such as
homogeneous functions of two variables) is not in general soluble, at
least in ordinary algebra.
To establish this it is sufficient to show the insolubility of
the problem in a particular case.
Let us suppose that x 2 + y 2 + 1 is resolvable into a product of factors which
are integral in x and y, that is, that
x" + y 2 + 1 = (px + qy + r) (p'x + q'y + r'),
then x 2 + y 2 + l =pp'x 2 + qq"y 2 + rr'
+ (Pi' +P'l)xy + (pr' +p'r)x
+ (qr' + q'r)y.
Since this is, by hypothesis, an identity, we have
pp' = 1
qq' = \
rr' = 1
First, we observe that, on account of the equations (1) (2) (3), none of the six
quantities p q r p' q' r' can be zero ; and further, p' = , <?' = , r' — . Hence,
(1)
Pi' + v'i ~ o
(4)
(2)
pr' + p'r =
(5)
(3)
q?''+q'r =
(6)
as
logical
consequences
of
our
hypothesis, we
^ + 2 =
q p
r p
o* n
have
from
(4) (5)
and
(6V
(7)
(8)
(9);
T q
and, from these again, if we multiply by pq, rp, and qr respectively, we get
p 2 + q 2 =0 (10)
^2 + r 2 = ( H )
q 2 + r 2 =0 (12).
Now from (11) and (12) by subtraction we derive
p 2 q 2 =0 (13);
140 QUADRATIC FUNCTION chap.
and from (10) and (13) by addition
2p 3 =0;
from this it follows that ^ = 0, which is in contradiction with the equation (1).
Hence the resolution in this case is impossible.
§ 13.] Nevertheless, it may happen in particular cases that
the resolution spoken of in last article is possible, even when the
function is not homogeneous. This is obvious from the truth
of the inverse statement that, if we multiply together two
integral functions, no matter of how many variables, the result
is integral.
One case is so important in the applications of algebra to
geometry, that we give an investigation of the necessary and
sufficient condition for the resolvability.
Consider the general function of x and y of the 2nd degree, and write it
F = ax 2 + 2hxy + by 2 + 2gx + 2/y + c.
We observe, in the first place, that, if it be possible to resolve F into two
linear factors, then we must have
F = ( \Jax + ly + m) (\Jax + I'y + to'),
= [sjax + >J(Z + Z') + h(l  I')} y + h(m + to') +£(m »»')]
x[sjctx+{\{l+l')l{ll')}y + ii{m + m')\{mm')],
= { sjax + J(J + V)y + \{m + m')} 2  {h(l  l')y + i(m  to')} 2 
Hence, when F is resolvable into two linear factors, it must be expressible in
the form L 2 M 2 , where L is a linear function of x and y, and M a linear
function of y alone ; and, conversely, when F is expressible in this form, it
is resolvable, namely, into (L + M) (L M).
Let us now seek for the relation among the coefficients of F which is
necessary and sufficient to secure that F be expressible in the form L 2  M 2 .
1st. Let a 4=0, then
F = a[x* + 2(hy + g)x/a +(by" + 2/y + c)/a],
= a[ {x + (hy + g)/a} 2  {(hy + gf  a(by* + 2/y + c)} /a"],
= a[ {x + (hy + g)/a} 2  {(^ 2  a% 2 +2(gh  af)y + (f  ac)} /«'].
Hence the necessary and sufficient condition that F be expressible in the form
L 2  M 2 is that (K 2  ab)y" + 2(gh  af)y + (g 2  ac) be a complete square as regards
y. For this, by § 7, it is necessary and sufficient that
l{ghaf)*4(h*db)(g*ac) = 0;
that is,  a {abc + 2fgh  a/ 2  bg 2  ch 2 ) — 0.
Now, since a 4=0, this condition reduces to
abc + 2/gh  a/ 2 bg 2 ch 2 = { 1 ).
2nd. If a = 0, but b 4=0, we may arrive at the same result by first arranging
F according to powers of y, and proceeding as before.
vii OF TWO VARIABLES HI
3rd. If a = 0, 6 = 0, and A#0, the present method fails altogether, but F
now reduces to
F = 2hxy + 2gx + 2fy + a,
and it is evident, since x 2 and y 2 do not occur, that if this be resolvable into
linear factors the result must be of the form 2h(x+p)(y + q). We must
therefore have 2g=2kq,
2f=Zhp,
c = 2hpq.
Now the first two of these give fg = hpq, that is, 2hpq=~ ; whence
using the third,
ch=2fg,
or, since h * 0, 2fgh  cK 2 = (2) ;
but this is precisely what (1) reduces to when a = 0, 6 = 0, so that in this third
case the condition is still the same.
Moreover, it is easy to see that when (2) is satisfied the resolution is
possible, being in fact
2hxy + 2gx + 2fy + c = 2hfx + fVy + ^\ (3),
which is obviously an identity if c = 2/g/h.
4th. If a = 0, 6 = 0, h = 0, F reduces to 2gx + 2/y + c. In this case we may
hold that F is resolvable, it being now in fact itself a linear factor. It is
interesting to observe that in this case also the condition (1) is satisfied.
Returning to the most general case, where a does not vanish, we observe
that, when the condition (1) is satisfied, we have, provided /i' 2 «6 + 0,
V{(#  ab),f + 2(gh  afjy + (g*  ac)} = VF^L+ ? p^Q,
so that the required resolution is
f h + \/l)?~ab g ghaf /r „ , "
Y=a\x+ y + + 7£5 — ^.\/h 2 ab [
I a a a(h 2  ab) )
( h\/h 2 ab q qhaf /.« " r a\
I. a a a{k  ab) )
To the coefficients in the factors various forms may be given by using the
relation (1) ; but they will not be rational functions unless h 2 ab be a com
plete square, and they will be imaginary unless h 2 ab is positive.
If h 2 ab = 0, then (1) gives (ghaf ') 2 = 0, that is, ghaf— 0; and the
required resolution is
( h g \'q ac\  h g \fg 2 ac\ ,>
Y = a\ .T + 2/ + + — * M x+y+  h (5).
I a a a J I a fa a )
The distinction between these cases is of fundamental importance in the
analytical theory of curves of the 2nd degree.
The function abc + 2fghaf 2 bg 2 ch 2 , whose vanishing is the condition
for the resolvability of the function of the 2nd degree, is called the Discrimi
nant of that function.
142
EXERCISES XI
CHAP.
It should be noticed that, if
F = ax 2 + 2hxy + by 2 + 2gx + 2fy + c
=(Ise+my+n)(l'x+m'y+n') (6),
then
ax 2 + 2hxy + by 2 =(lx + my) (Vx + m'y),
so that the terms of the 1st degree in the factors of F are simply the factors
of ax 2 + 2h xy + b y". We have therefore merely to find, if possible, values for
n and n' which will make the identity (6) complete.
Example. To factorise 3a; 2 + 2xy  y + 2x  2y  1 . We have 3a; 2 + 2xy  y 2
= (3xy) (x + y). Hence, if the factorisation be possible, we must have
3x 2 + 2xyy 2 + 2x2yl = (3xy + n){x + y + n') (7).
Therefore, we must have
»+3»'=2 (8),
nn'=2 (9),
«/;'= 1 (10).
Now, from (8) and (9), we get n=  1, and n = +1. Since these values
also satisfy (10), the factorisation is possible, and we have
3x 2 + 2xy  y 2 + 2x  2y  1 = (3x  y  1 ) (x + y + 1 ).
It should be noticed that the resolvability of
F = ax 2 + 2hxy + by 2 + 2gx + 2fy + c
carries with it the resolvability of the homogeneous function of
three variables having the same coefficients, namely,
F = ax 5 + by 2 + cz 2 + 2fyz + 2gzx + 2hxy,
as is at once seen by writing xfz, y/z, in place of x and y.
Exercises XL
Factorise the following functions : —
(1.
(3.
(5.
(7.
(10
(13.
(16.
(18.
(20.
(22.
(24.
(23.
(27.
(28.
(29.
(30.
(a + b) 2 + [a + c) 2  (c+ d) 2 {b + d) 2 . (2. ) ia 2 b 2  {a 2 + b 2  c 2 ) 2 .
(a 2  2¥  c 2 ) 2  i(b 2  c 2 ) 2 . (4. ) (5a 2  11a; + 12) a  (4.C 2  15a; + 6) 3 .
{x 2  03 + y)x + fa} 2 (x yT(x  a) 2 . (6. ) a*  y 6 .
x$y\ (8.) x 2 + Qxy + 9y 2 ±. (9.) 2a^ + 3a;2.
x 2 +6x16. (11.) a; 2 10a; + 18. (12.) x 2 + a;30.
a; 2 + 14a; + 56. (14.) ar + 4a;+7. (15.) 2a; 2 + 5a;12.
x 2 + 2xs/( 2 J + q) + 2q. (17.) x*2bx/(b + c) + (bc)/(b + c).
[x 2 +pq) 2 (p + q) 2 x 2 . (19.) ab(x 2 y 2 )+xy(a 2 b 2 ).
pq[x + y?{p + q){x 2 y 2 ) + {xy) 2 . (21.) a; 3 15a; 2 + 71a; 105.
a, J 14x 2 +148a;. (23.) a 3  13.^ + 54a; 72.
3?  8.c 2 + x  8. (25. ) x 3 + Zpx 2 + (3p 2  q 2 )x +p(p 2  q 2 ).
(p + q)x 3 + (pq)x 2 (p + q)x(pq).
x 3  (1 +p +p 2 )x 2 + (p +p 2 +p 3 )x p 3 .
x*  (a + byx 3 + (a 2 b + ab 2 )x  a 2 b 2 .
x 6 + x i a + xta 2  x 2 ^  xa s  a\
(l+x) 2 (l+i/ 2 )(l+y) 2 (l+ar ! ). (31.) x 4 + x 2 y 2 + y*.
vii EXERCISES XI 143
(32. ) Assuming x l + y i =(x~ +pxy + y 2 ) (x 2 + qxy + y 2 ), determine p and q.
(33.) Factorise a 4 + y 4 2(a 2 + ?/ 2 ) + l.
(34.) Determine r and s in terms of a, p, and q in order that x~ a 2 may
be a factor in x* +px* + qx 2 + rx + s.
Factorise
(35. ) (x m +») 2  {x'"a n ) 2  (x"a m ) 2 + («'»+»).
(36. ) (x 2 + a 2 ) 2 ^ + a 2 x 2 + a 4 )  (x* + a; 4 a 4 + a 8 ).
(37. ) xy 2  2xy y 2 + x + 2yl. (38. ) 2a 2 + xy + 7x + 3y + 3.
(39.) 2.)j 2 + a2/32/ 2 a;47/l. (40.) xy + 7x + 3y + 21.
(41.) a: 2 2!/ 2 3z 2 + 7yz + 2za; + a;#.
(42.) Determine X so that (x + 6y l){6x + y l)+\(3x + 2y + l)(2x + Sy+l)
may be resolvable into two linear factors.
(43.) Find an equation to determine X so that ax 2 + by 2 + 2hxy + 2gx + 2/y
+ c + \xy may be resolvable into two linear factors ; and find the value of X
when c = 0.
(44.) Find the condition that (ax + Py + yz) {a'x + p'y + y'z)  (a"x + p"y
+ y"z) 2 break up into two linear factors.
(45.) If (x+p) {x + 2q) + (x + 2p) (x + q) be a complete square in a', then
9p 2 Upq + 9q 2 =0.
(46. ) If (x 4 b) (x + c) + (x + c) (x + a) + (x + a) (x + b) be a complete square in
x, show that a = b = c.
Factorise
(47.) a 3 + b 3 + c?3abc. (48.) x 3 + 3axy + ?/ 3  a 3 .
(49.) (xx 2 ) 3 + {x 2 l) 3 + (lx) 3 .
Factorise the following functions of a:, y, z: — *
(50.) ^(y 2 + x 2 )(z 2 + x 2 )(yz).
(51.) 2(^ + ^)^2/). (52.) 2,x i (y 2 z 2 ). (53.) (2a) 3 Zx 3 .
(54.) Simplify {Z(x 2 + y 2 z 2 )(x 2 + z 2 y 2 )} /U{x±y±z).
(55.) Show that 2(1/"^"  y n z m ) and 2,x n (y m zt>  yPz m ) are each exactly
divisible by (y z)(z x) (x  y).
(56.) Show that nx n+1  (n + 1 )x n + 1 is exactly divisible by (a*!) 2 .
(57.) Show that ~Zx 2 (y + z  a:) 3 is exactly divisible by 2a; 2 22j/z.
(58.) Show that (x + y + z) 2n+1 x 2n + 1 ~y 2rl + 1 z 2n + 1 is exactly divisible by
{y + z){z + x)[x + y).
(59. ) (y  z) 2 ^ 1 + {z a) 2 "+ 1 + (x  y) 2 '* 1 is exactly divisible by {y  z) (z  x)
(x ~ !/)■
(60. ) If n be of the form 6m  1, then (,y  z) n + (z  x) n + (x  y)» is exactly
divisible by 2.x 2 Zxy ; and, if n be of the form 6»i + l, the same function is
exactly divisible by (2a: 2  2a*i/) 2 .
(61.) Prove directly that xy\ cannot be resolved into a product of two
linear factors.
(62.) If a and b be not zero, it is impossible so to determine p and q that
x +py + qz shall be a factor of x 3 + ay 3 + bz 3 .
* Regarding the meaning of 2 in (50), (51), &c, see the footnote on p. 84.
CHAPTER VIII.
Rational Fractions.
§ 1.] By a rational algebraical fraction is meant simply the
quotient of any integral function by any other integral function.
Unless it is otherwise stated it is to be understood that we
are dealing with functions of a single variable x.
If in the rational fraction A/B the degree of the numerator
is greater than or equal to the degree of the denominator, the
fraction is called an improper fraction, if less, a proper fraction.
GENERAL PROPOSITIONS REGARDING PROPER AND
IMPROPER FRACTIONS.
§ 2.] Every improper fraction can be expressed as the sum of
an integral function and a proper fraction ; and, conversely, the sum
of an integral function and a proper fraction may be exhibited as an
improper fraction.
For if in '— the degree m of A m be greater than the
degree n of B n , then, by the divisiontransformation (chap, v.),
we obtain
A w _ pj J*
n Dn
which proves the first part of our statement, since Q m . n is
integral, and the degree of II is < n.
Again, if F p be any integral function whatever, and A TO /B n a
proper fraction (that is, m<n), then
CHAP, nil PROPER AND IMPROPER FRACTIONS 145
A P P, + A
P "Ml _ P n "I
p K~~ "bT '
which is an improper fraction, since the degree of the numerator,
namely, n + p, is > n.
Examples of these transformations have already been given
under division.
It is important to remark that, if two improper fractions be
equal, then the integral parts and the properly fractional parts must be
equal separately.
For let ^• = Q M + ^
A' / "R'
am ' t>, — H m'n' + T3' '
t> ,i' d n '
by the above transformation.
A A' ,
I hen, it
B„ B' tt < '
R ~ R'
we have Q m _„ + == Q' m < _ M  +
B„ ^" l " BV
'/i
nence h»/i« V»h')i' — t> tv
*> » & ri
Now, since the degrees of R' and R are less than n' and n
respectively, the degree of the numerator on the righthand
side of this last equation is less than n + n' . Hence, unless
Qmn — Q'm'ri — Q) we nave an integral function equal to a
proper fraction, which is impossible (see chap, v., § 1). We must
therefore have
T> TV
Qmn = Q'm'n'j and consequently ~ = ™~.
D/i ° ri
X.B. — From this of course it follows that m n= m'  it'.
As an example, consider the improper fraction [a? + 2x + Zx + 4)/(ar + x + 1),
and let us multiply both numerator and denominator by ar + 2x+ 1 ; we thus
obtain the fraction
{a? + Ax 4 + 8x* + 12x 2 + 11«+ 4)/(* 4 + Zx 3 + 4.? 2 + 3a?+ 1),
which, by chap, iii., § 2, must be equal to the former fraction. Now transform
each of these by the longdivision transformation, and we obtain respectively
£+3
■ +1 +?+.+r
VOL. I
146 DIRECT OPERATIONS chap.
x 3 + 5x 2 + 7x + 3
and X + 1 +
x 4 + 3x 3 + ^ + 3x + Y
The integral parts of these are equal ; and the fractional parts are also equal
(see next section).
The sum of two 'proper algebraical fractions is a proper algebraical
fraction.
After what has been given above, the proof of this proposition will present
no difficulty. The proposition is interesting as an instance, if any were needed,
that fraction in the algebraical sense is a totally different conception from
fraction in the arithmetical sense ; for it is not true in arithmetic that the
sum of two proper fractions is always a proper fraction ; for example, f + i = ,
which is an improper fraction.
§ 3.] Since by chap, iii., § 2, we may divide both numerator
and denominator of a fraction by the same divisor, if the nu
merator and denominator of a rational fraction have any common
factors, we can remove them. Hence every rational fraction can
be so simplified that its numerator and denominator are algebraically
prime to each other ; when thus simplified the fraction is said to be at
"its lotvest terms."
The common factors, when they exist, may be determined by
inspection (for example, by completely factorising both numerator
and denominator by any of the processes described in chap, vii.) ;
or, in the last resort, by the process for finding the G.C.M., which
will either give us the common factor required, or prove that
there is none.
Example 1.
a? +53?+ 7s +3
se*+3a 8 +4a! 2 +3a;+l'
By either of the processes of chap. vi. the G. C. M. will be found to be x + 2.r + 1.
Dividing both numerator and denominator by this factor, we get, for the
lowest terms of the given fraction,
a+3
a, a +ss+r
The simplification might have been effected thus. Observing that both
numerator and denominator vanish when x=  1, we see that x+ 1 is a com
mon factor. Removing this factor we get
.•} ■!..• + 3
a*+2ar 1 +2a+l'
Here numerator and denominator both vanish when x=  1, hence there is the
common factor as+1. Removing this we get
vin WITH RATIONAL FRACTIONS 147
x+3
aP + x + l'
It is now obvious that numerator and denominator are prime to each
other ; for the only possible common factor is x + 3, and this does not divide
the denominator, which does not vanish when x =  3.
§ 4.] The student should note the following conclusion from
the above theory, partly on account of its practical usefulness,
partly on account of its analogy with a similar proposition in
arithmetic.
If two rational fractions, P/Q, P'/Q', be equal, and P/Q be at its
lowest terms, then P' = AP, Q' = AQ, where A is an integral function
of x, which will reduce to a constant if P'/Q' be also at its lowest
terms.
To prove this, we observe that
p p
Q'~Q'
whence r = — ~,
Q
that is, Q'P/'Q must be integral, that is, Q'P must be divisible
by Q ; but P is prime to Q, therefore by chap, vi., § 1 2, Q' = AQ,
where A is an integral function of x. We now have
Q
so that P' = AP, Q' = AQ.
If P'/Q' be at its lowest terms, P' and Q' can have no com
mon factor ; so that in this case A must be a constant, which
may of course happen to be unity.
DIRECT OPERATIONS WITH RATIONAL FRACTIONS.
§ 5.] The general principles of operation with fractions
have already been laid down ; all that the student has now to
learn is the application of his knowledge of the properties of
integral functions to facilitate such operation in the case of
rational fractions. The most important of these applications is
the use of the G.C.M. and the L.C.M., and of the dissection of
functions by factorisation.
148 EXAMPLES OF DIRECT OPERATIONS chap.
No general rules can be laid down for such transformations as we proceed
to exemplify in this paragraph. But the following pieces of general advice
will be found useful.
Never make a step that you cannot justify by reference to the fundamental
laws of algebra. Subject to this restriction, make the freest use of your judg
ment as to the order and arrangement of steps.
Take the earliest opportunity of getting rid of redundant members of a
function, unless you see some direct reason to the contrary.
Cultivate the use of brackets as a means of keeping composite parts of a
function together, and do not expand such brackets until you see that some
thing is likely to be gained thereby, inasmuch as it may turn out that the
whole bracket is a redundant member, in which case the labour of expanding
is thrown away, and merely increases the risk of error.
Take a good look at each part of a composite expression, and be guided in
your treatment by its construction, for example, by the factors you can per
ceive it to contain, by its degree, and so on.
Avoid the unthinking use of mere rules, such as that for long division,
that for finding the G.C.M., &c. , as much as possible ; and use instead pro
cesses of inspection, such as dissection into factors ; and general principles,
such as the theorem of remainders. In other words, use the head rather than
the fingers. But, if you do use a rule involving mechanical calculation, be
patient, accurate, and systematically neat in the working. It is well known
to mathematical teachers that quite half the failures in algebraical exercises
arise from arithmetical inaccuracy and slovenly arrangement.
Make every use you can of general ideas, such as homogeneity and sym
metry, to shorten work, to foretell results without labour, and to control
results and avoid errors of the grosser kind.
Example 1. Express as a single fraction in its simplest form —
2x s + 4x + 3x + A 2ar>+4x 2 3.c2 „
— &n ^i — =F say 
Transform each fraction by division, then
F = (2x + 4) +£, ■ (2x+ 4)  £±?,
x + 1 a; — l
_ x{a?1)(x + 2){j?+l)
.> 4 l
_ 2x?2xr1
~ x*l '
_ 2(3 3 a 8 l)
x A l '
Example 2. Express as a single fraction
111 1
F=
tfZxt+Zxl x*a~x+l a A 2x 3 + 2x~l x i  2x* + 2x : ~ 2x+ 1"
vin WITH RATIONAL FRACTIONS H9
AVe have
x^ofix+l =x 3 + 1x(x+1) = (j: + 1)(x 2 x + 1x)
= (x+l) (xlf;
x 4  2:< 3 + 2x  1 =x*  1  2x{x 2  1),
= (* 2 l)(.>'l),
= (xl)*(x+l);
x*  2X 3 + 2x 2  2x + 1 = (a? + 1 ) 2  2x(x + 1 ),
= (x*+l)(xl)*.
Whence
ill 1
F =
(x1? (jc+l)(a;l) a (xlf(x+l) (ar + lU:cl)'
_ (a;+l)(.r~l) _ (a; 2 + l) + (al)(a;+l )
[x + 1 ) (x  1 ) 3 [x  l) s (aj+ 1) (ar> + 1) '
2 2^
(x + l)(a;l) 3 (a}l) s (se+l)(a?+l)'
x^+lx*
(x + l)(x 2 + l){xlf
2
"(x*l)(xl) 2 '
2
x*2x i + x i  x 2 + 2x  1
Example 3.
/ .k  y x 3  y 3 \ / x + y x s + y 3 \
\x + y ar' + y 3 / \xy x^y 3 /'
( X V\ (i x* + x*/ + y \ (x+y\ / t  x 2 xy + y 2 \
\x+yj \ xPxy + y 2 / \xy) \ x 2 + xy + y 2 )'
=( ~ 2xy \ x ( 2 ^ + y z )\
\x 2 xy + y 2 J \^ + xij + y)'
ixy(x + y")
x 4 + x 2 y 2 + y f
Example 4.
^2 bc 2 ca 2 ab
* =T~:+ i \t~ m + r— : + r— 5T7T — , + — ; +
bc (ca)(ab) ca (ab)(bc) ab (bc)(ca)'
^ 2(ca)(ab) + (bc) 2 + 2(ab)(bc) + (ca) 2 + 2{bc){ca) + (ab) 2
(bc){c~a)(ab)
_ {(bc) + (ca) + (ab)} 2
&c.
2
&C. (bc)(ca){ab) '
it being of course supposed that the denominator does not vanish.
150 DIRECT OPERATIONS WITH RATIONAL FRACTIONS CHAP.
Example 5.
« s b 3 3
■C =7 tt~, ; +7^ m r +
(ab){ac) (bc)(b~a) (ca)(cb)
_  a 3 {b  c)b 3 {c a)  <?{ab)
~ (bc)(c a) (a  b)
Now we observe that when b = c the numerator of F becomes 0, hence bc
is a factor; by symmetry ca and ab must also be factors. Hence the
numerator is divisible by (b c){c a) (a  b). Since the degree of the numer
ator is the 4th, the remaining factor, owing to the symmetry of the expression,
must be Pa + PS + Pc. Comparing the coefficients of a s b in
 a 3 (b  c)  b 3 (c  a)  <?{a  b)
and F{a + b + c){bc){ca){ab),
we see that P= +1.
Hence, finally, ¥ = a + b + c.
Example 6.
^_ a 2 +pa + q b 2 +pb + q c 2 +pc + q
~ (ab)(ac)(xa) {ba) {bc) (xb) {ca)(cb){xc)'
■p_ (hc) (a 2 +pa + q)(x b) (x  c) + &c. + &c.
(b  c) (c a) (a b) (x  a) (x  b) (x  c) '
_ (b  c) (a? + pa + q) {x 2  (b + c)x + bc} +&c. +&c.
&c.
Now, collect the coefficients of x 2 , x, and the absolute term in the numerator,
observing that the two &c.'s stand for the result of exchanging a and b and a
and c respectively in the first term. We have in the coefficient of x 2 a part
independent of p and q, namely,
a 2 (b c) + b 2 (c a)+c*(a b)=  (J  c) (c  a) (a  b) (1).
The parts containing p and q respectively are
{a(b c) + b(c a)+c{ab))p = Q
and {(bc) + (ca) + (ab)}q = 0.
The coefficient of a? therefore reduces to (1).
Next, in the coefficient of x we have the three parts,
 {a"(b 2  c 2 ) + b 2 (c 2  a 2 ) + cV  b 2 )} = 0,
 {a(b 2 c 2 ) + b(c 2  a 2 ) + c(a 2  b 2 ))p
= (bc)(ca)(ab)p (2),
and  {(b 2 c 2 ) + (c 2 a 2 ) + {a 2 b 2 )}q = 0.
Finally, in the absolute term,
abc {a(b c) + b(c a)+ c(a  b)} = 0.
abc {{b c) + (c a) + (a b)}p  0,
{bc(b  c) + ca(c a)+ ab(a b)}q
= (bc)(cu)(ab)q (3).
vin INVERSE METHODS 151
Hence, removing the common factor (bc) (ca)(a b), which now appears
both in numerator and denominator, and changing the sign on both sides, we
have
,r +px + q
F =
(xa) (x b) (x — e)
The student should observe here the constant use of the identities on pp.
81S3, and the abbreviation of the work by twothirds, effected by taking
advantage of the principle of symmetry. In actual practice the greater part
of the reasoning above written would of course be conducted mentally.
INVERSE METHOD OF PARTIAL FRACTIONS.
§ 6.] Since we have seen that a sum of rational fractions can
always be exhibited as a single rational fraction, it is naturally
suggested to inquire how far we can decompose a given rational
fraction into others (usually called "partial fractions") having
denominators of lower degrees.
O
Since we can always, by ordinary division, represent (and that
in one way only) an improper fraction as the sum of an integral
function and a proper fraction, we need only consider the latter
kind of fraction.
The fundamental theorem on which the operation of dissec
tion into " partial fractions " depends is the following : —
If A/PQ be a rational proper fraction whose denominator contains
two integral factors, P and Q, which are algebraically prime to each
other, then we can always decompose A/PQ into the sum of two proper
fractions, P'/P + Q'/Q.
Proof. — Since P and Q are prime to each other, we can (see
chap, vi., § 11) always find two integral functions, L and M,
such that
LP + MQ=1 (1).
Multiply this identity by A/PQ, and we obtain
A _AL AM
PQ " Q + "P" ( ~ } '
In general, of course, the degrees of AL and AM will be higher
than those of Q and P respectively. If this be so, transform
AL/Q and AM/P by division into S + Q' Q and T + P'/P, so that
152 PARTIAL FRACTIONS chap.
S, T, Q', and P' are integral, and the degrees of P' and Q' less
than those of P and Q respectively. We now have
PQ P Q v ; '
where S + T is integral, and P'/P + Q Q a proper fraction. But
the lefthand side of (3) is a proper fraction. Hence S + T must
vanish identically, and the result of our operations will be simply
A^ + g (4).
PQ P Q w
which is the transformation required.
To give the student a better hold of the above reasoning, we
work out a particular case.
Consider the fraction
x*+l
F=:
" (x 3 + &E 2 + 2x + 1 ) (X s + x + 1) '
Here A=a*»+1 > F = x 3 + Bx 2 + 2x + l, Q=x" + x + l.
Carrying out the process for finding the G. C. M. of P and Q, we have
11 + 1)1 + 3 + 2 + 1(1 + 2
2 + 11
11)1 + 1 + l(l +
+ 1
+ 1
whence, denoting the remainders by Ri and Ro,
P = (x + 2)Q + Ri, Q= x&i + R*
From these successively we get
Ri = P(a + 2)Q,
l = R, = Q + a ,R 1 ,
= Q + xPx{x + 2)Q,
= (x n 2x+l)Q + .,;!> (1).
In this case, therefore,
M=z*2x+l, L=x.
Multiplying now by A/PQ on both sides of (1), we obtain (putting in the
actual values of P and Q in the present case)
(af' + l)(a 3 2g + l) (z* + l)x .
x 2 + Zx? + 2x+\ x*+x+l'
_ x e 2x 5 + x i x 2 2x+l j~  x
x 3 + 3x? + 2x+l + a? + x+l '
or, carrying out the two divisions,
viii SPECIAL CASES 153
or, seeing that the integral part vanishes, as it ought to do,
which is the required decomposition of F into partial fractions.
Cor. If P, Q, R, S, . . . be integral functions of x which are
prime to each other, then any proper rational fraction A/PQRS . . .
can be decomposed into a sum of proper fractions, P'/P + Q' Q +
R'/R + sys + . . .
This can be proved by repeated applications of the main
theorem.
§7.] Having shown a priori the possibility of decomposition
into partial fractions, we have now to examine the special cases
that occur, and to indicate briefer methods of obtaining results
which we know must exist.
"We have already stated that it may be shown that every
integral function B may be resolved into prime factors with real
coefficients, which belong to one or other of the types (x  a)'',
(/ + 0b + 7 y.
1st. Take the case where there is a single, not repeated,
factor, X  a. Then the fraction F = A B may be written
F *
(•!•  a)Q
say, where X  a and Q are prime to each other. Hence, by our
general theorem, we may write
f F + Q ' m
each member being a proper fraction.
In this case the degree of P' must be zero, that is, P' is a
constant.
It may be determined by methods similar to those used in
chap, v., § 21. See below, Example 1.
P' determined, we go on to decompose the proper fraction
Q Qi by considering the other factors in its denominator.
154 SPECIAL CASES chap.
2nd. Suppose there is a repeated factor (x  a) r ; say B =
(.r  a) r Q, where Q does not contain the factor x  a. "We may,
by the general principle, write
r, P' + Q'
(a;  a)'" Q*
P' is now an integral function, whose degree is less than r\
hence, by chap, v., § 21, Ave may put it into the form
P' = a + di(z  a) + . . . +a r _i(a;a) r " 1 ,
and therefore write
" + ...+^+g (2),
(xaf (a; a)'" 1 xa Q
where a , a lf . . ., a,._i are constants to be determined. See
below, Example 2.
3rd. Let there be a factor (x 2 + fix + y) g , so that
b = (x 2 + fix + y yq,
Q being prime to x 2 + fix + y. Now, we have
P' Q'
(x 2 + px + 7 y q
P' is in this case an integral function of degree 2s  1 at most.
We may therefore write, see chap, v., § 21,
P' = (a + b x) + (</, + b x x) (x 2 + fix + y)
t (a 2 + kjx) (x 2 + fix + y) 2
+ (a g „ l + b 8  1 z)(x* + px + y y 1 .
We thus have
_ a + h„r a, + b x x at^ + bgtf Q'
(x 2 + /3x+ y y> (x 2 + fix + y)* 1 x 2 + (3x + y Q { ' '
where the 2s constants a m b , &c, have to be determined by any
appropriate methods. See Examples 3 and 4.
In the particular case where s = 1, we have, of course, merely
_ a + &Q.T Q'
Viil EXPANSION THEOREM 155
By operating successively in the way indicated we can decompose
every rational fraction into a sum of partial fractions, each of which
belongs to one or other of the two types p r j{x  a) r , (a s + b s x)/(x 2 + ftx
+ y) s , where a, ft, y, p r , a s , b 8 are all real constants, and r and s
positive integers.
It is important to remark that each such partial fraction
has a separate and independent existence, and that if necessary
or convenient the constant or constants belonging to it can be
determined quite independently of the others.
Cor. If P be an integral function of x of the nth degree, and
a, a, . . ., a; ft, (3, . . . , ft ; y, y, . . . , y, . . . constants not less
than n + 1 in number, r of which are equal to a, s equal to ft, t equal
to y, . . . , then we can ahoays express P in the form
P = 2{a + a,(x «)+...+ ctr.^x  a) r " 1 }(a:  ft)%x  y) ( . . .,
where a , a u . . . , a r _ u . . . are constants. In particular, if r = 1.
s = 1 , t = 1, . . . , we have
P = ?a (xft)(xy) ...
These theorems follow at once, if we consider the fraction
p/(x  a y(x fty(x  7 y...
There is obviously a corresponding theorem where x  a,
x ft, x  y are replaced by any integral functions which are
prime to each other, and the sum of whose degrees is not less
than n + 1.
§ 8.] We now proceed to exemplify the practical carrying
out of the above theoretical process ; and we recommend the
student to study carefully the examples given, as they afford a
capital illustration of the superior power of general principles as
contrasted with "rule of thumb" in Akebra.
o
Example 1. It is required to determine the partial fraction, corresponding
to x 1, in the decomposition of
(Iv 4  16a»+17a?  8.r + 7)/(.r  1) (a  2) 2 (a 2 + 1).
We have
r _ 4a^16^ + 17x 2 8a + 7 _ p Q'
(a;l)(a;2) 2 (a; 2 +l) al + (aj2) 2 (£B 2 + l) ( >'
and we have to find the constant p.
156 EXAMPLES CHAP.
From the identity (1), multiplying both sides by (x1) (j;2) 2 (a; 2 + l), we
deduce the identity
±x A  16x 3 + 1 7 x s  8.r + 7 =p{x  2) 2 (« 2 + 1) + Q'(x  1 ) (2).
Now (2) being true for all values of x, must hold when o=l ; in this case it
becomes
4 = 2^), that is,p=2.
Hence the required partial fraction is 2/(x 1).
If it be required to determine also the integral function Q', this can be
done at once by putting ^> = 2 in (2), and subtracting 2(.c2) 2 (.£ 2 + l) from
both sides. We thus obtain
2x*  8a? + 7,<"  1 = Q'(x  1 ) (3).
This being an identity, the lefthand side must be divisible by x1.* It is
so in point of fact ; and, after carrying out the division, we get
2,r 3 6a: 2 + a:+l = Q' (4),
which determines Q'.
The student may verify for practice that we do actually have
4^16^ + 1 7.x 2  8x + 7 _ _2_ 2 a 3  6x ° + x + l
(xl){x2)*(x* + l) x  1 + (x  2) s (x 2 + 1 ) *
Example 2. Taking the same fraction as in Example 1, to determine the
group of partial fractions corresponding to (a;  2) 2 .
1°. "VVe have now
4x *  1 6^ + 1 7a; 2  8a: + 7 _ a n a x Q' ..
(a;l)(a;2) 2 (a; 2 +l) ~ {x2) + (x^2) + (x1) (x 2 +l) ^ "
whence
ix i 16x 3 + l7x' 2 8x + 7 = a (xl) (z 2 + 1) + <n(x 2) (aj1) (x 2 + 1)
+ Q'(*2) 2 (2).
In the identity (2) put x = 2, and we get
 5 = 5«oi that is, « =  1 .
Putting now a = 1 in (2), subtracting (  1) (.« 1) (.c 2 + l) from both sides
and dividing both sides by x  2, we have
4r i 7v i + 2x3 = a 1 (xl)(x + l) + Q'{x2) (3).
Put x = 2 in this last identity, and there results
+ 5 = 5cti, that is, a,\ = + 1.
The group of partial fractions required is therefore
l/(x2T + l/(x2).
If required, Q' may be determined as in Example 1 by means of (3).
2°. Another good method for determining a and a\ depends on the use of
"continued division."
If we put x = y + 2 on both sides of (1), we have the identity
4(y + 2) 4 16(y + 2) 3 + 17(y + 2) 2 8(?/ + 2) + 7 = a a x Q"
(2/+l)2/ 2 l(y + 2) 2 +l} y 2 y + (y + l){{y + 2f + \\'
* If it is not, then there lias been a mistake in the working.
vin EXAMPLES 157
that is,
54y + &c. _fl ffl _ Q"
5y + 9f + kc. y 2+ y (l+y)(3 + iy + y)
Now, by chap, v., § 20, the expansion of a rational fraction in descending
powers of 1/y and ascending powers of y is unique. Hence, if we perform
the operation of ascending continued division on the left, the first two terms
must be identical with a /if + ai/y ; for Q"/(l +y) (3 + 4y + ?/ 2 ) will obviously
furnish powers of y merely.
AVe have
54+... 5 + 9+ . , . .
+5+... 1+1+...
therefore a = 1, ai= +1.
The number of coefficients which we must calculate in the numerator and
denominator on the left depends of course on the number of coefficients to be
determined on the right.
Example 3. Lastly, let us determine the partial fraction corresponding to
7? + 1 in the above fraction.
We must now write
4.7;* I6x? + l7x 2 8x + 7 _ ax + b Q' .
(xl)(x2)*(x 2 +lj~~x 2 + l + (xl)(x2f ( '•
1°. Whence, multiplying by (x 1) (x 2) 2 ,
4ar 1 16x 3 + 17x 2 8^ + 7 _ jax + b) (x1) {x 2) 2
x*+l ~ x+l
whence
■ + Q' (2)
4 ,,.2 _ 1 6i , + 13 + 8 4_J = {nx + b) ( x _ 5 + i£±l\ + Q'
X'+l \ X 1 + 1 /
, ,,, rN 7ax + (7b + a)x+b _.
= (ax + b) (x 5)4 \, + 1  + < x >\
... r , „ (7b + a)x + (b7a) _, .
= (ax + b) .r5 +7a + v ^—. \ ' + Q (3).
ar + 1
Now the proper fractions on the two sides of (3) must be equal — that u, we
must have the identity
( 7 b + a)x + (b  7a) = Sx  6,
therefore 7b + a — 8, b7a=  6.
Multiplying these two equations by 7 and by 1 and adding, we get
506 = 50, that is, 6=1.
Either of them then gives «=1, heuce the required partial fraction is
(x+l)l(x*+l).
2". Another method for obtaining this result is as follows.
Remembering that x 2 + 1 = [x + i) (x  I) (see chap, vii.), we see that a; 2 +1
vanishes when x—i.
Now we have
lx i 16x"+l7x 2 8x + 7 = (ax+b){xl)(x2) 2 + Q'{x n +l)
= [ax + b) (a?  5x 2 + 8x  4) + Q'(.>" + 1 ) (4).
158 EXAMPLES
CHAP.
Put in tins identity x=i, and observe that
i i = Pxi 2 = (l)x(l)=+l,
^—■Px i—( l)xi=  i ;
and we have 8iG(ai + b) (7i+l),
= (7b + a)i+{b7a) ;
whence (7b + a  8)i=  b + 7a 6,
an equality which is impossible * unless both sides are zero, hence
7b + a8 = 0, b + 7a60,
from which a and b may be determined as before.
3°. Another method of finding a and b might be used in the present case.
We suppose that the partial fractions corresponding to all the factors
except ic 2 + 1 have already been determined. We can then write
„ 2 1 1 ax + b /c .
From this we obtain the identity
4.i 4 16j 3 + 17a; 2 8a; + 7
= 2(x2) 2 (x 2 + l)(xl)(x°+l) + (xl)(x2)(x 2 +l)
+ (ax+b){xrl){x 2) 8 ;
whence
% A  Ao? + Sx* + ix  4 = (ax + b) [x  1) [x  2) 2 ;
and, dividing by (x  1) (x  2) 2 ,
x+l=ax + b.
This being of course an identity, we must have
a = l, b = l.
Another process for finding the constants in all the partial fractions depends
on the method of equating coefficients (see chap, v., § 16), and leads to their
determination by the solution of an equal number of simultaneous equations
of the 1st degree.
The following simple case will sufficiently illustrate this method.
Example 4.
To decompose (3xi)/(x l)(„e2) into partial fractions.
We have
Zx  4 a b
{x\)[x~2)~x^l Jr x^~V
therefore 3x l = a(x2) + b(x 1),
= (a + b)x  (2a + b).
Hence, since this last equation is an identity, we have
a + b = 3, 2a + b=i.
Hence, solving these equations for a and b (see chap. xvi. ), we find «=i,
b = 2.
* For no real multiple (differing from zero) of the imaginary unit can be a
real quantity. See above, chap, vii., § 6. The student should recur to this
case again after reading the chapter on Complex Numbers.
VIII
EXERCISES XII
159
Example 5. We give another instructive example. To decompose
F=,
x+px + q
we may write
x +px + q
(x a)(x b) (xc)'
C
A B
+ 
+ ■
(1),
(x  a) (x  b) {x  c) xa xb xc
where A, B, C are constants.
Now
x 2 +px + q = A(x b)(xc) + B(x  e) {x a) + C{x  a) (x  b) (2).
Herein put x = a, and there results
a 2 +pa + q= A(a b)(ac) ;
whence
By symmetry
We have therefore
x^+px + q
A= a2+ P a + 9
(a b)(ac)'
b 2 + pb + q
B=
(x a)(x b)(x c)
a" +pa + q
+
(ba)(bc)'
c 2 +pc + q
(c~a)(c b)'
b 2 +pb + q
+ .
c 2 +pc + q
(a b)(ac)(x a) (bc){ba)(xb) (ca)(cb)(xc)
(3),
an identity already established above, § 5, Example 6. It may strike the
student as noteworthy that it is more easily established by the inverse than
by the direct process. The method of partial fractions is in point of fact a
fruitful source of complicated algebraical identities.
Exercises XII.
Express the following as rational fractions at their lowest terms.
(1.
(2.
(3.
(4.
(5.
(6.
(7
(8.
(9.
(10.
{x 3 + 2x 2 x + 6)/(x*  x 2 + ix  4).
(9ar» + 53.T 2  9a J 8)/(4x 2 + 44« + 120).
z*+2x*2xl _ x*_+ ,< a  3<"  5x2
x* + a*3x*5x2 a? 4 + 2x*  2x  1 "
{3x*x*z 1 )/(8a? + 5x 2 + Zx + 1 ) + (se* + 3a? + 5x + 3)/{x 3 + x 2 + x  3).
(a 6  2a? + l)/(x 2 2x + l) + {a? + 2a? + 1 )!(x 2 + 2x + 1 ).
(6a? + IZax 2  9a 2 x  10« 3 )/(9x 3 + \2ax 2  Ua 2 x  10a 3 ).
(la)l{(l+ax) 2 (a + x) 2 }.
{(w + x + z)(w + x)y(y + z)}/{(w + x + z)(w + z)y(x + y)}.
(\x)(\x 2 fl { (T^p " (lx)(lx 2 ) + (l^tf) 2 ) '
{ (al + bm) 2 + (am  bl) 2 } / { (ap + bq) 2 + (aq  bpf } .
160
EXERCISES XII
CHAP.
(11.
(12.
(14,
(15.
(16.
(17.
(18.
(19.
(20.
(21.
(22.
(23.
(24.
(25.
(26.
(27.
(28.
(29.
(30.
(31.
(32.
(33.
(34.
(35.
{px 2 + (ks)x + r} 2  {px 2 +(k + s)x + r} 2
\p3? + {k + t)x + r} 2  {px 2 + (kt)x + r} 2 '
x %y
2x 2y 2y2x
(13.)
■ + ■
1
ab(ab)x a + b + (a + b)x
l/(a 2b lj(a 2b l/(a  2b) ) ).
1/(6* + 6)  l/(2a3  2) + 4/(3  3a; 2 ).
x^y 3
x*y*
xy _ , / x + y 1_ )
x 2 y 2 t \x 2 + y 2 x + y]'
/ x lx \ I / x \x \
\\+x x )J \l+a: x )'
6a;
3a;2 9a, 2 + 4
30a~+4a Ax
2x +1
24(aY^l) ' 8(a;+l) ' 4(a + l) 2 2(a;+iy 3 "3(a; 2 + a;+l)•
_1 1_ _2_ _2_
(a;+l) 2 (a; + 2) 2 (x + 2f + x + l x+2'
{a + b)j{x + a) + {a  b)/(x a) 2a{x + ty/ix 2 + a").
{(xy)l{x+y)} + {{xy)l{x + y)} 2 + {(xy)/(x+y)}*.
/ a 3 3a+2 \ /a 2 + 2a;+l \
\.r i + 2x 2 + 2x+l) X Wbx+i)'
'a 2 + x 2
/a 2 + x 2 \
(r^x +1 ) x
2ax
x + y
+
+ 1
xy
ax
5a; +4,
4a(a + x,
a 2  ax + x
2'
2
x'  %f
aPy 3 ar 5 + y 3 x i + x 2 y 2 + y 4 '
x 2 y 2
1 1
 + .
x y<
I
Ht Hfi' t ts)( t .)
x° y
+ ;
+ •
2a(ac)(xa) 2a(a + c)(x + a) (c 2  a 2 ) (x + c)
( _20__180 420__280\ /i__20_ 2?2 _ i 2 ^ 280 \
(. + a+l a + 2 + a; + 3 a'+4j X t xl + x2~ x^S + x 4j
{(xyl) 2 + (x + y2)(x + y2xy)}/{(xy + l) 2 (x + y) 2 }.
{l+y*+*Syz)/(l+y+z).
{a{a + 2b) + b(b + 2c) + c(c + 2a)} / [a 2 b 2 c 2  2bc) .
(a + b) s +(b + c) s (a + 2b + c) i
(a + b)(b + c)(a + 2b + c)
x 6 + a 6
■+
orx*
(I c + a 6 ) (a?  a 2 ) + a 2 x 2 (x*  a 4 ) X* « G  « V V s  a 2 )'
a 2 + (2ac  b 2 )x 2 + c 2 x A a 2 + (ac  b 2 )x 2  bcx*
a 2 + 2abx + {2ac+b a ' •zhcjF+c 2 x A * a + \ac >> 2 + bex '■'
VIII
EXERCISES XII
161
X 2 + y 2 +x + yxy + l  x 2 + y 2 + x  y + xy + 1
xy1 x + y1
(37.)
(38.)
(■r 5 10,cy + 5.CT/ 4 ) 2 + (5afy  lOieV + //•/'
(b + c) + 2(b 2 c°) + (bcf
(39.) 2(6 2 + c 2 a 2 )/(a&)(<w).
(41.) 2(6 + c)/(co)(a&).
(43.) 2(& 2 + &c + c 2 )/(a&)(«c).
(44.) {II(la?) + n(a;y 2 )}/(laj^).
(45.) {Z(J + c) s 3II(& + c)}/{2a 3 3a&c}.
(46.) ]~ x i g ~* y i y1 1 ( 1  x )( x y)(y 1 )
(40.) (Zx)CZx 2 )/xyzZ(y + z)/x.
(42.) 2,bc{a + h)/{ab){ac).
(47.
1+x x + y y + 1 (l+a:)(a: + ?/)(y + iy
(y 2 ) 2 +( z a;) 2 + (.ry) 2  £ / 1 [ 1 1\
(y2)(»aj)(ajy) Vyz sa: a;y/
ras \ &c , c« , aft , (bc)(ca)(ab)
148.) 1 \ 17 rj re, ,.
x  a xb xc (xa)(xb)(xc)
(49.) 2(a+p)(a + q)f{ab)(ac)[a + h).
(50.) 2a 2 /(ab){ac)(ha). (51.) 2a 2 /(a 2 b°)(o 2 c 2 )(h 2 + a 2 ).
(52. ) Z(y 2 + 0 x 2 )/yz(x  y) (x  z).
a(bc) 3 + b(ca) 3 + c(ab) s + {b 2 c 2 )(bc) + (c 2 a 2 )(ca) + (a 2 b 2 )(ab)
a 2 (bc) + b\ca) + c 2 (ab)
{U) {{x + yf + {y + zf) {( z + x y> + (x + w) 2 }
{(z + y){z + x) + (y + z)(x + w)} 2 +{(x+~y)(x + w){y + z){z + x)} 2
(53.)
Prove the following identities : —
(55.) 2a 3 /(ab){ac) = Za.
(56.) c(ic 2 v) = au(l uv), c{v 2 u) = bv{l uv),
where tc = {abc 2 )j{bca 2 ), v=(abc 2 )/(cab 2 ).
(57. ) 2 (a + a) (a + /3) (a + y)(a{a b){ac){ad)=  aPyfabcd.
abed
. (bc) 3 + (ca) 3 + (abf K '
(59.)
(60.)
(ab  erf) (a 2  6 2 + c 2  rf 2 ) + («c  bd) (a 2 + b 2 c 2  d 2 )
(a 2  b + c  W) (a 2 + b 2  c 2  d 2 ) + 4.{ab  cd) (fflc  bd)
(b + c)(a + d)
~(b + c) 2 +(a + df
a 5 (c b) + b\a  c) + c 5 (6  a)
(c b)(ac)(ba)
(61.) {2{y8)8}/{2(y^}4n(y Z )a={2 a ?2^}».
Decompose the following into sums of partial fractions : —
(62.) 0r 2 l)/(x2)( ; «3). (63.) x 2 /(x l)(x2)(x 3).
(64.) 30a*/(a?l)(a?4). (65.) (.c 2 + 4)/(a; + l) 2 (x2)(x + 3).
VOL. I M
1G2
EXERCISES XII
CHAP. VIII
(66.) (&2)l(a?l).
(68.) {2x3)/(xl)(x 2 + iy.
(67.) (z 2 + * + l)/(a; + l)(z 2 + l).
(69.) l/{xa)(xb)(x 2 2px + q), p*<q.
(70.) (l + aj + a?)/(l »»* + a 5 ). (71.) 18/(ar* + 4;e+8).
(72.) (* + 3)/(^l). (73.) lKafi + tfrfa*).
(74.) Express (3a 2 + ic + l)/(a; 8  1) as the sum of two rational fractions
whose denominators are x i \ and x 4 + l.
(75.) Expand 1/(3 x) (2 + a?) in a series of ascending powers of x, using
partial fractions and continued division.
(76.) Expand in like manner 1/(1 x) 2 (l+x 2 ).
(77.) Show that
2 (b + c + d)/(ba)(cct)(da)(x  a) = (xabcd)J(x  a)(x  b)(x  c)(x  d).
abed
CHAPTER IX.
Further Application to the Theory of Numbers.
ON THE VARIOUS WAYS OF REPRESENTING INTEGRAL AND
FRACTIONAL NUMBERS.
§ 1.] The following general theorem lies at the root of the
theory of the representation of numbers by means of a systematic
scale of notation : —
Let r l} r.,, r 3 , . . ., r n , r n+1 , . . . denote an infinite series of
integers* restricted in no way except that each is to be greater than 1,
then any integer N may be expressed in the finite form —
N =p +p lTl + p 2 r 1 r. 2 +iW 8 r a + • • • +iW a . . . r n ,
where p <r u p x <r s , p. 2 <r 3 , . . ., p n <r n+l . When r u r. 2 , r a , . . .
are given, this can be done in one way only.
For, divide N by r lt the quotient being N, and the remainder
p ; divide N\ by r 2 , the quotient being N 2 and the remainder
p l} and so on until the last quotient, say p n , is less than the next
number in the series which falls to be taken as divisor. Then, of
course, the process stops. We now have
N =p + T$ 1 r l (j? <0 (1),
N t = Pl + N 2 r 2 (p x <r 3 ) (2),
N 3 =^ 2 + N 3 r 3 (p t <r a ) (3),
N„ _ , = p n _ ! + p n r n { p n _ !<r„) (n).
* In this chapter, unless the contrary is distinctly implied, every letter
used denotes a positive integral number.
164 FACTORIAL SERIES FOR AN INTEGER chap.
From (1), using (2), we get
N=Po + r 1 (p 1 + 'N,r 2 ),
=Po + PiTi +rjrJS r
Thence, using (3),
N =p +P{>\ + PJV 2 + W.N,,
and so on.
Thus we obtain finally
N =p +PS\ +iW g + pj\r 2 r 3 + . . . + p n r 1 r 2 . . . r n (A).
Again, the resolution is possible in one way only. For suppose
we also had
N =p ' +p 1 'r 1 + p 2 'r 1 r 2 + p 3 'r 1 r 2 r 3 + . . . +p n 'r 1 r 2 . . . r n (B),
then, equating (A) and (B), and dividing both sides by r lt we
should have
p
 + (Pi +P*r a +p a r 2 r a + . . . +p n r 3 r a . . . r n )
' i
= 7 + (ft'+ft''i+A'Vi + . • • +Pn'r a r a . . . r n ) (C).
' i
But the two brackets on the right and left of (C) contain integers,
and p fr } and p '/r l are, by hypothesis, each a proper fraction.
Hence we must have^,/^ pdj^'i ', that is,
Po*=Po,
p x +p 2 r e + p 3 r 2 r 3 + . . .4 p n r 2 r 3 . . . r n
= Pi +P*r* + p a 'r a r a + . . . + p n 'r 2 r 3 . . . r n (D).
Proceeding now with (D) as we did before with (C), we shall
prove Pi—pi'; and so on. In other words, the two expressions
(A) and (B) are identical.
Example. Let N = 719, and let the numbers r\, r%, r s , . . . be the natural
series 2, 3, 4, 5, . . . Carrying out the divisions indicated above, we have
2 )719
3 )359 ... 1
4 )119 ... 2
5)29 ... 3
5 ... 4.
ix FACTORIAL SERIES FOR A FRACTION 165
Hence Po = l, 2*1=2, #s=3, 2>s=4, Pi=5 ;
and we have 719 = 1 + 2x2 + 3x2.3 + 4x2.3.4 + 5x2.3.4.5.
§ 2.] There is a corresponding proposition for resolving a
fraction, namely, r u r.,, . . ., r n , &c, being as before,
Any proper fraction A/B can be expressed in the form
± = Pi + r±. + J!± + . . . + *>» +F
B r, i\r 2 r,r, 2 r 3 i\r 2 . . . r n
where p^r^ p. 2 <r 2 , . . ., p n <r n ; and F is either zero or can be made
as small as we please by taking a sufficient number of the integers
r u r 2 , . . ., r n . When o\, r 2 , . . ., r n , . . . are given, this resolution
can be effected in one way only.
The reader will have no difficulty in deducing this proposition
from that of last paragraph. It may also be proved thus : —
A _ Ar, _ Ar,/B
b^bt;  "^  '
Now we may put A^/B into the form p l + qJB, where g\<B.
We then have
A p x + g,/B
B~ r> '
where pi<r Xi since, by hypothesis, A<B.
Hence
 = ^ + ^ (1)
Treating the proper fraction ^,/B in the same way as we treated
A/B, we have
(.2).
B r 2 + r 2
°2
B'
where
2>*<r ai &<B
Similarly,
I 2 = il + I
B r 3 r a
ft
B'
"'"•
Pa<r a , 23<~B, &c
(3).
166 FACTORIAL SERIES FOR A FRACTION chap.
And, finally,
gni = Pn 2_ . ?»
B 7 M r„ B'
where Ihi^n, 8»<B (w).
Now, using equations (1), (2), . . ., (n) in turn, we deduce
successively
B r, r x r 2 r 1 r a B f
r, r,r 2 v/, W.B'
^' + A + _J?i_ + ■ ^
+ q — p (A),
r,r 8 . . . r n B
where p^r,, p 2 <r. 2 , . . ., p?i<r n , q n <B.
It appears therefore that F = q n \r{r % . . . r n B, which can
clearly be made as small as we please by sufficiently increasing
the number of factors in its denominator. This of course in
volves a corresponding increase in the number of the terms of
the preceding series.
It may happen, of course, that q n vanishes, and then F = 0.
We leave it as an exercise for the student to prove that this case
occurs when rfa . . . r n is a multiple of B, and that if A/B be
at its lowest terms it cannot occur otherwise. He ought also to
find little difficulty in proving that the resolution is unique when
fii r„ • • ., r n , . . . are given.
Example 1. Let A/B = 444/576, and let the numbers r\, r%, &c. , be 2, 4,
6, 8, &c.
We find 444 1 2 1_
57«~2 + 2.4 + 2.4.6'
Example 2. A/B = 11/13, r 1; r«, &c., being 2, 3, 4, 5, 6, . . ., &c.
11 1 2 1 3 3
16 22,32. 8. 42. 3. 4. 52. 3. 4. 5. 62. 3. 4. 5. 6x13
Since T\, r 2 , &c, are arbitrary, we may so choose them that the numer
ators p it ju, &c, shall each be unity. We thus have a process for decompos
ing any fraction into a sum of others with unit numerators.
IX EXAMPLES 1G7
Example 3. 11 5 1
2 x 3x 13 x
13)22(1 13)15(1 13~)l3(l
13 13 13
9 2
2x 7 x
13)18(1 13)14(1
13 13
Whence
5 1
n_i _i_ l _i i
13~2 + 2.2 + 2.2.3 + 2.2.3.7 + 2.2.3.7.13"
Here we have chosen at each step the least multiplier possible. When
this is done, it may be shown that the successive remainders diminish down
to zero, the successive multipliers increase, and the process may be brought
to an end. If this restriction on the multiplier be not attended to, the reso
lution may be varied in most cases to a considerable extent. Since, however,
we always divide by the sa me divisor B, there are only B possible remainders,
namely, 0, 1, 2, . . ., B  1 ; hence after B  1 operations at most the remainder
must recur if the operation has not terminated by the occurrence of a zero.
Example 4. Thus we have
2_1 1
3~2 + 2.3 '
. Ill li
also =o+?r 7 + ^r75+. • .+s— r„ + ;
2 2.4 2. 4 2 2.4" 2. 4". 3'
Example 5.
U + JU
29 5 5 . 5 5 . 5 . 29 '
als ° 4nV^ + ^ + CT + ^ + 5^ + 5T^ + &C  ;
, 111 1
also =  + t—z ■ + , „ „ + 
66. 3 6. 3. 36. 3. 3. 29'
and so on.
§ 3.] The most important practical case of the proposition in
§ 1 is that where r„ r 2 , . . . are all equal, say each =r. Then
we have this result —
Every integer N can be expressed, and that in one tvay only, in
the form
P*r n +PnS n ~ 1 + • • • +2hr+p ,
where _p , p iy . . ., p n are each < r.
In other words, detaching the coefficients, and agreeing that
their position shall indicate the power of r which they multiply,
and that apposition shall indicate addition (and not multiplica
tion as usual), we see that, r being any integer whatever chosen
168 SCALES OF NOTATION, INTEGERS CHAP.
as the radix of a scale of notation, any integer whatever may be
represented in the form p n p n i • • • PiPol where each of the
letters or digits p , p lf . . ., p n must have some one of the integral
values 0, 1, 2, 3, . . ., r  1.
For example, if r = 1 0, any integer may be represented b.y
PnJPni ■ • • P\Po where p ,Pi, • • ,p n have each some one of
the values 0, 1, 2, 3, 4, 5, G, 7, 8, 9.
The process of § 1 at once furnishes us with a rule for finding
successively the digits p ,P\,p 2 , . . ., namely, Divide the given integer
N by the chosen radix r, the remainder will be p ; divide the integral
quotient of last division by r, the remainder will be p t , and so on.
Usually, of course, the integer N will be given expressed in
some particular scale, say the ordinary one whose radix is 10;
and it will be required to express it in some other scale whose
radix is given. In that case the operations will be carried on in
the given scale.
The student will of course perceive that all the rules of ordi
nary decimal arithmetic are applicable to arithmetic in any scale,
the only difference being that, in the scale of 7 say, there are
only 7 digits, 0, 1, 2, 3, 4, 5, 6, and that the "carriages" go by
7's and not by 10's.
If the radix of the scale exceeds 10, new symbols must of
course be invented to represent the digits. In the scale of 12,
for example, digits must be used for 10 and 11, say t for 10
and e for 11.
Example 1. To convert 136991 (radix 10) into the scale of 12.
12)136991
12)11415
. .
. e
12)951
. .
. 3
12)79
. 3
6
. 7
The result is
6733e.
Example
2. To convert G733e (radix
12)
into the scale of r.
r)6733e
T)7el7
. .
. 1
t)9i;i
. 9
r)e4
. .
. 9
r)ll
. 6
1
. 3
The result is
136991.
IX EXAMPLES OF ARITHMETICAL OPERATION 1G9
Although this method is good practice, the student may very probably
prefer the following : —
6733e (radix 12) means
6xl2 4 + 7xl2 3 + 3xl2 a + 3xl2 + ll.
Using the process of chap, v., § 13, Example 1, we have
6+ 7 + 3+ 3+ 11
+ 72 + 948 + 11412 + 136980
6 + 79 + 951 + 11415 + 136991.
§ 4.] From one point of view the simplest scale of notation
would be that which involves the fewest digits. In this respect
the binary scale possesses great advantages, for in it every digit
is either or 1. For example, 365 expressed in this scale is
101101101. All arithmetical operations then reduce to the
addition of units. The counterbalancing disadvantage is the
enormous length of the notation when the numbers are at all
large.
"With any radix whatever we can dispense with the latter
part of the digits allowable in that scale provided we allow the
use of negative digits. For let the radix be r, then whenever,
on dividing by r, the positive remainder p is greater than r/2, we
can add unity to the quotient and take  (r  p) for a negative
remainder, where of course rp<r/2. For example, 3978362
(radix 10) might be written 4022442, where 2 stands for 2;
so that in fact 4022442 stands for 410° + O'lO 5  2"10 4  210 3
+ 410 2 410 + 2.
Example 1. Work out the product of 1698 and 314 in the binary scale.
1698 = 11010100010
314= 100111010
11010100010
11010100010
11010100010
11010100010
110101000100
10000010001010110100 ( = 533172 radix 10).
Example 2. Express 1698 and 314 in the scale of 5, using no digit greater
than 3, and work out the product of the two transformed numbers.
170 EXAMPLES OF ARITHMETICAL OPERATION chap.
5)1698 5)314
5)339 . .
5)68 . .
5)13 .
. 3
. 1
. 3
5)63 . .
5)12 . .
2 .
. 1
. 3
. 2
2 .
. 3
23313
2231
23313
131111
102111
102111
121121303 *
The student may verify that 121121303 (radix 5) = 533172 (radix 10).
Example 3. Show how to weigh a weight of 315 lbs. : first, with a series
of weights of 1 lb. , 2 lbs., 2 2 lbs., 2 3 lbs., &c, there being one of each kind ;
second, with a series of weights of 1 lb., 3 lbs., 3 2 lbs., 3 3 lbs., &c, there being
one only of each kind.
First. Express 315 in the binary scale. We have
315 = 100111011,
315 = l+2 + 2 3 + 2 4 + 2 5 + 2 8 .
Hence we must put in one of the scales of the balance the weights 1 lb., 2 lbs.,
2 3 lbs., 2 1 lbs., 2 5 lbs., and 2 8 lbs.
Second. Express 315 in the ternary scale, using no digit greater than
unity. We have
315 = 110100.
Hence over against the given weight we must put the weights 3 4 lbs. and 3 5
lbs. ; and on the same side as the given weight the weight 3 2 lbs.
§ 5.] If we specialise the proposition of § 2 by making
r, = r 2  . . . = r n , each = r say, we have the following : — Every proper
fraction A/B can be expressed, and that in one way only, in the form —
A Pi , p* , p a , , p n
+ "
B r r 2 r
4 . 3 v n >
where p u p 2 , . . ., p n are each<r, and F either is zero, or can be
made as small as we please by sufficiently increasing n.
If r be the radix of any particular scale of notation, the fraction
r r» r n
* The arrangement of the multiplication in Examples 1 and 2 is purposely
varied, because, although it is of no consequence here, sometimes the one order
is more convenient, sometimes the other. A similar variety is introduced in
§ 6, Examples 1 and 2.
IX IN VARIOUS SCALES 171
is usually called a radix fraction. We may detach the coeffi
cients and place them in apposition, just as in the case of
integers, a point being placed first to indicate fractionality. *
Thus we may write
A
where p x in the first place after the radix point stands for p x /r,
p 2 in the second place stands for p a /r*, and so' on.
Since the digits p x p. z p 3 . . . p n are the integral part of the
quotient obtained by dividing Ar n by B, the radix fraction can
not terminate unless Ar n is a multiple of B for some value of n.
Hence, if we suppose A/B reduced to its lowest terms, so that A
is prime to B, we see that the radix fraction cannot terminate
unless the prime factors of B (see chap, iii., § 10) be powers of
prime factors which occur in r. For example, since r=10 = 2x5,
no vulgar fraction can reduce to a terminating decimal fraction
unless its denominator be of the form 2 1 "5'\
In all cases, however, where the radix fraction does not
terminate, its digits must repeat in a cycle of not more than
B  1 figures ; for in the course of the division no more than B  1
different remainders can occur (if we exclude 0), and as soon as
one of the remainders recurs the figures in the quotient begin
to recur.
Example 1. To express 2/3 as a radix fraction in the scale of 10 to within
l/100000th—
2_ 200000 _ 66666+j
3~3xl0 5 ~ L0 5 '
_ 6xl0 4 + 6xl0 3 + 6xl0 2 + 6xl0 + 6 2/3
10 5 + 100000'
6 6 6 6 6^
10 10 2 10 J 10 4 10 3 '
where F =
2/3 ^ 1
1000U0 < 100000'
* Napier of Merchiston was apparently the first who used the modern form
of the notation for decimal fractions. The idea of the regular progression of
decimals is older. Stevin fully explains its advantages in his ArithmMique
(1585); and germs of the idea may be traced much farther back. According
to those best qualified to judge, Napier was the first who fully appreciated the
172 EXAMPLES OF RADIX FRACTIONS CHAP.
In other words, we have to the required degree of accuracy
~= 66666.
o
It is ohvious from the repetition of the figures that if we take n 6's after the
point we shall have the value of 2/3 correct within l/10"th of its value.
Example 2. Let the fraction be 5/64. Since 64 = 2 s this fraction ought
to be expressible as a terminating decimal. "We have in fact
5 5000000 78125
64
64 x 10 s
= •078125
10 6
5
Example 3
. To
express
2/3 as
a radix fraction
in
the
scale
of
2
to
within l/2 3 th.
2
3 :
2x2 6
~3x2 B_
128/3
2 6
42 + 2/3
2 6 *
Neglecting
2 / 3 i • i, ■
p". which is < ^ and
expressing 42 in
the scale of 2,
we
have
2
3 :
101010
2 6
= 101010 (radix 2).
§ 6.] When a fraction is given expressed as a radix fraction
in any scale, and it is required to express it as a radix fraction
in some other scale, the following process is convenient.
Let <f> be the fraction expressed in the old scale, r the new
radix, and suppose
. .Pi .iVPa ,
^ r r 2 r
then r <b = p , + +^ + . . .
= [\ + <£i say.
Now $, is a proper fraction, hence p x is the integral part of r<f>.
A • Pa
Again ?•</>! =p a + — + . . .
=jp a + & say.
So that p a is the integral part of r$ u and so on.
It is obvious that a vulgar fraction in any scale of notation
must transform into a vulgar fraction in any other ; and we shall
operational use of the decimal point ; and in his Constructio (written long
before his death, although not published till 1619) it is frequently used. See
Glaisher, Art. " Napier, " Encyclopaedia Britannka, 9th ed.; also Eae's recent
translation of the Constructio, p. 89.
ix EXAMPLES 173
show in a later chapter (see Geometrical Progression) that every
repeating radix fraction can be represented by a vulgar fraction.
Hence it is clear that every fraction which is a terminating or a
repeating radix fraction in any scale can be represented in any
other scale by a radix fraction which either terminates or else
repeats. It is not, however, true that a terminating radix fraction
always transforms into a terminating radix fraction or a repeater
into a repeater. Nonterminating nonrejieating radix fractions
transform, of course, into nonterminating nonrepeating radix
fractions, otherwise we should have the absurdity that a vulgar
fraction can be transformed into a nonterminating nonrepeating
radix fraction.
It is obvious that all the rules for operating with decimal
fractions apply to radix fractions generally.
Example 1. Reduce 3*168 and 11 "346 to the scale of 7, and multiply the
latter by the former in that scale ; the work to be accurate to l/1000th
throughout.
The required degree of accuracy involves the 5th place after the radical
point in the scale of 7.
•346
7
•168
_7
1)176
_7
1)232
_7
l)624
_7
4)368
_7
2V576
2)422
_7
2) 954
_7
6)678
_7
4)746
_7
5) 222
3168 = 311142 (radix 7). 11346 = 1422645.
1422645
311142
4601601
142265
14227
1423
632
32
5064146
174 REMAINDER ON DIVIDING BYf1 chap.
On account of the duodecimal division of the English foot into 12 inches,
the duodecimal scale is sometimes convenient in mensuration.
Example 2. Find the number of square feet and inches in a rectangular
carpet, whose dimensions are 21' 3" by 13' llf". Expressing these lengths
in feet and duodecimals of a foot, we have
21' 3" = 1936.
13' ll"=ll'e9.
If, following Oughtred's arrangement, we reverse the multiplier, and put the
unit figure under the last decimal place which is to be regarded, the
calculation runs thus —
1936
9ell
19360
1936
1763
13e
20978
209 (radix 12) = 288 + 9 = 297 (radix 10) feet.
•78 (radix 12) = 7 x 12 + 8 = 92 square inches.
Hence the area is 297 feet 92 inches.
§ 7.] If a number N be expressed in the scale of r, and if we
divide N and the sum of its digits by r  1 , or by any factor of r— 1 ,
the remainder is the same in both cases.
Let N = p + p t r + p 2 r* + . . . + p n r n .
Hence N  (p +i> x + • •  + Pn) = Pi{r  I) + p. 2 (r 2  1) + . . .
+ Pn(r n ~l) (1).
Now, m being an integer, r m  1 is divisible by r — 1 (see
chap, v., § 17). Hence every term on the right is divisible by
r  1, and therefore by any factor of r  1. Hence, p being r — 1,
or any factor of it, and /x some integer, we have
N0'o +;>, + . • +pn) = W (2).
Suppose now that the remainder, when N is divided by p, is <r,
so that N = vp + <t. Then (2) gives
p + p, + . . . + p n = (v  fl)p + o (3),
which shows that when p +Pi + • • . +p n is divided by p the
remainder is <r.
IX CASTING OUT THE NINES 175
Cor 1. In the ordinary scale, if we divide any integer by 9 or by
3, the remainder is the same as the remainder we obtain by dividing
the sum of its digits by 9 or by 3.
For example, 31692^9 gives for remainder 3, and so does
(3 + 1 + G + 9 + 2) + 9.
Cor. 2. It also follows that the sum of the digits of every midtiple
of 9 or 3 must be a multiple of 9 or 3. For example,
2x918 1 + 8 = 9 ■
13 x 9117 1 + 1 + 7 = 9
128x9 = 1152 1 + 1+5 + 2 = 9
128x3 = 384 3 + 8 + 4 = 15 = 5x3.
§ 8.] On Cor. 1 of § 7 is founded the wellknown method
of checking arithmetical calculations called " casting out the
nines."
Let L = MN ; then, if L = 19 + L', M = m9 + M', N = nd + N',
so that L', M', N' are the remainders when L, M, N are divided
by 9, we have —
19 + L' = (m9 + M') (nd + N'),
= mra81 + (M'w + N'm)9 + M'N',
= (mn9 + Wn + N'm)9 + M'N' ;
whence it appears that L' and M'N' must have the same re
mainder when divided by 9. L', M', N' are obtained in accord
ance with Cor. 1 of § 7 by dividing the sums of the digits in the
respective numbers by 9.
Example 1. Suppose we wish to test the multiplication
47923x568 = 27220264.
To get the remainder when 47923 is divided by 9, proceed thus: 4 + 7 = 11,
cast out 9 and 2 is left ; 2 + 9 = 11, cast out 9; 2 + 2 + 3 = 7. The remainder
is 7. Similarly from 563 the remainder is 1, and from 27220264, 7. Now
7x1 + 9 gives of course the same remainder as 7+^9. There is therefore a
strong presumption that the above multiplication is correct. It should be
observed, however, that there are errors which this test would not detect ; if
we replaced the product by 27319624, for instance, the test would still be
satisfied, but the result would be wrong.
In applying this test to division, say to the case L/M = N + P/M, since
we have L = MN + P, and therefore L P = MN, we have to cast out the nines
from L, P, M, and N, and so obtain L', P', M', and N' say. Then the test is
that L'  P' shall be the same as the result of casting out the nines from M'N'.
176 LAMBERT'S THEOREM CHAF.
Example 2. Let us test —
27220662+568 = 47923 + 398 + 568,
or 27220662 = 47923x568 + 398.
Here L'  P' = 0 2= 2,
M'.N'=7x 1=92.
The test is therefore satisfied.
§ 9.]* The following is another interesting method for ex
panding any proper fraction A/B in a series of fractions with
unit numerators : —
Let (?!, q,, q a , . . ., q n , and r x , r 2 , ? 3 , . . ., r n , be the quotients and
remainders respectively when B is divided by A, r lt r 2 , . . ., r n _, re
spectively, then
 = + . . . + v i — + F (1),
B ft q,q 2 q,q^ 3 q,q 2 ...q n
where F = (  1 ) n i " n JMi • • • o n ~B, that is, F is numerically less than
1/qfa . . . q n .
For we have by hypothesis
B = Aq x + r u therefore A/B = l/g 1  rjqfi (2),
B = i\q 2 + r s , therefore rjB = l/q 2  r 2 /q 2 B (3),
B = r 2 q 3 + r 3 , therefore r„/B = l/q 3  r 3 Jq 3 ~B (4),
and so on.
From (2), (3), (4), we have successively
A.l J_ J_AY\
B ~?i Mi 2&W'
= ±± + l LfcV
2i Mi MA Mzl^W'
and so on.
Since r n r 2 , . . ., r n go on diminishing, it is obvious that, if
A and B be integers as above supposed, the process of successive
division must come to a stop, the last remainder being 0. Hence
* In Lis Essai d' Analyse Numirique sur la Transformation dcs Fractions
(CEuvres, t. vii. p. 313), on which the present chapter is founded, Lagrange
attributes the theorem of § 9 to Lambert (17281777). Heis, Sammlung von
Bcispielen und Au/gaber. aits dcr allgemeinen Arithmetik und Algebra (1882), p.
322, has applied series of this character to express incommensurable numbers
such as logarithms, square roots, &c. In the same connection see also Syl
vester, American Jour. Math., 1880. Sec also Cyp. Stephanos, Bull. Soc.
Math. Fr. 7 (1879), p. 81 ; G. Cantor, Zeitsch. f. Math. 14 (1869), p. 124 ;
J. Liiroth, Math. Ann. 21 (1883), p. 411.
IX
EXERCISES XIII 177
every vulgar fraction can be converted into a terminating series
of the form
1 JL JL
Example.
113 1 1 1
: +
244 2 2.13 2.13.24 2.13.24.61
From this resolution we conclude that 1/21/2.13 represents 113/244 within
l/26th, and that 1/21/2.13 + 1/2.13.24 represents 113/244 within l/624th.
Exercises XIII.
(1.) Express 16935 (scale of 10) in the scale of 7.
(2.) Express 16935 (scale of 10) in the scale of 7.
(3.) Express 315 "34 (scale of 10) in the scale of 11.
(4.) Express r7e9ee (scale of 12) in the scale of 10.
(5.) Express Ir8e54 (scale of 12) in the scale of 9.
(6.) Express 345"361 (scale of 7) in the scale of 3.
(7.) Express 112/315 (scale of 10) as a radix fraction in the scale of 6.
(8.) Express 3169 in the form ^ + g3 + r3.5+s3.5.7 + &c, where ]i<3,
2<5, r<7, &c.
(9.) Express 7/11 in the form pl2 + q/2.S + r/2.ZA + kc., where p<2,
q<3, r<4, &c.
(10.) Express 113/304 in the form ^3 + ?/3.5 + r/3 2 .5 + s/3 2 .5 2 M/3 3 .5 2 f &r.,
where p< 3, q<5, r<3, &c.
(11.) Multiply 31263 by 56341 in the scale of 7.
(12.) Find correct to 4 places 31 3432 x 150323, both numbers being in
the scale of 6.
(13.) Find to 5 places 31 343272 67312, both numbers being in the scale
of 12.
(14.) Extract the square root of 365738 (scale of 9) to 3 places.
(15.) Express 887/1103 in the form l/qi  l/qiq» + l/qiq«q3  &c
(16.) Show how to make up a weight of 35 lbs. by taking single weights
of the series 1 lb., 2 lbs., 2 2 lbs., &c.
(17.) With a set of weights of 1 lb., 5 lbs., 5 2 lbs., &c, how can 7 cwt. be
weighed ? First, by putting weights in one scale only and using any number
of equal weights not exceeding four. Second, by putting weights in either
scale but not using more than two equal weights.
(18.) Find the area of a rectangle 35 ft. 3* in. by 23 ft. 6 in.
(19.) Find the area of a square whose side is 17 ft. 4 in.
(20.) Find the volume of a cube whose edge is 3 ft. 9} in.
(21.) Find the side of a square whose area is 139 sq. ft. 130 sq. in.
(22.) Expressed in a certain scale of notation, 79 (scale of 10) becomes 142 ;
find the radix of that scale.
VOL. I N
178 EXERCISES XIII
CHAP.
(23.) In what scale of notation does 301 represent a square integer ?
(24.) A number of 3 digits in the scale of 7 lias its digits reversed when
expressed in the scale of 9 ; find the digits.
(25.) If 1 be added to the product of four consecutive integers the result
is always a square integer ; and in four cases out of five the last digit (in the
common scale) is 1, and in the remaining case 5.
(26.) Any integer of four digits in the scale of 10 is divisible by 7, pro
vided its first and last digits be equal, and the hundreds digit twice the. tens
digit.
(27. ) If any integer be expressed in the scale of r, the difference between
the sums of the integers in the odd and even places respectively gives the
same remainder when divided by r + 1 as does the integer itself when so
divided. Deduce a test of multiplication by "casting out the elevens."
(28.) The difference of any two integers which are expressed in the scale
of 10 by the same digits differently arranged is always divisible by 9.
(29.) If a number expressed in the ordinary scale consist of an even
number of digits so arranged that those equidistant from the beginning and
end are equal, it is divisible by 11.
(30.) Two integers expressed in the ordinary scale are such that one has
zeros in all the odd places, the other zeros in all the even places, the remaining
digits being the same in both, but not necessarily arranged in the same order.
Show that the sum of the two integers is divisible by 11.
(31.) The rule for identifying leap year is that the number formed by the
two last digits of the year must be divisible by 4. Show that this is a
general criterion for divisibility by 4, and state the corresponding criterion
for divisibility by 2".
(32.) If the last three digits of an integer be^o^'o, show that the integer
will be exactly divisible by 8, provided p$ + 2_£>i + 4p2 be exactly divisible by 8.
(33.) Show that the sum of all the numbers which can be formed with the
digits 3, 4, 5 is divisible by the sum of these digits, and generalise the theorem.
(34.) Itp/n and (nj^/n, p<n, be converted into circulating decimals, find
the relation between the figures in their periods.
(35.) If, in converting the proper fraction ajb into a decimal, a remainder
equal to ba occurs, show that half the circulating period has been found,
and that the rest of it will be found by subtracting in order from 9 the digits
already found. Generalise this theorem.
(36.) In the scale of 11 every integer which is a perfect 5th power ends in
one or other of the three digits 0, 1, t.
(37.) In the scale cf 10 the dilference between the square of every number
of two digits and the square of the number formed by reversing the digits, is
divisible by 99.
(38.) A number of six digits whose 1st and 4th, 2nd and 5th, 3rd and 6th
digits are respectively the same is divisible by 7, by 11, and by 13.
(39.) Show that the units digit of every integral cube is either the same
as that of the cube root or else is the complementary digit. (By the comple
mentary digit to 3 is meant 10  3, that is, 7.)
(40.) If in the scale of 12 a square integer (not a multiple of 12) ends
ix EXERCISES XIII 179
with 0, the preceding digit is 3, and the cube of the square root ends with
60.
(41.) If a be such that a m + a — r, then any number is divisible by a m ,
provided the first m integers po, pi, ■ ■ ■ , p m i of its expression in the scale of
r are such that^ +2'i^+ • • • +Pmi^ m ~ 1 is divisible by a" 1 .
(42.) The digits of a are added, the digits of this sum added, and so on,
till a single digit is arrived at. This last is denoted by <p(u). Show that
<p(a + b) = <f> {</){a) + </>{b)} ; and that the values of <p(8n) for ft = 1, 2, . . ., a,
successively consist of the nine digits continually repeated in descending
order.
(43.) A number of 3 digits is doubled by reversing its digits : show that
the same holds for the number formed by the first and last digit, and that
such a number can be found in only one scale out of three.
CHAPTEK X.
Irrational Functions.
GENERALISATION OF THE CONCEPTION OF AN INDEX.
INTERPRETATION OF z\ X^l, X~ m .
§ 1.] The definition of an index given in chap, ii., § 1, be
comes meaningless if the index be other than a positive integer.
In accordance with the generalising spirit of algebra we
agree, however, that the use of indices shall not be restricted to
this particular case. We agree, in fact, that no restriction is to
be put upon the value of the index, and lay down merely that
the use of the indices shall in every case be subject to the laws
already derived for positive integral indices. Less than this we
cannot do, since these laws were derived from the fundamental
laws of algebra themselves, to which every algebraical symbol
must be subject.
The question now arises, What signification shall we attri
bute to x m in these new cases ? We are not at liberty to proceed
arbitrarily, and give any meaning we please, for we have already
by implication defined x m , inasmuch as it has been made subject
to the general laws laid down for indices.
§ 2 ] Case of x p! ? where p and q are any positive integers. Let
z denote the value of x p to, whatever it may be ; then, since x^i is
to be subject to the first law of indices, we must have —
zV = zxzxzx . . . a factors,
= xvli x xrti x xPlv x . . . q factors,
 3;P/?+P/?+.P/9+ • • • 1 terms,
= XP.
chap, x INTERPRET ATION OF X p,q 181
Iii other words, z is such that its qt\\ power is x* } that is, z
is what is called a qth root of xP, which is usually denoted
by tfxP.
Hence x.p'9 = *J&.
In particular, if p  1,
We have now to consider how far an algebraical value of
a 5th root of every algebraical quantity can be found.
In the case of a real positive quantity k, since zi passes con
tinuously* through all positive values between and + x> as z
passes through all positive values between and + 00 , it is clear
that, for some value of z between and + 00 , we must have
z?  k. In other words, there exists a real positive value of %Jk.
Unless the contrary is stated we shall, when k is positive,
take k 1! i as standing for this real positive value.
The student should, however, remark that when q is even,
= 2r say, there is at least one other real value of 0/k ; for, since
(  z) 2r = z 2r , if we have found a positive value of z such that
z 2r = k, that value with its sign changed will also satisfy the re
quirements of the problem.
Next let k be a negative quantity. If q be odd, then, since
z? passes through all values from  00 to as z passes through
all values from  00 to 0, there must be some one real negative
value of z, such that & = k. In other words, if q be odd, there
is a real negative value of {/k.
If q be even, then, since every even power of a real quantity
(no matter Avhether + or  ) is positive, there is no real value
of z. Hence, if k be negative and q even, %/k is imaginary. This
case must be left for future discussion.
It will be useful, however, for the student to know that
ultimately it will be proved that */k has in every case q different
values, expressions for which, in the form of complex numbers,
can be found. Of these values one, or at most two, may be real,
as indicated above (see chap, xii.)
* For a fuller discussion of the point here involved see chaps, xv. and xxv.
182 VERIFICATION OF THE LAWS FOR X rlq chap.
Only iii the case where h is the pth power of a rational
quantity can %/k be rational.
Example.
ltk=+h&>,
2 £>/k has two real values, +h and h.
If k=+h*P+\
ip t}/k lias one real value, +h.
HJc= J&+ 1 ,
v+Vk has one real value,  h.
In all that follows in this chapter, we shall restrict the radicand,
I; to be positive ; we shall regard only the real positive value of the qth
root of k ; and this (ivhich is called the PRINCIPAL value of the
root) is what ice understand to be the meaning of £ 1/ ?
The theory of fractional indices could (as in the first edition of this volume)
he extended so as to cover the case of a negative radicand, hut only so far as the
order of the root is odd. The practical advantage gained by this extension is
not worth the trouble which it causes by complicating the demonstrations.
We think it better also, from a scientific and educational, as well as from a
practical point of view, to consider the radication of negative radicands as a
particular case of the radication of complex radicands (see chap, xii., § 19).
§ 3.] We have now to show that the meaning just suggested
for x p l q is consistent with all the Laws of Indices laid down in
chap. ii. The simplest way of doing this is to reprove these
laws for the newly denned symbol x^i.
We remark in the first place that it is necessary to prove
only I. (a), II, and III. (a) ; because, as has been shown in chap.
ii., I. (/8) can be deduced from I. (a), and III. (J3) from III. (a),
without any appeal to the definition of x m .
To prove I. (a), consider x pl ? and x r/s , where p, q, r, s are
positive integers, and let
z = xPb x r ' s .
Then, since x? 1 * and x r ' s are, by hypothesis, each real and
positive, z is also real and positive. Also
z<i» = (xPlq x rl *)i s ,
X VERIFICATION OF THE LAWS FOR X p,q 183
all by the laws for positive integral indices, regarding which
there is no question.
Now, by the meanings assigned to aP^ and x rls , we have
(a^/«)9 = %* and (x rls ) a = x r . Hence
= xP s xP,
= K*"+« r )
hy the laws for positive integral indices.
It now follows that is the qsth root of a^ JS +?'' ; and, since z
is real and positive, it must be that qsth root which we denote
by ofp*+9r)lv. Therefore
z = z(2«+ <?'•)/</■',
that is to say,
z = ?pl<i+ r l s .
The proof is easily extended to any number of factors.
To prove Law II., consider (x^) r ' s , where p, q, r, s are posi
tive integers,
and let z = (x^) r ' s .
Then, since, by hypothesis, x*ti is real and positive, therefore
(xPtey 1 *, that is z, is real and positive. Also
z v = [(xPlqyisy^
= [{(ajP/s)*"/*} 8 ]?,
by laws for positive integral indices ;
= [(xPte) r ]i,
by definition of a fractional index ;
= (xvl*)9 r ,
by laws of positive integral indices ;
by definition of a fractional index ;
= xv,
by laws of positive integral indices.
Hence z is a qsth. root of »*"", and, since z is real and positive, we
must have
Z = oVrlqs,
that is, z = a^/9)( r /»).
184 PARADOXES chap.
Lastly, to prove Law III. (a), let
Then, since, by hypothesis, x^i and yvli are each real and
positive, z is real and positive. Also
z q = ( x plq yplqy^
by laws for positive integral indices ;
= xPf,
by definition for a fractional index ;
by laws for positive integral indices.
Hence z is a qi\\ root of (%y) p ; and, since z is real and positive,
we must have
a = (.vy)pli.
The proof is obviously applicable where there is any number of
factors, x, y, . . .
§ 4.] Although it is not logically necessary to give separate
proofs of Laws I. (/3) and III. (/?), the reader should as an
exercise construct independent proofs of these laws for himself.
It should be noticed that in last paragraph we have supposed
both the indices pjq and r/s to be fractions. The case where
either is an integer is met by supposing either q — 1 or s = 1 ; the
only effect on the above demonstrations is to simplify some of
the steps.
§ 5.] Before passing on to another case it may be well to
call attention to paradoxes that arise if the strict limitation as to
sign of xPfo be departed from.
By the interpretation of a fractional index
x*l 2 = Z/x* = ± x\
But x^ = x\
which is right if we take x 4/2 to stand for the positive value of
A 2 /V ; but leads to the paradox x* =  x~ if we admit the negative
value.
A similar difficulty would arise in the application of the law,
(x m ) n = x mn = (.'■")"' ;
x INTERPRETATION OF X° 185
for example, (4*) 9 = (4*)'
would lead to ( ± 2f = ± 4,
that is, 4 = ± 4,
if both values were admitted. Such difficulties are always apt
to arise 'with x^i where the fraction pfq is not at its lowest terms.
The true way out of all such difficulties is to define and
discuss x 11 as a continuously varying function of n, which is called
the exponential function. In the meantime fractional indices are
introduced merely as a convenient notation in dealing with
quantities which are (either in form or in essence) irrational ;
and for such purposes the limited view we have given will be
sufficient.
§ 6.] Case of x°. This case arises naturally as the extreme
case of Law I. (/3), when n = m; for, if we are to maintain that
law intact, we must have, provided x 4= 0,*
that is, X° = 1.
This interpretation is clearly consistent with Law I. (a), for
x m n° = x m +°
simply means
x m x 1 = x m ,
which is true, whatever the interpretation of x m may be.
Again, a;'" = (x ,rt )°,
that is x° = (x m )°,
simply means 1 = 1 by our interpretation ;
and x m0 = (x ) m ,
or x° = (x°) m ,
gives 1 = 1™
which is right, even if m be a positive fraction, provided we
adopt the properly restricted interpretation of a fractional index
given above. The interpretation is therefore consistent with II.
The interpretation a; = 1 is also consistent with III. (a), for
a?>f = (xyf
simply means 1x1 = 1,
* This provision is important since the form 0° is indeterminate (see chap,
xxv.)
186 NEGATIVE INDICES chap.
§ 7.] Case of x~ m , where m is any real positive (or signless)
number, and x 4= 0.
Let z = x~ m , then, since x m ^0, Ave have
z = x~ m x Z m 7X m ,
if Law I. (u) is to hold for negative indices. Whence
,0 /•>»>'
z = ar/x
/a*"»
by last paragraph. In other words, x~ m is the reciprocal of x m .
As an example of the reconciliation of this with the other
laws, let us prove I. (a), say that
By
our
definition,
X
we
~ m x
have
 n = (\fo: m )(ljx n ),
= l/x m x n ,
s= l/x m + n ,
the last step by
the laws
already
established for all positive
indices ;
by definition of a
negative index.
Hei
ice
X'
■m x
n _ r*  vi  n
In like manner we could show that
rwm v  n _ g.m  n
The verification of the other laws may be left as an exercise.
§ 8.] The student should render himself familiar with the
expression of the results of the laws of indices in the equivalent
forms with radicals ; and should also, as an exercise, work out
demonstrations of these results without using fractional indices
at all.
For example, he should prove directly that
Vx*/z= P !/xP+* (1);
V{ V xP Y = V xPr = Vi V&Y ( 2 ) i
yx&yyz=y(xyz) (3);
y x mj y y m = "/(^jy)™ (4).
EXAMPLES 187
EXAMPLES OF OPERATION WITH IRRATIONAL FORMS.
§ 9.] Beyond the interpretations x pl ^, x°, z~ m , the student has
nothing new to learn, so far as mere manipulation is concerned,
regarding fractional indices and irrational expressions in general.
Still some practice will he found necessary to acquire the requisite
facility. "We therefore work out a few examples of the more
commonly occurring transformations. In some cases we quote
at each step the laws of algebra which are appealed to ; in others
we leave it as an exercise for the student to supply the omission
of such references.
Example 1.
To express A v B in the form V P.
A v / B = AB 1 /™ = (A m ) , /"'B 1/m , by law of indices II.,
= (A"'B) 1; »', by law of indices III. (a),
= X /(A"'B).
Example 2.
'v/a=7a
for v^A = A 1/m = AP /m P,
= m VAP.
Example 3.
sJxP m +<l = x(P m +tf'' m
= x p+q!m i
= xPx xn' m , by law of indices I. (a),
m /
Example 4.
To express \/x»\ y/y as the root of a rational function of x and y.
s/xp\ \/y r = xP ! v/y r !» = xP^v/yW'Q',
= {xP'^lvftyvryiV,
= {xP'/yiry/q^
= V{xP'ly r )
Example 5.
V32=V(16x2),
= Vl6xV2,
= 4x V2.
188
EXAMPLES
Example 6.
2 x V2 x v"2 x \A
= 2x2"x2 } x2',
= 22,
= 4.
CHAP
Example 7.
= (« 2 )'"/ 2 »(la2/a2)m/2» )
= a mln (lx 2 /a 2 ) m ' 2n .
Example 8.
\/{yx + x") x ^(yz + ex)
= V Hy + *)} x V {2(2/ + a)} ,
= VZ X V(y + *) X \/z X \/(2/ +^);
= V(^)x{V(2/ + ^)} 2 ,
=(y+«)x V(a*)
Example 9.
V240 + V40
= V(16x3x5) + V(4x2x5),
= V16V3V5+ VW2V5,
= V5(4V3 + 2V2).
Example 10.
(V3 + 2V2 + 3V6)(V32 N /2 + 3V6)
= (V3 + 3V6) 2 (2V2) 2 ,
= (V3) 2 + 6V3V6 + (3V6) 2 (2V2) 2 .
= 3 + 6V(3x6) + 3 2 x62 2 x2,
= 49 + 6\/18,
= 49 + 18V2.
Example 11.
{V(la) + V(l+.T)} 4
= {(lx) i + (l+x) i }*,
= (lxf + {l+ x ) n 
+ i(lx)*(l+x) i +4(lx) i (l+x)i
+ 6(lx)(l+x),
= 8  4x 2 + 4(1 x) h (l + x)\l  x+l+x)
= 84.t 2 +8V(ltf 2 ).
Example 12.
V \xy) V \x+y)
_W(x + y)\ 2 +{s/(xy)} 2
\/{{xy)(x + y)}
RATIONALISING FACTORS 189
_x+y+x~y
2a;
Example 13.
V(* 2 ~2/ 2 )'
(a^a^+a: * a;  *) x (;*:*+ 1+a: *)
=x*7? +a:*aT*
+ ar  a; 1 + x *  x *
Tit "tC V X «C y
5 1 _ 1 _ 5
sx'+aTa; a; *
Example 14.
Show that
We have
g2 = a+s/Wb) + ayW5) + 2 /f {a + V(a 2  6)} (a  V(a 2  5)} ~
= a + V[« 2 {V(« 2 &)}''],
= re + \/6.
Hence, extracting the square root, we have
S=V(a+V & )
RATIONALISING FACTORS.
§ 10.] Given certain irrationals, say Jp, *Jq, y/r, we may
consider rational, and it may be also integral, functions of these.
For example, I *Jp + m \/q + n yV, and /( *Jp) s + m \/{pq) + n{ s/qf y
are integral functions of Jp, *Jq, Jr, of the 1st and 2nd
degrees respectively, provided /, m, n do not contain Jp, Jq, Jr.
Again, (I Jp + m Jq)/(l \/g + m Jr) is a rational, but not integral,
function of these irrationals. J (I \Jp + in \/q), on the other
hand, is an irrational function of Jp and Jq.
The same ideas may also be applied to higher irrationals,
such us p llm , q lln , &c.
§ 11.] Confining onrselves for the present to quadratic
irrationals, we shall show that every rational function of a
given set of quadratic irrationals, Jp, Jq, ^/r, &c, can be
190 RATIONALISING FACTORS chap.
reduced to a linear integral function of the square roots of p, q, r,
and of their products, pq, pr, gr, pgr, &c.
This reduction is effected mainly by means of rationalising
factors, whose nature and use we proceed to explain.
If P be any integral function of certain given irrationals, and Q
another integral function of the same, such that the product QP is
rational so far as the given irrationals are concerned, then Q is called
a rationalising factor ofP with respect to the given irrationals.
It is, of course, obvious that, if one rationalising factor, Q,
has been obtained, we may obtain as many others as we please
by multiplying Q by any rational factor.
§ 12.] Case of Monomials.
1°. Suppose we have only quadratic irrational forms to deal
with, say two such, namely, pi and qK
Then the most general monomial integral function of these
is
i = A(pi) 2m +\r) 2n+ \
where A is rational. There is no need to consider even indices,
since (jpi) Zm =p m is rational.
Now I reduces to
I = (Ap'YOiM
where the part within brackets is rational.
Hence a rationalising factor is jj'qK for we have
Ipigi = (Aj> m q n )pq,
which is rational.
Example. A rationalising factor of 16 . 2 ? . 3* . 5* is 2*35', that is, (30) .
2°. Suppose we have the irrationals p v *, q l!t , r^ u , say, and
consider
I = ApV s q m ^ ?'"/" *
which is the most general monomial integral function of these.
A rationalising factor clearly is
„ .1  l/s f ,\  in It r l  n/u
p q ' i
or 4*  > (ft  "0. 't >•(«  »)/«
* Where of course l<s, m<t, «<«, for if they were not they could be
reduced by a preliminary process like that in case 1°.
x RATIONALISING FACTORS 191
Example.
1=81. 8*. 6*. 7*,
=31.3 1+ *.5*.7 i+i ,
=(31.3.7 4 ).3*.5 f .7*.
5 2 1
A rationalising factor is 3 . 5 . 7 .
§ 13.] Case of Binomials.
1°. The most general form when only quadratic irrationals
are concerned is a \/p + b \/q, where a and b are rational ; for, if
we suppose p a complete square, this reduces to the more special
form A + B \Aj, where A and B are rational.
A rationalising factor clearly is a s/p  b \/q. For, if
I = a */p + b x /q,
I(a \ >  b y/q) = (a ^p) 2  (b s /q)\
= ap  b'q,
which is rational
The two forms a \/p + b \/q and a \/p  b *Jq are said to be con
jugate to each other with reference to y/q, and we see that any binomial
integral function of quadratic irrationals is rationalised by multiplying
it by its conjugate.
2°. Let us consider the forms ap*l* ± bqW, to which binomial
integral functions of given irrationals can always be reduced.* Let
x = ap aly , y = bqffl,
I = ap*b  hf r \
= «  y
Let m be the L.C.M. of the two integers, y, 8. Now, using
the formula established in chap, iv., § 16, we have
(z m  1 + x m ~ 2 y+. . . + xy m ~ 2 + y m " 1V I = x m  y m .
Here x m  y>" = (a'" p mx ^  b"< tf&P), where ma/y and m/3/8 are
integers, since m is divisible by both y and S, that is, x™  y'" is
rational.
A rationalising factor is therefore x 1 " ~ 1 + x m ~ y + . . .
+ xy m ~ 2 + y m ~ l , in which x is to be replaced by ap , and y by b<f'
* Tartaglia's problem. See Cossali Storia dell' Abjebra (1797), vol. ii. p. 266.
192 RATIONALISING FACTORS chap.
The form ap" ly + bq m may be treated in like manner by
means of formulas (4) or (5) of chap, iv., § 16.
Example.
1 = 3.2* 4.3*.
Here m = 6, z=3.2*, 27 = 4.3*;
and a rationalising factor is
a 5 + xHj + x*if + xhf + xy i + rf
= 3 5 .2 S + 3 l .4.2 l .3* + 3 3 .4 2 .2.3* + 3 2 .4 3 .2 § .3* + 3.4 4 .2*.3 § + 4 s .3 i ,
= 3 5 .2.2 3 + 3 4 .8.2*.3* + 3 3 .32.3* + 3 2 .4 3 .2 S .3* + 3.4 4 .2*.3 § + 4 5 .3 f .
§ 14.] Trinomials with Quadratic Irrationals. This case is
somewhat more complicated. Let
I = \/p + \/q+ Jr ; *
and let us first attempt to get rid of the irrational Jr. This
Ave can do by multiplying by the conjugate of \/p + sfq + s/r
with respect to *Jr, namely, sip + \'q  sir. We then have
( Jp + Vg  v/r)I = ( Jp + JqY  ( vV) 2 ,
= p + qr+2j(pq) (1).
To get rid of nj(pq) we must multiply by the conjugate of
p + q  r + 2 s/(p<j) with respect to \/(pq). Thus finally
{^ + 2r2 N \pq)}{ s/p + Jq ~ Jr)l = (p + q  r) 2  {2 v /( pq)}\
=f + ? 2 + r 2  2j?2  2p  2jr.
Hence a rationalising factor of I is
{p + qr 2 \/(j>g)}( «/p + slq  Jr),
or
(Jp  Jq + */r)(vi>  s'q Jr)(Jp + v'2  v'r) (2).
Ey considering attentively the factor (2) the student will see
that its constituent factors arc obtained by taking every possible
arrangement of the signs + and  in
+ *Jp ± \fq± \ V,
except the arrangement + + + , which occurs in the given trinomial.
* This is really the most general form, for a\/]) + b\Jq + c\/r may be
written V(^) + V(& 8 2) + vW
x REDUCTION TO LINEAR FORM 193
Example 1. A rationalising factor of
\/2  V3 + V5
is (V2  \ft  V») (V2 + V3 + V5) ( v'2 + \/3 yf5).
Example 2. A rationalising factor of
1+2V33V2
is (1 + 2V3+3V2)(12V3 + 3V2)(12V33V2).
In actual practice it is often more convenient to work out
the rationalisation by successive steps, instead of using at once
the factor as given by the rule. But the rule is important,
because it is general, and will furnish a rationalising factor for a
sum of any number of quadratic irrationals.
Example 3. A rationalising factor of
1 + V2V3 + V4
is (1 + V2 V3 \/4)(l+V2+V3 + V4)(l+ V2 + V3 V*)
x (1  V2  \/3 + \/4) (1  V2  V3  V4) (1  V2+ V3 + V4)
x (1  sJ2 + V3  V4).
Before giving a formal proof of the general truth of this
rule, it will be convenient to enunciate one or two general pro
positions which are of considerable importance, both for future
application and for making clear the general character of the
operations which we are now discussing.
§ 15.] Ever)/ integral function of a series of square roots,
\/p, Jq, \/r, &c, can be expressed as the sum of a rational term
and rational multiples of \tp, \/q, \/r, &c, and of their products
*J(P<l), J(pr)> v^Fi''). & i*
First, let there be only one square root, say \/p, and consider
any rational integral function of \jp, say <£( \''p). Every term
of even degree in *Jp will be rational, and every term of odd
degree, such as A( v//>) 2w+1 may be reduced to (Kp m ) \/p, that
is, will be a rational multiple of s/p. Hence, collecting all the
even terms together, and all the odd terms together, we have
</>(^) = P + Qv> (l),
where P and Q are rational.
* Such a sum is called a " Linear Form."
VOL. I
194 RATIONALISATION OF ANY INTEGRAL FUNCTION chap.
Next, suppose the function to contain two square roots, say
4>( JPi *Jq)' First of all, proceeding as before, and attending
to Jp alone, we get
«/>( <Sp, \ /( L) = P + Qv / P,
where P and Q are rational so far as p is concerned, but are
irrational as regards q, being each rational integral functions of
\/q. Reducing now each of these with reference to Jq we shall
obtain, as in (1),
P = F + QV2, Q = P" + Q"vA;,
and, finally,
<K Jp, Jq) = F + QVg + (P" + Q" Jq) y/p,
= F + P" s/p + Q' Jq + Q" J(pq) (2),
which proves the proposition for two irrationals.
If there be three, we have now to treat P', P", Q', Q" by means
of (1), and we shall evidently thereby arrive at the form
A+B Jp + C Jq+T> vV+E^r) +F J(rp)+G J(pq) + li y/{pgr),
and so on.
Cor. It follows at once from the process by which we arrived
at (1) that
4>( Jp) = P  Q yip.
Hence if <f>( Jp) he any integral function of sip, <£(  Jp) is a
rationalising factor of <£( \/p) ; and, more generally, if <f>( sjp, Jq,
s/r, . . .) be an integral function of \/p, Jq, Jr, . . ., then, if we
take any one of them, say Jq, and change its sign, the product
</>( Jp> Jq> J r > • • •) x <K J Pi ~ J°> J r > • • •) ?s rational, so far
as Jq is concerned.
Example 1. If <p(.r)=:x 3 + .r 2 + x+l, find the values of 0(l + \/3) and
0(1  V3) and 0(1 + \J3) x 0(1  \JZ).
0(1 + V3) = (l + V3) 3 + (l + V3) 2 + (1 + V3)+ 1,
= l + 3\/3 + 3.3 + 3\/3
+1+2V3+3
+ 1 + V3
+ 1,
= 16 + 9V3
0(1  \/3) is deduced by writing  \JZ in place of + \/3 everywhere in the
above calculation. Hence
0(1V3) = 169V3;
0(1 + V3) x 0(1  V3) = (16) 2  (9 V3) 2 ,
= 256243,
= 13.
X LINEAR FUNCTION OF SQUARE ROOTS RATIONALISED 195
Example 2. Find the value of a?+y i \7? — xyz i when x=sjq \/r,
y=sjr \/P> z = s/p ~ \'<1
Since x + y + z=\/q \Jr+\Jr \/p+s/p\/q
= 0,
we have (chap. iv. , § 25, IX.)
Sb*  3xyz = 2a(Za: 2  2xy),
= 0.
Therefore Zx 3  xyz = 2xyz,
= 2( V?  \Jr) ( V ~ Vi>) ( Vl>  VsOi
= 2(2  r)s/p + 2(r ^) >/? + %(p  ?)V r 
Example 3. Evaluate (l+y+is)(l+2+a:)(l+as+y) when a:= V 2 > 2/ = V 3 >
z=V5
(l+y+8)(l + «+a;)(l+a!+y)
= 1 + 2(se + y + 2) + x 2 + (y + z)x + yz + &c. + &c.
+ x{y + z 2 ) + &c. + &c. + 2xyz,
= 1 + x + y 2 + z 2 + (2 + y 2 + z 2 )x + (2 + z 2 + x 2 )y
+ (2 + a; 2 + y 2 )z + %yz + 3zx + Bxy + 2xyz,
= 11 + 10 V2 + 9 V3 + 7 V5 + 3V!5 + 3\/10 + 3V6 + 2V30.
§ 16.] We can now prove very easily the general proposition
indicated above in § 14.
If P be the sum of any number of square roots, say s/p, \/q,
sfr, . . ., a rationalising factor Q is obtained for P by multiplying
together all the different factors that can be obtained from P as
follows : — Keep the sign of the first term unchanged, and tale every
possible arrangement of sign for the following terms, except that which
occurs in P itself.
For the factors in the product Q x P contain every possible
arrangement of the signs of all but the first term. Hence along
with the + sign before any term, say that containing sjq, there
will occur every possible variety of arrangement of all the other
variable signs ; and the same is true for the  sign before \/q.
Hence, if we denote the product of all the factors containing
+ s!q by 4>( \/q), the product of all those factors that contain
 s/q will differ from <£( \fa) only in having  \/q in place of
+ s/q, that is, may be denoted by <£(  s/q). Hence we may
write Q x P = </)( s/q) x <£(  s/q), which, by § 15, Cor. 1, is rational
so far as v^ is concerned. The like may of course be proved for
every one of the irrationals \/q, \ f r, . . . Also, for every factor
in Q x P of the form sip + k there is evidently another of the form
196 RATIONAL FUNCTION REDUCED TO LINEAR FORM chap.
sip k; so that Q x P is rational as regards Jp. Hence Q x P
is entirely rational, as was to be shown.
§ 17.] Every rational function, whether integral or not, of any
number of square roots, s/p, sjq, s/r, . . ., can be expressed as the sum
of a rational part and rational midtiples of \/p>, *Jq, Jr, &c, and of
their products \/(pq), y/(pr), \/(qr), J(pqr), dr.*
For every rational function is the quotient of two rational
integral functions, say E/P. Let Q be a rationalising factor of
P (which we have seen how to find), then
R RQ
P = PQ'
But PQ is now rational, and RQ is a rational integral function of
sip, sjq, \/r, . . ., and can therefore be expressed in the required
form. Hence the proposition is established*
Example 1. To express 1/(1 + \J2 + \J3) as a sum of rational multiples of
square roots. Rationalising the denominator we obtain by successive steps,
1 1 + V2j!_\/3
1 + V2 + V3 (1 + \/2) 2  (V3) 2 '
_ i+V2ya
2V2
_ V2(l + y2V3 )
2x2
= 4 l (V2 + 2V6),
Example 2. Evaluate {x 2 x+1 )/(« 2 + x + 1 ), where x = y3 + \/5.
xx + l _ 9 + 2V15 V3 V 5
x 2 + x + 1 ~ 9 + 2 V15 + \J3 + v'5'
_ (9 + 2Vl5) 2  2(9 + 2V15)(V3 + V5) + (V3 + V5) 2
(9 + 2Vl5) 2 (V3 + V5) a
149  38 V3  30\/5 + 38 y/15
133 + 34V15
(14938V330V5 + 88y i5)(13334yi5)
133 2 34xl5 '
_ + 437 + 46 y3  114y5  12yi5
849
* Besides its theoretical interest, the process of reducing a rational func
tion of quadratic irrationals to a linear function of such irrationals is important
from an arithmetical point of view ; inasmuch as the linear function is in
general the most convenient form for calculation. Thus, if it be required to
calculate the value of l/(l + y2 + y3) to six places of decimals, it will be
found more convenient to deal with the equivalent form ^ + jy2  \\J6.
THEORY FOR IRRATIONALS OF ANY ORDER 197
GENERALISATION OF THE FOREGOING THEORY.
§ 18.] It may be of use to the student who has already
made some progress in algebra to sketch here a generalisation of
the theory of §§ 1317. It is contained in the following pro
positions : —
I. Every integral function of p 1/n can he reduced to the form
X + A 1 p l!n + A 2 p 2 l' l + . . . + A n . l p^^ n , where A u , A,, . . .,
A n _, are rational, so far as j)^ ,n is concerned.
After what has been done this is obvious.
II. Every integral function of p 111 , q 1,m , r lln , &c, can he ex
pressed as a linear function of p 1 ' 1 , p 2/l , . . ., jflW • q 1 *' 1 ", q /m , . . .,
£mi)lm ; yl/w ,.2/n _ _ ^ ,<ni)/w ; £ c ^ am i f t j ie products of these
quantities, two, three, &c, at a time, namely, p 111 q 1/m , p 2 ' 1 q llm , &c,
the coefficients of the linear function heing rational, so far as p 111 , q llm ,
r l!n , &c, are concerned.
Proved (as in § 15 above) by successive applications of I.
III. A rationalising factor of A + A,^ 1/rt + A. 2 p 2ln + . . .
+ A n _ jj/'ii)/" am always be found.
We shall prove this for the case n = 3, but it will be seen
that the process is general.
Let P = A + A^ + A 2 p% ( 1 ),
then p$P =pA 2 + A p^ + A^ (2),
and p%P =joAi + pA. 2 p* + A p ?s (3).
Let us now put x for jp«, and y for p%, on the righthand sides of
(1), (2), and (3) ; we may then write them
(A o P) + A,*; + A 2 y = (1'),
(M 2 /P) + A^ + A 1 ^0 (2'),
(M. i»*P) +M^ + A y = (3'),
whence, eliminating x and y, we must have (see chap, xvi., § 8)
(A  P) (A 2 pA.A.,) + (pA. 2 pi?) (pAf  A A,)
+ ( P A l piF)(A°A u A 2 ) = (4).
198
THEORY FOR IRRATIONALS OF ANY ORDER
C1IAV.
Whence
{(A 2 ^AA) + (pV  A A t )/ + (A x 2  AA^JP
= A (A 2 j?AA) +pA 2 (pA 2 2  A A,) + Mi(A, 2  A A 2 )
= A 3 + Mi 3 + /A/  3MAA,, (5).
Hence a. rationalising factor of P is
(A 2 MA) + (M/  AA>* + (A, 2  A A. 2 )/ (6),
and the rationalised result is
A 3 + M. 3 + p 2 A. 2 3  3;>Ao A, A 2 (7).
The reader who is familiar with the elements of the theory
of determinants will see from the way we have obtained them
that (6) and (7) are the expansions of
and
1
A,
A 2
1
A
A,
2
M«
A
A
A,
A 2
pK
A
A,
Mi
pA,
A
(6'),
(n
and will have no difficulty in writing down the rationalising
factor and the result of rationalisation in the general case.
IV. A rationalising factor can be found for any rational integral
function of p 1/l , q llm , r lln , . . . , &c, by first rationalising with
respect to p 111 , then rationalising the result with respect to q l,m ,
and so on.
V. Every rational function of p 1/l , q Vm t r^*, whether integral or
not, can be expressed as a linear function of p 1!l , pr 11 , • • • , p^ 1 ' 1 ^ 1 ;
q Vm ,q 2lm , . . ., (fi™**!™, &C ; and of the products p*' 1 qt /n r* lm . . .,
the coefficients of the function being rational, so far as p 1!l , q 1!m , r 1/n
are concerned.
For every such function has the form P/Q where P and Q
are rational and integral functions of the given irrationals ; and,
if R be a rationalising factor of Q, PR QR will be of the form
required.
X EXERCISES XIV 199
Example 1. Show that a rationalising factor of a; + y + z is
(a; 3 + y 3 + z 8  y V  s 3 a; 3  a: y 3 )
/I 3 i ,, 4 J 1 i 4 4>
x (a + y + z + 2y 3 3 3  z 3 x 3  x 6 y )
, i 3 9 4 4^44 i is
x (ar + 1/ 3 + ; 3  y V + 2z 3 ar  ary 3 )
,3 9 3 44 4Jo4Sn
x (a; 3 + y 3 + 2 3  y V  z 3 a; 5 + 2x 6 if) ;
and that the result of the rationalisation is (x + y + z) 3  27 xyz.
Exercises XIV.
Express as roots of rational numbers — ■
(1.) 2 ! x3*x4 i x9*. (2.) {!/(**)}*+ {S/tf)}*
(3.) 3V(Z/7)2W?/7). (4.) {1/{W*)} x #(1S»)}.
Simplify the following —
(5.) {(a a /6 2 )VW 6 )} x ^ / (« 1; <" 1, i 1/( " 1) )
(6.) (/fl/IHWM) x (a;(c+«)'(<»»))l/(»c) X (x(»+*)/(»e))V(«a).
(7.) Showthat n/("%/ ,!+ n/(''%= ^v' + ^x"^
Simplify —
(S.) [x^y'^ixtyh (9.) (ar» 2 + *»)/(* 2 4ar 1 ).
(10.) {x + x'^^x'^f.
(11.) (a^a^+l) (ar*+af*+ l){xx h + l).
(12.) (x i  x ~ V+ 3 *V ) (■**  2T )•
(13.) (x l + 4*/ 2 )/(x 3 + 2^ 3 + 2y).
(14. ) 2 3  1 + 2 3 /(2 3  1) + 1/(2* + 1/(2 3 + 1)).
(15. ) {x? "  2./ 1 " //' " + y 2/ ") (x 1 " + x l ''" y 1 "" + y 1/n ) (aj 1 '"  y 1 ")•
(16.) Show that x/{x i l)x i /{x i + l)l/(x^l) + l/(x h +l)=x i + 2.
(17.) If 0(a) = (a x «"*)/(«* + «"*)> F(a) = 2/(a* + «T*),
then tf>(a; +y)= {<fi(x) + <p(y) } / {1 + £(a)0(y) } ;
F(* + y)= F(*)F(y)/ {1 + tfaMy) } .
(18.) UxP/y=l, then xP m /y9 n =x n J''i m =y" m '''''.
(19.) If ma x , n = ay, m«n x = a il , then xy; = l.
Transform the following into sums of simple irrational terms : —
(20. ) V«/( \Ja + \/b) + V&/( V« ~ V 6 )
(21.) (2V53V'^ + W6) 2 .
(22. ) {x + 1  V2 + \/3) (a: + 1 + y/2  V 3 ) (a: + 1  V2  V 3 ), arranging ac
cording to powers of x.
200 EXERCISES XIV chap.
(23. ) (1/ V»+ 1/ sja) (x i  a h )l {( V« + \/xf  ( V« " V*) 2 } ■
r24 v V(« 2 + {«(!  »0/3V»t} 8 ) + o ( to 1)/2V ot
K '' \/(a*+{a(\m)lZsJm,y)a{m\)l2\/vi
(C} r v \/(p/a + x) + * s /(p/az) 2pq
(■to ) „ t— — r — ,> —, {, where a,x'=: = — „.
s/(p/a + x)  \/{p/a x)' 1 + cf
ton \ Pa pb p + a p + b
{to) 1 r— — r, where x\/(nb).
xa xb x + a x + b vv '
(27.) 1/ { X /( P q) + y/ p + V<?} + 1/ ! Mp q)\/p \/?} + 1/(VJP  V?)
(28) /( l*^(2x + x*) ) / / 1 x+ V(2.r + ^) ) .
v '' V Ua+V(2»+a*)/" r V llajV^+a: 2 )/
(29. ) ( V(« + & + c) + V(«  6 + '0)7( V(« + * + c )  V(«  6 + c)) 2 .
(30.) {U(pq) + */p<s/q)Ksf(pq)y/p\/q)}.
(31.) (2.r 5 6.r + 5)/(4 / 2x'+ vA + l).
(32. ) (a? + 3 ^23 + 1 )/(.r + ^2  1).
(33. ) { l/{a + b) l/(a  b) } { [ y/{a + b)f + [ l/{u  i)] 2  $/{<#  b) \ .
Show that
<"•> (r^r# Gt^)^^ ve*»
(35.) (V(y + i) + V(p 2 i)) 3 + (V(^ 2 + i)V(yi)) 3 =(yi)V(r+i)
(36.) V[\/{« 2 + \V& 2 )} + VF+ ^V& 4 )}] = (« § + *<¥
Express in linear form —
(37.) {Z/xZ/y)/(l/x+i/y).
(38. ) (1 + V3 + V5 + V7)/(l  V3  \ 7 5 + \/7)
(39.) 2(v&+ Vc)/(V & + Vc  V«)  42«(Vft + Vc)/n(\/ft + V< V«)
Rationalise the following : —
(40.) 3.5* 4^. (41.) 2V(Hc«).
(42.) V5 + V3 + V4  V6 (43.) 3.2* + 4.2 4 l.
(44.) a* + & J + c*. (45.) 2*+2* + l.
(46.) If« = ; rV(l+i/ 2 ) + 2/\/(l+* 2 ).tlienV(l+«) = ''2/ + \/((l+» 2 )( 1 +2/ 2 ))
(47.) Show that V(y .. ) + >/0 ,l, ) + V(a> . y j
= ( »/  z) % + (z  aQ j + (a  y) 1 + (y  z)\z  x)\x  y Y
■'" + y 2 + z 2 — yzzx xy
(48. ) If jr = l/( Vi + \/c  \fa), y = l/(\/c + s/a  \/£), z=lf(s/a + V*
 /y/c), U = l/( V« + Vft + V c )>
then Tl(x+y + z + m)/(Ssb  ?t) 3 = n ( ^ + c  a)/8abc.
xyz
Historical Note. — The use of exponents began in the works of the German
"Cossists," Rudolf! (1525) and Stifel (1544), who wrote over the contractions
x HISTORICAL NOTE 201
for the names of tbe 1st, 2nd, 3rd, . . . powers of the variable, which had been
used in the syncopated algebra, the numbers 1, 2, 3, . . . Stifel even states
expressly the laws for multiplying and dividing powers by adding and subtracting
the exponents, and indicates the use of negative exponents tor the reciprocals of
positive integral powers. Bombelli (1579) writes _., ^, „£,, 3> • • •> where
we should write .v°,.r, ./;, .*.•■'', . . . Stevin (1585) uses in a similar way @, ©,
©i ©>•••> an d suggests, although he does not practically use, fractional
powers such as ©, ©, which are equivalent to the x , x , of the present day.
Viete (1591) and Oughtred (1631), who were in lull possession of a literal
calculus, used contractions for the names of the powers, thus, Aq, Ac, Aq</, to
signify A' 2 , A 3 , A 4 . Harriot (1631) simply repeated the letter, thus, aa, aaa,
acuta, for ft 2 , « :i , a 4 . Herigone (1634) used numbers written after the letter,
thus, A, A2, A3, . . . Descartes introduced the modern forms A, A 2 , A 3 , . . .
The final development of the general idea of an index unrestricted in magnitude,
that is to say, of an exponential function a x , is due to Newton, and came in
company with his discovery of the general form of the binomial coefficients as
functions of the index. He says, in the letter to Oldenburg of 13th June 1676,
"Since algebraists write a 2 , ft 3 , «', &c, for aa, aaa, aaaa, &c. , so I write
a , a", a 3 , for \Ja, \/a s , \/c.a s : and I write a 1 , or, ar", &c, for , — ,
1 « aa
—,kr."
aaa
The sign \/ was first used by Rudolff.; both he and Scheubel (1551) used
/fj to denote 4th root, and wJ to denote cube root. Stifel used both •sj'fr.
and \J to denote square root, \/%%. to denote 4th root, and so on. Girard
(1633) uses the notation of the present day, \J, \/, &c. Other authors of the
17th century wrote y/2 :, \/3 : , &c. So late as 1722, in the second edition of
Newton's Arithmetica Universalis, the usage fluctuates, the three forms \/3:,
/3 3/
V : , v' all occurring.
In an incomplete mathematical treatise, entitled 1)e Arte Logistica, &c,
which was found among the papers of Napier of Merchiston (15501617 ; pub
lished by Mark Napier, Edinburgh, 1839), and shows in every line the firm grasp
of the great inventor of logarithms, a remarkable system of notation for irrationals
is described. Napier takes the figure " [_, and divides it thus 4J _5j [6.
' •' t m it
He then uses J, \_J, I , &c. , which are in effect a new set of symbols for the
nine digits 1, 2, 3, &c, as radical signs. Thus UlO stands for \A0, I 10 for
^10, _i°10 for ^/lO, J! 10 for ^10, _] ^j or =] for "Ao ; and so on.
Many of the rules for operating with irrationals at present in use have come,
in form at least, from the German mathematicians of the 16th centuiy, more
particularly from Scheubel, in whose Algebra Compendiosa Fo.cilisqiie Descriptio
(1551) is given the rule of chap, xi , § 9, for extracting the square root of a
binomial surd. Iu substance many of these rules are doubtless much older (as
old as Book X. of Euclid's Elements, at least) ; they were at all events more or
less familiar to the contemporary mathematicians of the Italian school, who did
so much for the solution of equations by means of radicals, although in symbol
ism they were far behind their transalpine rivals. See Hutton's Mathematical
Dictionary, Art. "Algebra."
The process explained at the end of next chapter for extracting the square or
202 HISTORICAL NOTE chap, x
cube root by successive steps is found in the works of the earliest European
writers on algebra, for example, Leonardo Fibonacci (c 1200) and Luca Pacioli (c.
1500). The first indication of a general method appears in Stifel's Arithmetica
Integra, where the necessary table of binomial coefficients (see p. 81) is given.
It is not quite clear from Stifel's work that he fully understood the nature of the
process and clearly saw its connection with the binomial theorem. The general
method of root extraction, together with the triangle of binomial coefficients, is
given in Napier's De Arte Logistica. He indicates along the two sides of his
triangle the powers of the two variables (prsecedens and succedens) with which
each coefficient is associated, and thus gives the binomial theorem in diagram
matic form. His statement for the cube is — " Supplementum triplicationis tribus
constat numeris : primus est, duplicati prsecedentis triplum multiplicatum per
succedens ; secundus est, prsecedentis triplum multiplicatum per duplicatum suc
cedentis ; tertius est, ipsum triplication succedentis." In modern notation,
CHAPTER XL .
Arithmetical Theory of Surds.
ALGEBRAICAL AND ARITHMETICAL IRRATIONALITY.
§ 1.] In last chapter we discussed the properties of irra
tional functions in so far as they depend merely on outward
form ; in other words, we considered them merely from the
algebraical point of view. We have now to consider certain
peculiarities of a purely arithmetical nature. Let p denote any
commensurable number ; that is, either an integer, or a proper or
improper vulgar fraction with a finite number of digits in its
numerator and denominator ; or, what comes to the same thing,
letp denote a number which is either a terminating or repeating
decimal. Then, if a be any positive integer, H/p will not be
commensurable unless p be the nth power of a commensurable
number ;* for if %/p = k, where k is commensurable, then, by the
definition of %/p, p = k n , that is, p is the nth power of a commen
surable number.
If therefore p be not a perfect nth power, %/p is incommensur
able. For distinction's sake %/p is then called a surd number.
In other words, we define a surd number as the incommensurable
root of a commensurable number.
Surds are classified according to the index, n, of the root to
be extracted, as quadratic, cubic, biquadratic or quartic, quintic,
. , . wtic surds.
The student should attend to the last phrase of the definition of a surd ;
because incommensurable roots might be conceived which do not come under
;; This is briefly put by saying that p is a perfect «th power.
204 CLASSIFICATION OF SURDS chap.
the above definition ; and to them the demonstrations of at least some of the
propositions in this chapter would not apply. For example, the number e (see
the chapter on the Exponential Theorem in Part II. of this work) is incom
mensurable, and \/c is incommensurable ; hence \/e is not a surd in the exact
sense of the definition. Neither is V( V 2 + 1), for \/2 + 1 is incommensurable.
On the other hand, \/( V 2 )> which can be expressed in the form tyl, does come
under that definition, although not as a quadratic but as a biquadratic surd.
He should also observe that an algebraically irrational function, say \/x,
may or may not be arithmetically irrational, that is, surd, strictly so called,
according to the value of the variable x. Thus \J1 is not a surd, but V 2 is 
CLASSIFICATION OF SURDS.
§ 2.] A single surd number, or, what comes to the same, a
rational multiple of a single surd, is spoken of as a simple mono
mial surd number ; the sum of two snch surds, or of a rational
number and a simple monomial surd number, as a simple binomial
surd number, and so on.
The propositions stated in last chapter amount to a proof of
the statement that every rational function of surd numbers can
be expressed as a simple surd number, monomial, binomial,
trinomial, &c, as the case may be.
§ 3 ] Two surds are said to be similar when they can be expressed
as rational multiples of one and the same surd ; dissimilar when this
is not the case.
For example, >/18 and N /8 can be expressed respectively in
the forms 3 N /2 and 2 J 2, and are therefore similar ; but J 6
and s/2 are dissimilar.
Again, ^/54 and £/16, being expressible in the forms 3^2
and 2 ^/2, are each similar to 1/2.
All the surds that arise from the extraction of the same nth root
are said to be equiradical.
Thus p*, p*, p*, P*° are all equiradical with p*.
There are n  1 distinct surds equiradical with 2> v '\ namely, p l/n ,
P 2!n , . . ., £>("■ D/'^ and no more.
For, if we consider p mln where m>n, then Ave have p m l n =
^*+»/n where /x and v are integers, and v < n. Hence p m,n =
pft pin _ a rational multiple of one of the above series.
All the surds equiradical with p l l* are rational functions (namely,
xi CLASSIFICATION OF SURDS 205
positive integral powers) of p lln ; and every rational function of p l,n
or of surds equiradieal with p l,n may be expressed as a linear function
of the n  1 distinct surds which are equiradieal with p l,n , that is, in
the form A + A^ 11 ' 1 + A.,p 2ln + . . . + A M _ 1 ^ n_1 ) /n , where A , A l5
. . . , A n _ i are rational so far as p lln is concerned.
This is merely a restatement of § 18 of chap. x.
§ 4.] The p'oducl or quotient of two similar quadratic surds is
rational, and if the product or quotient of the two quadratic surds is
rational they are similar.
For, if the surds are similar, they are expressible in the
forms A s/p and B s/p, where A and B are rational ; therefore
A \/p x B s/p  ABp ; and A \/p/B s/p = A/B, which proves the
proposition, since ABp and A/B are rational.
Again, if s/p x \/q = A, or \/p/ \/q = B, Avhere A and B are
rational, then in the one case s/p  (AJq) s ^q, in the other s/p
= B \'q. But A/q and B are rational. Hence s/p and \/q are
similar in both cases.
The same is not true for surds of higher index than 2, but
the product of two similar or of two equiradieal surds is either rational
or an equiradieal surd.
INDEPENDENCE OF SURD NUMBERS.
§ 5.] If p, q, A, B be all commensurable, and none of them zero,
and s/p and sjq incommensurable, then we cannot have
s/p = A + B sjq.
For, squaring, we should have as a consequence,
p = A 2 + B 2 q + 2AB s/q ;
whence
s/q = (p  A 3  BV)/2AB,
which asserts, contrary to our hypothesis, that Jq is commen
surable.
Since every rational function of s/q may (chap, x., § 15) be
expressed in the form A + B Jq, the above theorem is equivalent
to the following : —
One quadratic surd cannot be expressed as a rational function of
another which is dissimilar to it.
206 INDEPENDENCE OF SURD NUMBERS chap.
Since every rational equation between <Jp and Jq which is
not a mere equation between commensurables (for example,
( \ / 3) 2 + ( \/2) 2 = 5) is reducible to the form
A sj(pq) + B x/jp + C s/q + D = 0,
where A, B, C, D are rational ; and, since this equation may im
mediately be reduced to another of the form
Jp = L + M */q,
where L and M are rational, it follows that
No rational relation,, which is not a mere equation between rational
numbers, can subsist between two dissimilar quadratic surds.
§6.] 7/" the quadratic surds >Jp, \/q, \/r, \/{qr) be dissimilar to
each other, then >Jp cannot be a rational function of Jq and ijr.
For, if this were so, then we should have
s/p = A + B *Jq + C Jr + D \/(gr),
where A, B, C, D are all rational.
Now we cannot, by our hypotheses, have three of the four
A, B, C, D equal to zero.
In any other case, we should get on squaring
p = {A + Bjq + C Jr + D J(qr)}°,
which would either be a rational equation connecting two dis
similar quadratic surds, which is impossible, as we have just seen ;
or else an equation asserting the rationality of one of the surds,
which is equally impossible.
An important particular case of the above is the following : —
A quadratic surd cannot be the sum of two dissimilar quadratic
surds.
It will be a good exercise for the student to prove this
directly.
§ 7.] The theory which we have established so far fur
quadratic surds may be generalised, and also extended to surds
whose index exceeds 2. This is not the place to pursue the
matter farther, but the reader who has followed so far will find
the ideas gained useful in paving the way to an understanding
of the delicate researches of Lagrange, Abel, and Galois regarding
xi INDEPENDENCE OF SURD NUMBERS 207
the algebraical solution of equations whose degree exceeds the
4 th.
§ 8.] It follows as a necessary consequence of §§ 5 and 6
that, if we are led to any equation such as
A + Bs'p + C v/g + D ^(pq) = 0,
where Jp and *Jq are dissimilar surds, then we must have
A = 0, B = 0, C = 0, D = 0.
One case of this is so important that we enunciate and prove it
separately.
If x, y, z, u be all commensurable, and Jy and sju incommen
surable, and if x + Jy = z + Ju, then must xz and y = u.
For if x =# z, but = a + z say, where a 4= 0, then by hypothesis
a + z + \/y = z + \''u,
whence a + \/y — *Ju,
a 3 + y + 2a s/y  u,
s/y = («  a 2  y)/2a,
which asserts that Jy is commensurable. But this is not so.
Hence Ave must have x = z ; and, that being so, we must also
have Jy — \ f u, that is, y = u.
SQUARE ROOTS OF SIMPLE SURD NUMBERS.
§ 9.] Since the square of every simple binomial surd number
takes the form p + Jq, it is natural to inquire whether J(j> + *Jq)
can always be expressed as a simple binomial surd number, that
is, in the form *Jx + \'y, where x and y are rational numbers.
Let us suppose that such an expression exists ; then
*J(jp+ s 'q)= \/x+ >Jy,
whence p + \'q = x + y + 2 *J(xy).
If this equation be true, we must have, by § 8,
x + y=p (1),
2^)= s'q (2);
and, from (1) and (2), squaring and subtracting, we get
(x + y) 3 4xy=p*  q,
that is, (zy) 2 =p 3 ~q ( 3 >
208 LINEAR EXPRESSION FOR </(« + \^) chap.
Xow (3) gives either
xy = + s!{/q) (4),
or xy=  *S(p'q) (4*).
Taking, meantime, (4) and combining it with (1), we have
(x + y) + (xy)=p + s/{p*q) (5),
(x + y)  (x y)=p s '{f q) (6) ;
Avhence 2x =p + \,'(p*  q),
2y=p s/(p*q);
that is, x = h{p+ s'dfq)} (7),
V = HP  n ; (/ " ?)} (8).
If we take (4*) instead of (4), we simply interchange the values
of x and y, which leads to nothing new in the end.
Using the values of (7) and (8) we obtain the following
result : —
Since, by (2), 2 s/x x v /y = + ^> we must take either the two
upper signs together or the two lower.
If we had started with \/(p  */q), it would have been
necessary to choose \/x and \fy with opposite signs.
Finally therefore we have
(9),
(9*).
The identities (9) and (9*) are certainly true ; we have in fact
already verified one of them (see chap, x., § 9, Example 14). They
will not, however, furnish a solution of our problem, unless the
values of x and y are rational. For this it is necessary and
sufficient that p*  q be a positive perfect square, and that p be
XI
EXAMPLES 209
positive. Hence the square root of p + *Jq am he expressed as a
simple binomial surd number, provided p be positive and p*  a be a
positive perfect square.
Example 1. Simplify >/(194V21).
Let V( 19  4 V 21 )=V / a ; + Vy
Then aj+y=19,
(ccy) 2 = 361336
= 25,
xy= +5 say,
x + y19 ;
whence # = 12, y=7,
*Jx  ± VI 2, \ // = T \''7,
so that V(l 9 W21)=±(Vl2\/7)
Example 2. To find the condition that \Z(VP+ V?) ma } r De expressible
4 /
in the form (\/x + \ f y) *Jp we have
•s/NP + V?) = fa x V U + V(?/l>)} •
Now V{1 + V((?/i°)} "ill t> e expressible in the form \/x+\Jy, provided
1  q/p be a positive perfect square ; this, therefore, is the required condition.
For example,
V(5V7 + 2V42)= \/7x V(5 + 2v'6)
= ± 4/7K/3 + V2).
Example 3. It is obvious that in certain cases V(.P + V?+ V r + N/' s ) roust
be expressible in the form \'x + \/y+ \ ~, where a 1 , y, z are rational. To find
the condition that this may be so, and to determine the values of x, y, z, let
V(P + \fl + V'' + V s ) = \ /x + \'y + V s C 1 ).
then }i+\/q+\/r+\Js = x + y + z + 2\/(yz) + 2s/(zx) + 2\/(xy) (2).
Now let us suppose that
2V(yz)=V? (3),
2sJ{zx) = s /r (4),
2V(«y)=V* (5).
From (4) and (5) we have by multiplication
4x\/(yz)=*/(rs);
whence, by using (3), x = hy/{>'s/q) (6).
Proceeding in like manner with y and r, we obtain
y=W(q*M (7),
~~=W(<7>7*) («)•
It is further necessary, in order that (2) may hold, that the values (6), (7),
(8) shall satisfy the equation
x + y + z=p (9),
VOL. I P
210 ARITHMETICAL SQUARE ROOT chap.
that is, we must have
V(W?) + VWO + s/iqr/s) = 2p (10),
where the signs throughout must be positive, since x, y, z must all be positive.
Also, since x, y, z must all be rational, we must have
rs o 9 s as W
q r s
where a, /3, y are positive rational numbers, such that
a + p + y = 2p,
whence, in turn, we obtain
q = §y, r = ya, s = ap.
ARITHMETICAL METHODS FOR FINDING APPROXIMATE RATIONAL
VALUES FOR SURD NUMBERS.
§ 10.] It has already been stated that a rational approxima
tion, as close as we please, can always be found for every surd
number. It will be well to give here one method at least by
which such approximations can be obtained. We begin with
the approximation to a quadratic surd ; and we shall afterwards
show that all other cases might be made to depend on this.
§ 11.] First of all, we may point out that in every case we
may reduce our problem to the finding of the integral part of
the square root of an integer. Suppose, for example, we wish
to find the square root of 3*689 correct to five places of decimals.
Then, since ^3689 = ^36890000000/10*, we have merely to
find the square root of the integer 36890000000 correct to the
last integral place, and then count off five decimal places.
§ 12.] The following propositions are all that are required
for the present purpose : —
I. The result of subtracting (A + B) 2 from N is the same as the
result of first subtracting A 2 , then 2AB, and finally B 2 .
This is obvious, since (A + B) 2 = A 2 + 2AB + B 2 .
II. If the first p out of the n digits of the square root of the
integer N have been found, so that P10 n_ ^ is a first approximation
to VN, then the next p  1 digits ivill be the first p  1 digits of the
integral part of the quotient {N  (Pl0 n ^) 2 }/2Pl0' 1 ^ with a possible
error in excess of 1 in the last digit,
xi ARITHMETICAL SQUARE ROOT 211
Let the whole of the rest of the square root be Q. Then
x /N = P10»" + Q,
where 1 0*  x < P < 1 0*, Q < 1 n * ;
whence N = (P10 w *>) 2 + 2PQ10** + Q 2 ,
N(P10"^) 2 _ Q 2 (1).
— \°l +
2P10""^ ^ 2P10 n "^
Now
Q72PlO n ^<10 2 ( ,l ^/2 x 10J ,  1 10 n *'<10 n * +1 /2.
Hence Q 2 /2P10 n P will at most affect the (n  2p + l)th place,
and the error in that place will be at the utmost 5 in excess.
Therefore, since Q contains np digits, the first p  1 of these
will be given by the first p  1 digits of {N  (P10 n ^) 2 }/2P10 n "P
with a possible error in excess of 1 in the last digit.*
§ 13.] In the actual calculation of the square root the first
few figures may be found singly by successive trials, Proposi
tion I. being used to find the residues, which must, of course,
always be positive. Then Proposition II. may be used to find
the succeeding digits in larger and larger groups. The approxi
mation can thus be carried out with great rapidity, as will be
seen by the following example : —
Let it be required to find the square root of N = 680100000000000000,
which, for shortness, we write 6801(14).
Obviously 8(8)<\/N<9(8) ; in other words, \/N contains 9 digits, and the
first is 8.
Now N (8(8)) 2 = 401(14), which is the first residue. We have now to
find the greatest digit x which can stand in the second place, and still leave
the square of the part found less than N, that is (by Proposition I.), leave the
residue 401(14) 2 x 8(8) xx(7)  {x(7)} 2 positive. It is found by inspection
that»  = 2. Carrying out the subtractions indicated, that is, subtracting
16(8) + 2(7)} x2(7) = 162x2(14) from 401(14), we have now as residue
7700(12).
* The effect of such an error would be to give a negative residue in
the process of § 13 ; so that in practice it would be immediately discovered
and rectified. As an example of a case where the error actually occurs, the
reader might take the square root of 5558(12), namely, 74551995, and attempt
to deduce from 745 the two following digits. He will find by the above rule
52 instead of 51. If it be a question of the best approximation, the rule gives
here, as always, the best result ; but this is not always what is wanted.
212
EXAMPLES
CHAP.
The double of the whole of the part of \J~N now found is 164(7) ; and
we have next to find y as large as may be, so that 7700(12) {164(7) + 2/(6)}
x ?/(6) shall remain positive. This value of y is seen to be 4. It might, of
course, be found (by Proposition II.) by dividing 7700(12) by 164(7), and
taking the first figure of the quotient.
The residue is now 112400(10). The process of finding the first four
digits in this way may be arranged thus : —
8(8) 6801(14) 8(8)
16(8) 6400(14)
162(7) 401(14) +2(7)
164(7) 324(14)
1644(6) 7700(12) +4(6)
1648(6) 6576(12)
16486(5) 112400(10) +6(5)
16492(5) 98916(10)
134840(9)
We might, of course, continue in the same way, figure by figure, as long as
we please ; and we might omit the records in brackets of the zeros in each line.
Havino, however, already found four figures, we can find three more by
dividing the residue 134840(9) by 16492(5), which is the double of 8246(5),
the part of V N already found.
16492(5) 134840(9)
131936
817(2)
29040
16492
125480
115444
10036000(4)
The next three digits are therefore 817. 10036000(4) is not the residue ;
for we have only subtracted from V N as yet {8246(5)} 2 and 2x8246(5)
x 817(2). Subtracting also {817(2)} 2 we get the true residue, namely,
93685110000. We may now divide this by 2x8246817(2), that is, by
1649363400, and thus get the last two figures. We have then
10036000(4)
667489(4)
1649363400 I
93685110000
8246817000
11216940000
9896180400
1320759600
56
We have now found the whole of the integral part of V 68 01(14), namely,
824681756.
XI
RATIONAL APPROXIMATION TO ANY SURD
213
If it were desired to carry the approximation farther, 8 places after the
decimal point could at once be found by dividing the true residue
(1320759600  56 2 ) by 2 x 824681756. If we require no more places than those
8 places, then the residue is of no importance, and we may save labour by
adopting the abbreviated method of long division (see Brook Smith's Arith
metic, chap. vi. , § 153). Thus
nftiW$$V$
1320759600
3136
1320756464
1319490810
1265654
1154554
80076736
111100
98962
12138
11545
593
495
"98
98
We thus find V6801(14) = 824681756 '80076736.
will find that in point of fact
On verifying, the reader
(824681756 80076736) 2 = 680100000000000001 82 . . .
It will be a good exercise for him to find out how many decimal places of
the square root of a given integer must be found before the square of the
approximation ceases to be incorrect in the last integral place.
§ 14.] By continually extracting the square root (that is to
say, hy extracting the square root, then extracting the square
root of the square root, and so on), we may bring any number
greater than unity as near unity as we please. In other words,
by making n sufficiently great, W 1 * may be made to differ from 1
by less than any assignable quantity.
For let it be required to make N I/2 less than 1 + a, where a
is any positive quantity. This will be done if 2 n be made such
that (1 + af l > N. Now (chap, iv., § 1 1) (1 + a) 2 " = 1 + 2' l a + a
series of terms, which are all positive. Hence it will be sufficient
if we make 1 + 2"u> N, that is, if we make 2 n a>N  1, thut is,
214 EXAMPLES chap.
if we make 2 n >(N l)/a, which can always be done, since by
making n sufficiently great 2 n may be made to exceed any
quantity, however great.
Example. How many times must we extract the square root beginning
with 51 in order that the final result may differ from 1 by less than  001 ?
We must have
2» > (511)/ 001,
2" > 50000.
Now 2 15 = 32768, 2 16 = 65536,
hence we must make ?i = 16.
In other words, if we extract the square root sixteen times, beginning
with 51, the result will be less than 1 "001.
§ 15.] It follows from § 14 that we can approximate to any
surd whatever, say p lln , by the process of extracting the square
root. For (see chap, ix., § 2) let 1/n be expressed in the binary
scale, then we shall have
1/n = a/2 + /3/2 2 + y/2' + . . ,+fjL,
where each of the numerators a, ft, y, . . . is either or 1, and
ix is either absolutely or < l/2 r , where r is as great as we
choose.
Hence
= pt/2 x ^/S/22 x ^/23 x x ^» ^y
Now, excepting the last, each of these factors is either ] , or of
the form p* 1 "' , which can be approximated to as closely as we
please by continued extraction of the square root. If /i = 0, the
last factor is 1 ; and if ll< \/2 r , since r may be as great as we
choose, we can make it differ from 1 by as small a fraction as
we choose. It follows therefore that the product on the right
hand of (1) may be found in rational terms as accurately as Ave
please.
Example. To find an approximate value of 5V 9 .
We have
i i_ i i i L L L L L L
3 2' 2 2* 2 6 2 8 "*"2 10 "*~2 12 "^ 2 U "^2 16 + 2 18 "^2""^' U '
where fi < 1/2 20 .
XI ALGEBRAICAL SQUARE ROOT 215
Now we have, correct to the fourth decimal place, the following values :—
5V*= 267234, 51^"= 100096,
51 1/24 = 1 27857, 51 1 "' 4 = 1 00023,
51^=106336, 51' 2,6 = 100006,
51 1  8 = 1 01548, 51 1 "' 8 = 1 00002,
5H/2">  1 00385, 51 1 / 220 < 1 00001.
Hence, multiplying the first nine numbers together, we get
5V 3 = 3 70841
The correct value is 3708429 . . .
§ 16.] The method just explained, although interesting in
theory, would be very troublesome in practice.
The method given in § 13 for extracting the square root may
be easily generalised into a method for extracting an ?it\\ root
directly, figure by figure, and group by group of figures. The
student will be able to establish for himself two propositions,
counterparts of I. and II., § 12, and to arrange a process for
the economical calculation of the residues. A method of this
kind is given in most arithmetical textbooks for extracting the
cube root, but it is needless to reproduce it here, as the extrac
tion of cube and higher roots, and even of square roots, is now
accomplished in practice by means of logarithmic or other tables
(see chap, xxi.) Moreover, the extraction of the nth root of a
given number is merely a particular case of the numerical solu
tion of an equation of the nth degree, a process for which, called
Homer's Method, will be given in a later chapter.
Our reason for dwelling on the more elementary methods of
this chapter is a desire to cultivate in the mind of the learner
exact ideas regarding the nature of approximate calculation —
a process which lies at the root of many useful applications of
mathematics.
SQUARE BOOT OF AN INTEGRAL FUNCTION OF X.
§ 17.] When an integral function of x is a complete square
as regards .r, its square root can be found by a method analogous
216 ALGEBRAICAL SQUARE ROOT chap.
to that explained in § 12, for finding the square root of a number.
Although the method is of little interest, either theoretically or
practically, we give a brief sketch of it here, because it illustrates
at once the analogy and the fundamental difference between
arithmetical and algebraical operations.*
I. We may restate Proposition I. of § 1 2, understanding now
A and B to mean integral functions of x.
II. IfF =p x 2n + p 1 x 2n ~ 1 + . . . +p, n , and if JF = {q„x n + q x x n ~ x
+ . . . + q n  p + x x n  p+1 ) + {q n p* n ~ p + • • • + q ) = p + Q,say; and if
we suppose the first p terms, namely, P = q x n + q x x n ~ l + . . . + q n  P +i
x n ~P +1 , of \/F to be biown, then the next p terms will be the first p
terms in the integral part of (F  P 2 )/2P.
for we have
F = F + 2PQ + Q 2 ;
, FF Q 2
hence —^ = Q + ^
Now the degree of the integral part of Q 2 /2P is 2(np)n
= n 2p. Hence Q 2 /2P will at most affect the term in x n ~ 2p .
Hence (F  P 2 )/2P will be identical with Q down to the term in
x n2p+i inclusive. In other words, the first np  (»  2p) =p
terms obtained by dividing F  P 2 by 2P will be the p terms of
the square root which follow P.
We may use this rule to obtain the whole of the terms one
at a time, the highest being of course found by inspection as the
square root of the highest term of the radicand ; or we may ob
tain a certain number in this way, and then obtain the rest by
division. t
The process will be understood from the following example,
* The method was probably obtained by analogy from the arithmetical
process. It was first given by Recorde in The Whetstone of Witte (black
letter, 1557), the earliest English work on algebra.
t Just as in division, we may, if we please, arrange the radicand according
to ascending powers of x. The final result will be the same whichever arrange
ment be adopted, provided the radicand is a complete square. If this is not
the case the operation may be prolonged indefinitely just as in continued
division. We leave the learner to discover the meaning of the result obtained
in such cases. The full discussion of the matter would require some refer
ence to the theory of infinite series.
XI
ALGEBRAICAL SQUARE ROOT
217
in which we first find three of the terms of the root singly, and
then deduce the remaining two by division : —
Exam] ile.
To find the square root of
a;io + 6^9 + 13Z 8 + 4a; 7  18.x 6  12a: 5 + 14ar*  12k 3 + 9a 2  2x + 1.
1
2 + 3
2 + 6 + 2
2+6+44
2+6+48
1 + 6 + 13+ 41812 + 1412 + 92 + 1
1
6 + 13+ 41812 + 1412 + 92 + 1
6+ 9
4+ 41812 + 1412 + 92 + 1
4 + 12+ 4
 82212 + 1412 + 92 + 1
 82416 + 16
2+ 4 212+92+1
2+ 6+ 4 8
+ 1
+ 3
+ 2
+ 11
 2 6 4 + 9^2 + 1
 2 6 4 + 8
12 + 1
Hence the square root is x 5 + 3a; 4 + 2a* 3  4a.* 2 + x  1 ; and, since the residue
x*2x+\ is the square of the two last terms, namely, a;— 1, we see that the
radicand is an exact square. Of course, we obtain another value of the square
root by changing the sign of every coefficient in the above result.
A similar process can be arranged for the extraction of the
cube root ; but it is needless to pursue the matter further.
§ 18.] The student should observe that in the simpler cases
the root can be obtained by inspection ; and that in all cases the
method of indeterminate coefficients renders any special process
for the extraction of roots superfluous. This will be understood
from the following example.
Example.
To extract the square root of
,.io
+ 6a; 9 + 13a: 8 + 4a; 7  18a: 6  12a; 5 + Ux 4  12a; 3 + 9a: 2  2x+ 1
(1).
If the radicand be a complete square, its square root must be of the form
ar 5 +px i + qx* + rx" +sx+ 1 (2 ).
The square of (2) is
x it) + '2px 9 +(p 2 + 2q)x s + (22)q + 2r)x 7 + (2pr + q ii + 2s)x e + . . . (3).
Now this must be identical with (1) ; hence we must have
2p = G, p 2 + 27=13, 2pq+2r=4, 2pr+g a +2s= 18.
218 EXERCISES XV
CHAP.
The first of these equations gives p = S ; p being thus known, the second
gives q=2 ; p and q being known, the third gives r— 4 ; andp, q, r being
known, the last gives 8=1. We could now find I in like manner ; but it is
obvious from the coefficient of £ that t= — 1.
Hence one value of the square root is
x 5 + 3X 4 + 2x*  Ax 2 + x  1 .
N.B. — The equating of the coefficients of the remaining terms of (1) and
(3) will simply give equations that are satisfied by the values of p, <7, r, s
already found, always supposing that the given radicand is an exact square.
A process exactly similar to the above will furnish the root of an exact
cube, an exact 4th power, and so on.
Exercises XV.
Express the following as linear functions of the irrationals involved.
(1.) l/(VH + V3 + \/14). (2.) x /12/(l + V2)(\ / 6V3).
(3.) (1  V2 + V3)/(l + V2+ V3) + (1  V2  V3)/(l + \/2  V3).
(4.) (3V5)/(V3 + V5) 2 + (3 + \/5)/(V3\/5) 2 .
(5.) V5/(V3 + V52V2)V2/(\/3 + V2\/5).
(6.) (7  2 V5)(5 + V7)(31 + 13 v '5)/(6  2V7)(3 + V5)(H + 4 V")
(7.) V(25 + 10V6). (8.) V(3/2 + v'2).
(9.) V(12322V2). (10.) V(44V2 + 12V26).
(11.) V{(8 + 4VlO)/(84VlO)}.
(12.) V(7 + 4\/3)+V(52V6).
(13.) V(154V14) + 1/V(15+4V14).
(14. ) 1/V(16 + 2 V63) + l/\/(16  2 V63).
(15.) l/V(16V3 + 6V21) + V(16v'36V21).
(16.) Calculate to five places of decimals the value of {\/(5 + 2\/6)
V(52V6)}/{V(5 + 2\/6)+V(52V<5)}
(17.) Calculate to seven places of decimals the value of \/(\/15 + \/13)
+ VW15V13).
Simplify—
(IS.) V{3 + V(9 i ; 2 )} + V{3V(9^ 2 )}.
(19.) ^{a + bc+2^(b(ac))}.
(20.) % /{« 2 2 + «V(« 2 4)}.
<U/{(>^Xr^)}
(22.) Show that V{2 + V(2  V2)} = ./ { 2 + V( ^ +n/2) }
(23. ) Express in a linear form \/( 5 + V 6 + V^O + V 15 )
(24.) „ ,, V(254V312 x /2 + 6V6).
XI
EXERCISES XV 219
(25.) If ad = bc, then \/{a + \Jb + \Jc + \Jd) can always be expressed in
the form {\/x + sfy){\JX + sJY). Show that this will be advantageous if
a 2  b and crc are perfect squares.
(26.) If f/(a + \Jb) = z+\/y, where a, b, x, y are rational, and \Jb and
s/y irrational, then f/{a  \Jb) = x  \/y. Hence show that, if a 2 b = z 3 i
where z is rational, and if x be such that 4x z Zxz = a, then ^/(a + sjb)
=x+»/(x*z).
(27.) Express in linear form <i/{99  35\/8).
(28.) ,, „ 4/(395 + 93 V18).
(29.) „ „ 4/(117V2 + 74V5). '
(30.) Show that 4/(90 + 34V7) ^(90Mv7) = 2>/7.
(31.) If x= J/{p + q)+ </{pq), and fq^r 3 , show that ar 3 3ra;2^
= 0.
(32.) If py* + qi/ + r = 0, wherep, q, r, y are all rational, and y irrational,
then p=0, q = Q, r=0. Hence show that, if x, y, z be all rational, and
x , y , z all irrational, then neither of the equations x+y=z, x+y=z is
possible.
(33.) Find, by the full use of the ordinary rule, the value of \/10 to 5
places of decimals ; and find as many more figures as you can by division
alone. Use the value of \/10 thus found to obtain \/  004.
Extract the square root of the following : —
(34.) (yz + zx + xy) 2 4xyz(z + x).
(35. ) 25a; 2 + 9r/ 2 + z 2 + 6yz  lOzx  SOxy.
(36.) 9ar l + 24ar 3 +10a: 2 8a;+l. (37.) af 4  4ar 3 + 2a: 2 +4a; + l.
(38.) 4x i l2x i y + 25xY2ixy 3 + l6y i .
(39.) ar s 6ar 4 + 4ar 3 + 9a; 2 12a; + 4.
(40. ) 4a; 6  12a* + 5ar* + 22x?  23a; 2  8a; +16.
(41. ) 27(;a + qf(p" + q 2 ) 2  2( 2 r + ipq + q 2 ?.
(42.) a;" 3 2a:Va; + 3a;2v'a;+l
(43.) Extract the cube root of
8a; 9  1 2a; 8 + 6a; 7  37a; 6 + 36;^  9a, 1 + 54a 3  27a; 2  27.
(44.) Extract the cube root of
18(p 3 +p 2 q +pq 2 + y 5 ) ±2 V3(5/ + 3p 2 q  Zpq 2  5? 3 ).
(45.) Show that X can be determined so that a 4 + 6a? + 7a; 2  6a; + X shall be
an exact square.
(46.) Find a, b, c, so that x R  8z s + ax 4 + bx* + ex 2  Ux + i shall be an
exact square.
(47.) If ax 4 + bx 3 + ca 2 be subtracted from (x 2 + 2a + 4) 3 the remainder is an
exact square ; find a, b, c.
(48. ) If a^ + ax 5 + tar 4 + car 3 + dx 2 + ex +/ be an exact square, show mat
d = ^\« 4  %a 2 b + \ b 2 + \ac,
e =  ^a 5 + \a?b  \a 2 c  \ab 2 + \be,
<
220 EXERCISES XV
CHAP. XI
And that the square root is
ar* + iaa? + (  «. 2 + \b)x + (&a? \db+ c).
(49.) 4a5 6 +12.r i + 5,r 4 2.*: :, are the first four terms of an exact square ; find
the remaining three terms.
(50.) If x^ + Sdx^ + ex^fx^ + gx^ + hx + P be a perfect cube, find its cube
root ; and determine the coefficients e,f, g, h, in terms of d and k.
(51.) Show that
b%a b)(c b){(a 6) 2 + (c6) 2 }  ffJM^ + O + **(«• & + c)
is an exact cube.
(52.) Express \/{l+x + a?+a?+. . . ad »} in the form a + bx + cx +
... as far as the 4th power of x. To how many terms does the square of
your result agree with 1 + x + x 2 + x 3 +. . .?
(53.) Express, by means of the ordinary rule for extracting the square
root, xAl ~ ■*') "s an ascending series of integral powers of x, as far as the
4 th power.
(54.) Express \J(x+l) as a descending series of powers of x, calculating
six terms of the series.
(55.) Show that Lambert's theorem (chap, ix., § 9) can be used to find
rational approximations to surd numbers. Apply it to show that \/'2 = \ + 1/2
 1/2.5 + 1/2.5.7  1/2.5.7.197 approximately ; and estimate the error.
CHAPTEE XII.
Complex Numbers.
ON THE FUNDAMENTAL NATURE OF COMPLEX NUMBERS.
§ 1.] The attempt to make certain formulae for factorisation
as general as possible has already shown us the necessity of in
troducing into algebra an imaginary unit i, having the property
i" = — 1. It is obvious from its definition that i cannot be equal to
any real quantity, for the squares of all real quantities are positive.
The properties of i as a subject of operation are therefore to be
deduced entirely from its definition, and from the general laws
of algebra to which, like every other algebraical quantity, it
must be subject.
Since i must, when taken along with other algebraical quan
tities, obey all the laws of algebra, we may consider any real
multiples of i, say yi and y'i, where y and y' are positive or
negative, and we must have yi = iy, yi + y'i  (y + y')i = i(y + y'),
and so on ; exactly as if i were a real quantity.
By taking all real multiples of i from  go i to + go i, we have
a continuous series of purely imaginary quantity,
 oo i . . .  i . . . Oi . . . + i . . . + cc i I.,
whose unit is i, and which corresponds to the series of real
quantity,
co .. .1. . .0. . . + 1.. . + oo II.,
whose unit is 1.
No quantity of the series I. (except Oi) can be equal to any
quantity of the series II., for the square of any real multiple of ?',
say yi, is y 2 i 2 = y 2 (  1) = if, that is, is a negative quantity.
222 FUNDAMENTAL CHAKACTER OF COMPLEX NUMBERS chap.
Hence no purely imaginary quantity except Oi can be equal to a real
quantity. Since Oi = ( + a  a)i = + (ai)  (ai) = 0, if the same
laws are to apply to imaginary as to real quantity, Ave infer that
Oi = 0. Hence is the middle value of the series of purely
imaginary, just as it is of the series of real quantity ; it is, in
fact, the only quantity common to the two series.
Conversely, if yi = 0, we infer that y = 0. For, since yi = 0,
yixyi = 0, that is,  y* = ; hence y = 0.
§ 2.1 If we combine, by addition, any real quantity x with a
purely imaginary quantity yi, there arises a mixed quantity x + yi,
to which the name complex number is applied.
We may consider the infinite series of complex numbers
formed by giving all possible real values to x, and all possible
real values to y. We thus have a doubly infinite series of com
plex quantity. The student should note at the outset this double
character of complex quantity, on account of the contrast which
thus arises between purely real or purely imaginary quantity
on the one hand, and complex quantity on the other. Thus there
is only one way of varying z continuously (without repetition of
intermediate values) from  1 to + 2, say, if z is to be always
real ; and only one way of varying z in like manner from  i to
+ 2i, if z is to be always purely imaginary. But there are an
infinite number of ways of varying z continuously from 1+4
to 2 + 3i, say, if there be no restriction upon the nature of z,
except that it is to be a complex number.
This will be best un
derstood if we adopt
the diagrammatic method
of representing complex
numbers introduced by
Argand.
Let XOX', YOY' be
two rectangular axes. We
shall call XOX' the axis of
Fig. i. real quantity, or zaxis ;
and YOY' the axis of purely imaginary quantity, or yaxis. To
xil argand's diagram 223
represent any complex number x + yi we measure from (called
the origin) a distance OM, containing x units of length, to the
right or left according as x is positive or negative ; and we draw
MP, containing y units of length, upwards or downwards accord
ing as y is positive or negative. The point P, or, as is more
convenient from some points of view, the " radius vector " OP, is
then said to represent the complex number x + yi. It is obvious
that to every conceivable complex number there corresponds one
and only one point in the plane of XX' and YY' ; and, conversely,
that to every one of the doubly infinite series of points in that
plane there corresponds one and only one complex number. P
is often called the ajfixe of x + yi, or simply the " Point x + yi."
If P lie on the axis XX', then y = 0, and the number x + yi is
wholly real. If P lie on the axis YY', then x = 0, and x + yi is
wholly imaginary. Now there is only one way of passing from
any point on XX' to any other point, if we are not to leave the
axis, namely, we must pass
along the rraxis ; and the
same is true for the axis YY'.
If, however, we are not re
stricted as to our path, there
are an infinity of ways of
passing from one point in the X' O
plane of XX' and YY' to any
other point in the same plane. Y
If we draw any continuous Fi o 
curve whatever from P to Q, and imagine a point to travel along
it from P to Q, the value of x corresponding to the moving point
will vary continuously from the value OM to the value ON, and
the value of y in like manner from MP to NQ. Hence there are
as many ways of varying x + yi from OM + MPi to ON + NQi as
there are ways of drawing a continuous curve from P to Q.
Similar remarks apply when P and Q happen, as they may
in particular cases, to be both on the xaxis, or both on the
yaxis, provided that there is no restriction that the varying
quantity shall be always real or always imaginary. There are
Q
PM QN
224 RATIONAL OPERATIONS WITH COMPLEX NUMBERS CHAi'.
many other properties of complex numbers, which are best under
stood by studying Argand's diagram, and we shall return to it
again in this chapter. In the meantime, however, to prevent
confusion in the mind of the reader, we shall confine ourselves
for a little to purely analytical considerations.
§ 3.] If x + yi = 0, then x = 0, y = 0.* For it follows from
x + yi = that x =  yi. Hence, if y did not vanish, we should
have a real quantity x equal to a purely imaginary quantity  yi,
which is impossible. We must therefore have y = ; and conse
quently x—  Oi = 0.
Cor. Hence if x + yi = x + y'i, then must x = x and y = y .
For x + yi = x' + y'i gives, if we subtract x + y'i from both sides,
(x — x) + (y — y')i = 0.
Hence x — x' — 0, y — jf = 0,
that is, x = x', y = y'.
RATIONAL FUNCTIONS OF COMPLEX NUMBERS.
§ 4.] We have seen that so long as we operate upon real
quantities, provided we confine ourselves to the rational opera
tions — addition, subtraction, multiplication, and division, we
reproduce real quantities and real quantities only. On the
other hand, if we use the irrational operation of root extraction,
it becomes necessary, if we are to keep up the generality of
algebraical operations, to introduce the imaginary unit i. We are
thus led to the consideration of complex numbers. The ques
tion now naturally presents itself, " If we operate, rationally or
irrationally, in accordance with the general laws of algebra on
quantities real or complex as now defined, shall we always re
produce quantities real or complex as now defined ; or may it
happen that at some stage it will be necessary in the interest of
algebraic generality to introduce some new kind of imaginary
quantity not as yet imagined 1 " The answer to this question
is that, so far at least as the algebraical operations of addition,
* Here and hereafter in this chapter, when we write the form x + yi, it is
understood that this denotes a complex number in its simplest form, so that
x and y are real.
XII RATIONAL FUNCTIONS OF COMPLEX NUMBERS 225
subtraction, multiplication, division, and root extraction are
concerned, no further extension of the conception of algebraic
quantity is needed. It is, in fact, one of the main objects of the
present chapter to prove that algebraic operations on complex
numbers reproduce only complex numbers.
§ 5.] The sum or product of any number of complex numbers, and
the quotient of two complex numbers, may be expressed as a complex
number.
Suppose we have, say, three complex numbers, a;, + yj,, x 2 + y 2 i,
x 3 + y 3 i, then
(«i + ffii) + (x 2 + y 2 i)  (x 3 + y 3 i) = (.r, + x 2  x 3 ) + (y, + y 2  y 3 )i,
by the laws of algebra as already established.
But x l + x 2  x 3 and y x + y 2  y 3 are real, since x x , x 2 , x 3 , y x , y 2 , y 3
are so. Hence (#, + x, 2  x 3 ) + (y x +y 2  y 3 )i is in the standard form
of a complex number. The conclusion obviously holds, however
many terms there may be in the algebraic sum.
Again, consider the product of two complex numbers, x x + yj,
and x 2 + yj. We have, by the law of distribution,
(«i + yd) («» + yd) = x&i + y<ij£ + w + <w
Hence, bearing in mind the definition of i, we have
(a + yd) ( r 2 + yj) = to  ysj») + to + «Wi)i,
which proves that the product of two complex numbers can be
expressed as a complex number.
To prove the proposition for a product of three complex
numbers, say for
P = (.r, + yj) (x, + yj) (x 3 + y 3 i),
Me have merely to apply the law of association, and write
P = {(»i + yj) («s + yd)} («3 + yd).
We have already shown that the function within the crooked
brackets reduces to a complex number ; hence P is the product
of two complex numbers. Hence, again, by what we proved
above, P reduces to a complex number. In this way we can
extend the theorem to a product of any number of complex
numbers.
VOL. I Q
226 RATIONAL FUNCTIONS OF COMPLEX NUMBERS chap.
Lastty, consider the quotient of two complex numbers. We
have
Si + ffii _ («i + yj) («»  yJ) *
x 3 + y 2 i (x 2 f  (y 2 if '
= (^gg + vm)  fay»  %2!ti)i
a* + y*
x x x 2 + y^/nX f^iVi ~
2
Xo t y^ / \ *^2 "■"
— S~ ]'
y 2 /
This proves that the quotient of two complex numbers can
always be reduced to a complex number.
Cor. 1 . Since every rational function involves only the opera
tions of addition, subtraction, multiplication, and division, it
follows from what has just been shown that every rational
function of one or more complex numbers can be reduced to a com
plex number.
Cor. 2. If <f>(x + yi) be any rational function of x + yi, having all
its coefficients real, and if
<j)(z + yi) = X + Yi,
then
<f>(x  yi) = X  Yi,
X and Y being of course real.
Cor. 3. Still more generally, if <£(», + yj, x 2 + y 2 i, . . . , x n + y n i)
be any rational function of n complex numbers, having all its coefficients
real, and if
4>{x, + y t i, x 2 + y 2 i, . . . , x n + y n i) = X + Yi,
then
^{x.y.i, x 2 y 2 i, . . . , x n y, l i) = XYi.
Cor. 4. If all the coefficients of the integral function <j>(z) be
reed, and if cj>(z) vanish when z = x + yi, then <f>(z) vanishes wlien
z = xyi.
' Here we perform an operation which we might describe as "realising"
the denominator ; it is analogous to the process of rationalising described in
chap. x.
xir EXAMPLES 227
For, by Cor. 1, <f>(x + yi) = X + Yi where X and Y are real.
Hence, if cf>(x + yi) = 0, we have X + Yi = 0. Hence, by § 3, X =
and Y = 0. Therefore <j>(x  yi) = X  Yi =  0/ = 0.
Cor. 5. If all the coefficients of the integral function <£(2„
z s , . . . , z n ) be real, and if the function vanish when z u z.,, . . . , z n
are equal to x x + yj,, x 2 + y 2 i, . . . , x n + y n i respectively, then the
function will also vanish when s u z. 2 , . . . , z n are equal to x x  y x i,
x s  yi 1 , ' • ■ j x n  Vni respectively.
Example 1.
3(3 + 20  2(2  3i) + (6 + 80 = 9 + 6i 4 + 6J+6+ Si,
til + 2Qi.
Example 2.
(2 + 30 (2  50 (3 + 20 = (2  50 (6  6 + 9i + 40,
= (2 5013;',
= 26i + 65,
= 65 + 26i.
Example 3.
(b + c  ai) [c + a  bi) (a + b ci)
= {U(b + c)2bc(b + c)\ + {abcZa{a + b){a + c)}i,
 2abc + {abc  2a 3  2a 2 (6 + c)  Zabc} i,
— 2abc  {a 3 + b 3 + c 3 + (b + c) [c + a) {a + b)} i.
Example 4.
To show that the values of the powers of i recur in a cycle of 4.
We have i=i, i 2 =l, i 3 = i 2 xi= i, i i =(i~)=+l !
i s =i 4 x i = i, i 6 = i 4 x i=  1, i 7 = i 4 x i 3 —  i, i 8 = { 4 x i 4 = + 1 ;
and, in general,
i4n+l = i } i*H2=l, $*"+•= t, l' 4 ("+ 1 )=+l.
Example 5.
3 + 5i _ (3 + 5Q(2 + 3Q _ 615 + 19i_ 9 19.
23i~ 4 + 9 13 13 + 13*'
Example 6.
(x + yi)" = x" + „Ci x n ~Hyi) + „C 2 8» 2 (yi) a + • • • .
=(.?;«, a z"v 2 +»c.iz"y. • ■)
+ (nC 1 x» 1 y n C 3 x n  3 y 3 + „C 5 x n hj 5 . . .)*■
In particular
(jb + yt)*={a?  Sxy° + 2/ 4 ) + (4afy  to*/ 8 )*.
Example 7.
If <p(z) = Z " '
then «/>(2 + 30 :
'z + z + V
(2 + 3Q 2 (2 + 3Q + l
: "(2 + 30 2 + (2 + 30 + l'
5 + 12t23i + l
5 + 12i + 2 + 3i+l'
228 CONJUGATE COMPLEX NUMBERS CHAP.
_j6 + 9i _ 3(2 + 3 Q(215Q
2 + 15i~ 229
= 2§9 {4 + 45_6 ' +30 ^'
_U7 _72 .
229 229 *'
From this we infer that
,., ... 147 72 .
^ (2  3l) = 229229 i;
a conclusion which the student should verify by direct calculation.
CONJUGATE COMPLEX NUMBERS, NORMS, AND MODULI.
§ 6.] Two complex numbers which differ only in the sign
of their imaginary part are said to be conjugate. Thus — 3 — 2i
and  3 + 2t are conjugate, so are  ii and + ii ; and, generally,
x + yi and x  yi.
Using this nomenclature we may enunciate Cor. 3 of § 5
as follows : —
If the coefficients of the rational function <£ be real, then
the values of
4fa + yj, x 2 + y.j, . . ., x n + y n i)
and </>(' r i*/A «»yA ■ • •,  r n ~ ?/n0>
where the values of the variables are conjugate, are conjugate
complex numbers.
The reader will readily establish the following : —
The sum and the product of two conjugate complex numbers are real.
Conversely, if both the sum and the product of two complex numbers
be real, then either both are real or they are conjugate.
§ 7.] By the modulus of the complex number x + yi is meant
+ J(x 2 + y 2 ). This is usually denoted by  x + yi \*
It is obvious that a complex number and its conjugate have the
same modulus; and thai this modulus is the positive value of the
square root of their product.
Examples.
3 + 4i=+ N /{(3) 2 M 2 }=5.
34;= + v /{(3) 2 + (4) 2 }=5.
l+i= + v /(l 2 +l 2 )=v/2.
* Formerly by mod {x + yi).
XII
MODULI 229
It should be noticed that if y = 0, that is, if the complex
number be wholly real, then the modulus reduces to + Jx 2 ,
which is simply the value of x taken with the positive sign, or,
say, the numerical value of x. For example,   3  = + \/(  3) 2
= + 3. For this reason Continental writers frequently use  x 
where x is a real quantity, as an abbreviation for " the numerical
value of SB." We shall occasionally make use of this convenient
contraction.
For reasons that will be understood by referring once more to § 2, the
ordinary algebraical ideas of greater and less which apply to real quantities
cannot be attached to complex numbers. The reader will, however, find that
for many purposes the measure of the " magnitude " of a complex number is
its modulus. We cannot at the present stage explain precisely how "magni
tude" is here to be understood, but we may remark that, in Argand's diagram,
the representative points of all complex numbers whose moduli are less than
p lie within a circle whose centre is at the origin and whose radius is p.
§ 8.] If a complex number vanish its modulus vanishes ; and,
conversely, if the modulus vanish the complex number vanishes.
For if x + yi = 0, then by § 3, x = and y = 0. Hence
*J(x* + f) = 0. '
Again, if J(x" + y 2 ) = 0, then x 2 + y 2  ; but, since both x
and y are real, both x 2 and y 2 are positive, hence their sum cannot
be zero unless each be zero. Therefore x = and y = 0.
If two complex numbers are equal their moduli are equal ; but the
converse is not true.
For, if x + yi = x' + y'i, then, by § 3, x = x\ y = y. Hence
J(x 2 + f) = V(x' 3 + y' 2 ).
On the other hand, it does not follow from J(x 2 + y 2 )
= J(x' 2 + y' 2 ) that x = x', y= y. Hence the converse is not true.
§ 9.] Provided all the coefficients in <£(x' + yi) be real, we have
seen (§ 5, Cor. 2) that if
<p(x + yi) = X + Yi,
where X and Y are real, then
c/,(.c  yi) = X  Yi.
Now \cf>(x + yi)\= v/(X 2 + Y 2 )= J{(X + Yi)(XYi)},
= J{<f>(x + yi) <}>(xyi)},
230 MODULI
CHAP.
In like manner it follows from § 5, Cor. 3, that
I <t>fa + yj, %a + thh • ■ ■, x n + yd) 
= I <M«i  Vih x,  yj, . . ., x n  y n i) \
= + s /[<i>{x x + y x i, x 2 + y 2 i, . . ., x n + yj)
x <f>( Xl  y,i, x 2  yj, . . ., x n  y n i)] (2).
The theorems expressed by (1) and (2) are very useful in
practice, as will be seen in the examples worked below.
It should be observed that (1) contains certain remarkable
particular cases. For example,
I («! + yj) fa + yd) • • • fa + yj) I
= + ^[(jc, + yj) (x 2 + yj) . . . fa + y n i)
* ( g i  yj) fa  y a i) • • • fa  y j)\
= + Vfa + y?) fa + y 2 ) . . . (x n 2 + y n 2 ),
= I fc + yj) I x I fa + yd) I x • • • x I fa + yj) I (3).
In other words, the modulus of the 'product of n complex numbers
is equal to the product of their moduli.
Also
g, + yj>
x 2 + y 2 i
= \/fa* + y. 2 ) _ ( g. + yj 1
" \/(x 2 +y 2 2 ) \x 2 + y 2 i\ (4).
In other words, the modulus of the quotient of two complex numbers
is the quotient of their moduli.
§ 10.] The reader should establish the results (3) and (4) of
last paragraph directly.
It may be noted that we are led to the following identities : —
fa + y?) fa + y.f) = fac,  yfo)' + {x,y 2 + ay/,) 2 .
If we give to x„ y u x 2 , y 2 positive integral values, this gives us
the proposition that the product of hoo integers, each of which is the
sum of two square integers, is itself the sum of two square integers ;
and the formula indicates how one pair of values of the two
integers last mentioned can be found.
Also
fa 2 + y?) fa 2 + y 2 ) fa + y.*) = zflp,  x,yjj 3  xjj,y,  x^y,)*
+ (yiWe + y&&i + y<FiX e  2W 3 ) 2 
XII
MODULI 231
This shows that the product of three sums of two integral squares
is the sum of two integral squares, and shows one way at least of
finding the two lastmentioned integers.
Similar results may of course be obtained for a product of
any number of factors.
Example 1.
Find the modulus of (2 + 3i) (3  2i) (6  4i).
(2 + 8i)(32i)(64t)
=(2+3*)[x(32*)[x(64i) >
= s f\lZ)x ^(13) x ^(52),
=26^/(13).
Example 2.
Find the modulus of ( ^2 + ^3) (s/3 + » V 5 )/(V2 + W&)
( v /2 + rV3)(V3 + tV5) 
/r /( v / 2 + i v /3)(V3 + ^ v /5) \ r (N/2W3)( x /3i\/5) \
= VLl" >/2 + iV5 J l V2W5 <U'
 V[ (?± l^]= 7(f)
Example 3.
Find the modulus of {(/3 + 7) + (/3  7)1} {(7 + a) + (7  a) i} {(a + /3) + (a  /3)i} .
The modulus is J({((HyT + (Py)} {(y + *? + {ya.f\ {(a + (3f + (a p)*})
= v/{8(^ + 7 2 )(7 2 +a 2 )(a 2 + ^)}.
Example 4.
To represent 26 x 20 x 34 as the sum of two integral squares.
Using the formula of § 10 we have
26 x 20 x 34 = (l 2 +5 2 ) (2 2 + 4 2 ) (3 2 + 5 2 ),
= (1.2.3 1.4.5 2.5.5 3.5.4) 2 + (5.2.3 + 4.3.1 + 5.1.2 5.4.5) 2 ,
= 124 2 + 48 2 .
§11.] The modulus of the sum of n complex numbers is never
greater than the sum of their moduli, and is in general less.
This may be established directly ; but an intuitive proof will
be obtained immediately from Argand's diagram.
§ 12.] We have seen already that, when PQ = 0, then either
P = or Q = 0, provided P and Q be real quantities. It is
natural now to inquire whether the same will hold if P and Q
be complex numbers.
If P and Q be complex numbers then PQ is a complex
number. Also, since PQ = 0, by § 8,  PQ  = 0. But  PQ 
=  p i*IQI» b y § 10  Hence P x IQI = 0. Now P and
232 argand's DIAGRAM chap.
I Q  are both real, hence either  P  = or  Q  = 0. Hence, by
§ 8, either P = or Q = 0.
We conclude, therefore, that if PQ = 0, then either P = or
Q = 0, whether P and Q be real quantities or complex numbers.
discussion of complex numbers by means of
argand's diagram.
§ 13.] Returning now to Argand's diagram, let us consider
the complex number x + yi, which is represented by the radius
vector OP (Fig. 1). Let OP, which is regarded as a signless
magnitude, or, what comes to the same thing, as always having
the positive sign, be denoted by r, and let the angle XOP,
measured counterclockwise, be denoted by 6.
We have seen that if OP represent x + yi, then x and y are
the projections of r on X'OX and Y'OY respectively. Hence
we have, by the geometrical definitions of cos 6 and sin 9,
r= + V(r + /) (1),
x/rcos$, y/r = sin d, (2).
From (1) it appears that r, that is OP, is the modulus of the
complex number. The equations (2) uniquely determine the
angle 6, provided we restrict it to be less than 2ir, and agree
that it is always to be measured counterclock wise from OX.*
We call 6 the amplitude of the complex number. It follows
from (2) that every complex number can be expressed in terms
of its modulus and amplitude ; for we have
x + yi = r(cos 6 + i sin 6) (3).
This new form, which we may call the normal form, possesses
many important advantages.
* Sometimes it is more convenient to allow 8 to increase from  ir to + w ;
that is, to suppose the radius OP to revolve counterclockwise from OX' to
OX' again. In either way, the amplitude is uniquely determined when the
coefficients x and y of the complex number are given, except in the case of a
real negative number, where the amplitude apart from external considera
tions is obviously ambiguous.
XII
COMPOSITION OF VECTORS
233
Since two conjugate complex numbers differ only in the sign
of the coefficient of i, it follows
that the radii vectores which re
present them are the images of
each other in the axis of x (Fig. 3).
Hence two such have the same
modulus, as we have already shown
analytically ; and, if the amplitude
of the one be 6, the amplitude of
the other will be 2tt  6. In other
words, the amplitudes of two con
jugate complex numbers are con
jugate angles.
Example.
 1  i= V 2 ("^2^') = V2 ( C0S T + l ' sin T)
§ 14.] If OP, OQ' (Fig. 4) represent the complex numbers
Y
Q ,Q
x + yi and x' + y'i, and if
— >
PQ be drawn parallel and equal
to OQ', then OQ will represent
the sum of x + yi and x' + y'i.
For the projection of OQ on the
zaxis is the algebraic sum of the
Yi
I'lii. 4.
projections of OP and PQ on the
— >
same axis, that is to say, the projection of OQ on the aaxis is
— >
x + x. Also the projection of OQ on the yaxis is, by the same
reasoning, y + y'. Hence OQ represents the complex number
(x + x) + (y + y')i, which is equal to (x + yi) + (x' + y'i).
By similar reasoning we may show that if OP„ OP 2 , OP 3 , OP 4 ,
OP 5 , say, represent five complex numbers, and if P,Q, be parallel
234
 2i + Z 3 +
+ Zn\<
+
+ \Zm
CHAP.
and equal to 0P 2 , Q 2 Q 3 , parallel and equal to 0P 3 , and so on, then
0Q 5 represents the complex number which is the sum of the
complex numbers represented by OP,, OP„, OP 3 , OP 4 , OP 5 .
This is precisely what is known as the polygon law for com
pounding vectors. Since OQ 5 is never greater than the peri
meter OPjQ.jQaQ.jQs, and is in general less, Fig. 5 gives us an
intuitive geometrical proof that the modulus of a sum of complex
numbers is in general less than the sum of their moduli. It is
equally obvious from Fig. 4 that the modulus of the sum of two com
Fig. 5.
plex numbers is in general greater than the difference of the moduli.
" Sum of complex numbers " in these theorems means, of course,
algebraic sum.
§ 15.] If we employ the normal form for a complex
number, and work out the product of two complex numbers,
r,(cos 0, + i sin : ) and r t (cos 9 2 + i sin 2 ), we have
r^cos 0, + i sin 0])r»(cos 9 2 + i sin B 2 )
= r,r a {(cos 0, cos 6. 2  sin 0, sin $,) + (sin 0, cos 6. 2 + cos 9^ sin 9 2 )i),
= r,r 2 {cos (0, + 2 ) + i sin (0, + 9. 2 )} (1).
We thus prove that the product of two complex numbers is a
complex number, whose modulus r{r 2 is the product of the moduli
xii demoivre's theorem 235
of the two numbers, a result already established ; and we have
the new theorem that the amplitude of the product is, to a multiple
of 2tt, the sum of the amplitudes of the factors. For we can always
find an angle <£ lying between and 2rr, such that cos <f> =
cos (0, + 2 ) and sin (f>  sin {d x + 2 ), and we then have 6 Y + 6 3
= 2mr + cf>.
This last result is clearly general ; for, if we multiply both
sides of (1) by an additional factor, r 3 (cos 3 +• i sin 3 ), we have
r^cos Q x + i sin 1 )r 2 (cos 0„ + i sin 2 )r 3 (cos 3 + i sin 3 )
= r^cos (0, + 6 2 ) + i sin (0 t + 6 2 )} r 3 (cos 6 3 + i sin $ 3 ),
= r t r 2 r 3 {cos ($ x + d 2 + d 3 ) + i sin (0 1 + 6 2 + 3 )},
by the case already proved,
= ?W 3 {cos (0, + B 2 + 9 3 ) + i sin (0 : + 2 + 3 )}.
Proceeding in this way we ultimately prove that
r,(cos $! + i sin tf^r^cos 2 + i sin 2 ) . . . r n (cos 9 n + i sin 6 n )
= r{r 2 . . . ? n {cos(#! + 2 + . . . + 9 n ) + i sin(#, + 6 2 + . . . + 6 n )} (2).
This result may be expressed in words thus —
The pvduct of n complex numbers is a complex number whose
modulus is the product of the moduli, and whose amplitude is, to a
multiple of 2ir, the sum of the amplitudes of the n complex numbers.
If we put r l = r 2 = . . . = r n , each = 1 say, we have
(cos #! + i sin 0j) (cos # 3 + i sin 6 2 ) . . . (cos 6 n + i sin d n )
= cos (6, + 6 2 + . . . + tl ) + i sin (0, + 2 + . . . + 6 n ) (3).
This is the most general form of what is known as Demoivre's
Theorem.
If we put 0, = 6 2  . . .  6 n , each = 0, then (3) becomes
(cos + i sin 0) n = cos nd + i sin nd (4),
which is the usual form of Demoivre's Theorem.* It is an analy
tical result of the highest importance, as we shall see presently.
* This theorem was first given in Demoivre's Miscellanea Analytica
(Lond. 1730), p. 1, in the form
^ = i\ / {^ + \/(^l)}+i/\/{^+v'(^l)}, where ar = cos0, f = cos«0.
236 QUOTIENT OF COMPLEX NUMBERS CHAP.
Since cos  i sin  cos (2r  0) + i sin (2ir  0),
we have, by (3) and (4),
lT(cos 0,  i sin 0,) = cos (20,)  i sin (20,) (3') ;
and (cos  i sin 9) n = cos n0  i sin «0 (4').
The theorem for a quotient corresponding to (1) may be
obtained thus —
r(cos + i sin 0) _ r(cos + i sin ) (cos 0'  i sin 0')
/(cos 0' + i sin 0') = r'(cos 2 0' + sin a 0') '
=  {(cos cos 0' + sin sin 0')
+ (sin cos 0'  cos sin 0')i},
= , {cos (0  0) + * sin (0  $')} (5).
Hence $e quotient of two complex numbers is a complex number
ichose modulus is the quotient of the moduli, and whose amplitude is
to a multiple of 2ir the difference of the amplitude of the two complex
numbers.
IRRATIONAL OPERATIONS WITH COMPLEX NUMBERS.
§ 16.] Since every irrational algebraical function involves
only root extraction in addition to the four rational operations,
and since we have shown that rational operations with complex
numbers reproduce complex numbers and such only, if we could
prove that the nth. root of a complex number has for its value,
or values, a complex number, or complex numbers and such only,
then we should have established that all algebraical operations
with complex numbers reproduce complex numbers and such
only.
The chief means of arriving at this result is Demoivre's
Theorem ; but, before resorting to this powerful analytical engine,
we shall show how to treat the particular case of the square root
without its aid.
Let us suppose that
J(x + yi) = X + Yi (1).
xii SQUARE ROOTS OP A COMPLEX NUMBER 237
Then x + yi = X 2 Y 2 +2XYi.
Hence, since X and Y are real, we must have, by § 3,
X 2 Y 2 = .x (2),
2XY = y (3).
Squaring both sides of (2) and (3), and adding, we deduce
(X 2 + Y 2 ) 2 = x 2 + f •
whence, since X 2 + Y 2 is necessarily positive, we deduce
X 2 + Y 2 = + V'(/ + /) (4).
From (2) and (4), by addition, we derive
2X 2 = + >J(x 2 + y 2 ) + x,
u. v ■ Y* + \ (x 2 + y 2 ) + x
that is, A =  •
We therefore have X= ± y j 2 ) ^'
In like manner we derive from (2) and (4), by subtraction, &c,
T .;y{±i^b'} (6 ).
Since x 2 + y 2 is numerically greater than x 1 , + ^(x 2 + y 2 ) is
numerically greater than x. Hence the quantities under the sign
of the square root in (5) and (6) are both real and positive. The
values of X and Y assigned by these equations are therefore real.
Since 2XY = y, like signs must be taken in (5) and (6), or
unlike signs, according as y is positive or negative.
We thus have finally
(7).
(8),
if y be negative.
Example 1.
Express \/(8 + 6i) as a complex number.
Let ^{8 + 6i) = x + yi.
Then a~i/ 2 = 8, 2xy=6.
Hence (* 2 + ?/ 2 ) 2 = 64 + 36 = 100
Hence x ! + if=\0;
and x~y 2 = 8;
\
if y be positive ;
= ±
238 EXAMPLES chap.
therefore 2x 2 = 18, 2y° = 2.
Hence x=±3, y==hl.
Since 2a*y = 6, we must have either x — +3 and y=+l, or ar=3 and
y=l.
Finally, therefore, we have
V(8 + 6i)=±(3 + i);
the correctness of which can be immediately verified by squaring.
Example 2.
v<37o=±{y(»>^/(»)}.
Example 3.
Express \/( + i) and \/( ~ *) as complex numbers.
Let \/{+i)=3:+#*;
then i = ar  2/ 2 + 2a;?/i.
Hence a. 2 2/ 2 = (a), 2asy=l (0).
From (a) we have (x + y)(xy) = ; that is, either ?/= a or y = x. The
former alternative is inconsistent with (/3) ; hence the latter must be accepted.
We then have, from (/3), 2a?=l, whence £ 2 = l/2 and a;=±l/\/2. Since
!/=£, we have, finally,
V+*=±^" (7)
Similarly we show that
Example 4.
To express the 4th roots of + 1 and  1 as complex numbers.
^/ + 1 = V(V + 1) = V±l = V + l or «s/l = ±l or dbi.
Hence we obtain four 4th roots of + 1, namely, +1,  1, + i,  i.
Again v  1 = V(V  l)= V^
Hence, by Example 3,
§ 17.] We now proceed to the general case of the rcth root
of any complex number, r(cos 6 + i sin 6).
Since r is a positive number, Vr has (see chap, x., § 2)
one real positive value, which we may denote by r 1/n .
Consider the n complex numbers —
r Vn [ cos  + i sin  ) (1),
\ n ii/ x
2rr+e . . 2tt + 0N
A In I
( 2tt + 9 . . 2tt + 9\ /t . x
cos h i sin ) (2),
\ n n J
xii MTH BOOTS OF ANY COMPLEX NUMBER 239
.. / Att + . . 4 + 0\ ...
i ' " cos + t sin ) (o),
\ n n /
2stt + 6 . . 2sir + 8\ , . .
r 1 '" cos  + * sin   ) (s + 1 ),
n n J
r Vn (
2(>il) + e . . 2(n l)ir+0\ . ,
cos — + % sin — hi).
n n /
i r Vnf
No two of these are equal, since the amplitudes of any two
differ by less than 2ir. The ?ith power of any one of them is
r(cos 6 + i sin 6) ; for take the (s + l)th, for example, and we have
2$ir + 6 . . 2stt + 6
cos + ^ sm
n a
( ^.\ n ( 2STT+0 . . 2S7T+8\ n
= I r xjn ) [ cos + i sin
\ / \ n n
( 2sw + 6 . . 2S7T + 8
= r [ cos n + i sin n
\ n n
by Demoivre's Theorem,
= r(cos (25tt + 6) + i sin (2stt + 6) ),
= r(cos + i sin 6).
Hence the complex numbers (1), (2), . . ., (n) are n different
nth roots of r(cos 6 + i sin 6).
We cannot, by giving values to 5 exceeding n— 1, obtain
any new values of the ?tth root, for the values of the series
(1), (2), . . ., (h) repeat, owing to the periodicity of the
trigonometrical functions involved. We have, for example,
r ,/H (cos.(2?i7r + 6)1 n + i sin.(2?i + 6)jn) = r ln (cos.6!n + i sin. 6jn) ;
and so on.
We can, in fact, prove that there cannot be more than n
values of the ?ith root. Let us denote the complex number
r(cos + i sin 6) by a, for shortness ; and let z stand for any
mth root of a. Then must z n = a, and therefore z n  a = 0.
Hence every »th root of a, when substituted for z in z n  a,
causes this integral function of z to vanish. Hence, if «„ z s , . . .,
240 JITH ROOTS OF ± 1 CHAP.
~ s be s nth roots of a, zz u zz 2 > • • ■, % — %* will all be
factors of z n  a. Now z 11  a is of the ?(th degree in z, and
cannot have more than n factors (see chap, v., § 10). Hence s
cannot exceed n ; that is to say, there cannot be more than n
nth roots of a.
We conclude therefore that every complex number has n nth
roots and no more ; and each of these nth roots can be expressed as a
complex number.
Cor. 1. Since every real number is merely a complex num
ber whose imaginary part vanishes, it follows that every real
number, whether positive or negative, has n nth roots and no more,
each of which is expressible as a complex number.
Cor. 2. The imaginary nth roots of any real number can be
arranged in conjugate pairs. For we have seen that, if x + yi be
any nth root of a, then (x + yi) n  a = 0. Hence, if a be real (but
not otherwise), it follows, by § 5, Cor. 4, that (x  yi) n  a = ;
that is, x  yi is also an nth root of a.
N.B. — This does not hold for the roots of a complex number
generally.
§ IS.] Every real positive quantity can be written in the
form
r(cos + i sin 0) (A) ;
and every real negative quantity in the form
r(cos 7r + i sin 7r) (B) ;
where r is a real positive quantity. Hence, if we know the n
nth roots of cos + ism 0, that is, of +1, and the n nth roots of
cos 7r + isin7r, that is, of  1, the problem of finding the n nt\i
roots of any real quantity, whether positive or negative, is
reduced to finding the real positive value of the nth root of a
real positive quantity r (see chap, xi., § 15).
By means of the nth roots of + 1 we can, therefore, com
pletely fill the lacuna left in chap, x., § 2. In addition to their
use in this respect, the nth. roots of ± 1 play an exceedingly im
XII WTH ROOTS OF ±1 241
portant part in the theory of equations, and in higher algebra
generally. We therefore give their fundamental properties here,
leaving the student to extend his knowledge of this part of
algebra as he finds need for it.
Putting r1 and = in 1, . . ., n of § 17, and remem
bering that V ln = 1, we obtain for the n nth roots of + 1,
cos + i sin 0, cos — + % sin — , ,
n n
2(/i1)tt . . 2(?i1)tt
cos — J  — v i sin — — •
n n
Putting r=l, 6 = tt, we obtain for the nth roots of  l,
TV . . 7T StT . . StT
cos  + i sin , cos — + i sin — , ...,..,
n n n n
(2w 1W . . (2711W
cos — + i sin •
n n
Cor. 1. Since cos . 2(?i  l)ir/w = cos . 27r/«, sin . 2(n  l)irjn =
 sin . iTrjn; cos . 2(w  2)7r , ?i = cos . iir/n, sin . 2(u  2)7rjn =
 sin . 47r/», and so on, we can arrange the roots of +1 as
follows: —
nth roots of + 1, n even, =2m say,
2tt . . 2tt 4ir . . 47T
+ 1, cos — ± x sin — , cos — ± % sin — , . . .,
n n n n
^ a 2(^iy 2(>ulV .
cos ±isin , 1 (L) ;
7ith roots of + 1, n odd, = 2m + 1 say,
, 2ir . . 2tt 4:tt . . 4tt
+ 1, cos — ±zsin — , cos — ±tsin — , . . .,
n n n n
2nnr . . 2nnr tir ..
cos ± i sin (D).
n n v '
vol. I R
242 WTH ROOTS OF ± 1 chap.
Similarly we can arrange the roots of  1 as follows: —
reth roots of  1, « even, = 2m say,
cos  ± i sm , cos — ± i sin — , . . . ,
n n n n
(2m 1W . . (2m l>r
cos '— ± i sin ^ (E) :
n n v '
?i th roots of  1, n odd, = 2m + 1 say,
Ti . . 7T 37T . . 3~
cos  ± i sin , cos — ± i sin — , ....
n n n n
(2m 1W L . . (2m 1W
cos v '— ± t sin v — — .  1 (F).
11 n v '
From (C), (D), (E), (F) we see, in accordance with chap, x.,
§ 2, that the ??th root of + 1 has one real value if n be odd, and
two real values if n be even ; and that the »th root of  1 has
one real value if n be odd, and no real value if n be even.
We have also a verification of the theorem of § 17, Cor. 2,
that the imaginary roots of a real quantity consist of a set of
pairs of conjugate complex numbers.
Cor. 2. The first of the imaginary roots of + 1 in the series
(1), . . .,(»), namely, cos. 2wjn + i sin. 2ir/n, is called a primitive*
rtth root of + 1. Let us denote this root by w.
Then since, by Demoivre's Theorem,
•2tt . . 2ttY 2stt . . 2S7T
cos — + i sin —  cos — + % sm — ,
n n / it n
and, in particular,
1% n
it
2tt . . 2ir\
(o n = ( cos — + i sin — I == cos 2n + i sin 2tt,
1,
* By a primitive imaginary ?tth root of + 1 in general is meant an ?;th root
which is not also a root of lower order. For example, cos.27r/3 + isin.27r/3
is a 6th root of +1, but it is also a cube root of +1, therefore cos.27r/3f
»sin.2r/3 is not a primitive 6th root of +1. It is obvious that cos.27r/« +
■i sir\.2ir/n is a primitive «th root ; but there are in general others, and it may
be shown that any one of these has the property of Cor. 2.
XII FACTORISATION OF x n ± A 243
we see that, if <o be a primitive imaginary nth root of + 1, then the
n nth roots of + 1 are
to , u) , u) , . . . , w n (Gr).
Similarly, i/" w'  cos.ir/n + i sin.7r/?t, w/tic/i we may call a primitive
imaginary nth root of  I, then the n nth roots of  1 are
,./2nl /fj\
OJ
§ 19.] The results of last paragraph, taken in conjunction
with the remainder theorem (see chap, v., § 15), show that
Every binomial integral function, x n ± A, can be resolved into n
factors of the 1st degree, whose coefficients may or may not be wholly
real, or into at most two real factors of the 1st degree, and a number
of real factors of the 2nd degree*
Take, for example, x 2m  a 2m . This function vanishes whenever we sub
stitute for x any 2»ith root of a 2m ; that is, it vanishes whenever x has any
of the values aw, aw 2 , . . . , aw 2 '" where w stands for a primitive 2?)ith root
of +1.
Hence the resolution into linear factors is given by
To obtain the resolution into real factors, we observe that, corresponding
to the roots +a and —a, we have the factors xa, x + a; and that, corre
sponding to the roots a(cos.STr/m±i sin .sir/m), we have the factors
(St . . sir\ /
xa cos ai sin — II
m m J \
sir . . sw\
xa cos \ai sin — I,
vi m J
( SW\" „ . S7T
= I x  a cos — ) + a sin" — ,
\ to / TO
o Sir „
=x  2ax cos h a~.
m
Hence the resolution into real factors is given by
x 2m  a 2m =(x a)(x + a) (x 2  lax cos — + a 2 ) lx 2  lax cos ha 2 )...
m m
We may treat x 2m + a im , x 2m + l  a~ m+1 , and x 2m + 1 + a 2m+l in a similar way.
Example 1.
To find the cube roots of + 1 and  1. We have +1 =1 {cos O + i sin 0}.
Hence the cube roots of+1 are
cos + i sin 0, cos . 27r/3 ± i sin . 27r/3,
that is to say , + 1 ,  1/2± i*j3/2.
* The solution of this problem was first found in a geometrical form by
Cotes ; it was published without demonstration in the Harmonia Mensurarum
(1722), p. 113. Demoivre (Misc. Anal., p. 17) gave a demonstration, and also
found the real quadratic factors of the trinomial 1 + 2 cos Ox" + x' 2 ".
244 EXAMPLES chap.
Again 1 = 1 {cos ir + i sin jt} . Hence the cube roots of  1 are
cos . ir/3 ± i sin . ir/B, cos it + i sin w,
that is to say, l/2±i\j3f2,  1.
Example 2.
To find the cube roots of 1+i. We have l+t=V 2 (Vv^+ ll /V 2 )
= \/2(cos 45° + i sin 45°). Hence the cube roots of 1 + i are
2 J (cos 15° + i sin 15°), 2*(cos 135° + i sin 135°), 2*(cos 255° + i sin 255°),
that is,
2*(cos 15° + i sin 1 5°), 2 h {  cos 45° + i sin 45°), 2*(  cos 75°  i sin 75°),
that is,
i/V3 + l V3l\ 1/ 1 . 1 \ 2 1/ V31 \/3 + l\
2 r2V2 _+ l "2V2~> H"V2 + V2> U~~2^ l 2V2>
«,„*;. (V3 + l) + (V3l)f 1+t (V8l) + CV3 + l)t
mat is, j , j — , 2 .
93 93 9?
Here it will be observed that the roots are not arranged in conjugate pairs, as
they would necessarily have been had the radicand been real.
Example 3.
To find approximately one of the imaginary 7th roots of + 1. One of the
imaginary roots is
cos 51°25'43" + i sin 51°25'43".
By the table of natural sines and cosines, this gives
•6234893+ 7818318?;
as one approximate value for the 7th root of + 1.
Example 4.
If a be one of the imaginary cube roots of + 1, to show that 1 + w + co 2 =
and that {ax + a 2 y){a 2 x + ay) is real.
AVe have 1 +w + w 2 = (l  w 3 )/(l  w) = 0, since co 3 =l and lw#0.
Again,
{ax + a 2 y) {a 2 x + ay) = a*x 2 + {a 4 + a 2 )xy + uPy' 2 .
Now u 3 =l ; and w 4 + w 2 =w 3 u> + w a =« + a 2 — 1, since l+« + w 2 = 0.
Hence
( ax + a 2 y) {a 2 x + ay) = x 2 — xy + y 2 .
FUNDAMENTAL PROPOSITION IN THE THEORY OF EQUATIONS.
§ 20.] If /(2)sA + A,« + A/+ • • • +A„; n be an integral
function of z of the nth degree, whose coefficients A , A,, . . . , A w
are given complex numbers, or, in particular, real numbers, where, of
course, A a + 0, then f(z) can always be expressed as the product of n
factors, each of the 1st degree in z, say z  z l} zz.,, z  z 3 , . . .,
z  z n , ;.',, z 2 , . . ., z n being in general complex numbers.
xii FACTORISATION OF ANY INTEGRAL FUNCTION 245
It is obvious that this proposition can be deduced from the
following subsidiary theorem : —
One value of z, in general a complex number, can always he found
which causes f{z) to vanish.
For, let us suppose that /(s,) = 0, then, by the remainder
theorem, f(z) =f(z) (z  «,), where f(z) is an integral function of z
of the (n  l)th degree. Now, by our theorem, one value of z at
least, say z,, can be found for which f(z) vanishes. We have,
therefore, /,(&) = ; and therefore /,(*) =f a (z) (z  z 2 ), where f 2 (z)
is now of the (»  2)th degree ; and so on. Hence we prove
finally that
f(z) s k(z  *,) (z  z 2 ) ...(z z tl ),
where A is a constant.
§21.] We shall now prove that there is always at least one
finite value of z, say z = a, such that by taking z sufficiently near
to a, that is by making \z  a\ small enough, we can make
\f(z)  as small as we please. So that in this sense every
integral equation/^) = has at least one finite root.
Let  z I = R. Then, since
\f(z)  =  A.  R" 1 1 + A n _ JA n z + . . . + A jA n z n  ,
we have, by § 14,
\f(z)  >  A n  R»{1   A n _,/A n z + . . . + A /A,,*"  },
provided  z , (i.e. R), be large enough ; therefore
/(*)>A*R»{1C(1/R+ . . . + 1/R»)},
where C is the greatest of A n _,/A n , . . ., [A /A w . There
fore, taking provisionally R>1, we have
\f(z)  >  A.  R»{1  CXI  1/R»)/R(1  1/R)},
>A n R»{lC/(Rl)} (1),
provided R > C + 1 .
Hence, by taking \z\ sufficiently large, we can make \f(z)\ as large
as we please ; and we also see that there can be no root of f(z) 
whose modulus exceeds C + 1 .
Let now w be the value of z at any finite point in the
Argand Plane, so that \f(ic)  is finite. It follows from what has
246 EXISTENCE OF A ROOT CHAP.
just been proved that we can describe about the origin a circle
S of finite radius, such that, at all points on and outside S,
/(~) I > \f( w )  • Then, since \f(w) ] is real and positive, if we
consider all points within S, we see that there must be a finite
lower limit L to the value of \f(w) \ ; that is to say, a quantity
L which is not greater than any of the values of \f(w)  within S,
and such that by properly choosing u we can make \f(w)  = L + e,
where e is a real positive quantity as small as we please.
We shall show that L must be zero. For, suppose L > 0,
and choose w so that \f(w)  = L + e. Let h be a complex number,
say r(cos + i sin 6). Then
f(w + h)  A + A (w + h) + . . . + A n (w + h) n ,
=f(w) + BJi + BJf+ . . .+A n h n (2),
where A n is independent both of w and h, and by hypothesis
cannot vanish, but B„ . . ., B n _! are functions of w, one or more
of which may vanish. Suppose that B m is the first of the B's
that does not vanish, and let b m (cos a m + i sin <x m ), etc., be the
normal forms of the complex numbers B m /f(w), etc. Then,
since \f{w)  is not zero, b mi etc., are all finite. Also we have,
by Demoivre's Theorem,
f(w + h)ff(w) = 1 + b m r™e m + b m+1 r™+iQ m+i + . . . + 6 n r»6 n (3),
where m = cos {md + a m ) + i sin (rnd + a m ), etc.
We have h, and therefore both r and 6 at our disposal. Let
us first determine so that cos {m6 + a m ) =  1, sin (mO + a m ) = ;
that is, give 6 any one of the m values {it  a m )/m, (Stt — a m )/w, . . .,
(2m  1. 7r  a m )/m, say the first. Then we have Q m = 1;
and Q m+l , etc., assume definite values, say, G' m+1> etc. We now
have
f(w + h)/f(w) = 1  b m r™ + b m+1 f m + 1 0' m+1 + . . . + b n r"Q' n (A).
Considering the right hand side of (4) as the sum of 1  b m r m
and b m+l r m+i Q' m+l + . . . + b n r n Q' n , we see, by § 14, that the
modulus of f(w + h)jf(w) lies between the difference and the sum
of the moduli of these two. Also
XII
EXISTENCE OF A ROOT 247
\b m+l r™+ l Q , m+l + . . . + b n O' n r n \<b m+l r m+1 + • • • +&»'*
<b(r m+1 + . . . +r n ),
where b is the greatest of b m+l , . . ., b n ,
< («  m)br m +\
provided we take r< 1, so that r m+i >r m+2 > _ # § >r n
Therefore we have
I  b m r m (n m)b? m+l < \f(w + h)/f(w) 
< 1  6 TO r m +{n  m)br m+l (5) ;
provided r be so chosen that 1  b m r m and 1  b m r m  (n  m)br m+1
are both positive. Let us further choose r so that (n  m)h m+1
<b m r m . All these conditions will obviously be satisfied if we
give a finite value to r less than the least of the three,
1, l/(26 m ) lM . b m /(nm)b. (6).
When r is thus chosen, \f(w + h)/f(w)  will lie between two
positive proper fractions, so that \f(w + h)jf(w)  = 1 — /x, where /x
is a positive proper fraction ; and we have
/(w + A) = (l/t)/HI = (l/*)(L + € ) = L + 6  /t (L + e)(7).
which, since e may be as small as we please, is less than L by a
finite amount. L is therefore not a finite lower limit as sup
posed; in other words, L must be zero. Our fundamental
theorem is thus established.
By reasoning as above we can easily show that, if
f(z) = A + A il z*+ . . . +A n z n ,
then
\Ao\{l(ns+l)d\z\»}<\f(z)\
<\A \{l+(ns+l)d\z\°} (8),
where d is the greatest of  A«/A 1 , . . .,  A n /A 1 , provided [ z \
is less than the lesser of the two quantities 1, l/{(n  s + l)d} l > s .
Combining this result with one obtained incidentally above,
we have the following useful theorem on the delimitation of the
roots (real or imaginary) of an equation.
Cor. 1, The equation A + AgZ 8 + . . . + A n z n = can have no root
whose modulus exceeds the greatest of the quantities 1 +  A /A w  ,
248 GRADIENT AND EQUIMODULAR CURVES CHAP.
1 +  A s jA n  , . . ., 1 +  A n _JA n  , or whose modulus is less than the
least of 1, l/{(n  s + 1)  A s /A 1 }V°, . . ., l/{(«  s + 1)  A n /A 1 }**.
Cor. 2. JFe ca» always assign a positive quantity ?;, such that, if
 A<?7, /(2 + h) f(z) <e, w/tere e is a positive quantity as small as
we please.
This is expressed by saying that the integral function f(z) is
continuous for all complex values of its argument which have a
finite modulus. The proof is obvious after what has already
been done.
The above demonstration is merely a version of the proof given by
Argand in his famous Essai, * amplified to meet some criticisms on the briefer
statement in earlier editions of this work. The criticisms in question touch
the formulation, but not the essential principle of Argand's proof, which is
both ingenious and profound. As some of the critics appear to me to have
missed the real point involved, perhaps the following remarks, which the
student will appreciate more fully after reading chap. xv. §§ 1719, and
chap, xxix., may be useful.
Taking any value of z=x + yi, let f(z) — u + vi, where u and v are real
functions of the real variables x and y. Plot u v and x y Argand
diagrams. Then to each point (x, y) there corresponds one point (u, v),
although it may happen that to one (u, v) point there correspond more than
one (x, y) point. If we start with any given point (x, y), and consider
\/( z ) I = \f( u2 + v2 )> it is obvious that the direction in which \f(z) \ varies most
rapidly is obtained by causing (x, y) to move in the x y plane, so that
(u, v) moves along the radius vector towards the origin in the u v plane ;
also that the rate of variation in the perpendicular direction is zero. If,
therefore, we trace one of the curves v\u = constant in the x y plane, the
tangent to this curve at every point z on it is the direction of most rapid
variation. We may call these curves the Gradient Curves of/(s).t
* See in particular his amplified demonstration given in a note in Ger
gonne's Ann. de Math. t. v. pp. 197209 (181415).
t These curves, together with the curves u^ + v 2 — constant, which we
may call the Equimodular Curves of /(c), possess a number of interesting
properties. Since the Equimodulars and Gradients are orthomorphosed (see
chap. xxix. § 36) from a series of concentic circles and their pencil of
common radii, they form two mutually orthogonal systems. Through
every given point (x' t y'), which is not the affixe of a root of /(s) = 0, there
passes one equimodular u 2 + v 2 = u" 1 + v' 2 and one gradient u'vv'u — 0.
Every gradient u'vv'u — passes through all the intersections of u — 0,
v — 0, i.e. through the affixes of all the roots of /(z) = 0. Near the root
points the equimodulars take the form of small ovals enclosing these points.
XII ARGAND'S PROGRESSION TO A ROOT 249
It may readily be shown that the process by which we pass from v to
w + h (—w v say) in the above demonstration simply amounts to passing a
certain distance along a tangent to the gradient curve through the point w.
We may repeat this process, starting from w x and passing along the tangent
to the gradient through w 1} and so on. We shall thus ha.vef(to)>f(w 1 )>
f{w ) > . . . This process we may call Argand's Progression towards the root
of an equation.
Since b n <b, b m r m + (n m)br m+l <\, and (nm)br m+1 <b m r'», it follows
that
r < { \A W ) I / 2 (' 1  m ) I A » I !• ' /(m+1) and r < ! B ™ I A* _ m ) I A » I •
The first of these conditions shows that the longest admissible steps of
Argand's Progression become smaller and smaller as we approach a root.
This is expressed by saying that the Progression becomes asymptotic as we
approach a root.
But the second condition shows that Progression also becomes asymptotic
as we approach a point at which Bj = ; or Bi = 0, B. 2 =:0 ; or B 1 = 0, B 2 =0,
B 3 = 0, etc. Such points are stationary points for the variation of \J\z)  on
any path which passes through them. They are also multiple points on the
gradient curves which pass through them ; so that at them we have 2, 3, 4,
etc., directions of most rapid variation of \f{z) \ .
If the original progression leads towards one of these points (for which of
course \f{iv) 4=0), we must infer its existence from the asymptotic approach,
and start afresh from that point along one of the tangents to the gradient
that passes through it ; or we may avoid the point altogether by starting
afresh on a path which docs not lead to it.
An interesting example is to take z 2  z + 1 = 0, and start from a point on
the real axis in the x y diagram. Argand's Progression will lead first to
the minimum point (1/2, 0) on the x axis ; then along a line parallel to the
y axis to the points, (1/2, ± *j3/2), which are the affixes of the two imaginary
roots. The diagram of chap. xv. § 19 will also furnish a curious illustration
by taking initial points on one or other of the two dotted lines.
It should be noted that the question as to whether /(a)  actually reaches
its lower limit is not essential in Argand's proof, if we merely propose to
show that a value of z can be found such that \f(z)  is less than any assigned
positive quantity, however small. Nor do we raise the question whether
the root is rational or irrational, which would involve the subtle question of
the ultimate logical definition of an irrational number (see vol. ii. (ed. 1900)
chap. xxv. §§ 2841).
§ 22.] We have now shown that in all cases
f(z) = A(z  2,) (z  gg) . . . (z  z n ),
where A is a constant.
z lf z,, . . ., Zn may be real, or they may be complex numbers
of the general form x + yi. They may be all different, or two
250 EQUATION OF WTA DEGREE HAS n ROOTS CHAP.
or more of them may be identical, as may be easily seen by
considering the above demonstration.
The general proposition thus established is equivalent to the
following : —
If f(z) be an integral function of z of the nth degree, there are
n values of z for which f(z) vanishes. These values may be real or
complex numbers, and may or may not be all unequal.
We have already seen in chap, v., § 16, that there cannot be
more than n values of z for which f(z) vanishes, otherwise all its
coefficients would vanish, that is, the function would vanish for
all values of z. We have also seen that the constant A is equal
to the coefficient A n . We have therefore the unique resolution
f(z) = An(z  z t ) (z  z 2 ) . . . (z  z n ).
§ 23.] If the coefficients of f(z) be all real, then we have
seen that if f(x + yi) vanish f(x  yi) will also vanish. In this
case the imaginary values among z u z 2 , . . ., z n will occur in
conjugate pairs.
If a + fii, a  (3i be such a conjugate pair, then, correspond
ing to them, we have the factor
(Z  a  (3i) (za + (3i) = (z af + ft,
that is to say, a real factor of the 2nd degree.
It may of course happen that the conjugate pair a ± fti is
repeated, say s times, among the values z„ z,, . . ., z n . In that
case we should have the factor (z  a) 2 + (3 s repeated s times ; so
that there would be a factor {(z  a) 2 + (3 2 } s in the function f(z).
Hence, every integral function of z, wlwse coefficients are all real,
can be resolved into a product of real factors, each of which is either
a positive integral power of a real integral f miction of the 1st degree,
or a positive integral power of a real integral function of the 2nd
degree.
This is the general proposition of which the theorem of § 19
is a particular case.
XII
EXERCISES XVI 951
Exercises XVI.
Express as complex numbers —
(1.) (a + bi) 8 + (abif.
(2) 1+i l 1 ~ i
2 + 36i f 726i
{4m) 6 + 8i + 34i'
u) (p±si\\(p^iY
\pqij \p + v i J
697V(15) + (V36V5)i
{ ' 3(V33V5)i
(6.) Show that
if n be any integer which is not a multiple of 3.
(7.) Expand and arrange according to the powers of a;
(as  1  i\J2) (x\ + i\/2){x2 + i^/3) (a;  2  i\/3).
(8. ) Show that
\(2abc)+i{bc)^B} 3 ={{2bca) + i(caWS} 3 .
(9.) Show that
{(V3 + 1) + (V3 W=16(l + i).
(10.) If £ + J?t be a value of x for which ax 1 + bx + c = 0, a, 6, c being all
real, then 2a£rj + br) = 0, a?? 2 = a£ 2 + &£ + c.
(11.) If #(a?+2/*)=X.+Y*, show that 4(X 2 Y 2 ) = x/X + y/Y.
(12.) If n be a multiple of 4, show that
l+2i + 3i 2 + . . . + (n + l)i n =%(n + 2ni).
(13.) Showthato + o 1 5 + . . . +a n z n \^\a n \ \z\ n (lnc/\z\), provided \z\
exceeds the greater of 1 and nc, and c is the greatest of \a Q [a„\, . . .,
I ««]/«« .
(14.) Find the modulus of
(23i)(3 + 4t )
(6+4i)(158i)"
(15.) Find the modulus of
{x+ J&+W
(16.) Find the modulus of
bc(b  ci) + ca(c  ai) + ab(a  bi).
(17.) Show that
1 1 + ix + i~x + iV + . . . ad »  = 1/ S !{1 + x 2 )', where as<l.
(18.) Find the moduli of (x + yi) n and (x + yi) n /(xyi) n .
Express the following as complex numbers : —
(19.) \/7 + 24f. (20.) \ f Q + iJ13.
(21.) *J  7/36 + 2t/3. (22.) s' lab + 2(o a  "Fjl
(23.) s!\+2x*]{x 2 \)i. (24.) Vl + i ^(x 4  1 ).
(25.) Find the 4th roots of  119 + 120;'.
252 EXERCISES XVI CHAP.
(26.) Resolve x* a 6 into factors of the 1st degree.
(27.) Resolve a; 5 +l into real factors of the 1st or of the 2nd degree.
(28.) Resolve x G + x 5 + x 4 +x 3 + x 2 +x + l into real factors of the 2nd
degree.
(29.) Resolve x 2m  2 cos 0a m x m + a 2m into real factors of the 1st or of the
2nd degree.
(30.) If w be an imaginary nth root of +1, show that l + w + w 2 + . . . +
co" 1 = 0.
(31.) Show that, if w be an imaginary cube root of +1, then
x 3 + y 3 + z 3  Sxyz = (x + y + z) (x + wy + w 2 z) (x + uPy + uz),
and
(x + wy + uhf + (x + w 2 y + wz) 3 = (2xy z) {2y z~x) {2zx y).
(32. ) Show that (x + y) m x m y m is divisible by x 2 + xy + y 1 for every odd
value of m which is greater than 3 and not a multiple of 3.
(33.) Show that
U„ 2/7T  T . 2nr\ („ . 2/7T „ 2rir\ . ^ n , v „..
Xcos Ysin ) + (Xsin + \ cos )i V = (\+\i) n .
n n J \ n n J J
(34.) Simplify
(cos 2d  i sin 20) (cos <£ + i sin 4>) 2 (cos 20 + i sin 20) (co s < p  i sin <f>) 2
cos (0 + <p) + i sin {0 + <p) cos (0 + 0) i sin (0 + 0)
(35.) If s /(a + bi)+ s /(c + di)= s /(x + yi), show that
(xac) 2 +(ybd) 2 = 4^J{(a 2 + b 2 )(e + d 2 )}.
(36.) Prove that one of the values of *J(a + bi) + £/(a  bi) is
v /[ J \2a + 2V(« 2 + b 2 )} + 2 Z'(a 2 + b 2 )\
(37.) If w = cos tt/7 + i sin tt/7, prove that {x  w) {x + w 2 ) (x  it?) (x + w 4 )
(x  w 5 ) (x + w 6 ) = x 6  x 5 + x 4  x 3 + x 2  x + 1 .
(38.) Find the value of w r 1 + w;'' 2 4 . . . +w r n ; w lt w 2 . . . w„ being the
nth roots of 1, and r a positive integer. What modification of the result is
necessary if r is a negative integer ?
Prove that 1/(1 +«)!«) f 1/(1 + «>2 a; ) +  • • +l/(l+w„a:)=M/(l a;").
(39.) Decompose 1/(1 +x + x 2 ) into partial fractions of the form aj(bx + c).
Hence show that
a:/(l + x + x 1 ) x  x 2 + x 4  x 5 + x 1  x 8 + . . . + x 3n + l  x 3n+2 + R,
where a^, x e , etc. , are wanting ; and find R.
(40.) Find the equation of least degree, having real rational coefficients,
one of whose roots is ^2 + i.
One root of x 4 + 3a; 3  30a; 2 + 366.*: 340 = is 3 + 5?:, find the other three
roots.
(41.) If a be a given complex number, and z a complex number whose
affixe lies on a given straight line, find the locus of the affixe of a + z.
(42.) Show that the area of the triangle whose vertices are the affixes of
ssii s 2 , z 3 is 2 {(z 2  z 8 )  z x  2 /Uzl} .
(43.) If 2 = (a + 7 cos 0) + i((i + y sin 0), where a, /3, y are constant and
variable, find the maximum and minimum values of  z  ; and of amp z
when such values exist.
XII
HISTORICAL NOTE 253
(44. ) If the affixe of x + yi move on the line Sx + iy + 5 = 0, prove that the
minimum value of\x + yi\ is 1.
(45.) If u and v are two complex numbers such that u = v+l/v, show
that, if the affixe of v describes a circle about the origin in Argand's diagram,
then the aflixe of u describes an ellipse (x 2 /a 2 + ?/ 2 /& 2 = l) ; and, if the affixe of
u describes a circle about the origin, then the affixe of v describes a quartic
curve, which, in the particular case where the radius of the circle described by
the affixe of u is 2, breaks up into two circles whose centres are on the iaxis.
(46.) If a; and y be real, and x + y = \, show that the affixe of xz l + yz 2 lies
on the line joining the affixes of z x and ».,. Hence show that the affixe of
xz x + yz 2 lies on a fixed straight line provided lx + my—l, I and m being
constants.
(47.) If £ + ij£ be an imaginary root of iE 3 + 2cc + 1 = 0, prove that (£, ij) is
one of the intersections of the graphs of ?j 2 = 3£ 2 +2 and ?? 2 = l/2 + 3/8£.
Draw the graphs : and mark the intersections which correspond to the roots
of the equation.
If o be the real root of this cubic, show that the imaginary roots are
Ha+*V(23/a)}.
(48.) If £±r)i be a pair of imaginary roots of o?px + q = 0, show that
(if, tj) are coordinates of the real intersections of 3£ 2  i) 2 p = 0, 8^rj + 2p^
3q = 0. Hence prove that the roots of the cubic are all real, or one real and
two imaginary, according as 4p 3 < >27g a . What happens if 4p s =272 a ?
(49.) If x 3 + qx + r = Q has imaginary roots, the real part of each is posi
tive or negative according as r is positive or negative.
(50.) The cubic x 3  9a; 2 + 33a: 65 = has an imaginary root whose
modulus is N /13 ; find all its roots.
(51.) Find the real quadratic factors of x 2n + x 2 " 1 + ... +1 ; and hence
prove that
o„ • if ■ 27r . nir ,,_ , .
2 ' Sm 2„ + 1 Sin S+l ' " ' Sm 2^Tl= ^ 2 " + 1 >
(52.) Find in rational integral form the equation which results by
eliminating 6 from the equations x — a cos + b cos 30, y = a smd+b sin 36.
(Use Demoivre's Theorem.) Give a geometrical interpretation of your
analysis.
Historical Note. — Imaginary quantities appear for the first time in the works
of the Italian mathematicians of the 16th century. Cardano, in his Artie Magna
sive de Rcgvlis Algebraicis Liber Units (1545), points out (cap. xxxvii., p. 66)
that, if we solve in the usual way the problem to divide 10 into two parts whose
product shall be 40, we arrive at two formula? which, in modern notation, may
be written 5 + *J  15, 5  N '  15. He leaves his reader to imagine the meaning
of these "sophistic" numbers, but shows that, if we add and multiply them
in formal accordance with the ordinary algebraic rules, their sum and product
do come out as required in the evidently impossible problem ; and he adds
" hucusque progreditur Arithmetics subtilitas, cujus hoc extremum ut dixi adeo
est subtile, ut sit inutile." Bombelli in his Algebra (1522), following Cardano,
devoted considerable attention to the theory of complex numbers, more especially
in connection with the solution of cubic equations.
254 HISTORICAL NOTE
CHAP XII.
There is clear indication in the fragment De Arte Logistica (see above, p. 201)
that Napier was in possession to some extent at least of the theory. He was
fully cognisant of the independent existence of negative quantity ("quantitates
defectives minores nihilo"), and draws a clear distinction between the roots of
positive and of negative numbers. He points out (Napier's Ed., p. 85) that
roots of even order have no real value, either positive or negative, when the
radicand is negative. Such roots he calls " nugacia " ; and expressly warns
against the error of supposing that J_  9=  M 9. In this passage there occurs
the curious sentence, " Hujus arcani magni algebraici fundamentum superius,
Lib. i. cap. 6, jecimus : quod (quamvis a nemine quod sciam revelatum sit) quan
tum tamen emolumeuti adferat huic arti, et cseteris mathematicis postea patebit."
There is nothing farther in the fragment De Arte Logistica to show how deeply he
had penetrated the secret which was to be hidden from mathematicians for 200
years.
The theory of imaginaries received little notice \mtil attention was drawn to
it by the brilliant results to which the use of them led Euler (17071783) and his
contemporaries and followers. Notwithstanding the use made by Euler and
others of complex numbers in many important investigations, the fundamental
principles of their logic were little attended to, if not entirely misunderstood.
To Argand belongs the honour of first clearing up the matter in his Essai
sur une maniere de representer les quantites imaginaires dans les constructions
geometriques (1806). He there gives geometrical constructions for the sum and
product of two complex numbers, and deduces a variety of conclusions therefrom.
He also was one of the first to thoroughly understand and answer the question of
§ 21 regarding the existence of a root of every integral function. Argand was an
ticipated to a considerable extent by a Danish mathematician, Caspar Wessel, who
in 1797 presented to the Royal Academy of Denmark a remarkable memoir Om
Direktionens ancdytiske Betegning, et Forslig, anvendt fwncmmelig til plane og
sphaeriske Polygoners Opliisning , which was published by the Academy in 1799,
but lay absolutely unknown to mathematicians, till it was republished by the
same body in 1897. See an interesting address by Beman to Section A of the
American Association for the Promotion of Science (1897). Even Argand's results
appear to have been at first little noticed ; and, as a matter of history, it was
Gauss who first initiated mathematicians into the true theory of the imaginaries of
ordinary algebra. He first used the phrase com2>lex number, and introduced the
use of the symbol i for the imaginary unit. He illustrated the twofold nature of
a complex number by means of a diagram, as Argand had done ; gave a masterly
discussion of the fundamental principles of the subject in his memoir on Bi
quadratic Residues (1831) (see his Works, vol. ii., pp. 101 and 171) ; and furnished
three distinct proofs (the first published in 1799) of the proposition that every
equation has a root.
From the researches of Cauchy (17891857) and Riemann (18261866) on
complex numbers has sprung a great branch of modern pure mathematics, called
on the Continent function  theory. The student who wishes to attain a full
comprehension of the generality of even the more elementary theorems of algebraic
analysis will find a knowledge of the theory of complex quantity indispensable ;
and without it he will find entrance into many parts of the higher mathematics
impossible.
For further information we may refer the reader to Peacock's Algebra, vol. ii.
(1845) ; to De Morgan's Trigonometry and Double Algebra (1849), where a list
of most of the English writings on the subject is given ; and to Hankel's
Vorlesungen ilber die complexen Zahlen (1867), where a full historical account
of Continental researches will be found. It may not be amiss to add that the
theory of complex numbers is closely allied to Hamilton's theory of Quaternions,
Grassmami's Ausdehuuugslehre, and their modern developments.
CHAPTER XIII.
Ratio and Proportion.
RATIO AND PROPORTION OF ABSTRACT QUANTITIES.
§ 1.] The ratio of the abstract quantity a to the abstract quantity
b is simply the quotient of a by b.
When the quotient a s b, or a/h, or y is spoken of as a ratio,
it is often written a : b ; a is called the antecedent and b the con
sequent of the ratio.
There is a certain convenience in introducing this new name,
and even the new fourth notation, for a quotient. So far, how
ever, as mere abstract quantity is concerned, the propositions
which we proceed to develop are simply results in the theory
of algebraical quotients, arising from certain conditions to which
we subject the quantities considered.
If a > b, that is, if a  b be positive, a : b is said to be a ratio
of greater inequality.
If a < b, that is, if a  b be negative, a : b is said to be a ratio
of less inequality.
When two ratios are multiplied together, they arc said to be
compounded. Thus, the ratio aa' : bb' is said to be compounded of
the ratios a : b and a' : V.
The compound of two equal ratios, a : b and a : b, namely,
a 2 : b 2 , is called the duplicate of the ratio a : b.
Similarly, of : b 3 is the triplicate of the ratio a : 5.*
* Formerly a 2 : b was spoken of as the double of the ratio a : b. Similarly
.i *
\Ja : \/b was called the half or subduplicate of a : b, and a 2 : b* the sesquipli
cate of a : b.
256 PROPERTIES OF A RATIO chap.
§ 2.] Four abstract numbers, a, b, c, d, are said to be proportional
when the ratio a:b is equal to the ratio c : d.
We then write
a : b = c : d*
a and d are called the extremes, and b and c the means, of the pro
portion, a and c are said to be homologues, and b and d to be
homologu.es.
If a, b, c, d, e, f, &c, be such that a : b = b : c = i : d = d : e = e :f
= &c, a, b, c, d, e, f, &c, are said to be in continued proportion.
If a, b, c be in continued proportion, b is said to be a mean
proportional between a and c.
If a, b, c, d be in continued proportion, b and c are said to be
two mean proportionals between a and d ; and so on.
§ 3.] If b be positive, and a>b, the ratio a:b is diminished by
adding the same positive quantity to both antecedent and consequent ;
and increased by subtracting the same positive quantity (<b) from
both antecedent and consequent.
If a<b, the words "increased" and "diminished" must be inter
changed in the above statement.
a + x a b(a + x)  a(b + x)
For,
b + x b b(b + x)
x(b  a)
b(b + x)'
Now, if a > b, ba is negative ; and x, b, b + x are all positive
by the conditions imposed ; hence x(b  a)jb(b + x) is negative.
TT a + x a .
Hence 7 — ■ 7 is negative,
b + x b
, a + x a
that is, t < r.
' b + x b
a x a x(a  b)
Ac;ain, ■? t = 777 r
' b  x b b(b  x)
But, since a > b, ab is positive, and x and b are positive, and,
since x<b, bx is positive. Hence x(a  b)/b(b  x) is positive.
* Formerly in writing proportions the sign : : (originally introduced by
Oughtred) was used instead of the ordinary sign of equality.
Xlll PERMUTATIONS OF A PROPORTION 257
TT a — x a
Hence . > .
b — x b
The rest of the proposition may be established in like manner.
The reader will obtain an instructive view of this proposition
by comparing it with Exercise 7, p. 267.
§ 4.] Permutations of a Proportion.
(1),
(2),
(3),
(4).
If
a:b = c:d
then
b:a = d:c
a :c = b:d
and
c :a = d :b
For, from (1), we have
a c
b = d'
Hence
'b V
that is,
b_d
a c '
that is,
b : ad: c,
which establishes (2).
Again, from (1),
a c
1 = 7v
multiplying both sides by , we have
a b c b
b c d c
that is,  =  , ;
c d
that is, a:c = b: d,
which proves (3).
(4) follows from (3) in the same way as (2) from (1).
§ 5.] The product of the extremes of a proportion is equal to the
product of the means ; and, conversely, if the product of two quantities
be equal to the product of two others, the four form a proportion, the
extremes being tlie constituents of one of the products, the means the
constituents of the other.
VOL. J S
258
RULE OF THREE
For, if
a : b = c : d,
that is,
a c
b = d'
then
 x bd =  : x bd,
b d
whence
ad = be.
Again, if
ad = be,
then
ad/bd = bc/bd,
whence
a c
b~d'
CHAP.
Cor. If three of the terms of a proportion be given, the remaining
one is uniquely determined.
For, when three of the quantities a, b, c, d are given, the equation
ad = be,
which results by the above from their being in proportion, be
comes an equation of the 1st degree (see chap, xvi.) to deter
mine the remaining one.
Suppose, for example, that the 1st, 3rd, and 4th terms of the proportion
are I, %, and f ; and let x denote the unknown 2nd term.
Then *:<*=*:*;
whence %y.x = \x%.
Multiplying hy f , we have x=xf xf,
_ o
— 3TT
§ 6.] Relations connecting quantities in continued proportion.
If three quantities, a, b, c, be in continued proportion, then
2 72 72 2
a : c = a : b = b : c ;
and b = \Z(ac).
If four quantities, a, b, c, d, be in continued proportion, then
a:d = a a : b' = b' : c 3 = c 3 : d 3 ,
and b = 'i/(a\l), c = t/(ad*).
For the general proposition, see Exercise 12, p. 267.
XIII
DETERMINATION OF MEAN PROPORTIONALS
259
For, if
then
Therefore
whence
a : b
b :c,
a _b
b~~c
a b b b
~ X  =  X .
(!)•
c c c
c~ c 2 ~ Ir
Also ac = b 2 ,
whence b = \/(ac) (2).
Equations (1) and (2) establish the first of the two proposi
tions above stated.
Again, if
a\b = b:c = c:d,
then
abac
b c b d"
Also
a a
b = b'
hence
a a a a b c
that is,
a 3 a
b 3 = d'
therefore
a a 3 b 3 c a
d~b 3 ~c 3 ~d 3
Further, since
a a 3
d = b 3 '
b 3 — ci'd ;
whence
b = l/(a*d).
Also, since
a c 3
d d 3 '
c 3 = ad 2 ;
whence
c = l/{ a d 2 )
(3).
(4>
(5).
It should be noticed that the result (2) shows that the finding of a mean
proportional between two given quantities a and c depends on the extraction
of a square root. For example, the mean proportional between 1 and 2 is
V(l x 2) = V2 = 14142 . . .
260 DELTAN PROBLEM
CHAP.
Again, (i) and (5) enable us to insert two mean proportionals between two
given quantities by extracting certain cube roots. For example, the two
mean proportionals between 1 and 2 are
^(lx 2) =4/2 = 1*2599 . . .
and V / (lx2 2 ) = ^ =15874 . . .
Conversely, of course, the finding of the cube root of 2, which again corre
sponds to the famous Delian problem of antiquity, the duplication of the
cube, could be made to depend on the finding of two mean proportionals, a
result well known to the Greek geometers of Plato's time.
§ 7.] After what has been done, the student will have no
difficulty in showing that
(2).
if
a :b = c:d,
then
ma :mb = nc: nd
and
ma : nb = me : nd
§8.]
if
Also that
a l :b i = c l :d 1 ,
a ± \ o 2 = c 2 : « 2 >
a n '. o n = e n : a n ,
then
a,<i, 2 . . . a n
: 6,6 S . . . b n = ca . . . c n
■ d/h
Cor.
If
a • b = c : d,
then
a n . ^ = c n : d n .
.dn (1).
(Here n, see chap, x., may he positive or negative, integral or
fractional, provided a n , &c, be real, and of the same sign as a,
&c.)
§9.] // a:b = c:d,
then a±b :b = c±d:d (1),
a + b: a b = c + d : c  d (2),
la + mb : pa + qb = lc+ md \pc + qd (3),
la r + mb r : pa r + qb r = lc r + md r :pc r + qd" (4),
where I, m, p, q, r are any quantities, positive or negative.
xiil CONSEQUENCES OF PROPORTION 261
Also, if a, : b, = a 2 : b a = a z : b 3 = . . . = a n : i n ,
a, + a., + . . . + a n : 6 a + b 3 + . . . + &„ (5) ;
and afeo fo
;/(/,< + W + • • • + W) : W + U>/ + • ■ ■ + W) (6).
Though outwardly somewhat different in appearance, these
six results are in reality very much allied. Two different
methods of proof are usually given.
FIRST METHOD.
Let us take, for example, (1) and (2).
„. a c
bince T = ,
b d
in a i c
therefore = ± 1 = •= ± 1 :
b a
a±b c ± d
whence — = — = — 5— ;
b d
this establishes the two results in (1).
Writing these separately we have
a + b c + d
b d '
a  b c  d
b d '
whence
(a + b),
h I
' (ab) (c + d)/ (c d)
b d 1 d '
that is,
a+b c+d
ab c — d'
which establishes (2).
Similar
treatment
may be applied to the rest of the six
results.
2G2 EXPRESSION IN TERMS OF FEWEST VARIABLES chap.
SECOND METHOD.
Let us take, for example, (2).
Since a/b = c/d, we may denote each of these ratios by the
same symbol, p, say. We then have
a
c
whence
a = pb,
c = pd
Now,
using (a),
Ave have
a + b
pb + b
ab
" pbb'
b( P +l)
~b(piy
p + i
 p v
In exactly
the i
same way, we
! have
c + d
pd + d
cd
' pd d'
p+l
pV
Hence
a + b p +
ab p
1 c + d
1 " c  d'
Again
, let
us take (5).
We h:
ive
a,
a 2 a 3
Z h~b 3 ~ • •
. = — , each
i = P
say,
hence
(
h = p
K <k = ph,
. . ., a n
= pbi
l 5
therefore
«i
+ do
+ . . . +a n
pb l + pb. 2 +
b x + b. 2 +
• • •
+ pbn
h
+ K
+ . . . + b n
+ b n '
p(K + h +
. . .
+ b n )
b l + b i + .
+ b» '
hence
= P,
a,
6,
a 2
 &C  n  ffl
+ a 2 + . . .
+ a n
— iVL, — p — ,
+ b 2 + . . .
+ b n
(a).
XIII
GENERAL THEOREM 2G3
Finally, let us take (G).
Since a/ = ( t AY = p'V,
a/ = (ph 2 y = p >V, &c.
we have
y(ifi* + W+ ■ • • + W) = V(p r (iA r + W+ ■ ■ ■ +inb/)),
(see chap, x., § 4). It follows that
y(/,o, r + W + • ■ ■ + l n (hi r ) = _ a, = a* _ &(j
;/(ZA r + w + ■ ■ • + W« r ) p ^ ^
Of the two methods there can be no doubt that the second
is the clearer and more effective. The secret of its power lies
in the following principle : —
In establishing an equation, between conditioned quantities, if ice
first express all the quantities involved in the equation in terms of the
fewest quantities possible under the conditions, then the verification of
the equation involves merely the establishment of an algebraical identity.
In establishing (2), for instance, we expressed all the quantities
involved in terms of the three b, d, p, so many being necessary,
by § 5, to determine a proportion.
A good deal of the art of algebraical manipulation consists
in adroitly taking advantage of this principle, without at the
same time destroying the symmetry of the functions involved.
§ 10.] The following general theorem contains, directly or
indirectly, all the results of last article as particular cases ; and
will be found to be a compendium of a very large class of
favourite exercises on the present subject, some of which will
be found at the end of the present paragraph.
If <M''i> ■''■_•> • • ., x„) be any homogeneous integral function of the
variables it,, x. 2 , . . ., x a of the rth degree, or a homogeneous function
of degree r, according to the extended notion of homogeneity and
degree give<i at the foot of p. 73, and if
«! : b l = a.,: b.,= . . . = a n : b n ,
then each of these ratios is equal to
264 EXAMPLES chap.
This theorem is an immediate consequence of the property
of homogeneous functions given in chap, iv., p. 73.
Example 1.
AV Inch is the greater ratio, x 2 + y 2 : x + y, or xy":xy, x and y being
each positive ?
x 2 + y 2 x 2 y 2 _ (x 2 + y' 2 ) (xy) {x 2  y 2 ) (x + y)
x + y ~ xy ~ (x + y)(xy)
2xy 2  2x 2 y
~{x + y)(xy)'
_ 2xy(x  y)
{x + y)(xy)'
_ 2xy ^
x+y
Now, if x and y be each positive, 2xyj(x + y) is essentially negative.
Hence
x 2 + y 2 : x + y < x 2  y 2 : x  y.
Example 2.
If a:b=c;d, and A:B = C:D, then ci\/A  b\/B : c> s /Cd s /D = asJA
+ 6VB:cVC+dVD
Let each of the ratios a : b and c :d = p, and each of the two A : B and
C : D = <r, then a = pb, c — pd; A = <rB, C = crD. A\ 7 e then have
as / A  b VB _ pb V(<rB)  b VB
(pV<rl)&VB _ K/B
(p\/<rl)d\/D dsJV
In the same way we get
a\/A + b K /B = ( P sJ<r + l)bsJB K/B
cVC + ^V^ (pV<r + l)rfv'D rfV^
From (a) and (j8) the required result follows.
(a).
(/3).
Example 3.
If b be a mean proportional between a and c, show that
(a+b + c)(ab + c) = a 2 + b 2 + c 2 (a),
and {a + b + c) 2 + d + b 2 + c 3 = 2(a + b + c){a + c) (j8).
Taking (a) we have
(a+i+c)(a6+c)=(a+c+6)(a+c6),
= (a+c) 2 6 2 .
Now, by data, a/b = b/c, and therefore b' 2 = ac ; hence
(« + c) 2  b 2 = (a + c) 2  ac,
= a 2 + OC + C 1 ,
= a + b + c,
since b 2 = ac. Hence (a) is proved.
XIII
EXAMPLES 265
Taking now (/3), and, for variety, adopting the second method of § 9, let
us put
a b
b = C =P '
Hence a — pb, b — pc; so that a = pipe) = p~c.
We have to verify the identity
(pc + pc + cf + (j?cf + {pcf + c 2 = 2(pc + pc + c){pc + c) ;
that is to say,
{(ffi + p +l)* + (j,* + l ? + l)}<* = 2(? + p + l)((?+l)c> (7).
Now
{(p + P + l) 2 + (^ + r^l)^• 2 =(p 2 +/» + l){(p 2 + p + l) + ( / )•• ! p + l)}c'• ! ,
= 2(p 2 + p + I)(p 2 + l)c 2 ,
which proves the truth of (7), and therefore establishes (j8).
Example 4.*
If x/(b +ea) = yl(c + b) = zfta + bc), then (bc)x + (ca)y + (ab)z = 0.
Let us put
a _ V _ z _
b+ca c+ab a+b c
then x = (b + ca)p,
y = (c + ab)p,
z = {a + b c)p.
Now, from the last three equations, we have —
{b  c)x + (c  a)y + (a  b)z
= (b c)(b + c a)p + (ca)(c + a  b)p+ (ab)(a + bc)p,
= {(S 8  c 2 + c  a 2 + a 2  b' 2 )  (a(b c) + b(c  a) + e(a b))}p,
= {00}*
= 0.
Example 5.
If bz + cy _cx + az _ay + bx
bc ca a b
then (a + b + c)(x + y + z)=ax + by + cz (2).
Let each of the ratios in (1) be equal to p, then
bz + cy = p(bc) (3),
cx + az = p(ca) (4),
ay + bx — p(ab) (5).
From (3), (4), (5), by addition,
(b + c)x+(c + a)y+ (a + b)z = p{ (b  c) + (ca) + (« b)},
= p0,
= (6).
If now we add ax + by + cz to both sides of (6) we obtain equation (2).
* Examples 4, 5, and 6 illustrate a species of algebraical transformation
which is very common in geometrical applications. In reality they are ex
amples of a process which is considered more fully in chap. xiv.
(1),
266 EXAMPLES
Example 6.
If cy + bz az + cx bx + ay
qb + rc pa re +}M  qb pa + qb re
show that
x
a {pa(a + b + c)  qb(a + bc) rc(a  b + c) }
y
CHAI\
(1),
b{qb(a + b + c) pa(a + b  c) rc( a + b + c)}
z
(2).
c {rc(a + b + c)qb( a + b + c) pa(a  b + c) }
Let each of the fractions of (1) be = p; and observe that the three
equations,
cy + bz=(qb +rc 'pa)p (a) ~
az + ex = [re +}>a  qb)p (/3) J (3),
be + ay — {pa + qb  rc)p (7) J
fa**)
which thus arise are symmetrical in the triple set J abc Y, so that the simul
[pqr)
taneous interchange of the letters in two of the vertical columns simply
changes each of the equations (3) into another of the same set. It follows,
then, that a similar interchange made in any equation derived from (3) will
derive therefrom another equation also derivable from (3).
Now, if we multiply both sides of (j3) by b, and both sides of (7) by c, we
obtain, by addition from the two equations thus derived,
2bcx + a{cy + bz} =p{b(rc+2>aqb) + c{pa + qbrc)} (4).
Now, using the value of cy + bz given by (a), we have
2bcx + pa(qb + re  2m) = p {pa(b + c)  qb(b  c) rc( b + c)} (5).
Subtracting pa(qb + rcpa) from both sides of (5), we have
Ibex = p {pa{a + b + c) qb(a + bc) rc(a  b + e) } (6).
From (6), we have
? =P. (7) .
a{pa{a + b + c)qb[a + bc) re(a b + c)} 2abc
/xap\ /xap\
We may in (7) make the interchange I into I, or I into I, and we shall
\ ybq / \ zcr '
obtain two other equations derivable from (3) by a process like that used to
derive (7) itself. These interchanges leave the righthand side of (7) un
altered, but change the lefthand side into the second and third members of
(2) respectively. Hence the three members of (2) are all equal, each being in
fact equal to pj2abc.
This is a good example of the use of the principle of symmetry in compli
cated algebraical calculations.
xiii EXERCISES XVII 2G7
Exercises XVII.
(1.) Which is the greater ratio, 5 : 7 or 151 : 208 ?
(2. ) If the ratio 3 : 4 be duplicated by subtracting x from both antecedent
and consequent, show that a' = lf.
(3.) What quantity x added to the antecedent and to the consequent of
a : b will convert this ratio into c : d ?
(4.) Find the fourth proportional to 3£, 5, 6 ; also the third proportional
to 1 + V2 and 3 + 2\/2.
(5.) Insert a mean proportional between 11 anil 19 ; and also two mean
proportionals between the same two numbers.
(6.) Find a simple surd number which shall be a mean proportional be
tween \J7  \/5 and 11V7+ 13\/5
(7.) If x and y be such that when they are added to the antecedent and
consequent respectively of the ratio a : b its value is unaltered, show that
x:y = a :b.
(8. ) If x and y be such that when they are added respectively to the ante
cedent and consequent and to the consequent and antecedent of a : b the two
resulting ratios are equal, show that either x = y or x + y =  a—b.
(9.) Find a quantity x such that when it is added to the four given quan
tities a, b, c, d the result is four quantities in proportion. Exemplify with
3, 4, 9, 13 ; and with 3, 4, 1£, 2.
(10.) If four quantities be proportional, the sum of the greatest and least
is always greater than the sum of the other two.
(11.) If the ratio of the difference of the antecedents of two ratios to the
sum of their consequents is equal to the difference of the two ratios, then the
antecedents are in the duplicate ratio of the consequents.
(12.) If the n quantities d\, a. 2 , . . ., o n be in continued proportion, then
m : a n = a 1 " 1 : a a n1 =aa"~ 1 : ai 11 ' 1 = &c. ; and
a,= "'^(a 1 "a n ), a s = VWV,,), . . ., « P = "v / ("i"'Vr, 1 ' 1 ).
(13.) If (pa+qb+rc+sd)(paqbrc+sd)
= {paqb + rc sd) (pa + qbrc sd),
then be : ad =ps : qr ;
and, if either of the two sets a, b, c, d or p, q, r, s form a proportion, the
other will also.
(14.) If a: b=c:d=e:f,
then a 3 + 3a n b + b*:c 3 + Zcd + cP = a* + &' : c 3 + d 3 (a) ;
//«V «V eV\ f(bd b"P df\
= a\lf+ c s b/+ c*bd : b 3 cc + tPae +J*ac (/3) ;
pa  qc + re:pbqd+ rf= ^/ace : y/bdf
= vV  c ° + e " + 2ac ) : V(& 2  cp +f +  hd ) M
268 EXERCISES XVII chap.
t
(15.) Ua:a'=b:b',
then «'"+" + a'"b" + &"•+" : «'"'+" + a' m b'" + b >mr < "
= («. + ft)»'+": («' + ft')" !+ ".
(16.) If «:& = c : rf, and a 1/3 = 7: 5,
then « 3 a 2 + (a 2 ft + aft 2 )a/3 + ft 3 /3 2 : (a 3 + ft 3 ) (a 2 + /3 2 )
= cV + (c 2 ^ + (5^)75 + d 3 8 2 : (c 3 + d 3 ) (y~ + 8 s ).
(17.) If a: b = b :c = c:d, then
[a? + ft 2 + e 2 ) (ft 2 + c 2 + d 2 ) = (ab + be + cdf (a) ;
(ft  c) 2 + (c  a) 2 + (d  ft) 2 = (a  d) 2 (j8) ;
«ft + crf + ft^=(« + ft + c)(ftc + f?) (7);
a + 6ctf = (a + ft)(ftf0/ft (0);
(a + 6 + c + a") (a ft  6 + 0") = 2(aft  erf) (ac  bd)/(ad + ftc) (e).
(IS.) If a, ft, c be in continued proportion,
then a 2 + aft + ft 2 : b 2 + be + c 2 = a : c (a);
a 2 («ft + c)(a + ft + c) = « 4 + a 2 ft 2 + ft 4 (/3) ;
(ft + c) 2 /(ftf) + (c + «) 2 /(c«) + (a + ft) 2 /(«ft) = 4ft(a + ft + c)/(«c) (7).
(19.) If a, ft, c, d he in continued proportion,
then («c)(ftrf)(«^)(?'c) = (&c) 2 (a);
V(a*)W(k)+V(«9=V{(« +*+«)(* +«+/>} (£)■
(20.) Uab=cd=ef, then
(ac + ce + ea)/dbf(d + ft +/) = (a 2 + c 2 + e 2 )/(ft 2 d 2 + d 2 / 2 +/'ft').
(21.) If (aJ)/(rfe)=(6c)/(e/) 1 then each of them
= {b(fd) + (cdaf)\/c(fd).
(22.) Ul;/yx=vli?=tfyz, thenx/^ = y/e = ^vi.
(23.) If 2a!+3y:3i/+4z:43+5aj=4a56:3fto:263a J
then 7a; + 6y+8«=0.
(24.) If ax + cy :by + dz = ay + cz:bz + dx = az + ex : bx + dy, and if
x + y + z + 0, abcd + 0, adbc + 0, then each of these ratios =a + c : ft + d ;
and x 2 + y 2 + z 2 = yz + zx + xy.
(25.) If (a  ny + mz)jl' = (b7z + nx)/m'=(c  <mx + ly)/n', then
/ m'cn'b \n_f ii'al'c \, _f /'ft  m'a \ , »
V X " K' + mm' + MwV ' ^ y ' U'+mm'+nn'J ' m ~[ z ~U' + m m' + n n'J' 71 '
RATIO AND PROPORTION OF CONCRETE QUANTITIES.
§ 11.] AVe have now to consider how the theorems Ave have
established regarding the ratio and proportion of abstract num
bers are to be applied to concrete quantities. We shall base
* Important in the theory of the central axis of a system of forces, &c.
xill CONCRETE RATIO AND PROPORTION 2G9
this application on the theory of units. This, for practical pur
poses, is the most convenient course, but the student is not to
suppose that it is the only one open to us. It may be well to
recall once more that any theory may be expressed in algebraical
symbols, provided the fundamental principles of its logic are in
agreement with the fundamental laws of algebraical operation.
§ 12.] If A and B be two concrete quantities of the same hind,
which are expressible in terms of one and the same unit by the com
mensurable numbers a and b respectively, then the ratio of A to B is
defined to be the ratio or quotient of these abstract numbers, namely,
a : b, or ajb.
It should be observed that, by properly choosing the unit, the ratio of
two concrete quantities which are each commensurable with any finite unit at
all can always be expressed as the ratio of two integral numbers. For ex
ample, if the quantities be lengths of Z\ feet and 4f feet respectively, then,
by taking for unit th of a foot, the quantities are expressible by 26 and 35
respectively ; and the ratio is 26 : 35. This follows also from the algebraical
theorem that (3 + £)/(4 + ) = 26/35.
If A, B be two concrete quantities of the same kind, whose ratio is
a : b, and C, D two other concrete quantities of the same kind {but not
necessarily of the same kind as A and B), whose ratio is c:d, then
A, B, C, D are said to be proportional when the ratio of A to B is
equal to the ratio of C to D, that is, wht n
a:b = c: d.
We may speak of the ratio A : B, of the concrete magnitudes
themselves, and of the proportion A : B = C : D, without alluding
explicitly to the abstract numbers which measure the ratios ; but
all conclusions regarding these ratios will, in our present manner
of treating them, be interpretations of algebraical results such as
we have been developing in the earlier part of this chapter,
obtained by operating with a, b, c, d. The theory of the ratio
and proportion of concrete quantity is thus brought under the
theory of the ratio and proportion of abstract quantities.
There are, however, several points which require a nearer
examination.
§ 13.] In the first place, it must be noticed that in a concrete
270
SPECIAL POINTS IN CONCRETE PROPORTION
CHAP.
ratio the antecedent and the consequent must be quantities of the
same kind ; and in a concrete proportion the two first terms must
be alike in kind, and the two last alike in kind. Thus, from the
present point of view at least, there is no sense in speaking of
the ratio of an area to a line, or of a ton of coals to a sum of
money. Accordingly, some of the propositions proved above —
those regarding the permutations of a proportion, for instance —
could not be immediately cited as true regarding a proportion
among four concrete magnitudes, unless all the four were of the
same kind.
This, however, is a mere matter of the interpretation of
algebraical formulae — a matter, in short, regarding the putting of
a problem into, and the removing of it from, the algebraical
machine.
§ 14.] A more important question arises from the considera
tion that, if we take two concrete
magnitudes of the same kind at random,
there is no reason to expect that there
exists any unit in terms of which each
is exactly expressible by means of com
mensurable numbers.
Let us consider, for example, the
historically famous case of the side AB
and diagonal AC of a square ABCD. On the diagonal AC lay
off AF = AB, and draw FE perpendicular to AC. It may be
readily shown that
3
A
E
/
\
\
SF
\
c
\
D
Hence
BE = EF = FC.
CF = AC  AB
CE = CB  CF
(2).
Now, if AB and AC were each commensurably expressible in
terms of any finite unit, each would, by the remark in § 1 2, be an
integral multiple of a certain finite unit. But from (1) it follows
that if this were so, CF would be an integral multiple of the
same unit j and, again, from (2), that CE would be an integral
multiple of the same unit. Now CF and CE are the side and
diagonal of a square, CFEG, whose side is less than half the side
xiu COMMENSURABLES AND INCOMMENSURABLES 271
of ABCD; and from CFEG could in turn be derived a still
smaller square whose side and diagonal would be integral mul
tiples of our supposed unit ; and so on, until we had a square
as small as we please, whose side and diagonal are integral
multiples of a finite unit; which is absurd. Hence the side
and diagonal of a square are not magnitudes such as A and B
are supposed to be in our definition of concrete ratio.
§ 15.] The difficulty which thus arises in the theory of con
crete ratio is surmounted as follows : —
We assume, as axiomatic regarding concrete ratio, that if
A' and A" be two quantities respectively less and greater than
A, then the ratio A : B is greater than A' ; B and less than
A" : B ; and we show that A' and A" can be found such that,
while each is commensurable with B, they differ from each other,
and therefore each differs from A by as little as we please.
Suppose, in fact, that we take for our unit the nth part of B,
then there will be two consecutive integral multiples of Bjn, say
mB/n and (m + l)Bfn, between which A will lie. Take these
for our values of A' and A" ; then
A"  A' = (m + 1)B/tc  mBjn,
= B>.
Hence A"  A' can, by sufficiently increasing n, be made as small
as Ave please.
We thus obtain, in accordance with the definition of § 12,
two ratios, m\n and (m + l)/n, between which the ratio A : B lies,
each of which may be made to differ from A : B by as little as
we please.
Practically speaking, then, we can find for the ratio of two
incommensurables an expression which shall be as accurate as
we please. Regarding this matter, see vol. ii., chap, xxv., §§ 2641.
Example.
If B be the side and A the diagonal of a square, to find a rational value
of A : B which shall be correct to l/1000th.
If we take for unit the l/1000th part of B, then B = 1000, and A 2 =
2,000,000. Now 1414''= 1999396, and 1415 2 = 2002225. Hence 1414/I00O
<A/B< 1415/1 000. But 1415/10001414/1000 = 1/1000. Hence we have
A/B = 1414, the error being < 1/1000.
272 EUCLIDIAN THEORY OF PROPORTION chap.
§ 16.] The theory of proportion given in Euclid's Elements gets over the
difficulty of incommensurables in a very ingenious although indirect manner.
No working definition of a ratio is attempted, but the proportionality of four
magnitudes is defined substantially as follows : —
If there be four magnitudes A, B, C, D, such that, always,
mk>, = , or <«B,
according as mC>, =, or <?;D,
m and n being any integral numbers whatsoever, then A, B, C, D are said to
be proportional.
Here no use is made of the notion of a unit, so that the difficulty of in
commensurability is not raised. On the other hand, there is substituted a
somewhat indirect and complicated method for testing the subsistence or non
subsistence of proportion ality.
It is easy to see that, if A, B, 0, D be proportional according to the
algebraical definition, they have the property of Euclid's definition. For, if
a:b and c : d be the numerical measures of the ratios A :B and"C:D, we
have
a _c
b~d'
, ma mc
hence — r = — ,,
no nd
from which it follows that ma>, —, or <nb, according as mc> , =, or <nd.
The converse, namely, that, if A, B, C, D be proportional according to
Euclid's definition, then
a _c
can be proved by means of the following lemma.
Given any commensurable quantity ajb, another commensurable quantity
can be found which shall exceed or fall short of a/b by as little as we please.
Let n be an integral number, and let mb be the least multiple of b which
exceeds na, so that
na—mbr,
where r < b.
Dividing both sides of this equation by rib, we have
a _ in r
b u nb '
, m a r
whence T = _ i;»
n b vb
so that m/n exceeds a/b by r/nb. Now, since r never exceeds the given
quantity b, by making n sufficiently great, we can make r/nb as small as we
plea.se ; that is to say, we can make m/n exceed a/b by as little as we please.
Similarly we may show that another commensurable quantity may be
found falling short of a/b by as little as we please.
From this it follows that, if two commensurable quantities differ by ever so
little, we can always find another commensurable quantity which lies between
xui COMPARISON OF THEORIES 273
them ; for we can find another commensurable quantity which exceeds tlic
less of the two by less than the difference between it and the greater.
Suppose now that
ma >, =, or <nb,
according as mo, —, or <nd,
m and n being any integers whatever, then we must have
a _ c
b~d'
For, if these fractions (which we may suppose to be commensurable by
virtue of § 15) dilfer by ever so little, it will be possible to find another
fraction, n/m say, where n and m are integers, which lies between them.
Hence, if a/b be the less of the two, we must have
t< — , that is, ma < nb ;
b m
> — . that is, mond.
d m
In other words we have found two integers, m and n, such that we have
at once
ma < nb
and mond.
But, by hypothesis, when ma<nb, we must have mc<nd. Hence the
fractions a/b and c/d cannot be unequal.
VARIATION.
§ 17.] There are an infinite number of ways in which we
may conceive one quantity y to depend upon, be calculable from,
or, in technical mathematical language, be a function of, another
quantity x. Thus we may have, for example,
y = 3&,
y =17/,
y = ax + b,
y = ax 2 + bx + c,
y='2 \ f x,
and so on.
For convenience x is called the independent variable, and y the
dependent variable ; because we imagine that any value we please
is given to x, and the corresponding value of y derived from it
by means of the functional relation. All the other symbols of
quantity that occur in the above equations, such as 3, 17, a, b, c,
VOL. I T
274 INDEPENDENT AND DEPENDENT VARIABLE chap.
2, &c, are supposed to remain fixed, and are therefore called
constants.
Here we attach meanings to the words variable and constant
more in accordance with their use in popular language than
those given above (chap, ii., § 6).
The justification of the double usage, if not already apparent,
will be more fully understood when Ave come to discuss the
theory of equations, and to consider more fully the variations of
functions of various kinds (see chaps, xv.xviii.)
§ 18.] In the meantime, we propose to discuss very briefly
the simplest of all cases of the functional dependence of one
quantity upon another, that, namely, which is characterised by
the following property.
Let the following scheme
Values of
the Independent Variable.
Corresponding Values of the
Dependent Variable.
X
x'
V
y
denote any two corresponding pairs whatever of values of the
independent and dependent variables, then the dependence is to
be such that always
y:y' = x:x' (1).
It is obvious that this property completely determines the
nature of the dependence of y upon x, as soon as any single cor
responding pair of values are given. Suppose, in fact, that,
when x has the value x , y has the value y , then, by (1),
y x
y ~x '
whence y = ( — )x.
Now we may keep x u and y„ as a fixed standard pair, for
reference as it were ; their ratio y /x is therefore a given con
Xlll SIMPLEST CASE OF FUNCTIONAL DEPENDENCE 275
stant quantity, which we may denote by a, say. We therefore
have
V = ax (2),
that is to say, y is a given constant multiple of x ; or, in the
language of chap, iv., § 17, a homogeneous integral function of x
of the 1st degree.
Example. Let us suppose that we have for any two corresponding pairs
y, x and y', x' the relation y :x = y' :x' ; and that when x=Z, y = 6 Then
since 6 and 3 are corresponding pairs y :x=6 :3. Hence y/x — 6/B = 2.
Hence y = 2x.
Conversely, of course, the property (2) leads to the property
(1). For, from (2),
y = ax ;
hence, if x' and y' be other two corresponding values,
y' = ax'.
y ax x
Hence '—~ — , = — •
y ax x
When y depends on x in the manner just explained it is said
to vary directly as x, or, more shortly, to vary as x.
A better * phrase, which is also in use, is " y is proportional
to x."
This particular connection between y and x is sometimes
expressed by writing
y<^x.
§ 19.] In place of x, we might write in equation (2) x 2 , l/x,
1/x 2 , x + b, and so on ; we should then have
y = ax 2 (a),
y = ajx ((3),
y = a\i (y),
y = a(x + b) (8).
* The use of the word " Variation " in the present connection is unfortunate,
because the qualifying particle " as " is all that indicates that we are here
concerned not with variation in general, as explained in § 17, but merely with
the simplest of all the possible kinds of it. There is a tendency in uneducated
minds to suppose that this simplest of all kinds of functionality is the only
one ; and this tendency is encouraged by the retention of the above piece of
antiquated nomenclature.
276 OTHER SIMPLE CASES chap.
The corresponding forms of equation (1) would then be
V : :'/ = X 2 : X" (a'),
y:y' = llx:l/x' (ft'),
T .y = 1/aM/i" (/),
y.y' = x + b: x' + b (8').
y is then said to vary as, or be proportioned to, x 2 , l/'x, Ijx 2 ,
x + b. In cases (ft) and (y) y is sometimes said to vary inversely
as x, and inversely as the square of x respectively.
Still more generally, instead of supposing the dependent
variable to depend on one independent variable, we may suppose
the dependent variable u to depend on two or more independent
variables, x, y, z, &c.
For example, we may have, corresponding to (2),
and, corresponding to (1),
u = axy
(4
u = axyz
(0,
u = a(x + y)
(v),
u = ax/y
{0);
u : to' = xy : x'y'
(0,
u : m' = xyz : x'y'z'
(0,
u:u' = x + y. x' + y
(v')>
to : u' = xjy : x'jy'
(d'\
In case (e) u is sometimes said to vary as x and y jointly ;
in case (6) directly as x and inversely as y.
§ 20.] The whole matter we are now discussing is to a large
extent an affair of nomenclature and notation, and a little
attention to these points is all that the student will require to
prove the following propositions. We give the demonstrations
in one or two specimen cases.
(1 .) If z°=y and y ^ x, then z<=>=x.
Proof. — By data z = ay, y = bx, where a and b are constants ;
therefore z = dbx. Hence z^x, since ah is constant.
(2.) If y i ' x x l and y 2 ^x 2 , then y x y 2 °= ;<•/.,.
Proof. — By data y t = a x x lt y 2 = a^, where a, and a 2 are con
XIII
PROPOSITIONS REGARDING VARIATION
277
stants. Hence y$ 9 = a^v^, which proves the proposition, since
ai« 2 is constant.
In general if y,^x„ y,^x,, . . ., y n <=^x n , then y l y 2 ...y n
oc #,£ 2 . . . x w And, in particular, if y cc x, then y n <^ x n .
(3.) If y°zx, then zy°czx, whether z be variable or constant.
(4.) If zozxy, then x^zjy, and y^zjx.
(5.) If z depend on x and y, and on these alone, and if z^x
when y is constant, and z°^y when x is constant, then z °= xy when
both x and y wry.
Proof. — Consider the following system of corresponding
values of the variables involved.
Dependent Variable.
Independent Variables.
Z
z'
x, y.
x', y.
x', y'.
Then, since y lias the same value for both z and z u we have,
by data,
z x
2, X
Again, since x' is the same for both z l
and z', we have, by
data,
»i y
z' y
From these two equations we have
z z x x y
7 x I 7 = v x 7? '
Zi z x y
1.1. l • z xy
thatls ' ?*?■
which proves that z^xy.
A good example of this case is the dependence of the area of a triangle
upon its base and altitude.
278 EXAMPLES chap.
"We have
Area oc base (altitude constant) ;
Area oc altitude (base constant).
Hence area oc base x altitude, when both vary.
(6.) In a similar manner we may prove that if z depend on
x l , x 2 , . . . , x n , and on these alone, and vary as any one of these when
the rest remain constant, then z oc x x x 2 . . . x n when all vary.
(7.) If zee x (y constant) and z<^l/y (x constant), then z<>=x/y
when both vary.
For example, if V, P, T denote the volume, pressure, and absolute tem
perature of a given mass of a perfect gas, then
V oc 1/P (T constant), V oc T (P constant).
Hence in general V oc T/P.
Example 1.
If s oc t 2 when/ is constant, and s oc/ when t is constant, and 2s=/ when
t = \, find the relation connecting s, f, t.
It follows by a slight extension of § 20 (5) that, when/ and t both vary,
s <xft 2 . Hence s = aft", where a is a constant, which we have to determine.
Now, when t = l, s = \f, hence \f=afl 2 , that is, \f—af\ in other words,
we must have a—h. The relation required is, therefore, s — \ffi.
Example 2.
The thickness of a grindstone is unaltered in the using, but its radius
gradually diminishes. By how much must its radius diminish before the
half of its mass is worn away ? Given that the mass varies directly as the
square of the radius when the thickness remains unaltered.
Let m denote the mass, r the radius, then by data, m = ar, where a is
constant.
Let now r become ?•', and, in consequence, m become Jm, then b>i = ar' 2 ,
hence
ar' 2
_hn
:
ar 2
~ m'
r >2
r
=h;
r'
l
—
^
r
\A
that is,
whence
It follows, therefore, that the radius of the stone must be diminished in
the ratio 1 : V2.
Example 3.
A ami B are partners in a business in which their interests are in the
ratio a : b. They admit C to the partnership, without altering the whole
amount of capital, in such a way that the interests of the three partners in
the business are then equal. C contributes £c to the capital of the firm.
XIII
EXERCISES XVIII 279
How is the sum £c which is withdrawn from the capital to be divided between
A and B ? and what capital had each in the business originally ?
Solution. — Since what C pays in is his share of the capital, they each have
finally £c in the business ; let now £x be A's share of C's payment, so that
£{cx) is B's share of the same. In effect, A takes £x and B £(cx) out
of the business. Hence they had originally £{c + x) and £(c + cx) in the
business. By data, then, we must have
c + x _a
2cx~b'
hence b(c + x) = a(2cx) ;
we have, therefore, bc + bx — 2ac  ax.
From this last equation we derive, by adding ax  be to both sides,
{a + b)x=(2ab)c.
Heuce, dividing by a + b, we have
x J2aV)c
a + b v '
Hence cx=c
a + b
(2ba)c
a + b
(2).
It appears, then, that A and B take £(2a  b)c/{a + b) and £{2b  a)c/(a + b)
respectively out of the business. C's payment must be divided between them
in the ratio of these sums, that is, in the ratio 2a b :2b a. They had in
the business originally £3ac/(a + b) and £3bc/(a + b) respectively.
Exercises XVIII.
(1.) Ifyccx, and if y = 3§ when a; = 6i, find the value of y when a=g.
(2. ) y varies inversely as x 2 ; and z varies directly as ,<:. When x= 2, y + z
= 340 ; when ar=l, ?/c = 1275. For what value of a; is y = zl
(3.) zeeuv; uxx; vecxr. When x = 2, s = 48 ; when x = 5, z = 30.
For what values of x is c = ?
(4.) If xy oc x 2 + y' 2 , and x = Z when y = 4, find the equation connecting
y and x.
(5.) If x + y<xxy, then x 2 + y 2 <x xy and oP + y 3 <x .ri/(.>±y).
(6.) If {x + y + z)(x + yz)(xy + z)( x + y + z) oc x'hf, then either x* + y 2
oc zr or xr ■> y — z oc xy.
(7.) Hxxy, then x 2 + y 2 cc xy.
(8. ) If a* + \ « x 3  i, then y <x. Ifx.
(9. ) If x oc y 2 , y 3 oc z 4 , z 5 oc u e , tt 7 oc ^ then (x/v) (y/v) (z/v) (ujv) is constant.
(10.) Two trains take 3 seconds to clear each other when passing in
opposite directions, and 35 seconds when passing in the same direction : find
the ratio of their velocities.
280 EXERCISES XVIII chap, xiii
(11.) A watch loses 2\ minutes per day. It is set right on the 15th March
at 1 p.m. : what will the proper time be when it indicates 9 A.M. on the 20th
April 1
(12.) A small disc is placed between two infinitely small sources of radiant
heat of equal intensity, at a point on the line joining them equidistant from
the two. It is then moved parallel to itself through a distance aj2\/S towards
one of the two sources, a being the distance between them : show that the
whole radiation falling on the disc is trebled.
(The radiation falling on the disc varies inversely as the square of the dis
tance from the source, when the disc is moved parallel to itself towards or
from the source.)
(13.) The radius of a cylinder is ?•, and its height h. It is found that by
increasing either its radius or its height by x its volume is increased by the
same amount. Show that x = r(r2h),'h. "What condition is there upon r
and h in order that the problem may be possible ?
(Given that the volume of a cylinder varies directly as its height when
its radius is constant, and directly as the square of its radius when its height
is constant.)
(14.) A solid spherical mass of glass, 1 inch in diameter, is blown into a
shell bounded by two concentric spheres, the diameter of the outer one being
3 inches. Calculate the thickness of the shell. (The volume of a sphere
varies directly as the cube of its diameter. )
(15.) Find, the radius of a sphere whose volume is the sum of the volumes
of two spheres whose radii are 3^ feet and 6 feet respectively.
(16.) Two equal vessels contain spirits and water, the ratios of the amount
of spirit to the amount of water being p : 1 and p : 1 respectively. The con
tents of the two are mixed : show that the ratio of the amount of spirit to the
amount of water in the mixture is p + p' + 2pp' :2 + p + p'.
CHAPTER XIV.
On Conditional Equations in General.
DEFINITIONS AND GENERAL NOTIONS.
§ 1 .] It will be useful for the student at this stage to attempt
to form a wider conception than we have hitherto presupposed of
what is meant by an analytical function in general. Dividing the
subjects of operation into variables (x, y, z, . . .) and constants
(a, b, c, . . .), we have already seen what is meant by a rational
integral algebraical function of the variables x, y, z, . . .; and we
have also had occasion to consider rational fractional algebraical
functions of x, y, z, . . . We saw that in distinguishing the
nature of such functions attention was paid to the way in which
the variables alone were involved in the function. We have already
been led to consider functions like s/(x + Jy), or %/(x + >/y),
or ax* + bx + c, where the variables are involved by way of root
extraction. Such functions as these are called irrational alge
braical functions. These varieties exhaust the category of what
are usually called Ordinary * Algebraical Functions, In short, any
intelligible concatenation of operations, in which the operands selected
for notice and called the variables are involved in no other ways than
by addition, subtraction, multiplication, division, and root extraction, is
called an Ordinary Algebraical Function of these variables.
Although Ave have thus exhausted the category of ordinary
algebraical functions, we have by no means exhausted the possi
* The adjective " Ordinary " is introduced to distinguish the class of func
tions here defined from algebraical functions as more generally denned in
chap, xxx., § 10. The word "Synthetic" is often used for "Ordinary" in
the present connection.
282
CONDITIONAL AND IDENTICAL EQUATIONS
CHAP.
bilities of analytical expression. Consider for example a* where,
as usual, x denotes a variable and a a constant. Here x is not
involved in any of the Avays recognised in the definition of an
algebraical function, but appears as an index or exponent. a x is
therefore called an exponential function of x. It should be care
fully noted that the discrimination turns solely on the way in
which the variable enters. Thus, while a x is an exponential
function of x, x 0, is an algebraical function of x. There are other
functions in ordinary use, — for example, sin a', logic, — and an
infinity besides that might be imagined, which do not come
under the category of algebraical ; all such, for the present, we
class under the general title of transcendental functions, so that
transcendental simply means nonalgebraical. We use the term
analytical function, or simply function, to include all functions,
Avhether algebraical or transcendental, and we denote a function
of the variables x, y, z, . . ., in which the constants a, b, c, . . .
are also involved, by
4>(x, y, z, . . . a, b, c, . . .);
or, if explicit mention of the constants is unnecessary, by
<f>(x, y, z, . . .).
§ 2.] Consider any two functions whatever, say cf>(x, y, z, . . .
a, b, c, . . .), and if(x, y, z, . . . a, b, c, . . .), of the variables
x, y, z, . . ., involving the constants a, b, c, . . .
If the equation
<f>(z, y,z, . . . a, b, c, . . .) = f(z, y, z, . . . a, b, c, . . .) (1)
be such that the lefthand side can, for all values of the variables
x, y, z, . . ., be transformed into the right by merely apply
ing the fundamental laws of algebra, it is called an identity.
With equations of this kind the student is already very familiar.
If, on the other hand, the lefthand side of the equation (1)
can be transformed into the right only when x, y, z, . . . have
certain values, or are conditioned in some way, then it is said to be
a Conditioned Equation, or an Equation of Condition.* Examples
* "When it is necessary to distinguish between an equation of identity and
an equation of condition, the sign = is used for the former, and the sign =
for the latter. Thus, we should write [x + 1) (» l)=u?  1 j but 1c + 2 = 2.
xiv CLASSIFICATION OF EQUATIONS 283
of such equations have already occurred, more especially in chap,
xiii. One of the earliest may be seen in chap, iv., § 24, where,
inter alia, it was required to determine B so that we should
have 2B + 2 = 2 ; in other words, to find a value of x to satisfy
the equation
2x+2 = 2 (2).
Here 2x + 2 can be transformed into 2 when (and, as we
shall hereafter see, only when) x = 0.
Every determinate problem, wherein it is required to deter
mine certain unknown quantities in terms of certain other given
or known quantities by means of certain given conditions, leads,
when expressed in analytical language, to one or more equations
of condition ; to as many equations, in fact, as there are condi
tions. The quantities involved are therefore divided into two
classes, known and unknown. The known quantities are denoted
by the socalled constant letters ; the unknown by the variable
letters. Hence, in the present chapter, constant and known are
convertible terms ; and so are variable and unknown. The con
stants may be actual numerical quantities, real and positive or
negative (4, \, 0, +1, + f , &c), or imaginary or complex
numbers (  i, 1 + 2i, &c); or they may be letters standing for
any such quantities in general.
§ 3.] Equations are classified according to their form, and
according to the number of variables that occur in them.
If transcendental functions appear, as, for example, in
2* = 2> x + 2, the equation is said to be transcendental. With
such for the present we shall have little to do.
If only the ordinary algebraical functions appear, as, for
example, in «/(jc + y) + \/(x  y) = 1 , the equation is called an
algebraical equation. Such an equation may, of course, be
rational or irrational, and, if rational, either fractional or
integral, according to circumstances.
It will be shown presently that every algebraical equation
can be connected with, or made to depend upon, an equation
of the form
284 MEANING OF SOLUTION chap.
where <t> is a rational integral function. Such equations are
therefore of great analytical importance ; and it is to them that
the " Theory of Equations," as ordinarily developed, mainly
applies. An integral equation of this kind is described by
assigning its degree and the number of its variables. The degree
of the equation is simply the degree of the function <f>. Thus,
of + Ixy + if  2 = is said to be an equation of the 2nd degree
in two variables.
§ 4.] Equations of condition may occur in sets of one or of
more than one. In the latter case we speak of the set as a set
or system of simultaneous equations.
The main problem which arises in connection with every system of
equations of condition is to find a set or sets of values of the variables
which shall render every equation of the system an identity literal oi
numerical.
Such a set of values of the variables is said to satisfy the
system, and is called a solution of the system of equations. If
there be only one equation, and only one variable, a value of
that variable which satisfies the equation is called a root. We
also say that a solution of a system of equations satisfies the system,
meaning that it renders each equation of the system an identity.
It is important to distinguish between two very different
kinds of solution. When the values of the variables which con
stitute the solution are closed expressions, that is, functions of
known form of the constants in the given equations, we have
what may be called a formal solution of the system of equations.
In particular, if these values be ordinary algebraical functions
of the constants, we have an algebraical solution. Such solutions
cannot in general be found. In the case of integral algebraical
equations of one variable, for example, if the degree exceed the 4th,
it has been shown by Abel and others that algebraical solutions
do not exist except in special cases, so that the formal solution,
if it could be found, would involve transcendental functions.
When the values of the variables which constitute the solution
are given approximately as numbers, real or complex, the solution
is said to be an approximate numerical solution. In this case the
xiv EXAMPLES OF SOLUTION 285
words " render the equation a numerical identity " are understood
to mean "reduce the two sides of the equation to values
which shall differ by less than some quantity which is assigned."
For example, if real values of the two sides, say P and P',
are in question, then these must be made to differ by less than
some given small quantity, say 1/100,000; if complex values
are in question, say P + Qi and P' + Q'i, then these must
be so reduced that the modulus of their difference, namely,
\/{(P  P') 2 + (Q  Q') 2 }> shall be less than some given small
quantity, say 1/100,000. (Cf. chap, xii., § 21.)
As a matter of fact, numerical solutions can often be obtained
where formal solutions are out of the question. Integral alge
braical equations, for example, can always be solved numerically
to any desired approximation, no matter what their degree.
Example 1.
2x + 2 = 2.
jb=0 is a solution, for this value of x reduces the equation to
2x0 + 2 = 2,
which is a numerical identity. Strictly speaking, this is a case of algebraical
solution.
Example 2.
ax — b 2 = 0.
x = bja reduces the equation to
a — b = 0,
a
which is a literal identity ; hence x = b 2 /a is an algebraical solution.
Example 3.
x"  2 = 0.
Here x— + \/2 and x=  \J2 each reduce the equation to the identity
22=0 ;
these therefore are two algebraical solutions.
On the other hand, a;=+l  4142 and x = — 1*4142 are approximate
numerical solutions, for each of them reduces x~2 to  "00003836. which
differs from by less than '00004.
Example 4.
0'l) 2 + 2 = 0.
x = \ + \/2i and x—l\/2i are algebraical solutions, as the student will
easily verify.
.T = 10001 + l  4142i and a;= 1*0001  1  4142i are approximate numerical
solutions, for they reduce (.r l) 2 + 2 to "00003837+ 00028284;:' and "00003837
 , 00028284t respectively, complex numbers whose moduli are each less than
"0003.
286 CONDITIONAL EQUATION A HYPOTHETICAL IDENTITY chap.
Example 5.
xy1.
Here se=l, y=0, is a solution ; so is a;=l"5, y = 5 ; so is x = 2, y — 1 ; and,
in fact, so is x = a+ 1, y = a, where a is any quantity whatsoever.
Here, then, there are an infinite number of solutions.
Example 6. Consider the following system of two equations : —
x — y=\, 2x + y—5.
Here x=2, y = l is a solution ; and, as we shall show in chap, xvi., there is
no other.
The definition of the solution of a conditional equation
suggests two remarks of some importance.
1st. Every conditional equation is a hypothetical identity. In all
operations with the equation ive suppose the variables to have such
values as will render it an identity.
2nd. The ultimate test of every solution is that the values which it
assigns to the variables shall satisfy the equations when substituted therein.
No matter how elaborate or ingenious the process by which
the solution has been obtained, if it do not stand this test, it is
no solution ; and, on the other hand, no matter how simply
obtained, provided it do stand this test, it is a solution.* In
fact, as good a Avay of solving equations as any other is to guess
a solution and test its accuracy by substitution.!
§ 5.] The consideration of particular cases, such as Examples
16 of § 4, teaches us that the number of solutions of a system of
one or more equations may be finite or infinite. If the number
be finite, we say that the solution is determinate (singly determin
ate, or multiply determinate according as there are one or more
solutions) ; if there be a continuous infinity of solutions, we say
that the solution is indeterminate.
The question thus arises, Under what circumstances is the
solution of a system of equations determinate 1 Part at least of
the answer is given by the following fundamental propositions.
Proposition I. The solution of a system of equations is in general
determinate (singly, or multiply according to circumstances) when the
number of the equations is equal to the number of the variables.
* A little attention to these selfevident truths would save the beginner
from many a needless blunder.
t This is called solving by "inspection."
xiv PROPOSITIONS AS TO DETERMINATENESS OF SOLUTION 287
Rightly considered, this is an ultimate logical principle which
may be discussed, but not in any strictly general sense proved.
Let us illustrate by a concrete example. The reader is aware
that a rectilinear triangle is determinable in a variety of ways
by means of three elements, and that consequently three condi
tions will in general determine the figure. To translate this into
analytical language, let us take for the three determining elements
the three sides, whose lengths, at present unknown, we denote
by x, y, z respectively. Any three conditions upon the triangle
may be translated into three equations connecting x, y, z with
certain given or constant quantities ; and these three equations
will in general be sufficient to determine the three variables,
x, y, z. The general principle :VL common to this and like cases is
simply Proposition I. The truth is that this proposition stands
less in need of proof than of limitation. What is wanted is an
indication of the circumstances under which it is liable to excep
tion. To return to our particular case : What would happen,
for example, if one of the conditions imposed upon our triangle
were that the sum of two of the sides should fall short of the
third by a given positive quantity % This condition could be
expressed quite well by an equation (namely, x + y = z  q, say),
but it is fulfilled by no real triangle, f Again, it might chance
that the last of the three given conditions was merely a con
sequence of the two first. We should then have in reality only two
conditions — that is to say, analytically speaking, it might chance
that the last of the three equations was merely one derivable from
the two first, and then there would be an infinite number of
solutions of the system of three variables. Such a system is
x + y + z = 6,
3.c+ 2y + z = 10,
2x + y= 4,
for example, for, as the reader may easily verify, it is satisfied
by x = a  2, y = 8  2a, z = a, where a is any quantity whatsoever.
* A name seems to be required for thi.s allpervading logical principle :
the Law of Determinate Manifoldness might be suggested.
+ See below, chap. xix.
288 PROPOSITIONS AS TO DETEEMINATENESS OF SOLUTION chap.
It will be seen in following chapters how these difficulties are
met in particular cases. Meantime, let us observe that, if we
admit Proposition I., two others follow very readily.
Proposition II. If the number of equations be less than the number
of variables, the solution is in general indeterminate.
Proposition III. If the number of independent equations be
greater than the number of variables, there is in general no solution,
and the system of equations is said to be inconsistent.
For, let the number of variables be n, and the number of
equations to, say, where to < n. Let us assign to the first n  m
variables any set of values we please, and regard these as constant.
This we may do in an infinity of ways. If we substitute any such
set of values in the to equations, we have now a set of to equa
tions to determine the last m variables ; and this, by Proposition
I., they will do determinately. In other words, for every set of
values we like to give to the first n  to variables, the to equations
give us a determinate set of values for the last to. We thus get an
infinite number of solutions ; that is, the solution is indeterminate.
Next, let to be > n. If we take the first n equations, these
will in general, by Proposition I., give a determinate set, or a
finite number of determinate sets of values for all the n
variables. If we now take one of these sets of values, and
substitute it in one of the remaining to  n equations, that
equation will not in general be satisfied ; for, if we take an
equation at random, and a solution at random, the latter will
not in general fit the former. The system of to equations will
therefore in general be inconsistent.
It may, of course, happen, in exceptional cases, that this
proposition does not hold ; witness the following system of three
equations in two variables : —
SBy=l, 2x + y=5, 3x"+2«/=8,
which has the common solution x = 2, y = 1.
§ 6.] We have also the further question, When the system
is determinate, how many solutions are there ? The answer
to this, in the case of integral equations, is furnished by the
two following propositions : —
xiv MULTIPLICITY OF DETERMINATE SOLUTIONS 289
Proposition I. An integral equation of the nth degree in one
variable has n roots and no more, which may be real or complex, and
all unequal or not all unequal, according to circumstances.
Proposition II. A determinate system ofm integral equations with
m variables, whose degrees in these variables are p, q, r, . . . respect
ively, has, at most, pqr . . . solutions, and has, in general, just that
number.
Proposition I. was proved in the chapter on complex num
bers, where it was shown that for any given integral function
of x of the nth degree there are just n values of x and no more
that reduce that function to zero, these values being real or
complex, and all unequal or not as the case may be.
Proposition II. will not be proved in this work, except in
particular cases which occur in chapters to follow. General
proofs will be found in special treatises on the theory of equa
tions. We set it down here because it is a useful guide to the
learner in teaching him how many solutions he is to expect.
It will also enable him, occasionally, to detect when a system
is indeterminate, for, if a number of solutions be found exceed
ing that indicated by Proposition II., then the system is certainly
indeterminate, that is to say, has an infinite number of solu
tions.
Example. The system x 2 + y 2 =l, xy = \ has, by Proposition II. , 2x1 = 2
solutions. As a matter of fact, these solutions are x = 0, y — — 1, and x=l,
y=o.
EQUIVALENCE OF SYSTEMS OF EQUATIONS.
§ 7.] Two systems of equations, A and B (each of which may con
sist of one or more equations), are said to be equivalent when every
solution of A is a solution of B, and every solution of B a solution
of A.
From any given system, A, of equations, we may in an in
finity of ways deduce another system, B; but it will not
necessarily be the case that the two systems are equivalent.
In other words, we may find in an infinity of ways a system,
B, of equations which will be satisfied by all values of the
VOL. I U
290 DEFINITION OF EQUIVALENCE chap.
variables for which A is satisfied ; but it will not follow con
versely that A will be satisfied for all values for which B is
satisfied. To take a very simple example, x  1 = is satisfied by
the value x=l, and by no other; £(.t1) = is satisfied by
as=l, that is to say, x(x  1) = is satisfied when x  1 = is
satisfied. On the other hand, x(x  1) = is satisfied either by
x = or by x = 1, therefore x — 1 = is not always satisfied when
x(x  1) = is so ; for x = reduces x  1 to — 1, and not to 0.
Briefly, x(x  1) = may be derived from x 1 = 0, but is not
equivalent to x — 1 = 0.
x(x  1 ) = is, in fact, more than equivalent to x — 1 = 0,
for it involves x  1 = and x = as alternatives. It will be
convenient in such cases to say that x(x  1) = is equivalent to
I 01
\;cl = 0j
When by any step we derive from one system another which
is exactly equivalent, we may call that step a reversible deriva
tion, because we can make it backwards without fallacy. If
the derived system is not equivalent, we may call the step
irreversible, meaning thereby that the backward step requires
examination.
There are few parts of algebra more important than the
logic of the derivation of equations, and few, unhappily, that
are treated in more slovenly fashion in elementary teaching.
No mere blind adherence to set rules will avail in this matter ;
while a little attention to a few simple principles will readily
remove all difficulty.
It must be borne in mind that in operating with conditional
equations we always suppose, the variables to have such values
as will render the equations identities, although we may not at
the moment actually substitute such values, or even know them.
We are therefore at every step, hypothetically at least, applying the
fundamental laws of algebraical transformation just as in chap. i.
The following general principle, already laid down for real
quantities, and carefully discussed in chap, xii., § 12, for com
plex quantities, may be taken as the root of the whole matter.
xiv ADDITION AND TRANSPOSITION OF TERMS 291
Let P and Q be two functions of the variables x, y, z, . . ., vjhich
do not become infinite * for any values of those variables that we have
to consider. J/PxQ=0 and Q 4= 0, then mil P = 0, and ifP x Q =
and P =t= 0, then will Q = 0.
Otherwise, the only values of the variables which make P x Q =
are such as make either P = 0, or Q = 0, or both P = and Q = 0.
§ 8.] It follows by the fundamental laws of algebra that if
P = Q • (l),
then P ± R = Q ± R (2),
where R is either constant or any function of the variables.
We shall show that this derivation is reversible.
For, if P ± R = Q ± R,
then P±RtR = Q±RtR,
that is, P = Q ;
in other words, if (2) holds so does (1).
Cor. 1 . If we transfer any term in an equation from the one side
to the other, at the same time reversing its sign of addition or subtrac
tion, 07' if toe reverse all the signs on both sides of an equation, we
deduce in each case an equivalent equation.
For, if P + Q = R + S, say,
then P + QS = R + SS,
that is, P + Q  S = R.
Again, if P + Q = R + S,
then P + QPQRS = R + SPQRS,
that is,  R  S =  P  Q,
or PQ=RS.
Cor. 2. Every equation can be reduced to an equivalent equation of
the form —
R=0.
For, if the equation be P = Q,
* In all that follows all functions of the variables that appear are supposed
not to become infinite for any values of the variables contemplated. Cases
where this understanding is violated must be considered separately.
292 MULTIPLICATION BY A FACTOR chap.
we have P  Q = Q  Q,
that is P  Q = 0,
which is of the form R = 0.
Example.
 Za? + 2x 2 + Sx = or  x  3.
Subtracting x  x  3 from both sides, we have the equivalent equation
3a; 3 + ar + 4a: + 3 = 0.
Changing all the signs, we have
3.« 3  x 2  4x  3 = 0.
In this way an integral equation can always be arranged with all its terms on
one side, so that the coefficient of the highest term is positive.
§ 9.] It follows from the fundamental laws of algebra that
if P = Q (1),
then PR = QR (2),
the step being reversible if R is a constant differing from 0, but not if
R be a function of the variables*
For, if PR = QR,
an equivalent equation is, by § 8,
PR  QR = (3),
that is, (P  Q)R = (4).
Now, if R be a constant 4= 0, it will follow from (4), by the
general principle of § 7, that
PQ = o,
which is equivalent to P = Q.
But, if R be a function of the variables, (4) may also be satisfied
by values of the variables that satisfy
R = (5) ;
and such values Avill not in general satisfy (1).
In fact, (2) is equivalent, not to (1), but to (1) and (5) as
alternatives.
* This is spoken of as "multiplying the equation by R." Similarly the
process of § 8 is spoken of as "adding or subtracting R to or from the equa
tion." This language is not strictly correct, but is so convenient that we
shall use it where no confusion is to be feared.
XIV DEDUCTION OF INTEGRAL FROM RATIONAL EQUATION 293
Cor. 1. From the above it follows that dividing both sides of
an equation by any function other than a constant not equal to zero is not
a legitimate process of derivation, since we may thereby lose solutions.
Thus PR = QR is equivalent to J _ Y ;
whereas PR/R = QR/R*
gives P = Q,
which is equivalent merely to
PQ = 0.
Example. If we divide both sides of the equation
(x~l)x 2 = i{xl) (a)
by £ 1, we reduce it to x = 4 (/3),
which is equivalent to (x  2) (x + 2) = 0.
(a), on the other hand, is equivalent to
(xl)(w2)(a;+2)=0.
Hence (a) has the three solutions x1, x = 2, x=  2 ; while (/3) has only the
two x = 2, x=  2.
Cor. 2. To multiply or divide both sides of an equation by
any constant quantity differing from zero is a reversible process
of derivation. Hence, if the coefficients of an integral equation be
fractional either in the algebraical or in the arithmetical sense, we can
always find an equivalent equation in which the coefficients are all
integral, and have no common measure.
Also, we can always so arrange an integral equation that the co
efficient of any term we please, say the highest, shall be + 1 .
Example 1.
3a + 2 6x + S_2x+_i
~~4~~ + 5  8
gives, on multiplying both sides by 40,
10(3x + 2) + 8(6i + 3) = 5(2x+4),
that is, 30^ + 20 + 48^ + 24 = 10x + 20,
whence, after subtracting lO.e + 20 from both sides,
68a; + 24 = ;
* As we are here merely establishing a negative proposition, the reader
may, to fix his ideas, assume that all the letters stand for integral functions
of a single variable.
294 DEDUCTION OF INTEGRAL FROM RATIONAL EQUATION chap.
whence again, after division of both sides by 68,
Example 2.
X+ Tf=°
K q p~q' J\ p P +
 q y)=^y
If we multiply both sides by pq(p  q) {p + q), that is, by pq(2>~ ~ <T)i we derive
the equivalent equation
{{p 2 q 2 )x+pqy} {{2rq)x+pqy)=2pq(pq)xy,
that is, ( p 2  q 2 ) 2 x 2 + 2pq{ p 2  q 2 )xy +p 2 qy 2 = 2pq{ p 2  q 2 )xy,
which is equivalent to [p 2 q 2 ) 2 x 2j rp 2 q 2 y 2 = Q.
Cor. 3. From every rational algebraical equation an integral equa
tion can be deduced ; but it is possible that extraneous solutions may
be introduced in the process.
Suppose we heave P = Q (a),
where P and Q are rational, but not integral. Let L he the
L.C.M. of the denominators of all the fractions that occur either
in P or in Q, then LP and LQ are both integral. Hence, if we
multiply both sides of (a) by L, we deduce the integral equation
LP = LQ (/?).
Since, however, the multiplier L contains the variables, it is
possible that some of the solutions of L = may satisfy (ft), and
such solutions would in general be extraneous to (a). We say
possible ; in general, however, this will not hajjpp n, because P
and Q contain fractions whose denominators are factors in L.
Hence the solutions of L = will in general make either P or Q
infinite, and therefore (P  Q)L not necessarily zero. The point
at issue will be best understood by studying the two following
examples : —
Example 1.
. z 2 6a; + 8 x2
2 *' 3+ x2 = x=S (a) 
If we multiply both sides by (x 2)(x 3), we deduce the equation
(2x3)(x2)(x3) + (x 2 6x+8)(x3) = (x2) 2 (/3),
which is integral, and is satisfied by any solution of (a). "We must, however,
examine whether any of the solutions of (x 2)(x 3) = satisfy (/3). These
solutions are x=2 and x = 3. The second of these obviously does not satisfy
(/3), and need not be considered; but x=2 does satisfy (/3), and we must
examine (a) to see whether it satisfies that equation also.
XIV RAISING BOTH SIDES TO SAME POWER 295
Now, since x 2  6a: + 8 = (as— 2) (a;  4), (a) may be written in the equivalent
form
x — 2
2x  3 + x  4 :
P = Q
(1)
P  Q =
(2).
+ P» 2 Q + P' l " 3 Q 2 + . .
. +Q n \ ■
x3
which is obviously not satisfied by x = 2.
It appears, therefore, that in the process of integralisation we have intro
duced the extraneous solution x=2.
Example 2.
2xB + 2 ^^ 8 = X ~l («<).
x  2 x  3
Proceeding as before, we deduce
{2x  3) (a;  2) (x  3) + (2a: 2  6x + 8) (a;  3) = (x  2f iff).
It will be found that neither of the values 33=2, a; = 3 satisfies (/3').
Hence no extraneous solutions have been introduced in this case.
N.B. — The reason why a? = 2 satisfies (/3) in Example 1 is that the numer
ator a, 2  6a: + 8 of the fraction on the left contains the factor x2 which
occurs in the denominator.
Cor. 4. Raising both sides of an equation to the same integral
power is a legitimate, but not a reversible, process of derivation.
The equation
is equivalent to
If we multiply by P' 1 " 1 + P n ~ 2 Q + P» 3 Q 2 + . . . + Q M ~\ we
deduce from (2)
P* _ Q» = (3),
which is satisfied by any solution of (1); (3), however, is not
equivalent to (1), but to
P = Q\
P»1 + P»2Q + . . > + Q»l = oj
It will be observed that, if we start with an equation in the
standard form P  Q = 0, transfer the part Q to the righthand
side, and then raise both sides to the wth power, the result is the
same as if we had multiplied both sides of the equation in its
original form by a certain factor. To make the introduction of
extraneous factors more evident we chose the latter process ; but
in practice the former may happen to be the more convenient.*
If the reader will reflect on the nature of the process described
in chap. x. for rationalising an algebraical function by means
of a rationalising factor, he will see that by repeated operations of
this kind every algebraical equation can be reduced to a rational
* See below, § 12, Example 3.
296 EVERY ALGEBRAICAL EQUATION CAN BE INTEGRALISED chap.
form ; but at each step extraneous solutions may be introduced.
Hence
Cor. 5. From every algebraical equation we can derive a rational
integral equation, which will be satisfied by any solution of the given
equation ; but it does not follow that every solution^ or even that any
solution, of the derived equation will satisfy the original one.
Example 1. Consider the equation
V(z+l) + V(*l) = l (a),
where the radicands are supposed to be real and the square root to have the
positive sign. *
(a) is equivalent to \J(x + l) = l  \J{x 1),
whence we derive, by squaring,
x+l = \+xl2'sJ{x\),
which is equivalent to 1 =  2\J(x  1).
From this last again, by squaring, we derive
1 = 4(351),
which is equivalent to the integral equation
Ax 5 = (/3),
the only solution of which, as Ave shall see hereafter, is x = \.
It happens here that x=\ is not a solution of (a), for V(f + 1) + V(i"  1)
— S i 1 _o
— T + V — ■
Example 2.
V(.c+i)V( >, i) = i (o).
Proceeding exactly as before we have
a; + l = l + a:l + 2V(*l)i '
1=+2V(*1),
1 = 4(31),
4a 5 = (/3'),
Here (/?') gives a:=f, which happens this time to be a solution of the
original equation.
We conclude this discussion by giving two propositions
applicable to systems of equations containing more than one
equation. These by no means exhaust the subject ; but, as our
object here is merely to awaken the intelligence of the student,
the rest may be left to himself in the meantime.
§ 10.] From the system
P, = 0, P 2 =0, . . ., P n = (A)
we derive
L 1 P 1 .+ IJP f +. . .+LnPn= 0, P 8 = 0, . . ., P n = (B),
and the two will be equivalent if J Jl be a constant differing from 0.
f When \Jx is imaginary, its "principal value" (see chap, xxix.) ought
to be taken, unless it is otherwise indicated.
XIV
EXAMPLES OF DERIVATION 297
Any solution of the system (A) reduces P 15 P 2 , . . ., P n all
to 0, and therefore reduces L^ + LP, + . . . + L n P n to 0, and
hence satisfies (B).
Again, any solution of (B) reduces P 2 , P 3 , . . ., P n all to 0,
and therefore reduces L,P, + L 2 P 2 + . . . + L n P„ = to L,P, = 0,
that is to say, if L, he a constant 4= 0, to Pj = 0. Hence, in this
case, any solution of (B) satisfies (A).
If L, contain the variables, then (B) is equivalent, not to (A)
simply, but to
f P, = 0, P 2 = 0, . . ., P w = 0]
tL.O, P s = 0, . . , P w = 0j
As a particular case of the above, we have that the two
systems
P = Q, R = S j
and P + R = Q + S, R = S
are equivalent. For these may be written
PQ = 0, R~^S=0;
PQ + RS = 0, RS = 0.
If I, V, m, m! he constants, any one of which may he zero, hut
which are such that lm'  I'm =t= 0, then the two systems
U = 0, U' = 0,
and IV + IV = 0, mV + niV =
are equivalent.
The proof is left to the reader. A special case is used and
demonstrated in chap, xvi., § 4.
§ 11.] Any solution of the system
P = Q, R = S (A)
is a solution of the system
PR = QS, R = S (B);
but the two systems are not equivalent.
From P = Q, we derive
PR = QR,
which, since R = S, is equivalent to
PR = QS.
It follows therefore that any solution of (A) satisfies (B).
298 EXAMPLES OF DERIVATION CHAP.
Starting now with (B), we have
PR = QS (1),
R = S (2).
Since II = S, (1) becomes
PR = QR,
which is equivalent to
(P  Q)R = 0,
that is, equivalent to
fPQ = 0\
Hence the system (B) is equivalent to
fP = Q, R = S\
R = 0, R = SJ"
that is to say, to
/P = Q,R = S\
\R = 0, S = 0J"
In other words, (B) involves, besides (A), the alternative system,
R = 0, S = 0.
Example. From x2 = ly, x = l+y,
a system which has the single solution x=2, y==l, we derive the system
x(x2) = ly\ x=l+y,
which, in addition to the solution x = 2, y—l, has also the solution x =Q,y= 1
belonging to the system
2=0, l + y = 0.
§ 12.] In the process of solving systems of equations, one of
the most commonlyoccurring requirements is to deduce from two
or more of the equations another that shall not contain certain
assigned variables. This is called " eliminating the variables in
question between the equations used for the purpose." In per
forming the elimination Ave may, of course, use any legitimate
process of derivation, but strict attention must always be paid to
the question of equivalence.
Example. Given the system
* 2 +r=i (i),
x+y=l (2),
it is required to eliminate y, that is, to deduce from (1) and (2) an equation
involving x alone.
xiv EXAMPLE OF ELIMINATION 299
(2) is equivalent to
y = lx.
Hence (1) is equivalent to
a?+(lxf=l,
that is to sav, to
2x 2 2x=0,
or, if we please, to
x*x = 0;
and thus we have eliminated y, and obtained an equation in x alone.
The method we have employed (simply substitution) is, of course, only
one among many that might have been selected.
Observe that, as a result of our reasoning, we have that the system (1) and
(2) is equivalent to the system
x 2 x=0 (3),
x + y=l (4),
from which the reader will have no difficulty in deducing the solution of the
given system.
§ 13.] Although, as we have said, the solution of a system
of equations is the main problem, yet the reader will learn,
especially when he comes to apply algebra to geometry, that
much information — very often indeed all the information that is
required— may be derived from a system without solving it,
but merely by throwing it into various equivalent forms. The
derivation of equivalent systems, elimination, and other general
operations with equations of condition have therefore an im
portance quite apart from their bearing on ultimate solution.
We have appended to this chapter a large number of exercises
in this branch of algebra, keeping exercises on actual solution for
later chapters, which deal more particularly with that part of
the subject. The student should work a sufficient number of the
following sets to impress upon his memory the general principles
of the foregoing chapter, and reserve such as he finds difficult for
occasional future practice.
The following are worked out as specimens of various artifices
for saving labour in calculations of the present kind : —
Example 1. Reduce the following equation to an integral form : —
ax + bx + c ax + b , .
(a).
px~ + qx + r px + q
"We may write (a) in the form
• x(ax + b) + c _ ax + b
x{px + q) + r ~px + q
(0).
300 EXAMPLES OF INTEGRALISATION chap.
Multiplying (/3) by (px + q){x(px + q) + r}, we obtain
x(ax + b) (px + q)+ c{px + q) = x(ax + b) (px + q) + r(ax + b) (7).
Now, (7) is equivalent to
c(px + q) = r(ax + b) (5),
which again is equivalent to
(cp  ra)x + (cq  rb) = (e).
The only possibly irreversible step here is that from (/3) to (7).
Observe the use of the brackets in (/3) and (7) to save useless detail.
Example 2.
Integralise
(a x)(x + m)_(a + x) (x  m)
x + n xn
(a).
Since x + m = (x + n) + (m  n), xm = (xn) (mn), (a) may be written
in the equivalent form,
. ./, , mn\ , , /, mv\ ,_,
whence the equivalent form
{ax)(a + x) + (mn)t + =0,
\x+n xn)
that is,
2x + 2{m y {n + a)x =0 (7).
x  n
Multiplying by £(a; 2 ?i 2 ), we deduce from (7) the integral equation
x{x 1 n' i  (m  n) (n + a)) = (5).
In this case the only extraneous solutions that could be introduced are
those of x 1  n = 0.
Note the preliminary transformation in (/3) ; and observe that the order
in which the operations of collecting and distributing and of using any
legitimate processes of derivation that may be necessary is quite unrestricted,
and should be determined by considerations of analytical simplicity. Note
also that, although we can remove the numerical factor 2 in (7), it is not
legitimate to remove the factor x ; x = Q is, in fact, as the student will see by
inspection, one of the solutions of (a).
Example 3.
X, Y, Z, U denoting rational functions, it is required to rationalise the
equation
X /X ± \/ Y ± y'Z ± VU = (a).
We shall take + signs throughout ; but the reader will see, on looking
through the work, that the final result would be the same whatever arrange
ment of signs be taken.
From (a),
VX+vy=vz\/u,
whence, by squaring,
X + Y + 2V(XY)=Z+U+2V(ZTJ) (/3).
xiv EXAMPLES OF RATIONALISATION 301
From (/3),
X + YZU= 2 V(XY) + 2 V(ZU),
whence, by squaring,
(X+YZU) 2 =4XY+4ZU8V(XYZU) (7).
We get from (7),
X 2 + Y 3 + Z + U 2  2XY  2XZ  2XU  2YZ  2YU  2ZU=  8 V(XYZU),
whence, by squaring,
J X s + Y 2 + Z 2 + U 2  2XY  2XZ  2XU  2YZ  2YU  2ZU j 2 = 64X YZU (5).
Since X, Y, Z, U are, by hypothesis, all rational, (5) is the required result.
As a particular instance, consider the equation
V(2k+3) + V(&g+2)  V(2as+5)  \/{&c)=0 [a').
Here X = 2a: + 3, Y=3a;+2, Z = 2a; + 5, U = Zx; and the student will find,
from (5) above, as the rationalised equation,
(48^ + 112a, + 24) 2 =64(2a;+3)(3a: + 2)(2a; + 5)3a; (8').
After some reduction (5') will be found to be equivalent to
(z3) 2 = (Y).
It may be verified that a; = 3 is a common solution of (a') and (e').
Although, for the sake of the theoretical insight it gives, we have worked
out the general formula (5), and although, as a matter of fact, it contains as
particular cases very many of the elementary examples usually given, yet
it is by no means advisable that the student should work particular cases by
merely substituting in (5) ; for, apart from the disciplinary advantage, it
often happens that direct treatment is less laborious, owing to intervening
simplifications. Witness the following treatment of the particular case (a')
above given.
From (a'), by transposition,
V(2aJ+8)+V(3a!+2)=V(2a!+5) + 's/(3a5),
whence, by squaring,
5x + 5 + 2\/(6ar + 13a: + 6) = 5x + 5 + 2 y/(6x + 15x),
which reduces to the equivalent equation
^(6x 2 + lZx+6)=s/(6x i +l5x) (£').
From (/3'), by squaring,
6X 2 + lBx + 6 = 6a; 2 + 15s,
which is equivalent to
a;3 = (5").
Thus, not only is the labour less than that involved in reducing (5), but
(5") is itself somewhat simpler than (5').
Example 4. If
x + y + z=0 (o),
show that
2(2/ 2 + yz + z*f = 3n(y 2 + yz + z") (/3j.
302 EXAMPLES OF TRANSFORMATION chap.
We have
vf + yz + z^if + ziy + z),
=(zx)*+z{x),by(a),
= Z 2 + ZX + X 2 ,
= ir + xy + y 2 , by symmetry.
It follows then that
2(f + yz + z 2 ) 3 = 3(</ 2 + yz + z*)* (7),
and m(y + ijz + z'>) = 3(f + yz + z*) 3 (5).
From (7) and (5), (/3) follows at once.
Example 5. If
x + y+z = (a),
show that
(y + z)(z + x) J
From (a), y + z=x (7),
whence, squaring and then transposing, we have
y*+z*=a?2yz (5).
Similarly z + x=y (7'),
z n  + x = y" i 2zx (5').
From the last four equations we have
2 (y + s 2 ) (z 2 + * 2 ) = s (a 2  2yz) (f  2sc)
(y+2)(«+a:) a;?/
_ aPy 2  2x?z  2y 3 z + ixyz 1
Z\xy2
xy
(3? + y 3 ls
3 + 4; 2 
xyz
= Xvy + 42a, 2  J 2.r% 3 + ,3) (e) .
Now, from (a), by squaring and transposing,
2x' ! =2?xy (f).
Also 2x 2 (y 3 + z 3 )EE2a:y 2 (a: + y),
=  ^xrfz, by (a),
=  xyzZxy (17).
If we use (f) and (r?), (e) reduces to
&+*)(*+*) = _ 5
(y + z)(z + x) J '
which is equivalent to (/3).
The use of the principles of symmetry in conjunction with the 2 notation
in shortening the calculations in this example caunot fail to strike the
reader.
Example 6. If
yzx^zxf
y+z z+x
xiv EXAMPLES OF TRANSFORMATION 303
and if x, y, z be all unequal, show that each of these expressions is equal to
(xy  z 2 )/(x + y), and also to x + y + z.
Denote each of the sides of (a) by U. Then we have
yzx 2
y + z
zxy
~z + x
U (/3),
= U ( 7 ).
Since y + z = and z + x = would render the two sides of (a) infinite, we
may assume that values of a, y, z fulfilling these conditions are not in ques
tion, and multiply (/3) and (7) by y + z and z+x respectively. We then
deduce
yex*(y+z)\J=0 (5),
zxy n (z + x)U = (e).
From (5) and (e), by subtraction, we have
z(xy) + (x 2 y 2 )(xy)V = 0,
that is, (x + y + zV)(xy) = (f).
Now xy — is excluded by our data ; hence, by (f), we must have
x + y + zU = 0, (,),
that is, V=x+y+z (6).
We have thus established one of the desired conclusions. To obtain the
other it is sufficient to observe that (77) is symmetrical in x, y, z. For, if we
start with (77) and multiply by x  z (which, by hypothesis, 4= 0), we obtain
y(x  z) + (x 2  s«)  (x  Z )U = ;
and, combining this by addition with (5),
xyi?(z+y)U=Q;
which gives (since x + y + 0)
jj_ xyz*
x + y
The reader should notice here the convenient artifice of introducing an
auxiliary variable U. He should also study closely the logic of the process,
and be sure that he sees clearly the necessity for the restrictions xy + Q,
x + y + 0.
Example 7. To eliminate x, y, z between the equations
y*+z*=ayz (a),
z 2 + x 2 = bzx (£),
x t +y*=cxy (7),
where x + 0, w + 0, z + 0.
In the first place, we observe that, although there are three variables, yet,
since the equations are homogeneous, we are only concerned with the ratios
of the three. We might, for example, divide each of the equations by x 2 ;
we should then have to do merely with y/x and z/x, each of which might be
regarded as a single variable. There are therefore enough equations for the
purpose of the elimination.
304 EXAMPLES OF ELIMINATION chap.
From (a) and (j3) we deduce, by subtraction,
x*y 2 ={bxay)z (8).
We remark that it follows from this equation that bx ay + 0; for bxay —
would give x 2 = y 2 , and hence, by (7), x = (at least if we suppose c=t=±2).
This being so, we may multiply (/3) by (bx  ay) 2 . "We thus obtain
z(bx  ay) 2 + x 2 (bx — ay) 2 = bxz(bx  ay) 2 ,
whence, using (5), we have
(a; 2  y 2 ) 2 + x 2 (bx — ay) 2 = bx(bx ay) (x 2  y 2 ),
which reduces, after transposition, to
(x 2  y 2 ) 2 = xy(ax  by) (bx  ay),
that is to say, (x 2 + y 2 ) 2  ±x 2 y 2 = xy(axby) (bxay) (e).
Using (7), we deduce from (e)
(c 2  i)x 2 y 2 = xy(ax  by) (bx  ay),
whence, bearing in mind that xy + 0, we get
(c 2 i)xy = ab(x 2 + y 2 )  (a 2 + b 2 )xy,
which is equivalent to
(a 2 + b 2 + c 2  i)xy = ab(x 2 + y 2 ) (f).
Using (7) once more, and transposing, we reach finally
(a 2 + b 2 + c 2  4  abc)xy  0,
whence, since xy + 0, we conclude that
a 2 + b 2 + c 2 4:abc = ( v ),
so that (77) is the required result of eliminating x, y, z between the equations
(a), (/3), (7). Such an equation as (77) is often called the eliminant (or re
sultant) of the given system of equations.
Example 8. Show that, if the two first of the following three equations be
given, the third can be deduced, it being supposed that x =t= y + z + 0.
a 2 (y 2 + yz + z 2 )  ayz(y + z) + y 2 z 2 = (a),
a 2 (z 2 + zx + x 2 )  azx(z + x) + z 2 x 2 = (p),
a 2 (x 2 + xy + y")  axy(x + y)+ x 2 y 2 = (7).
This is equivalent to showing that, if we eliminate z between (a) and (/3), the
result is (7).
Arranging (a) and (/3) according to powers of z, we have
aY a(ay + y 2 )z + (a 2  ay + y 2 )z 2 = (5),
a 2 x 2 a(ax + x 2 )z + (a 2 ax + a?)z 2 = (e ).
Multiplying (5) and (e) by x 2 and y 2 respectively, and subtracting, we get
a 2 xy(x  y)z + {a 2 (x + y)  axy} (x  y)z 2 = 0,
whence, rejecting the factor a(xy)z, which is permissible since x + y, z#=0,
axy+ {a(x + y)xy}z = (f).
Again, multiplying (5) and (e) by a 2 ax + x 2 and a 2 ay + y 2 respectively,
and subtracting, we get, after rejecting the factor a 2 ,
a(x + y)xy+ \a(x + y)}z=0 (77).
XIV
EXERCISES XIX 305
Finally, multiplying (f) and (v) by a{x + y)xy and axy respectively, and
subtracting, we get, since z=$=0,
. {a{x + y) xy) 2  axy {a  (x + y)} = 0,
which gives a 2 (x 2 + xy + y 2 )  axy(x + y) + x 2 y 2 = 0,
the required result.
Exercises XIX.
(On the Reduction of Equations to an Integral Form)
Solve by inspection the following systems of equations : —
(1.) ar43(;r + 2) = 0.
(2.) £±j ;=«.
x x—bx—a
(3.) (ab)xa 2 + b = 0.
(4.) x(bc)+y(ca) + (ab)=0,
ax(b c) + by(c a) + c(ab) = 0.
(5.) x + y + z = a + b + c,
ax + by + cz = a 2 + b' + c 2 ,
bx + cy + az = be + ca + ab.
(6.) For what values of a and b does the equation
(a:  a) (3a: 2)= Bx 2 + bx + 10
become an identity ?
Integralise the following equations ; and discuss in each case the equiva
lence of the final equation to the given one.
(7.) «±« + ?»±" 14
x  4 a;  2
(8.)
l8/ ( g+l) l + 3/(a;l)
a; + 1 + l/fc+l) as  1 + l/(as  1)'
/n X 1111
(9. ) + = +
x+a xc xa x+c
,,~ x flf 6" ar a + b
( 10 ) : — =+:ri=;r; +
(11.)
(12.)
xa x—b xa xb
(3 a) (a; + 10) (8 + x) (x  10)
a;+ll a 11
x 2 +px + q _ x 2 +px + 1
x 2 + rx + 2q ~ x 2 + rx + 2t
(131 (xa) 3 (xb) 3 (xc) 3
K '' (ca)(ab) (ab)(bc) (bc)(ca)
(u) x + T + U a; + TU
1 x 2 + (2t)x + s(2st) x 2 + (2s)x + t(2s<) A
when 2T=s + ts 2 stt 2 .
VOL. I
(22.)
(23.)
(24.)
(25.)
306 EXERCISES XIX, XX chap.
,,_ . x^ + ax + b x 2 + cx + d x" + ax + b' x" + cx + d'
(15.) 1 — = 1 .
x + a x + e x + a x + c
Rationalise the following equations and reduce the resulting equation to
as simple a form as possible : —
(16.) \/X + \/Y + \JZ = 0, where X, Y, Z are rational functions of the
variables.
(17.) \/{x + a) + \/(x + b) + \J(x + c) = 0.
(18.) \/Q.+x) + *J(i + x)y/(9 + x) = 0.
(19.) [xc+ {(xc) n  + f' i i ]/[x + c+ {(xc) + y°} h ] = m.
(20. ) x  a = V {a  vV* 2  <*)} .
(21.) VaB+V(a»7) = 21/V(*7).
V» + 3" _ Va; + 29'
\/{2+x) V(2a;)
V2 + V( 2 + ^O V2  V( 2  *) '
(24. ) \J{x + a) + V(se  a) + V(& + a") + V( 6 «) = 0.
V(l+a; + a; i! ) + V(la: + a; 2 ) _
V(i+*) + V(i*)
(26. ) (y  s) (OSB4 &)* + («») (ay + ft) 4 + (a  ?/) (az + &)* = 0.
(27.) 2V(y~) = 0; and show that 2x= V( 32 2/~) (three variables a, y, z).
<28 » v{»w(»i)( + v^v^i)r vl ^ +1)1 
(29.) sb* +53!* 22=0.
, Qn v v/(a + a;) \/(a + a:) Va;
^ou. ; (.
a; a c
(31. ) v/(a + V») + \/(a  vfc) = #fc
(32.) a£+y*+z*=0,
where a; + 2/ + 2=0.
Exercises XX.
(On t%e Transformation of Systems of Equations.)
[In working this set the student should examine carefully the logic of every
step he takes, and satisfy himself that it is consistent with his data. He
should also make clear to himself whether each step is or is not reversible.]
(1.) If y + s + z J^ + x ±y + 2 = 0, x + y + z*0,
x y z
41 1111
then H ( = ■
x y z x+y+z
xiv EXERCISES XX 307
(2.) If a* + y*=i?,
then {{a? + z*)y\  + {(a*  y s )z) 3 = {(if + a 8 )*} s ( Tait).
(3.) If x, y, z be real, ami it' x\y  z) + y i (z x) +z 4 {xy) = 0, then two at
least of the three must be equal.
(4.) If (x + y + z) 3 = x* + y 3 + z\
then (x + y + z)^ 1 = x 2n + l + y»+ 1 + z 2 "+ ] .
(5.) If
(2> 2 x + 2pry + r 2 z) (q 2 x + 2qsy + s"z) — [pqx + (ps + qr)y + rsz) 2 ,
then either y zx = or ps  qr = 0.
(6.) if jp^+*pr + *> =0i
where r 2 = a; 2 + y + z 2 ,
.1 « 2 V 2 z 2
then 3., 5 + ^ — a + 5Ki 5 = 0,
bcr  p" era  p a~b~  p
where p 2 = a 2 x + b 2 y 2 + c 2 z 2 .
(Important in the theory of the wave surface. — Tait.
(7.) If ±* = !±!! = *±£, andas+y+*=0i
bc ca ab
show that each of them is equal to \/{2a; 2 /2(2<( 2  S6c)}.
a(&y + cz  ckc) = b(cz + ax by) = c(oa; + by cz),
a + b + c = 0,
x + y + z = 0.
x + 2y _ y + 2z _ z + 2x
2a + b~2b + c~2c~+a
(8.)
If
and if
then
(9.)
If
then
(10.)
If
then
(11.)
If
then
/ZxY _Xxy _2x 2
\Za) _ 2a6~2ft 2 "
2ab + b 2 a 2 b
X ~a 2 + ab + b 2 ' V ~a' + ab + b 2
x 3 + y = y + x.
(a b) 2 a + b ab
x = a + b + y. — —, y = —— + —,
4(« + o) 4 a + b
(xa) 2 (yb) 2 = b 2 .
(12.) If a — ax + by + cz + dw,
(3 = bx + ay + dz + cw,
y = ex + dy + az + bio,
d = dx + cy + bz + aw,
and if
/{a, ft 7, 5) = (a + /3 + 7 + 5)(a i 3 + 75)(a/37 + 5)(a + /375),
then
/(«, /3, y, 5) =/(«, 6, c, d)f{x, y, z, w).
(13.) If x+ y +2=0, then Zl/* 2 =(Zl/a;) 2 .
308 EXERCISES XX, XXI chap.
and
Qin+n fyni\n gvi+n
show that (2aj™ n/ (" , +»)) (2a,"" 2 /< m +">) = d m .
(15.) If a _^ =& _J =c _^ a . +0} y=#O jZ *0 J
., , w  z , z  xr x*  y
then a + J . — b] = c + f.
bc c a ab
(16.) If x + {yz  x 2 )j(x 2 + y 2 + z 2 ) be unaltered by interchanging x and y, it
will be unaltered by interchanging x and z, provided x, y, z be all unequal ;
and it will vanish if x + y + z=l.
(17.) If zxy/(y 4 2)  7? = xyz/(z + x)  y 2 , and x + y, then each of these is
equal to xyz/(x + y)  z 2 and also to yz + zx + xy.
(18.) Of the three equations
__x y + z
x 2  w 2 ~ (m + 1 )w 2  (n + 1 )yz '
V _ s + jg
y 2 w 2 (m+l)iv 2 (7i+l)zx'
z _ % + y
z 2 w 2 {m + l)vr~(n + l)xy'
where x #=2/ + z, any two imply the third (Cayley).
(19.) Given
 1  V + ^— = 1,
1 + x + xz l + y + xy l+z + yz
x xy 1
. + , ■ ... + , ■ . ... =h
1+x + xz l+y + xy l+z + yz
none of the denominators being zero, then x = y = z.
(20.) Given ■2(y + z) 2 /x = 3Sx, 2a; *0, prove Z(y + zx) 3 + II{y + zx) = 0.
(21.) Given 2a; =0, prove 2(a? + y 3 )/(a: + ?/) + 5a7/~2(l/ai) = 0.
(22.) Given 2a; = 0, prove that Ske'Sa^/XB 6 is independent of a;, y, z,
(23. ) If 2a; 5 =  5xyzSxy, then 2a = 0, or 2a, 4  Safy + 1.x 2 y 2 + 2Zx 2 yz = 0.
(24.) If II(ar + l) = a 2 +l, n(a^l)=a 2 l, and 2ay = 0, then x+y+z=0
or = ±«.
(25.) If x + y + z + u=0, then 42x* + 32,(y + z)(u + y)(u + z) = 0, where the
2 refers to the four variables a;, y, z, u.
Exercises XXI.
(On Elimination.)
(1.) Eliminate x between the equations
x + l/x = y, x n +l/x 5 — z.
(2.) If z= \/{ayd 2 ly), y = \J(ax 2 a 2 /x), express \J(az 2  a 2 [z) in terms
of a;.
XIV
(3.)
If
the i
I
(i.)
Given
prove that
EXERCISES XXI 309
(p(x) = (a x  ar x )l(a x + ar x )\
F(x)=2/{a*+ar*),
4>{x + y) = (<p(x) + <p(y) )/(l + <p{x)4>{y) ),
¥{x + y) = F(x)¥(y)/(l+<p(x)cl>(y)).
x(y + zx) _ y(z + xy) _ z(x + y z)
a b c
a(b + ca) _b{c + a b) _c(a + bc)
x y z
(5.) Given bz + cy = cx + az — ay + bx, x 2 + y 2 + z 2 =2yz + 2zx + 2xy, prove
that one of the functions «±£>±e = 0.
(6.) Show that the result of eliminating x and y between the equations
x y , x 2 y" „ „
a + b = 1 > S + ^= 1 ' Xy =r>
is (b"c" + a"cr 2 )p* + 2abc 2 cf(bc" + a"d n   2a 2 b")p 2 + arb^d^a 2  c 2 ) (b 2  d?) = 0.
(7.) Eliminate x, y, x', y' from
ax + by = c 2 , x + y 2 = c~, , , _
a'x' + by = c' 2 , x' 2 + y' 2 = c' 2 , Xy + X V ~
(8.) If l/(x + a) + l/(y + a) + l/(z + a) = l/a, with two similar equations it
which b and c take the place of a, show that Z(l/«) = 0, provided a, b, c be all
different.
(9.) Show that any two of the following equations can be deduced from
the other three : —
ax + be = zu, by + ca = nv, cz + db = vx, du + ec = xy, ev + ad = yz.
(10.) Eliminate x, y, z from the three equations
(z + xy){x + yz) = ay~,
(■'' s yz){y + zx) = bzx, [y + z x) {z + x y) = cxy ;
and show that the result is abc — {a + b + c 4) 2 .
CHAPTER XV.
Variation of Functions.
§ 1.] The view which we took of the theory of conditional
equations in last chapter led us to the problem of finding a set
of values of the variables which should render a given conditional
equation an identity. There is another order of ideas of at least
equal analytical importance, and of wider practical utility, which
we now proceed to explain. Instead of looking merely at the
values of the variables x, y, z, . . . which satisfy the equation
f(x, y, z, . . .) = 0,
that is, which render tlie function /(x, y, z, . . .) zero, we consider
all possible values of the variables, and all possible corresponding
values of the function ; or, at least, we consider a number of such
values sufficient to give us a clear idea of the whole ; then, among
the rest, we discover those values of the variables which render
the function zero. The two methods might be illustrated by the
two possible ways of finding a particular man in a line of soldiers.
We might either go straight to some part of the ranks where
a preconceived theory would indicate his presence ; or we might
walk along from one end of the line to the other looking till we
found him. In this new Avay of looking at analytical functions,
the graphical method, as it is called, is of great importance.
This consists in representing the properties of the function in
some way by means of a geometrical figure, so that we can with
the bodily eye take a comprehensive view of the peculiarities of
any individual case.
CHAP. XV
THE GRAPHICAL METHOD
311
GENERAL PROPOSITIONS REGARDING FUNCTIONS OF ONE
REAL VARIABLE.
§ 2.] For the present we confine ourselves to the case of a
function of a single variable, fix)', and we suppose that all the
constants in the function are real numbers, and that only real
values are given to the variable x. "We denote, as in chap,
xiii., § 1 7, f(x) by y, so that
V =/(■'■) (IX
and we shall, as in the place alluded to, speak of x and y as
the independent and dependent variables ; we are now, in fact,
merely following out more generally the ideas broached there.
Y
Fig. 1.
To obtain a graphical representation of the variation of the
function /(.»•) we take two lines X'OX, Y'OY, at right angles to
each other (coordinate axes). To represent the values of x we
measure x units of length, according to any convenient scale,
from the intersection along X'OX to the right if x have a
positive value, to the left if a negative value. To represent the
values of y we measure lengths of as many units, according to
the same or, it may be, some other fixed scale, from X'OX
parallel to Y'OY, upwards or downwards according as these
values are positive or negative.
For example, suppose that, when Ave put x =  20, x = — 7,
x= + IS, x= + 37, the corresponding values of
312
are
CONTINUITY OF FUNCTION AND OF GRAPH
/(20), /(7), /( + 18), /(+37)
+ 4,  10, +7,  6
CHAP.
respectively ; so that we have the following scheme of corre
sponding values : —
X
y
20
 7
+ 18
+ 37
+ 4
 10
+ 7
 6
l
Then we measure off OM 2 (left) =20, OM 4 (left) = 7, OM 7 (right)
= 18, OM 9 (right) = 37 ; and M 2 P 2 (up) = 4, M 4 P 4 (down) = 10,
M 7 P 7 (up) = 7, M 9 P 9 (down) = 6.
To every value of the function, therefore, corresponds a re
presentative point, P, whose abscissa (OM) and ordinate (MP)
represent the values of the independent and dependent variables;
that is to say, the value of x and the corresponding value of /(./:).
Now, when we give x in succession all real values from — oo to
+ qo , y will in general * pass through a succession of real values
without at any stage making a sudden jump, or, as it is put, without
becoming discontinuous. The representative point will therefore
trace out a continuous curve, such as Ave have drawn in Fig. 1.
This curve Ave may call the graph of the function.
§ 3.] It is obvious that when Ave Icuoav the graph of a
function we may find the value of the function corresponding to any
value of the independent variable x Avith an accuracy that depends
merely on the scale of our diagram and on the precision of our
drawing instruments. All Ave have to do is to measure off the
value of x in the proper direction, OM 7 say ; then draw a
parallel through M 7 to the axis of y, and find the point P 7 Avhere
this parallel meets the graph ; then apply the compasses to M 7 P 7 ,
and read off the number of units in M 7 P 7 by means of the scale
of ordinates. This number, taken positive if P 7 be above the
We shall retina to the exceptional eases immediately.
xv GRAPHICAL SOLUTION OF AX EQUATION 313
axis of x, negative if below, will be the required value of the
function.
The graph also enables us to the same extent to solve the
converse problem, Given the value of the function, to find the corre
sponding value or values of the independent variable.
Suppose, for example, that Fig. 1 gives the graph of f(x),
and Ave wish to find the values of x for which f(x) = + 7. All
we have to do is to measure ON 7 = 7 upwards from on the axis
of y ; then draw aline (dotted in the figure) through N 7 parallel
to the axis of x, and mark the points where this line meets the
graph. If P 7 be one of them, we measure N 7 P 7 (obviously = OM 7 )
by means of the scale of abscissae, and the number thus read off
is one of the values of x for which f(x) = + 7 ; the others are
found by taking the other points of intersection, if such there be.
Observe that the process we have just described is equivalent
to solving the equation
f{x) = + 7.
In particular we might look for the values of x for which /(./•)
reduces to zero. When/(a;) becomes zero, that is, when the ordin
ate of the graphic point is zero, the graph meets the axis of x.
The axis of x, then, in this case acts the part formerly played by
the dotted parallel, and the values of x required are  OMj,
OM 3 , +OM, +OM 8 , +OM 10 , where OM„ OM 3 , Sec, stand
merely for the respective numbers of units in these lengths when
read off upon the scale of abscissae. Hence
By means of the graph of the function /(.c) we can solve the
equation
/(■*) = (2).
The roots of this equation are, in point of fact, simply the values
of x which render the function f(x) zero ; we may therefore,
when it is convenient to do so, speak of them as the roots of the
function itself.
§ 4.] The connection between the general discussion of a
function by means of the graphical or any other method and the
problem of solving a conditional equation will now be apparent
to the reader, and he will naturally ask himself how the graph
314
EXAMPLE
CHAP.
is to be obtained. We cannot, of course, lay down all the
infinity of points on the graph, but we can in various ways infer
its form. In particular, we can assume as many values of the
independent variable as we please, and, from the known form of
the function f(x), calculate the corresponding values of y. We
can thus lay down as many graphic points as we please. If care
be taken to get these points close enough where the form of the
curve appears to be changing rapidly, we can draw with a free
hand a curve through the isolated points which will approach
the actual graph sufficiently closely for most practical purposes.
When the form of the function is unknown, and has to be
determined by observation — as, for example, in the case of the
curve which represents the height of the barometer at different
times during the day — the course we have described is the one
actually followed, only that the value of y is observed and not
calculated.
Before going further into details it will be well to illustrate
by a simple example the above process, Avhich may be unfamiliar
to many readers.
Example.
Let the function to be discussed be lx", then the equation (1) which
determines the graph is y= 1  a?.
We shall assume, for the present without proof, what will probably be at
once admitted by the reader, that, as x increases without break from up to
+ oo, x 2 increases without break from up to + oo ; and that a ,2 > = <1,
according as x> = < 1.
Consider, in the first place, merely positive values of x. When x = 0,
y = l ; and, so long as x<l, lx is positive. When x=l, y = l 1 = 0.
When x>l, then » ,2 >1 and 1 x 2 is negative. Hence from x = until x=l,
lx" continually decreases numerically, but remains always positive. When
x = l, lx 2 becomes zero, and when x is further increased lx 2 becomes
negative, and remains so, but continually increases in numerical value.
We may represent these results by the following scheme of corresponding
values : —
X
y
i
<+l
+
+ 1
>+l

+ 0O
— CO
XV
EXAMPLE
315
The general form of the graph, so far as the righthand side of the axis of y
is concerned, will be as in Fig. 2.
As regards negative values of
x and the lefthand side of the
axis of }i, in the present case,
it is merely necessary to notice
that, if we put a; = a, the re
sult, so far as 1  x 2 is concerned,
is the same as if we put x = +a;
for 1  (  af= 1  ( + a). Hence
for every point P on the curve,
whose abscissa and ordinate are
+ OM and + MP, there will be
a point P', whose abscissa and
ordinate are  OM and + M P. P
and P' are the images of each
other with respect to Y'Y ; and the part AP'B' of the graph is merely an
image of the part APB with respect to the line Y'Y.
Let us see what the graph tells us regarding the function 1 x".
First we see that the graph crosses the araxis at two points and no more,
those, namely, for which x= + 1 and x=  1. Hence the function 1 x has
only two roots, +1 and  1 ; in other words, the equation
1^ =
has two real roots, x — hi, x=  1, and no more.
Secondly. Since the part BAB' of the graph lies wholly above, and the
parts C'B', CB wholly below the a:axis, we see that, for all real values of x
lying between 1 and +1, the function 1 ar is positive, and for all other
real values of x negative.
Thirdly. We see that the greatest positive value of 1 x 2 is 1, correspond
ing to x—0 ; and that, by making x sufficiently great (numerically), we can
give 1 x a negative value as large, numerically, as we please.
All these results could be obtained by direct discussion of the function,
but the graph indicates them all to the eye at a glance.
§ 5.] Hitherto we have assumed that there are no breaks or
discontinuities in the graph of the function. Such may, how
ever, occur ; and, as it is necessary, when we set to work to
discuss by considering all possible cases, above all to be sure
that no possible case has escaped our notice, we proceed now to
consider the exceptions to the statement that the graph is in
general a continuous curve.
f.
The function f(x) may become infinite for a finite value of x.
316
DIFFERENT KINDS OF DISCONTINUITY
CHAP.
Example 1.
Consider the function 1/(1 a;). "When a; is a very little less than + 1,
say x= 99999, then y = \j(\x) gives y = + 1/00001 = +100000 ; that is to
say, y is positive and very large ; and it is obvious that, by bringing x suffi
ciently nearly up to + 1, we can give y as large a positive value as we please.
On the other hand, if a: be a very little greater than +1, say x= +1 '00001,
then 2/=l/( "00001)= 100000 ; and it is obvious that, by making x ex
ceed 1 by a sufficiently small quantity, y can be made as large a negative
quantity as we please.
The graph of the function 1/(1  x) for values of x near + 1 is therefore as
follows : —
The branch BC ascends to
an infinite distance along
KAK' (a line parallel to the
2/axis at a distance from it
= +1), continually coming
nearer to KAK', but never
reaching it at any finite dis
tance from the a;axis. The
branch DE comes up from an
infinite distance along the
other side of KAK' in a similar
manner.
Here, if we cause x to in
crease from a value OL very
little less than + 1 to a value
OM very little greater, the
value of y will jump from a very large positive value + LC to a very large
negative value  WD ; and, in fact, the smaller we make the increase of x,
provided always we pass from the one side of + 1 to the other, the larger will
be the jump in the value of y.
It appears then that, for x= +\, 1/(1— a) is both infinite and discon
tinuous.
Example 2.
y= 1/(1 x)~.
We leave the discussion to the
reader. The graph is as in Fig. 4.
The function becomes infinite
when x= +1 ; and, for a very small
increment of x near this value, the
increment of y is very large. In fact,
if we increase or diminish x from the
value + 1 by an infinitely small
amount, y will diminish by an in
finitely great amount.
Here again we have infinite value of the function, and accompanying
discontinuity.
Y
B
K
M
O
I
k'
\
X
r
Fig. 3.
Fio. 1.
XV
DIFFERENT KINDS OF DISCONTINUITY
31
Fio. 5.
II. The value of the function may make a jump without becoming
infinite.
The graph for the neighbourhood of such a value would be
of the nature indicated in Fig. 5, where, while x passes through
the value (DM, y jumps from MP
to MQ.
Such a case cannot, as we shall
immediately prove, occur with in
tegral functions of x. In fact it ol M
cannot occur with any algebraical
function, so that we need not
further consider it here.
The cases we have just considered lead us to give the follow
ing formal definition.
A function is said to he continuous when for an infinitely small
change in the value of the independent variable the change in the value
of the function is also infinitely small ; and to be discontinuous when
for an infinitely small change of the independent variable the change in
the value of the function is either finite or infinitely great.
III. It may happen that the value of a function, all of whose con
stants are real, becomes imaginary for a real value of its variable.
Example.
This happens with the function + \J{1  x"). If we confine ourselves to the
positive value of the square root, so that we have a singlevalued function to
deal with, the graph is as in Fig. 6 : —
a semicircle, in fact, whose centre is at the
origin.
For all values of a;>+l, or <— 1, the
value of y= + \/(l  ar) is imaginary ; and the
graphic points for them cannot be constructed
 in the kind of diagram we are now using.
The continuity of the function at A can
not, strictly speaking, be tested ; since, if we
attempt to increase x beyond + 1, y becomes
imaginary, and there can be no question of the
magnitude of the increment, from our present point of view at least.*
No such case as this can arise so long as /(.?•) is a rational
algebraical function.
Fio. 6.
See below, § 18.
318 LIMITING CASES chap.
We have now enumerated the exceptional cases of functional
variation, so far at least as is necessary for present purposes.
Graphic points, at which any of the peculiarities just discussed
occur, may be generally referred to as critical points.
ON CERTAIN LIMITING CASES OF ALGEBRAICAL OPERATION.
§ 6.] We next lay down systematically the following propo
sitions, some of which we have incidentally used already. The
reader may, if he choose, take them as axiomatic, although, as
we shall see, they are not all independent. The important
matter is that they be thoroughly understood. To secure that
they be so we shall illustrate some of them by examples. In
the meantime we caution the reader that by " infinitely small "
or "infinitely great" we mean, in mathematics, " smaller than
any assignable fraction of unity," or " as small as we please," and
"greater than any assignable multiple of unity," or "as great as
Ave please." He must be specially on his guard against treating
the symbol <x> , which is simply an abbreviation for " greater
than any assignable magnitude," as a definite quantity. There
is no justification for applying to it any of the laws of algebra,
or for operating with it as we do with an ordinary symbol of
quantity.
I. If P be constant or variable, provided it does not become
infinitely great when Q becomes infinitely small, then when Q becomes
infinitely small PQ becomes infinitely small.
Observe that nothing can be inferred without further examin
ation in the case where P becomes infinitely great when Q
becomes infinitely small. This case leads to the socalled inde
terminate form oo x 0.*
Example 1.
Let us suppose, for example, that P is constant, =100000, say. Then, if
we make Q = l/100000, we reduce PQ to 1 ; if we make Q = 1/100000000000,
we reduce PQ to 1/1000000 ; and so on. It is abundantly evident, therefore,
that by making Q sufficiently small PQ can be made as small as we please.
* Indeterminate forms are discussed in chap, xxv
ZV LIMITING CASES 319
Example 2.
LetP = x + l, Q = a:1.
Here, when x is made to approach the value + 1, P approaches the finite
value + 2, while Q approaches the value 0. Suppose, for example, we put x — 1
+ 1/100000, then
PQ = (2 + 1/100000) x 1/100000,
^2/100000 + 1/10 10 ,
and so on. Obviously, therefore, by sufficiently diminishing Q, we can make
PQ as small as we please.
Example 3.
Y = l/(xl), Q=jb1.
Here we have the peculiarity that, when Q is made infinitely small, P (see
below, Proposition III.) becomes infinitely great. We can therefore no longer
infer that PQ becomes infinitely small because Q does so. In point of fact,
PQ=(« l)/(x l)=l/(aj+l), which becomes 1/2 when x=l.
II. If ¥ be either constant or variable, provided it do not become
infinitely small when Q becomes infinitely great, then when, Q becomes
infinitely great PQ becomes infinitely great.
The case where P becomes infinitely small when Q becomes
infinitely great must be further examined ; it is usually referred
to as the indeterminate form x oo .
Example 1.
Suppose P = l/100000. Then, by making Q = 100000, we reduce TQ to 1 ;
by making Q = 100000000000 we reduce PQ to 1000000 ; and so on. It is
clear, therefore, that by sufficiently increasing Q we could make PQ exceed
any number, however great.
The student should discuss the following for himself : —
Example 2.
Y=x + \, Q = l/(zl).
PQ = oo whena;=l.
Example 3.
P=(a;l) a , Q = l/(.*l).
PQ = 0when x=l.
III. If P be either constant or variable, provided it do not become
infinitely small when Q becomes infinitely small, then when Q becomes
infinitely small P/Q becomes infinitely great.
The case where P and Q become infinitely small for the same
value of the variable requires further examination. This gives
the socalled indeterminate form •
320 LIMITING CASES
CHAP.
Example 1.
Suppose P constant = 1/100000. If we make Q = 1/100000, P/Q becomes
1 ; if we make Q = 1/100000000000, P/Q becomes 1000000 ; and so on. Hence
we see that, if only we make Q small enough, we can make P/Q as large as
we please.
The student should examine arithmetically the two following cases : —
Example 2.
T=x + 1, Q=a1.
P/Q = oo when x—1.
Example 3.
P=*l, Q=«l.
P/Q=l whencc=l.
IV. If P be either constant or variable, provided it do not become
infinitely great when Q becomes infinitely great, then when Q becomes
infinitely great P/Q becomes infinitely small.
The case where P and Q become infinitely great together re
quires further examination. This gives the indeterminate form
00
00
Example 1.
Suppose P constant = 100000. If we make Q = 100000, P/Q becomes 1;
if we make Q = 100000000000, P/Q becomes 1/1000000 ; and so on. Hence by
sufficiently increasing Q we can make P/Q less than any assignable quantity.
Example 2.
P=aj+1, Q=l/(asl).
p/Q = when x—1.
Example 3.
P = l/(a;l) a , Q = l/(.zl).
P/Q = oo when sc=l.
V. If P and Q each become infinitely small, then P + Q becomes
infinitely small.
For, let P be the numerically greater of the two for any
value of the variable. Then, if the two have the same sign, and,
a fortiori, if they have opposite signs, numerically
P + Q < 2P.
Now 2 is finite, and, by hypothesis, P can be made as small as
we please. Hence, by I. above, 2P can be made as small as
we please. Hence P + Q can be made as small as we please.
VI. If either P or Q become infinitely great, or if P and Q each
XV
LIMITING CASES 321
become infinitely great and both have finally the same sign, then P + Q
becomes infinitely great.
Proof similar to last.
The inference is not certain if the two have not ultimately
the same sign. In this case there arises the indeterminate
form oo  co .
Example 1.
?=x*/(xl)*, Q = (2x\)/(xir.
When x= 1, we ha ve P = 1/0 = + oo , Q , = 1 /0 = + oo . Also
F + Q = rA^+ 2X ~ l 3S + 2X ~ l
(xi)'(ziy (xiy
Example 2.
2 i i
= =oo, when x=l.
P=a?/(a!l) 2 , q=(2xl)J(xlf.
Here x=l makes P=+oo, Q=oo, so that we cannot infer r + Q = <».
In fact, in this case,
p. __£! 2*l_( a :l) a
L+K (;clr< (xlfixl)* 1
for all values of x, or, say, for any value of x as nearly = + 1 as we please. In
this case, therefore, by bringing x as near to +1 as we please, we cause the
value of P + Q to approach as near to +1 as we please.
§ 7.] The propositions stated in last paragraph are the funda
mental principles of the theory of the limiting cases of algebraical
operation. This subject will be further developed in the chapter
on Limits in the second part of this work.
In the meantime we draw the following conclusions, which
will be found useful in what follows : —
I. If P = P^ . . . P Jt , then P will remain finite if P,, P 2 , . . .,
P n all remain finite.
P will become infinitely small if one or more of the functions
P n P 2 , . . ., ~P n become infinitely small, provided none of the remain
ing ones become infinitely great.
P will become infinitely great if one or more of the functions P n
P 2 , . . ., P n become infinitely great, provided none of the remaining
ones become infinitely small.
II. If S = P, + P, + . . . + P n , then S will remain finite if P,,
P 2 , . . ., P,j, each remain finite.
VOL. I Y
322 LIMITING CASES chap.
S will become infinitely small if P u P 2 , . . ., P n each become in
finitely small.
S vMl become infinitely great if one or more of the functions P,,
P 3 , . . ., P n become infinitely great, provided all those that become
infinitely great have the same sign.
III. Consider the quotient P/Q.
p
j xoill certainly be finite if both P and Q be finite,
may be finite if P = 0, Q = 0,
or if P — oo , Q = oo .
P
pr icill certainly = if P = 0, Q. 4= 0,
or ?/ P 4= co , Q = oo •
may =0i/P = 0, Q = 0,
or if P = co , Q = co ,
P
pr will certainly = co (/* P = oo , Q 4= oo ,
^ on/P + O, Q = ;
may = oo if P = 0, Q = 0,
or ?/ P = oo , Q = co .
ON THE CONTINUITY OF FUNCTIONS, MORE ESPECIALLY OF
RATIONAL FUNCTIONS.
§ 8.] We return now to the question of the continuity of
functions.
By the increment of a function f(x) corresponding to an increment
h of the independent variable x we mean f(x + h)  f {•>')■
For example, HfixJ—x 1 , the increment is (x + h) 2 x i —2xh + Jr.
lff(x) = l/x, the increment is l/{x + h)l/x= h/x(x + h).
The increments may be either positive or negative, according
partly to choice and partly to circumstance. The increment of
the independent variable x is of course entirely at our disposal ;
but when any value is given to it, and when x itself is also
assigned, the increment of the function or dependent variable
is determined.
xv CONTINUITY OF A SUM OR PRODUCT 323
Example.
Let the function be 1/x, then if x=l, h = 3, the corresponding increment
ofl/.ris 3/1(1+3)= 3/4. Ifz=2, h = 3, the increment of 1/x is 3/2(2 + 3)
~  3/10, and so on.
If P be a function of x, and p denote its increment when x
is increased from x to x + h, then, by the definition of p, P +p is
the value of P when x is altered from x to x + A.
"We can now prove the following propositions : —
I. The algebraic sum of any finite number of continuous functions
is a continuous function.
Let us consider S = P  Q + R, say. If the increments of P,
Q, R, when x is increased by h, be p, q, r, then the value of S,
when x is changed to x + h, is (P + p)  (Q + q) + (R + r) ; and the
increment of S corresponding to h is p  q + r. Now, since P, Q,
R are continuous functions, each of the increments, p, q, r, becomes
infinitely small when h becomes infinitely small. Hence, by § 7,
I., p — q + r becomes infinitely small when h becomes infinitely
small. Hence S is a continuous function.
The argument evidently holds for a sum of any number of
terms, provided there be not an infinite number of terms.
II. The poduct of a finite number of continuous functions is a
continuous function so long as all factors remain finite.
Consider, in the first place, PQ. Let the increments of P
and Q, corresponding to the increment h of the independent vari
able x, be p and q respective^. Then when x is changed to x + h
PQ is changed to (P + p) (Q, + q), that is, to VQ, + pQ, + q~P + pq.
Hence the increment of PQ corresponding to h is
jjQ + ^P +pq.
Xow, since P and Q are continuous, p and q each become in
finitely small when h becomes infinitely smalL Hence by § 7, I.
and II., it follows that pQ + qP + pq becomes infinitely small
when h is made infinitely small ; at least this will certainly be
so, provided P and Q remain finite for the value of x in question,
which we assume to be the case.
It follows then that PQ is a continuous function.
Consider now a product of three continuous functions, say
PQR. By what has just been established, PQ is a continuous
324 ANY INTEGRAL FUNCTION CONTINUOUS chap.
function, which we may denote by the single letter S ; then
PQR = SR where S and R are continuous. But, by last case, SR
is a continuous function. Hence PQR is a continuous function.
Proceeding in this way, we establish the proposition for any
finite number of factors.
Cor. 1. If A be constant, and P a continuous function, then AP
is a continuous function.
This can either be established independently, or considered
as a particular case of the main proposition, it being remembered
that the increment of a constant is zero under all circumstances.
Cor 2. Ax m , where A is constant, and in a positive integer, is a
continuous function.
For x m = x x xx . . . x x (m factors), and x is continuous, being
the independent variable itself. Hence, by the main proposi
tion, x m is continuous. Hence, by Cor. 1, At'" is a continuous
function.
Cor. 3. Every integral function of x is continuous ; and cannot
become infinite for a finite value of x.
For every integral function of x is a sum of a finite number
of terms such as Ax" 1 . Now each of these terms is a continuous
function by Cor. 2. Hence, by Proposition I., the integral func
tion is continuous. That an integral function is always finite
for a finite value of its variable follows at once from § 7, I.
III. If P and Q be integral functions of x, then P/Q is finite and
continuous for all finite values of x, except such as render Q = 0.
In the first place, if Q 4= 0, then (see § 7, III.) P/Q can only
become infinite if either P, or both P and Q, become infinite ; but
neither P nor Q can become infinite for a finite value of x, because
both are integral functions of x. Hence P/Q can only become
infinite, if at all, for values of x which make Q = 0.
If a value which makes Q = makes P 4= 0, then P/Q certainly
becomes infinite for that Aalue. But, if such a value makes both
Q = and also P = 0, then the matter requires further investi
gation.
Next, as to continuity, let the increments of P and Q corre
XV CONTINUITY OF A RATIONAL FUNCTION 325
sponding to h, the increment of x, be p and q as heretofore.
Then the increment of P Q is
Y+p _ P _ pQ  gP
Q + 'l Q~Q(Q + ?)"
Now, by hypothesis, ^ and q each become infinitely small
when h does so. Also P and Q remain finite. Hence pQ  qP
becomes infinitely small. It follows then that (pQ  #P)/Q(Q + q)
also becomes infinitely small when h does so, provided always
(see § 6) that Q does not vanish for the value of x in question.
Example.
The increment of l/(se— 1) corresponding to the increment, h, of a: is
l/(x + hl)l/(xl)=  h/(x  1) (a*+ h  1). N ow, if x = 2, say, this becomes
 hftl+h), which clearly becomes infinitely small when h is made infinitely
small. On the other hand, if x—1, the increment is h/Oh, which is
infinitely great so long as h has any value differing from by ever so little.
§ 9.] When a function is finite and continuous between two
values of its independent variable x = a and x = b, its graph forms
a continuous curve between the two graphic points whose
abscissas are a and b ; that is to say, the graph passes from the
one point to the other without break, and without passing any
where to an infinite distance.
From this we can deduce the following important pro
position : — 
Jff(x) be continuous from x  a to x = b, and iff(a) =p, f(b) = q,
then, as x passes through every algebraical value between a and b, fix)
passes at least once, and, if more than once, an odd number of times
through every algebraical value between p and q.
Let P and Q be the graphic points corresponding to x = a and
x = b, AP and BQ their ordinates ; then AP = p, BQ = q. We
have supposed p and q both positive ; but, if either were negative,
we should simply have the graphic point below the a'axis, and
the student will easily see by drawing the corresponding figure
that this would alter nothing in the following reasoning.
Suppose now r to be any number between p and q, and
draw a parallel UV to the zaxis at a distance from it equal
to r units of the scale of ordinates, above the axis if r be
positive, below if r be negative. The analytical fact that r is
326
LIMITS FOR THE ROOTS OF AN EQUATION
CHAP.
intermediate to p and q is represented by the geometrical fact
that the points P and Q lie on opposite sides of UV.
Y
Q
u
f?
R R
V
p
A M, M 2 Nlj B X
Fig. 7.
Now, since the graph passes continuously from P to Q, it
must cross the intermediate line UV ; and, since it begins on one
side and ends on the other, it must do so either once, or thrice,
or five times, or some odd number of times.
Every time the graph crosses UV the ordinate becomes equal
to r ; hence the proposition is proved.
Cor. 1. If f(a) be negative andf(b) be positive, or vice versa, then
f(x) has at least one root, and, if more than one, an odd number of
roots, between x = a and x = b, provided f{x) be continuous from x = a
to xb.
This is merely a particular case of the main proposition, for
is intermediate to any two values, one of which is positive and
the other negative. Hence as x passes from a to bf(.r) must pass
at least once, and, if more than once, an odd number of times
through the value 0.
In fact, in this case,
the axis of x plays the
/ \ part of the parallel UV.
Observe, however, in
regard to the converse of
12
Fig. 8.
Fio. 9.
this proposition, that a
function may pass through the value without changing its sign.
For the graph may just graze the reaxis as in Figs. 8 and 9.
XV VALUES FOR WHICH A FUNCTION CHANGES SIGN 327
Cor. 2. Iff((i) and f(b) have like signs, then, if there be any real
roots of f(x) between x = a and x — b, there must be an even number,
provided f(x) be continuous between x = a and x = b.
Since an integral function is always finite and continuous for
a finite value of its variable, the restriction in Cor. 1 is always
satisfied, and we see that
Cor. 3. An integral function can change sign only by passing
through the value 0.
Cor. 4. If P and Q be integral functions of x algebraically prime
to each other, P/Q can only change sign by passing through the values
or oo .
"With the hint that the theorem of remainders will enable
him to exclude the ambiguous case 0/0, we leave the reader to
deduce Cor. 4 from Cor. 3.
Example 1.
When sb=0, la?=+l; and when x= +2, lar= 3. Hence, since
1 a; 2 is continuous, for some value of x lying between and +2 1 — a? must
become 0; for is between +1 and  3. In point of fact, it becomes once
between the limits in question.
Example 2.
y=a?fa?+llx6.
Whenai=0, y — 6; and when x= + 4, y= +6. Hence, between x = and
x= +4 there must lie an odd number of roots of the equation
a 3 6a; 2 +lla;6 = 0.
It is easy to verify in the present case that this is really so; for x 3 6x i
+ llx  6 = (x  1) (x  2) (x  3) ; so that the roots in question are x — 1, x — 2,
se=3.
The general form of the graph in the present case is as follows : —
Fig. io.
Example 3.
When x = 0, l/(la;)=+l; and when x= +2, l/(la:)=] : but since
1/(1 — x) becomes infinite and discontinuous between x = and x= +2, namely,
when X=l, we cannot infer that, for some value of x between and +2,
328 SIGN OF /(0) AND /( ± oo ) chap.
1/(1 x) will become 0, although is intermediate to + 1 and 1. In fact,
1/(1 — x) does not pass through the value between x=0 and x= +2.
§ 10.] It will be convenient to give here the following pro
position, which is often useful in connection with the methods
Ave are now explaining.
If fix) be an integral function of x, then by making x small
enough we can always cause f(x) to have the same sign as its lowest
term, and by making x large enough we can always cause f(x) to have
the same sign as its highest term.
Let us take, for simplicity, a function of the 3rd degree, say
y = px 3 + qx 2 + rx + s.
If we suppose s 4= 0, then it is clear, since by making x small
enough we can (see § 7, II.) make px 3 + qx 2 + rx as small as we
please, that we can, by making x small enough, cause y to have
the same sign as s.
If s = 0,
then we have y =px* + qx 2 + rx,
= (px 2 + qx + r)x.
Here by making x small enough we can cause px 2 + qx + r to have
the same sign as r, and hence y to have the same sign as rx,
which is the lowest existing term in y.
Again, we may write
ir
Here by making x large enough we may make q/x + rjx 2 + s/x 3 as
small as we please (see § 6, IV., and § 7, II.), that is to say,
cause p + q/x + rjx 2 + s/x 3 to have the same sign as p. Hence by
making x large enough we can cause y to have the same sign
as px 3
If we observe that, by chap, xiv., § 9, we can reduce every
integral equation to the equivalent form
f{x) = x n + p n _ , x" * + . . .+^ = 0,
and further notice that, in this case, if n be odd,
/( + oo ) = + oo , /(  oo ) =  oo ,
ami, if n be even,
/( + oo ) = + oo , /(  oo ) = + oo ,
we have the following important conclusions.
XV
MINIMUM NUMBER OF REAL ROOTS
329
Cor. 1. Every integral equation of odd degree with real co
efficients has at least one real root, and if it has more than one it
has an odd number.
Cor. 2. If an integral equation of even degree with real coefficients
has any real roots at all, it lias an even number of such.
Cor. 3. Every integral equation with real coefficients, if it has any
complex roots, has an even number of such.
The student should see that he recognises what are the cor
responding peculiarities in the graphs of integral functions of
odd or of even degree.
Example.
„ Show that the equation
x i 6, 3 + llxrl = ()
has at least two real roots.
Let y = x i 6x' i + llxx4.
We have the following scheme of corresponding values : —
X
— OO
+ oo
V
+ oo
 4
+ oo
Hence one root at least lies between  oo and 0, and one at least between
and + oo . In other words, there are at least two real roots, one negative
the other positive.
We can also infer that, if the remaining two of the possible four be also
real, then they must be either both positive or both negative.
When the real roots of an integral equation are not very
close together the propositions we have just established enable
us very readily to assign upper and lower limits for each of
them ; and in fact to calculate them by successive approxima
tion. The reader will thus see that the numerical solution of
integral equations rests merely on considerations regarding con
tinuity, and may be considered quite apart from the question
of their formal solution by means of algebraical functions or
otherwise. The application of this idea to the approximate
determination of the real roots of an integral equation will be
found at the end of the present chapter.
330
MAXIMA AND MINIMA ALTERNATE
CHAP.
GENERAL PROPOSITIONS REGARDING MAXIMA AND MINIMA
VALUES OF FUNCTIONS OF ONE VARIABLE.
§11.] When fix) in passing through any value, f(a) say, ceases to
increase and begins to decrease, f(a) is called a maximum value of f{x).
When fix) in passing through the value f (a) ceases to decrease and
begins to increase, f(a) is called a minimum value of fix).
The points corresponding to maxima and minima values of
the function are obviously superior and inferior culminating
points on its graph, such as P 2 and P 9 in Fig. 1. They are
also points where, in general, the tangent to the graph is parallel
to the axis of x. It should be noticed, however, that points
such as P and Q in Fig. 1 1 are maxima and minima points,
according to our present definition, although it is not true in
any proper sense that at them the tangent is parallel to OX. It
Fig. 11.
Fio. 12.
should also be observed that the tangent may be parallel to OX
and yet the point may not be a true maximum or minimum
point. Witness Fig. 12.
We shall include both maximum and minimum values as at
present defined under the obviously appropriate name of turning
values.
§ 12.] By considering an unbroken curve having maxima
and minima points (see Fig. 1) the reader will convince himself
graphically of the truth of the following propositions : —
I. So long as f(x) remains continuous its maxima and minima
values succeed each other alternately.
II. If x = a, x = b be two roots of f(x) (a alg.<b), then, if f(x) be
not constant, but vary continuously between xa and x = b, there must
XV
CRITERION FOR TURNING VALUES 331
be either at least one maximum or at least one minimum value of f(x)
between x  a and x = b.
In particular, if f(x) become positive immediately after x
passes through the value a, then there must he at least one
maximum before x reaches the value b ; and, in like manner, if
f(x) become negative, at least one minimum.
§ 1 3.] It is obvious, from the definition of a turning value,
and also from the nature of the graph in the neighbourhood of
a culminating point, that we can always find two values of the func
tion, on opposite sides of a turning value, which shall be as nearly
equal as we please. These two values will be each less or each greater
than the turning value according as the turning value is a maximum
or minimum.
Hence, if p be infinitely near a turning value of fix) (less in
the case of a maximum, greater in the case of a minimum), then
two roots oi fix) p will be infinitely nearly equal to one another.
It follows, therefore, that, if p be actually equal to a turning value
of f(x), the function f(x) p will have two of its roots equal. This
criterion may be used for finding turning values, as will be seen
in a later chapter.
CONTINUITY AND GRAPHICAL REPRESENTATION OF A
FUNCTION OF TWO INDEPENDENT VARIABLES.
§ 14.] Let the function be denoted by f(x, y), and let us
denote the dependent variable b} r z ; so that
*=/(«, y)
We confine ourselves entirely to the case where fix, y) is an integral
function, and we suppose all the constants to be real, and consider mily
real values of x and y. The value of z will therefore be always real.
Since there are now two independent variables, x and //,
there are two independent increments, say h and k, to consider.
Hence the increment of z, that is, f(x + h, y + k) f(x, y), now
depends on four quantities, x, y, h, k. Since, however, f(x, y)
consists of a sum of terms such as Ax m f n , it can easily be shown
by reasoning, like that used in the case oif(x), that the increment
of z always becomes infinitely small when h and k are made infinitely
332
GRAPHIC SURFACE FOR Z =/(», y)
CHAP.
small. Hence, as x and y pass continuously from one given pair of
values, say (a, b), to another given pair, say (a', b'), z passes continu
ously from one value, say c, to another, say c'.
§ 15.] There is, however a distinct peculiarity in the case
now in hand, inasmuch as there are an infinity of different ways
in which (x, y) may pass from (a, b) to (a', b'). In fact we re
quire now a twodimensional diagram to represent the variations of
the independent variables. Let X'OX, Y'OY be two lines in a
Fio. 13.
horizontal plane drawn from west to east and from south to
north respectively. Consider any point P in that plane, whose
abscissa and ordinate, with the usual understanding as to sign,
are x and y. Then P, which we may call the variable point, gives
us a graphic representation of the variables (x, y).
Let us suppose that for P x = a, y = b, and that for another
point P' x = a', y = b'. Then it is obvious that, if Ave pass
along any continuous curve whatever from P to P', x will vary
continuously from a to a', and y will vary continuously from
b to 1/ ; and, conversely, that any imaginable combination of a
continuous variation of x from a to a' with a continuous varia
tion of y from b to V will correspond to the passage of a point
from P to P' along some continuous curve.
It is obvious, therefore, that the continuous variation of
XV CONTOUR LINES 333
(x, y) from (a, h) to (a\ b') may be accomplished in an infinity
of ways. We may call the path in which the point which repre
sents the variables travels the graph of the variables.
To represent the value of the function z = f(x, y) we draw
through P, the variable point representing (x, y), a vertical line
PQ, containing .: units of any fixed scale of length that may be
convenient, upwards if z be positive, downwards if z be negative.
Q is then the graphic point which represents the value of the
function z =/(•'", y).
To every variable point in the plane XOY there corresponds
a graphic point, such as Q ; and the assemblage of graphic points
constitutes a surface which we call the graphic surface of the
function f{x, y)
When the variable point travels along any particular curve S
in the plane XOY, the graphic point of the function travels along
a particular curve 2 on the graphic surface ; and it is obvious
that S is the orthogonal projection of 2 on the plane XOY.
§ 1 6.] If we seek for values of the variables which correspond
to a given value c of the function, we have to draw a horizontal
plane U, c units above or below XOY according as c is positive
or negative ; and find the curve 2 where this plane U meets the
graphic surface. This line 2 is what is usually called a contour
line of the graphic surface. In this case the orthogonal projection
S of 2 upon XOY will be simply 2 itself transferred to XOY,
and may be called the contour line of the function for the value c.
All the variable points upon 8 correspond to pairs of values of
(x, y), for which f(x, y) has the given value c.
If we take a number of different values, c„ c 2 , c 3 , . . ., c,„ we
get a system of as many contour lines. Suppose, for example,
that the graph of the function were a rounded conical peak, then
the system of contour lines would be like Fig. 14, where the
successive curves narrow in towards a point which corresponds
to a maximum value of the function.
Any reader who possesses a one inch contoured Ordnance
Survey map has to hand an excellent example of the graphic
representation of a function. In this case x and y are the dis
334
f(x, y)=0 REPRESENTS A PLANE CURVE
CHAP.
tances east from the lefthand side of the map, and north from
the lower side ; and the function z is the elevation of the land
Fig. 14.
at any point above the sea level. The study of such a map from
the present point of view will be an excellent exercise both in
geometry and in analysis.
An important particular case is that where we seek the
values of x and y which make f(x, y) = 0. In this case the plane
U is the plane XOY. This plane cuts the graphic surface in a
continuous curve S {zero contour line), every point on which
has for its abscissa and ordinate a pair of values that satisfy
f{x, y) = 0.
The curve S in this case divides the plane into regions, such that
in any region f(x, y) has always either the sign + or the sign — ,
and S always forms the boundary between two regions in which f(x, y)
has opposite signs.
If we draw a continuous curve from a point in a + region
to a point in a  region, it must cross the boundary S an odd
number of times. This corresponds to the analytical statement
that iff(a, b) be positive and f(a, b') be negative, then, if (x, y) vary
continuously from {a, b) to (a', b'), f(x, y) will pass through the value
an odd number of times.
The fact just established, that all the " variable points " for
which f(x, y) — lie on a continuous curve, gives us a beautiful
geometrical illustration of the fact established in last chapter, that
the equation f(x, y) = has an infinite number of solutions, and
gives us the fundamental idea of coordinate geometry, namely, that
XV
EXAMFLE
335
a plane curve can be analytically represented by means of a single
equation connecting two variables.
Example.
Consider the function z = x 2 + y 2  1. If we describe, with as centre, a
circle whose radius is unity, it will be seen that for all points inside this circle
z is negative, and for all points outside z is positive. Hence this circle is the
zero contour line, and for all points on it we have
x 2 + ifl=0.
Y
Fig. 15.
INTEGRAL FUNCTIONS OF A SINGLE COMPLEX VARIABLE.
§ 17.] Here we confine ourselves to integral functions, but no longer
restrict either the constants of the function or its independent variable
x to be real.
Let us suppose that x = £ + ?ii, and let us adopt Argand's
method of representing £ + yi H
graphically, so that, if 12 M = £,
MP = ij, in the diagram of Fig.
1 6, then P represents £ + tji.
If P move continuously from
any position P to another P', the
complex variable is said to vary
continuously. If the values of
(£, rj) at P and P' be (a, ft) and
(a', ft') respectively, this is the Fro. 16.
same as saying that £ + iji is said to vary continuously from the
value a + fti to the value a + /3'i, when £ varies continuously from a
to a, and tj varies continuously from ft to ft'. There are of course
P
Q
M
M' 2
336
CONTINUITY OF COMPLEX FUNCTION
CHAP.
an infinite number of ways in which this variation may be
accomplished.
§ 1 8.] Suppose now we have any integral function of x whose
constants may or may not be real. Then we have f(x) =f($ + rji) ;
but this last can, by the rules of chap, xii., always be reduced
to the form £' + rji, Avhere £' and rj are integral functions of £
and ?; whose constants are real (say real integral functions of
(£, rj) ). Now, by § 14, £' and rj are finite and continuous so
long as (£, ?/) are finite. Hence /(£ + rfi) varies continuously when
£ + r/i varies continuously.
A graphic representation of the function /(£ + rji) can be
obtained by constructing another diagram for the complex
number £' + iji. Then the continuity of /(£ + rji) is ex
pressed by saying that, when the graph of the independent vari
able is a continuous curve S, the graph of the dependent variable is
another continuous curve S'.
Example.
Let
2/=+V(l^).
H
12
H
$
1
J
B'
C,
0'
A'
c
»— *
Fig. 17.
Fig. 18.
For simplicity, we shall confine ourselves to a variation of x which admits
only real values ; in other words, we suppose r\ always =0.
The path of the independent variable is then IACBJ, the whole extent
of the £axis. In the diagram we have taken CA = CB = 1 ; so that A and
B mark the points in the path for which the function begins to have, and
ceases to have, a real value.
Let Fig. 18 be the diagram of the dependent variable, ;/ = £' + t?7 I f
A'C'l (A', B', and W are all coincident), then the path of the dependent
XV
COMTLEX ROOTS DETERMINED GRAPHICALLY
337
variable is the whole of the 77' axis above fi, together with A'C, each
reckoned twice over. The pieces of the two paths correspond as follows : —
Independent
Variable.
Dependent Variable.
IA
AC
CB
BJ
FA'
A'C
CB'
B'J'
§ 19.] £' and rj being functions of £ and >;, we may represent
this fact to the eye by writing
r = </>(£ v), v' = M v)
If we seek for values of (£, rj) that make £' = 0, that is the same
as seeking for values of (£, rj) that make <£(£, rj) = 0. All the
points in the diagram of the independent variable corresponding
to these will lie (by § 16) on a curve S.
Similarly all the points that correspond to rj = 0, that is, to
^(£, rj) = 0, lie on another curve T.
The points for which both £' = and ?/ = 0, — in other words,
the points corresponding to roots of /(£ + ?;?), — must therefore be
the intersections of the two curves S and T.
Example.
If we jmtx=a+rri, and y=s' + 7 ?'i wu have
=m$v) + (ev)i.
Hence £'=2(4 £»;), v ' = ^ v ".
Hence the S and T curves, above spoken of, are given by the equations
2(4 fr) = (S),
?v=Q (T).
These are equivalent to
VOL. I
4
(S),
(T).
338
HORNER S METHOD
CHAP.
The student should have no difficulty in constructing these. The diagram
that results is
H
\
N
M
Fir,. 19.
The S curve (a rectangular hyperbola as it happens) is drawn thick. The
T curve (two straight lines bisecting the angles between the axes) is dotted.
The intersections are P and Q.
Corresponding to P we have if = 2, ?? = 2; corresponding to Q, £= 2,
V
_ o
It appears therefore that the roots of the function are + 2 + 2i and  2  2i.
The student may verify that these values do in fact satisfy the equation
ix 2 + 8 = 0.
HORNER'S METHOD FOR APPROXIMATING TO THE VALUES OF THE
REAL ROOTS OF AN INTEGRAL EQUATION.
§ 20.] In the following paragraphs we shall show how the
ideas of §§ 810 lead to a method for calculating digit by digit
the numerical value of any real root of an integral equation. It
will be convenient in the first place to clear the way by estab
lishing a few preliminary results upon which the method more
immediately depends.
§ 21.] To deduce from the equation
p^ n + p x X n ~ l + . . . + p n _ iX + p n =0 (1 )
another equation each of wlwse roots is m times a corresponding root
xv DIMINUTION OF ROOTS 339
of (1). Let x be any root of (1) ; and let £ = mz. Then x = gjm.
Hence, from (1), we have
Pod m) n +p 1 (g/m) n  1 + . . . +p n  l ($/m) +p n = 0.
If we multiply by the constant in' 1 , we deduce the equivalent
equation
pg 1 + pM n  1 + ■ ■ ■+ Pn  .»" " *£ +JW»" = (2),
which is the equation required.
Cor. The equation whose roots are those of (1) with the signs
changed is
Po$" Pi? 1 + • • ■ + ( r^Pn^ + (  )"Pn = (3).
This follows at once by putting m =  1 in (2). We thus
see that the calculation of a negative real root of any equation
can always be reduced to the calculation of a positive real root
of a slightly different equation.
Example. The equation whose roots are 10 times the respective roots of
Sa?15x i +5x + 6 =
is 3^ 150x 2 + 500.c+ 6000 = 0.
§ 22.] To deduce from the equation (1) of § 21 another, each of
xohose roots is less by a than a corresponding root of (I).
Let x denote any root of (1); £ the corresponding root of
the required equation ; so that g = x — a, and x = £ + a. Then
we deduce at once from (1)
P& + a) n +;>,(£ + a)"" 1 + . . . +p n i($ + «) +Pn = (4).
If we arrange (4) according to powers of £, we get
ft£* + $£*"* + : •  + q n i€ + qn = o (5),
which is the equation required.
It is important to have a simple systematic process for
calculating the coefficients of (5). This may be obtained as
follows.
Since £ = x  a, we have, by comparing the lefthand sides of
(1) and (5),
p x n +p l x n ~ 1 + . . . +p n _ l x +p n
=p Q (xa) n + q l (za) n  1 + . . . + ?»i(a:a) + ?»•
340 APPROXIMATION TO ROOT chap.
The problem before us is, therefore, simply to expand the
function/(.T) =p x n +p 1 x n ~' 1 + . . . +p n iX +p n m powers of (x  a).
Hence, as we have already seen in chap, v., § 21, q n is the
remainder when f{x) is divided by x  a ; q n _^ the remainder
when the integral quotient of the last division is divided by
x a ; and so on. The calculation of the remainder in any
particular case is always carried out by means of the synthetic
process of chap, v., § 13.
It should be observed that the last coefficient, q n , is the value
of f(a).
Example. To diminish the roots of
by 1.
AVc simply reproduce the calculation of § 21, Example 1, in a slightly
modified form ; thus —
5 11
+ 10
2 (1
5
6
4
6
4
1 2
5
1
1
1 3
5
1 4
Hence the rerpiired equation is
5f + 4£ 2 +3£ + 2 = 0.
§ 23.] If one of the roots of the equation (1) of § 21 be small,
say between and + 1, then an approximate value of that root is
Pn/Pnv For, if x denote the root in question, we have,
by (i),
x = 
Pn
(Pn2+Pn 3 V +  ■ .+iV'"" 2 ) (6).
Pa i Pni
Hence, if x be small, we have approximately z =  p n ;Pn\
It is easy to assign an upper limit to the error. We have,
in fact,
x = PnfPni 5
■II _/•
where mode<^— (1 + x + . . . +o: n ~"),
l'llX
p r being the numerically greatest among the coefficients p . . .,
Pni, Pn2 Hence, since x^>l, we have
mode<(?i l)p,/p n i
XV iiokxek's PROCESS 341
It would be easy to assign a closer limit for the error ; hut
in the applications which we shall make of the theorem we have
indirect means of estimating the sufficiency of the approxima
tion ; all that is really wanted for our purpose is a suggestion of
the approximation.
Example. The equation
x" + 8192ar + 16036288a;  5969856 =
has a root between and 1, find a first approximation to that root.
By the above rule, we have for the root in question
a: = 59S9856/160362S8e
= "37227  e
where e<2 x 8192/16036288 < 00103.
Hence o;= *372, with an error of not more than 1 in the last digit.
In point of fact, since x< '4, we have
e< (4) 3 + 8192 x (4) 2 }/16036288,
< 1311/16036288 < 1600/16000000,
<0001 ;
so that the approximation is really correct to the 4th place of decimals.
§ 24.] Horner s Method. Suppose that we have an equation
/(.'•) = 0, having a positive root 235  67 . . . This root would be
calculated, according to Horner's method, as follows : — First we
determine, by examining the sign of f(x) } that /(.<•) = has one
root, and only one,* lying between 200 and 300 : the first digit
is therefore 2. Then we diminish the roots of f(z) = by 200,
and thus obtain the first subsidiary equation, say /,(.>•) = 0. Then
fi(x)0 has a root lying between and 100. Also since the
absolute term of f t (x) is /(200), and no root of /(.») = lies
between and 200, the absolute term of f(x) (that is, /(0)) and
the absolute term oif^x) must have the same sign. By examin
ing the sign of /,(«) for x = 0, 10, 20, . . ., 90, we determine
that this root lies between 30 and 40 : the next digit of the
root of the original equation is therefore 3. The labour of this
last process is, in practice, shortened by using the rule of § 23.
Let us suppose that 30 is thus suggested ; to test whether this
* For a discussion of the precautions necessary when an equation has two
roots which commence with one or more like digits, see Burnside and Pcmton's
Theory of Equations, § 104.
342 HORNER'S PROCESS chap.
is correct we proceed to diminish the roots of /,(•>') = by 30, —
to deduce, in fact, the second subsidiary equation f s (x) = 0. Since
the roots of /(.?:) have now been diminished by 230, the absolute
term off a (x) is/(230). Hence the absolute term of f,(.r) must
have the same sign as the absolute term of /,(#), unless the digit
3 is too large. In other words, if the digit 3 is too large, we
shall be made aware of the fact by a change of sign in the
absolute term. In practice, it does not usually occur (at all
events in the later stages of the calculation) that the digit
suggested by the rule of § 23 is too small ; but, if that were so,
we should become aware of the error on proceeding to calculate
the next digit which would exceed 9.
The second subsidiary equation is now used as before to
find the third digit 5.
The third subsidiary equation would give "6. To avoid the
trouble and possible confusion arising from decimal points, we
multiply the roots of the third (and of every following) subsidiary
equation by 10; or, what is equivalent, we multiply the second
coefficient of the equation in question by 10, the third coefficient
by 100, and so on; and then proceed as before, observing, how
ever, that, if the trial division for the next digit be made after
this modification of the subsidiary equation, that digit will
appear as 6, and not as "6, because the last coefficient has been
multiplied by 10 n and the second last by 10 B1 .
The fundamental idea of Horner's method is therefore simply
to deduce a series of subsidiary equations, each of which is used
to determine one digit of the root. The calculation of the
coefficients of each of these subsidiary equations is accomplished
by the method of § 22. After a certain number of the digits of
the root have been found, a number more may be obtained by a
contraction of the process above described, the nature of which
will be easily understood from the following particular case.
Example. Find an approximation to the least positive root of
/(.•) = .," + 2.v 2  5z7 = (1).
Since /(0)=7, /(l)=9, /(2)=l, /(3)=+23,
the root in cjuestion lies between 2 and 3. The first digit is therefore 2.
XV EXAMPLE 34.°,
We now diminish the roots of (1) by 2. The calculation of the coefficients
runs thus : —
1 + 2 5 7 (2
_2 _8 _6
4 3 "\  1
2 12
~6 Tu
2
where the prefix  is used to mark coefficients of the first subsidiary equation.
The first subsidiary equation is therefore
a? + 8x* + l5xl = 0.
Since the next digit follows the decimal point, we multiply the roots of
this equation by 10. The resuming equation is then
a? + 8(k 2 + 1500.*! 1000 = (2).
Since 1000/1500 <1, it is suggested that the next digit is 0. "We there
fore multiply the roots of (2) by 10, and deduce
ar ? + 800x 2 + 150000.>; 1000000 = (3).
Since 1000000/150000 = 6 . . ., the next digit suggested is C. We now
diminish the roots of (3) by 6.
+ 800
+ 150000
1000000 (06
6
4836
929016
806
154836
4 I 70984000
6
4872
812
4 i 15970800
6
8180
The resulting subsidiary equation, after the multiplication of its roots
by 10, is
a? + 8180a 3 + 15970800a  70984000 (4).
Since 70984000/15970800 = 4 . . ., the next digit suggested is 4. The
reader should notice that, owing to the continual multiplication of the roots
by 10, the coefficients towards the right increase in magnitude much more
rapidly than those towards the left : it is for this reason that the rule of § 23
becomes more and more accurate as the operation goes on. Thus, even at the
present stage, the quotient 70984000/15970800 would give correctly more than
one of the following digits, as may be readily verified.
We now diminish the roots of (4) by 4 ; and add the zeros to the coefficients
as before.
344 EXAMPLE chap
EXAMPLE
+ 8180
4
+ 15970800
32736
70984000 (4
64014144
8184
4
8188
4
81920
16003536
32752
s l 1603628800
b  6969856000
Then we have the subsidiary equation
a; 3 + 81920a 2 + 1603628800a  6969856000 (5).
It will be observed that throughout the operation, so far as it has gone,
the two essential conditions for its accuracy have been fulfilled, namely, that
the last coefficient shall retain the same sign, and that each digit shall come
out not greater than 9. It will also be observed that the number of the
figures in the working columns increases much more rapidly than their utility
in determining the digits of the root. All that is actually necessary for the
suggestion of the next digit at any step, and to make sure of the accuracy of
the suggestion, is to know the first two or three figures of the last two
coefficients.
Unless, therefore, a very large number of additional digits of the root is
required, we may shorten the operation by neglecting some of the figures in
(5). If, for example, we divide all the coefficients of (5) by 1000, we get the
equivalent equation *
•001a 3 + 81 92a 2 + 1603628 "8a  6969S56 = (5').
Hence, retaining only the integral parts of the coefficients, we have
0a 3 + 81a 2 + 1603628a  6969856 = (5").
It will be noticed that the result is the same as if, instead of adding zeros,
as heretofore, we had cut off one figure from the second last coefficient, two
from the third last, and so on.t
Since 6969856/1603628 = 4 . . ., we have for the next digit 4. We then
diminish the roots of (5") by 4. In the necessary calculation the first working
column now disappears owing to the disappearance of the coefficient of a 3 ; we
have in its place simply 81 standing unaltered. It is advisable, however, in
multiplying the contracted coefficients by 4 to carry the nearest number of
tens from the last figure cut off (just as in ordinary contracted multiplication
and division and for the same reason).
* If the reader find any difficulty in following the above explanation of
the contracted process, he can satisfy himself of its validity by working out
the above calculation to the end in full and then running his pen through the
unnecessary figures.
t In many cases it may not be advisable to carry the contraction so far at
each step as is here done. We might, for instance, divide the coefficients of
5 by 100 only. The resulting subsidiary equation would then be
Oa 3 + 81 9a 2 + 1 6036288a  09698560,
with which we should proceed as before.
XV
EXAMPLE
345
The next step, therefore, runs thus :
81 +1603628
328
1603957
328
6969856
64 15828
 554028
(4
(6);
(6').
"I 1604285
The corresponding subsidiary equation is
81a 2 + 16042850a  55402800 =
or, contracted, Ox + 160428.1  554028 =
The next digit is 3 ; and, as the coefficient of x 2 , namely  81, still has a
slight effect on the second working column, the calculation runs thus: —
+160428 554028 (3
2 481293
71
160431
2
72735
'I 160433
The resulting subsidiary equation after contraction is
16043a; 72735 = (7).
The rest of the operation now coincides with the ordinary process of
contracted division ; it represents, in fact, the solution of the linear equation
(7), that is (see chap, xvi., § 1), the division of 72735 by 16043.
The whole calculation may be arranged in practice as below. But the
prefixes 2 , 3 , &c, which indicate the coefficients of the various equations,
may be omitted. Also the record may be still farther shortened by performing
the multiplications and additions or subtractions mentally, and only recording
the figures immediately below the horizontal lines in the following scheme.
The advisability of this last contraction depends of course on the arithmetical
power of the calculator.
+ 2
5
7 (2064434533
2
8
6
4
3
8   1000000
2
12
929016
6
" 3  150000
4  70984000
2
4836
64014144
3  800
154836
1  6969856
6
4872
6415828
806
1 15970800
"1  554028
6
32736
481293
812
16003536
'172735
6
32752
64173
4  8180
b l 1603628^
"1  8562
4
328
8022
8184
1603957
9 5l0
4
328
481
8188
"1 160428$
10 l59
4
2
48
'I^N
160431
2
11
16fc(^Sf
346 EXERCISES XXII
i'H U\
The number of additional digits obtained by the contracted process is less
by two than the number of digits in the second last coefficient at the beginning
of the contraction. Owing to the uncertainty of the carriages the last digit
is uncertain, but the next last will in such a case as the present be abso
lutely correct. In fact, by substituting in the original equation, it is easily
verified that the root lies between 2064434534 and 2064434535 ; so that the
last digit given above errs in defect by 1 only. The number of accurate
figures obtained by the contracted process will occasionally be considerably
less than in this example ; and the calculator must be on his guard against
error in this respect (see Horner's Memoir, cited below).
§ 25.] Since the extraction of the square, cube, fourth, . . .
roots of any number, say 7, is equivalent to finding the positive
real root of the equations, x 2 + Ox  7 = 0, x a + Ox 2 + Ox  7 = 0,
x* + Ox 3 + (bf + Oo:  7 = 0, . . . respectively, it is obvious that
by Horner's method we can find to any desired degree of
approximation the root of any order of any given number
"whatsoever. In fact, the process, given in chap, xi., § 13, for
extracting the square root, and the process, very commonly
given in arithmetical textbooks, for extracting the cube root will
be found to be contained in the scheme of calculation described
in § 24.*
* Horner's method was first published in the Transactions of the Philoso
phical Society of London for 1819. Considering the remarkable elegance,
generality, and simplicity of the method, it is not a little surprising that it
has not taken a more prominent place in current mathematical textbooks.
Although it has been well expounded by several English writers (for example,
De Morgan, Todhunter, Burnside and Panton), it has scarcely as yet found a
recognised place in English curricula. Out of five standard Continental text
books where one would have expected to find it we found it mentioned in only
one, and there it was expounded in a way that showed little insight into its
true character. This probably arises from the mistaken notion that there is
in the method some algebraic profundity. As a matter of fact, its spirit is
purely arithmetical ; and its beauty, which can only be appreciated after one
has used it in particular cases, is of that indescribably simple kind which
distinguishes the use of position in the decimal notation and the arrangement
of the simple rules of arithmetic. It is, in short, one of those things whose
invention was the creation of a commonplace. For interesting historical
details on this subject, see De Morgan — Companion to British Almanack, for
1839; Article "Involution and Evolution," Penny Cyclopsedia ; and Budget
of Paradoxes, pp. 292, 374.
XV
EXERCISES XXII 34"
Exercises XXII.
[The student should trace some at least of the curves required in the
following graphic exercises by laying them down correctly to some convenient
scale. He will find this process much facilitated by using paper ruled into
small squares, which is sold under the name of Plotting Paper.]
Discuss graphically the following functions : —
(40 y=rz 1 Tv i  (5.) y=^i (6.) y= x
"(sb1) 3 ' v ' J se2 v "' * J' 2 9
(7.) Construct to scale the graph of y =  x + 8x  9 ; and obtain graphic
ally the roots of the equation a 2  8a: + 9 = to at least three places of
decimals.
(8.) Solve graphically the equation
a?  16a; 2 + 7Lc 129 = 0.
(9.) Discuss graphically the following question. Given that y is a con
tinuous function of x, does it follow that a; is a continuous function of y ?
(10.) Show that when h, the increment of x, is very small, the increment of
p„x" +p n ix" 1 + . . . +pxx +p
is (npnX" 1 + (n  1 )p n ix" 2 + . . . + 1 .pi)h.
(11. ) If h be very small, and x=l, find the increment of 2a* 3  9ar + 12a 1 + 5.
(12.) If an equation of even degree have its last term negative, it has at
least two real roots which are of opposite signs.
(13.) Indicate roughly the values of the real roots of
10a. 3 17ar + a; + 3 = 0.
(14.) What can you infer regarding the roots of
a, 3 5a; + 8 = 0?
(15.) Show by considerations of continuity alone that a;" 1=0 cannot
have more than one real root, if n be odd.
(16.) If f{x) be an integral function of a;, and if/(a)= p,f{b)= +q, where
p and q are both small, show that x — (qa+2^b)/{p + q) is an approximation to
a root of the equation /(a:) = 0.
Draw a series of contour lines for the following functions, including in
each case the zero contour line : —
(17.) z = xy. (18.) z = ~. (19.) z = xf. (20.) z = ^±£ '.
y x
Is the proposition of § 16 true for the last of these ?
Draw the Argand diagram of the dependent variable in the following
cases, the path of the independent variable being in each case a circle of radius
unity whose centre is fi : —
<21.) y=. (22.)y= + sjx. (23.) y= ^x. (24.) y = \ a 2 .
348
EXERCISES XXII
CHAP. XV
Find by Horner's method the positive real roots of the following equations
in each case to at least seven places of decimals : — *
(25.) a? 2 = 0.
(27.) a 3 + x 1000 = 0.
(29.) a 4 + x 3 + x" + x 127694 = 0.
(30. ) a 4  80a 3 + 24a: 2  6a:  80379639 = 0.
(26.) x 3  2a 5 = 0.
(28. ) x 3  46a; 2  36a + 18 = 0.
(31.) x 5  4a 4 + 7a 3 863 = 0.
(33.) x 5 4a 2000 = 0.
(32.) a 5 7 = 0.
(34. ) 4a 6 + 7a 5 + 9a: 4 + 6a 3 + 5a 2 + 3a  792 = 0.
(35.) Find to twenty decimal places the negative root of 2a 4 + 3a 3  6x  8
= 0.
(36.) Continue the calculation on p. 344 two stages farther in its uncon
tracted form ; and then estimate how many more digits of the root could
be obtained by means of the trial division alone.
* Most of these exercises are taken from a large selection given in De
Morgan's Elements of Arithmetic (1854).
CHAPTEE XVI.
Equations and Functions of the First Degree.
EQUATIONS WITH ONE VARIABLE.
§ 1.] It follows by the principles of chap. xiv. that every
integral equation of the 1st degree can be reduced to an equiva
lent equation of the form
ax + b = (I);
this may therefore be regarded as a general form, including all
such equations. As a particular case b may be zero ; but we
suppose, for the present at least, that a is neither infinitely great
nor infinitely small.
Since a 4= 0, we may write (1) in the form
"{— ("«)}"° (2>:
whence we see that one solution is x =  bja. We know already,
by the principles of chap, xiv., § 6, that an integral equation of
the 1st degree in one variable has one and only one solution. Hence
we have completely solved the given equation (1).
It may be well to add another proof that the solution is unique.
Let us suppose that there are two distinct solutions, x — a and x = §, of (1).
Then we must have
aa + b = 0,
ap + b=0.
From these, by subtraction, we derive
a(a  jS) = 0.
Now, by hypothesis, «=t=0, therefore we must have a/S = 0, that is, a = /3;
in other words, the two solutions are not distinct.
350 TWO LINEAR EQUATIONS IN ONE VARIABLE chap.
§ 2.] Two equations of the 1st degree in one variable will in
general be inconsistent.
If the equations be ax + b = (1),
a'x + b' = (2),
the necessary and sufficient condition for consistency is
aV  a'b = (3).
The solution of (1) is x = b/a, and the solution of (2) is
x =  b'/a'. These will not in general be the same ; hence the
equations (1) and (2) will in general be inconsistent.
The necessary and sufficient condition that (1) and (2) be
consistent is
b V
Since a 4= 0, a 4= 0, (4) is equivalent to
a'b  ab',
or ab'  a'b = 0.
Obs. 1. If b = and b' = 0, then the condition of consistency
is satisfied. In this case the equations become ax  0, a'x = ;
and these have in fact the common solution x = 0.
Obs. 2. When two equations of the 1st degree in one vari
able are consistent, the one is derivable by multiplying the other
by a constant. In fact, since a 4= 0, if we also suppose b 4= 0, we
derive from (3), by dividing by ab and then transposing,
— = r, each = I; say ;
a b ' J '
hence a' = ka, b' = kb,
so that a'x + b' = hax + kb,
= k(ax + b).
If, then, (3) be satisfied, (2) is nothing more or less than
k(ax + b) =
where k is a constant.
This might have been expected, for, transpositions apart, the
only way of deriving from a single equation another perfectly
equivalent is to multiply the given equation by a constant.
XVI
EXERCISES XXIII 35 i
Exercises XXIII.
Solve the following equations : —
l + (lg)/2
(2.) 3 =1
(3.)
51 62
5 +
3a; 1 12a; + 5
2 29
(4.) 3 4 i~ 5 "
iX
(5.) 68(32a;5) + p—=3694.r,
find x to three places of decimals.
(6.) a/(lbx)=b/(lax).
(7. ) (" + x) (b + x)  a{b +c) = (ca? + bx 2 )jb.
xa xc _
(8) i, +i = 2 
K ' b a bc
x 2 a" xb 2 %*<?
(9.)  + r + = a + b + c3x.
N ' xa x  o xc
(10.) (a 3 + b 3 )x + a 3  b x = ft 4 b* + a 6(« 3 + b").
O 9 O 9
a;^  or ar  c
(11.) i i =ca.
^ > b+a b+c
(12.) (a?l)(aJ+2)(2jj2)=(2a;l)(2a5+l)(a!/2+I).
(13.)
(14.)
(15.)
(16.)
(17.)
(18.)
12 3 4
as+1
X + 2
x + 3
a;+4*
x
1
xB
x2
a; 4
x
2
x4
x 3
a;5"
11
12B + 11
5
6a; +~5~
7
'4a; + 7
3
1
x
X
5a:
'T^x~
ar>2
1Zx + x 2
2
; +
14
10
r +
6
aH2 a+10 a; + 6 a+14'
a; 2  4a; 4 5
a; + 6a; +10
\x+Zj
352 ax + hj + c = HAS do 1 solutions chap.
rim J J 3(^+5) 6.r+i7
1 ; sc + l .r + 2 (s+l)(as+2) tf + 2 '
a; + 2a a:2a_ 4a&
(20) 2b~^x + 2b + x~ ib 2 x 2 '
(a + b)x + c ( ab)x + e _ iab
( ' (ab)x + d {a + b)x+f~(a + b){ab)
a+b ab a b
(22.) + ,= ;.
v ' xa xb xa xb
(23.) 5.+ * " + *
xa xb x(xab) + ab'
(24) r^+ ~rr = 2 
' x+abx+bc
(85 ) X _ ? 1
(a;  a) (x  b) (x a){x c) (x b)(x c)
1 2
+ ;
{x + a)(x + b) (x + a)(x + c) (x + b)(x + c)
EQUATIONS WITH TWO VARIABLES.
§ 3.] A single equation of the 1st degree in two variables has a
onefold infinity of solutions.
Consider the equation
ax + by + c = (1).
Assign to y any constant value we please, say (3, then (1)
becomes
ax + b/3 + c = (2).
We have now an equation of the 1st degree in one variable,
which, as we have seen, has one and only one solution, namely,
x =  (bp + c)ja.
We have thus obtained for (1) the solution x=  (b(3 + c)ja,
y = f3, where ft may have any value we please. In other words,
we have found an infinite number of solutions of (1).
Since the solution involves the one arbitrary constant /?,
we say that the equation (1) has a onefold infinity (sometimes
symbolised by oo x ) of solutions.
Example.
the solutions are given by
2/31 .
XVI
TWO LINEAR EQUATIONS IN TWO VARIABLES
353
we have, for example, for /3= 2, 0=  1, j3 = 0, /3= +, /3= + 1, /3= + 2, the
following solutions : —
/3
_2
1
1
+ 2
+ 1
+ 2
5
1
1
a;
~3
1
~3
+ 3
+ 1
y
2
'
1
+ 2
+ 1
+ 2
And so on.
§ 4.] We should expect, in accordance with the principles of
chap, xiv., § 5, that a system of two equations each of the 1st
degree in two variables admits of definite solution.
The process of solution consists in deducing from the given
system an equivalent system of two equations in which the
variables are separated ; that is to say, a system such that x
alone appears in one of the equations and y alone in the
other.
We may arrive at this result by any method logically con
sistent with the general principles we have laid down in chap
xiv., for the derivation of equations. The following proposition
affords one such method : —
If I, V, m, m' be constants, any one of ichich may be zero, but
which are such that lm!  I'm 4= 0, then the two systems
ax + by + c = (1),
a'x + b'y + c' = Q (2),
and
J(ax + by + c)+ l'(a'x + b'y + c') = (3),
m(ax + by + c) + m'ia'x + b'y + c) = (4),
are equivalent.
It is obvious that any solution of (1) and (2) will satisfy (3)
and (4) ; for any such solution reduces both ax + by + c and
a'x + b'y + c' to zero, and therefore also reduces the lefthand
sides of both (3) and (4) to zero.
Again, any solution of (3) and (4) is obviously a solution of
VOL. I 2 a
354 SOLUTION BY CEOSS MULTIPLICATION chap.
m'{ l(ax + by + c) + l'(a'x + b'y + c')}
 I' {m(ax + by + c) + m'(a'x + b'y + c')} = (5),
 m { l(ax + by + c) + l'(a'x + b'y + c')}
+ 1 {m(ax + by + c) + m'(a'x + b'y + c')} = (6).
Now (5) and (6) reduce to
(lm'  I'm) (ax + by + c ) = (7),
{lm!  I'm) {ax + b'y + c') = (8),
and, provided lm'  l'm^¥ 0, (7) and (8) are equivalent to
ax +by + c  0,
a'x + b'y + c'  0.
We have therefore shown that every solution of (1) and (2) is a
solution of (3) and (4) ; and that every solution of (3) and (4) is
a solution of (1) and (2).
All we have now to do is to give such values to I, V, m, m'
as shall cause y to disappear from (3), and x to disappear from
(4). This will be accomplished if we make
I = + V, V = b,
m =  a', m' = + a ;
so that lm' — I'm — ab'  a'b.
The system (3) and (4) then reduces to
(ab'  a'h)x + cb'  c'b = (3'),
(ab'  a'b)y + c'a  ca' = (4') ;
and this new system (3'), (4') will be equivalent to (1), (2),
provided
ah' a'b^O (9).
But (3') and (4') are each equations of the 1st degree in one
variable, and, since ab'  a'b 4= 0, they each have one and only
one solution, namely —
cb'  c'b "
X ab'  a'b
y
ac'  a'c
" ab' a'b J
(10).
XVI
MEMORIA TECHNICA 355
It therefore follows that the system
ax + by + c = (1),
a'x + b'y + c'=0 (2)
has one and only one definite solution, namely, (10), provided
ab'  a'b * (9).
The method of solution just discussed goes by the name of
cross multiplication, because it consists in taking the coefficient
of y from the second equation, multiplying the first equation
therewith ; then taking the coefficient of y from the first equation,
multiplying the second therewith ; and finally subtracting the
two equations, Avith the result that a new equation appears not
containing y.
The following memoria tcchnica for the values of x and y will enable the
student to recollect the values in (10).
The denominators are the same, namely, ab'  a'b, formed from the co
efficients of x and y thus
a\ yb
a'y \6'
the line sloping down from left to right indicating a positive product, that
from right to left a negative product.
The numerator of x is formed from its denominator by putting c and c' in
place of a and a' respectively.
The numerator of y by putting c and c' in place of b and b'.
Finally, negative signs must be affixed to the two fractions.
Another way which the reader may prefer is as follows : —
Observe that we may write (10) thus,
_bc'b'c ca'  c'a
X ~ab r a'b' y ~ab'a'b (11) '
where the common denominator and the two numerators are formed according
to the scheme
\
^a .
It is very important to remark that (1) and (2) are col
laterally symmetrical with respect to I a, b , see chap, iv., § 20.
V, v)
Hence, if we know the value of x, we can derive the value of y
by putting everywhere b for a, a for b, b' for a', and a for b'. In
356
EXAMPLES
CHAP.
fact the value of y thus derived from the value of x in (10) is
 (ca  c'a)/(ba'  b'a) ; and this is equal to  (ac'  a'c)/(ab'  a'b),
which is the value of y given in (10).
Example 1.
3x+22/3 =
9x + iy + 5 =
(a),
08).
Proceeding by direct application of (11), we have
+ 3\/+2 X /3\/+3
10 + 12 11
9
y=
2715 2
12 + 18 15' " 12 + 18 5'
Or thus : multiply (a) by 2, and we have the equivalent system
Sx + iy  6 = 0,
 9a; + 4y + 5 = ;
whence, by subtraction,
15a;ll=0,
which gives
11
* = I5
Again multiplying (a) by 3, and then adding Q3), we have
10y4 = 0,
which gives
4 2
Example 2.
a/3 *'
a'V 7
Multiplying the first of these equations by , and subtracting the second, we
obtain
i i\ _i i
a 2 /3 * J87 '
<x 2 (7/3)
that is,
whence
# =
7(«"/3)
Since the equations are symmetrical in f ' ' ) we get the value of y by
interchanging a and /3, namely,
y=
7 (/3a)
xvi SPECIAL CASES 357
Sometimes, before proceeding to apply the above method, it
is convenient to replace the given system by another which is
equivalent to it but simpler.
Example 3.
a?x+by = 2ab(a + b) (a),
b(2a + b)x + a(a + 2% = a 3 + a?b + ab 2 + b* (/3).
By adding, we deduce from (a) and (/3)
(a + b)°~x + (a + b) 2 y = (a + b) 3 ,
which is equivalent to
x + y = a + b (y).
It is obvious that (a) and (7) are equivalent to (a) and (/3). Multiplying
(7) by b' 2 and subtracting, we have
(a 2 b 2 )x = 2a 2 b + ab 2 b 3 ,
= b(2ab){a + b).
Hence #=— — =— .
a b
Since the original system is symmetrical in { '"' J, we have
a(2b  a)
§ 5.] Under the theory of last paragraph a variety of par
ticular cases in which one or more of the constants a, b, c, a\ b', c'
involved in the two equations
ax + by + c = 0,
a'x + b'y + c' =
become zero are admissible ; all cases, in short, which do not
violate the condition ab'  a'b =t= 0.
Thus we have the following admissible cases : —
'.>
a = (1), b' = (4),
b = (2), a = and b' = (5),
a'  (3), a' = and 6 = (G).
The following are exceptional cases, because they involve ab'  a'b
= 0:—
a = and a' = (I.), a, b, a', V all different
a = and b = (H.), from 0, but such that
V = and a' = (III.), aJ'  a'J = (V.)
b' = and 6 = (IV.),
358 HOMOGENEOUS SYSTEM chap.
We shall return again to the consideration of the exceptional
cases. In the meantime the reader should verify that the
formulae (10) do really give the correct solution in cases (1) to
(6), as by theory they ought to do.
Take case (1), for example. The equations in this case reduce to
bi/ + e0, a'x + b'y + c' = 0.
The first gives y —  c/b, and this value of y reduces the second to
a x  r + c = 0,
.  , • b'cbc'
which gives x= n — ■•
a b
It will be found that (10) gives the same result, if we put a = 0.
There is one special case that deserves particular notice, that,
namely, where c = and c' = ; so that the two equations are
homogeneous, namely,
ax + by = (a),
a'x + b'y = (ft).
If ab'  a'b 4= 0, these formula? (10) give x = 0, y = as the only
possible solution. If ab'  a'b = 0, these formulae are no longer
applicable ; what then happens will be understood if we reflect
that, provided y 4= 0, (a) and (/S) may be written
az + b = (a'),
a'z + b' = ((3'),
where z  x/y.
We now have two equations of the 1st degree in z, which
are consistent (see § 2), since ab' — a'b = 0. Each of them gives
the same value of z, namely, z =  b/a, or z =  b'/a' (these two
being equal by the condition ab'  a'b = 0).
If then ab' — a'b 4= 0, the only solution of (a) and ((3) is x = 0,
y = ; if ab'  a'b = 0, x and y may have any values such that the
ratio x/y = b/a=  b'/a'.
§ 6.] There is another way of arranging the process of solu
tion, commonly called Bezoitt's method* which is in reality merely
a variety of the method of § 4.
* For an account of Bezout's methods, properly so called, see Mini's
papers on the " History of Determinants ;" Proc. R.S.E., 1886.
&vi USE OF UNDETERMINED MULTIPLIER 359
If X be any finite constant quantity whatever* then any solution of the
system
ax + by + c=0, a'x + b'y + c' = (1)
is a solution of the equation
[ax + by + c) + \(a'x + b'y + c') = (2),
that is to say, of (a + \a')x + (b + \b')y + (e + \c') = (3).
Now, since X is at our disposal, we may so choose it that y shall disappear
from (3) ; then must
\b' + b = Q (4),
and (3) will reduce to (a + \a')x + (c + \c') = 0' (5).
From (4) we have \=  bjb', and, using this value of X, we deduce from (5)
c_+\c'_ b'c  be'
a + \a' ab'a'b'
which agrees with (10).
The value of y may next he obtained by so determining X that x shall
disappear from (3). We thus get
\a' + a = (6),
(b + \b')y + (c + \c') = (7),
and so on.
To make this method independent and complete, theoretically, it would
of course be necessary to add a proof that the values of x and y obtained do
in general actually satisfy (1) and (2); and to point out the exceptional case.
§ 7.] There is another way of proceeding, which is inter
esting and sometimes practically useful.
The systems
ax + by + c = ] . >
a'x + b'y +. d = J { '
i ax + c x
aml »— J" } (2)
a'x + b'y + c' = J
are equivalent, provided b 4= 0, for the first equation of (2) is
derived from the first of (1) by the reversible processes of trans
position and multiplication by a constant factor.
Also, since any solution of (2) makes y identically equal
to  (ax + c)/b, we may replace y by this value in the second
equation of (2). We thus deduce the equivalent system,
* So far as logic is concerned X might be a function of the variables, but
for present purposes it is taken to be constant. A letter introduced in this
way is usually called an "indeterminate multiplier " ; more properly it should
be called an "undetermined multiplier."
360 SOLUTION BY SUBSTITUTION CHAP.
ax + c '
b'iax + c) , „
a'a;  ^—7 — ' + c' =
Now, since b 4= 0, the second of the equations (3) gives
(a'b  aJ> + (6c'  b'c) = (4).
If a'b  aft' 4= 0, (4) has one and only one solution, namely,
be' b'c . .
x = aY^7b (5) '
this value of x reduces the first of the equations (3) to
1 f a(bc'  b'c) }
qj= ~nw^b +c \'
abc'  a'bc
b(ab'  a'b)'
that is, to y = ca ~ C f (6).
* ab  ao
The equations (5) and (6) are equivalent to the system (3),
and therefore to the original system (1). Hence we have proved
that, if ab'  a'b =1= and b 4= 0, the system (1) has one and only
one solution.
We can remove the restriction b 4= ; for if b = the first of
the equations (1) reduces to ax + c = 0. Hence (if a 4= 0, which
must be, since, if both a = and b = 0, then ab'  a'b = 0) we have
x =  c/a, and this value of x reduces the second of equations
(l)to
a 'c ,, , „
+ b'y + c = 0,
which gives (since V cannot in the present case be without
making ab'  a'b = 0) y = (ca'  c'a)/ab'. Now these values of x
and y are precisely those given by (5) and (6) when b = 0.
The excepted case b = is therefore included ; and the only
exceptional cases excluded are those that come under the condi
tion ab'  a'b = 0.
xvi SYSTEM OF THREE EQUATIONS IN X AND y 3G1
The method of this paragraph may be called solution by
substitution. The above discussion forms a complete and
independent logical treatment of the problem in hand. The
student may, on account of its apparent straightforwardness
and theoretical simplicity, prefer it to the method of § 4.
The defect of the method lies in its want of symmetry ; the
practical result of which is that it often introduces needless
detail into the calculations.
Example.
3«+2y 3 = (a),
9ai + 4i/ + 5 = (/3).
From (a) we have y= — ^ (y)
a
Using (y), we reduce (/3) to
9a!+2(3a;+3) + 5=0,
that is, 15b+11=0;
whence
This value of x reduces (7) to
15
3x^ + 3
15
y= — 2 —
2
= 5'
The solution of the system (a) and (J3) is therefore
11 2
X= 15' y = 5
§ 8.] Three equations of the 1st degree in two variables, say
ax + by + c = 0, a'x + b'y + c'  0, a"x + b"y + c" = (1 ),
will not be consistent unless
a" {be'  b'c) + b"(ca'  c'a) + c"(ab'  a'b) = (2) ;
and they will in general be consistent if this condition be satisfied.
We suppose, for the present, that none of the three functions
ab' — a'b, a'bab", a'b"  a'b' vanishes.* This is equivalent to
supposing that every pair of the three equations has a deter
minate finite solution.
If we take the first two equations as a system, they have
the definite solution
* See below, § 25.
362 CONDITION OF CONSISTENCY chap.
be'  b'c ca'  c'a
X = ^b'~^ r b' V = ab'  a'b'
The necessary and sufficient condition for the consistency of the
three equations is that this solution should satisfy the third
equation ; in other words, that
be' b' c erf ca
a n  + b j, + e = 0.
ab  ab ab  ab
Since ab'  a'b 4= 0, this is equivalent to
a" {be'  b'c) + b"(ca'  c'a) + c"(ab'  a'b) = (3).
The reader should notice that this condition may be written in
any one of the following forms by merely rearranging the
terms : —
a(b'c"  b"c') + b(c'a"  c"a') + c(a'b"  a'b') = (4),
a'(bc"  b"c) + b'(ca"  c"a) + c'(ab"  a"b) = (5),
a(b'c"  b"c') + a'(b"c  be") + a"(bc'  b'c) = (6),
b(c'a"  c'a) + b'(c"a  ca") + b"(ca'  c'a) = (7),
c(a'b"  a"b') + c'(a"b  ab") + c"(ab'  a'b) = (8),
ab'c"ab"c' + bc'a" bc"a' + ca'b" ca"b' = (9).
The forms (4) and (5) could have been obtained directly by
taking the solution of the two last equations and substituting in
the first, and by solving the first and last and substituting in the
second, respectively. Each of these processes is obviously logic
ally equivalent to the one actually adopted above.
The forms (6), (7), (8) would result as the condition of the
consistency of the three equations
ax + a'y + a" = 0, bx + b'y+b" = Q, cx + c'y + c" = (10).
We have therefore the following interesting side result : —
Cor. If the three equations (I) be consistent, then the three equa
tions (10) are consistent.
If the reader will now compare the present paragraph with § 2, he will
see that the function
ab'  a'b
plays the same part for the system
ax + b = 0, a'.r + b' =
XVI
DETERMINANT OF THE SYSTEM
363
as does the function
a(b'c"  b"c') + b{c'a"  c"a') + c(a'b"  a"b')
for the system
ax + by + c = 0, a'x + b'y + c' = 0, a"x + b"y + c" = 0.
These functions are called the determinants of the respective systems of equa
tions. The}' are often denoted by the notations
a b
a! V
for ab'  a'b
\a b c
\a' V c'
for ab'c"  ab"c' + bc'a"  bc"a' + ca'b"  ca"b'
(ii);
(12).
The reader should notice—
1st. That the determinant is of the 1st degree in the constituents of any
one row or of any one column of the square symbol above introduced.
2nd. That, if all the constituents be considered, its degree is equal to the
number of equations in the system.
A special branch of algebra is nowadays devoted to the theory of deter
minants, so that it is unnecessary to pursue the matter in this treatise. For
the sake of more advanced students we have here and there introduced results
of this theory, but always in such a way as not to interfere with the progress
of such as may be unacquainted with them.
The reader may find the following memoriae technicse, useful in enabling
him to remember the determinant of a system of three equations : —
For the form (4),
a b c
b'
b"
V
b\
to be interpreted like the similar scheme in § 4.
For the form (9),
b
or &"/ no
where the letters in the diagonal lines are to form products with the signs f
or  , according as the diagonals slope downwards from left to right or from
right to left.
Example.
To show that the equations
3.r + 5t/2 = 0, 4ic + 6?/l=0, 2.r + 4?/  3 =
are consistent.
Solving the first two equations, we have x= 7/2, j/ = 5/2. These values
364
EXERCISES XXIV
CHAP.
reduce 2x + iyB to 7 + 103, which is zero. Hence the solution of the
first two equations satisfies the third ; that is, the three are consistent.
We might also use the general results of the above paragraph.
Since 3x65x4= 2, 5x23x4=2, 4 x 4 2 x 6= +4, each pair
of equations has by itself a definite solution. Again, calculating the deter
minant of the system by the rule given above, we have, for the value of the
determinant,  54  10  32 + 24 + 12 + 60 = 0. Hence the system is consistent.
+ 3+5 2+3+5
+ 2 +
Exercises XXIV.
Solve the following : —
(1.) £*+&=«, aj+Jy=6J.
(2.) &B+3y=18, 3x2y = 9.
(3.) 123.?+ 685?/ = 3 34, 893a! '59% =8 '71,
find x and y to five places of decimals.
(4.)
(5.)
(6.)
(7.)
(8.)
(9.)
(10.)
(11.)
(12.)
(13.)
x + y:xy = 5 : 3, x + 5y=Z6.
Sx + l = 2y + l = Zy + 2x.
{x+S)(y+S) = (xl)(y+2), 8x + 5 = 9y + 2.
x + y = a + b, (x + a)l(y + b) = b/a.
x y
la mb
2via~ r Zlb
ax + by = 0, (ab)x+(a + b)y = 2c
(a + b)x (ab)y~c, (ab)x + (a + h)y=e.
(a + b)x + (a b)y = a + 2ab  b 2 , (a  by + [a  b)y = a 2 + b 2 .
y
ab,
V
a b 2 a 2 + ab + b 2 ~""'' a 2 +b 2 ' a 2 ab + b 2
(ap m + bq'")x + {ii]i mJ r l + bq m + x )y = «/>"'+ + bq m+2 ,
[np n + bq")x + {«2>"+ l + bq"+^ )y = «^"+ + bq"+ 2 .
a{2a+b).
(14.) Find \ and p. so that x 3 + Xx 2 + /xx + abc may be exactly divisible by
x  b and by x  c.
(15.) If X 4=0, and if xy = ab, ^— + ^—=1,  X —+!L=l, be con
J a + X b + X ' a  X b\
sistent, show that X= ±\/aL
(16.) If the system (b + c)x+{c + a)y + (a + b) = 0, (c + a)x+(a + b)y+ (b + c)
— 0, (a + b)x+{b + c)y + (c + a)~0, be consistent, then os 8 + &*+<? 3a&c=0.
(17.) Find the condition that ux\by = c, a 2 x+bi/ = c', a*x + b 3 y = c 3 be
consistent.
xvi SYSTEMS OF ONE AND OF TWO EQUATIONS IN X, 1J, Z 365
(18.) Find an integral function of a; of the 1st degree whose values shall
be +9 and +10 when x has the values 3 and + 2 respectively.
(19.) Find an integral function of x of the 2nd degree, such that the
coefficient of its highest term is 1, and that it vanishes when x=2 and when
x= 3.
(20.) Find an integral function of x of the 2nd degree which vanishes
when .r=0, and has the values 1 and + 2 when x—  1 and x =+3
respectively.
EQUATIONS WITH THREE OR MORE VARIABLES.
§ 9.] A single equation of the 1st degree in three variables admits
of a twofold infinity of solutions.
For in any such equation, say
ax + by + cz + d = 0,
we may assign to two of the variables any constant values we
please, say y = ft, z — y, then the equation becomes an equation
of the 1st degree in one variable, which has one and only one
solution, namely,
b(3 + cy + d
x = — •
a
We thus have the solution
bB + cy + d
x = f — ■. y = P, * = y
Since there are here two arbitrary constants, to each of
which an infinity of values may be given, we say that there is
a twofold infinity (oo 2 ) of solutions. A symmetric form is given
for this doubly indeterminate solution in Exercises xxv., 27.
§ 10.] A system of two equations of the 1st degree in three vari
ables admits in general of a onefold infinity of solutions.
Consider the equations
ax + by + cz + d = 0, a'x + b'y + c'z + d' = (1).
We suppose that the functions be'  b'c, ca  c'a, ab'  a'b do not
all vanish, say ab' — a'b 4= 0.
If we give to z any arbitrary constant value whatever, say
z = y, then the two given equations give definite values for x and
y. We thus obtain the solution
366 HOMOGENEOUS SYSTEM OF TWO EQUATIONS chap.
(bc'b'c)y + (bd' b'd) (ca'  c'a)y + (da'  d'a) _
X = ~~; ab r ^7b ,V ~ ab'ab »* 7 W
Since we have here one arbitrary constant, there is a onefold
infinity of solutions.
Cor. There is an important particular case of the above that
often occurs in practice, that, namely, where d = and d' = 0.
We then have, from (2),
be'  b'c ca'  c'a
ab  ab ' * ah  ab' '
This result can be written as follows : —
x y
be'  b'c ab'  aV
V = 7
ca'  c'a ab'  a'b'
z = 7
ab'  a'b ab'  a'b
Now, y being entirely at our disposal, we can so determine
it that y/(ab'  a'b) shall have any value we please, say p. Hence,
p being entirely arbitrary, we have, as the solution of the system,
ix +by + cz =0 ) . v
i'x + b'v + c'z = 0) ^ '*
ax
ax + by
x  p(bc'  b'c), y = p(ca'  c'a), z = p(ab'  a'b) (4).
It will be observed that, although the individual values of
x, y, z depend on the arbitrary constant p, the ratios of x, y, z
are perfectly determined, namely, we have from (4)
x : y : z = (be'  b'c) : (ca'  c'a) : (ab'  a'b).
Example 1.
2x + 3y + 4z=0,
3x2y6z0,
x y z
2 6 3 2
give x:y:z 10 : 24 : 13;
xvi SYSTEM OF THREE EQUATIONS IN X, If, Z 367
or, which is the same thing,
a;=10p, ?/ = 24/>, z —  ISp,
p being any quantity whatsoever.
Example 2.
ax + by + cz — 0,
a 2 x + b 2 y + c 2 z=0,
give x = {be 2  b 2 c)p =  bcp(b  c),
y = {ca 2 c 2 a)p =  cap(c  a) ,
z= {ab 2  a 2 b)p —  abp(a  b):
If we choose, we may replace  abep by <r, say, and we then have
xa(bc)/a, y = <r(ca)/b, z — a{ab)\c,
where a is arbitrary.
In other words, we have
x :y :z = (b c)/a : (c  a)/b : (a  b)/c.
§ 11.] A system of three equations of the 1st degree in three
variables, say
ax +by + cz + d =0 (1),
a'x + b'y + c'z + 6! = (2),
a"x + b"y + c"z + d" = (3),
has one and only one solution, provided
ab'c"  ab"c' + be' a"  bc"a' + ca'b"  cab' * (4).
The three coefficients c, c, c" cannot all vanish, otherwise we
should have a system of three equations in two variables, x and
y, a case already considered in § 8.
Let us suppose that c =t= 0, then the following system
ax +by + cz + d =0 (5),
c' (ax + by + cz + d)  c(a'x + b'y + c'z + d') = (6),
c"(ax + by + cz + d) c(a"x + b"y + c"z + d") = (7),
is obviously equivalent to (1), (2), and (3). Matters are so
arranged that z disappears from (6) and (7) ; and if, for short
ness, we put
A  ac'  a'c, B = be'  b'c, C = dc'  d'e,
A'  ac"  a"c, B' = be"  b"c, C = dc"  d"c,
we may write the system (5), (6), (7) as follows : —
ax + by + cz + d = (5'),
Ax + By + C  (6'),
Ax + B'y + 0' = (7').
368
GENERAL SOLUTION
CHAP.
(8),
(9).
(10).
(11).
Now, provided AB'  A'B 4=
(6') and (7') have the unique solution
, _ BC  B'C
X ~ AB'  A'B
_ CA  CA
y ~ AB'  A'B
These values of x and y enable us to derive from (5')
g(BC  B'C) + b(CA'  CA) + d(AB'  A'B)
Z ~ f (AB'  A'B)
(9), (10), and (11) being equivalent to (5'), (6'), (7'), that is,
to (1), (2), (3), constitute a unique solution of the three given
equations.
It only remains to show that the condition (8) is equivalent
to (4).
We have
AB'  A'B = {ad  a'c) (be"  b"c)  (ac"  a"c) (be'  b'c),
= c(ab'c"  ab"c' + be' a"  be" a' + ca'b"  ca"b') (12).
Hence, since c 4= 0, (8) is equivalent to (4).
Although, in practice, the general formulae are very rarely used, } r et it may
interest the student to see the values of x, y, z given hy (9), (10), (11) ex
panded in terms of the coefficients. We have
 (BC  B'C) = {dc'  d'e) (Id'  b"c)  (dc"  d"c) (be'  b'c).
Comparing with (12), we see that (BC'B'C) differs from AB'A'B
merely in having d written everywhere in place of a (the dashes being
imagined to stand unaltered). Hence
 (BC  B'C)=c(db'c"  db"c' + be'd"  be'd' + cd'b"  cd"b').
So that we may write
d(b'c"  b"c') + d'(b"c  be") + d"(bc'  b'c)
X= 7
(13).
a(b'c"  b"c') + a'(b"c  be") + a"(bc'  b'c)
We obtain the values of y and 2 by interchanging a and b and a and c
respectively, namely,
d(a'c"  a"c') + d'(a"c  ac") + d"(ac'  a'c)
?/= 
b(a'c"  a"c') + b'(a"c  ac") + b"(ac'  a'c)
d(b'a"  b"a') + d'(b"a  ba") + d"(ba'^ b'a)
c(b'a"  b"a') + c'(b"aba") + c"(ba'  b'a)
(14),
(15).
XVI
HOMOGENEOUS SYSTEM OF THREE EQUATIONS
3G9
Written in determinant notation these would become
V =
d
b
c
a
b
c
d'
V
c
^
a'
V
e
(13'),
d"
b"
c
a"
b"
c
a
d
c
a
b
c
a'
d'
c'
4
a'
V
c'
an
a"
d"
c"
a"
b"
It
c
a
h
d
a
b
c
a'
V
d'
4
a'
b'
c'
(15').
a"
b"
d"
a"
b"
c"
§ 12.] In the special case where d — 0, d' = 0, d" = 0, the
equations (1), (2), (3) of last paragraph become
ax +by + cz =0 (1),
a'x + b'y + c'z = (2),
a"x + b"y + c'z  (3),
which are homogeneous in x, y, z.
If the determinant of the system, namely, a" (be  b'c) + b"(ca'  c'a)
+ c"(ab'  a'b), do not vanish, tee see from § 11 (9), (10), (11) (or
more easily from (13), (14), and (15) of the same section) that
x = 0, y  0, 3 = 0.
If the determinant does vanish, this conclusion does not necessarily
follow.
In fact, if we write (1), (2), (3) in the form
a  + b  + c
z z
«' + ^ + c' =
z z
a' ,X  + b" V  + c" =
z z
0'),
(2'),
(3'),
and regard xjz and yjz as variables, these equations are con
sistent, since
a" {be'  b'c) + b"(ca'  c'a) + c"(ab'  a'b) = (4),
and any two of them determine the ratios xjz, yjz ; so that we
have
x:y:zbc' b'c : at'  c'a : ab'  a'b,
= be"  b"c : ca"  c"a : ab"  ab,
= b : c"  b"d : c'a"  c"a' : a'b"  a"b'.
VOL. I 2 B
370 EXAMPLES
CHAP.
These different values of the ratios are in agreement, by virtue
of (4), as the student should verify by actual calculation.
Hence, if the determinant of a system of three homogeneous equa
tions of the 1st degree in x, y, z vanish, the values of x, y, z are inde
terminate {there being a onefold infinity of solutions), but their ratios
are determinate.
§ 13.] Knowing, as we now do, that a system of three equa
tions of the 1st degree in x, y, z has in general one definite
solution and no more, we may take any logically admissible
method of obtaining the solution that happens to be convenient.
(1) We may guess the solution, or, as it is put, solve by inspec
tion, verifying if necessary. (2) We may carry out, in the
special case, the process of § 11; this is perhaps the most gene
rally useful plan. (3) We may solve by substitution. (4) We
may use Bezout's method. (5) We may derive from the given
system another which happens to be simpler, and then solve
the derived system. The following examples illustrate these
different methods : —
Example 1.
x + y + z = a + b + c, (bc)x + {ca)y + {ab)z = 0,  + \ + —3.
' v '" ' a b c
A glance shows ns that this system is satisfied by x = a, y = b, z — c; and,
since the system has only one solution, nothing more is required.
Example 2.
Bx + 5y 7; 2 = (a),
4af+8y14z+3=0 (jS),
Sx + 6y 8~3 = (7).
Multiplying (a) by 4 and (/3) by 3, and subtracting, we obtain
4^143+17 = (3).
From (a) and (7), by subtraction,
2/~l = (e).
Multiplying (e) by 4, and subtracting (5), we have, finally,
l(b21 = 0;
whence z=2'l.
Using this value of z in (e), we find
V = 31;
and, putting 3/ =8 1, z = 2'\ in (a), we find
x =4.
The solution of the system (a), (/3), (7) is therefore
x=i, t/ = 31, ss=21.
XVI
EXAMPLES 371
Example 3.
Taking the equations (a), (/3), (7) of last example, we might proceed by
substitution, as follows : —
From (a)
5 7 2
X =3 y + 3 Z + 3
20 28 8
This value of x reduces (/3) to
which is equivalent to
4y14z + 17 = (5')
Substituting the same value of as as before in (7), we deduce
2/~l=0 (e').
Now (e') gives
y=z+l,
and this value of y reduces (5') to
102 + 21 = 0,
hence ~ = 21.
The values of y and x can now be obtained by using first (e') and then (a).
Example 4.
Taking once more the equations (a), (^), (7) of Example 2, Ave might pro
ceed by Bezout's method.
If X and fi be two arbitrary multipliers, we derive from (a), (/3), (7),
(3x + 5y7z2) + \(ix + 8yliz + 3) + /jL(3z + 6y8z3) = (5').
Suppose that we wish to find the value of x. We determine X and /j. so that
(5') shall contain neither y nor z. We thus have
8X + 6/*+5 = (e'),
14X8/x7 = (D,
(3 + 4X + 3/4£2 + 3X3m=0 (if).
If we solve (e') and (f), we obtain
X=l, ^=7.
The last equation (rf) thus becomes
(3421).b2 3 + 21 = 0,
that is, 'ox # 2 = ;
whence x= 2/*5 = 4.
The values of y and z may be obtained by a similar process.
Example 5.
ax + by + cz = (a),
(b + c)x + {c + a)y + (a + b)z=0 (3),
a 2 x + b*y + ch = a 2 {b  c) + b{c  a) + c(a  b) (7).
From (a) and (/3) we derive, by addition,
(a + b + c)(x + y + z) = 0,
which, provided « + i + c + 0, is equivalent to
x + y + z = (5).
We can now, if we please, replace (a) and (/3) by the equivalent simpler pair
(a) and (5).
372 EXAMPLES chap.
Now (see § 10), by virtue of (a) and (5), we have
x y z . .
bc ca ab
If none of the three, bc, ca, ab, vanish, we may write (7) in the form
a 2 (b  c),— + b(c  a)^— + c(a  b) ^ = a 2 (b c) + b 2 (c a) + c"(a  b).
b — c c — a a — b
Using (e) we can replace y/{ca) and zftab) by x/(bc), and the last equa
tion becomes
{a°(bc) + b(ca) + <?(ab)},— = a"(bc) + b*(ca) + c 2 (ab);
c
and, since a"(bc) + b{ca)+c(a b) = (bc) {ca) (a b), which does not
vanish, if our previous assumptions be granted, it follows that
bc
Hence x = bc, and, by symmetry, y = ca, z = ab.
This solution might of course have been obtained at once by inspection.
Example 6.
x + ay + a 2 z + a 3 =0\
x + by + b 2 z + ¥ = ol (a).
x + cy + cz+c 3 0J
From the identity
Z 3 +p£ z +qZ+r={£a)$b){Zc),
(see chap, iv., § 9), where
2)= a bc, q — bc + ca + ab, r=abc,
we have
r + aq + a 2 p + a? = 0\
r + bq+b 2 j,+ b 3 =0l (£).
r + cq+cp+ (^0)
It appears, therefore, from (,3) that
xr, y = q, z=p
is a solution of (a). Hence, since (a) has only one solution, that solution is
x=abc, y=bc + ca + ab, z='—a — b — c
This result may be generalised and extended in various obvious ways.
§ 14.] A system of more than three equations of the 1st degree
in three variables will in general be inconsistent. To secure consistency
one condition must in general be satisfied for every equation beyond
three. This may be seen by reflecting that the first three equa
tions will in general uniquely determine the variables, and that
the values thus found must satisfy each of the remaining equa
tions. Thus, in the case of four equations, there will be one
condition for consistency. The equation expressing this condition
could easily be found in its most general form ; but its expression
would be cumbrous and practically useless without the use of
xvi GENERAL THEORY FOR A LINEAR SYSTEM 373
determinantal or other abbreviative notation. There is, how
ever, no difficulty in working out the required result directly in
any special case.
Example.
Determine the numerical constant p, so that the four equations
2x~Zy+5z=18, 3xy + lz = 2Q, 4x + 2yz=5,
{p+l)x+{p+2)y+(p + S)z=76
shall he consistent.
If we take the first three equations, they determine the values of x, y, z,
namely, x=l, y = B, 2 = 5.
These values must satisfy the last equation ; hence we must have
(p + l) + (p + 2)3 + (p + 3)5 = 76,
which is equivalent to
9p = 5i.
Hence p = 6.
§ 15.] If the reader will now reconsider the course of reason
ing through which we have led him in the cases of equations of
the 1st degree in one, two, and three variables respectively, he
will see that the spirit of that reasoning is general ; and that,
by pursuing the same course step by step, we should arrive at
the following general conclusions : —
I. A system of n  r equations of the 1st degree in n variables
has in general a solution involving r arbitrary constants; in other
words, has an rfold infinity of different solutions.
II. A system of n equations of the 1st degree in n variables has
a unique determinate solution, provided a certain function of the co
efficients of the system, which we may call the determinant of the
system, does not vanish.
III. A system of n + r equations of the 1st degree in n variables
will in general be inconsistent. To secure consistency r different con
ditions must in general be satisfied.
There would he no great difficulty in laying down a rule for calculating
step hy step the function spoken of above as the determinant of a system of
n equations of the 1st degree in n variables ; hut the final form in which it
would thus be obtained would be neither elegant nor luminous. Experience
has shown that it is better to establish independently the theory of a certain
class of functions called determinants, and then to apply the properties of
these functions to the general theory of equations of the 1st degree. A
brief sketch of this way of proceeding is given in the next paragraph, and
will be quite intelligible to those acquainted with the elements of the theory
of determinants.
374
GENERAL SOLUTION OF LINEAR SYSTEM
CHAP.
GENERAL SOLUTION OF A SYSTEM OF LINEAR EQUATIONS
BY MEANS OF DETERMINANTS.
§16.] Consider the system
"D  ! =a n x x + a 12 .r 2 + . .
. + a m x n + c,
=
(1),
KJ 2 == ^'21 1 ^22* 2 ' * *
• + CWEji + C2
=
(2),
U„ * ffljuJB, + rt, !2 a 2 + . . . + fl^Bn + C n = (»),
where there are n variables, 2,, x 2 , . . ., #«, and ra equations to
determine them.
Let
A =
a n a ]2
"«2
and let A M A 2 , . . ., A n denote the determinants obtained from A
by replacing the constituents of the 1st, 2nd . . . nth columns
respectively, by the set r,, c 2 , . . ., c n .
Also let the cofactors of a,„ a I2 , . . ., a m , a 21 , a 22 , . . ., a m ,
&c., A be denoted by A u , A I2) . . ., A m , A 21 , A 22 , . . ., A 2n , &c,
as usual.
Then, by the theory of determinants, we have
OuA,, + OnAa + . . . + a m A m = A \
a IS A„ + rt 23 A 21 + . . . + a n2 A m =
(a),
a, n A u + a. m A 2i + . . . + a nn A m =
c iAi + £3^1 + . . . + <" n A, u = A, J
and so on.
If the determinant A, which we call the determinant of the
system of equations, does not vanish, then A u , A 21 , . . ., A, n
cannot all vanish. Let us suppose that A„ 4= 0. Then, by
chap, xiv., § 10, the system
A U U, + A 21 U 2 + . . . + A /U U„ = 0,
U 2 =0, u, = o, . . ., u n =o,
is equivalent to the system (1), (2) . . . (»). If we collect the
XVI
BY MEANS OF DETERMINANTS
375
coefficients of the variables ar, , x 2 , . . ., x n in the first of these
equations, and attend to the relations (a), that equation reduces to
A», + A t = 0.
Since A + 0, this is equivalent to
*,= A/A.
By exactly similar reasoning we coidd show that x 2 =  A.,/ A, . . .
x n =  A n /A. Hence the solution, and the only solution, of (1),
(2), . . ., (n) is
k, =  A,/A, x 2 =  A 2 /A, . . ., x n =  A,,./ A (ft).
Although, from the way we have conducted the demonstration,
it is not necessary to verify that (ft) does in fact satisfy (1),
(2), . . ., (a), yet the reader should satisfy himself by substitu
tion that this is really the case.
We have thus shoivn that a system of n equations of the 1st degree
in n variables has a unique determinate solution, provided its determin
ant does not vanish.
Next, let us suppose that, in addition to the equations (1),
(2), . . ., (n) above, we had another, namely,
0»+i,i x i + a »+i, 2 x 2 + • • • + a w +i,« %n + Cn+i = Oj (a + 1 ),
the system of n + 1 equations thus obtained will in general be
inconsistent.
The necessary and sufficient condition for consistency is that the
solution of the first n shall satisfy the n+lth, namely, ^/A=t=0,
— #71+1,1 A,
tf»+i >2 A 2 
a
u
a 2l
a™,
a n+i,n Aji + C,!)!
... i
A = 0, that is,
(12,1
II
rt+i,i #n+2,s
" nil ' II
a n+2,n ("n+i
=
(y)
Lastly, let us consider the particular case of n homogeneous
equations of the 1st degree in n variables. In other words, let
us suppose that, in equations (1), (2), . . ., (/<) above, we have
c, = 0, c 2 = 0, .
., Cn=0.
376 CONDITION OF CONSISTENCY HOMOGENEOUS SYSTEM chap.
1st. Suppose A 4= 0, then, since now A l = 0, A.,  0,
A n = 0, (/3) gives Xy = 0, x 2 = 0, . . ., x n = 0.
2nd. Suppose A = 0.
We may write the equations in the form
x x
x,
On— + «m— + • • • + a m = 0,
■' a
X~
« 2 i— + «»— + . . .+a m = 0,
"21 T •*«!
X\ x%
Q"M ^ ^W2 I"
+ a nn = 0.
These may be regarded as a system of n equations of the 1st
degree in the n — 1 variables xfx m x.,/x n , . . ., x n ^\x n \ and, since
A = 0, they are a consistent system. Using only the last n\
of them, we find
x 1
Xn
"'in ( *22 • • • ®2,n  1
^371 ^32 • • • Q'3,n  1
™"im ™na
l n,n\
Clo, do
(I.,
J 2\ "'22 • • • "B,n  1
^'31 ^32 • " * "3,71—1
0"K\ Q1V2 . • • tt
n,ni
= "("I) 2
A A
\2n3 tt!L  n
A A *
In a similar way we prove that
X 2 /X n = A 12  A m . . ., X nl jX n = Ai jn _ 1 /A ln .
iZe/tce we /«we
3/ t : £ 2 • • • • • ^n ~ "li • "is • • • • ■ A lrt J
and, fo/ 'parity of reasoning,
x l : x. 2 : . . . : x n = A rl : A r2 : . . . : A m
where r = 2, = 3, . . ., = n, as we please. In other words, the ratios
of the variables are determinate, but their actual values are in
determinate, there being a onefold infin iiy of different solutions.
Exercises XXV
Solve the following systems : —
(1.)
x y z
a , v
3 + 4 + 6^' i5 + 20 + 9 10, 2 + 10 + 4 _4,J 
xvr
EXERCISES XXV 377
(2.) 2x + %y + \z = 2§, 3x+2y+5z=S2, 4.z + 3?/ + 2i = 25.
(3.) '8x+l2y+68z=l, SSx25yZ'82z=5,
•Ola: 003y 301;= 013 ;
calculate as, y, z to four places of decimals.
(4.) x + y + z = 26, xy = <±, xz=6.
(5.) If (x+l) a _ A  B:r+C
(x + 2)(ar + x+l)~x + 2 a? + x+V
determine the numerical constants A, B, C.
(6.) Find a linear function of x and y, which shall vanish when x=x',
y = y', and also when x = x", y = y", and which shall have the value +1 when
x = x'", y = y'".
(7.) An integral function of x of the 2nd degree vanishes when x = 2,
and when x=3, and has the value  1 when x=  2 ; find the function.
Solve the following systems : —
(8.) y + z = a, z + x = b, x + y — c.
(9.) JL+ * =2a, ^ + — = 26,  X .+ y ,=2c.
b + c bc c+a ca a+b a b
(10.) An integral function of a; of the 2nd degree takes the values A, B, C,
when x has the values a, b, c respectively ; find the function.
Solve the following systems : —
(11.) bc{b  c)x + ca(c  a)y + ab(a  b)z=0,
(a + bc)x + (b + ca)y+(c + ab)z = a 2 + b 2 + c 2 ,
b 2 c 2 x + c 2 a 2 y + a 2 b 2 z = abc(bc + ea + ab).
(12.) If _^_ + ^_ + _f_=l >
a + a b+a c + a
x y z .
« + /3 6 + (3 c + /3
x y z .
then
a + y b + y c+y
I V ■ z  Iz£.
(a + a)(a + p) 2 "(b + a)(b + p) 2 (c + a) (c + /3) 2 (a + j8) (b + p) (c + /3)
(13.) aa; + fa/ + c3 = « + & + c,
a 3 x + fe 2 2/ + c^ = (a + 6 + c),
bcx + cay + abz = 0.
(14.) ax + cy + bz = ex + by + az = bx + ay + cz = a 3 + b* + c 3  Zabc.
(15.) lx + my + nz = mn + nl + hn,
x + y + z=l + m + n,
(mn)x + (n l)y + (l m)z—0.
(16.) h + my + nz = 0,
(ra + n)x + (7i + ?)y + {l + m)z=l + m + n,
Px + m 2 y + n 2 z =p 2 .
(17.) Show that (b c)x + bycz = 0, (ca)y + czax=Q, (ab)z + axby
= 0, are consistent.
378 EXERCISES XXV chap.
(18.) Show that the system cybz=f, ass — cz—g, bxay = h has no finite
solution unless af+bg + ch = 0, in which case it has an infinite number of
solutions.
Find a symmetrical form for the indeterminate solution involving one
arbitrary constant.
Solve the following systems : —
(19.) 3x2y + 3u = 0, xy+z=0, 3y + Sz2u = 0, x + 2y + Bz + 4u = 8.
(20.) Uax=pr, by—ps, cz = rs, d{y + z) = sq, e(z + x) = qr,f(X + y)
= qp + g, find z in terms of a, b, c, d, e,f, g.
iz + x 3u + x 5v + x
(21.) 2y + x =
3 4 8
x + y + z + u + v  1 _ 5x + ly + Bz + 2u + v + 2
4 9
(22.) ax + by=l, cx + dz=l, ez+fu = l, gu + hv=l, z + y + z + u + v = 0.
(23.) Prove that, with a certain exception, the system U = 0, V = 0, W = 0,
and \XJ + fiY + vW = 0, \'U + /x'V + k'W = 0, X"U + /j."V + p"W = are equivalent.
(24.) If x=by + cz + du, y = ax + cz + du,
z = ax + by + du, u = ax + by + cz,
„ a b c d
then  + i — H 7 +5 — =r=l
a+l 6+1 c+1 d+l
(25.) Show that the system ax + by + cz + d = 0, a'x + b'y + c'z + d' = 0,
a"x + b"y + c"z + d" = will be equivalent to only two equations if the system
ax + a'y + a"Q, bx + b'y + b" = Q, cx + c'y + e" = Q, dx + d'y + d" — be con
sistent, that is, if
b'c"  b"c' _ b"c  be" _ be'  b'c
a'd"  a"d' ~ a"d  ad" ~~ad' a'd'
Show that in the case of the system
7 x y z . x y z .
x + y + z = a + b + c,  + f + =l, v. + pj + ^ = 0,
a b c a 6 b f c 8
the above two conditions reduce to one only, namely,
bc + ca + ab = Q.
(26.) Show that the three equations
x = A + A'u + A"v, y = B + B'u + B"v, z = C + C'tc + C"v,
where u and v are variable, are equivalent to a single linear equation con
necting x, y, z; and find that equation.
(27.) If ax + by + cz + d—0, show that
y= (£ +q y e a)£
where p and q are arbitrary constants.
xvi EQUATIONS REDUCIBLE TO LINEAR SYSTEMS 379
(28. ) If ax + by + cz + d = 0, a'x + b'y + c'z + d' = 0, show that
x=p(bc!  b'c) + \{b'  c')d (b c)d'}/{a(b'  c') + b(c'a') +c(a' b')},
y=p(ca' ~ c'«) +{(e' a')d  (c  a)d'\/{a(b'  c') + b(c'  a') + c(a' &')},
z=p(ab'  a'b) + { (a'  b')d (a b)d'}/{a{b'  c') + b(c'  a') + c(a' b')},
where p is an arbitrary constant.
EXAMPLES OF EQUATIONS WHOSE SOLUTION IS EFFECTED BY
MEANS OF LINEAR EQUATIONS.
§ 17.] We have seen in chap. xiv. that every system of
algebraical equations can be reduced to a system of rational
integral equations such that every solution of the given system
will be a solution of the derived system, although the derived
system may admit of solutions, called " extraneous," which do not
satisfy the original system. It may happen that the derived
system is linear, or that it can, by the process of factorisation,
be replaced by equivalent alternative linear systems. In such
cases all we have to do is to solve these linear systems, and then
satisfy ourselves, either by substitution or by examining the
reversibility of the steps of the process, which, if any, of the
solutions obtained are extraneous. The student should now
reexamine the examples worked out in chap, xiv., find, wher
ever he can, all the solutions of the derived equations, and
examine their admissibility as solutions of the original system.
"We give two more instances here.
Example 1.
• + ./;.„ .L 2 1U =V{2(^ + D} («)■
(Positive values to be taken for all tbe square roots.)
If we rationalise the two denominators on the left, we deduce from (a) the
equivalent equation,
V{^V0« 2 i)}+V{* + V(^i)}=V{2(^+i)} (/s).
From (3) we derive, by squaring both sides,
2x + 2\f{x 2 (x°~l)} =2(a* + l),
that is, 2.e + 2 = 2x 3 + 2 (y).
Now (7) is equivalent to x 3 x = 0,
that is, to x(x  1) (se+ 1) =0 (8).
Again (5) is equivalent to the alternatives
that is to say, its solutions are # = 0, x=l, x =
380
EXAMPLES
niAP.
Since, however, the step from (fi) to (7) is irreversible, it is necessary to
examine which of these solutions actually satisfy (a).
Now x = Ogives \f  i+ \J \i=\j2,
that is (see chap, xii., § 17, Example 3),
which is correct.
Also, £=1 obviously satisfies (a).
But x=  1 gives 2i = 0, which is not true, hence x=  1 is not a solution
of (a).
Remark that x= \ is a solution of the slightly different equation,
V{z+V(z 2 D} " VteVC* 8 !)} = ^'^ + 1) '
Example 2.
x 2 ~y 2 = xy, 2x+3yl = (a).
Since the first of these equations is equivalent to (x~y)(x + yl) = 0, the
system (a) is equivalent to
/ xy = 0, and 2x + 3i/l=0 N
\x + yl = 0, an.l 2z + 32/l = 0,
now the solution of xy = 0, 2x+Byl~0 is a: =1/5, 2/ = 1/5 ; and the solu
tion of a: + 2/l = 0, 2»+3yl = is a;=2, y— 1. Hence the solutions of
(a) are
a;
y
1/5
2
1/5
1
§ 18.] The solution of linear systems is sometimes facilitated
by the introduction of Auxiliary Variables, or, as it is sometimes
put, by changing the variables. This artifice sometimes enables us
to abridge the labour of solving linear systems, and occasionally
to use methods appropriate to linear systems in solving systems
which are not themselves linear. The following are examples : —
Example 1.
(a).
{xaf _x2ab
{x + b'f x + a + 2b
Let x + b = z, so that x = zb ; and, for shortness, let c = a + b.
Then (a) may be written
(zc) 3 _z2c
2 s ~ z + c
From (/3) we derive
{zc)\z + c) = z 3 ^z2c\
(£)■
xvi CHANGE OF VARIABLES 381
that is,
z*  2z»c+ 2z<?  c 4 = 2 4  2rV,
which is equivalent to
2c s 3c 4 =0 (7).
Now (7) has the unique solution z=e/2, which evidently satisfies (a)
Hence x = c/2  b, that is, x — (ab)j2, is the only finite solution of (a).
Example 2.
a(x + y) + b(xy)+c = 0, a'(x + y) + b'(xy) + c' = (a).
Let £ = x + ?/, q = xy, then the system (a) may be written
a^ + bij + c0, a's + b'T] + e' = Q (a').
Now (a') is a linear system in £ and 7;, and we have, by § 4,
be' b'c _ca'  c'a
*~ab'a'b' V ~ab'a'b &>'
Replacing f and 7; by their values, we have
be'  b'c ca' c'a , .
X + y = aV^b' X ~ y= W^Tb iy) 
From (7), by first adding and then subtracting, we obtain
_ be'  b'c + ca'  c'a _ be'  b'c  ca' + c'a
X ~ 2{ab'a'b) ' V ~ 2(ab'  a'b) '
Example 3.
cy + bz — az + cx — bx + ay — abc.
Dividing by be, by ca, and by ab, we may write the given system in the
following equivalent form
!+: »
x y , .
a + l= C W
Now, if we add the equations (/3) and (7), and subtract (a), we have
(MM;+s)(K)»+^
x
that is, 2 — b + ca\
a
, a(b + ca)
wlience x= — — .
By symmetry, we have
y = b{c + ab)/2, z = c{a + bc)/2.
Here we virtually regard x/a, yjb, z/c as the variables, although we have
not taken the trouble to replace them by new letters.
382
EXAMPLES
CHAP.
Example 4.
x—ay—b z c
r
(a),
03).
have
(7).
(*)•
P 1
Ix + my + nz = d
Represent each of the three equal functions in (a) by p. Then we
(xa)/p = p, (yb)/q = p, (zc)/r = p,
which are equivalent to
x = a+pp, y = b + qp, zc + rp
Using (7), we reduce (j8) to
l(a +pp) + m[b + qp) + n(c + rp) = d,
for which we obtain, for the value of the auxiliary p,
_dla mb  nc
lp + mq + nr
From (7) and (5) we have, finally,
d—la — mb  nc
x = a + v 7 ,
lp + mq + nr
_ m{aq  bp) + n(ar  cp) \})d
lp + mq + nr
The values of y and z can be similarly found, or they can be written
down at once by considering the symmetry of the original system.
Example 5.
x2y + Zz = (a),
2a;3y + 4t = Q3),
ia? + By 3 + z 3 xyz = 216 (7).
From (a) and (/3) we have (see § 10 above)
x/l=y/2=z/l=p, say.
Hence & = P, y — %P> ~ = / } (5).
By means of (5) we deduce from (7)
27 p 3 = 216,
which is equivalent to p* = 8. (e).
Now the three cube roots of 8 are (see chap, xii., § 20, Example 1)
2, 2(  1 + V3i). 2(  1  \/30
Hence the solutions of (e) are
P = 2, p=2(l + V3i). p=2(lV8i)
Hence, by (5), we obtain the three following solutions of (a), (/3), (7) ; —
X
V
z
2
 1 + \J%i
 1  V3i
4
2(  1 + V30
2(  1  V30
2
 1 + V3*
 1  V3i
xvi EXERCISES XXVI 383
Since, by chap. xiv. , § 6, the system in question has only three solutions,
we have obtained the complete solution.
N.B. — In general, if Ui, w 2 , • • •, « n i be homogeneous functions of the
1st degree in n variables, and v a homogeneous function of the nth degree in
the same variables, the solution of the system
!/l = 0, W2 = 0, . . ., « n i = 0, v =
may be effected by solving a system of n  1 linear equations in n  1 variables,
and then extracting an »th root. See in this connection § 16 above.
Example 6.
ax + by 2 + c = 0, a'x? + b'y + c'=0.
If we regard x? and y' 2 as the variables, we have to do with a linear system,
and we obtain as heretofore,
x 2 ={bc'  b'c)/(ab'  a'b), y" = (ca'  c'a)/(ab'  a'b).
Hence
x=±\/{M b'c)/(ab' a'b), y=±\/(ca' c'a)/{ab' a'b).
Since either of the one pair of double signs may go with either of the other
pair, we thus obtain the full number of 2 x 2 = 4 solutions.
Example 7.
ay + bx + cxy = 0, a'y + b'x + c'xy=0.
These two equations evidently have the solution x=0, y = 0.
Setting these values aside, we may divide each of the two equations by
xy. We thus deduce the system
a+b + c = 0, a' + b' + c' = 0,
x y x y
which is linear, if we regard 1/x and 1/y as the variables. Solving from this
point of view, we obtain
1 _ be'  b'c 1 _ ca'  c'a _
x~ab'a'b' y~ab'a'b '
for which we have
x = {aV  a'b)/(bc'  b'c), y = {aV  a'b)/(ca'  c'a\
"We have thus found two out of the four solutions of the given system. There
are no more finite solutions.
Exercises XXVI.
Solve the following by means of linear systems :
,, > \Jox +\/b _ \/a + \/b
\/ax  yjb ~ \/b
,„ , Va;+4?>i _ \/x + 27H
\/x + 3)i ~ \Jx + n '
(3.) ^/( x + 22)s/(x+ 11) = 1.
(4.) V*+V(*+3) = 12/V(* + 3).
384 EXERCISES XXVI
(5.) V(^ + 2 ) + V(«2) = 5/V(a + 2).
(6.) V aj +V( a + / 3 )V(aJ 8 )=V(*+2/3).
(7. ) \J{x + q r) + V(a; + rp) + s/(x +p q) = 0.
*j{xp)y/ P + V(»p)+Vp = ^ (x ~ p) ~
(9.) z=v> 2 aV(& 3 + a; 2 a 2 )}+a.
(10. ) V( V' c + V«) + V(V*  V«) = V(2 V® + 2 \ /b )
(11.) \?x+^/{B^(2x + x 2 )} = ^3.
( 12 ) V^'V^)^' V(*2)=v(*3).
(13.) \/{ax)^J{yx) = sJy, \J{bx) + *J(yx)~s/y.
1 17
\H.) Va ; \ / 2/=4. ^y^^g
(15.) (xa) 2 (y6) 2 = 0, (a; 6) (y «) = «( 26 a).
(16.) a2/ = 3, z 2 7/ 2 = 45.
(17.) xfy = a/b, x^if — d.
(18.) a; + ay + a 2 2 + a 3 ii + a 4 =0,
x + by + bh + b s u + b* = 0,
x + cy + <?z + <?u + c 4 = 0,
re + tf y + cPz + d?u + d 4 = 0.
(19.) x+y+z=0, ax + by + czQ,
bcx + cay + abz = (b  c 2 ) (c 2  a 2 ) (a 2  b").
mx ly _ ny mz _ Iznx 1
^ "' lm{ab) mn(b — c) nl(ca) hnn'
CHAP.
(21.)
mnx + nly + hnz — a + b + c.
x + 7/ — z y + zx z + xy ,
b+c c+a a+b
(22.) Z(&c)a;=0; 2a(6 2 c 2 )x=0, 2a(bc)x=JI(bc).
(23.) 2sc=l, 2te/(6c)=0 J 2.<:/(i 2 c 2 ) = 0.
(24.) ax + k(y + z + ic)0, by + k(z + u + x) = 0,
cz + k(u + x + y)0, du + k(x + y + z) = 0.
(25.) x + y\z—a, y + z + u — b, z + u + x = c, u + x + y=d.
(26.) 3/»2/y=i, 4»+7y=lay.
1 1 _12 ay + 2s + 3y + 2
1 ' sb+5 y + 7 35' rry+1
< 28 > w+J 11  L + h 16 
X L M 2 \2 + M 2 X 2_ M 2
y x + y 2 a: 2 y ar + y
(29) ^ +■ M 1 ^ "^ =1
XVI
EXERCISES XXVI
385
(30.)
(31.)
(32.)
(33.)
a(b y) + b(a  x) = c(a x)(b y),
a\b y) + b\a x) = c 2 {a x)(b y).
b « 2 ft 2
+ . , . =1, — +
c+a+x c + b + y
b 2 c 2 , 6
— + ~i — 1,
I/ 2 z 2
c+a+x c+b+y
a + b.
+ * 2=] '
a 2 b 2 ,
— +— =1,
yz zx
z 2
+— =1,
2a; xy
a 2 6 2 ,
3 +  = l.
x 2 y
^2/ 2/2
(34. ) ayz  bzx  cxy =  ayz + bzx  cry —  ayz  bzx + cxy = xyz.
(35.) Show that (l + lx) (l + ay) = l + lz. {l+mx)[\ + by) = l+mz,
(1 + nx) (1 +cy) = l \nz are not consistent unless
(6  c) ajl + {c a)b/m + (ab) cfn = 0.
If this condition be satisfied, then x = (c/nb/m)/(b  c) ; and particular
solutions for y and z are y—  \ja, z=  l/l.
GRAPHICAL DISCUSSION OF LINEAR FUNCTIONS OF ONE AND
OF TWO VARIABLES.
§ 19.] The graph of a linear function of one variable is a
straight line.
Consider the function
y = ax+ b. To find the
point where its graph cuts
Y
B
p
N
sk
o i\
ft X
Fio. 1.
OY, that is, to find the
point for which X = 0,
we have to measure
OB = b upwards op
downwards, according
as b is positive or
negative (Figs. 1 and 2). Through B draw a line parallel to the
xaxis.
Let OM represent any positive value of x, and MP the cor
responding value of if.
vol. I 2 c
Fio. 2.
386
THE GRAPH OF CLV + b
CHAP.
By the equation to the graph, we have (y  b)]x = a. Now, since
b = + OB = + MN in Fig. 1,   OB =  MN in Fig. 2, we have
yb = PM  MN  PN in Fig. 1,
y  b = PM + MN = PN in Fig. 2.
Hence we have in both cases
PN_PN_y6
BN ~ OM ~ "
= a.
In other words, the ratio of PN to BN is constant ; hence,
by elementary geometry, the locus of P is a straight line. If a
be positive, then PN and BN must have the same sign, and the
line will slope upwards, from left to right, as in Figs. 1 and 2 ;
if a be negative, the line will slope downwards, from left to
right, as in Figs. 3 and 4. The student will easily complete the
discussion by considering negative values of x.
Y Y
M X
Fig. 3.
Fio. 4.
§ 20.] So long as the graphic line is not parallel to the axis
of x, that is, so long as a =# 0, it will meet the axis in one point,
A, and in one only. In analytical language, the equation
ax + b = has one root, and one only.
Also, since a straight line has no turning points, a linear
function can have no turning values. In other words, if we
increase x continuously from  oo to + oo , ax + b either increases
continuously from  oo to + oo , or decreases continuously from
+ oo to  oo ; the former happens when a is positive, the latter
when a is negative.
Since ax + b passes only once through every value between
XVI
THE GRAPH OF OX + b 387
+ oo and  oo , it can pass only once through the value 0. We have
thus another proof that the equation ax + b = has only one root.
A purely analytical proof that ax + b has no turning values
may be given as follows : — Let the increment of x be h, then the
increment of ax + b is
{a(x + h) + b)  {ax + b) = ah.
Now ah is independent of x, and, if h be positive, is always posi
tive or always negative, according as a is positive or negative.
Hence, if a be positive, ax + b always increases as x is increased ;
and if a be negative, ax + b always decreases as x is increased.
§ 21.] We may investigate graphically the condition that the
two functions ax + b, a'x + b' shall have the same root ; in other
words, that the equations, ax + b = 0, a'x + b' = 0, shall be con
sistent. Denote ax + b and a'x + V by y and y' respectively, so
that the equations of the two graphs
are y = ax + b, y' = a'x + b'. If both
functions have the same root, the
graphs must meet OX in the same
point A. Now, if P'M PM be ordi
nates of the two graphs corresponding
to the same abscissa OM, and if the
graphs meet OX in the same point A,
it is obvious that the ratio P'M/PM
is constant. Conversely, if P'M/PM FlQ  5 
is constant, then P'M must vanish when PM vanishes ; that is,
the graphs must meet OX in the same point. Hence the neces
sary and sufficient analytical condition is that (a'x + b')j(ax + b)
shall be constant, = k say. In other words, we must have
a'x + V = k(ax + b).
From this it follows that
a' = ka, V = kb,
and ab'  a'b = 0.
These agree with the results obtained above in § 2.
§ 22.] By means of the graph we can illustrate various limiting
cases, some of which have hitherto been excluded from con
sideration.
388
GRAPHICAL DISCUSSION OF LIMITING CASES
CHAP.
I. Let b = 0, a 4= 0. In this case OB = 0, and B coincides
with ; that is to say, the graph passes through (see Figs.
Y
Fig. 6. Fig. 7.
6 and 7). Here the graph meets OX at 0, and the root of
ax = is x = 0, as it should be.
II. Let b =¥ and a = 0. In this case the equation to the
graph is y = b, which represents a line parallel to the aaxis (see
Figs. 8 and 9). In this case the point of intersection of the
Y
B
O
X
o
Fig. 8.
B
Fig. 9.
graph with OX is at an infinite distance, and OA = oo . If we
agree that the solution of the equation ax + b = shall in all
cases be x —  b/a, then, when b 4= 0, a = 0, this will give x = oo ,
in agreement with the conclusion just
derived by considering the graph.
This case will be best understood
^ by approaching it, both geometric
— r ally and analytically, as a limit.
Let us suppose that 6= — 1, and
that a is very small, =1/100000,
say. Then the graph correspond
ing to y = x/l 00000  1 is something like Fig. 10, where
B
Fio. 10.
:cvi INFINITE ROOT 389
the intersection of BL with the axis of x is very far to the right
of ; that is to say, BL is nearly parallel to OX.
On the other hand, the equation
1=0
100000
gives x = 100000, a very large value of x. The smaller we
make a the more nearly will BL become parallel to OX, and the
greater will be the root of the equation ax + b = 0.
If, therefore, in any case where an equation of the 1st degree in
x was to be expected, we obtain the paradoxical equation
6 = 0,
where b is a constant, this indicates that the root of the equation has
become infinite.
III. If a = 0, b = 0, the equation to the graph becomes y  0,
which represents the axis of x itself. The graph in this case
coincides with OX, and its point of intersection with OX becomes
indeterminate. If we take the analytical solution of ax + b =
to be x =  bja in all cases, it gives us, in the present instance,
x = 0/0, an indeterminate form, as it ought to do, in accordance
with the graphical result.
§ 23.] Tho graphic surface of a linear function of two inde
pendent variables x and y, say z = ax + by + c, is a plane. It
would not be difficult to prove this, but, for our present pur
poses, it is unnecessary to do so. We shall confine ourselves to
a discussion of the contour lines of the function.
The contour lines of the function z = ax + by + c are a series of
parallel straight lines.
For, if k be any constant value of z, the corresponding con
tour line has for its equation (see chap, xv., § 1 6)
ax + by + ck (1).
Now (1) is equivalent to
/ a\ k c
r(j>+T < 2) 
But (2), as we have seen in § 19 above, represents a straight
line, which meets the axes of x and y in A and B, so that
390
CONTOUR LINES OF ClX + by + C
CHAP.
b b I b a
Fig. 11.
Let h' be any other value of z, then the equation to the corre
sponding contour line is
ax + by + c = k' (3),
el
s'
a\ k'  c
y x+ b
(4).
Hence, if this second contour line meet the axes in A' and B'
respectively, we have
OB' =
Hence
Jc'c
IT'
OA =
OB~
OA' =
OA'
k'c
OB'
which proves that AB is parallel to A'B'.
The zero contour line of z = ax + by + c is given by the equa
tion
ax + by + c = (5).
This straight line divides the plane XOY into two regions, such
that the values of x and y corresponding to any point in one of
them render ax + by + c positive, and the values of x and y cor
responding to any point in the other render ax + by + c negative.
§ 24.] Let us consider the zero contour lines, L and I/, of
two linear functions, z = ax + by + c and z' = a'x + b'y + c'. Since
the coordinates of every point on L satisfy the equation
ax + by + c = (1),
XVI GRAPHIC ILLUSTRATION OF INFINITE SOLUTION 391
and the coordinates of every point on L' satisfy the equation
a'x + Vy + ti = (2),
it follows that the coordinates of the point of intersection of L
and L' will satisfy both (1) and (2); in other words, the co
ordinates of the intersection will be a solution of the system
(1), (2).
Now, any two straight lines L and L' in the same plane have
one and only one finite point of intersection, provided L and L'
be neither parallel nor coincident. Hence we infer that the
linear system (1), (2) has in general one and only one solution.
It remains to examine the two exceptional cases.
I. Let L and L' (Fig. 1 1 ) be parallel, and let them meet the
axes of X and Y in A, B and in A', B' respectively. In this case
the point of intersection passes to an infinite distance, and both
its coordinates become infinite.
The necessary and sufficient condition that L and L' be
parallel is OA/OB = OA'/OB'. Now, OA =  c/a, OB =  c/b ;
and OA' =  c'fa', OB' =  c'/b'. Hence the necessary and suffi
cient condition for parallelism is bja  b'/a', that is, ab'  a'b  0.
We have thus fallen upon the excepted case of §§ 4 and 5.
If we assume that the results of the general formulae obtained
for the case ab'  a'b # 0, namely,
be'  b'c ca'  c'a
x =zr> — I7i> y
n i
ab'  a'b' J ab'  a'b
hold also when ab'  a'b = 0, we see that in the present case
neither of the numerators be'  b'c, ca'  c'a, can vanish. For if,
say, be'  b'c = 0, then  c/b =  c'/b', that is, OB = OB' ; and the
two lines AB, A'B', already parallel, would coincide, which is not
supposed.
It follows, then, that
be'  b'c ca'  c'a
Z = — ^ = 00, y = _ — =co:
and the analytical result agrees with the graphical.
II. Let L and L' be coincident, then the intersection becomes
indeterminate. The conditions for coincidence are
392 INDETERMINATE SOLUTION CHAP.
OA = OA', OB = OB',
whence
c/a = c'/a', c/b= c'/b'.
These give
a b c
a' = b' = c"
which again give
be'  b'c = 0, ca'  c'a = 0, ab'  a'b = 0.
We thus have once more the excepted case of §§ 4 and 5, but
this time with the additional peculiarit\ r that be'  b'c = and
ca'  c'a ~ 0.
If we assert the truth of the general analytical solution in
this case also, we have
* = o' ?/ = 0'
that is, the values of x and y are indeterminate, as they ought to
be, in accordance with the graphical result.
§ 25.] Since three straight lines taken at random in a plane
have not in general a common point of intersection, it follows
that the three equations,
ax + by + c = 0, a'x + b'y + c' = 0, a"x + b"y + c" = (1),
have not in general a common solution. When these have a
common solution their three graphic lines, L, L', L", will have a
common intersection. We found the analytical condition for
this to be
ab'c"  ab"c + be a"  be" a + ca'b"  ca'b' = (2).
In our investigation of this condition we left out of account the
cases where any one of the three functions, ab' — a'b, a"b  ab",
a'b"  a'b', vanishes.
We propose now to examine graphically the excepted cases.
First, we remark that if two of the functions vanish, the
third will also vanish ; so that we need only consider (I.) the
case where two vanish, (II.) the case where only one vanishes.
I. ab'  a'b = 0, a'b  ab" = 0.
This involves that L and L' are parallel, and that L and L" are
parallel ; so that all three, L, L', L", are parallel ; and we have,
in addition to the two given conditions, also a'b"  a'b' = 0.
xvi EXCEPTIONAL SYSTEMS OF THREE EQUATIONS 393
Hence, since the condition (2) may be written
c(a'b"  a"b') + c'(a"b  ab") + c"(ab'  a'b) = 0,
it appears that the general analytical condition for a common
solution is satisfied.
This agrees with the graphical result, for three parallel
straight lines may be regarded as having a common intersection
at infinity.
In the present case is of course included the two cases where
two of the lines coincide, or all three coincide. The corre
sponding analytical peculiarities in the equations will be obvious
to the reader.
II. ab'  a'b = 0.
Here two of the graphic lines, L and L', are parallel, and the
third, L", is supposed to be neither coincident with nor parallel
to either.
Looking at the matter graphically, we see that in this case
the three lines cannot have a common intersection unless L and
L' coincide, that is, unless
a' = ka, b' = kb, c' = kc,
where k is some constant.
Let us see whether the condition (2) also brings out this
result, as it ought to do.
Since ab'  a'b = 0,
we have  = _ = /• say.
a b J
Hence a' = ka, b' = kb.
Now, by virtue of these results, (2) reduces to
a"(be'  b'c) + b"(ca  c'a) = 0,
that is, to
a"(bc'  kbc) + b"(cka  c'a) = 0,
that is, to
(a"b  ab") (c'  kc) = 0,
which gives, since a"b  ab" 4= 0,
c  kc = 0,
that is, c' = kc.
394 EXERCISES XXVII chap.
Hence the agreement between the analysis and the geometry is
complete.*
§ 26.] It would lead us too far if we were to attempt here
to take up the graphical discussion of linear functions of three
variables. We should have, in fact, to go into a discussion of
the disposition of planes and lines in space of three dimensions.
We consider the subject, so far as we have pursued it, an
essential part of the algebraic training of the student. It will
help to give him clear ideas regarding the generality and
coherency of analytical expression, and will enable him at the
same time to grasp the fundamental principles of the application
of algebra to geometry. The two sciences mutually illuminate
each other, just as two men each with a lantern have more light
when they walk together than when each goes a separate way.
Exercises XXVII.
Draw to scale the graphs of the following linear functions of x : —
(1.) y=x + l. (4.) y = 2x + 3.
(2.) y=x+\. (5.) y=\x\.
(3.) y=2El. (6.) y=Z[x\).
(7.) Draw the graphs of the two functions, 3a; 5 and 5a; +7 ; and by
n>eans of them solve the equation Sx 5 = 5»+7.
(8.) Draw to scale the contour lines of z = 2x3y + l, corresponding to
g=2, z=l, z = 0, z=+l, z=+2.
(9. ) Draw the zero contour lines of z = 5x + 6y  3 and z' = 8x  9y + 1 ; and
by means of them solve the system
5x + 6y3 = 0, 8a:9y+l=0.
* It may be well to warn the reader explicitly that he must be careful to
use the limiting cases which we have now introduced into the theory of
equations with a proper regard to accompanying circumstances. Take, for
instance, the case of the paradoxical equation b = 0, out of which we manu
factured a linear equation by writing it in the form 0x + & = 0; and to which,
accordingly, we assigned one infinite root. Nothing in the equation itself
prevents us from converting it in the same way into a quadratic equation,
for we might write it Ox 2 + Ox + b = 0, and say (see chap, xviii., § 5) that it
has two infinite roots. Before we make any such assertion we must be sure
beforehand whether a linear, or a quadratic or other equation was, generally
speaking, to be expected. This must, of course, be decided by the circum
stances of each particular case.
XVI
EXERCISES XXVII 395
Also show that the two contour lines divide the plane into four regions,
such that in two of them (5x + 6yZ) (8x9y + l) is always positive, and in
the other two the same function is always negative.
(10.) Is the system
3x4y + 2 = 0, 6x8y + 3 = 0, x$y+l=0
consistent or inconsistent ?
(11.) Determine the value of c in order that the system
2se+yl=0, 4.r + 2y + 3 = 0, {c + 1)x + (c + 2)y + 5 = Q
may be consistent.
(12.) Prove graphically that, if ab' a'b — 0, then the infinite values of x
and y, which constitute the solution of
ax + by + c0, a'x + b'y + c' = 0,
have a finite ratio, namely,
x/y=(bc'  b'c)j(ca'  c'a).
(13. ) If (ax + by + c)/{a'x + b'y + c') be independent of x and y, show that
ab'a'bO, ca'c'a = 0, bc'b'c = 0;
and that two of these conditions are sufficient.
(14.) Illustrate graphically the reasoning in the latter part of § 5 of the
preceding chapter.
(15.) Explain graphically the leading proposition in § 6.
CHAPTER XVII.
Equations of the Second Degree.
EQUATIONS OF THE SECOND DEGREE IN ONE VARIABLE.
§ 1.] Every equation of the 2nd degree (Quadratic Equation)
in one variable, can be reduced to an equivalent equation of
the form
ax 2 + bx + c = (1).
Either or both of the coefficients b and c may vanish ; but
we cannot (except as a limiting case, which we shall consider
presently) suppose a = without reducing the degree of the
equation.
By the general proposition of chap, xii., § 23, when a, b, c
are given, two values of x and no more can be found which
shall make the function ax 2 + bx + c vanish ; that is, the equation
(1) has always two roots and no more. The roots may be equal or
unequal, real or imaginary, according to circumstances.
The general theory of the solution of quadratic equations is
thus to a large extent already in our hands. It happens,
however, that the formal solution of a quadratic equation is
always obtainable ; so that we can verify the general proposition
by actually finding the roots as closed functions of the coefficients
a, b, c.
§ 2.] We consider first the following particular cases : —
I. c = 0.
The equation (1) reduces to
ax 2 + bx = 0,
chap, xvn BOOTS EQUAL AND OF OPPOSITE SIGN 397
that is, since a 4= 0,
az(z + ]) = 0,
which is equivalent to
a: =
x +  b =
a
Hence the roots are x = 0, x =  bfa.
II. & = 0, e = 0.
The equation (1) now reduces to
ax x x = 0,
which, since a 4 0, is equivalent to
{:::}■
Hence the roots are x = 0, x = 0. This might also be deduced
from I.
Here the roots are equal. "We might of course say that there
is only one root, but it is more convenient, in order to maintain
the generality of the proposition regarding the number of the
roots of an integral equation of the nth degree in one variable,
to say that there are two equal roots.
III. b = 0.
The equation (1) reduces to
ax 2 + c = 0,
that is, since a =t= 0, to
WIX*v/;H
which is equivalent to
V a
i
 = o
a J
Hence the roots are x=  s/(c/a), x= + ^(cja); that is,
the roots are equal, but of opposite sign. If cja be negative,
398
GENERAL CASE
CHAP.
both roots will be real ; if c/a be positive, both roots will be
imaginary, and we may write them in the more appropriate form
x =  i \/(cJa), x = + i \/(cJa).
§ 3.] The general case, where all the three coefficients are
different from zero, may be treated in various ways ; but a little
examination will show the student that all the methods amount
to reducing the equation
ax 2 + bx + c =
(1)
to an equivalent form, a(x + A) 2 + jm = 0, which is treated like the
particular case III. of last paragraph.
1st Method. — The most direct method is to take advantage
of the identity of chap, vii., § 5. We have
f b + J(b 2 Aac)) ( b J(l/iac))
hence the equation (1) is equivalent to
b+ J( b 2 4:ac) ) (  b  y/(6 2  iac) }
6+ ^(b'Aac)
alx
that is, to
x
\
X
b
2a
sl(b 2  Aac)
2a
=
=
>■
The roots of (1) are therefore {b+ s /(b 2  iac)}/2a, and
{h J(b 2 iac)}/2a.
2nd Method. — We may also adopt the ordinary process of
" completing the square." We may write (1) in the equivalent form
x + ~x = —
a a
(2),
and render the lefthand side of (2) a complete square by adding
(b/2a) 2 to both sides. We thus deduce the equivalent equation
\ X + 2a) ~ia* a'
b 2  4«c
4a*
(3).
xvn VARIOUS METHODS OF SOLUTION 399
The equation (3) is obviously equivalent to
/ fb 2 Aac\
b
2a
i
b I (b 2  iac
x + — = 
y.
2a V \ 4 a*
from which we deduce
x = {  b + */(!?  iac) }/2a, x={b K /(b 2  iac) } /2a,
as before.
3rd Method. — By changing the variable, we can always make
(1) depend on an equation of the form az 2 + d = 0. Let us
assume that x = z + h, where h is entirely at our disposal, and z
is to be determined by means of the derived equation. Then,
by (1), we have
a(z + Kf + b(z + h) + c = (4).
It is obvious that this equation is equivalent to (1), provided x
be determined in terms of z by the equation x = z + h.
Now (4) may be written
az 2 + (2ah + b)z + (ah 2 + bh + c) = (5).
Since h is at our disposal, we may so determine it that 2ah
+ b0 ; that is, we may put h =  bj2a. The equation (5)
then becomes
'♦■(sMa)*
that is, az 2 — = (6).
From (6) we deduce z = + J(b 2  iac)/2a, z =  ^(b 2  4ac)/2a.
Hence, since x = z + h =  b/2a + z, we have
x = {b+ s/(b 2 iac)}j2a, x={b >J(b 2  iac)} /2a,
as before.
In solving any particular equation the student may either
quote the forms {  b ± *J(b 2  iac)} /2a, which give the roots in
all cases, and substitute the values which a, b, c happen to have
in the particular case, or he may work through the process of
az
400 DISCRIMINATION OF THE ROOTS chap.
the 2nd method in the particular case. The latter alternative
will often be found the more conducive to accuracy.
§ 4.] In distinguishing the various cases that may arise when
the coefficients a, b, c are real rational numbers, we have merely
to repeat the discussion of chap, vii., § 7, on the nature of the
factors of an integral quadratic function.
We thus see that the roots of
ax* + bx + c = 0,
(1) Will be real and unequal if b 2  lac be positive.
(2) Will be real and equal if b 2  lac = 0.
(3) Will be two conjugate complex numbers if b 2  lac
be negative. The appropriate expressions in this case are
{b + i s /(lac  b 2 )}/2a, {bi ^(lac  b 2 )}j2a.
(4) The roots will be rational if b 2  lac be positive and the
square of a rational number.
(5) The roots Avill be conjugate surds of the form A ± JB
in the case where b 1  lac is positive, but not the square of a
rational number.
(6) If the coefficients a, b, c be rational functions of any
given quantities p, q, r, s, . . . then the roots will or will not
be rational functions of p, q, r, s, . . . according as b 2  lac is or
is not the square of a rational function of p, q, r, .?,...
It should be noticed that the conditions given as characterising
the above cases are not only sufficient but also necessary.
The cases where a, b, c are either irrational real numbers, or
complex numbers of the general form a + a'i, are not of sufficient
importance to require discussion here.
Example 1.
2jt 2 3.7 = 0.
By inspection we see that the roots are x = 0, x=3j2.
Example 2.
This equation is equivalent to # 2 + 4 = 0, whose roots are x = 2i, x= 2i.
Example 3.
35x 2 2u;l = 0.
The equation is equivalent to
2 _ 1
X '35 X ~T$'
XVII
EXAMPLES, EXERCISES XXVIII
401
that is, to
Henca
Hence
(&
L\ 2 _1_ 1__36_
: 35 2 + 35~35 2 '
1 +1
* 35 35"
1±6
W
x=
The roots are, therefore, + 1/5 and  1/7.
Example 4.
a? ! 2a!2=0.
The roots are 1 + v'3 and 1  \/3.
Example 5.
3a; 2 + 2ix + 48 = 0.
The given equation is equivalent to
a? + 8^ + 16 = 0,
that is, to (a; + 4) 2 = 0.
Hence x=  4±0 ; that is to say, the two roots are each equal to  4,
Example 6.
a; 2 4a;+7 = 0.
This is equivalent to
a; 2 4a; + 4=3 )
that is, to (as2) a =3**.
Hence the roots are 2 + \jZi, 2  \JZi.
Example 7.
a; 2  2{p + qfx + 2p* + 1 Iff + 2q i = 0.
This equation is equivalent to
{x  (p + q)*\ 9 = (p +q)*  2p*  12pY  2q*,
= (pq)\
= {pq)H\
Hence the roots are (p + qf + {pqfi, (p + q^ipqfi
Exercises XXVIII.
(1.) a: 2 + a; = 0.
(2.)
(3.) (a:+l)(asl)+l=0.
(4.)
(5.) (a;l) 2 + (a2) 2 = 0.
(6.)
(7.) p(x + a) 2 q(x + p)* = 0.
(8.)
(9.) 2a, 2 + 3a:+5 = 3a; 2 + 4a: + l.
(10.)
(11.) 255a; 2  431a: + 182 = 0.
(12.)
(13.) x* 22a; +170 = 0.
(14.)
(15.) a? + 102a; + 2597 = 0.
(16.)
(17.) a? + 6\/7* + 55 = 0.
(18.)
(19. ) a; 2 + (23 + 12i)x + 97 + 137*' = 0.
(20.)
VOL. I
(2a:l)(3aj2) = 0.
(a:l) 2 + 3(a;l) = 0.
.3(xl) 2 2(a:2) 2 =0.
{px + q) 2 +{qx+p) 2 = 0.
ar> + 8a: 2 + 16a:l = (a: + 3) 3 .
4a, 2 40a; + 107 = 0.
a; 2 201a; + 200 = 0.
a; 2 4a; 2597 = 0.
a. 2 2(l + V2)a; + 2V2 = 0.
a. 2 (82i)a:=38i31.
2 D
402 EQUATIONS REDUCIBLE TO QUADRATICS chap.
(21.) («l)(aj2) + (ajl)(a!3) + (a!2)(aJ3)=0.
(22.) (xl) 3 + {xl) 2 (x2)2{x + l) 3 = 0.
(23.) (xi)(xi) + (xi)(xl) = 0.
(24.) {xa) 2 + (xb) 2 = a 2 + b 2 . (25.) a 2 + 4aa;=(&c) 2 + 4(&ca 2 ).
(26.) x 2 +(bc)x=a? + bc + ca + ab.
w^iK^ + y>
(28.) (a + b)(abx 2 2) = (a 2 + b 2 )x.
(29.) (ab)x 2 (a 2 + ab + b 2 )x + ab(2a+b) = 0.
(30.) (c + «2Z>);z 2 +(a + &2c),?+(& + c2«) = 0.
(31. ) {a 2  ax + c 2 ) (a 2 + ax + c 2 ) = a 4 + aV + c 4 .
(32. ) x  2(a 2 + b 2 + c 2 )x + a i + b i + c 4 + b 2 c 2 + <?a* + a 2 b 2 = 2abc(a + b + c).
(33.) (bc){xa) 3 +{ca)(xb) 3 + (ab){xc) 3 = 0.
(34.) Evaluate V(7 + V(7 + V(7 + V(7... ad oo... )))).
EQUATIONS "WHOSE SOLUTION CAN BE EFFECTED BY MEANS OF
QUADRATIC EQUATIONS.
§ 5.] Reduction by Factorisation. — If we know one root of an
integral equation
/(a) = (1),
say x = a, then, by the remainder theorem, we know that f(x) =
(x  a)cj>(x), where <j>(x) is lower in degree by one than f(x).
Hence (1) is equivalent to
(2).
tip)
The solution of (1) now depends on the solution of <f>(x) = 0. It
may happen that <f>(x) = is a quadratic equation, in which case
it may be solved as usual ; or, if not, Ave may be able to reduce
the equation <f>(x) — by guessing another root ; and so on.
Example 1.
To find the cube roots of  1.
Let x be any cube root of  1, then, by the definition of a cube root, we
must have x 3 —  1. We have therefore to solve the equation
3^ + 1 = 0.
We know one root of this equation, namely, x= 1 ; the equation, in fact, is
equivalent to
(x + l)(x 2 x+l) = 0,
that is, to { . X + } = °A.
{ x 2  x + 1 = J
The quadratic x 2 x + l = 0, solved as usual, gives a;=(l±t\/3)/2.
xvii INTEGKALISATION AND RATIONALISATION 403
Hence the three cube roots of  1 are  1, (1 + i\/S)/2, (1  i\jZ)j2, which
agrees with the result already obtained in chap. xii. by means of Demoivre's
Theorem.
Example 2.
This equation is obviously satisfied by x=l. Hence it is equivalent to
{7x 2 6x3)(xl) = 0.
The roots of the quadratic 7x 2  6x 3 = are (3±V30)/7. Hence the three
roots of the original cubic are 1, (3 + V30)/7, (3  V3lr)/7.
It may happen that we are able by some artifice to throw an
integral equation into the form
PQR . . . = 0,
where P, Q, R, . . . are all integral functions of x of the 2nd
degree. The roots of the equation in question are then found
by solving the quadratics
P = 0, Q=0, R = 0, ...
Example 3.
p(ax 2 + bx + r) 2  q(dx 2 + ex +/ ) 2 = 0.
This equation is obviously equivalent to
{^p(ax 2 + bx + c) + \Jq{dx 2 + ex+f)} {^p(ax*+bx + c)  *Jq(dx 2 + ex+f)} = 0.
Hence its roots are the four roots of the two quadratics
(a sjp + dsjq)x 2 + (b\Jp + c\Jq) x + {csjp +f\Jq) = 0,
(a \/p  d\/q)x 2 + {b\Jp  esjq) x + (c\/p f\/q) = 0,
which can be solved in the usual way.
§ 6.] Integralisation and Rationalisation. — We have seen in
chap. xiv. that every algebraical equation can be reduced to an
integral equation, which will be satisfied by all the finite roots of
the given equation, but some of whose roots may happen to be
extraneous to the given equation. The student should recur to
the principles of chap, xiv., and work out the full solutions of
as many of the exercises of that chapter as he can. In the exer
cises that follow in the present chapter particular attention
should be paid to the distinction between solutions which are
and solutions which are not extraneous to the given equation.
The following additional examples will serve to illustrate the
point just alluded to, and to exemplify some of the artifices that
are used in the reduction of equations having special peculiarities.
404 EXAMPLES chap.
Example 1.
1 =0.
a b
If we combine the first and last terms, and also the two middle terms, we
derive the equivalent equation
2x 2x _
>+ .» ,_ ra =0
x 2 [a + bf x'{abf
If we now multiply by {x~  (a + b) 2 } {x 2 (a b) 2 } we deduce the equation
2x{2x 2 2(a 2 + b 2 )}=0;
and it may be that we introduce extraneous solutions, since the multiplier
used is a function of x.
The equation last derived is equivalent to
\x 2 {a 2 + b) = 0j
Hence the roots of the last derived equation are 0, + \/{a 2 + b 2 ),  \/{a 2 + b 2 ).
Now, the roots, if any, introduced by the factor \x 2 (a + b) 2 ) {x 2 (a b) 2 }
must be ±(a + b) or ±(ab). Hence none of the three roots obtained from
the last derived equation are, in the present case, extraneous.
Example 2.
ax a + x . i \
\/a+\J{ax) \/a+^/{a + x)
If we rationalise the denominators on the left, we have
(gar){VaV(ag)} + { a + x){^a s/ja + x)} _ ,
x x ' ''
From (^), after multiplying both sides by x, and transposing all the terms
that are rational in x, we obtain
[a + as)t (««)?= 3a; \/a (7).
From (7), by squaring and transposing, we deduce
2a 3 3ax 2 = 2(a 2 x 2 )i (5).
From (5), by squaring and transposing, we have finally the integral equation
(4^3« 2 )a^=0 (e).
The roots of (e) are (repeated four times, but that does not concern us so far
as the original irrational equation * (a) is concerned) and ±a\/3/2.
It is at once obvious that x = is a root of (a).
If we observe that s/(l±s/Z/2) = (s/3±l)/2, we see that ±«V 3 / 2 are roots
of (a), provided
2:pV3 2±V3
xl + \/3 2±l + \/3 '
that is, provided
2^1 + V^ 2±l + \/3
2 V3 , 2 + V3 =
l + Vs^s + vs '
which is not true.
Hence the only root of (a) is x = 0.
* For we have established no theory regarding the number of the roots of
an irrational equation as such.
xvil EXAMPLES 405
Example 3.
V(ffl + a:) \/(ax)
V« + V( re + x ) V a _ V(« ~ a:)
By a process almost identical with that followed in last example, we deduce
from (a) the equation
4z*3ah?=0 (/3).
The roots of (/3) are 0, and ±a\/3/2 ; but it will be found that none of these
satisfy the original equation (a).
Example 4.
V(2a; 2  4aT+ 1)+ V(*' 2 5x+2)= V(2a?  2a; + 3) + V(« 2  3a: + 4) (a).
The given equation is equivalent to
V(2a?  4«+ 1)  V(>» 2 ~ 3a: + 4) = s/(2x 2  2x + 3)  V(a*  Bas + 2).
From this last, by squaring, we deduce
3a; 2  7x + 5  2V(2a: 2  4a; + 1) (a; 2  3a; + 4)
= 3a; 2  7x + 5  2V(2ar !  2a; + 3) (a 2  5a; + 2),
which is equivalent to
V(2a, 4  10a? + 21a?  19a; + 4) = yj(2x*  12a? + 17a?  19a + 6) (j8).
From (/3), by squaring and transposing and rejecting the factor 2, we deduce
a?+2a?l=Q (7).
One root of (7) is x —  1, and (7) is equivalent to
(«+l)(aj 2 +a!l)=0.
Hence the roots of (7) are  1 and (  1 ±\/5)/2.
Now x= 1 obviously satisfies (a). We can show that the other two
roots of (7) are extraneous to (a) ; for, if x have either of the values
(l±\/5)/2, then x* + xl = 0, therefore a; 2 = a; + l. Using this value of
a; 2 , we reduce (a) to \/{  6x + 3)=\J{  4a; + 5). This last equation involves
the truth of the equation  6a; + 3=  4ar+ 5, which is satisfied by x=  1, and
not by either of the values x— (  1 ± \J5)/2.
N.B. — An interesting point in this example is the way the terms of (a)
are disposed before we square for the first time.
Example 5.
1  \/(l  a; 2 ) __ 27 V(l+aQ + V(lg ) (a)<
1 + V(l  * 2 ) V(l +*)  V(i  x)
Multiply the numerator and denominator on the left by 1  \/(l a: 2 ), and the
numerator and denominator on the right by y/(l+x)  \/(l x), and we ob
tain the equivalent equation
(i_vr^ ) 2 _ x
a?  1V(I^ 2 )'
Multiply both sides of the last equation by a^l  Vl  x% and we deduce
{lsj(lx*)} 3 = 27x 3 08).
406
EXERCISES XXIX, XXX
CHAP.
If 1, w, u 2 (see chap, xii., § 20) be the three cube roots of 41, then (/3) is
equivalent to
1 \/(lx 2 ) = 3x
lV(lar»)=3«a! !
l^/(lx 2 ) = Z<J i x j
By rationalisation we deduce from (7) the three integral equations
10x 2 6.r = (f
{l + 9w 2 )a?6wx =  (5).
^l+QoO^ear.x^O,
The roots of these equations (5) are 0, 3/5 ; 0, 6w/(l +9w 2 ) ; 0, 6w 2 /(l +9w).
The student will have no difficulty in settling which of them satisfy the
original equation (a).
Exercises XXIX.
(1.) 1 X' 6 1 = (X 2 + ) 2 (X 2 1).
(2.) x 3 (a + b + c)x 2 {a~ + b + cbccaab)x + a 3 + P+c $ Babc=0.
(3.) x 4  40a; + 39 = 0.
(4.) x A + 2(a2)x s +(a2) n x 2 + 2a 2 (a2)x + a i =0.
(5.) 2a 3 a 2 2a; 8 = 0. (6.) ax 3 + x + a + l = 0.
(7.) ar*3a^ + 4a; 2 3a; + l = 0.
x 2 . x . 1 P p 2
p p 2 p 3 X 2 X
(9.) a; 4 6a, 3 + 10a^8x + 16 = 0. (10.) x i 6 = 5x(x 2 xl).
(11.) (x 2 + 6x + 9) (x 2 + 8x + 1 6) = (a; 2 + 4a; + 4) (x 2  12a; + 36).
(1.) ax = 2(l + l
Exercises XXX.
1 !\
x +  T .
a 0/
(2.)
b+x a+x
, n ,2x 2 xl 2a?3a;8 8a; 2 8
(3.) ST+
a;2
a;3
2a; 3
9a; + 5 4a;2 _ 12a; + 3 4a; + 3 11
( ' 12 7xl _ 16 7z + 9 48'
(5.) a;3 =
a? 27
a^ + 8 '
,. . ax + b ax + b 2ax + d b
(6 ° r + <^+b = ~2r + c
ax + b bx + a_(a + b)(x + 2)
cx + b cx + a
(8 .) 5±5 + "±»5±?
x  a xb xc
cx + a + b
n r
.,+
bx
lex
, n . xa xb b
9.) =— + = 
b a xa xb
a?a 2 ^x 2 b 2 x*c 2
a
XVII
EXERCISES XXX, XXXI 407
ac bc a + blc
^ l0 '' 2b + x + 2a + x a + b + :>:'
x + a xa _ x 2 + a? 3 2 « 2
(1L) x~^a + x + a~x li a + x 2 + a^
(x  a) (xb) _ (x c)(x d)
(12.)
xa—b x c—d
a + 2x
/ gt + ax + x
^ \a 2 ax + x'
a2x
( ' 2z 2 + 2a: + 3 x + 1
x 74
( 15 ') a 2 2a 15 cc 2 + 2x35 a; 2 + 10a; + 2l'
2x+3a 2xSa_ a + b ab
^ ' 2x3a + 2x+Sa~ab a + b'
K ''' (xb){xc) (xc){xa) {xa){xb)
(1.)
Exercises XXXI.
x+\/x_ x(xl)
x \/x 4
^ v^T2) = ^ +2) + 2 ^
(4.) (a 2 + bx)s/{a? + c) = (a 2 + be) V(a 2 + * ! ).
(5.) 6xtxl)2 s /{Z{x2)(x + l)2(x5)}=±{x+Z).
(6. ) Viz + Vz) + V(*  V*) = aV*/V(* + Va).
(7.) (l+x)V(la; 2 ) + (a :  1 ) = 
(8.) (aj8)M^6* + 8 6 ) = ( aj  4 )M^8» + 64).
(9. ) (2a  a)/ V(«* ~ «* + « 2 ) = ( 2a; " & V V(* 2 " hx + ^
wV(.4.Hy^)y(£iO}
(ii.) vV 2 + 6a + 1)  V(3' 2 + to + 4 ) + V(* 2 + 6a "  s)=o.
(12.) V(ff 2 + &x) + V(& 2 + a*) = 3(« + &).
(13.) V {a(te a 2 )/6} + V{ ft («*  b ')l a ) = a ~ h 
(14.) ^/(a + x) + s/{b + x) = 2s/(a + b + x).
Consider more especially the case where a=b.
(15.) \/(x + i)^(xi) = ^(xl).
(16.) 2x s /{x i + a' 2 ) + 2x'sJ{x i + V i ) = a?b.
(17. ) V(* 2 + 4x + 3)  V(* 2 + 3 * + 2) = 2(x+ 1).
Two solutions, x  1 and another.
(18. ) a 2 + a 2 + V(<' 4 + « 4 )  2 * V {■<■" + V(* 4 + « 4 ) } •
408 CHANGE OF VARIABLE chaP.
(19.) x=\J{ax + x 2 a\/(ax + x 2 )}.
(oqn 7 12 1 6
y ~ '' V(a:6) + 4 V(«6) + 9 V(a'6)4 + V( a;  6 )9 _
(21.) V(« 2 + a' 2 ) + V(2a») = V(« 2 + 3oaj) + VC* 2 + So*).
(22.j_._l 1
\/(a + x) \Ja \Ja + \/(a + x) \/(a + x)  \/{a  x)
(23. ) V« + V(« + «)  V(«  *) = ^/(« 2  * 2 )
(24. ) BiV(a+a!) + » V(« «)= V(™ 2 + « 2 ) #(«*  * 2 )
(25. ) Rationalise and solve 2,\J{x bc) = \Jx.
(26.) V{(^ 3 + a 2 )(^ + & 2 )}+x{V(a; 2 + a 2 )\/( a;2 + &2 )}= : ^ 2 + a;2 
(27.) a + (a; + J)V{^ 2 + « 2 )/(a; 2 + i 2 )}=6 + (a: + a)v'{(a; 2 + i 2 )/(a; 2 + a 2 )}.
§ 7.] Reduction of Equations by change of Variable. If we have
an equation which is reducible to the form
{_W +*{/(*)} +2«0 (a),
then, if we put £ =/(%), we have the quadratic equation
to determine £. Solving (/3), we obtain for £ the values
{ p ± J(p 2  4q)}/2. Hence (a) is equivalent to
<
pJ(p 2 4q)
m
1
(?)•
If the function f{x) be of the 1st or 2nd degree in x, the
equations (y) can be solved at once ; and all the roots obtained
will be roots of (a).
Even when the equations (y) are not, as they stand, linear
or quadratic equations, it may happen that they are reducible to
such, or that solutions can in some way be obtained, and thus
one or more solutions will be found for the original equation (a).
In practice it is unnecessary to actually introduce the
auxiliary variable £. We should simply speak of (a) as a
quadratic in f(x), and proceed to solve for f(x) accordingly.
Example 1.
<_*>/ + _»*' 12=0.
We may write this equation in the form
(a5* , .)_+4(_:_'/e)_12__0.
XVII
EXAMPLES 409
It may therefore be regarded as a quadratic equation in o'J'i. Solving, we find
xp ;, j=+2, xi>''J=6.
From the first of these we have
.rP = 2'l.
Hence, if 1, w, or, . . ., w^ 1 be the pth roots of +1, we find the following p
values for x : —
2t'i\ oj2">'p, w 2 2i'p, . . ., wP 1 2i'p.
In like manner, from xp : '< =  6, we obtain, if q be even, the p values
&P, w6«* w 2 6i'p, . . ., w* 1 6*'8;
and, if q be odd, the p values
w'6«*. w' 3 6** s w' 5 6^ ( . . ., uPPi&l*,
where &>', w' 3 , . . . , w' 2 ^ 1 are the ^?th roots of  1 .
Example 2.
x 2 + 3 = 2 V(^' 2 ~ 2a + 2) + 2z.
This equation may be written
a?2sB+22VO^2a!+2) + l=0 ;
that is,
{ VCe 2  2a + 2) } 2  2 { VC* 3  2x + 2)} + 1 =0,
which is a quadratic in \f(3? 2x + 2).
Solving this quadratic we have
x /(x 2 2x + 2) = l.
Whence ^2* + 2 = 1,
that is, (xl) 2 = 0.
The roots of this last equation are 1, 1, and x=l satisfies the original equation.
Example 3.
22* _ 32*42 + 32 = 0.
We may write this equation as follows,
(2*)2_ 12(2*) + 32 = 0;
that is, (2*4)(2*8)=0.
Hence the given equation is equivalent to
\2*=8J
The first of these has for one real solution x = 2; the second has the real
solution x=3.
Example 4.
(x + a) (x + a + b) (x + a + 2b) {x + a + 3b) = c 4 .
Associating the two extreme and the two intermediate factors on the left,
we may write this equation as follows,
{a; 2 + (2« + 3b)x + a(a + 36) } {sc* + {2a + %b)x + {a + b) (a + 26)} = c\
410 RECIPROCAL BIQUADRATIC chap.
If ^ = x i + (2a + 3b)x + {w i + dab), the last equation may be written
^+26 a )=c*;
that is, ? + 2b 2 Z + b i = b i + c i .
Hence £= &±s/(&+ct).
The original equation is therefore equivalent to the two quadratics
x + (2a + 3b)x + a 2 + 3ab + b i =±\J{b i + c i ).
§ 8.] Reciprocal Equations. — A very important class of equa
tions of the 4th degree (biquadratics) can be reduced to
quadratics by the method we are now illustrating.
Consider the equations
ax* + bx 3 + ex 2 + bx + a = (1),
ax 4 + bx 3 + ex 2  bx + a = (I.),
where the coefficients equidistant from the ends are either equal,
or, in the case of the second and fourth coefficients, equal or
numerically equal with opposite signs. Such equations are
called reciprocal*
If we divide by x 2 , we reduce (1) and (I.) to the forms
B (* + ?)
+ b(x + j + c =
(2),
a ( x2+ ?J
+ b(xJ +c =
(II.)
r alent to
a(x + l) + b{
x + J +c2a =
(3),
«(«!) + b[
x — ) + c+ 2a =
\ xJ
(III.)
3 and III. are quadratics in x + 1 jx and x  l/x respectively. If
their roots be a, /?, and y, 8 respectively, then (3) is equivalent to
* If in equation (1) we write l/£ for x, we get an equation which is equiva
lent to a? + b? + c? + bS + a = Q. Hence, if f be any root of (1), l/£ is also a
root. In other words, two of the four roots of (1) are the reciprocals of the
remaining two. In like manner it may be shown that two of the roots of (I.)
are the reciprocals of the remaining two with the sign changed.
XViI GENERALISED RECIPROCAL BIQUADRATIC 411
that is, to
j X  aX + 1 = (^ ,..
//fc+l=0j W "
Similarly, III. is equivalent to
The four roots of the two quadratics (4) or (IV.) are the roots
of the biquadratic (1) or (I.)
Generalisation of the Reciprocal Equation. — If we treat the
general biquadratic
ax* + bx 3 + ex 2 + dx + e =
in the same way as we treated equations (1) and (I.), we reduce
it to the form
a { x 2 + — 5 ) r b(x + j) + c = 0.
\ ax"/ \ ox/
Now, if e/a = cPjb 2 , this last equation may be written
( d\ 2 J d\ n ad n
a(x + E ) + b(x + ) + c2 T = 0,
which is a quadratic in x + djbx.
Cor. It should be noticed that the following reciprocal equations
of the 5th degree can be reduced to reciprocal biquadratics, and can
therefore be solved by means of quadratics, namely,
ax* + bx* + ex 3 ± ex 2 ± bx ± a = 0,
where, in the ambiguities, the upper signs go together and the
lower signs together.
For the above may be written
a(x> ± 1) + bx(x 3 ± 1) + cx\x ± 1) =0,
from which it appears that either x + 1 orz1 is a factor on
the lefthand side. After this factor is removed, the equation
becomes a reciprocal biquadratic, which may be solved in the
manner already explained. The roots of the quintic are either
+ 1 or  1, and the four roots of this biquadratic.
412 EXAMPLES CHaP.
Ill an appendix to this volume is given a discussion of the
general solution of the cubic and biquadratic, and of the cases
where they can be solved by means of quadratics.
Example 1.
To find the fifth roots of +1. Let x be any fifth root of + 1 ; then x 5 = l.
Hence we have to solve the equation
^1 = 0.
This is equivalent to
Xx^ + a^ + x^ + x + lOJ'
The latter equation is a reciprocal biquadratic, and may be written
H) 2+ H) 1=0 
After solving this equation for x + \jx, we, find
1 1 + V5 1 1  \/5
x 2 x 2
These give the two quadratics
a?+ 1 J^6 a . +1=0j ^ + 11^ + 1 = 0.
These again give the following four values for x : —
 (1 + V5)/4±»V(10  2V5)/4,  (1  V5)/4±*V(10 + 2V5)/4,
these, together with 1, are the five fifth roots of + 1. This will be found to
agree with the result obtained by using chap, xii., § 19.
Example 2.
(cc + «.) 4 + (a: + Z>) 4 =17(aZ>) 4 .
This equation may be written
(z + «) 4 + (a + &) 4 =17{(a; + a)(.r + &)} 4 ,
from which, by dividing by (x + b)\ we deduce
or £ 4 + l = 17(Sl) 4 ,
where £ = (» + a)/(x + b).
This equation in £ is reciprocal, and may be written thus —
Hence * + F = 2'
, 1 7 •
From this lust pair we deduce
{=2, or £ ; and £ = 
±iV(15)
8
xvii RATIONALISATION BY MEANS OF AUXILIARY VARIABLES 413
Hence we have the four equations
x + a_ x + a_, x + a_ 7±i\/(15)
x~+b~' J ' aT+l> _ *' x~+~b~ < 8 '
From these, four values of x can at once be deduced. The real values are
x = a2b and x = b 2a.
§ 9.] By introducing auxiliary variables, we can always make
any irrational equation in one variable depend on a system of
rational equations in one or more variables. For example, if we
have
s'(x + a)+ J(x + b)+ s/(x + c) = d,
and we put u = \/(x + a), v= sj(x + b), w= J(x + c), then we
deduce the rational system
u + v + w = d, n 2 = x + a, v 2 = x + b, w 2 = x + c.
Whether such a transformation will facilitate the solution de
pends on the special circumstances of any particular case. The
following is an example of the success of the artifice in question.
Example.
(a +aj)* + («»)=&.
We may write the given equation thus —
(re + «)* + (re  xy j {(re +z) + (a x)} *.
Hence we deduce
\axj /t) a \t la — x )
(2a)
Let now y={(a + x)l(ax)}* t
we then have y + 1 = Ay* + 1 )*.
(2a)*
From the last equation we deduce
2a(y+l) i =V(y*+l),
which is a reciprocal biquadratic, and can therefore be solved by means of
quadratics. Having thus determined y, we deduce the value of x by means
of the equation (re + x)/(a x) = y i .
Exercises XXXII.
(1. ) a; 2 " 1  x'"(b m + c m ) + b m c m = 0.
(2.) e?*' 2 + qe~ 3xl ' 2 =p ; show that the sum of the two real values of x is
° 7 2
(3.) 2xH^)^= C ^f~(x 1 'i> + x 1 ^).
4H EXERCISES XXXII chap.
(4.) (9*)*  2(3*)*3*+ 1 = B 2 *^.
(5.) **£l+*+ '£*=*•
(6.) (x + \) + l/(x + \) = f i.
(7.) (lx + x ! )l(l + a 2 x) = (l + a 2 + x)/(l+x + x i ).
(8.) 6x i + 5x 3 Z8x 2 + 5x + 6 = 0. (9.) 6a^31x 3 + 51a; 2 31x + 6 = 0.
(10.) 2x i 7x 3 + 7x 2 7x + 2 = Q. (11.) 8x 4  42ar* + 29a: 2 + 42a; + 8 = 0.
(12.) ax 3 + bx 2 + bx + a0. (13.) rta^ + i>x 2 6xa=0.
(14.) aa^ + 6ar + c = 0. (15.) ax* + bx 3  bx  a = 0.
(16.) a 2 x i + 2abx 3 + b 2 x 2 c 2 = 0. (17.) a: 5 + l = 0.
(18.) x 5 + 7x i + 9x 3 9x 2 7xl = 0.
(19.) 12a 5 + x 4 + 13a 3  13a; 2 x 12 = 0.
(20.) Show that the biquadratic ax i + bx 3 + cx 2 + dx + e = can be solved
by means of quadratics, provided b/2a = iad/(iac b 2 ).
(21.) a; 1 + 10ar 3 + 22a; 2 15a;+2 = 0.
(22. ) x i + 2(p q)x 3 + (p 2 + q 2 )x 2 + 2pq{p  q)x +pq(p 2 +pq + q 2 ) = 0.
(23.) 3/(x 2 7x + B)2/(x 2 + 7x + 2) = 5.
(24.) at(l+lX{to?+x)=7Q. (25.) VU o?) = \ + *J(l+x 2 )/x.
(26.) N /(a?+l) + 4=5/V(« 2 +l) (27) (x + 5) i + (x + 5)~ h = 2.
™ {fc:M^{(i*:)*'»W
(29.) 2ar + 2 v / (z 2 + 4a;5) = 4x 2 + 8a;5.
(30.) x 2 + 7x3 = s/(2x 2 +Ux + 2).
(31.) (,z7) i +(z + 9)* + 2(ar + 2:c63) i =702.r.
(32. ) ^(x 2 +px + a) + \J(x 2 +px + b) + s/(x 2 +px + c) = 0.
(33.) Show that the imaginary 7th roots of + 1 are the roots of
x 2  ax + 1 = 0, a; 2  fix + 1 = 0, x 2  yx + 1 = 0, where a, §, y are the roots
of the cubic x 3 + x 2  2x  1 = 0.
(34.) ^ + J = a;V2\/(^ 4 i) (35.) 5(l + ar>)/(l x 3 )= {(1 +x)/(l x)} 3 .
(36.) {ax) 5 + (xb) 5 = (ab) 5 . (37.) v'a; + v^a;  1 ) = \/(x + 1 ).
(38.) (as + 3)(»+8)(a! + 13)(aj + 18) = 51.
SYSTEMS WITH MORE THAN ONE VARIABLE WHICH CAN BE
SOLVED BY MEANS OF QUADRATICS.
§ 10.] According to the rule stated without proof in chap,
xiv., § 6, if we have a system of two equations of the Zth and
with degrees respectively in two variables, x and y, that system
has in general Im solutions. Hence, if we eliminate y and
deduce from the given system an equation in x alone, that equa
tion will in general be of the Imth degree, since there must in
XVII MOST GENERAL SYSTEM HAYING TWO SOLUTIONS 415
general be as many different values of x as there are solutions of
the original systems. We shall speak of this equation as the
Resultant Equation in x.
In like manner, if we have a system of three equations of
the /th, wth, and nth degrees respectively, in three variables
x, y, z, the sj r stem has in general Imn solutions ; and the re
sultant equation in x obtained by eliminating y and z will be of
the Imnth degree ; and so on.
From this it appears that the only perfectly general case in which
the solution of a system of equations will depend on a quadratic equation
is that in which all the equations hit one are of the 1st degree, and
that one is of the 2nd.
It is quite easy to obtain the solution in this case, and thus
verify in a particular instance the general rule from which we
have been arguing. All we have to do is to solve the n\
linear equations, and thereby determine n  1 of the variables as
linear functions of the nth. variable. On substituting these
values in the nth. equation, which we suppose of the 2nd
degree in all the n variables, it becomes an equation of the
2nd degree in the nth variable. We thus obtain two values
of the nth variable, and hence two corresponding values for each
of the other n  1 variables ; that is to say, we obtain two solu
tions of the system.
Example 1.
lx + my + n = (I),
ax 2 + 2hxy + by + 2gx + 2fy + c=0 (2).
(1) is equivalent to
Ix+n
y = m  (3);
and this value of y reduces (2) to
am 2 x*  2hmx{lx + n) + b(lx + rif + 2gmx  2fm(lx + n) + cm = 0,
that is,
{am 2  2hlm + bP)x 2 + 2{gm' 1  hmn + bnl flm)x + {bn~  2fmn + c??i 2 ) = (4).
The original system (1), (2) is therefore equivalent to (3), (4). Now (4) gives
two values for x, and for each of these (3) gives a corresponding value of y.
For example, the two equations
3.c + 2y + l=0, x 2 + 2xy + yx + y + Z = 0,
will be found to be equivalent to
y=?x\, ar8;c+ll0.
416 SYSTEM IN TWO VARIABLES chap.
Hence the two solutions of the system are
x= 4 + V5, 4 s/5 ;
2/=¥iV5, V+fs/6
Example 2.
3a;+2yi8=l > x + y3z = 2, x 2 + y 2 + z 2 =l.
The system is equivalent to
x=5z3, y = 8z + 5, 90z 2 + 1102 + 33 = 0.
The solutions are
the upper signs going together and the lower together.
§ 11.] For the sake of contrast with the case last considered,
and as an illustration of an important method in elimination, let
us consider the most general system of two equations of the
2nd degree in two variables, namely —
ax* + 2hxy + by 2 + 2gx + 2fy + c = (1),
ax 2 + 2h'xy + b'tf + 2g'x + 2fy + c' = (2).
We may write these equations in the forms —
bif + 2(hx +f)y + (ax 2 + 2gx + c) = 0,
b'y 2 + 2(h'x +f')y + (a'x 2 + 2g'x + c') = 0,
say by 2 +py + q = (1'),
Vtf+ft + jQ (2'),
Avhere p = 2(hx +/), q = ax 2 + 2gx + c, &c.
If we multiply (1') and (2') by b' and by b respectively, and
subtract, and also multiply them by q' and by q respectively, and
subtract, we deduce
(pb'p'b)y + (b'qbq') = (3),
(b'qbq')y 2 +(p'qpq')y = (4);
and provided bq'  b'q^ 0, (3) and (4) will be equivalent to (1')
and (2'). In general, the values of x which make bq'b'q=0
will not belong to the solutions of (3) and (4), nor will the value
y  belong to those solutions. Hence we may say that, in
general, the system
{j>h'p'b)y + (b'qbq) = (3'),
(b'qbq')y + (p'qpq') = (4'),
is equivalent to (1') and (2').
Again, if we multiply (3') and (4') by b'q  bq and by pb'  p'b
respectively, and subtract, we deduce
(b'qb q y( P b'p'b)(p'q~pq') = (5),
XVII
OF THE ORDER (2, 2) 417
and, provided bq  b'q 4= 0, (4') and (5) will be equivalent to (3')
and (4').
Hence we finally arrive at the conclusion that, in general, the
system
{b'(ax 2 + 2gx + c)  b(a'a? + 2g'x + c')} 2  4={b'(hx +/)
 b(h'x +/)}{{h'x +/') (atc a + 2gx + c)  (hx +/) (a'af + 2g'x + c')}
0 (6),
{b'(ax 2 + 2gx + c)  b(a'x 2 + 2g'x + c')}y
+ 2{{h'x +/') (ax 2 + 2gx + c)  (Iix +/) (ax 2 + 2g'x + c')}
= (7),
is equivalent to (1) and (2).
The first of these is a biquadratic giving four values for x,
and, since (7) is of the 1st degree in y, for each value of x we
obtain one and only one value of y. We have therefore four
solutions, as the general rule requires.
In general, the resultant biquadratic (6) will not be reducible
to quadratics. It may, however, happen to be so reducible in
particular cases. The following are a few of the more im
portant : —
I. If, for example, b'/b =/'//= c'/c, then (6) reduces to
x 2 [{b'(ax + 2g)  b(a'x + 2g')} 2  i(b'h  bh'){(h'x + f) (ax + 2g)
 (hx+f) (a'x + 2g') + (h'c  he')}] = 0,
two of whose roots are zero, the other two being determinable
by means of a quadratic equation.
II. Again, if a' /a = b'/b = h'/h, it will be found that the two
highest terms disappear from (6). Hence in this case two of its
roots become infinite (see chap, xviii., § 6), and the remaining
two can be found by means of a quadratic equation.
III. If/=0, g = 0,f = 0, g' = 0, it will be found that only
even powers of x occur in (6). The resultant then becomes a
quadratic in x 2 .
IV. The resultant biquadratic may come under the reciprocal
class discussed in § 8 above.
Most of these exceptional cases are of interest in the theory
of conies, because they relate to cases where the intersection of two
conies can be constructed by means of the ruler and compasses
alone. The general theory is given in the Appendix to this vol.
VOL. I 2 E
418 EXAMPLES chap.
Example 1.
The system 3x 2 + 2xy + y 2 =l7, x 2 2xy + 5y 2 = 5,
is equivalent to 12.vy + 14a; 2  80 = 0, 73a; 4 692a; 2 + 1600 = 0.
The solutions of which are
x=+2, 2, +20/V(73), 20/V(73);
y=+l, 1, +1/V(73), 1/V(73).
Example 2.
n{x 2 + 2y) = 1(1 + 2xy), n(ij 2 + 2x) = m(l + 2xy).
Here the elimination is easy, because the first equation is of the 1st degree
in y. We deduce from it
?ia; 2  1
y ~2{lxn)'
This reduces the second equation to
n(nx 2  l) 2 + 8nx(lxn) 2 = irn(lxn) 2 +imx(lxn) (nx 2  I),
which is equivalent to
(n 2  4.lm)x i + 4(2P + mn)x*  18nlx 2 + i(2n 2 + lm)x + (I 2  4»m) = 0.
If n = l, this biquadratic is reciprocal, and its solution depends upon
(Z4m)f 2 + 4(2Z + m)£+ (8m 200=0,
where £ = a; + l/a;.
In general, if we have an equation of the 1st degree in x
and y together with an equation of the nth. degree in x and y,
the resultant equation in x will be of the wth degree. In par
ticular cases, owing to the existence of zero or infinite roots,
or for other special reasons, this equation may be reducible to
quadratics.
Example.
x + y = 18, a; 3 + y 3 =4914,
is equivalent to
y=lSx, ar ! + (18a;) 3 = 4914.
The second of these two last equations reduces, as it happens, to
a; 2  18a; +17 = 0.
Hence the finite solutions of the given system are
35=17,1;
y=l, 17.
§ 12.] A very important class of equations are the socalled
Homogeneous Systems. The kind that most commonly occurs is
that in which each equation consists of a homogeneous function
of the variables equated to a constant. The artifice usually em
ployed for solving such equations is to introduce as auxiliary
variables the ratios of all but one of the variables to that one.
Thus, for example, if the variables were x and y, we should put
y = vx, and then treat v and x as the new variables.
xvii HOMOGENEOUS SYSTEMS EXAMPLES 419
Example 1.
x 2 + xy = 12, xy 2y 2 = l.
Put y = vx, and the two equations become
z 2 (l + fl) = 12, x 2 (v2v 2 ) = l.
From these two we derive
x 2 (l + v)\2x 2 {v2ir) = 0,
that is,
a?{24t?llt>+l}=0.
Since x = evidently affords no solution of the given system, we see that the
original system is equivalent to
a?(l+v) = 12, 24b 2 11v + 1 = 0.
Solving the quadratic for v, we find v—l/S or 1/8.
Corresponding to v=l/3, the first of the last pair of equations gives x 2 = 9,
that is, x= ±3.
Corresponding to v—1/8, we find in like manner x= ±4n/(2/3).
Hence, bearing in mind that y is derived from the corresponding value of
x by using the corresponding value of v in the equation y — vx, we have, for
the complete set of solutious,
*=+3, 3, +4V(2/3), 4V(2/3);
y=+l, 1, +1/V6, VV6.
Example 2.
x 2 + 2yz=l, y 2 + 2zx = m, z 2 + 2xy = n.
Let x=uz, y — vz, then the equations become
(u 2 + 2v)z 2 = l, {v 2 + 2u)z 2 = m, (l + 2uv)z 2 = n.
Eliminating z, we have, since 2 = forms in general no part of any solution,
n ( v? + 2v) = I (1 + 2uv), n {v 2 + 2u) = m (1 + 2wi>).
"We have already seen how to treat this pair of equations (see § 11,
Example 2). The system has in general four different solutions, which can
be obtained by solving a biquadratic equation (reducible to quadratics when
n = I).
If we take any one of these solutions, the equation (1 +2icv)z 2 = n gives
two values of z. The relations x = uz, y = vz, then give one value of a; and one
value of y corresponding to each of the two values of z.
We thus obtain all the eight solutions of the given system.
There is another class of equations in the solution of which
the artifice just exemplified is sometimes successful, namely,
that in which each equation consists of a homogeneous function
of the variables equated to another homogeneous function of the
variables of the same or of different degree.
Example 3.
The system
ox 2 + bxy + cy 2 = dx + ey, a'x 2 + b'xy + c'y 2 = d'x + e'y (1)
is equivalent to
(a + bv + cv 2 )x 2 ={d + ev)x, (a' + b'v + c'v 2 )x 2 = (d' + e'v)x (2)
where y = vx.
420 EXAMPLES chap.
From this last system we derive the system
x*{(a + bv + cv°) {d' + e'v) (a' + b'v + c'v>)(d + ev)} =0\ ,„.
£{(« + &u + cr 2 )a;(d + er)}=0/ (6 >>
which is equivalent (see chap, xiv., § 11) to (2), along with
(a + bv + cv 2 )a? = (4),
{d + ev)x = (5).
If we observe that x=0, y — is a solution of the system (1), and keep
account of it separately, and observe further that values of v which satisfy both
(1) and (5) do not in general exist, we see that the system (1) is equivalent to
(a + bv + cv 2 ) (d' + e'v)  {a' + b'v + c'v 2 ) (d + ev) = (6)
along with (a + bv + ci?)x (d + ev) = (7)
and x = 0, y = 0.
The solution of the given system now depends on the cubic (6). The
three roots of this cubic substituted iu (7) give us three values of x, and y = vx
gives three corresponding solutions of (1). Thus, counting x = 0, y = 0, we
have obtained all the four solutions of (1).
The cubic (6) will not be reducible to quadratics except in particular cases,
as, for example, when ad' a'd = Q or ce'  c'e — 0.
For example, the system
3a; 2  2xy + 3y 2 = x + 1 2y, 6x 2 + 2xy  2y 2 2x + 29y,
is equivalent to x=0, y = 0, together with
7>(llli> 2  86*;+ 8) = 0, (3  2v + 3v)x= 1 + 12t>.
The values of v are 2/3, 4/37, and 0. Hence the solutions of the system are
x=0, 3, 185/227, 1/3;
y=0, 2, 20/227, 0.
§ 13.] Symmetrical Systems. — A system of equations is said to
be symmetrical when the interchange of any pair of the variables
derives from the given system an identical system. For example,
x + y = a, x 2 + y 2 = b ; z 3 + y = a, y 3 + x a ;
x + y + z = a, x 2 + y 2 + z* — b, yz + zx + xy = c,
are all symmetrical systems.
There is a peculiarity in the solutions of such systems, which
can be foreseen from their nature. Let us suppose in the first place
that the system is such that it would in general have an even
number of solutions, four say. If Ave take half the solutions, say
X = a 1 , a 2 ,
V = Pn Am
then, since the equations are still satisfied when the values of x
and y are interchanged, the remaining half of the solutions are
x = (3 n ft,
y = a u a 2 .
xvn SYMMETRICAL SYSTEMS 421
If the whole number of solutions were odd, five say, then
four of the solutions would be arranged as above, and the fifth
(if finite, which in many cases it would not be) must be such
that the values of x and y are equal ; otherwise the interchanges
of the two would produce a sixth solution, which is inadmissible,
if the system have only five solutions.*
These considerations suggest two methods of solving such
equations.
1st Method. — Replace the variables by a new system of vari
ables, consisting of one, say x, of the former, and the ratios to it
of the others, u, v, . . . say. Eliminate x, v, . . . and obtain
an equation in u alone ; then this equation will be a reciprocal
equation ; for the values of u are
a l Pi a 2 P2 o / 1 • 1 i \
u = — , — , r, — , &c. (and, it may be, u = 1 ),
Pi «i Pa a 2
that is to say, along with each root there is another, which is its
reciprocal. The degree of this resultant equation can therefore
in all cases be reduced by adjoining a certain quadratic, just as
in the case of a reciprocal biquadratic.
2nd Method. — Replace the variables x, y, z, . . . by an equal
number of symmetric functions of x, y, z, . . ., say by Sic, i>//,
~2xyz, . . ., ifec, and solve for these.
The nature of the method, its details, and the reason of its
success, will be best understood by taking the case of two
variables, x and y.
Let us put u = x + y, v = xy. After separating the solutions,
if any, for which x = y, Ave may replace the given system by a
system each equation of which is symmetrical. We know, by
the general theory of symmetric functions (see chap, xviii., § 4),
that every integral symmetric function can be expressed as an
* We have supposed that for all the solutions (except one in the case of
an odd system) x + y. It may, however, happen that x = y for one or more
solutions. Such solutions cannot be paired with others, since an interchange
of values does not produce a new solution. This peculiarity must always
arise in systems which are symmetrical as a whole, but not symmetrical in the
individual equations. As an example, we may take the symmetrical system
x 3 + y=a, y 3 + xa, three of whose solutions are such that x = y.
422 SYMMETRICAL SYSTEMS chap.
integral function of u and v. Hence it will always be possible to
transform the given system into an equivalent system in u and v.
We observe further that, in general, u and v will each have
as many values as there are solutions of the given system, and
no more ; but that the values of u and v corresponding to two
solutions, such as x = a u y = ($„ and x = /3 l} y = a l} are equal.
Hence in the case of symmetrical equations the number of solu
tions of the system in u and v must in general be less than usual.
Corresponding to any particular values of u and v, say u = a,
v = (3, we have the quadratic system x + y = a, xy = fi, which gives
the two solutions
X = {a ± v/(a 2  4j8)}/2, y={a T */(a S  4/?)}/2.
If we had a system in three variables, x, y, z, then we should
assume u = x + y + z, v = yz + zx + xy, w = xyz, and attempt to solve
the system in u, v, w. Let u = a, v  j3, iv  y, be any solution of
this system ; then, since
(f*)tfy)tf*)^«f + «f«i
we see that the three roots of
eaf' + /3$y =
constitute a solution of the original system, and, since the
equations are symmetrical, any one of the six permutations of
these roots is also a solution. In this case, therefore, the number
of solutions of the system in u, v, to would, in general, be less
than the corresponding number for the system in x, y, z.
The student should study the following examples in the light
of these general remarks : —
Example 1. A (x + y 2 ) + Bxy +C (x f y) + T> =0 \
A'(x o  + y") + B'xy + C'{z + y) + D' = 0) ( )
If we put y = vx, and then eliminate x by the method employed in §11,
the resultant equation in v is
{ (D'A) + (D'B)» + (D'AH 2 } 2 = (D'C) (1 + vf { (C'A) + (C'B)« + (C'A)« 2 } (2),
where (D'A) stands for D'A  DA', (D'B) for D'B  DB', and so on.
The biquadratic (2) is obviously reciprocal, and can therefore be solved by
means of quadratics.
The solution can then be completed by means of the equation
{(D'A) + (D'B)v + (D'A)^}x + (D'C)(l + v) = (3).
i
XVII
EXAMPLES 423
As an instance of this method the student should work out in full the
solution of the system
2(x 2 + y)  3xy + 2(x + y)  39 = 0,
3(x 2 + y°)ixy + (x + i/)  50 = 0.
"We may treat the above example by the second method of the present
paragraph as follows. The system (1) may be written
A(x + y) 2 +{B 2A)xy + C(x + y) + T> = 0,
A'(x + y) 2 + (B'  2A!)xy + C'(x +y) + D'=0 ;
Am 2 +(B 2A)i + Cm+D=0\ ...
tnatis > AV+(B'2A> + C'«+D'=0j v h
Eliminating first u 2 and then v, we deduce the equivalent system
(A'B)i?+(A'C)M + (A'D)=0\
(A'B)u 2 + { (C'B)  2(C'A) J u + { (D'B)  2(D'A) } =0 J
where (A'B), &c, have the same meaning as above.
The system (5) has two solutions,
u = a, a',
say, corresponding to which we find for the original system
x
— I
;a±vV4/3)}/2, {a'±V("' 2 4/3')}/2,
in all four solutions.
This method should be tested on the numerical example given above.
Example 2. x i + y i = 82, x + y=4.
We have x i + y i =(x + ijf  4xy(x 2 + y 2 )  6x 2 y 2 ,
= (x + yf  ixy { (x + yf  2xy }  6x 2 if,
= u i 4u 2 v + 2v 2 .
Hence the given system is equivalent to
u 4 4u*v + 2v 2 = 82, u=i.
Using the value of u given by the second equation, we reduce the first to
f 2 32w + 87 = 0.
The roots of this quadratic are 3 and 29. Hence the solution of the u, v
system is u = 4, 4,
v=S, 29.
From x+y=i, xy = 29, we derive {'y)= 100, that is, xy±10i;
combining this with x + y=4, we have se=2±5i, y = 2:f 5i.
From x + y=i, xy = 2, we find x = S, y = l ; x=l, ?/ = 3.
All the four solutions have thus been found.
Example 3. x i =mx+ny, y 4 = nx + my (I}.
Let us put y = vx; then, removing the factor x in both equations, and
noting the corresponding solution, x = 0, y = 0, we have
x 3 = m + nv, v i a? = n + mv.
These are equivalent to
x 3 =m + nv, v*(m + nr) = mv + n (2).
424 EXAMPLES chap.
The second of these may be written
n(v°l) + mv{v?l) = (3),
and is therefore equivalent to
:" +i V + (5° i > + (5 +i > +i =°}'
The second of these is a reciprocal biquadratic. Hence all the five roots of (3)
can be found without solving any equation of higher degree than the 2nd.
To the root v—1 correspond the three solutions,
x = y = (m + m)"' 3 . u{ni + n) llS , to"(m + n) 1!3 ,
of the original system, where (ra + n) 1 ' 3 is the real value of the cube root, and
w, or are the imaginary cube roots of unity.
In like manner three solutions of (1) are obtained for each of the remain
ing four roots of (3). Hence, counting x = 0, y = 0, we obtain all the sixteen
solutions of (1).
The reader should work out the details of the numerical case
x i = 2x + By, y i 3x + 2y,
and calculate all the real roots, and all the coefficients in the complex roots,
to one or two places of decimals.
Example 4. yz + zx + xy = 26,
yz(y + z) + zx(z + x) + xy(x + y) = 1 62,
yz{if + z 2 ) + zx{z + u?) + xy{x + y 2 ) = 538.
If we put u = x + y + z, v = yz + zx + xy, w = xyz, the above system reduces to
i>=26, M03w=162, (u 2 2v)vuw = 538.
Hence 26t*3w=162, 26icmc = 1890.
Hence 26m 2 +81m 2835 = 0.
The roots of this quadratic are it = 9 and u= — 315/26.
We thus obtain for the values of u, v, w, 9, 26, 24, and  315/26, 26,
 159. Hence we have the two cubics
£ 3 9£ 2 + 26£24 = 0,
£ 3 + 3^£ 2 + 26f + 159 = 0.
Twelve of the roots of the original system consist of the six permutations
of the three roots of the first cubic, together with the six permutations of the
roots of the second cubic.
The first cubic evidently has the root £ = 2 ; and the other two are easily
found to be 3 and 4. Hence we have the following six solutions : —
x=2, 2, 3, 3, 4, 4;
y = 3, 4, 4, 2, 2, 3;
a = 4, 3, 2, 4, ?, 2.
Other six are to be found by solving the second cubic.
§ 14.] We conclude this chapter with a few miscellaneous
examples of artifices that are suggested merely by the peculi
xvii MISCELLANEOUS EXAMPLES 425
arities of the particular case. Some of them have a somewhat
more general character, as the student will find in working the
exercises in set xxxiv. A moderate amount of practice in solv
ing puzzles of this description is useful as a means of cultivating
manipulative skill ; hut he should beware of wasting his time
over what is after all merely a chapter of accidents.
Example 1.
ax by (a + b)c
a + x b + y a + b + c'
Let a + x = (a 4 b + c)£, b + y = {a + b + c)q ;
the system then reduces to
a 2 ^ + b"lr, = (a + b) 2 , £+ij=l.
This again is equivalent to
{(a + b)^a} 2 = 0, f+ij=l.
Hence we have the solution £ = a/(a + b), r) = b/(a + b) twice over.
The solutions of the original system are therefore x = ac/(a + b), y=bc/(a + b)
twice over.
Example 2.
ax 2 + bxy + cy 2 = bx 2 + cxy + ay 2 =d (1).
This system is equivalent to
(a  b)x 2 + (b  c)xy + (ca)y = (2),
ax 2 + bxy + cy 2 =d (3).
The equation (2) (see chap, xvi., § 9) is equivalent to
x 2 = (c<x + l) P , xy = (aa + l)p, y 2 =(ba + l) P (4),
where p and <r are undetermined.
Since x 2 y 2 = (xy) 2 , we must have
(C(r + l)(6(j + l) = (ffcr + l) 2 .
J) A n 0/t
Hence we deduce c = 0, a— — r 1 — r~ (5).
' a be
The first of these, taken in conjunction with (4), gives x = y ; and hence
4 / *~ .
x = y = ± A / i — >
J V a+b+c
that is to say, two solutions of (1). If we take the second value of a we find
, 2 _picaf p{ca){ai) „ p(«^) 2 (6)
x ~ a 2 bc' J ~ a? be ' T ~ a? be K >'
where it remains to determine p/(a 2  be). This can be done by substituting
in (3). We thus find
p/(a 2  be) = d/(a 3 + ac 2  ca 2 + ab 2  a 2 b  abc).
We now deduce from (6)
±(ca)<P
x = :
y=±, &c.
(a 3 + ac 2  ca + ab 2  a 2 b  abc) 1 ■'
two more solutions of the original system.
The system (2), (3) could also be solved very simply by putting y = vx,
as in § 12.
426 MISCELLANEOUS EXAMPLES chap.
Exam
These
pie 3.
equations
give
yz =
a 2 , zx =
zxxxy
■b 2 , xy
b 2 xc 2
~ a 2 '
b 2 c 2
~ a 2 '
c 2 .
t is,
yz
Hence x=±bc/a ; the two last equations of the original system then give
y=±ca/b, z=±ab/c. The upper signs go together and the lower together ;
so that we have only obtained two out of the possible eight solutions.
Example 4.
x{y + z) = a 2 , y(z + x) = b 2 , z(x + y)=c 2 .
This can be reduced to last by solving for yz, zx, xy.
Example 5.
x{x + y + z) = a 2 , y(x + y + z) = b 2 , z(x + y + z) = c 2 .
Let x + y + z — p. Then, if we add the three equations, we have
p 2 = a 2 + b 2 + c 2 .
Hence p = ± \/(« 2 + b 2 + c 2 ) ; and we have
±a 2 ±b 2 ±c 2
X — ///v  2 i 12 , »2\l V~
«J(a 2 + b 2 + c 2 )' J ^{a 2 + b 2 + c 2 Y V^' + ^ + c 2 )'
Example 6.
To find the real solutions of
a* + tf+p =t p (1), ${y+z)+tf=bc (4),
tf' + ? + e = b 2 (2), v{z + x) + t t =ca (5))
z 2 + e + T = c 2 (3), fte+y)+ij=a5 (6).
From (2), (3), and (4) we deduce
U(y+s)+i?f} 2  w+f+ewz'+e+v^^o ;
that is, (l 2 S») 8 +(^f2) a +(i?w) 3 =0 (7).
Every solution of the given system must satisfy (7). Now, since (£ 2 yz) 2 ,
(£77 zf) 2 , (ftyy) 2 are all positive, provided x, y, z, £, 77, f be all real, it
follows that for all real solutions we must have ^ 2 — yz, §J7=fo ff = 777/.
Hence, from the symmetry of the system, we must have
£ 2 = yz, ■ n 2 = zx, £ 2 =xy, (8),
— £ '?• 8 =T (9) 
By means of (8) we reduce (1), (2), (3) to
x(x + y + z) = a 2 , y(x + y + z) = b 2 , z(x + y + z) = c 2 .
Hence, by Example 5, we have
±a 2 _ ±b 2 _ _±f
X ~^(a 2 + b 2 + c 2 y y ~^(a 2 + b 2 + c 2 )' Z ~^/(a 2 + b 2 + c 2 Y
From (8) we now derive
zkbc zkca ,__ ±«£>
* "~ . //_2 1 M 1 2\' V ~~ . J7Z.2 , la . .2\' i —
V(a 2 + * 2 + c 2 )' ''^/(a? +&+<?)' s V(a 2 + ^ + 0'
xvil MISCELLANEOUS EXAMPLES 427
If we take account of (4), (5), (6) we see that the upper signs must go to
gether throughout, and the lower together throughout ; so that we find only
two real solutions.
Example 7.
x(xa)=yz, y(yb) = zx, z(zc) = xy (1).
From the first two equations we derive (x y) (x + y + z) = ax  by, which,
if we put p = x + y + z, may he written (pa)x=(pb)y. Hence, bearing in
mind the symmetry of the system, we have
*= — i y= — i> *= — ( 2 )>
pa " pb pc
where p and a have to he determined.
From the first equation of (1) we have
<x f cr
pa\pa ) {p~b)(pc)
Removing the factor <r, to which will correspond the solution x — y = z=0,
we find
o { (2a  b  c)p + (be  a") }=a(p~ a)(p b)(p c) (3).
Similarly we find
<r{(2bca)p+(cab")}=b(pa)(pb)(pc) (4).
From (3) and (4) we now eliminate <x, observing that in the process we
reject the factors <r, p  a, pb, pc, which correspond to three solutions,
namely,
x = a, 0, ;
y=0, b, 0;
2 = 0, 0, c.
We thus deduce p = . ,
r a+b+c
which gives one more solution. We have in fact pa = (bca")jZa,
pb = {ca b*)/2a, pc = (ab c 2 )IZa.
Hence (2) gives
<x _(cab)(abc)
pa Babe  2a 3 '
and, by symmetry, we have two corresponding values for y and z.
This example is worthy of notice on account of the symmetrical method
which is used for treating the given system of equations. The solution might
be obtained fully as readily by putting x = u~, y = vz, and proceeding as in
§ 13, Example 3.
Exercises XXXIII.
(1.) x + y = 30, xy = 2\Q. (2.) xy=3, x" + if = 65.
(3.) x+if = 5S, xy = 1\, (4.) x + y = 8, 3x"2xy + y' i =5i.
(5.) x + 2y = x\ 2x + y = y 2 .
(6.) x + y + 2(x + y) = U, Zxy = 2(x + y).
(7.) x~ + y~ — a~, x + y = b.
428
EXERCISES XXXIII
CHAP.
and
(8.
(9.
10.
11.
the
(12.
(14.
(15.
(16.
(17.
(18.
(19.
(20.
(21.
(22.
(23.
(24.
(25.
(26.
(27.
(28.
(29.
(30.
(32.
(33.
(35.
(36.
(37.
(38.
(39.
(40.
(41.
(42.
(43.
(44.
(45.
(46.
(47.
(48.
(49.
(50.
(51.
(52.
a(x + y 2 ) =px  qy, b(x 2 + y 2 ) = qx  py.
(x + y)/(l+xy)=a, {xy)/(l ~xy) = b.
ax + by = e, b/x + a/y = d.
It ax + by = 1, ex 2 + dy 2 = 1, have only one solution, then arjc + b 2 /d = 1,
solution in question is x = a/c, y—b/d.
2x 2 %xy = \, y 2 + 5xy = 34. (13.) x 2 +xy = 8i, xy + y 2 = 60.
x 3 +4xy + y 3 = 38, x + y = 2.
l/x 2 + l/xy = 1/a 2 , l/y 2 + l/xy = l/b 2 .
(px + qy) (x/p + y/q) = x 2 + y 2 +p 2 + q 2 , xjp + yjq = V5.
x 2 + a 2 = y 2 + b 2 = {x + yf + (a  bf.
(xyf = a 3 (x + y), {x + y) 2 = b 3 {xy).
(a 2  b 2 )/(x 2 + y 2 ) + (a 2 + b^/ix 2 y 2 ) = l,
3
x 2 jp 2 y 2 /q 2 = 0.
x+Z ,y3_
x3 + y+3~ '
xZ w 
; = 1.
2x+3 2y + 3
2(xy)+xy = Bxy(xy) = 7.
(x + y)/7 = 8/(x + y + l), xy=12.
x+l/y=10/x, y + l/x = l0x.
Z{x 2 + y 2 )  2xy=27, 4(x 2 + y 2 )  6xy = 16.
cc 3 ?/ = 208, xy=L
x 2 y + xy 2 = 162, a? + y 3 = 243.
x 2 y + xy 2 = 30, x i y 2 + x 2 y 4 =468.
x 3 + y 3 = (a + b)(xy), x 2 + xy + y 2 = ab.
x i + x 2 if + y*=741, x 2 xy + y 2 = 19
xy(x + ij) = 48, x 3 + y 3 = 72.
x i + y i = Q7, x + y=5.
(31.) x 4 + y 4 = a 4 , x + y = b.
x i + y i = {p 2 + 2)x 2 y 2 , x + y = a.
x 2 y 2 = 2xy + x + y, x 3 y 3 = 3xy(x + y).
(x + y) {x 3  y 3 ) = 819, (x  y) {x 3 + y 3 )= 399.
x 2 \y + y 2 \x = 2, x + y =5.
(34.) x i + y 5 = 33, x + y=Z.
x 2 y 2 (x i y i )=a'
x i x 2 + y i y 2 
xy(x i + y 4 ) (x 2  y 2 ) = a.
84, x 2 + x 2 y 2 + y 2 = 49.
y s /x=b 2 xy.
i/2
x 3 /y = a 2 xy,
x + y+ \J(xy) = 14, x 2 + y 2 + xy = 84.
V(l«A') + V(l«/2/)=\ / (l+«/^)> x + y = b.
x+y+ VO* 2  2/ 2 ) = a, 2ysJ{x 2  y 2 ) = V\
*J(x 2 + l2y)+sJ(y 2 + l2x) = 33, x + y = 23.
V(*/y) + V(y/*) = 5/2, V(*W + V(s/7*) = 9 V2/2.
\/{x + a) + V(2/  «) = W a > s/( x ~ a ) + \/(y a) = f \/a.
X s + y • =a~,
(x 2 + y 2 ) h + (2xyf = b.
a?+a?+y*+b*=>s l /2{z( L a+y)b{ay)},
x 2 a 2 y 2 + b 2 ='s/2{x(ay) + b{a + y)\.
(x 2 + a 2 ) {y 2 + b 2 ) = m(xy + ab) 2 , {x 2  a 2 ) (y 2  b 2 ) = mJifix  ay) 2 .
b b b
x= y + ^rr z , rr
a a a
y+ y+ y +
gmyn—^ymn
x z +y=y 4a , y
y = x +
x+ x+ x +
'=uy
*+: ' = :»•«
xvii EXERCISES XXXIV 429
Exercises XXXIV.
*(1.) 2e=0, Saaj=0, Zah?=ZIL{bc).
(2.) (ya){za) = bc, (zb)(xb) = ca, (xe){y c) = ab.
(3.) yz+2(y+z)=U, zx+2{z+x)=8, xy+2(x+y)=16.
(4.) ^ + ^+; = 4 > yz+xs+xy=tyxyz, 2zx + 3yz=2xy.
(5.) z(j/ + 2) = 24, y{z + x) 18, z(x + y) = 20.
, (6.) a<y+«)=y(«+a;)=a(a!+y)=l.
(7.) ( 2 + a:)(a; + 2/) = « 2 , (a; + y) {y + z) = b 2 , (y + z)(z + x) = c 2 .
(8.) ax + yz=ay + zx=az + xy=p 2 .
(9.) a; 2 + 2?/2 = 128, y 2 + 2zx=15B, z 2 + 2xy=128.
(10.) a 2 (j/ + 2) 2 = a 2 a; 2 + l, 6 2 (2 + cc) 2 = &V + l, c 2 (x + y) 2 = c 2 z 2 + 1 .
(11.) a(i/ + za;) = (a! + i/ + z) 2 2&2/, b(z + xy) = (x + y + zf2cz,
c(x + yz) = (x + y + z) 2  2ax.
(12. ) ZOc 2  v/2)  2(a; 2  yz) = a 2 , ^(x 2  yz)  2{y 2 zx) = b 2 ,
2(x 2 yz)2(z 2 xy) = c 2 .
(13.) , Zbcx=0, 2aijz = Q, 2a; 2 =1.
(14.) a(x + yz) = b(y + zx) = c(z + xy), x 2 + y 2 + z 2 + 2xyz = l.
(15.) x(a + y + z)=y(a + z + x) = z(a + x + y) = 3a(x + y + z).
(16.) a; 2 + ?/ 2 + 2 2 =a 2 + 2a;(?/42)a; 2 , and the two equations derived from
this one by interchanging { J .
(17.) aa?=+ t by 2 = ~~~, cz 2 = + .
v ' y z' " z x x y
(18.) y*z*+z*x*+xhj*=i9, x 2 + if + z 2 = li, x{y + z) = 9.
(19. ) (yz  **)/a?x=(zx  y 2 Wy= {xy  z 2 )/Sz = l/xyz.
(20.) « 2 .c 2 (2/ + 2) 2 =(a 2 + a; 2 )2/ 2 2 2 , and the two derived therefrom by inter
changing^}.
(21.) 2x 3 = a{2x2x) = b(2x2y) = c(2x2z).
(22.) (xl)(y+z5)=77, (y2)(z+x 4)=.72, (z3){x+y3)=65.
(23.) u{yx)/(zu)=a } z(yx)/{zu) = b, y(uz)j{x y) = c,
x(uz)/(xy) = d.
(24. ) If ar 3 + ^ + z 3 + faeyz = a, 3(y 2 z + z 2 x + x 2 y) = b, 3(yz 2 + zx 2 + xy 2 ) = c,
show that
x + y + z = (a + b + c) , x + wy + uPz = (a + wb + arc) ,
X + u 2 y + uz = (a + u 2 b + wc) ,
where w 2 + w + l = 0. Find all the real solutions when a— 72, b = 75, c=69.
t C\£ \ 9 O O JO o o
(25.) xryz — a, y zx = b", z xy = c.
* In this set of exercises 2 and II refer to three letters only ; and 11(6 c)
stands for (bc)(ca)(ab), and not for (b  c) (c  a) (a  b) (c  b)(a c)(b~a),
as, strictly speaking, it ought to do.
430 EXERCISES XXXV CHAP, xvn
Exercises XXXV.
Eliminate *
(1.) x from the system
ax + b _ a'x + V _ a"x + b"
cx + d~c'x + d'~<7x + d"'
(2. ) x and y from
Ix my a + b al bm _ab x 2 y 2 _
a b ~ a  b x y a+b a 2 b 2
(3.) x and y from
x 2 + xy = a 2 , y 2 + xy = b 2 , x 2 + y 2 — c 2 .
(4.) x, y, z from
x(y + z)=a 2 , y{z + x) = b 2 , z(x + y)=c 2 , xyz = d\
(5.) x, y, z from
x y z . a. . b, . c, .
x + y + z=0,  + +=0, (xp)=(yl)={*r).
(6.) x, y, z from
2Az 2 =0, 2A'x 2 = 0, 2aa:=0,
and show that the result is
21/ {b 2 (GA') + c 2 (AB')  « 2 (BC) } = 0,
where (CA')CA'  C'A, &c.
(7.) Show that the following system of equations in x, y, z are inconsistent
unless r 3  p 3 = Zrgp, and that they have an infinite number of solutions if this
condition be fulfilled.
Skc 3  Zxyz =p 3 , "Zyz = q 2 , 2z = r.
Eliminate
(8.) x and y from
{a  x) (a  y) =p, {bx){by) = q, (ax)(by)/{bx)(ay)=c.
(9.) x, y, z from
x + y z = a, x 2 + y 2 z 2 = b 2 , x 3 + y 3 z 3 — c 3 , xyz — a 3 .
(10.) x, y, z from
ax + yz = be, by + zx = ca, cz + xy — ab, xyz = abc.
(11.) x, y, z from
2a; 2 =p 2 , 2,3? = q 3 , 2x 4 = ? 4 , xyz = s i .
(12.) x, y, z from
(x + a)(y + b){z + c) = abc, (yc)(zb) = a 2 , (za){xc) — b 2 , {xb){ya) = c 2 .
(13.) The system
X1X2 + y 1 y 2 = h i , x& z + y 2 y 3 = kf, . . ., x„x 1 + y„y 1 = k, 2 ,
osi 2 + yi 2 = x 2 2 + y.2 2 = . . .=x n 2 + y n 2 = a 2 ,
either has no solution, or it has an infinite number of solutions.
* The eliminant is in all cases to be a rational integral equation. '
CHAPTEE XVIII.
General Theory of Integral Functions, more
particularly of Quadratic Functions.
RELATIONS BETWEEN THE COEFFICIENTS OF A FUNCTION AND ITS
ROOTS — SYMMETRICAL FUNCTIONS OF THE ROOTS.
§ 1.] By the remainder theorem (chap, v., § 15), it follows
that if a,, a 2 , . . ., a n be the n roots of the integral function
p X n +p l X n ~ 1 +p^ n ~ 2 + . . . +Pn\ X + Pn (1)>
that is to say, the n values of x for which its value becomes 0,
then we have the identity
PoX n + p t x n ~ l + p.p: n " L + . . . + p n
=jp (x  a x )(x  03) . . . (xa n ) (2).
Now we have (see chap, iv., § 10)
(X  a,)(x  og) . . . (X  a n ) = X n  P^" 1 + P 2 Z» 2 ... + ( l)»P n>
where P t , P 2 , . . ., P n denote the sums of the products of the n
quantities a,, a 2 , . . ., a n , taken 1, 2, . . ., n at a time re
spectively. Hence, if we divide both sides of (2) by p , we have
the identity
x n + Pl x nl + P* x n2 + , _ + Pn
Pa Pa Pa
=x n T l x n  1 + T^ n  2 . . . + (l)' l P n (3).
Since (3) is an identity, we must have
p l /p,= P„ Wl'o = P 2 , • • , Pn/Pa = ( ~ 1) B P« (4).
In particular, if p = 1, so that we have the function
x n +p l 37 l ~ 1 +p. 2 x n  2 ' + . . . + p,i (5),
then i» 1 =P lJ #, = P 8 , . • ., i?» = (l) n P« (6).
432 SYMMETRIC FUNCTIONS OF TWO VARIABLES chap.
Hence, if we consider the roots of the function
x n + p& n ~ l + p# n " 2 + . . .+p n _ lX + p n ,
or, ivhat comes to the same thing, the roots of the equation
x n +p 1 z n  1 +px n  2 + . . .+p n . 1 x+p n = 0,
then pi is the sum of the n roots; p a the sum of all the products of
the roots, taken two at a time; p 3 the sum of all the products, taken
three at a time, and so on.
Thus, if a and f3 be the roots of the quadratic function
ax 2 + bx + c, that is, the values of x which satisfy the quadratic
equation ax 2 + bx + c = 0, then
a + (3=  b/a, a/3 = cja (7).
Again, if a, /3, y be the roots of the cubic function ax 3 + bx 2
+ c x + d, then
a + (3 + y =  b/a, {3y + ya + af3 = c/a, a/3y=  d/a (8).
§ 2.] If s 1} s 2 , s 3 , . . ., s r stand f m the sums of the 1st, 2nd,
3rd, . . . , rth powers of the roots a and f3 of the quadratic equation
x 2 +p 1 x+p 2 = (1),
we can express s„ s,, . . ., s r as integral functions of p x and p 2 .
In the first place, we have, by § 1 (6),
s 1 = a + /3=  Pl (2).
Again
S 2 = a 2 + (3 2 = (a + (3) 2 2a/3,
=1K  2p 2 (3).
To find s 3 we may proceed as follows. Since a and (3 are
roots of (1), we have
a 2 +^ ia +^ = 0, {3 2 +p,{3+p 3 =0 (4).
Multiplying these equations by a and [3 respectively, and adding,
we obtain
s 3 +pA+p.A = (5).
Since s t and s 2 are integral functions of p x and p 2 , (5) determines
5 3 as an integral function of p l and p,. We have, in fact,
h = lh{p;2p 2 )+p 2 p„
= ~Px + 3p>P* (6).
XVI 1 1
SYMMETRIC FUNCTIONS OF. TWO VARIABLES 433
Similarly, multiplying the equations (4) by a 2 and (? respectively,
and adding, we deduce
S4+P1S3 +#&=() (7).
Hence s 4 may be expressed as an integral function of p x and p a ,
and so on.
We can now express any symmetric integral function whatever of
the roots of the quadratic (1) as an integral function of >, and p.,.
Since any symmetric integral function is a sum of sym
metrical integral homogeneous functions, it is sufficient to prove
this proposition for a homogeneous symmetric integral function
of the roots a and ft. Tbe most general such function of the
?  th degree may be written
A(a' + ft r ) + Ba/3(a' 2 + jS*" 2 ) + Ca 2 /3V~ 4 +^" 4 ) + . .
that is to say,
As r  Bp 2 s,._ 2 + Cp 2 2 s,._ 4 + . . . (8),
where A, B, C are coefficients independent of a and ft.
Hence the proposition follows at once, for we have already
shown that s r , s r .,, s r _ 4 , . . . can all be expressed as integral
functions of p y and p. z .
It is important to notice that, since a and ft may be any
two quantities whatsoever, the result just arrived at is really a
general proposition regarding any integral symmetric function of
two variables, namely, that any symmetric integral function of two
variables a, ft can be expressed as a rational integral function of the two
elementary symmetrical functions p x =  (a + ft) and p. z = aft.
There are two important remarks to be made regarding this
expression.
1st. If all the coefficients of the given integral symmetric function
be integers, then all the coefficients in the expression for it in terms of
l> i and p., will also be integers.
This is at once obvious if we remark that at ever)' step in
the successive calculation of s ly s 2 , s 3 , . . ., &c, we substitute
directly integral values previously obtained, so that the only
possibility of introducing fractions would be through the co
efficients A, B, C, ... in (8).
VOL. I 2 F
434 EXPRESSION IN TERMS OF p 1 AND p 2 chap.
2nd. Since all the equations above written become identities,
homogeneous throughout, when for p x and p g we substitute their
values  (a + /3) and a/3 respectively; and since ^, is of the 1st
and p 2 of the 2nd degree in a and /3, it follows that t?i ever^
term of any function of p x and p 2 which represents the value of
a homogeneous symmetric function, the sum of the suffixes * of the
p's must be equal to the degree of the symmetric function in a and ft.
Thus, for example, in the expression (6) for s 3 the sum of the
suffixes in the term p*, that is, piPiPi, is 3 ; and in the term
3p } p 2 also 3.
This last remark is important, because it enables us to write
down at once all the terms that can possibly occur in the ex
pression for any given homogeneous symmetric function. All
we have to do is to write down every product of p t and p s , or
of powers of these, in which the sum of the suffixes is equal to
the degree of the given function.
Example 1.
To calculate a 4 + /3 4 in terms of p\ and# 2 .
This is a homogeneous symmetric function of the 4th degree. Hence, by
the rule just stated, we must have
^ + 13* = Api* + Bj?^ + Cpr,
where A, B, C are coefficients to be determined.
In the first place, let /3 = 0, so that ]h=  a, #2=0. We must then have
the identity a 4 sAa'. Hence A = l.
We now have
a 4 + /3 4 s(a + /3) 4 + B(a + /3) 2 a/3 + Ca 2 /3 2 .
Observing that the term a :! /3 does not occur on the left, we see that B must
have the value  4.
Lastly, putting a= /3=1, so that;>i = 0, p 2 = 1, we see that C = 2.
Hence
a'+p=pSipip» + 2p.f.
The same result might also be obtained as follows. We have
Hence, using the values of s 2 and s 3 already calculated, we have
Si= ]h( pf + ZpiPt) pilh 2 ~ 2?a)i
=£i 4 4pi s .p»+2p8*.
Example 2.
Calculate a 6 + ^ + o s /3 s H a'' ( f : in terms ofjB] and jh.
* This is called the uxiylit of the symmetric function. See Salmon's
Higher Algebra, § 5tJ.
xvni NEWTON'S THEOREM 435
We have
a 5 + /3 5 + a 3 /3 2 + a 2 /3 3 = Kp x 5 + Efefa + Cpipj 1 .
Putting /3 = 0, we find A=  1 ; considering the term a 4 /i, we see that B = 5;
and, putting a = /3 = l, we find C=  6. Hence
a 3 + jS 5 + a*p 2 + a 2 ^ =  p? + bpjfa  Spiff.
Since any alternating integral function * of a, ft, say /(«, ft),
merely changes its sign when a and ft are interchanged, it follows
that we have /(a, ft) = /(/?, a). Hence, if we put ft = a, we
have/(a, a) = /(a, a); that is, 2/(a, a) = 0. Therefore /(a, a) = 0.
It follows from the remainder theorem that /(a, ft) is exactly
divisible by a  ft. Let the quotient be <j(a, ft). Then g(a, ft)
is a symmetric function of a, ft. For g(a, ft) =/(<*, /3)/(<*  /?)> an d
,/(/?, a) =/(/?, a)/(/3«)= f(a,ft)l(fta) = /(a,0)/(a  /3); that is,
;/((z, /?) = £(/?, a). Hence any alternating integral function of a and
/? can be expressed as the product of a  ft and some symmetric
function of a and ft. Hence any alternating function of a and ft can
be expressed without difficulty as the product of ± >J(p*  4j» s ), and
an integral function of p. and p. 2 .
Example 3.
To express a 5 /3 a/3 5 in terms of pi and ^2
We have
a 5 /3 a/3 5 =a/3(a 4  /3 4 ),
= (a/3)a/3(a + /3)(a 2 + /3 2 ),
= ± V( Pi 2  4ft) { Pi?' 2 ( Pi*  2i? 2 )} •
Every symmetric rational function of a and ft can be ex
pressed as the quotient of two integral symmetric functions of
a and ft, and can therefore be expressed as a rational function
of p x and p.,.
Example 4.
a :i + 2a 2 /3 + 2a/3 2 + ^ (a + /3) 3  a/3(a + /3)
a 2 /3 + a/3* a/3(a + /3)
7'i 3 + ;'i?'2
_ P?Vz
Pi
§ 3.] The general proposition established for symmetric
functions of two variables can be extended without difficulty to
symmetric functions of any number of variables.
* See p. 77, footnote.
436
newton's theorem
chap.
We shall first prove, in its most general form, Newton's
Theorem that the sums of the integral powers of the roots of any
integral equation,
x n + p^ x n  1 + p aX n  2 + _ _ _ + ^ _ q (\^
can be expressed as integral functions of p x , p s , . . ., p n , whose co
efficients are all integral numbers.
Let the n roots of (1) be a,, a 2 , . . ., a n , and let the equation
whose roots are the same as those of (1), with the exception of
a n be
y.n1
f'H ii?i« + iPi
.113
+
+ iPni =
(2);
also let the equation whose roots are the same as those of (1),
with the exception of a. 2 , be
X n ~ 1 + 2 p,X n ~ 2 + 2 p. 2 X n ' 3 + . . .+ s p n i = Q (3),
and so on.
Then
x n_1 + j^a' 12 + ,p 2 x n ~ 3 + . . . + x p n  x
= (x ?l +p 1 X n ~ 1 + p 2 X n ~ 2 + . . . + p n )K%  a,),
= X n  1 + (a l +p ] )x^ 2 + (a i 2 +p l a 1 +p.^X n  s . . .
+ (a 1 r +p 1 a 1 r  1 + . . . +p r )x n  r ~ 1 + . . .
by chap, v., § 13.
Hence, equating coefficients, we have
iP» = a i 2 +?i a i + P»>
rl
[ p r = a l 1 +p 1 a 1 ' " +
+ P.
iPni =< 1+ .2V*i
«2 +
+Pni
(3').
Similar values can be obtained for „j\, a p s , 2 p 3 , . . ., g J?ni
in terms of a, and p lt p„ . . ., p»j and so on.
Taking the (r l)th equation in the system (3'), and multi
plying by a,, we have
rl
Similarly
and so on.
i£Vi«i = <*i +j>W +
2 p r iOL 2 = a 2 r +p l a 2 r l +
+jpri a i
XVIII
NEWTON S THEOREM
43;
Adding the 11 equations thus obtained, we have
iPri*i+2Pri<h +  . . + „p r ia n = S r +2> l S r  l + . . . +iVl»j (4)
Now ,^ r _, is the sum of all the products r—\ at a time of the
n— 1 quantities  a,,  a 3 , . . .,  a n . Hence i/> r i a i is the sum,
with the negative sign, of all those products r at a time of the
n quantities  a 1?  a 2 , . . .,  a n which contain a,. Similarly
the next term contains all those products rat a time in which
a 2 occurs ; and so on. Hence on the left all the products r at
a time of the n quantities a,,  a 2 , . . .,a n occur, each as
often as there are letters in any such product, that is to say, r
times. Hence the equation (4) becomes
rp r = S r +jt> 1 S r L l + . . . +p r  i S n
Or Sr+iVri+ • • • + Pr~ i s i + r pr = 0.
This will hold for any value of r from 1 to n  1, both inclusive.
We have therefore the system
s i +p i =0
s 2 +p v s t + 2p 2 =
s 3 + p x s a +p a s l + 3p a =0 }■ (5).
Sni +PiS n 2 + ■ • • (n l)i>»_i =
Again, since a, is a root of (1), we have
a I B + ^,a, n  1 + . . . +p n = 0.
Similarly
a/ 1 + p t a 2
+
■ +Pn = 0;
and so on.
If we first add these n equations as they stand, then
multiply them by a,, a 2 , . . ., a a and add, then multiply them
by a, 2 , a/, . . ., a a 2 respectively, and add, and so on, we obtain
s n +PiS n i + • • • + np n =0
Sn+i+Pi*n +• ■ • +8 1 p n =0 I
Sn+a+PJn+i +■ ■  + S 2 p n = \
(6),
and so on.
J
The equations (5) and (G) constitute Newton's Formula for
438 INTEGRAL SYMMETRIC FUNCTION CHAP.
calculating s n s 2 , s 8 , . . ., &c, in terms ofpi,p a , . . ., p n . It is
obvious that s,, s,, s 3 , . . . are determined as rational integral
functions of p, , p 2 , . . . , p n , in which all the coefficients are
integral numbers.
A little consideration of the formula? will show that in the
expression for s r the sum of the suffixes of the p's in each term will
be r.
Hence to find all the terms that can possibly occur in s r we
have simply to write down all the products of powers of p n p 2 ,
. . ., p n in which the sum of the suffixes is r.
Example.
To find the sum of the cubes of the roots of the equation
a?  2a? + 3a: + 1 = 0.
We have
512 = 0, s 2 2s 1 + 2x3 = 0, 53 2s 2 + 3s 1 + 3 x 1 = 0.
Hence si = 2, s 2 =2, s 3 =13.
§ 4.] We can now show that evqry integral symmetric function
of the roots can be expressed as an integral function of p t , p 3 , . . ., p n .
The terms of every symmetric function can be grouped into
types, each term of a type being derivable from every other of
that type by merely interchanging the variables a,, a,, . . ., a n
(see chap, iv., § 22). All the terms belonging to the same type
have the same coefficient. It is sufficient, therefore, to prove the
above proposition for symmetric functions containing only one
type of terms. Such symmetric functions may be classed as
single, double, triple, &c, according as one, two, three, &c,
of the variables a u a 2 , . . ., a n appear in each term. Thus
^a,^, ^a,^a/, ^a^a^a/, &c, are single, double, triple, &c,
symmetric functions.
For the single functions, which are simply sums of powers,
the theorem has already been established. We can make the
double function depend on this case as follows : —
Consider the distribution of the product
(tt,P + a.,P + . . . + a H P) (a,? + aj + . . . + a^).
Terms of two different types, and of two only, can occur, namely,
XVIII
ORDER AND WEIGHT 439
terms derivable from a?a£, that is, af+% and terms derivable
from a^a 2 2. We have in fact
SpS q = SttjP+S + ^a.PaJ.
Hence Sa^cu? = s p s q  s p+q .
Now s p , s q , s p+q can all be expressed as integral functions of
PitPsi • ■ •> Pn Hence the same is true of lafa^.
Here we have supposed p 4= q. If p = q, then the term a^a/
will occur twice, and we have
s p 2 = 2 ai 2 * + S2*?aJ> ;
but this does not affect our reasoning.
The case of triple functions can be made to depend on that
of double and single functions in a similar way. In the distri
bution of
(ttj* + <%» + ... + a/) (a,? + a? + . . . + a,?) (a, r + a/ + . . . + a/)
every term is of the form a M p a/a w r , where m, v, w are, 1st, all
different ; 2nd, such that two are equal ; 3d, all equal. Any
particular term can occur only once if p, q, r be all unequal.
Hence we have
SpSgSr = SoAsW + 2af+W + Sa^+W + 2<+2V + W +9+r 
In the last equation every term, except Safafaf, can be ex
pressed as an integral function oip lf p 2 , . . .,p n . Hence 'Zafaja/
can be so expressed.
If two or more of the numbers p, q, r be equal, tlien each
term of lafa^a/ will occur a particular number of times ; and
the same is true of certain of the other terms in the equa
tion last written. But this does not affect the conclusion in
any way.
We can now make the case of quadruple symmetric functions
depend on the cases already established ; and so on. Hence the
proposition is generally true.
It is obvious, from the nature of each step in the above pro
cess, and from what has been already proved for s,, s.,, s 3 , . . .,
that in the expression for any homogeneous symmetric function of
degree r the sum of the suffixes of the p's will be r for each term ;
so that we can at once write down all the terms that can possibly
440 GENERAL STATEMENT OF THEORY CHAP.
occur in that expression, and then determine the coefficients by
any means that may happen to he convenient.
It is important to remark that the degree in p lt p.,, . . . , p n
of the expression for 'Eafa^a.f . . . in terms of p lt p s , . . .,p n
must be equal to the highest of the indices p, q, r . . . For, let
the term of highest degree he p^ipfo . . . p 7 ^ n , then, since
P\  \P\ ~ «d Pz = iPa  \P\0ii where $ lf ,j? 2 , &c, do not contain a 1} *
we see that p^ l p. 2 x * . . . p n Xn , when expressed in terms of
a,, a 2 , . . ., a n , will introduce the power a,^i + X 2+' ' + *» with
the coefficient ( — l) n iPi^p 3 3 . . . i^ni*" Now, since there
are no terms of higher degree than p^ipfo . . . p,^ n , if the
power a^i + x i +  ■ • +x » occur again, it must occur as the highest
power, resulting from a different term of the same degree ; that
is to say, it will occur with a different coefficient and cannot
destroy the former term. Hence the index of the highest
power of any letter in the symmetric function must be equal
to the degree of the highest term in its expression in terms of
PoPm, ■ ■ •, VnA
Although, in establishing the leading theorem of this para
graph, we have used the language of the theory of equations,
the result is really a fundamental principle in the calculus of
algebraical identities ; and it is for this reason that we have
introduced it here. We may state the result as follows : —
elementary symmetric functions of the system of n variables
x u x 2 , . . ., x n . Then we caii express any symmetric integral function
of sc, , o: 2 , . . ., x n as an integral function of the n elementary
symmetric functions ; and therefore any rational symmetric function
of these variables as a rational function of the n elementary symmetric
functions.
On account of its great importance we give a proof of this
* They are, in fact, the functions of a 2 , ag, . . ., a„ defined in § 3. See
Exercises xxxvi., 51.
t Salmon, Higher Algebra, § 58.
XVIII
PROOF OF GENERAL THEOREM 441
proposition not depending on Newton's Theorem (which is itself
merely a particular case).*
Let n q„ n q a , . . ., n <ln denote the n elementary symmetric
functions of the % variables x lt x 2 , . . ., x m that is to say,
2 n K,, 2 n .r,in,, • • , x l x a ...x n ; and let „_,&, „_,&, • • •» ntfni
denote the n~\ elementary symmetric functions of x x , x 2 , . . .,%_,,
that is, 2a;,, ? x,r 2 , . . ., sea . . . x n ^. It is obvious that,
?! — 1 ?(— 1
when x n = 0, ££,, n q 2 , . . ., n <lni become „_,£,, „_,&, . . ., nflni
respectively.
Let us now assume that all symmetric integral functions not
involving more than n  1 variables can be expressed as integral
functions of „_,?,, n i?8j . . ., ni?ni Let/(:r.,, a a , . . ., »„_„ « w )
be any symmetric integral function of the « variables x„ x„ . . ., £„,
Then /(a;,, a:*, . . ., ar n _,, 0) is a symmetric integral function of
x u x 2 , . . ., x n _„ and can therefore, by hypothesis, be expressed
integrally in terms of „_,?„ »,&, • • •, niini Let this ex
pression be <K»i?» »i2« ■ • •» nrfnOj so that <£ is a known
function.
Now assume
+ xj/(x l , x 2 , . . ., %_,, .r„) (7).
Then, since <£(»&, n </ 2 , . . ., n3Wi) is a symmetric integral
function of a,, x 2 , . . ., x m if/(x 1} x,, . . ., x n _„ a^) is obviously
a symmetric integral function of these variables.
If we put x n = on both sides of the identity (7), then
J\X X , X 2 , . . ., X n _ u 0) — <p{n\ ( lu n^121 • • '} niQni)
+ xf (./,, X s , . . ., Xn.\, 0) (8).
But/(.r,, ..',, . . ., ar B _„ 0) = <f>( n _ l q 1 , w _,2„ . . ., „_,£„_,)• Hence,
by (8), )/<(>,, x 2 , . . ., a^_ 19 0) = 0. Therefore the integral
function \f/(x l} x 2 , . . ., x n _ u x n ) is exactly divisible by some
* This proof is taken from a paper by Mr. R. E. Allardice, Pruc. Edinb.
Math. Soc. for 1889.
442 PROOF OF GENERAL THEOREM chap.
power of x n , say x n * ; hence, on account of its symmetry, also by
x*, x", . . ., x n _". We may therefore put
lf,(x 1} «b, . . ., X n  1 ,X n )= n qnfi(Vi,Xa, • ■ ■^ni^n),
where /, is a symmetric integral function of a,, K g , . . ., % of
lower degree than /
We can now deal with f x in the same way as we dealt with
f; and so on. We shall thus resolve f(x u x 2 , . . ., x n _ u x n )
into a closed expression of the form
<f> + n2»"V. + nq n ai+x % 2 + • • • + .fo" 1 W • • ' +K ' n *. (9),
where <£,, 4>,, . . ., <f> m are, like <f>, all known functions of n q x ,
n<l>, • • ., «?«>, or else constants.
If, therefore, the integral expression in question be possible
for n  1 variables, it is possible for n variables.
Now every integral function of a single variable, x n is a
symmetric function of that variable, and can be expressed
integrally in terms of //,, which is simply .t,. Hence it follows
by induction that every symmetric integral function of n variables
can be expressed as an integral function of the n elementary
symmetric functions.
Cor. It follows at once, by induction, from the form of (9) that the
coefficients of the expression for any symmetric integral function
/(»„ x, 2 , . . ., x n ) in terms of n q u n q a , . . ., n q n are integral
functions of the coefficients off In particular, if the coefficients of f
be integral numbers, the coefficients of its expression in terms of
n°i> w&j • • •> nQn w ^ a ^ so be integral numbers.
We now give a few examples of the calculation of symmetric
functions in terms of the elementary functions, and of the use of
this transformation in establishing identities and in elimination.
Example 1.
If a, /3, 7 be the roots of the equation
gXpja? +p 2 x ^3 = 0,
express /3 3 > + fiy 3 + /a + ya? + a?p + a/3 3 in terms of ;j 1; p 2 , ]h
Here we have;>i = Za, p i ='Lap, p s = af3y. Remembering that no term of
higher degree than the 3rd can occur in the value of 2a 3 /3, we see that
XVIII
EXAMPLES 443
2a 3 £ = Aprpi + Bp a & + Cpr ( 1 ),
where A, B, C are numbers which we have to determine.
Suppose 7 = ; then 2h = a + P, P2 = *P, J>s=0 ; and (1) becomes
a 3 (3 + ap s = A(o + /3) 2 a/3 + Ca 2 /5 2 ;
that is to say, a 2 + /3 2 = A(a + /3) 2 + Ca/3.
Hence A = l, C=2.
We now have 2a 3 /3 =^i 2 ^ 2 + Bpi p 3  2p 2 2 .
Let a =/3=7 = 1, so that pi = 3, ^a = 3, p» = 1 . We then have
6 = 27 + 3B18.
Hence B=  1.
Therefore, finally, 2a 3 /3 =£1^2  p^  2i> 2 2 .
In other words, we have the identity
2a 3 /3=(2a) 2 2a/3  a/3 7 2a  2(2a/3) 2 .
Example 2.
To show that
{yz  xu ) (zx  yu) (xy  zu) = (yzu + zux + uxy + xyzf  xyzu(x + y + z + u) 2 (2).
The lefthand side of (2) is a symmetric function of a, y, z, u. Let us calcu
late its value in terms of/?i = 2a:, /? 2 = 2xy, p 3 Y.xijz, p i = xyzu.
Since the degree of U(yz  xu) in x, y, z, u is 6, and the degree in z alone
is 3, we have
U(yz  xu) = Api 2 pi + Bpip^&t + Cp 2 3 + BpsPi + Ep 3 2 (3).
If we put u = 0, then pi = ~Z 3 x, p 2 = 2 3 xy, p 3 = xyz, p t =0, where the suffix
3 under the 2 means that only three variables, x, y, z, are to be considered. If
Pi, lh, Pz have for the moment these meanings, then (3) becomes the identity
^ 3 2 = BpiPsPa + Cpi + Ep 3 2 .
Hence B = 0, C=0, E=l.
Hence Jl(yz  xu) = Aprp* + VpiPi + P3* ( 4 )
Now let x = y = l, and z = u \, so thatj9i = 0, #2= 2, ^3=0, p t = l.
Then (4) becomes
0=2D.
Hence D = 0.
We now have II (yz  xu) = Apfp* +p 3 2 .
In this put x = y = zu = l, and we have
= 16A + 16.
Hence A=  1.
Hence, finally, H(yz  xu)=p 3 2  p^p*,
which establishes the identity (2).
Example 3.
If x + y + s = 0, show that
g n + y ii + 8 u _^ ± ^ ± j» x * + f + z* (xs + y^ + z 3 ) 3 x 2 + y 2 + z 2
11 ~ 3 " 2 9 2 ' ; '
( Wolstcnholme.)
444 EXAMPLES chap.
If^> 1 =Sar, p 2 = ?,xy, p 3 xyz, s 3 = 2£ 2 , s 3 =2.r 3 , &c, then we are required to
prove that
su s$s a s/s 2 ._,.
11 _ "6" ""18 ( ,
We know that s n is a rational function of^i, Pi, lh I' 1 the present case
Pi = 0, and we need only write down those terms which do not contain p x .
We thus have
9n = AfrY + Bp a p s * (6),
provided x + y + z=0.
A may be most simply determined by putting z = (x + y), writing out
both sides of (6) as functions of x and y, dividing by xy, and comparing the
coefficients of a; 9 . We thus find A = 11.
We have therefore
s n = llp 2 4 p 3 + B2 ] 2P3 S 
In this last equation we may give x, y, z any values consistent with
x + y+z=0, say x=2, y= 1, z= 1. We thus get B=  11. Hence
Sn = 1 lp 2 4 P3 ~ 1 l&&£ ( 7 )•
In like manner we have
s 8 = Ap2 i + Bp 2 ps 2 
Putting in this equation first x=l, y=  1, z — 0, and then x = 2, y —  1,
z= 1, we find A=2, B= 8.
Hence s 8 = Ipt  8p 2 p 3 2 ( 8 ).
We also find S3 = 3jt? 3 (9),
s,= 2p 2 (10).
From (8), (9), and (10) we deduce
T  IT =^ 4  4 ^ 2) + 3 ^'
"IV
which is the required equation.
Since we have four equations, (7), (8), (9), (10), and only two quantities,
P2, Ps, to eliminate, we can of course obtain an infinity of different relations,
such as (5) ; all these will, however, be equivalent to two independent equa
tions, say to (5), and
725 8 = 95 2 '' + 4.V:" (11).
Example 4.
Eliminate x, y, z from the equations x + y + z = 0, x i + y' i + z* = a, x^ + if + z 5
= 6, x 7 + y 7 + z 7 = c.
Using the same notation as in last example, we can show that
s 3 = 3;>3 , S S =  5;j 2 ;>3 , $7 = 7prj>:i ■
Our elimination problem is therefore reduced to the following : —
To eliminate p% andf7s from the equations
3y'3 =a,  5pa Pi = !>, "pfl '3 = <".
This can be done at once. The result is
2W 25oc=0.
xvin EXERCISES XXXVI 445
Exercises XXXVI.
a and /3 being the roots of the equation x 2 +px + q = 0, express the follow
ing in terms of p and q : —
(1.) a» + /3» (2.)(a« + /3«)/(a/3) 2 . (3.) a~ 5 + /3 3 . (1 ) a" 5  /3~ 5 .
(5. ) (a :1 + /3 s ) 1 + (a 3  /3»)i. (6. ) (1  a) 2 /3 2 + (1  /3)V.
(7.) If the sum of the roots of a quadratic be A, and the sum of their
cubes B 3 , find the equation.
(8.) If s„ denote the sum of the nth powers of the' roots of a quadratic,
then the equation is
(SnS»2  S„i 2 )x 2  (s„ + iS„_o  S n S n _i)x + (s n+1 S n i  S n 2 ) = 0.
(9.) If a and /3 be the roots of x 2 +px + q = 0, find the equation whose
roots are (ah) 2 , (/3A) 2 .
(10.) Prove that the roots of
x 2  {2pq)x+p 2 pq + q 2 =
are p + uq, p + uq, w and or being the imaginary cube roots of 1.
(11.) If a, /3 be the roots of x* + z + l, prove that a" + /3" = 2, or =  1,
according as n is or is not a multiple of 3.
(12.) Find the condition that the roots of ax 2 + bx + c = may be deduc
ible from those of a'x 2 + b'x + c' — by adding the same quantity to each root.
(13.) If the differences between the roots of x 3 +px + q = and x 2 + qx+p
— be the same, show that either p = q or p + q + 4 = 0. What peculiarity is
there when p> = q1
Calculate the following functions of a, j3, y in terms of^! = 2a, j? 2 = Sa/3,
p 3 = aj3y:—
(14.) a. 2 /py + p 2 /ya + y 2 /ap. (15.) a~ s + ^ 5 + y s .
(16.) (/ja+yxy+oW + Zn (17.) Z(a 2 + /3 7 )/(a 2 /3 7 ).
(18.) 2(/3 7 )'. (19.) Z(a0) 2 (/3 7 )2. (20.) 2(/3 + 7 )\
Calculate the following functions of a, /3, 7 , 5 in terms of the elementary
symmetric functions : —
(21.) Za\ (22.) Sa3. (23.) 2a 2 /3 2 . (24.) 2a 2 /3 7 . (25.) 2(a+/3)*
(26. ) If a, /3, 7 , 3 be the roots of the biquadratic ar 1 +p 1 x :i +p 2 x 2 +psx +p 4 = 0,
find the equation whose roots are /3 7 + a5, ya + {15, a^ + yd.
(27.) If the roots of x 2 2hr+p 2 — 0, x 2 q 1 x + q 2 = Q, x 2 rix + r 2 = 0, be /3,
y ; y, a ; a, /3 respectively, then a, /3, 7 are the roots of
K 3  Ki»i + ?i + ^i)* 2 + (#! + qi+ r 2 )x  hiPilin ~PiPa  q\qi  nr 2 ) = 0.
(28. ) If a, p, y be the roots of x 3 +px + q = 0, show that the equation whose
roots are a + /3 7 , /3 + ya, y + a/3, is X s px 2 + {p + 3q)x + q(p + q) 2 — 0.
(29.) If a, j8, 7 be the roots of
p/(a + x) + q/(b + x) + r/{c + x) = 1,
show that p = (a + a) (a + /3) (a + 7 )/(a  6) (a  c).
If a, /3 be the roots of x 2 +pix+p 2 = 0, and a', (3' the roots of x 2 +pi'x+p 2
= 0, express the following in terms of p\, p 2 , p{, p 2 : —
(30.) (a'a)(/3'/3)+V/3)(/3'«).
446 EXERCISES XXXVI CHAP.
(31. ) (a'a)' + (/?'  jB)« + (a'  /3) 2 + (/3'  a)*.
(32.) (a + a')(/3 + /3')(a + /3')(/3 + a').
(33.) 4(aa')(a/3')(/3a')(/3/3').
[The result in this case is
icfty*)" + ♦(piftO (pip2piP2) = (2 P 2 + 2p 2 ' PlPl r{pi' 2  if*) bi' 2  ^m
(34.) A, A' and B, B' are four points on a straight line whose distances,
from a fixed point on that line (right or left according as the algebraic
values are positive or negative), are the roots of the equations
ax 2 + bx + c = 0, a'x 2 + b'x + c' = Q.
If AA'.BB' + AB'.BA' = 0,
show that 1m' + 2c 'a W = 0;
and if AA'.BA' + AB'.BB'=0,
that 2c«' 2  2cW + ab' 2  a'W = 0.
(35.) a, j3 are the roots of x~2ax + b 2 = 0, and a', (3' the roots of
x 2  2cx + d = 0. If aa' + /3/3' = 4w 2 , show that
(a 2 b 2 )(c 2 d 2 ) = {ae2n) 2 .
(36.) a, /3, a', /3', being as in last exercise, form the equation whose roots
are aa' + /3/3', a/3' + a'/3.
(37.) If the roots of ax 2 + bx + c = be the squai'e roots of the roots of
a'x 2 + b'x + c' = 0, show that a'b 2 + ab' = 2aa'c.
(38.) Show that when two roots of a cubic are equal, its roots can always
be obtained by means of a quadratic equation.
Exemplify by solving the equation 12a. 13  56a; 2 + 87a;  45 = 0.
(39.) If one of the roots of the cubic a?+pia?+p^xs+ps=0 be equal to the
sum of the other two, solve the cubic. Show that in this case the coefficients
must satisfy the relation
Pi 8 4pj#s+8p8=0.
(40.) If the square of one of the roots of the cubic x s +piX 2 +p2X+2>3=0 be
equal to the product of the other two, show that one of the roots is JJg/pi ;
and that the other two are given by the quadratic
PiPix +2h(Pi" pi)x +2 } i 2 P3 = °
As an example of this case, solve the cubic
x 3 9a; 2  63a; + 343 = 0.
(41.) If two roots of the biquadratic x i +pix 3 +p<c 2 +p 3 x+p i = be equal,
show that the repeated root is a common root of the two equations
4X 3 + ZpiX 2 + 2p»z + ]>3 = 0, ./; 4 + p X 7? + p&p +2) 3 x + p 4 = 0.
(42.) If the three variables x, y, z be connected by the relation 2x—xyz,
show that 22a:/(l  x 2 ) = U2x/{1  x 2 ).
(43.) If 2z = 0, show that 22a. 7 = 1xijzZx\
(44.) If Saj=0, show that Zx s =2{2yz) i  8x 2 ifz 2 Zyz.
(45.) If Sa=0 (three variables), then (2(6»  e?)/a 3 ) (Za^ft 8  <?) ) =s
364(2a :, )(2a 3 ).
(46.) If 2ar* = Q, 2.t 4 = (three variables), show that Zx 5 + xyz{{yz)(zx)
xviii DISCRIMINATION OF THE ROOTS OF A QUADRATIC 447
(47.) Ifas+y+2+tt=0, show that (2a?y i =9{2xyzf=m(yzxu).
(48.) Under the hypothesis of last exercise, show that
ux{ u + xf + yziuxf + uy(u + yf + zx(u  yf + uz(u + zf + xy(ic  zf + ixyzu = 0.
Eliminate x, y, z between the equations
<»•><;)='. *(:)*="■ =©'""• KS)'* 1 .
(50.) Sa?=a a , Zxy = V, 2xhf = c\ Zx* = d\
(51.) Show that Pi = B p\a i , P2 = >P2~ >P\*s, • • ; Pr= tPr~ »Pr\**i ■ ■ ;
p„—  ,2hiio. s , where 2h> P2> • • •> tPi> iPz> • • • have the same meanings as
in § 3.
SPECIAL PROPERTIES OF QUADRATIC FUNCTIONS.
§ 5.] Discrimination of Boots. — We have already seen (chap,
xvii., § 4) how, without solving a quadratic equation, to dis
tinguish between cases where the roots are real, equal, or imag
inary. There are a variety of other cases that occur in practice
for which it is convenient to have criteria. These may be treated
by means of the relations between the roots and the coefficients
of the equation given in § 1 of the present chapter. If a, ft be
the roots of
az* + bx+c = (1),
then a + ft   b/a, aft = cja.
If both a and ft be positive, then both a + ft and aft are posi
tive. Conversely, if aft be positive, a and ft must have like
signs ; and if a + ft be also positive, each of the two signs must
be positive ; but if a + ft be negative, each of the two signs
must be negative. Hence the necessary and sufficient condition that
both roots of (1) be positive is that b/a be negative and cja positive;
and the necessary and sufficient condition that both roots be negative is
tin if hja be positive and cja positive. This presupposes, of course,
that the condition for the reality of the roots be fulfilled, namely,
b 2  iac > 0.
Reality being presupposed, the necessary and sufficient condition
that the roots have opposite signs is obviously that cja be negative.
The necessary and sufficient condition that the two roots be numeric
ally equal, but of opposite sign, is a + ft = 0, that is, b/a = 0.
If one root vanish, then aft = ; and, conversely, if aft = 0,
then at least one of the two, a, ft, must vanish. Hence the neces
448 INFINITE ROOTS
CHAP.
sary and sufficient condition for one zero root is cja  0, that is, c = 0,
a being supposed finite.
If both roots vanish, then a/3 = and a + f3 = ; and, con
versely, if a/3 = and a + (3 = 0, then both a = and (3 = 0; for
the first equation requires that either a = or (3 = 0, say a = ;
then the second gives + (3 = 0, that is, (3 = also. Hence the
necessary and sufficient condition for two zero roots is c/a = 0, b/a = 0,
that is, a being supposed finite, c — 0, b = 0.
The two last conclusions have already been arrived at in
chap, xvii., § 2. Perhaps they will be more fully understood by
considering the case as a limit. Let us suppose that the root a
remains finite, and that the root (3 becomes very small. Then
a/3 becomes very small, and approaches zero as its limit, while
a + (3 approaches a as its limit. In other words, c/a becomes
very small, and  b/a remains finite, becoming in the limit equal
to the finite root of the quadratic.
If both a and (3 become infinitely small, then both a + (3 and
a/3, that is to say, both  b/a and c/a, become infinitely small.
Infinite Boots. — If the quadratic (1) have no zero root, it is
equivalent to
that is, if £= 1/x, to
c£ 2 + b£ + a = (2).
The roots of (2) are 1/a and \/(3; and we have 1/a+ 1/(3
=  b/c, l/a/3 = a/c. Let us suppose that one of the two, a, f3,
say (3, becomes infinitely great, while the other, a, remains finite j
then \/f3 becomes infinitely small, and l/a/3, that is, a/c, becomes
infinitely small, while l/a+l//3, that is, b/c, approaches the
finite value 1/a. Hence the necessary and sufficient condition
that one root of (1) be infinite is a = 0, c being supposed finite.
In like manner, the condition that two roots of (1) become
infinite, that is, that two roots of (2) become zero, is a = 0, b = 0.
If therefore in any case where a quadratic equation is in
question we obtain an equation of the form bx + c = 0, or an equa
tion of the paradoxical form c = 0, ice conclude that one root of the
XVIII
TABLE GENERAL RESULTS CUBIC
449
quadratic lias become infinite in the one case, and that the two roofs
have become infinite in the other.
For convenience of reference we collect the criteria for dis
criminating the roots in the following table : —
Roots real
i
b"4ac>0.
Both roots negative
c/a + , b/a + .
Roots imaginary
b  4ac < 0.
Roots of opposite
Roots equal .
b4ac = 0.
signs .
c/a .
Roots equal with
One root =0
c = 0.
opposite signs
b = 0.
Two roots =
b = 0, c = 0.
Both roots positive
c/a + , bja .
One root = qo
a = 0.
Two roots = oc
b = 0, « = 0.
§ 6.] The reader should notice that some of the results em
bodied in the table of last paragraph can be easily generalised.
Thus, for example, it can be readily shown that if in the equation
p x n +p l x n  1 + . . . +p n = (1)
the last r coefficients all vanish, then the equation will have r zero
roots ; and if the first r coefficients all vanish it will have r infinite
roots.
Again, if p t = 0, the algebraic sum of the roots will be zero ;
and so on.
It is not difficult to find the condition that two roots of any
equation be equal. AVe have only to express, by the methods
already explained, the sjanmetric function TI(ai  a,) 2 of the roots
in terms of p , p u . . ., p n , and equate this to zero. For it is
obvious that if the product of the squares of all the differences
of the roots vanish, two roots at least must be equal, and con
versely.
For example, in the case of the cubic
a?+pix 2 +p 2 x +2^30 (2),
whose roots are a, /3, 7, we find
(/3  7 ) 2 (7  a) 2 (a  /3) 2 =  Wjh +2hW + 1 Zjhlh]h ~ W  2"i> 3 2 .
The condition lor equal roots is therefore
 4]h 3 p3+Pipr + 18piP2i'3 ~ *Pa 8  27j> 3 2 = 0.
Further, if all the roots of the cubic be real, (/3  7) 2 (7  a) 2 (a  /3) 2 will be
positive, and if two of them be imaginary, say /3=\H^i, y = \/xi, then
( i 37)2(7 a )2 i a^ 2 =4 M 2 {(Xa) 2 + ^} 2 , that is, (^  7)^(7  «) 2 (a  /3) 2 is
negative. Hence the roots of (2) are real and unequal, such that two at least
are equal, or such that two are imaginary, according as
VOL. 1 2 G
450 TWO QUADRATICS, CONDITION OF EQUIVALENCE CHAr.
 ipi 3 P3 +Pi~p? + IZpiPiPz ~ ±pi  27p 3 2
is positive, zero, or negative.
The further pursuit of this matter belongs to the higher theory of equa
tions.
§ 7.] If the two quadratic equations
ax 2 + bx + c  0, a'x 2 + b'x + c' =
be equivalent, then b/a = b'/'a' and c/a = c'/a'. For, if the roots of
each be a and /3, then
bja =  (a + /5) = b'/a', cfa = afl = c'/a' ;
and this condition is obviously sufficient.
The above proposition leads to the following : A quadratic
function of x is completely determined when its roots are given, and
also the value of the function corresponding to any value of x which is
not a root. This we may prove independently as follows. Let
the roots of the function y be a and (3 ; then y = A(x  a) (x  /5).
Now, if V be the value of y when x = X, say, then we must have
V = A(A  a) (A  /8).
This equation determines the value of A, and we have,
finally,
y  V (Aa)(A/?) W
The result thus arrived at is only a particular case of the
following : An integral function of the nth degree is uniquely deter
mined when its n + 1 values, V,, V 2 , . . ., V M+1 , corresponding re
spectively to the Ti+1 values A n A 2 , . . ., A rt+1 of its variable x, are
given. To prove this we may consider the case of a quadratic
function.
Let the required function be ax 2 + bx + c ; then, by the con
ditions of the problem, we have
a A, 2 + ?>A, + cY u akf + bk 2 + c = V 2 , a\.f + bX x + c = V 3 .
Tbese constitute a linear system to determine the unknown
coefficients a, b, c. This system cannot have more than one
definite solution. Moreover, there is in general one definite
solution, for we can construct synthetically a function to satisfy
the required conditions, namely,
xvnr lagkange's interpolation formula 451
= v C g  A *K C ~ K) v (* Ai)(sAg) v ( ■<•  a,) (x  a 2 )
' J \K\,^K  A 3 ) 2 (A,  A,)(A 2  A 3 ) 3 (A 3  A,)(A 3  A,)
(2).
The reasoning and the synthesis are obviously general. We
obtain, as the solution of the corresponding problem for an in
tegral function of x of the nth degree,
V y (■/'  A 2 ) (■'■•  A 3 ) . . . (X  A, t+1 ) ,gX
(A,  A.) (A!  A 3 ) . . . (A t  A )l+1 )
This result is called Lagrange's Interpolation Formula.
Example 1.
Find the quadratic equation with real coefficients one of whose roots is
5 + 6i.
Since the coefficients are real, the other root must be 5  6i. Hence the
required equation is
(x5 + 6i)(x56i) = 0,
that is, (x5f+6 2 = 0,
that is, x*10x + 61 = Q.
Example 2.
Find the quadratic equation with rational coefficients one of whose roots
is 3 + v7.
Since the coefficients are rational,* it follows that the other root must be
3  \J7. Hence the equation is
{x3 + ^/7)(x3^/7) = 0,
that is, a; 2  6^ + 2 = 0.
Example 3.
Find the equation of lowest degree with rational coefficients one of whose
roots is \J2 + \J3.
By the principles of chap, x.* it follows that each of the quantities
\/2 \J3,  \J2 + \J3,  \/2 \/3 must be a root of the required equation.
Hence the equation is
(x  V2  V3) {x  V2 + V3) (* + V2  V3) (x + s/2 + s/3) = 0,
that is, a?l0a?+l = 0.
Example 4.
Construct a quadratic function of x whose values shall be 4, 4, 6, when
the values of x are 1, 2, 3 respectively.
* This we have not explicitly proved ; but we can establish, by reasoning
similar to that employed in chap, xii., § 5, Cor. 4, that, if a + b\fp be a root of
f(x) = 0, and if a and b and also all the coefficients of f(x) be rational so far as
\/p is concerned, then a  b\/p is also a root of/(.t') = 0.
452 TWO QUADRATICS, CONDITION FOR COMMON ROOT chap.
The required function is
( 8 2)(g8) (■>:!) (a 3) (as1) (s2)
4 (l2)(l3) + 4 (2l)(23) + °(3l)(32)'
that is, a.' 2  3.r + 6.
§ 8.] The condition that the two equations
ax 2 + bx + c = 0, ft V + b'x + c' =
have one root in common is the same as the condition that the
two integral functions
y = ax 2 + bx + e, y = a'x 2 + b'x + c'
shall have a linear factor in common. Now any common factor
of y and y' is a common factor of
c'y  cy ', and ay' ay,
that is, if we denote ad  a'c by (ac'), Sec, a common factor of
(ac')x 2 + (bc')x, and (ab')x + (ac') ;
that is, since x is i ot a common factor of y and y' unless c =
and c' = 0, any common factor of y and y' is a common factor of
(uc')x + (be'), and (ab')x + (ac').
Now, if these two linear functions have a common factor of
the 1st degree in x, the one must be the other multiplied by a
constant factor.
Hence the required condition is
(ac') _ (be)
(ab')~(ac'y
or (ac'  a'c)* = (be  b'c) (ab'  a'b).
The common root of the two equations is, of course,
be'  b'c ac'  a'c
x = 
ac
a'c ab'  a'b'
By the process here employed we could find the r conditions
that two integral equations should have r roots in common.
It is important to notice that the process used in the demon
stration is simply that for finding the G.C.M. of two integral
functions — a process in which no irrational operations occur.
Hence
xvm EXERCISES XXXVII 453
Cor. 1. If two integral equations have r roots in common, these
roots are the roots of an integral equation of the rth degree, whose co
efficients are rational functions of the coefficients of the given equations.
In particular, if the coefficients of the two equations be real
rational numbers, the r common roots must be the roots of an
equation of the rth degree with rational coefficients.
For example, two quadratics whose coefficients are all rational
cannot have a single root in common unless it be a rational root.
Cor 2. We may also infer that if two integral equations whose
coefficients are rational have an odd number of roots in common, then
one at least of these must be real.
Exercises XXXVII.
Discriminate the roots of the following quadratic equations without solving
them : —
(1.) 4a; 2 8a; + 3 = 0. (2.) 9x  12* 1 = 0. (3.) ix*4x+6 = 0.
(4.) 9a; 2  36x436 = 0. (5.) la?  ix  3 = 0. (6.) 4a?+8x+8=0.
(7.) (a53)(aj+4) + (a;2)(a:+3)=0.
(8.) Show that the roots of {b 2  4ac)x 2 + 4(a + c)x  4 = are always real ;
and find the conditions— 1° that both he positive, 2° that they have opposite
signs, 3° that they be both negative, 4° that they be equal, 5° that they be
equal but of opposite sign.
(9.) Show that the roots of x 2 + 2(p + q)x + 2(jj 2 + q 2 ) = are imaginary.
(10.) Show that the roots of
{<?2bc + l 2 }x 2 2{c 2 {a + b)c + ab}x+{2a 2 2(b + e)a + b 2 + e 2 }=0
are imaginary.
(11.) Show that the roots of
(a;  b) [x c) + {x  e) [x  a) + (x  a) [x b) =0
are real ; and that they cannot be equal unless a=b = c.
(12. ) The roots of a/(x a) + b/(x b) + c/(x  c) = are real ; and cannot be
equal unless either two of the three, a, b, c, are zero, or else a = b = c.
(13.) Find the condition that the cubic x 3 + qx + r = have equal roots.
(14.) Show that the cubic 1 2x? 52a; 2 + 75a;  36 = has equal roots; and
solve it.
(15.) If two of the roots of a cubic be equal, and its coefficients be all
rational, show that all its roots must be rational.
(16.) Find the condition that two roots of the biquadratic ax i + dx + e=0
be equal.
(17.) If a/(x + a) + b/(x + b)c/(x + c) + d/(x + d) have a pair of equal roots..
then either one of the quantities a or b is equal to c or d, or else 1/a + l/b
= l/c + l/d. Prove also that the roots are then a. a, 0,  b, b, 0,
or 0, 0,  2ab/(a + b).
454 EXERCISES XXXVII chap.
Write down and simplify the equations whose roots are as follows: —
(18.) 1,0. (19.) h . (20.) 3 + V2, 3V2.
(21.) (a+Va 2 l)/(«\/« 2 l), (a Va 2  1 )/(« + Va 2  1 ).
Find the equations of lowest degree with real rational coefficients which
have respectively the following for one root : —
(22.) a+pi. (23.) 1 + V2V3 (21.) s/2 + i^3.
(25.) »/2 + fyi. (Result, x 3  6x 6 = 0.)
(26. ) 8/2 + f/3. ( Result, x 9  1 5x 6  8 Tx 3  1 25 = 0. )
(27.) V(?'0 + VM + V(i>9)
(28.) Find the equation of the 6th degree two of whose roots are
1 + y/2 and 1+^1.
(29.) Find an equation with rational coefficients one of whose roots is
ap 2 ' 3 + bp 1 ' 3 + c.
Hence show how to find the greatest integer in ap 2/3 + bp 113 + c without
extracting the cube roots.
(30.) Form the equation whose roots are p + atf, p + a 2 q, . . ., p + a 2n q,
where a 1; a 2 , . . ., a 2n are the imaginary (2?i + l)th roots of 1, showing that
the coefficients are all rational, and finding the general term of the equation.
(31.) Construct a quadratic function whose roots shall be equal with
opposite signs, and whose values shall be 23 and 67 when x=5 and when
x = 6 respectively.
(32.) Construct a cubic function y corresponding to the following table ol
values : —
a: = 25, 3, 35, 4;
y= 6, 8, 15, IS.
(33.) If x 3 +ax + bc = 0, x 2 + bx + ca = have a common root, then their
other roots satisfy x 2 + cx + ab = 0.
(34.) If 2{p + q + r) = a 2 + p 2 + y 2 , and the roots of a 2 + aa:  ;; = be/3 and y,
and the roots of x 2 + fix  q = be y and a, then the equation whose roots are
a and /3 is x 2 + yx r—0.
VARIATION OF QUADRATIC FUNCTIONS FOR REAL VALUES
OF THE VARIABLE.
§ 9.] The quadratic function
y = ax 2 + bx + c
may be put in one or other of the three forms
y = a{(x — I) 2  m] I.,
y = a{(;xiy} II.,
y = a{(x  I) 3 + vi] III.,
according as its roots « and ji are real (say a = l + Jin,
xvni VARIATION OF ca^ + bx + c 455
(3 = I  \/m), equal (say a = l,f$ = l), or imaginary (say a = I + i sjm,
(3 = 1  i x /m). I and m are both essentially real quantities, and
m is positive.
Each of these three cases may be farther divided into two,
according as a is positive or negative.
In all three cases when x is very great (x  I) 3 is very great
and positive. Hence, in all three cases, y is infinite when x is
infinite, and it has the same sign as a.
In all three cases the function within the crooked bracket
diminishes in algebraical value when x diminishes, so long as
x > I, and has an algebraically least value when x = I ; for (x  If,
the only variable part, being essentially positive, cannot be less
than zero. When x is diminished beyond the value xl, (x  If
continually increases in numerical value.
We conclude, therefore, that in all three cases the quadratic
function y has an algebraical minimum or maximum value when x = l,
according as a is positive or negative; and that the function has no
other turning value.
In Case I., where the roots are real and unequal, y will have the
same sign as a or not, according as the value of x does not or does lie
between the roots.
For y = a(x  a)(x  (3) ; and (x  a) (x  fS) will be positive if
x be algebraically greater than both a and (3, for then x  a and
x  (3 are both positive ; and the same will be true if x be alge
braically less than both a and (3, for then x  a and x (3 are
both negative. If x lie between a and (3, then one of the two,
x  a, x (3, is positive and the other negative.
In Cases II. and III, where the roots are either equal or imaginary,
the function y will have the same sign as a for all values of x.
For in these cases the function within the crooked brackets
has clearly a positive value for all real values of x.
§ 10.] The above conclusions may be reached by a different
but equally instructive method as follows : —
Let us trace the graph of the function
y = ax z + bx + c (1);
and, for the present, suppose a to be positive.
456
GRAPH OF A QUADRATIC FUNCTION
CHAP.
To find the general character of the graph, let us inquire
where it cuts a parallel to the axis of x, drawn at any given
distance y from that axis. In other words, let us seek for the
abscissa? of all points on the graph whose ordinatcs are equal to y.
We have
y = ax" + bx + c,
that is, ax 2 + bx + (c  y) = (2).
We have, therefore, a quadratic equation to determine the
abscissa? of points on the parallel. Hence the parallel cuts the
graph in two real distinct points, in two coincident real points,
or in no real point, according as the roots of (2) are real and
unequal, real and equal, or imaginary.
Since a is positive, it follows that when x=  <x> , y = +co;
and when x + ao , y = + oo . Moreover, the quadratic function
y is continuous, and can only become infinite when x becomes
infinite. Hence there must be one minimum turning point on
the graph. There cannot be more than one, for, if there were,
it would be possible to draw a parallel to the &axis to meet
the graph in more than two points.
The graph therefore consists of a single festoon, beginning
and ending at an infinite distance above the axis of x.
The main characteristic point to be determined is the
minimum point. To obtain this we
have only to diminish y until the
parallel UV (Fig. 1) just ceases to
meet the graph. At this stage it is
obvious that the two points U and
V run together ; that is to say, the
two abscissa? corresponding to y
become equal. Hence, to find //, we
have simply to express the condition
that the roots of (2) be equal. This
condition is
4a (c  y) = 0.
y= h lz^ (3).
J 4a v
V
/
V^f^M/
/
/
/
/
/
/
/
/
/
' A S
\
\
\
\
V
\
\
\
Fig. 1.
lr
Hence
XVIII
THREE FUNDAMENTAL CASES
457
The corresponding value of x is easily obtained, if we notice
that the sum of the roots of (2) is in all cases  b/a, and that
when the two are equal each must be equal to  b/2a. Hence
the abscissa of the minimum point is given by
b_
9. a.
X =
(0.
There are obviously three possible cases — ■
I. The value of y given by (3) may be negative. Since a is
supposed positive, this will happen when b 2  iac is positive.
In this case the minimum point A will lie below the axis of
.<■, and the graph will be like the fully drawn curve in Fig. 1.
Here the graph must cut the a'axis, hence the function y
must have two real and unequal roots, namely, x = OL, x = OM ;
and it is obvious that y is positive or negative, that is, has the
same sign as a or the opposite,
according as x does not or does
lie between OL and OM.
II. The value of y given
by (3) will be zero, provided
b 2  inc = 0.
Fig. 2.
In this case the minimum point
A falls on the axis of x, and the
graph will be like the fully drawn
curve in Fig. 2.
Here the two roots of the function
are equal, namely, each is equal to OA.
It is obvious that here y is always
positive, that is, has the same sign as a.
III. The value of y given by (3)
will be positive, provided b 2  iac be negative.
In this case the graph will be like the fully drawn curve in
Fig. 3.
Fig.
3.
458 EXAMPLES
CHAP.
Here the graph does not cut the axis of x, so that the
function has no real roots. Also y is always positive, that is,
has the same sign as a.
If we suppose a to be negative, the discussion proceeds
exactly as before, except that for positive we must say negative,
and for minimum maximum. The typical graphs in the three
cases will be obtained by taking the mirrorimages in the axis of
x of those already given. These graphs are indicated by dotted
lines in Figs. 1, 2, 3.
For simplicity we have supposed the abscissae of the points
L, M, N, A to be positive in all cases. It will of course happen in
certain cases that one or more of these are negative. The cor
responding figures are obtained in all cases simply by shifting
the axis of y through a proper distance to the right.
Example 1.
To find for what valuesof x the function y = 2x 2  12.T + 13 is negative, and
to find its turning value.
We have y = 2(.r 2  6ar + 9)  5,
= 2{(z3) 2 
4i
= 2{z(3Vf)}{a;(3+Vf)}
Hence ?/ will be negative if x lie between 3 V(°/2) and 3 + \/(5/2), and will
be positive for all other values of x.
Again, it is obvious, from the second form of the function, that y is
algebraically least when (x 3) 2 = 0. Hence y=  5 is a minimum value of y
corresponding to x=Z.
Example 2.
To find the turning values of (a;  8a; + 15)/a;.
15 „
We have y—x +
x
First, suppose x to be positive, then we may write
from which it appears that y has a minimum value, 8 + 2a/15, when
\Jx \/(15/.c) = 0, that is, when x=\J\5.
Next, let x be negative, = £ say, then we may write
» 15
2Vi5(vf V ^)*
win MAXIMA AND MINIMA 459
from which we see that  8  2\/15 is a maximum value of y corresponding to
£ = /^/l 5, that is, to x —  \/l 5.
Example 3.
If a and y be both positive, then —
If x + y be given, the greatest and least values of xy correspond to the
least and greatest values of (xy) 2 ; so that the maximum value of xy is
obtained by putting x=y, if that be possible under the circumstances of the
problem.
If xy be given, the greatest and least values of x + y correspond to the
greatest and least values of (xy) ; so that the minimum of x + y is obtained
by putting x = y, if that be possible under the circumstances of the problem.
These statements follow at once from the identity
(x + y) 2 (xy) 2 =4xy.
i,,y = c(xy) 2 .
For, iix + y = c, then
And, if xy=d a , then
(x + y) 2 =id 2 + (.ry) 2 .
Hence the conclusions follow immediately, provided x and y, and therefore
xy and x+y, be both positive.
These results might also be arrived at by eliminating the value of y by
means of the given relation. Thus, if x + y = c, then xy = x(cx) = cxx 2
= c 2 /4(c/2x)' 2 . Hence xy is made as large as possible by making x as
nearly =c/2 as possible, and so on.
Many important problems in geometry regarding maxima
and minima may be treated by the simple method illustrated in
Example 3.
Example 4.
To draw through a point A within a circle a chord such that the sum of
the squares of its segments shall be a maximum or a minimum.
Let r be the radius of the circle, d the distance of A from the centre, x and
y the lengths of the segments of the chords.
Then, by a wellknown geometrical proposition,
xy=i*d* (1).
Under this condition we have to make
u = x + y (2)
a maximum or minimum.
Now, if we denote x" and y by £ and 77, then £ and 77 are two positive
quantities ; and, by (1), we have
!;V=(i*d*)* (3).
Hence, by Example 3, { + 77 is a minimum when = 77, and is a maximum
when (£ tj) 2 is made as great as possible. If we diminish 77, it follows, by
(3), that I increases. Hence (f77) 2 will be made as great as possible by
making £ as great as possible.
460 GEOMETRICAL MAXIMA AND MINIMA chap.
Hence the sum of the squares on the segments of the chord is a minimum
when it is bisected, and a maximum when it passes through the centre of
the circle.
Example 5.
A and B are two points on the diameter of a circle, FQ a chord through
B. To find the positions of PQ for which the area APQ is a maximum or a
minimum.
Let be the centre of the circle. The area OPQ bears to the area APQ
the constant ratio OB : AB. Hence we have merely to find the turning
values of the area OPQ.
Let OB = a, and let x denote the perpendicular from on PQ. Then, if
u denote the area OPQ, u = x\J(r 2 x 2 ).
We have therefore to find the turning values of u. Since u is positive,
this is the same thing as finding the turning values of u 2 . Now
,.4 / r 2N
u = x(r 2  x) = —  f x  —
There are two cases to consider. First, suppose a>r/\J'2. Then, since
the least and greatest values of x allowable under the circumstances are and
a, we have, confining ourselves to half a revolution of the chord about A,
three turning values. If we put x=0 we give to (* 2 r 2 /2) 2 the greatest
value which we can give it by diminishing x below r(\j2. Hence x = gives
a minimum value of OPQ.
If we put x = rj\j2, we give (x 2 r 2 /2) 2 its least possible numerical value.
Hence, for x = r/\j2, OPQ is a maximum.
If we put x = a, we give (arr 2 /2) 2 the greatest value which we can give
it by increasing x beyond rj\j2. Hence to x = a corresponds a minimum
value of OPQ.
Next, suppose a<rj\j2. In this case we cannot make x= or >rj\j2.
Hence, corresponding to a' = 0, we have, as before, OPQ a minimum. But
now (x 2  r 2 /\/2j 2 diminishes continually as x increases up to a. Hence, for
x = a, OPQ is a maximum.
Remark. — This example has been chosen to illustrate a peculiarity that
very often arises in practical questions regarding maxima and minima,
namely, that all the theoretically possible values of the variable may not be
admissible under the circumstances of the problem.
Example 6.
Given the perimeter of a rightangled triangle, to show that the sum of
the sides containing the right angle is greatest when the triangle is isosceles.
Let x and y denote the two sides, p the given perimeter. Then the
hypotenuse i&pxy ; and we have, by the condition of the problem,
{p(x+y)}*=x*+y*.
2
Hence xyp{x+y) = ^.
This again may be written
{px){py)j£ (1).
XVIII
MAXIMA AND MINIMA, GENERAL METHOD
461
Under the condition (1) we have to make
u — x\y
a maximum.
If we put £ —p  x, f) —p  y, we have
to2
(2)
(3);
and we have to make u = 2p (£ + *?)
a maximum ; this is, to make £ + ?? a minimum. Now, under the condition
(3), i + v is a minimum when £ = y. Hence x + y is a maximum when v — y.
§ 11.] The method employed in § 10 for finding the turning
points of* a quadratic function is merely an example of the
Fm. 4.
general method indicated in chap, xv., § 13. Consider any
function whatever, say
?/=/('•) (!)■
Let A be a maximum turning point on its graph, whose
abscissa and ordinate are x and y. If we draw a parallel to OX
a little below A, it will intersect the graph in a certain number
of points, TUVW say. Two of these will be in the neighbour
hood of A, left and right of AL. If we move the parallel up
wards until it pass through A, the two points U and V will run
together at A, and their two abscissae will become equal. If we
move the parallel a little farther upwards, we lose two of the
real intersections altogether.
Hence to find y we have simply to express the condition that the
roots of the equation « v _ _ ( . / \
462
EXAMPLES
CHAP.
be equal, and then examine whether, if we increase y by a small
amount, we lose two real roots or not. If we do, then y is a maximum
value.
If it appears that two real roots are lost, not by increasing but by
diminishing y, then y is a minimum value.
Example 1.
To find the turning values of
y = x s 9x° + 24x + 3.
The values of x corresponding to a given ordinate y are given by
x 3 9x + 2ix + (3  y) = 0.
If D denote the product of the squares of the differences of the roots of this
cubic, then all its roots will be real, two roots will be ecpial or two imaginary,
according as D is positive, zero, or negative.
Using the value of D calculated in § 6, and putting pi=  9, ^ 2 =24,
p s = 3  y, we find
T)=2?(y19)(y2S).
Hence y = 19, ?/ = 23 are turning values of y. If we make y a little less than
19, D is negative, that is, two real roots of the cubic are lost. Hence 19 is a
minimum value of y. If we make y a little greater than 23, D is again
negative ; hence 23 is a maximum value of y.
It is easy to obtain
the corresponding val
ues of a;, if we remember
that two of the roots of
the cubic become equal
when there is a turning
value. In fact, if the
two equal roots be a, a,
and the third root y, we
have, by § 1,
2a + 7 = 9, a 2 + 2ct7 = 24.
Hence
a 2 6a 4 8 = 0,
which gives
a = 2, or a = 4.
Fig. 6.
It will be found that x = i corresponds to the minimum value y = 19 ; and
that x=2 corresponds to the maximum y = 23.
XVIII
EXAMPLES
463
Remark. — The above method is obviously applicable to any cubic integral
function whatsoever, and we see that such a function has in general two
turning values, which are the roots of a certain quadratic equation easily ob
tainable by means of the function D.
If the roots of this quadratic be real and unequal, there are two distinct
turning points, one a maximum, the other a minimum.
If the roots be equal, we have a point which may be regarded as an
amalgamation of a maximum point with a minimum, which is sometimes
called a maximumminimum point.
If the roots be imaginary, the function has no real turning point.
If the coefficient of a? be positive, the graphs in the first two cases have
the general characters shown in Figs. 5 and 6 respectively.
Example 2.
To discuss the turning values of
arS.T + 15 ....
y= — — (D
The equation for the values of x corresponding to any given value of y is
ai 2 (?/ + 8);r+15 = 0.
Let D be the function b 2  Aac of § 5, whose sign discriminates the roots of a
quadratic. In the present instance we have
D = (y + 8) 2 60={y(8VG0)}{2/(8 + V60)} (2).
Hence the turning values of y are
y = _8 V(60), and y= 8 + V(60).
If y has any value between
these, D is negative, and the
roots of (1) are imaginary.
Hence the algebraically less of
the two, namely,  8  V(60),
is a maximum ; and the
algebraically greater, namely,
 8 + V(60), a minimum.
The values of x correspond
iugtotheseare atonce obtained
from the equation x = {y + 8)/2.
They are x=  \J{15) and
x= + \/(15) respectively.
The reader should examine
carefully the graph of this
function, which lias a discon
tinuity when x=0 (see chnp. xv
corresponding values : —
»=«, 1, 0,
§ 5).
Fig. 7.
We have the following series of
+ 0,
y=—ao, 24, — 00, + oo ,
Hence the graph is represented by Fig. 7
+ 3,
0,
>3
<5'
>o,
0, +
+ 00,
+ 00.
464 GENERAL DISCUSSION OF chap.
Example 3.
To discuss generally the turning values of the function
_ a,r 2 + b.v + c
V ~a'x 2 + b'x + c' (1) '
The equation which gives the values of x corresponding to any given value
of y is
(a  a'y)x 2 + (b b'y)x + (c  c'y) = 0.
Let ~D = {bb'yf4{aa'i/){cc'y),
= (ft' 2  ia'c')y 2 + 2(2a'c + 2ac'  W)y + (Ir  lac),
= Ay 2 + By + C, say.
Then we have x=  (bb'y)±^D
2{aay) K '
The turning values of y are therefore given by the equation
A2/ 2 + By + C = (3).
I. If B 2 4AC>0, this equation will have real unequal roots, and there
will be two real turning values of y.
If A be positive, then, for real values of x, y cannot lie between the roots
of the equation (3). Hence the less root will be a maximum and the greater
a minimum value of y.
If A be negative, then, for real values of x, y must lie between the roots of
(3). Hence the less root will be a minimum and the greater a maximum
value of y.
II. IfB 2 4AC<0, the equation has no real root, and D has always the
same sign as A. In this case the sign of A must of necessity be positive ;
for, if it were not, there would be no real value of x corresponding to any
value of y whatever.
Hence there is a real value of x corresponding to any given value of y
whatever ; and y has no turning values.
III. If B 2  4AC = 0, we may apply the same general reasoning as in Case
II. The present case has, however, a special peculiarity, as we shall see im
mediately.
The criteria for distinguishing these three cases may be expressed in terms
of the roots a, /3 and a', 13' of the two functions axr + bxrC and a'x? + b'x + c',
and in this form they are very useful.
We have
B 2  4AC = 4(2«c' + 2a 'c  bb'f  4{b"  lac) (6' s  Aa'c'),
f2 i.^yCg4iUS^},
a' a a a J \a aj \a* a'
= 4aV 2 [2ct'/3' + 2a/3  (a + /S) (a' + /3') ] 2  (a  /3) 2 (a'  /3') 2 } ,
= 4a 2 a' 2 { 2a' 13' + 2a/3  (a + j8) (a' + /3')  (a  /3) (a'  /3') }
x {2a'/3' + 2ai8(o + j8)(a' + j8') + (aj8)(o'j8')},
= 16feV 2 (a  a') (a  /3') (/3  a') (/3  /3').
Hence it appears that the sign of B 2  4 AC depends merely on the sign of
E = (aa')(a/3')(/3a')(/3/3') (4)
xvi ir (ax~ + bx + c)l(a'r + b'x + c') 465
Since ft, b, c, a', b', c' are all real, the roots of an? + bx + c and of
a'x 2 + b'x + c', if imaginary, must be conjugate imaginaries. Hence, by
reasoning as in § 6, we see that, if the roots of ax 2 + bx + c, or of a'x" + b'x + c',
or of both, be imaginary, E is positive.
The same is true if the roots of either or of both of these functions be
equal.
Consider, next, the case where a, /3, a, /3' are all real and all unequal.
Since the sign of E is not altered if we interchange both a with a' and /3
with fi', or both a with /3 and a' with /3'. we may, without losing generality,
suppose that a is the algebraically least of the four, a, /3, a', /?', and that a' is
algebraically less than /3'. If we now arrange the four roots in ascending
order of magnitude, there are just three possible cases, namely, a, p, a', /3' ;
a, a', /3', /3 ; a, a', /3, /3'. In the first case, a a', a/3', /3a', /3/3' have
all negative signs ; in the second, two have negative signs, and two positive ;
in the third, three have negative signs, and one the positive sign. It is,
therefore, in the third case alone that E has the negative sign. The peculi
arity of this case is that each pair of roots is separated as to magnitude by
one of the other pair. We shall describe this by saying that the roots inter
lace.
Lastly, suppose E = 0. In this case one at least of the four factors, a  a',
/?  /3', (3  a', /3  j8', must vanish ; that is to say, the two functions ax 2 + bx + c
and a'x 2 + b'x + c' must have at least one root, and therefore at least one linear
factor in common.*
Hence, in this case, (1) reduces to
say. Hence we have
ft(*a)
J a'(xa') {0) >
a:a' + a'a a a(a'  a)
V = a ■«'(* a')" = a' + a^^') (6) "
From (6) it appears that y has a discontinuity when x = a', passing from
the value + co to  co , or the reverse, as X passes through that value ; but
that, for all other values of x, y either increases or decreases continuously as
x increases. Hence y has no real turning values in this case, unless we choose
to consider the value y = aja, which corresponds to x= ±°o, as a maximum
minimum value.
The graph in this case, supposing a/a', a, and a'  a to be both positive, is
like Fig. 8, where OA = a, OA' = a', OB=aJa',
To sum up —
Case I. occurs when the roots of either or of both of the functions
aa? + bx + c, a'x 2 + b'x + c' are imaginary or equal, and when all the roots
are real but not interlaced.
Case II. occurs when the roots of both quadratic functions are real and
interlaced.
* In the case where they have two linear factors in common, y reduces to
a constant, a case too simple to require any discussion.
VOL. I. 2 H
466
(ax* + bx + c)/(a'a? + b'x + c')
CHAP.
Case III. occurs when the two quadratic functions have one or both roots
in common. In this case y reduces to the quotient of two linear functions,
or to a constant, and has no maximum or minimum value properly so called.
In the above discussion we have assumed that neither a nor a' vanish ; in
other words, that neither of the two quadratic functions has an infinite root.
The cases where infinite roots occur are, however, really covered by the above
Y
B
^
X
M
A'
Fig. 8.
statements, as may be seen either by considering them as limits, or by work
ing out the expression for B 2  4AC in terms of the finite roots in the particular
instances in question.
In stating the above conclusions so generally as this, the student must
remember that one of the turning values may either become infinite or corre
spond to an infinite value of x ; otherwise he ma}' find himself at a loss in
certain cases to account for the apparent disappearance of a turning value.
A great variety of particular cases are included under the general case of
this example. If we put a' = 0, c' = 0, for instance, we have the special case
of Example 2.
As our object here is merely to illustrate methods, it will be sufficient to
give the results in two more particular cases.
Example 4.
To trace the variation of the function
_ a 2 7a: + 6
V ~x8x + lo
The quadratic for x in terms of y is
(lyy(78ij)x+(615y) = Q.
Hence
D = (78 2 /)24(l2/)(615 2 /) = 4{ 2 /(tV6)}{ya+V6)}.
Hence 7/2 \/6 and 7/2 + \/6 are maximum and minimum values of y re
spectively. The corresponding values of a; are given by
XVIII
GRAPHS FOR PARTICULAR CASES
467
X = is
r8y
12/
and are 9 + 2\/6 and 9  2\/6 respectively. We observe farther that y is dis
continuous when x = 3 and when x = 5 ; that when a; = + oo or = — oo , y = 1 ;
and that the other value of a for which ?/ = l is a:=9.
We have thus the following table of corresponding values : —
x=co, 0, +1, +30, +3 + 0, +41,
y=
+ 1, +"4, 0,

°°,
+ °°,
+ 59,
mill.
x=
+ 50, +5 + 0,
+ 6,
+ 9,
+ 139,
+ oo
y=
+ CO, CO,
o,
+ 1,
+ 105,
+ 1.
The graph has the
general form indicated in
Fig. 9, which is not
drawn to scale, but dis
torted in order to bring
out more clearly the
maximum point 15.
Example 5.
To trace the variation
of the function
The quadratic for x is
Y
y
i
i
i
max.
B
„ m "" ™
\
/ i x
M
1 "
x?5x+4
' x8x+ 15'
Fig. 9.
(ly)x 2 (58y)x + (i15y) = 0.
Here we find
D=4{(yi) 2 +2}.
Hence there are no real turning values.
The graph will be found to be as in Fig. 10.
Y
\
^__
1
X
Fig. 10.
468 MAXIMA AND MINIMA, METHOD OF INCREMENTS chap.
Example 6.
To find the turning values of z = x 2 + y 2 , given that cut?+bxy+cy 2 =l.
We have, since ax 2 + bxy + cy 2 — l,
x 2 + y 2 £ 2 + l
~ ~ ax' 2 + bxy + cy 2 a£ + b% + c'
where £ = ,r/y.
We have now to find the turning values of z considered as a function of f.
The quadratic for £ is
(azl)£ + bz£ + (czl) = 0.
Hence the turning values of z are given by
&V=4(oal)(csl),
that is, by
(b 2 4ac)z 2 +i(a + c)zi = 0.
The result thus arrived at constitutes an analytical solution of the well
known problem to find the greatest and least central radii (that is, the semi
axes) of the ellipse whose equation is ax 2 + bxy + cy 2 = 1 .
Remark. — The artifice used in this example will obviously enable us to
find the turning values of u=J{x, y), when <p(x, y) = c, provided /(.r, y) and
<p(x, y) be homogeneous functions of x and y whose degree does not exceed
the 2nd. Indeed it has a general application to all cases where f(x, y) and
<f>(x, y) are homogeneous functions ; the only difficulty is in discriminating
the roots of the resulting equation.
§ 12.] Examination of the Increment. — There is yet another
method which is very useful in discussing the variation of
integral functions. Suppose we give x any small increment,
h, then the corresponding increment of the function fix) is
f(x + h) f(x). If this is positive, the function increases when x
increases ; if it is negative, the function decreases when x increases.
The condition that x = a corresponds to a maximum value of f(x)
is therefore that, as x passes through the value a, f(x + h) f(x)
shall cease to he positive and begin to be negative, and for a
minimum shall cease to be negative and begin to be positive.
The practical application of the method will be best under
stood by studying the following example : —
Example.
To find the turning values of
y = x 3 9x 2 + 2ix + 3.
Let I denote the increment of y corresponding to a \ ery small increment,
h, of x ; then
I = (./• + hf  9(a! + h) 2 + 24(.c + h) + 8  a? + 9.r ;  24a  3,
= (3z 2 18a: + 24)A + (3x 9)/r + A :! .
XVIII
EXERCISES XXXVIII 469
Now, since for our present purpose it does not matter how small h may
be, we may make it so small that (3x  9)h" + h 3 is as small a fraction of
(3a; 2  18x+24)A as we please. Hence, so far as determining the sign of I is
concerned, we may write
I = (3x 2 18a; + 24)A.
Here h is supposed positive, hence the sign of I depends merely on the sign
of 3ar18a; + 24. Hence I will change sign when, and only when, x passes
through a root of the equation
3a; 2  18a: + 24 = 0.
Hence the turning values of y correspond to x = 2 and x = i.
Moreover, we have
l=8(x2)(x4)h.
Therefore, when a; is a little less than 2, I is positive ; and when x is a
little greater than 2, I is negative. Hence the value of y corresponding to
a' = 2 is a maximum.
In like manner we may show that the value of y corresponding to a;=4 is
a minimum.
Exercises XXXVIII.
(1. ) Find the limits within which x must lie in order that 8(.r 2  a 2 )  65xa
may be negative.
Trace the graphs of
(2.) y = x 2 5x + 6. (3.) y= Sx + l2x6.
(4.) y=  4a; 2 + 20a;  25.
Find the turning values of the following ; and discriminate between
maxima and minima : —
(5.) ae** + be~ kx . (6.) afx+a/(ax).
(7.) V(l+aO + V(l*) ( 8 ) xl + s/(x+l).
Trace the graphs of the following, and mark, in particular, the points
where the graph cuts the axes, and the points where y has a turning value : —
(9.) y = (a: 2 + 8a; + 16)/(a; 2 7a; + 12).
(10.) i/ = (a; 2 7a; + 12)/(a; 2 +8a' + 16).
(11.) y = (a; 2 + 8a; + 16)/(a; 2 6a;+9).
(12.) y = (a 2 10aj + 27)/(a; 2 8,r + 15).
(13.) y = (x 2 8x + 15)/(x 2 \0x + 27).
(14.) 2/ = (ar10a: + 27)/(a; 2 14a; + 52).
(15.) i/=(x 2 9a: + 14)/(a^ + 2a;15). (16.) y = (a; 2 +a; 6)/(se* 1).
(17.) y = (x 2 + 5x + Q)/{2x + 3). (18.) y = l/(x? + $x + 5).
(19.) y = (2a^ + a;6)/(2a; 2 +5a;12).
(20. ) Show that the algebraically greatest and least values of (a; 2 + 2x 2)/
(x 2 + 3a; + 5) are \/( 12 Al) an< i ~ V( 12 /H) J an( l n °d the corresponding values
of a;.
(21.) Show that (axb) (dzc)/(bxa)(cx d) may have all real values,
provided (a 2  b) (c  d 2 ) > 0.
(22.) Show that (ax 2 + bx + c)/(cx 2 + bx + a) is capable of all values if
470 EXERCISES XXXV1I1 CHAP, xvtn
b 2 >(a + c) 2 ; that there are two values between which it cannot lie if
(a + c) 2 >b 2 >lac; and that there are two values between which it must lie
if b 2 < iac (Wolstenholme).
(23. ) If ra >pb, tlien the turning value of [ax + b)/(2}x + r) 2 is ar/4p(ra ~pb).
Find the turning values of the following ; and discriminate maxima and
minima : —
(24.) (xl)(x3)jx 2 . (25.) (x3)/(x 2 + x3).
(26. ) l(ax + b) 2 + l'(a'x + b'f + l"(a"x + b") 2 .
(27. ) ax + by, given x 2 + y 2 = c 2 . (28. ) a 2 x 2 + b"y 2 , given x + y = a.
(29.) xy, given a 2 /x 2 + b 2 Jy 2 =l. (30.) x 3 y + x 2 y 2 + xy 3 , given xy = a 2 .
(31. ) ax 2 + 2hxy + by 2 , given Ax 2 + 2Hxy + By 2 = 1.
(32.) xy/^J(x 2 + y 2 ). (33.) (2x  1) (Bx  4) (x  3).
(34.) \l\Jx + \j\Jy, given x + y = c.
(35.) To inscribe in a given square the square of minimum area.
(36. ) To circumscribe about a given square the square of maximum area.
(37.) To inscribe in a triangle the rectangle of maximum area.
(38.) P and Q are two points on two given parallel straight lines. PQ
subtends a right angle at a fixed point 0. To find P and Q so that the area
POQ may be a minimum.
(39.) ABC is a rightangled triangle, P a movable point on its hypotenuse.
To find P so that the sum of the squares of the perpendiculars from P on the
two sides of the triangle may be a minimum.
(40.) To circumscribe about a circle the isosceles trapezium of minimum
area.
(41.) Two particles start from given points on two intersecting straight
lines, and move with uniform velocities u and v along the two straight lines.
Show how to find the instant at which the distance between the particles is
least.
(42.) OX, OY are two given straight lines ; A, B fixed points on OX ; P a
movable point on OY. To find P so that AP 2 + BP 2 shall be a minimum.
(43. ) To find the rectangle of greatest area inscribed in a given circle.
(44.) To draw a tangent to a given circle which shall form with two given
perpendicular tangents the triangle of minimum area.
(45.) Given the aperture and thickness of a biconvex lens, to find the radii
of its two surfaces when its volume is a maximum or a minimum.
(46. ) A box is made out of a square sheet of cardboard by cutting four
equal squares out of the corners of the sheet, and then turning up the flaps.
Show how to construct in this way the box of maximum capacity.
(47.) Find the cylinder of greatest volume inscribed in a given sphere.
(48.) Find the cylinders of greatest surface and of greatest volume in
scribed in a given right circular cone.
(49.) Find the cylinder of minimum surface, the volume being given.
(50.) Find the cylinder of maximum volume, the surface being given.
CHAPTEE XIX.
Solution of Arithmetical and Geometrical Problems
by means of Equations.
§ 1.] The solution of isolated arithmetical and geometrical
problems by means of conditional equations is one of the most
important parts of a mathematical training. This species of
exercise can be taken, and ought to be taken, before the student
commences the study of algebra in the most general sense. It
is chiefly in the applications of algebra to the systematic investi
gation of the properties of space that the full power of formal
algebra is seen. All that we need do here is to illustrate one or
two points which the reader will readily understand after what
has been explained in the foregoing chapters.
§ 2.] The two special points that require consideration in
solving problems by means of conditional equations are the
choice of variables, and the discussion or interpretation of the solution.
With regard to the choice of variables it should be remarked
that, while the selection of one set of variables in preference to
another will never alter the order of the system of equations on
whose solution any given problem depends, yet, as we have
already had occasion to see in foregoing chapters, a judicious
selection may very greatly diminish the complexity of the system,
and thus materially aid in suggesting special artifices for its
solution.
With regard to the interpretation of the solution, it is im
portant to notice that it is by no means necessarily true that
all the solutions, or even that any of the solutions, of the system
of equations to which any problem leads are solutions of the
472 INTERPRETATION OF THE SOLUTION chap.
problem. Every algebraical solution furnishes numbers which
satisfy certain abstract requirements ; but these numbers may
in themselves be such that they do not constitute a solution of
the concrete problem. They may, for example, be imaginary,
whereas real numbers are required by the conditions of the con
crete case ; they may be negative, whereas positive numbers are
demanded ; or (as constantly happens in arithmetical problems
involving discrete quantity) they may be fractional, whereas
integral solutions alone are admissible.
In every concrete case an examination is necessary to settle
the admissibility or inadmissibility of the algebraical solutions.
All that we can be sure of, a priori, is that, if the concrete
problem have any solution, it will be found among the algebraical
solutions ; and that, if none of these are admissible, there is no
solution of the concrete problem at all.
These points will be illustrated by the following examples.
For the sake of such as may not already have had a sufficiency
of this kind of mental gymnastic, we append to the present
chapter a collection of exercises for the most part of no great
difficulty.
Example 1.
There are three bottles, A, B, C, containing mixtures of three substances,
P, Q, 11, in the following proportions : — ■
A, aP + «'Q + «"R;
B, bY + b'Q + b"R ;
C, cP+c'Q + c"R.
It i3 required to find what proportions of a mixture must be taken from A,
B, C, in order that its constitution may be dP + d'Q + d"R (Newton, Arithmdica
Universalis).
Let x, y, z be the proportions in question ; then the constitution of the
mixture is
{ax + by + cz) P + (a'x + b'y + c'z)Q + [a"x + b"y ■* c'z)R.
Hence we must have
ax + by + czd, a'x + b'y + c'z = d', a"x + b"y + c"z = d".
The system of equations to which we are thus led is that discussed in
chap, xvi., § 11, with the sole difference that the signs of d, d', d" are re
versed.
If, therefore, ab'c"  ab"c' + bc'a"  be" a' + ca'b"  ca"b' 4= 0, we shall obtain a
unique finite solution. Unless, however, the values of x, y, z all come out
positive, there will be no proper solution of the concrete problem. It is in
XIX
EXAMPLES 473
fact obvious, a priori, that there are restrictions ; for it is clearly impossible,
for instance, to obtain, by mixing from A, P., C, any mixture which shall
contain one of the substances in a proportion greater than the greatest in which
it occurs in A, B, or C.
Example 2.
A farmer bought a certain number of oxen (of equal value) for £350. He
lost 5, and then sold the remainder at an advance of £6 a head on the original
price. He gained £365 by the transaction ; how many oxen did he buy ?
Let x be the number bought ; then the original price in pounds is 350/a:.
The selling price is therefore 350/a + 6. Since the number sold was as 5, we
must therefore have
(x5)(—+6j 350=365.
This equation is equivalent to
6a; 2  395a; 1750 = 0,
which has the two roots a; = 70 and x = 25/6. The latter number is in
admissible, both because it is negative and because it is fractional ; hence the
only solution is a; = 70.
Example 3.
A'OA is a limited straight line such that OA = OA' = a. P is a point in
OA, or in OA produced, such that OP=;?. To find a point Q in A'A such
that PQ 2 = AQ.QA'. Discuss the different positions of Q asp varies from
to its greatest admissible value.
Let OQ = a?, a* denoting a positive or negative quantity, according as Q is
right or left of O. Then PQ=±(ajj>), A'Q = a + x, AQ=az; and we have
in all cases
(xpf = (a + x)(ax) = a 2 x (1).
Hence x 2 px + h(p 2 a n ) = (2).
The roots of (2) are ^±V(i« 2 ~ lP%
These roots will be real if pr<2a ; that is to say, confining ourselves to
positive values ofp, \{p<\j2a.
From (1) we see that in all cases where x is real it must be numerically
less than a. Hence Q always lies between A' and A.
When p = 0, the roots of (2) are ±a\J2 ; that is to say, the two positions of
Q are equidistant from O.
So long as p is <a, ftp 3  a?) will be negative, and the roots of (2) will be
of opposite sign ; that is to say, the two positions of Q will lie on opposite
sides of O. Since the sum of the two roots is p( = OV), if QiQ 2 be the two
positions of Q, the relative positions of the points will be as in Fig. 1, where
OQ 2 = PQi.
i n m
A' Q 2 PQiA
Fig. 1.
When ,p = a, Qa moves up to O, and Qi up to A.
If p>a, then both roots are positive, and the points will be as in Fig. 2,
where OQ2=QiP.
474 EXAMPLES
CHAP.
I, I IMI II
A' O Q,>C<M PB
Fig. 2.
If OB = \/2a, then B is the limiting position of P for which a solution of
the problem is possible. When P moves up to B, Qi and Q 2 coincide at C
(OC being £OB).
Example 4.
To find four real positive numbers in continued proportion such that their
sum is a and the sum of their squares b 2 .
Let us take for variables the first of the four numbers, say x, and the
common value, say y, of the ratio of each number to the preceding. Then
the four numbers are x, xy, xy 2 , xy 3 . Hence, by our data,
x + xy + xy 2 + xy 3 = a,
x 2 + x 2 y 2 + xY + xhf = b 2 ;
that is to say, x{l+y){l+y 2 )a (1),
x 2 (l + y 2 )(l + i/) = b 2 (2).
From (1) we derive x 2 (l +y)\l + y 2 )" = ar (3),
and from (2) and (3), rejecting the factor y 2 +l, which is clearly irrelevant,
we derive
a 2 (l+7/) = b%l+y 2 )(l+y) 2 (4).
The equation (4) is a reciprocal biquadratic in y, which can be solved by the
methods of chap. xvii. , § 8.
For every value of y (1) gives a corresponding value of x.
The student will have no difficulty in showing that there will be two
proper solutions of the problem, provided a be > b. Since, however, the two
values of y are reciprocals, and since x(l +y) (1 + y 2 ) = xy 3 {l + l/y) (1 + l/i/ 2 ),
these two solutions consist merely of the same set of four numbers read for
wards and backwards. There is, therefore, never more than one distinct
solution.
Newton, in his Arithmctica Universalis, solves this problem by taking as
variables the sum of the two mean numbers, and the common value of the
product of the two means and of the two extremes. He expresses the four
numbers in terms of these and of a and b, then equates the product of the
second and fourth to the square of the third, and the product of the first and
third to the square of the second. It will be a good exercise to work out the
problem in this way.
Example 5.
In a circle of given radius a to inscribe an isosceles triangle the sum of
the squares of whose sides is 2b 2 .
Let x be the length of one of the two equal sides of the triangle, 2y the
length of the base.
If ABC be the triangle, and if AD, the diameter through A, meet BC in
E, then, since ABD is a right angle, we have AB 2 = AD . AE. Hence
x 2 = 2a s /(. l "y 2 ) (1).
XIX
EXAMPLES 475
Again, by the conditions of the problem, we have
2x 2 + 4y 2 =2i 2 ,
that is, x 2 + 2y 2 =b 2 (2).
From (1) and (2) we derive
x i 6aW + 2a 2 b' i = (3).
The roots of (3) are
*V{«y( l 5)};
and the corresponding values of y are given by (2).
The necessary and sufficient condition that the values of x and of y be real
is that b<3a/\/2. When this condition is satisfied, there are two real posi
tive values of ,r, and if b>2a there are two corresponding real positive values
ofy.
It follows from the above that, for the inscribed isosceles triangle the
sum of the squares of whose sides is a maximum, b = Ba/\/2. Corresponding
to this we have x=\J2>a, 2y=\/3a ; that is to say, the inscribed triangle, the
sum of the squares of whose sides is a maximum, is equilateral, as is well
known.
Example 6.
Find the isosceles triangle of given perimeter 2p inscribed in a circle of
radius a ; show that, if 2p be less than 3\/3, and greater than 2a, there are
two solutions of the problem ; and that the inscribed triangle of maximum
perimeter is equilateral.
Taking the variables as in last example, we find
x* = 2a^(x'>y") (1),
x + y=p (2).
Hence a 4  8a 2 px + idy" = (3).
We cannot reduce the biquadratic (3) to quadratics, as in last example ;
but we can easily show that, provided p be less than a certain value, it has
two real positive roots.
Let us consider the function
y = x*8a' i px+iay (4);
and let I be the increment of y corresponding to a very small positive incre
ment (h) of x. Then we find, as in chap, xviii., § 12, that
l = 4(.^2ap)h (5).
Hence, so long as x i <2a"p, I is negative ; and when x 3 >2a 2 p, I is positive.
Hence, observing that y— + <x> when x— ±oo , we see that the minimum value
ofy corresponds to x= ^/(2arp), and that the graph of (4) consists of a single
festoon. Hence (3) will have two real roots, provided the minimum point be
below the araxis ; that is, provided y be negative when x= ^/(2a 2 p) ; that is,
provided
4n])U — «s +p§ I
476 EXERCISES XXXIX chap.
be negative ; that is, provided 2p<Z\J%a. It is obvious that both the roots
are positive ; for when x = we have y = 4a 2 p 2 , which is positive ; hence the
graph does not descend below the axis of x until it reaches the righthand
side of the axis of y.
From the above reasoning it follows that the greatest admissible perimeter
is SsjSa. When 2p has this value, the minimum point of the graph lies on the
axis of x, and x— £/(2a 2 p>) = */(3\/3a 3 )= \JBa corresponds to two equal roots
of (3). The corresponding value of 2y is given by 2y = 2p  2x = B\/3a  2\/3a
= \jZa; in other words, the inscribed isosceles triangle of maximum peri
meter is equilateral.
Another interesting way of showing that (3) has two equal roots is to dis
cuss the graphs (referred to one and the same pair of axes) of the functions
y = x*, and y = 8apx  iarp 2 .
These can be easily constructed ; and it is obvious that the abscissae of their
intersections are the real roots of (3).
Exercises XXXIX.
(1.) How long will an up and a down train take to pass each other, each
being 44 yards long, and each travelling 30 miles an hour ?
(2.) Diophantus passed in infancy the sixth part of his life, in adolescence
a twelfth, then he married and in this state he passed a seventh of his life
and five years more. Then he had a son whom he survived four years and
who only reached the half of his father's age. How old was Diophantus when
he died ?
(3.) A man met several beggars and wished to give 25 pence to each ; but,
on counting his money, he found that he had 10 pence too little for that ; and
then made up his mind to give each 20 pence. After doing this he had 25
pence over. What had he at first, and how many beggars were there ?
(4.) Two bills on the same person are sent to a banker, the first for £580
payable in 7 months, the second for £730 payable in 4 months. The banker
gives £1300 for the two. What was the rate of discount, simple interest
being allowed in lieu of discount ?
(5. ) A basin containing 1 200 cubic metres of water is fed by three fountains,
and can be emptied by a discharging pipe in 4 hours. The basin is emptied
and the three fountains set on ; how long does it take to fill with the dis
charging pipe open ? — given that the three fountains each running alone
would fill the basin in 3, 6, and 7 hours respectively.
(6. ) If I subtract from the double of my present age the treble of my
age 6 years ago, the result is my present age. What is my age ?
(7.) A vessel is filled with a mixture of spirit and water, 70% of which is
spirit. After 9 gallons is taken out and the vessel filled up with water, there
remains 58 J°/ °' spirit : find the contents of the vessel.
(8. ) Find the time between 8 and 9 o'clock when the hour and minute
hands of a clock are perpendicular.
(9. ) A and B move on two paths intersecting at O. B is 500 yards short
XIX
EXERCISES XXXIX 477
of when A is at ; in two minutes they are equidistant from 0, and in eight
minutes more they are again equidistant from 0. Find the speeds of A
and B.
(10.) I have a sum to buy a certain number of nuts. If I buy at the rate
of 40 a penny, I shall spend 5d. too much, if at the rate of 50 a penny,
lOd. too little. How much have I to spend ?
(11.) If two numbers be increased by 1 and diminished by 1 respectively,
their product is diminished by 4. If they be diminished by 1 and increased
by 2 respectively, their product is increased by 16. Find the numbers.
(12.) A is faster than B by p miles an hour. He overtakes B, who has
a start of h miles, after a run of q miles. Required the speeds of A and B.
(13.) To divide a given number a into two parts whose squares shall be
in the ratio m: 1.
(14.) Four apples are worth as much as five plums ; three pears as much as
seven apples ; eight apricots as much as fifteen pears ; and five apples sell for
twopence. I wish to buy an equal number of each of the four fruits, and to
spend an exact number of pence ; find the least sum I can spend.
(15.) A man now living said he was x years of age in the year x 2 . What
is his age and when was he bom .'
Remark on the nature of this and the preceding problem.
(16.) OABCD are five points in order on a straight line. If OA = a,
OB = 6, OC = c, OB = d, find the distance of P from O in order that
PA : PD = FB : PC. (Assume P to lie between B and C.)
(17.) A man can walk from P to Q and back in a certain time at the rate
of 3£ miles an hour. If he walks 3 miles an hour to and 4 miles an hour back,
he takes 5 minutes longer ; find the distance PQ.
(18.) A starts to walk from P to Q half an hour alter B ; overtakes B mid
way between P and Q ; and arrives at Q at 2 P.M. After resting 7i minutes,
he starts back and meets B in 10 minutes more. When did each start
from P ?
(19.) At two stations, A and B, on a line of railway the prices of coals are
£p per ton and £q per ton respectively. If the distance between A and B be
d, and the rate for the carriage of coal be £r per ton per mile, find the distance
from A of a station on the line at which it is indifferent to a consumer whether
he buys coals from A or from B.
(20.) A merchant takes every year £1000 out of his income for personal
expenses. Nevertheless his capital increases every year by a third of what
remains ; and at the end of three years it is doubled. How much had he at
first?
(21.) A takes m times as long to do a piece of work as B and C together ;
B n times as long as C and A together ; C x times as long as A and B together.
Find a;; and show that l/(se+l) + l/(m+l)+l(»+l)=l.
(22.) The total increase in the number of patients in a certain hospital in
a certain year over the number in the preceding year was 2%. In the
number of outpatients there was an increase of 4% ; but in the number of
inpatients a decrease of 11 %• Find the ratio of the number of out to the
number of inpatients.
478 EXERCISES XXXIX CHAP.
(23.) The sum of the ages of A and B is now 60; 10 years ago their ages
were as 5 to 3. Find their ages now.
(24.) Divide 111 into three parts, so that onethird of the first part is
greater by 4 than onefourth of the second, and less by 5 than onefifth of the
third.
(25.) In a hundred yards' race A can beat B by \" ; but lie is handicapped
by 3 yards, and loses by 1 T V yards. Find the times of A and B.
(26.) A and B run a mile, and A beats B by 100 yards. A then runs with
C, and beats him by 200 yards. Finally, B runs with C ; by how much does
he beat him ?
(27. ) A person rows a miles down a river and back in t hours. He can
row b miles with the stream in the same time as c miles against. Find the
times of going and returning, and the velocity of the stream.
(28. ) A mixture of black and green tea sold at a certain price brings a
profit of 4°/ on the cost price. The teas sold separately at the same price
would bring 5% and 3% profit respectively. In what proportion were the
two mixed ?
(29. ) If a rectangle were made a feet longer and b feet narrower, or a' feet
longer and V feet narrower, its area would in each case be unaltered. Find
its area.
(30.) Two vessels, A and B, each contain 1 oz. of a mixture of spirit and
water. If 1/mth oz. of spirit be added to A and l/?ith oz. of spirit to B, or if
l/;ith oz. of water be added to A and 1/mth oz. of water to B, the percentages
of spirit in A and B in each case become equal. What percentage of spirit
is there in each ?
(31.) A winemerchant mixes wine at 10s. per gallon with spirit at 20s. per
gallon and with water, and makes 25% profit by selling the mixture at lis. 8d.
per gallon. If he had added twice as much spirit and twice as much water, he
would have made the same profit by selling at lis. 3d. per gallon. How much
spirit and how much water does he add to each 100 gallons of wine ?
(32. ) Find the points on the dial of a watch where the two hands cross.
(33.) Three gamesters agree that the loser shall always double the capital
of the two others. They play three games, and each loses one. At the end
they have each £«. What had they at first ?
(34. ) A cistern can be filled in 6 hours by one pipe, and in 8 hours by
another. It was filled in 5 hours by the two running partly together and
partly separately. The time they ran together was twothirds of the time
they ran separately. How long did each run ?
(35.) A horse is sold for £24, and the number expressing the profit per
cent expresses also the cost price. Find the cost price.
(36.) I spent £18 in cigars. If I had got one box more for the money,
each box would have been 5s. cheaper. How many boxes did I buy ?
(37.) A person about to invest in 3% consols observed that, if the price
had been £5 less, he would have received ^°/ more interest on his money.
Find the price of consols.
(38.) Out of a cask containing 360 quarts of pure alcohol a quantity is
drawn and replaced by water. Of the mixture a second quantity, 84 quarts
XIX
EXERCISES XXXIX 479
more than the first, is drawn and replaced by water. The cask now contains
as much alcohol as water. "What quantity was drawn out at first ?
(39.) Find four consecutive integers such that the product of two of them
may be a number which has the other two for digits.
(40.) The consumption of an important commodity is found to increase as
the square of the decrease of its price below a certain standard price (j>). If
the customs' duty be levied at a given percentage on the value (a) of the
commodity before the duty is paid, show that, provided the rate be below a
certain limit, there are two other rates which will yield the same total
revenue, and determine the rates which will yield the greatest and least
revenues.
(41.) A number has two digits, the sum of the squares of which is 130.
If the order of the digits be reversed the number is increased by 18. Find
the number.
(42.) Three numbers are in arithmetical progression. The square of the
first, together with the product of the second and third, is 16 ; and the square
of the second, together with the product of the first and third, is 14.
(43. ) To find three numbers in arithmetical progression such that their sum
is 2a, and the sum of their squares 46 2 .
(44.) The sides of a triangle are the roots of x 3 ax 2 + bxc = 0. Show
that its area is \ \J {a( 4ab a 3  8c) } .
(45. ) The. hypotenuse of a rightangled triangle is h, and the radius of
the inscribed circle r. Find the sides of the triangle. Find the greatest
admissible value of r for a given value of h.
The following are from Newton's Arithmetica Universalis, q.v., pp. 119
ct scq. : —
(46.) Given the sides of a triangle, to find the segments of any side made
by the foot of the perpendicular from the opposite vertex.
(47.) Given the perimeter and area of a rightangled triangle, to find the
hypotenuse.
(48.) Given the perimeter and altitude of a rightangled triangle, to find
its sides.
(49.) The same, given the hypotenuse and the sum of the altitude and the
two sides.
(50.) Find the sides of a triangle which is such that the three sides, a, b, c,
and the perpendicular on a form an arithmetical progression.
(51.) The same, the progression being geometric.
(52.) To find a point in a given straight line such that the difference of
its distances from two given points shall be a given length.
CHAPTER XX.
Arithmetic and Geometric Progressions and the
Series allied to them.
§ 1.] By a series is meant the sum of a number vf terms formed
according to some common law.
For example, if f(n) be any function of n whatsoever, the
function
/(l) +/(2) +/(3) + . . .+/(/•) + . . .+/(») (1)
is called a series.
/(I) is called the first term; /(2) the second term, &c; and/(r)
is called the rth, or general, term.
For the present we consider only series which have a finite
number n of terms.
As examples of this new kind of function, let f(n) — n, then we have the
series
1+2 + 3 + . . .+% (2) ;
let f(n) — I /(a + bn), and we have the series
a + b + ^+T2 + a + b3 + • • • + ^Ui (3) ;
let f(n) — \Jnl{2 \Jn), and we have the series
2 VI 2 V2 2V3 ' ' 2V»
and so on.
It is obvious that when the nth term of a series is given we
can write down all the terms by simply substituting for n 1, 2,
3, . . . successively.
Thus, if the ?>th term be n + 2n, the series is
(l' ! + 2.1) + (2 2 + 2.2) + (3 , + 2.3) + . . .+(» 8 +2tt),
or 3 + 8 + 15 + . . .+{n 2 + 2n).
It is not true, however, that when the first few terms are
CHAT\ xx MEANING OF SUMMATION 481
given we can in general find the nth term, if nothing is told us
regarding the form of that term. This is sufficiently obvious
from the second form in which the last series was written ; for
in the earlier terms all trace of the law of formation is lost.
If we have some general description of the nth term, it may
in certain cases be possible to find it from the values of a certain
number of particular terms. If, for example,, we were told that
the nth term is an integral function of n of the 2nd degree,
then, by chap, xviii., § 7, we could determine that function if the
values of three terms of the series of known order were given.
§ 2.] If we regard the series
/(l)+/(2)+. . .+/(»)
as a function of n, and call it <f>(n), it has a striking peculiarity,
shared by no function of n that we have as yet fully discussed,
namely, that the number of terms in the function <$>(n) depends on the
value of its variable. For example,
</>(l)=l, 0(2) = 1 + 2, <£(3)=l + 2 + 3,
and so on.
It happens in certain cases that an expression can be found
for (f>(n) which has not this peculiarity ; for example, we shall
show presently that
l2 2 02 a n (n + 1) (2n+ 1)
1 +2 +3 + . . . +n " = * '? '.
o
On the left of this equation the number of terms is n ; on the
right we have an ordinary integral function of n, the number of
terms in which is independent of n, and which is therefore called
a closed function of n.
When, as in the example quoted, toe can find for the sum of a
series an expression involving only known functions and constructed by
a fixed number of steps, then the series is said to admit of summation ;
and the closed expression in question is spoken of as the sum, par
excellence, of the series.
The property of having a sum in the sense just explained is
an exceptional one ; and the sum, where it exists, must always
be found by some artifice depending on the nature of the series.
What the student should endeavour to do is to group together,
VOL. I 2 I
482 ARITHMETIC SERIES chap.
and be sure that he can recognise, all the series that can be
summed by any given artifice. This is not so difficult as might
be supposed ; for the number of different artifices is by no means
very large.
In this chapter we shall discuss two very important cases,
leaving the consideration of several general principles and of
several interesting particular cases to the second part of this work.
SERIES WHOSE WTH TERM IS AN INTEGRAL FUNCTION OF n.
§ 3.] An Arithmetic Series, or an Arithmetic Progression, as it
is often called, is a series in which each term exceeds the pre
ceding by a fixed quantity, called the common difference. Let
a be the first term, and b the common difference ; then the terms
are a, a + b, a + lb, a + 3b, &c, the nth term being obviously
a + (n  1 )b.
Here a and b may be any algebraical quantities whatsoever,
the word " exceed " in the definition being taken in the algebraical
sense.
Since the nth. term may be written (a — b) + bn, where a  b
and b are constants, we see that the nth. term of an arithmetical
series is an integral function of n of the 1st degree. Such a
series is therefore the simplest of the general class to be con
sidered in this section.
The usual method of summing an A.P. is as follows. Let
2 denote the sum of n terms, then
2= a+ {a + b) + {a + 2b}+. . .+ {a + (nl)b}.
If we write the terms in the reverse order, we have
2= {a+(nl)b} + {a + {n2)b}+ {a + (n3)b} +. , . + a.
If we now add, taking the pairs of terms in the same vertical
line together, we find
2S= {&* + (» 1)6}+ {2a + {nl)b} + {2a + (nl)b}+. . . + {2a + (nl)b}.
Hence, since there are n terms,
^ = ~{2a + (nl)b} (1).
This gives 2 in terms of n, a, b.
xx EXAMPLES 483
If we denote the last term of the series by I, we have
l = a + (nl)b. Hence
^ a + 1 /«\
2 = »y (2).
That is to say, the sum of n terms of an A. P. is n times the
average of the first and last terms, a proposition which is con
venient in practice.
Example 1.
To sum the arithmetical series 5 + 3 + 113 . . . to 100 terms. Here
a — 5 and b=  2. Hence
S=a^{2x5+(1001)(2)} >
= 50(10198),
=  9400.
Example 2.
To find the sum of the first n odd integers.
The nth. odd integer is 2n  1. Hence
2 = 1 + 3 + 5+ . . . +(2»l),
l+(2nl)
= n 2 '
= n 2 .
It appears, therefore, that the sum of any number of consecutive odd integers,
beginning with unity, is the scpuare of their number. This proposition was
known to the Greek geometers.
Example 3.
Sum the series 12 + 34 + 5 . . . to n terms. First suppose n to be
even, —2m say. Then the series is
2 = 12 + 34+. . . + (2m I) 2m,
= 1+3+ . . . +(2to1)
 2  4  . . . 2m.
In each line there are m terms. The first line has for its sum m 2 , by Example 2.
The second gives m(2 + 2m)/2, that is, m(m + l). Hence
2 = wi 2 m(m + l)= m= — ■=.
Next suppose n to be odd, =2»i 1, say.
Then we have
2 = 12 + 34+ . . . +(2ml).
To find the sum in this case, all we have to do is to add 2m to our former
result. We thus find
2 = 2toto = wi,
_ n+\
2 •
This result might be obtained even more simply by associating the terms of
the given series in pairs.
484 GENERAL INTEGRAL SERIES chap.
§ 4.] The artifice of § 3 will not apply to the case where the
nth term of a series is an integral function of n of higher degree
than the 1st. We proceed, therefore, to develop a more general
method.
Let the nth term of the series be
p n r +p 1 n r ~ 1 +p a n r ~ 2 + . . . +p r (1),
where_p , p l} p. 2 , . . ., p r are independent of n.
And let us denote the sums of the first, second, third, . . .,
rth powers of the first n integral numbers by n s 1} n s 2 , n s a , . . .,
n s r ; so that
jjOj — x r U T <J T . . . T l*j
n * 2 =l 2 + 2 2 + 3 2 + . . . +n\
n . S 3=l 3 +2 3 +3 3 + . . . +n 3 ,
and so on.
If 2 denote the sum of n terms of the series whose rath term
is (1), we have,
2=p l r +p l V 1 +p 2 l r  2 + . . . +p r ,
+ 2V2 r +p l 2 r ~ 1 +p 2 2 r  2 + . . . +p r ,
+ p y+p l 3 r  l +p 2 3 r ~ 2 + . . . +p r ,
+ p n r +p i 7i r ~ 1 +p 2 n r ~ 2 + . . . +p r .
Hence, adding in vertical columns, we have
2 = 7WV + Pi n S r  i + P» n Sr» + ■ • ■ + Wr (2).
From this formula we see that we could sum the series whose
general term is (1) if oidy we knew the sums of the first, second,
third, . . ., 7th powers of the first n integers.
These sums can be calculated successively by a uniform
process, as we shall now show.
§ 5.] To calculate n s,.
If in the identity (,c + 1 )'  x' = 2x +• 1 we put successively
z = n, x  n  1 , . . ., x = 2, x  1 , we have the following equations —
(jh l) 2  n' = 2w +1,
n' (n~l) 2 = 2(n 1)41,
3 e 2 8 =2.2 t 1,
2 2  1 2 = 2 1 +1.
XX
SUMMATION OF l r + 2 r + . . . + n r 485
If we add all these equations, the terms on the left mutually
destroy each other, with the exception of two, which give
(ft + l) 2  1 ; and those on the right, added in vertical columns,
give 2 n s l + n. Hence
(n+l) 2 l = 2 n s 1 + ft (1).
From this we have
2 n s 1 = (ft+l) 2 >+l), .
= (ft + 1)«,
wOi
ft(ft + 1)
(2);
2
a result which we might have obtained by the method of § 3,
for 1 + 2 + . . . + n is an A.P.
Cor. The sum of the first powers of the first n integers is an
integral function of n of the 2nd degree.
§ 6.] To calculate n s 2 .
In the identity (x + l) 3  z* = 3x 2 + Sx + 1 put successively
x = n, x = nl, . . ., x = 2, x=l, and we have
(ft + l) 3  ft 3 = 3ft 2 +Sn +1,
ft 3 (?il) 3 = 3(ftl) 2 + 3(ftl)fl,
3 3 2 3 = 3.2 2 +3.2 +1,
2 3 l 3 =3.1 2 +3.1 +1.
Hence, adding all these equations, we have
(n+lfl = S n s 2 +S n s l + n (1).
Using in (1) the value already found for „*„ we have
3 n s 2 = (ft+l) 3 n(ft+l)(ft + l),
= ^{2(ft + l) 2 3, i 2},
Hence
= — — (2n* + n).
_n(n + l)(2ft+ 1)
~6~
(2)
Cor. The sum of the squares of the first n integers is an integral
function of n of the 3rd degree.
486 SUMMATION OF V + 2 r + . . . + n f chap.
§ 7.] To calculate n s 3 .
In the identity (x + l) 4  x* = 4x 3 + 6x 2 + Ax + 1 put success
ively x = n, x — n  1, . . ., x— 2, x= I; add the n equations so
obtained, and we find, as before,
(n + l) 4  1 = 4„s 8 + 6„s 2 + 4 n s, + »,
or (ra + l) 4 (%+ l) = 4 n s a + 6 n Sj + 4 n s, (1).
Using the values of n s 2 and n Sj already found, we have
4 w s 3 = n(n +l)(n 2 + 3n+ 3) n(n + 1) (2w + 1)  2»(w + 1 ),
= »i(w + 1) (n 2 + 3n + 3  2rc  1  2),
= n 2 (?i + If.
Hence
A= «^i)' (3) .
Cor. 1. w 5 3 zs aw integral function of n of the 4th degree.
Cor. 2. T/je swm o/ <Ae cubes of the first n integers is the square
of the sum of their first powers.
§ 8.] Exactly as in § 7 we can show that
(n + l) 5  (n + 1) = 5 n s t + 10 n s 3 + I0 n s, + 5 n s, (1);
and from this equation, knowing n Si, n s 2 , n s 3 , we can calculate
n s 4 . The result is
_ n(n + 1) (6n 3 + 9?i 2 + n  1) 9 v
wS< ~ " "30  W
§ 9.] This process may be continued indefinitely, and the
functions „s 1} n s 2 , . . ., n s r _ l . . . calculated one after the other.
Suppose, in fact, that n s u n s 2 , . . ., n s r . x had all been cal
culated. Then, just as in §§ 58, we deduce the equation
(n + l) r+1  (n + 1) = r+i C in s r + r+1 C 2n s r _, +. . . + r+1 C rn S, (1),
where r +iC,, r +iC 2 , &c, are the binomial coefficients of ther + 1th
order.
The equation (1) enables us to calculate n s r .
Cor. 1. n s r is an integral function of n of the r + \th degree, so
that we may write
n s r = q n r+1 + q^ r + qjf 1 + . . . + &■+,
and it is obvious from (1) that
i )
XX SUM OF ANY INTEGRAL SERIES 487
1 1
q ° ~ r+1 C,  r + 1 *
Cor. 2. n s r is divisible by n(n + 1), so that we may write
[n r  1 1
n s r = n(n + 1)J  — +p l n r * + p. 2 n r ~ s + . . . +^ r _, V ;
for this is true when r = 1, r = 2, r = 3, r = 4 ; hence it must be
true for r = 5, for we have
(n + 1)'  (n + 1) = 6 C in s 5 + 6 C 2n s 4 + 6 C 3n s s + 6 C 27i s a + AiAi
and (■« + l) 6  (/i + 1) is divisible by »(% + 1) ; and so on.
§ 10.] We can now sum any series whose Tith term is re
ducible to an integral function of n. By § 4 and § 9, Cor. 1,
we see that the sum of n terms of any series whose nth term is an
integral function of n of the rth degree, is an integral function of n of
the r + \th degree. We may, therefore, if we choose, in summing
any such series, assume the sum to be A?i ,,+1 + ~Bn r + . . . + K ;
and determine the coefficients A, B, . . ., K by giving particular
values to ?7. If S,, S 2 , . . ., S r+2 be the sums of 1, 2, . . ., r + 2
terms of the series, then it is obvious, by Lagrange's Theorem,
chap, xviii., § 7, that the sum is
r + 2 s (n  1 ) ( n  2) . . . (tt.  s + 1) (n  s  1) . . . (n  r  2)
\ ' \s~\)(s2) 1 ( 1) .. . (sr 2)'
The following are a few examples : —
Example 1.
To sum the series 1, = a + (a + b) + (a + 2b) + . . . + {a + {nl)b}.
The ?ith term is (ab) + nb.
The n  1th term is (a  b) + (n l)b.
The 2nd term is (ab) + lb.
The 1st term is (a  b) + lb.
Hence 2 = (a b)7i + b n si,
= (a  V) n + b~ — ,
rn
as was found in § 3.
488 EXAMPLES
CHAP.
Example 2. 2 = l 2 + 3 2 + 5 2 + to n
terms.
The nth. term is (2m  1 ) 2 = 4 m 2  in + 1 .
Hence 2 = 4?i 2  in
+ 1
+ 4(m1) 2 4(m
■1) + 1
+ 4.2 2 4.2
+ 1
+ 4.1 2 4.1
+1.
Hence, adding in vertical columns, we have
2 = 4„s 2  4„si + n,
?i(m + 1)(2m + 1)
6
4^ 2 +1 Vn,
_(2;i1)m(2m+1)
3
Example 3.
2 = 2.3.4 + 3.4.5 + 4.5.6+. . . to n terms.
The Mth term is (m + 1) (?i + 2) (m + 3) = m 3 + 6?i 2 + 11m + 6.
Hence 2 = „s 3 + 6„S2+11„Si + 6m,
=  (n 4 + 10m 3 + 35m 2 + 50m).
Example 4.
A wedgeshaped pile of shot stands on a rectangular base. There are m
and n shot respectively in the two sides of the lowest rectangular layer, m  1
and m  1 in the two sides of the next rectangular layer, and so on, the upper
most layer being a single line of shot. Find the whole number of shot in the
pile, m being greater than n.
The number in the lowest layer is mil ; in the next (m  1) (n  1) ; in the
next {in  2) (m  2), and so on ; the number in the last layer is {inn\)
[n  m  1 ), that is, (m n + 1 ).
Hence we have to sum the series
S=m?i+(j»l)(Ml) + (m2)(»2)+. . . +(mnl)(nn 1),
in which there are n terms.
The rth term of the series is (mrl)(nrl), that is, (m+lr)
(M + lr), that is, (wi + 1)(m + 1) (m + M + 2)? + r 2 .
Hence we may write the series as follows : —
2= (wi + 1)(m + 1) (?>i + m + 2)m +m 2
+(m+l)(»+l) (mi + m + 2)(?i1) +(m1) 2
+ (wt + l)(/i + l) (m + M + 2)2 +2 2
+(m+l)(m+l) (??i + m + 2)1 +1 2 ,
= n{m + l)(n + l)(m + n + 2)„si + n « 2 >
= {in + l)n(n+l)§{m + n+2)n{n+l) + l / t(n+~l)(2n + l),
= ln(n+l){Smn+l).
Remark. — In working examples by this method the student must be care
ful to see that the scries is complete ; in other words, that there are exactly
n terms, all formed according to the same law. If any terms are wanting, or
if there are redundant terms, allowance must be made by adding or subtract
ing terms, as the case may be.
XX
GEOMETRIC SERIES 489
SERIES WHOSE ftTH TERM IS THE PRODUCT OF AN INTEGRAL
FUNCTION OF ft AND A SIMPLE EXPONENTIAL FUNCTION OF ft.
§ 11.] The typical form of the ftth term in the class of series
now to be considered is
(p n g +p 1 n s ~ 1 + . • • + P& n ,
where p , p u . . ., p 8 , r are all independent of n, and s is any
positive integer.
The simplest case is that in which the integral function re
duces to a constant. The ftth term is then of the form p s r n , or
sa,jp s r . i*" 1 , that is, at*" 1 , where a =p 8 r is a constant.
The ratio of the ftth to the (ft  l)th term in this special case
is ar n /ar n ~ l = r, that is to say, is constant.
A series in which the ratio of each term to the preceding is con
stant is called a geometric series or geometric progression ;
and the constant ratio in question is called the common ratio.
If the first term be a and the common ratio r, the second
term is ar • the third (ar)r, that is, ar 2 ; the fourth (ar*)r, that is,
or 3 ; and so on. The ?ith term is ar n ~ l . A geometric series
is therefore neither more nor less general than that particular
case of the general class of series now under discussion which
introduced it to onr notice.
§ 12.] To sum a geometrical series.
Let 2 = a + ar + ar 2 + . . .+ar n ~ 1 (1).
Multiply both sides of (1) by 1  r and we have
(1  r)2 = a + ar + ar 2 + . . . + ar n ~ l
 ar  ar 2  ...  ar n ~ 1  ar' 1 ,
= a ar n (2).
1  r n
Hence 2 = a (3).
1 — r
Since the number of operations on the righthand side of (3)
is independent of n* we have thus obtained the sum of the
series (1).
Cor. If I be the last term of the series, then l = ar n  1 and
ar n = rl. Hence (3) may be written
* Here we regard the raising of r to the ttth power as a single operation.
490
EXAMPLES
a  rl
CHAP.
2 =
Example 1.
2=f+£+f+.
In this case « = $, r = £. Hence
1 r
. . to 10 terms.
(*>•
Example 2.
Here a=l, r
2=12+48+16
2. Hence
l( 2)"_l (l)"2"
S_1 * l(2)~ 3 '
=i(l  2 n ), if ?i. be even,
= 1(1+2"), if ?i be odd.
Example 3.
2 = (a; + y) + (a; ? + xy + ?/ 2 ) + (a? + a% + xy 2 + ■>/) + .
_x 2 y" x?y 3 x i y i
xy xy xy
to n terms.
+
'WH~1 _ i/«+l
a; 1/
= — (x 3 + £c 3 + . . .+a»+ 1 )^(?/ + 2 / 3 + .
xy ' ajy
to ?i terms,
+ 2/«+i),
xy
{l+x
y
Now
and
hnnce
h. . . +ic" 1 )^— (l + w + . . .+7/" 1 ).
xy
l+ai+. . .+x n ~ 1 = {lx n )/(lx),
1 + 1/ + . . .+2/ m_1 = (l2/")/(l2/).
_ a: 2 (l  x 71 ) y{\  y n )
~{xy){\x) {xy){ly)'
§ 13.] "We next proceed to consider the case where the integral
function which multiplies r n is of the 1st degree.
The general term in this case is
(a + bn)r n (1 .
where a and b are constants.
It will be observed that a term of this form would result if
we multiplied together the wth term of any arithmetic series by
the nt\\ term of any geometric series. For this reason a series
whose wth term has the form (1) is often called an arithmetico
geometric series.
The series may be summed by an extension of the artifice
employed to sum a G.P.
Let
2 = (a + b. 1>' + (a + b. 2)r 2 + (a + b. 3)/ s + . . . + (a + b. n)r n .
XX
ARITHMETICOGEOMETRIC SERIES 491
Multiply by 1  r, and we have
(1rJZ
= (a + b.iy + (a + b. 2)r 2 + (a + b.3)r 3 + . . . + (a + b n)r n
 (a + b. l)r 2  (a + b. 2)r 3  . . .  (o + bn  1 )r n
 (a + bn)r n +\
= (a + b.l)r + \br + br 3 +. . . + br n \ (a + bn)r n + l (1).
Looking merely at the terms within the two vertical lines,
we see that these constitute a geometric series. Hence, if we
multiply by 1  r a second time, there will be no series left on
the righthand side ; and we shall in effect have found the
required expression for 2. We have, in fact,
= (1  r) (a + b)r + h ? + br 3 + . . . + br n
br 3 . . .  br n  br n+1
 (1  r) (a + bn)r n +\
= (1  r) (a + b)r + br 2  br n + l  (1  r) (a + bn)r n +\
= (a + b)r  (a + b)r 2 + br 2 {<* + (»+ l)b}r n + l + (a + bn)r n + 2 (2).
Hence
v (a + b)r  (a + b)r 2 + br 2  {a + (n + I )6}r»+ 1 + (a + bn)r n + 2
2 = ~ (lr) s {6) 
§ 14.] If the reader has not already perceived that the
artifice of multiplying repeatedly by 1  r will sum any series of
the general form indicated in § 11, probably the following argu
ment will convince him that such is the case.
Let f s (n) denote an integral function of n of the 5th degree ;
then the degree of f g (n) f/n  1) is the (s  l)th, since the two
terms in n s destroy each other. Hence we may denote /,(n)
fin  1) by /,_,(»). Similarly, /,_,(«) /,_,(«  1) will be an
integral function of n of the (5  2)th degree, and may be denoted
by/,_ g (n), and so on.
Consider now the series
2=/s(l>'+/*(2> 2 + . • .+//»>" (I)
Multiply by 1  r, and we have
492 INTEGROGEOMETRIC SERIES chap.
(1rjS
=/ s (iy+ //ay+ / s (3> 3 +//»>•
 f s (iy /,(2)r 3 . . ./ g (»l)r» fJny»+\
=f s (\y + I /,_ 1 (2)r 2 +f a .pY + . . . +/.. 1 (n>  /»"+ 1 (2).
The series between the vertical lines in (2) is now simpler
than that in (1) ; since the integral function which multiplies r n
is now of the (s  l)th degree only.
If we multiply once more by 1  r we shall find on the right
certain terms at the beginning and end, together with a series
whose nth. term is now f s _ 2 (n)r n .
Each time we multiply by 1  r we reduce the degree of the
multiplier of r n by unity. Hence by multiplying by (1  r) s+1
we shall extirpate the series on the righthand side altogether,
and there will remain only a fixed number of terms.
It follows that any series whose nth term consists of an integral
function of n of the sth degree multiplied by r n can be summed by
simply multiplying by (1 —r) s+1 .
This simple proposition contains the whole theory of the sum
mation of the class of series now under discussion.
Example 1. 2 = IV + 2 V 2 + 3 V + . . . + n 2 r".
Here the degree of the multiplier of r n is 2. Hence, in order to effect the
summation, we must multiply by (1  r) 3 . We thus find
(1  r) 3 2
= 1V + 2V + 3V + 4V + . . .+ »V»
3.1V 2 3.2V 3 3. 3V 4 . . . 3(?il)V* 2,nr n + l
+ 3. IV + 3. 2V 4 . . .+3(7i2)V"l3(?il)V I + 1 + 3/iV"+ 2
 IV... (»3)V* (?i2)V"+ 1 (7(.iyV»+ 2
 m V+ 3 ,
= r + ,2 _ ( }l + 1 f r n+i + (2n + 2>i 1 >"+ 2  ?iV"+ 3 .
Hence
_r + r"  (n + 1 ) V"* 1 + (2?t 2 + 2ra  I )r n + 2  WV+ 3
S  (1r) 8
Example 2. 2=1 2r+3r 9  4r 3 + . . .  2?ir 2 "" 1 .
Multiply by (1 + r) 2 , and we. have
(l+r) 2 2 = l2/ + 3r 2  4r»+. . .  2»ir 2 " 1
+ 2;2.2r 2 + 2.3? 3 . . . +2(2?i l)? 2 " 1  2.2nr 2 "
+ r 2_ 2r 3 + . . . (2n,2)r 2 » 1 + (2wl>r 2 ' ! 2«r 2 ' , + 1 ,
= 1  (2n + l)r" n  27ir"" +1 .
Hence
^ l(2»,+ l)r 2 ' t 2?tr 2 "+ 1
2 ~ (1 + r) 2 ~~ "
xx CONVERGENCY OF GEOMETRIC SERIES 493
If we put r=l, we deduce
12 + 34 . . . 2«= n,
which agrees with § 3, Example 3, above.
CONVERGENCY AND DIVERGENCY OF THE ABOVE SERIES.
§ 15.] We have seen that the sum of n terms of a series
whose nth term is an integral function of n is an integral function
of n ; and we have seen that every integral function becomes
infinite for an infinite value of its variable. Hence the sum of n
terms of any series whose nth. term is an integral function of
n may be made to exceed (numerically) any quantity, however
great, by sufficiently increasing n.
This is expressed by saying that every such series is
divergent.
§ 16.] Consider the geometric series
2 = a + ar + ar 2 + . . . + ar n ~ \
If r = 1, the series becomes
"S = a + a + a + . . . + a = na.
Hence, by sufficiently increasing n, we may cause 2 to surpass
any value, however great.
If r be numerically greater than 1, the same is true, for we have
a(r n  1)
2 =
r1 '
ir n a
r1 r1
Now, since r> 1, we can, by sufficiently increasing n, make r",
and therefore ar n /(r  1), as great as we please. Hence, by suffi
ciently increasing n, we can cause 2 to surpass any value, how
ever great (see Ex. ix. 46).
In these two cases the geometric series is said to be divergent.
If r be numerically less than 1, we can, by sufficiently increas
ing n, make r n as small as we please, and therefore ar n /(l  r) as
small as we please. Hence, by sufficiently increasing n, we can
cause 2 to differ from «/(l  r) as little as we please. This is
often expressed by saying that when r is numerically less than 1,
the sum to infinity of the series a + ar + ar 2 + ... is a! (I  r).
494 EXAMPLE OF INFINITE GEOMETRIC SERIES chai\
In this case the series is said to be convergent, and to converge to
the value a/ (I r).
There is yet another case worthy of notice.
If r   1 , the series becomes
~2 = aa + aa + . . .
Hence the sum of an odd number of its terms is always a, and
the sum of an even number of them always 0. The sum, there
fore, does not become infinite when an infinite number of terms
are taken ; but neither does it converge to one definite value. A
series having this property is sometimes said to oscillate.
Example 1.
Find the limit of the sum of an infinite number of terms of the series
For n terms we have
1 1 1
s =*lzi£=i.
2 14 ' 2"'
Hence, when n is made infinitely great,
2=1.
This case may be illustrated geometrically as follows : —
Let AB be a line of unit length.
— j j j — —  Bisect AB in P x ; bisect PiB in P 2 ,
A Pi Po P3 P4B P 2 B in P 3 ; and so on indefinitely.
It is obvious that by a sufficient
number of these operations we can come nearer to B than any assigned dis
tance, however small. In other words, if we take a sufficient number of
terms of the series
AP 1 + P 1 P 2 + P 2 P 3 + P3P 4 +. . .,
we shall have a result differing from AB, that is, from unity, as little as we
please.
This is simply a geometrical way of saying that
111 ,
2 + 2i + 2» +  • * adco=1 
Example 2.
To evaluate the recurring decimal "34.
Let
^ S/ 34 34 34
s= ' 34= Too + io^ + ioo"3 +   ad ™'
Then 2 is obviously a geometric series, whose common ratio, 1/100, is less
than 1. Hence
34 1 34
2 =
100 1xfo 99
xx PROBLEMS ON ARITHMETIC PROGRESSION 495
PROPERTIES OF QUANTITIES WHICH ARE IN ARITHMETIC,
GEOMETRIC, OR HARMONIC PROGRESSION.
§ 17.] If a be the first term, b the common difference, n the
number of terms, and 2 the sum of an arithmetic progression,
we have
2 = ^{2a + (nl)b] (1).
This equation enables us to determine any one of the four quan
tities, 2, a, b, n, when the other three are given. The equation
is an integral equation of the 1st degree in all cases, except
when n is the unknown quantity, in which case the equation is a
quadratic. This last case presents some points of interest, which
we may illustrate by a couple of examples.
Example 1.
Given 2 = 36, a = 15, &=3, to find n. We have by the formula (1)
above
36={S0(»l)3}.
Hence n 2  lira + 24 = 0.
The roots of this equation are n = B and n = 8. It may seem strange that
there should be two different numbers of terms for which the sum is the same.
The mystery is explained by the fact that the common difference is negative.
The series is, in fact,
15 + 12 + 9 +6 + 3 + 036] 9. . .;
and, inasmuch as the sum of the part between the vertical lines is zero, the
sum of 8 terms of the series is the same as the sum of 3 terms.
Example 2.
2=14, a = Z, 6 = 2.
The equation for n in this case is
tt 2 + 2?i = 14.
Hence ji=  1± V(15)= +287 . . ., or 487 . . .
The second of the roots, being negative, has no immediate reference to our
problem. The first root is admissible so far as its sign is concerned, but it is
open to objection because it is fractional, for, from the nature ot the case, n
must be integral. It may be conjectured, therefore, that we have set our
selves an impossible problem. Analytically considered, the function n 2 + 2n
varies continuously, and there is in the abstract no difficulty in giving to it
any value whatsoever. The sum of an arithmetic series, on the other hand,
varies per saltum ; and it so happens that 14 is not one of the values that 2
can assume when a — B and b — 2. There are, however, two values which 2
496 DETERMINATION OF ARITHMETIC SERIES BY TWO DATA chap.
can assume between which 14 lies ; and we should expect that the integers
next lower and next higher than 2 "87 would correspond to these values of 2.
So, in fact, it is ; for, when w = 2 2 = 8, and when n = 3 2=15.
§ 18.] An arithmetic progression is determined when its first
term and common difference are given ; that is to say, when
these are given we can write down as many terms of the pro
gression as we please. An arithmetic progression is therefore
what mathematicians call a twofold manifoldness ; that is, it is
determined by any two independent data.
Bearing this in mind, we can write the most general arith
metic progressions of 3, 4, 5, &c. terms as follows : —
a  (3, a, a. + {3,
a  3/3, a  (3, a + fi, a + 3/?,
a 2(3, a ft, a, a + (3, a + 2/3,
&c,
where a and (3 are any quantities whatsoever. It will be
observed that in the cases where we have an odd number of
terms the common difference is (3, in the cases where we have
an even number 2(3. These formulas are sometimes useful in
establishing equations of condition between quantities in A.P.
Example 1.
Given that the ^th term of an A.P. is P, and that the qth term is Q, to
find the A.P. Let a be the first term and b the common difference ; then
the ^th and qth terms are a + (j»  1 )b and a + [q  l)b respectively. Hence
a + (pl)b = ~P, a + {ql)b = Q.
These are two equations of the 1st degree to determine a and b.
We find
& = (PQ)/(i>0). ct={(pl)Q(ql)?}l(pq).
Example 2.
If a, b, c be in A. P., show that
a\b + c) + b{c + a) + c\a + b) = y(a + b + cf.
We may put a = ap, b = a, c = a+/3.
The equation to be established is now
(a/3)(2a + /i) + a.2a + (a + i3) 2 (2a/3)=?(3c l ) 3 ,
— 6a s .
Since a and /3 are independent of one another, this equation must be an
identity. The lefthand side reduces to
XX ARITHMETIC MEANS 497
2a {(a  (3f + (a + j8) 2 } + j8 {(a  /3) 2  (a + 0)*} + 2a\
= 2a{2a 2 + 2/3'} + /3{4a/3} +2a 3 ,
= 6a :! .
Hence the required result is established.
§ 19.] If three quantities, a, b, c, be in A. P., we have
b  a = c  b by definition. Hence
b = (c + «)/2.
In this case b is spoken of as the arithmetic mean between a
and c. The arithmetic mean between two quantities is therefore
merely what is popularly called their average.
If a and c be any two quantities whatsoever, and A x , A 2 , . . ., A n
n others, such that a, A,, A 2 , . . ., A n , c form an A. P., then A n A 2 ,
. . ., A n are said to be n arithmetic means inserted between a and c.
There is no difficulty in finding A,, A 2 , . . ., A n when a and
c are given. For, if b be the common difference of the A.P.,
a, A u A 2 , . . ., A, i; c, then
A! = a + b, A 2 = a + 2b, . . ., A n = a + nb,
and 6 = a + (n + 1 )b.
From the last of these we deduce b — (c  a)j{n +1). Hence
we have
A C ~ a A O C ~ a S
A,=a + , A„ = a + 2 , 6zc.
n + 1 " n + 1
N.B. — By the arithmetic mean or average of n quantities a,, a 2 ,
. . ., a a is meant («, + a„ + . . . + a tl )/n.
In the particular case where two quantities only are in
question, the arithmetic mean in this sense agrees with the
definitions given above ; but in other cases the meanings of the
phrases have nothing in common.
Example 1.
Insert 30 arithmetic means between 5 and 90 ; and find the arithmetic
mean of these means.
Let b be the common difference of the A. P. 5, Ai, A 2 , . . ., A M , 90.
Then
6=(905)/(80+l)=85/31.
Hence the means are
,85 n 85 _ 85 c
5 + 3l' 5 + 2 3l' 5 + 3 3T &C ' ;
,. t . 240 325 410 .
that is, — , — , — , &c.
VOL. I 2 K
498 EXAMPLES chap.
We have A 1 + A 2+ . . . + A„ = 1 ( r A^+A. 
n n \ 2 J
_A! + A»
■~ 2~ '
_ 85 nn 85 \ /„
= (5 + 90)/2 = 95/2.
Remark. — It is true generally that the arithmetic mean of the n arith
metic means between a and c is the arithmetic mean between a and c.
Example 2.
The arithmetic mean of the squares of n quantities in A. P. exceeds the
square of their arithmetic mean by a quantity which depends only upon n
and upon their common difference.
Let the ?i quantities be
a + b, a + 2b, . . ., a + nb.
Then, by §§ 5 and 6,
{a + bT + {a + 2b?+. . . +(a + nbf
n
If. , , , , ., , ,.,« (n + 1 2?i + l)\
:■{ an + abnln + 1) + b 2 — ^ —  V,
n I 6 J
6 2
= a 2 + ab(n + l) + {2n 2 + 3n + l).
. . f(a + b) + (a + 2b) + . . . +(a + nb)\ 3 / ra + l,\ a
Again,  — ) a= ^+__»ji
TO
= « 2 + «&( ,i + 1 ) + ( w 2 + 2n + 1 ).
M 2  1
Hence A. M. of squares  square of A.M. = 5 2 ,
which proves the proposition.
§ 20.] If Z be the sum of n terms of a geometric progression
whose first term and common ratio are a and r respectively, we
have
r n _ 1
When any three of the four, 2, a, r, n, are given, this equation
determines the fourth. "When either i: or a is the unknown
quantity, we have to solve an equation of the 1st degree. When
r is the unknown quantity, we have to solve an integral equation
of the ?<th degree, which, if n exceeds 2, will in general be
effected by graphical or other approximative methods. If n be
the unknown quantity, we have to solve an exponential equation
of the form r 11 = s, where r and s are known. This may be
xx DETERMINATION OF GEOMETRIC SERIES BY TWO DATA 499
accomplished at once by means of a table of logarithms, as we
shall see in the next chapter.
§ 21.] Like an A.P., a G.P. is a twofold manifoldness, and
may be determined by means of its first term and common ratio,
or by any other two independent data.
In establishing any equation between quantities in G.P., it
is usual to express all the quantities involved in terms of the
first term and common ratio. Since these two are independent,
the equation in question must then become an identity.
Example 1.
The ^th term of a G.P. is P, and the qth. term is Q ; find the first term
and common ratio.
Let a be the first term, r the common ratio. Then we have, by our data,
Prom these, by division, we deduce
?*«=: P/Q, whence r=(?/Q) 1! (>><>K
Using this value of r in the first equation, we find
a = P/(P/Q)Ip 1 V(p«) = PPd /(/>«)Q(i*>)/(«*).
Hence we have
a = YViV(pq)Qj}p)!(<ip) } r = pi/(i>«)Qi/(«j>).
Example 2.
If a, b, c, d be four quantities in G. P., prove that
4(« 2 + b 2 + c 2 + d')  (a + b + c + df={a  bf + (c  df + 2(a  df.
If the common ratio be denoted by r, we may put b — ra, c = r 2 a, d = r i a.
The equation to be established then becomes
4cr(l + >~ + r 4 + I s )  a 2 {l + r + r 2 + r 3 ) 2 = a''(l  r) 2 + aV(l  r) 2 + 2a 2 (l  r 3 ) 2 ,
that is,
4(1 + r 2 +r*+ r«)  (1 + It + 3r 2 + 4r= + dr 4 + 2r 5 + r 6 )
= l2r + r 2 + r i 2r 5 + r K + 2 4r 3 + 2r 6 ,
which is obviously true.
§ 22.] When three quantities, a, b, c, are in G.P., b is called the
geometric mean between a and c.
We have, by definition, c/b = b/a. Hence b 2 = ac. Hence, if
we suppose a, b, c to be all positive real quantities, b = + \/(ac).
That is to say, the geometric mean between two real positive quantities
is the positive value of the square root of their product.
If a and c be two given positive quantities, and G 1; G 2 , . . ., G n
n quantities, such that a, G lt G 2 , . , ., G n , c form a G.P., then
G„ G 2 , . . ., G n are said to be n geometric means inserted between a
and c.
500 GEOMETRIC MEANS
CHAP.
Let r be the common ratio of the supposed progression.
Then we have G, = ar, G 2 = ar 2 , . . ., G M = ar n , c = ar n+1 . From
the last of these equations we deduce r = (c/a) 1/(  n+l \ the real
positive value of the root being, of course, taken. Since r is thus
determined, we can find the value of all the geometric means.
The geometric mean of n positive real quantities is the positive
value of the nth root of their product. This definition agrees with
the former definition when there are two quantities only.
Example.
The geometric mean of the n geometric means between a and c is the
geometric mean between a and c.
Let the n geometric means in question be ar, ar 2 , . . ., ar n , so that
c = ar n+1 . Then
{ar. ar" . . . ar n ) Vn — (a"r l ^ 2 + " ■ • +»)i«,
= {a 2 r"+ 1 ) 1 ' 2 ,
= {acr }
which proves the proposition.
§ 23.] A series of quantities which are such that their reciprocals
form an arithmetic progression are said to be in harmonic progression.
From this definition we can deduce the following, which is
sometimes given as the defining property : —
If a, b, c be three consecutive terms of a harmonic progression, then
ajc = {ab)l(bc) (1).
For, by definition, I /a, 1/6, 1/e are in A. P., therefore
1_1 1 _1
b a c b
ab b  c
Hence
Hence
ah be
ab ab a
b — c be c '
which proves the property in question.
§ 24.] A harmonic progression, like the arithmetic j)rogression,
from which it may be derived, is a twofold munifoldness. The
following is therefore a perfectly general form for a harmonic
xx HARMONIC PROGRESSION, HARMONIC MEANS 501
series, 1 J(a + b), l/(a + 2b), l/(a + 36), . . ., lfca + rib), . . .,
for it contains two independent constants a and b ; and the
reciprocals of the terms are in A. P.
The following forms (see § 18) are perfectly general for
harmonic progressions consisting of 3, 4, 5, . . . terms respect
ively : —
l/(a£), 1/a, l/(a + P);
l/(a3/3), l/(«/3), l/(a + /3), l/(a+3/3);
1/(0.2(3), l/(«/3), 1/a, l/(a + j8), l/(a+2/3);
&c.
The above formulae may he used like those in § 18.
§ 25.] If a, b, c be in IIP., b is called the harmonic mean
between a and c. We have, by definition, 1/c — 1/6 = 1/b — 1/a
Hence 2/b  1/a + 1/c, and b = 2ae/(a + c).
If a, H,, H B , . . ., H„, c form a harmonic progression, H,, H,,
. . ., H w are said to be n harmonic means inserted between a and c.
Since 1/a, 1/H,, 1/H 2 , . . ., 1/H n , 1/c in this case form an
A. P., whose common difference is d, say, we have
d = (l/c l/a)/(n + 1) = (a  c)j(n + \)ac.
Hence
1/H, = 1/a + (a  c)/(n + l)ac, 1/H g = 1/a * 2(a  c)j(n + l)ac, &c;
and H, = (n + l)ae/(a + nc), H 2 = (n + l)ac/(2a + (n  l)c), &c.
If a quantify H be such that its reciprocal is the arithmetic mean
of the reciprocals of n given quantities, H is said to be the harmonic
mean of the n quantities.
It is easy to see, from the corresponding proposition regard
ing arithmetic means, that the harmonic mean of the n harmonic
means between a and c is the harmonic mean of a and c.
§ 26.] The geometric mean between two real positive quantities a
and c is the geometric mean between the arithmetic and the harmonic
means between a and c ; and the arithmetic, geometric, and harmonic
means are in descending order of magnitude.
Let A, G, H be the arithmetic, geometric, and harmonic
means between a and c, then
A = (a + c)/2, G  + */(ac), H = 2ac/(a + c).
502 ARITHMETIC, GEOMETRIC, AND HARMONIC MEANS CHAP.
XT Ka a + c %ac 2
Hence AH = —— x = ac = G ,
2 a + c
which proves the first part of the proposition.
Again, A  G = — \/(oc) = h( y/a  \U) 2 .
GH= sl(ac) — =^ks/a *Jc)\
a + c a+c
Now, since a and c are both positive, Ja and N /c are both real,
therefore ( s/a  Jcf is an essentially positive quantity ; also
*J(ac) and a + c are both positive. Hence both A  G and G  H
are positive.
Therefore A >G>H.
The proposition of this paragraph (which was known to the
Greek geometers) is merely a particular case of a more general
proposition, which will be proved in chap. xxiv.
§ 27.] Notwithstanding the comparative simplicity of the
law of its formation, the harmonic series does not belong to the cate
gory of series that can be summed. Various expressions can be
found to represent the sum to n terms, but all of them partake
of the nature of a series in this respect, that the number of steps
in their synthesis is a function of n.
It will be a good exercise in algebraic logic to prove that
the sum of a harmonic series to n terms cannot be represented
by any rational algebraical function of n. The demonstration
will be found to require nothing beyond the elementary principles
of algebraic form laid down in the earlier chapters of this work.
Exercises XL.
Sum the following arithmetical progressions : —
(1.) 5 + 9 + 13+ ... to 15 terms. (2.) 3 + 3^ + 4 + ... to 30 terms.
(3.) 13 + 12 + 11+. . . to 24 terms. (4.) i + l+ ■ ■ • to 16 term*
,k 1 , n 1
(o.)  + — + ... to 7t terms.
n n
(6.) {a~ny t + {a? + n) + {a + n)+ . . . to to terms.
r) l + l, 41 7 +
(<•) T _ r7 + n772+ • • • to ( terms.
(8.) The 20th term of an A. P. is 100, and the sum of 30 terms is 500 ;
find the sum of 1000 terms of the progression.
XX
EXERCISES XL 503
(9.) The first term of an A. P. is 5, the number of its terms is 15, and the
sum is 390 ; find the common difference.
(10.) How many of the natural numbers, beginning with unity, amount
to 500500 ?
(11.) Show that an infinite number of A.P.'s can be found which have
the property that the sum of the first 2m terms is equal to the sum of the
next m terms, m being a given integer. Find that particular A. P. having
the above property whose first term is unity.
(12.) An author wished to buy up the whole 1000 copies of a work which
he had published. For the first copy he paid Is. ' But the demand raised
the price, and for each successive copy he had to pay Id. more, until the
whole had been bought up. What did it cost him ?
(13.) 100 stones are placed on the ground at intervals of 5 yards apart.
A runner has to start from a basket 5 yards from the first stone, pick up the
stones, and bring them back to the basket one by one. How many yards
has he to run altogether ?
(14.) AB is a straight line 100 yards long. At A and B are erected per
pendiculars, AL, BM, whose lengths are 4 yards and 46 yards respectively. At
intervals of a yard along AB perpendiculars are erected to meet the line LM.
Find the sum of the lengths of all these perpendiculars, including AL and BM.
(15.) Two travellers start together on the same road. One of them
travels uniformly 10 miles a day. The other travels 8 miles the first day,
and increases his pace by half a mile a day each succeeding day. After how
many days will the latter overtake the former ?
(16.) Two men set out from the two ends of a road which is I miles long.
The first travels a miles the first day, a + b the next, a + 2b the next, and so
on. The second travels at such a rate that the sum of the number of miles
travelled by him and the number travelled by the first is always the same for
any one day, namely c. After how many days will they meet ?
(17.) Insert 15 arithmetic means between 3 and 30.
(18.) Insert 10 arithmetic means between  3 and +3.
(19.) A certain even number of arithmetic means are inserted between 30
and 40, and it is found that the ratio of the sum of the first half of these
means to the second half is 137 : 157. Find the number of means inserted.
(20.) Find the number of terms of the A. P. 1 +8 + 15+ . . . the sum of
which approaches most closely to 1356.
(21.) If the common difference of an A. P. be double the first term, the
sum of m terms : the sum of n terms = m : n.
(22.) Find four numbers in A. P. such that the sum of the squares of the
means shall be 106, and the sum of the squares of the extremes 170.
(23.) If four quantities be in A. P., show that the sum of the squares of
the extremes is greater than the sum of the squares of the means, and that
the product of the extremes is less than the product of the means.
(24.) Find the sum of n terms of the series whose rth term is (3r + 1).
(25. ) Find the sum of n terms of the series obtained by taking the 1st, rth,
2rth, 3rth, &c. terms of the A. P. whose first term and common difference are
a and b respectively.
504 EXERCISES XL
CHAP.
(26.) If the sum of n terms of a series be always n(n + 2), show that the
series is an A. P. ; and find its first term and common difference.
(27.) Show by general reasoning regarding the form of the sum of an
A. P. that if the sum of p terms be P, and the sum of q terms Q, then the
sum of n terms is Pn(?i  q)/p{p q) + Qn(n p>)lq{<l p).
(28.) Any even square, (2n) 2 , is the sum of n terms of one arithmetic
series of integers ; and any odd square, (2ft + l) 2 , is the sum of n terms of
another arithmetic series increased by 1.
(29.) Find n consecutive odd numbers whose sum shall be nP.
Show that any integral cube is the difference of two integral squares.
(30.) Find the nth term and the sum of the series
13 + 610 + 1521+ ....
(31.) Sum the series 3 + 6+ . . . +3w.
(32.) If si, s 2 , . . ., s p be the sums of p arithmetical progressions, each
having n terms, the fhst terms of which are 1, 2, . . ., p, and the common
differences 1, 3, . . ., 2pl respectively, show that 5i + s L .+ . . . +g p is
equal to the sum of the first np integral numbers.
(33.) The series of integral numbers is divided into groups as follows : —
1,  2, 3,  4, 5, 6,  7, 8, 9, 10,  . . ., show that the sum of the nth group
is (?i 3 + ?i).
If the series of odd integers be divided in the same way, find the sum of
the ?ith group.
Sum the following series : —
(34.) 4 2 + 7 2 + . . . +(3u+l) 2 . (35.) 2„(?i 3 l)(?il).
(36.) Z n \p + q(nl)}{p + q(n2)}.
(37.) l 2 2 2 + 3 2  . . . + (2?il) 2 (2ra) 2 .
(38.) a s + (a + bf+. . . +( a + n~^lb)*.
(39.) (l3i) + ( 2 3_2) + (3 3 3)+. . . tow terms.
(40.) 1.2 2 +2.3 2 + 3.4 2 +. . . ton terms.
(41.) l + 2.3 2 + 3.5 2 + 4.7 2 + 5.9 2 +. . . to n terms.
(42.) 1.3.7 + 3.5.9 + 5.7.11+ . . . to n terms.
(43.) l 2 + (l 2 + 2 2 ) + (l 2 + 2 2 + 3 2 )+ . . . ton terms.
(44. ) A pyramid of shot stands on an equilateral triangular base having
30 shot in each side. How many shot are there in the pyramid ?
(45.) A pyramid of shot stands on a square base having m shot in each
side. How many shot in the pyramid ?
(46.) A symmetrical wedgeshaped pile of shot ends in a line of m shot
and consists of I layers. How many shot in the pile ?
(47.) If B S r =l' , +2'+ . . . +n',then n 8 r =pon<+ 1 +p 1 nr+. . . +p r+h where
Po, Pi, • . . can be calculated by means of the equations
r+lC 3 po + rCiJ?i = rQi,
HlQs p + pC2.Pl + rlCi ^2 = rC» ,
„C r denoting, as usual, the rth binomial coefficient of the nth rank.
XX
EXERCISES XLI
505
(48.) Show that
„S r =»(»+l)(nl)(H2)...(»r)
rlS r
H
r tn r
r(r+l)(»r)(rl)!
r nO r
(ri)r(nr+l)l\(r2)\ ' (r2)(rl)(nr + 2)2!(r3)!
( ~ ) r iS r 1
1.2(7il)(rl)!i '
where r ! stands for 1.2.3 . . . ?•.
(49.) If ? r denote the sum of the products r at a time of 1, 2, 3, . . ., n,
and S r denote l r + 2 r + . . . + W, show that ?P r =SiP r _rS 2 P,2 + S3Pr3~ . . .
Hence calculate P 2 and P3.
(50.) UJ[x) he an integral function of x of the (?• l)th degree, show that
f(x) r C 1 f(xl)+ r Q i f(x2) . . . () r f{nr)=0, r Ci, &
coefficients.
Exercises XLI.
Sum the following geometric progressions : —
(1.) 6 + 18 + 54+ . . . to 12 terms. (2.) 618 + 54
1 1
being binomial
(3.) 3333. . . to n terms.
(5.) 64+ . . . to 10 terms
^ V3 + 1 V3 + 2
( 4 ) Ji + P"
. to 12 terms,
to n terms.
to 20 terms.
('■) 1+3+32+ • •
(9.) lx + x 2 x 3 +
(10.) V2 + ^+. •
(12.) a ±? a ^ +
to n terms.
. . to oo, a;<l.
to 00. (11.)
, . to 00 .
(8.) 15+i
2 2'
\/3
V3
V3 + 1 ' V 3 + 3
+
to 00 .
+
to 00.
ax a+x
Sum, by means of the formula for a G.P., the following : —
(13.) l+xx 2 x 3 + x 4 + x 5 x 6 x 7 + . . . tooo,a;<l.
(14.) {x y)+(l^y(t 3 ty. . . to * terms.
(15.) l+(x + y) + (x 2 + o:y + y 2 ){{x i + x 2 y + xy 2 + y 5 )+. . . to n terms.
(16.) 33 +333+ 3333 + . . . to n terms.
Sum the series whose ?ith terms are as follows : —
(17.) 2"3"+ 1 . (18.) {x n + n)(x n n).
(19.) Ae»lVy».i). (20.) (jt»gr)(p*+r).
(21.) 2»(3" 1 + 3» 2 +. . .+1). (22.) (l)"a 3 ''.
(23. ) Sum to « terms the series (r n + l/>") 2 + (r^ 1 + l/r"+ 1 ) 2 + . . . .
(24.) Sum to ?i terms (l + l//) 2 + (l + l/r 2 )+. . . .
(25.) Show that {a + b) n b" = ab n ~ l + ab"~ 2 (a + b)+. . . +a(a + b) n \
(26.) If S, denote the sum of n terms of a G.P. beginning with the tth
term, sum the series S1 + S0+ . . . +S ( .
(27.) Show that £/('037)=3.
506 EXEECISES XLI chap.
Sum the following series to n terms, and, where admissible, to infinity : —
(28.) l2x + 3x 2 lx s +. . . . (29.) 1f + £•£+. . . .
9 S 2 4 2 12 2 3 3 4
(so.) 1I44+. • • < 31 > 1+ ir + 1r + ir +  ■ • ■
(32.) l 3 + 2 3 x + S s x 2 +. . . .
12 3 12 3
(33.) = + =5 + =; + =3 + ss + =a+. • . to oo, where the numerators recur
7 7 i J i r r
with the period 1, 2, 3.
(34. )  + — , + = + : + —. + a + . . . to 3)i terms, where the numerators recur
with the period a, J, c.
(35.) A servant agrees to serve his master for twelve months, his wages to
be one farthing for the first month, a penny for the second, fourpence for the
third, and so on. What did he receive for the year's service ?
(36. ) A precipitate at the bottom of a beaker of volume V always retains
about it a volume v of liquid. It was originally precipitated in an alkaline
solution ; find what percentage of this solution remains about it after it has
been washed n times by filling the beaker with distilled water and emptying
it. Neglect the volume of the precipitate itself.
(37.) The middle points of the sides of a triangle of area Ai form the
vertices of a second triangle of area A 2 ; from A 2 a third triangle of area A 3 is
derived in the same way ; and so on, ad infinitum. Find the sum of the
areas of all the triangles thus formed.
(38.) OX, OY are two given straight lines. From a point in OX a perpend
icular is drawn to OY ; from the foot of that perpendicular a perpendicular
on OX ; and so on, ad infinitum. If the lengths of the first and second per
pendicular be a and b respectively, find the sum of the lengths of all the per
pendiculars ; and also the sum of the areas of all the rightangled triangles in
the figure whose vertices lie on OY and whose bases lie on OX.
(39.) The population of a certain town is P at a certain epoch. Annually
it loses d per cent by deaths, and gains b per cent by births, and annually a
fixed number E emigrate. Find the population after the lapse of n years.
2, a, r, n, I, having the meanings assigned to them in § 12, solve the
following problems : —
(40. ) Express 2 in terms of a, n, I ; and also in terms of r, n, I.
(41.) 2= 4400, « = 11, »=4,.findr.
(42.) 2 = 180, r = 3, n = 5, find a.
(43.) 2 = 95, a = 20, n = S, find r.
(44.) 2 = 155, « = 5, »=3, findr.
(45.) 2 = 605, a=5, r = 3, find n.
(46.) If the second term of a G.P. be 40, and the fourth term 1000, find
the sum of 10 terms.
XX
EXEKCISES XLI, XLII 507
(47.) Insert one geometric mean between \/3/\/2 and 3V3/2V2
(48.) Insert three geometric means between 27/8 and 2/3.
(49.) Insert four geometric means between 2 and 64.
(50.) Find the geometric mean of 4, 48, and 405.
(51.) The geometric mean between two numbers is 12, and the arithmetic
mean is 25§ : find the numbers.
(52.) Four numbers are in G.P., the sum of the first two is 44, and of the
last two 396 : find them.
(53.) Find what common quantity must be added to a, b, c to bring them
into G.P.
(54.) To each of the first two of the four numbers 3, 35, 190, 990 is added
SB, and to each of the last two y. The numbers then form a G.P. : find x
and y.
(55.) Given the sum to infinity of a convergent G.P., and also the sum to
infinity of the squares of its terms, find the first term and the common ratio.
(56.) IfS=Oi + a2+. . . + «„be a G.P., then S' = l/a 1 + l/a 2 + . . . +l/a n
is a G.P., and S/S' = «!«„.
(57.) If four quantities be in G.P., the sum of the squares of the extremes
is greater than the sum of the squares of the means.
(58.) Sum In terms of a series in which every even term is a times the
term before it, and every odd term c times the term before it, tire first term
being 1.
(59.) If x = a + a/r + a/r 2 + . . . ad oo ,
y = b b/r+ bjr 2  . . . ad oo ,
z = c + clr 2 +cjr i + . . . adoo,
then xylz = ab/c.
(60. ) Find the sum of all the products three and three of the terms of an
infinite G.P., and if this be onethird the sum of the cubes of the terms, show
that r=\.
Exercises XLII.
(1.) Insert two harmonic means between 1 and 3, and five between 6 and 8.
(2.) Find the harmonic mean of 1 and 10, and also the harmonic mean of
1, 2, 3, 4, 5.
(3.) Show that 4, 6, 12 are in H.P., and continue the progression both
ways.
(4.) Find the H.P. whose 3rd term is 5 and whose 5th term is 9.
(5.) Find the H.P. whose joth term is P and whose gill term is Q.
(6. ) Show that the harmonic mean between the arithmetic and geometric
means of a and b is 2(a + b)/ {{a/b^ + (b/a) i } n :
(7.) Four numbers are proportionals ; show that, if the first three are in
G.P., the last three are in G.P.
(8.) Three numbers are in G.P. ; if each be increased by 15, they are in
H.P. : find them.
(9. ) Between two quantities a harmonic mean is inserted ; and between
each adjacent pair of the three thus obtained is inserted a geometric mean.
508 EXERCISES XLII chap, xx
It is now found that the three inserted means are in A. P. : show that the ratio
of the two quantities is unity.
(10.) The sides of a rightangled triangle are in A. P. : show that they are
proportional to 3, 4, 5.
(11.) a, b, c are in A.P., and a, b, d in H.P. : show that c/d =
1  2(o  bflab.
(12.) If x be any term in an A. P. whose two first terms are a, b, y, the
term of the same order in a H.P. commencing with the same two terms, then
(xa)/(yx) = b/(yb).
(13.) If a 2 , b\ c 2 be in A. P., then 1/(5 + c), l/(c + a), lfta + b) are in A. P.
(14.) If P be the product of n quantities in G.P., S their sum, and S' the
sum of their reciprocals, then P 2 =(S/S') n .
(15.) If a, b, c be the pth, qth, and rth terms both of an A.P. and of a
G.P., then a*" &*»«»» =1.
(16.) If P, Q, R be the .pth, qth., rth terms of a H.P., then 2{PQ(j3 q)}
= 0, and {S(g 2 r 2 )/P 2 } 2 = 4{2tor)/P 2 }{2 ? r( ? r)/F}.
/17.) If the sum of m terms of an A.P. be equal to the sum of the next n,
and also to the sum of the next^J, then (m + n) (1/n  l/p) = (n +p) (1/m  I/n).
(18.) If the squared differences of p, <?> r be in A.P., then the differences
taken in cyclical order are in H.P.
(19.) If a + b + c, a 2 + & 2 + c 2 , a? + b z + <? be in G.P., prove that the common
ratio is ^2a  3a&c/226c.
(20.) If x, y, z be in A. P., ax, by, cz in G.P., and a, b, c in H.P., then
H.M. of a, c:G.M. of a, c = H.M. oix, z:G.M. of a, z.
(21.) Ifa 2 + 6 2 c 2 , P + ca", c 2 ia 2 & 2 be in G.P., then a 2 /c 2 + c 2 /i 2 ,
& 2 /c 2 + c 2 /6 2 , a 2 /6 2 + 6 2 /c 2 are in A.P.
(22.) If a, b, c, d be in G.P., then abcdl 2  ] = (2a) 2 , and
V(^ + ^) + V(6* + c*) + V(c« + * )_&«■ ^ „ 4 m
(23.) The sum of the n geometric means between a and k is
{aVWik  aWl»W)l(tM n + 1 )  a 1 ( n +V).
(24.) If Ai, A 2 , . . ., A„ be the n arithmetic means, and H^ H 2 , . . .,
H„ the n harmonic means, between a and c, sum to n terms the series whose
rth term is (A r a)(H r «)/H r .
(25.) If«i,a 2 , . . ., a n be in G. P. , then
(^ + a, 2 + 03) + {d2 + a 3 + en) 2 + . . . + (a„_n + «.„_;, + a n f
= («! 2 + «!«, + «.o 2 ) 2 (a! 2 « 4  «.2 2 " 4 )/(«l 2  «2>1 5 " 4 .
(26.) If a r , b r be the arithmetic and geometric means respectively between
« r _i and b r i, show that
a„ 2  {a n *±(a„ 2 &„ 2 ) } 2 ,
Z>„_ 2 = {aj 2f{a n 2 b n ) } 2 .
(27.) If «j, a 2 , . . ., a n be real, and if
(ai 2 + 02 2 +. . . +a„i 2 )(a 2 2 + (T 3 2 + . . . + a„ 2 ) = {aiao + a 2 a i + . . . + a„ia„), 2
then ai, a 2 , • ■ , cin are in G.P.
CHAPTEE XXI.
Logarithms.
§ 1.] It is necessary for the purposes of this chapter to define
and discuss more closely than we have yet done the properties
of the exponential function a x . For the present we shall sup
pose that a is a positive real quantity greater than 1. AY hat
ever positive value, commensurable or incommensurable, we give
to x, we can always find two commensurable values, mjn and
(m + l)/n (where m and n are positive integers), between which
x lies, and which differ from one another as little as we please,
see chap, xiii., § 15. In defining a x for positive values of x, we
suppose x replaced by one (say m/n) of these two values, which
we may suppose chosen so close together that, for the purpose
in hand, it is indifferent which we use. We thus have merely
to consider a'" ,ln ; and the understanding is that, as in the
chapter on fractional indices, we regard only the real positive
value of the nth root ; so that a m,n may be read indifferently as
( Z/a) m , or as ya m .
For negative values of x we define a x by the equation
a x = lja~ x , in accordance with the laws of negative indices.
§ 2.] We shall now show that a x , defined as above, is a continu
ous function of x susceptible of all positive values between and + oo .
1st. Let y be any positive quantity greater than 1, and let
n be any positive integer. Since a > 1 , a lln > 1 ; but, by suffi
ciently increasing n, we may make a 1!n exceed 1 by as little as
we please. Also, when n is given, we can, by sufficiently in
creasing w, make a m!n as great as we please.* Hence, whatever
* See chap, xi., § 14.
510
CC CONTINUOUS ITS GRAPH
CHAP.
may be the value of y, we can so choose n that a 1/n <y ; and then
y will lie between two consecutive integral powers of a l!n ; say
a min < y < a (»»+iy» Now the difference between these two values
of a x is a mln (a 1/n  1); and this, by sufficiently increasing n, we
may make as small as we please. Hence, given any positive
quantity y > \, we can find a value of x such that a x shall be as
nearly equal to y as we please.
2nd. Let y be positive and < 1 ; then 1/y is positive and
greater than 1. Hence we can find a value of x, say x', such
that a x ' = \jy as nearly as we please. Hence a~ x ' = y.
We may make y pass continuously through all possible values
from to + oo . Hence a x is susceptible of all positive values
from to + oo . It is obviously a continuous function, since
the difference of two finite values corresponding to x = m 'n and
x = (wi + 1 )/n is a m!n (a 1/n  1 ), which can be made as small as
we please by sufficiently increasing n.
Cor. We have the following set of corresponding values : —
X= oo, , 1, 0, +, 1, +co;
y = a x = 0. < 1, 1/a, 1, > 1, a, + oo .
Fio. 1.
In Fig. 1 the fulldrawn curve is the graph of the function
y ~ lO* ; the dotted curve is the graph of y = 100*.
xxi DEFINITION OF LOGARITHMIC FUNCTION 511
It will be observed that the two curves cross the axis of y
at the same point B, whose ordinate is + 1 ; and that for one
and the same value of y the abscissa of the one curve is double
that of the other.
§ 3.] The reasoning by Avhich we showed that the equation
y = a x , certain restrictions being understood, determines y as a
continuous function of x, shows that the same equation, under
the same restrictions, determines a; as a contiimous function of y.
This point will perhaps be made clearer by graphical considera
tions. If we obtain the graph of y as a function of x from the
equation y = a x , the curve so obtained enables us to calculate x
when y is given ; that is to say, is the graph of x regarded as a
function of y. Thus, if we look at the matter from a graphical
point of view, we see that the continuity of the graph means the
continuity of y as a function of x, and also the continuity of x as
a function of y.
When we determine x as a function of y by means of the
equation y = a x , we obviously introduce a new kind of transcend
ental function into algebra, and some additional nomenclature
becomes necessary to enable us to speak of it with brevity and
clearness.
The constant quantity a is called the base.
y is called the exponential of x to base a (and is sometimes
written exp a ;r).*
x is called the logarithm of y to base a, and is usually written
log a y
The two equations
y = a* (1),
z = log a y (2),
are thus merely different ways of writing the same functional
relation. It follows, therefore, that all the properties of our new
logarithmic function must be derivable from the properties of our old
exponential function, that is to say, from the laws of indices.
* This notation is little used in elementary textbooks, but it is con
venient when in place of x we have some complicated function of x. Thus
exp a (l +X + X 2 ) is easier to print than a 1+x+x2 .
512 FUNDAMENTAL PEOPERTIES OF LOGARITHMS chap.
The student should also notice that it follows from (1) and
(2) that the equation
y = a 1 **** (3)
is an identity.
§ 4.] If the same hase a be understood throughout, we have
the following leading properties of the logarithmic function : —
I. The logarithm of a product of positive numbers is the sum of
the logarithms of the separate factors.
II. The logarithm of the quotient of hvo positive numbers is the
excess of the logarithm of the dividend over the logarithm of the divisor.
III. The logarithm of any power {positive or negative, integral
or fractional) of a positive number is equal to the logarithm of the
number multiplied by the power.
Let ?/,, y 2 , . . ., y n be n positive numbers, x lt x 3 , . . ., x n their
respective logarithms to base a, so that
«i = log a ?/ 1) x 2 = \og a y 2 , . . ., x n = log a y n .
By the definition of a logarithm we have
y x = a x \ y 2 = a x \ . . ., y n = a x \
Hence y^Ji ■ ■ II n = a Xl a x  . . . a Xn  a x i +x *+ • ■ • + Xn ,
by the laws of indices.
Hence, by the definition of a logarithm,
x x + x. 2 + . . . + x n = logafay, . . . y n ),
that is, log^ + log a ?/ 2 + . . . + log a y n = log (y,y 9 . . . y n ).
We have thus established I.
Again, y 1 /y 2 = a x i/a x ~ = a x ^ _a ' 2 ,
by the laws of indices.
Hence, by the definition of a logarithm,
x x x a = logaiyM,
that is, log a y,  \og a y, = logatyjyj),
which is the analytical expression of II.
Again, y* = (a x i) r = a rx \
by the laws of indices.
xxi EXAMPLES 513
Hence, by the definition of a logarithm,
ra, = \og a y x \
that is, r log a y 1 = log*^,
which is the analytical expression of III.
Example 1.
log 21 = log (7x3) = log 7 + log 3.
As the equation is true for any base, provided all the logarithms have the
same base, it is needless to indicate the base by writing the suffix.
Example 2.
log (113/29)= log 113 log 29.
Example 3.
log (540/539) = log (2 2 . 3 3 . 5/7 2 . 11),
= 2 log2 + 3 log3 + log52 log7logll.
Example 4.
log 4/(49)/ 7/(21) = log (7 2 '*/3^7n
= #log7f log3f log 7,
=Hlog7f log 3.
COMPUTATION AND TABULATION OF LOGARITHMS.
§ 5.] If the base of a system of logarithms be greater than
unity, we have seen that the logarithm of any positive number
greater than unity is positive ; and the logarithm of any positive
number less than unity is negative.
The logarithm of unity itself is always zero, whatever the base
may be.
The logarithm of the base itself is of course unity, since a = a 1 .
The logarithm of any power of the base, say a r , is r ; and, in
particular, the logarithm of the reciprocal of the base is  1.
The logarithm of + oo is + cc , and the logarithm of is  oo .
It is further obvious that the logarithm of a negative number
could not (with our present understanding) be any real quantity.
With such, however, we are not at present concerned.
The logarithm of any number which is not an integral power
of the base will be some fractional number, positive or negative,
as the case may be. For reasons that will appear presently, it is
usual so to arrange a logarithm that it consists of a positive
VOL. I 2 L
514
DETERMINATION OF CHARACTERISTIC
CHAP.
fractional part less than unity, and an integral part, positive or
negative, as the case may be.
The positive fractional part is called the mantissa.
The integral part is called the cliaracteristic.
For example, the logarithm of "0451 to base 10 is the negative
number  1 '3458235. In accordance with the above understand
ing, we should write
log 10 0451
=  13458235 =  2 + (1  3458235),
  2 + '6541765.
For the sake of compactness, and at the same time to prevent
confusion, this is usually written
log 10 0451 = 26541765.
In this case then the characteristic is 2 (that is,  2), and the
mantissa is 6541765.
§ 6.] To find the logarithm of a given number y to a given
base a is the same problem as to solve the equation
a* = y,
where a and y are given and x is the unknown quantity.
There are various ways in which this may be done ; and it
will be instructive to describe here some of the more elementary,
although at the same time more laborious, approximative methods
that might be used.
In the first place, it is always easy to find the characteristic
or integral part of the logarithm of any given number y. We
have simply to find by trial hvo consecutive integral powers of the
base between which the given number y lies. The algebraically less of
these two is the characteristic.
Example 1.
To find the characteristic of log 3 451.
We have the following values for consecutive integral powers of 3 : —
Power
1
2
3
4
5
6
Value
3
9
27
81
243
729
XXI
LOGARITHMS TO BASK 10
515
Hence 3 5 <451<3 fi . Hence log 3 451 lies between 5 am! 6.
log 3 451 = 5+a proper fraction.
Hence char. log 3 451=5.
Example 2.
Find the characteristic of log3 , 0451. We have
Therefore
Powers of Base
1
2
3
Values
1
•333 . . .
•111. . .
•037 . . .
Hence 3~ 3 < '0451 <3~ 2
Hence
; that is to say,
log3  0451 =  3 + a proper fraction,
char. log 3 0451=3.
When the base of the system of logarithms is the radix of
the scale of numerical notation, the characteristic can always be
obtained by merely counting the digits.
For example, if the radix and base be each 10, then
If the number have an integral part, the characteristic of its
logarithm is + {one less than the number of digits in the integral
part).
If the number have no integral part, the characteristic is  {one
more than the number of zeros that follow the decimal point).
The proof of these rules consists simply in the fact that, if a
number lie between 10 M and 10 n+1 , the number of digits by
which it is expressed is n + 1 ; and, if a number lie between
10 (*+!) and 10 _ri , the number of zeros after the decimal point
is n.
For example, 351 lies between 10 2 andl0 3 . Heucechar. logi 351 = 2 = 3  1,
according to the rule.
Again, 0351 lies between '01 and T, that is, between 10 2 and 10' 1 .
Hence char, log w '0351=  2~  (1 + 1), which agrees with the rule.
The rule suggests at once that, if 1 be adopted as the base of
our system of logarithms, then the characteristic of a logarithm depends
merely on the position of the decimal point ; and the mantissa is in
dependent of the position of the decimal point, but depends merely on
the succession of digits.
516 COMPUTATION OF LOGARITHMS BY SOLUTION OF a x = y chap.
We may formally prove this important proposition as
follows : —
Let N be any number formed by a given succession of digits,
c the characteristic, and m the mantissa of its logarithm. Then
any other number which has the same succession of digits as N,
but has the decimal point placed differently, will have the form
10*N, where i is an integer, positive or negative, as the case may be.
But log 10 10*N = log.JO 1 ' + log 10 N, by § 4, = i + log 10 N = (i + c) + m.
Now, since by hypothesis m is a positive proper fraction, and c
and i are integers, the mantissa of log I0 10 f N is m, and the
characteristic is i + c. In other words, the characteristic alone
is altered by shifting the decimal point.
§ 7.] The process used in § 6 for finding the characteristic
of a logarithm can be extended into a method for finding the
mantissa digit by digit.
Example.
To calculate log 10 4217 to three places of decimals.
The characteristic is obviously 0. Let the three first digits of the mantissa
be xyz. Then we have
4217 = 10°^, hence (4217) 10 =10*3«.
We must now calculate the 10th power of 4217. In so doing, however,
there is no need to find all the significant figures— a few of the highest will
suffice. We thus find
1778400 = 10^^.
We now see that x is the characteristic of logi 1778400. Hencea:=6.
Dividing by 10 6 , and raising both sides of the resulting equation to the 10th
power, we find
(l77S) in = 10J' 2 ; hence 315'7 = 10^.
Hence y = 2. Dividing by 10 2 , and raising to the 10th power, we now find
(316) 10 =10 2 ; hence 99280 = 10*.
Hence z = 5 very nearly.
Wc conclude, therefore, that
log 10 4217=625 nearly.
This method of computing logarithms is far too laborious to
be of any practical use, even if it were made complete by the
addition of a test to ascertain what effect the figures neglected
in the calculation of the 10th powers produce on a given decimal
place of the logarithm ; it has, however, a certain theoretical
xxi COMPUTATION BY INSERTING GEOMETRIC MEANS 517
interest on account of its direct connection with the definition of
a logarithm.
By a somewhat similar process a logarithm can he expressed
as a continued fraction.
§ 8.] If a series of numbers be in geometric progression, their
logarithms are in arithmetic progression.
Let the numbers in question be y 1} y. 2 , y a , . . ., y n . Let the
logarithm of the first to base a be a, and the logarithm of the
common ratio of the G.P. y l} y 2 , y.,, . . ., y n to the same base be
(3. Then we have the following series of corresponding values : —
Vi, y2, y*, • • •, Vu,
ii ii ii ii
a a , a x +?, a x+ <\ a *+(».i)^
from which the truth of the proposition is manifest.
As a matter of history, it was this idea of comparing two
series of numbers, one in geometric, the other in arithmetic pro
gression, that led to the invention of logarithms ; and it was on
this comparison that most of the early methods of computing
them were founded.
The following may be taken as an example. Let us suppose
that we know the logarithms x x and a" 9 of two given numbers, y,
and y 9 ; then we can find the logarithms of as many numbers
lying between y x and y 9 as we please. We have
Vi = a Xl , ft = a* 3 
Let us insert a geometric mean, y 6 , between y l and y g , then
% = (Mi.)* = « ( * 1+a " 9)/2 = «*, say,
where x b is the arithmetic mean between x^ and x a . We have
now the following system : —
Logarithm x l x s x ti ,
Number y x y, y 9 .
Next insert geometric means between y } , y B and y u y 9 . The
logarithms of the corresponding numbers will be the arithmetic
means between x u x b and x b , x y . We thus have the system —
Logarithms x 1 , x s , x b , x 7 , x 9
Numbers y u y a , y s , y 7 , y 9 .
^a >
518
COMPUTATION BY INSERTING GEOMETRIC MEANS chai\
Proceeding in like manner, we derive the system —
Logarithms
wo
/y* /y rv* /■>• iy iy* <y* •
[q, a, 2? «^ 3 j w 4 , WfiJ rt 6 , .(' 7 , «^g, «*/g J
Numbers y„ y 9J y 3J y i} y„ y 6 , y 7 , y 8 , y 9 ;
and so on. Each step in this calculation requires merely a multi
plication, the extraction of a square root, an addition accompanied
by division by 2, and each step furnishes us with a new number
and the corresponding logarithm.
Since x 1} x 2 , . . ., x n form an A.P., the logarithms are spaced
out equally, but the same is not true of the corresponding num
bers which are in G.P. It is therefore a table of antilogarithms *
that we should calculate most readily by this method. It will
be observed, however, that by inserting a sufficient number of
means we can make the successive numbers differ from each other
as little as we please ; and by means of the method of interpola
tion by first differences, explained in the last section of this
chapter, we could space out the numbers equally, and thus con
vert our table of antilogarithms into a table of logarithms of the
ordinary kind.
As a numerical example we may put a = 10, y 1 = l,y s = 10; thena^=0, a'9=l.
Proceeding as above indicated, we should arrive at the following table : —
Number.
Logarithm.
Number.
Logarithm.
10000
ooooo
42170
06250
13336
01250
56235
07500
17783
02500
74990
08750
23714
03750
100000
1 oooo
31622
05000
§ 9.] In computing logarithms, by whatever method, it is
obvious that it is not necessary to calculate independently the
logarithms of composite integers after we have found to a suffi
cient degree of accuracy the logarithms of all primes up to a
certain magnitude. Thus, for example, log 4851 = log 3 2 .7 2 11
= 2 log 3 + 2 log 7 + log 11. Hence log 4851 can be found when
the logarithms of 3, 7, and 1 1 are known.
* By the antilogarithm of any number N is meant the number of which
X is the logarithm.
XXI
CHANGE OF BASE 519
Again, having computed a system of logarithms to any one base
a, we can without difficulty deduce therefrom a system to any other base
b. All we have to do is to multiply all the logarithms of the former
system by the number fx = l/\og a b.
For, if x = log b y, then y = L x .
Hence log rt y = log a ^,
= ajlog a &, by § 4.
Hence \og b y = x = log a ?//log a & (1).
The number //, is often called the modulus of the system
whose base is b with respect to the system whose base is a.
Cor. 1. If in the equation (1) we put y=a, we get the
following equation, which could easily be deduced more directly
from the definition of a logarithm : —
log & «= l/log a 6,
or log a Hog 6 a=l (2).
Cor. 2. The equation y = b x may be written
y = a xlos « b or y = a x!l0Bba .
Hence the graph of the exponential b x can be deduced from the
graph of the exponential a x by shortening or lengthening all the abscissce
of the latter in the same ratio 1 : log a b.
This is the general theorem corresponding to a remark in § 2.
We may also express this result as follows : —
Given any two exponential graphs A and B, then either A is the
orthogonal projection of B, or B is the orthogonal projection of A, on a
plane passing through the axis of y.
USE OF LOGARITHMS IN ARITHMETICAL CALCULATIONS.
§ 10.] We have seen that, if we use the ordinary decimal
notation, the system of logarithms to base 10 possesses the im
portant advantages that the characteristic can be determined by
inspection, and that the mantissa is independent of the position
of the decimal point. On account of these advantages this
system is used in practical calculations to the exclusion of all
others.
520
SPECIMEN OF LOGARITHMIC TABLE.
Oil A P.
No.
12 3 4
5 6 7 8 9
Diff.
3050
484 2998 3141 32S3 342G 3568
3710 3853 3995 4137 4280
51
4422 4564 4707 4849 4991
5134 5276 5418 5561 5703
52
5845 5988 6130 6272 6414
6557 6699 6841 6984 7126
53
7268 7410 7553 7695 7837
7979 8121 8264 8406 8548
54
8690 8833 8975 9117 9259
9401 9543 9686 9828 9970
55
485 0112 0254 0396 0539 0GS1
0823 0965 1107 1249 1391
56
1533 1676 1818 1960 2102
2244 2386 2528 2670 2812
57
2954 3096 3239 33S1 3523
3665 3807 3949 4091 4233
142
58
4375 4517 4659 4801 4913
5085 5227 5369 5511 5053
1 14
59
5795 5937 6079 6221 6363
6505 6647 6788 6930 7072
2 28
3 43
4 57
60
7214 7356 7498 7640 7782
7924 8066 8208 8350 8491
3061
8633 8775 8917 9059 9201
9343 9484 9626 9768 9910
5 71
62
4860052 0194 0336 0477 0619
0761 0903 1045 1186 1328
6 85
63
1470 1612 1754 1895 2037
2179 2321 2462 2604 2746
7 99
64
2888 3029 3171 3313 3455
3596 3738 3880 4021 4163
8 114
9 128
65
4305 4446 4588 4730 4872
5013 5155 5297 5438 5580
66
5722 5863 6005 6146 62S8
6430 6571 6713 6855 6996
67
7138 7279 7421 7563 7704
7846 7987 8129 8270 8412
68
8554 8695 8837 8978 9120
9261 9403 9544 9686 9827
69
9969 0110 0252 0393 0535
0676 0818 0959 1101 1242
70
4871384 1525 1667 1808 1950
2091 2232 2374 2515 2657
3071
2798 2940 3081 3222 3364
3505 3647 3788 3929 4071
72
4212 4353 4495 4636 4778
4919 5060 5202 5343 5484
73
5626 5767 5908 6050 6191
6332 6473 6615 6756 6897
74
7039 7180 7321 7462 7604
7745 7886 8027 8169 8310
75
76
8451 8592 8734 8875 9016
9863 0004 0146 0287 042S
9157 9299 9440 9581 9722
0569 0710 0852 0993 1134
77
488 1275 1416 1557 1698 1839
1981 2122 2263 2404 2545
78
2686 2827 2968 3109 3251
3392 3533 3674 3815 3956
79
4097 4238 4379 4520 4661
4802 4943 5084 5225 5366
80
5507 5648 5789 5930 6071
6212 6353 6494 6635 6776
141
1 14
2 28
3081
6917 7058 7199 7340 7481
7622 7763 7904 8045 8185
82
8326 8467 860S 8749 8890
9031 9172 9313 9454 9594
3 42
83
9735 9876 0017 0158 0299
0440 0580 0721 0862 1003
4 56
84
4891144 1285 1425 1566 1707
1848 1989 2129 2270 2411
5 71
85
2552 2692 2833 2974 3115
3256 3396 3537 3678 3818
6 85
7 99
8 113
86
3959 4100 4241 4381 4522
4663 4804 4944 5085 5226
87
5366 5507 5648 5788 5929
6070 6210 6351 6492 6632
9 127
88
6773 6914 7054 7195 7335
7476 7617 7757 7898 8038
89
8179 8320 8460 8C01 8741
8882 9023 9163 9304 9444
90
9585 9725 9866 0006 0147
0287 0428 0569 0709 0850
3091
4900990 1131 1271 1412 1552
1693 1833 1973 2114 2254
92
2395 2535 2676 2816 2957
3097 3238 3378 3518 3659
93
3799 3940 4080 4220 4361
4501 4642 4782 4922 5063
94
5203 5343 5484 5624 5765
5905 6045 6186 6326 6466
95
6G07 6747 6887 7027 7168
7308 7448 7589 7729 7869
96
8010 8150 8290 8430 8571
8711 8851 8991 9132 9272
97
9412 9552 9693 9833 9973
0113 0253 0394 0534 0674
98
4910S14 0954 1094 1235 1375
1515 1655 1795 1935 2076
99
2216 2356 2496 2636 2776
2916 3057 3197 3337 3477
3100
3617 3757 3897 4037 4177
4317 4457 4597 4738 4878
xxl USE OF LOGARITHMIC TABLE 521
la printing a table of logarithms to base 10 it is quite un
necessary, even if it were practicable, to print characteristics.
The mantissa? alone are given, corresponding to a succession of
five digits, ranging usually from 10000 to 99999.*
A glance at p. 520, which is a facsimile of a page of the
logarithmic table in Chambers's Mathematical Tables, will show
the arrangement of such a table. To take out the logarithm of
30715 from the table, we run down the column headed "No."
until we come to 3071 ; the first three figures of the mantissa
are 487 (standing over the blank in the first half column) ; the
last four are found by running along the line till Ave reach the
column headed 5, they are 3505. The characteristic is seen by
inspection to be 4. Hence log 30715 = 44873505.
To find the number corresponding to any given logarithm
we have of course simply to reverse the process.
To find the logarithm of '030715 Ave have to proceed exactly
as before, only a different characteristic, namely 2, must be pre
fixed to the mantissa. We thus find log 030715 = 24873505.
If we Avish to find the logarithm of a number, say 3  083279,
Avhere we have more digits than are given in the table, then we
must take the nearest number whose logarithm can be found
by means of the table, that is to sa} r , 30833. "We thus find
log 30833 = 04890158f nearly. Greater accuracy can be at
tained by using the column headed c 'Diff," as will be explained
presently.
Conversely, if a logarithm be given Avhich is not exactly
coincident with one given in the table, Ave take the one in the
table that is nearest to it, and take the corresponding number as
an approximation to the number required. Greater accuracy
can be obtained by using the difference column. Thus the
number whose logarithm is 1 4872191 has for its first five
* For some purposes an extension of the table is required, and such ex
tensions are supplied in various ways, which need not be described here. For
rapidity of reference in calculations that require no great exactness a short
table for a succession of 3 digits, ranging from 100 to 999, is also usually
given.
+ The bar over 0158 indicates that these digits follow 489, and not 48S.
522 NUMBER OF FIGURES REQUIRED chap.
significant digits 30705 ; but, if we wish the best approximation
with five digits, we ought to take 30706. Since the character
istic is 1, the actual number in question has two integral digits.
Hence the required number is 30'706, the error being certainly
less than '0005.
§ 11.] The principle which underlies the application of
logarithms to arithmetical calculation is the very simple one
that, since to any number there corresponds one and only one
logarithm, a number can be identified by means of its logarithm.
It is this principle which settles how many digits of the
mantissa of a logarithm it is necessary to use in calculations
which require a given degree of accuracy.
Suppose, for example, that it is necessary to be accurate
down to the fifth significant figure ; and let us inquire whether
a table of logarithms in which the mantissae are given to four
places would be sufficient. In such a table we should find log
30701 =04871, log 30702 = 04871 ; the table is therefore
not sufficiently extended to distinguish numbers to the degree
of accuracy required. Five places in the mantissa would, in the
present instance, be sufficient for the purpose; for log 30701
= 048715, log 30702 = 048716. Towards the end of the
table, however, five places would scarcely be sufficient ; for log
94910 = 097731 and log 94911 = 097731.
§ 12.] The great advantage of using in any calculation
logarithms instead of the actual numbers is that we can, in
accordance with the rules of § 4, replace every multiplication by
an addition, every division by a subtraction, and every operation
of raising to a power or extracting a root by a multiplication or
division.
The following examples will illustrate some of the leading
cases. We suppose that the student has a table of the loga
rithms of all numbers from 10000 to 100000, giving mantissa?
to seven places.
Example 1.
Calculate the value of 16843 x 00132 J '3692.
If A = 16843 x 00132H3692,
log A = log 1 6843 + log 00132  log 3692,
XXI
EXAMPLES 523
log 16843 = 2264194
log 00132 = 31205739
3*3469933
log 3692 =15672617
logA = 37797316.
Hence A= 0060219.
Observe that the negative characteristics must be dealt with according to
algebraic rules.
Example 2.
To extract the cube root of 016843.
Let A = (016843) 1 / 3 , then
log A=  log 016843,
= i(2 2264194),
= 4(3 + 12264194),
= 14088065.
A= 25633.
Example 3.
Calculate the value of A = (368)"' 3 /(439) 5;9 .
Log A =1 log 368 flog 439.
I log 368 = (2 5658478) = 59869782
$ log 439 = f (2 6424645) = 1 "4680358
log A = 45189424
A= 33033.
Example 4.
Find how many digits there are in A = (l01) 10000 .
log A= 10000 log 101,
= 10000 x 0043214,
= 43214.
Hence the number of digits in A is 44.
Example 5.
To solve the exponential equation 12*=11 by means of logarithms.
AVe have log 1 2* = log 1 • 1 .
Therefore a?logl'2=logl , L
„ _ log 11 _ 0413927
Ce ""log l2~ 0791812 "
Hence log.r = l_og 0413927 log 0791812,
= 17183059.
Therefore x — 52276.
Remark.— It is obvious that we can solve any such equation as a* 2 *"** = &,
where;, q, a, b are all given. For, taking logarithms of both sides, we have
(x 2 px + q) log a = log b.
We can now obtain the value of a: by solving a quadratic equation.
524 INTERPOLATION BY FIRST DIFFERENCES chap.
INTERPOLATION BY FIRST DIFFERENCES.
§ 13.] The method by which it is usual to find (or "interpo
late ") the value of the logarithm of a number which does not
happen to occur in the table is one which is applicable to any
function whose values have been tabulated for a series of equi
different values of its independent variable (or "argument").
The general subject of interpolation belongs to the calculus
of finite differences, but the special case where first differences
alone are used can be explained in an elementary way by means
of graphical considerations.
We have already seen that the increment of an integral
function of x of the 1st degree, y = A.?; + B say, is proportional
to the increment of its argument ; or, what comes to the same
thing, if we give to the argument x a series of ecpiidifferent
values, a, a + h, a + 2h, a + 3/j, &c, the function y will assume a
series of equidifferent values Aa + B, Aa + B + Ah, Aa + B + 2 Ah,
Aa + B + 3Ah, &c.
If, therefore, we were to tabulate the values of Ax + B for a
series of equidifferent values of x, the differences between suc
cessive values of y (" tabular differences ") would be constant, no
matter to how many places we calculated y.
Conversely, a function of x which has this property, that the
differences between the successive values of y corresponding to
equidifferent values of x are absolutely constant, must be an in
tegral function of x of the 1st degree.
If, however, we take the difference, h, of the argument small
enough, and do not insist on accuracy in the value of y beyond
a certain significant figure, then, for a limited extent of the table
of any function, it will be found that the tabular differences are
constant.
Bef erring, for example, to p. 520, it will be seen that the
difference of two consecutive logarithms is constant, and equal
to 0000141, from log 30660 up to log 30S99, or that there is
merely an accidental difference of a unit in the last place ; that
is to say, the difference remains constant for about 240 entries.
XXI
LIMITS OF THE METHOD
525
A similar phenomenon will be seen in the following extract
from Barlow's Tables, provided we do not go beyond the 7th
significant figure :
Number.
Cube Root.
Diff.
2301
132019740
19122
2302
132038862
19117
2303
132057979
19111
2304
132077090
19105
2305
132096195
Let us now look at the matter graphically. Let ACSDQB
be a portion of the graph of a function y = /(•>') ; and let us
suppose that up to the nth significant figure the differences of y
are constant for equidifferent
values of x, lying between
OE and OH. This means
that in calculating (up to the
nth significant figure) values
of y corresponding to values
of x between OE and OH we
may replace the graph by the
straight line AB. Thus, for
example, if x  OM, then
/(OM) = MQ ; and PM is the
value calculated by means of the straight line AB. Our state
ment then is that PM  QM, that is PQ, is less than a unit in
the n significant place.
If this be so, then, a fortiori, it will be so if Ave replace a
portion of the graph, say CD, lying between A and B by a
straight line joining C and D.
In other words, if up to the nth place the increment of the func
tion for eqiudifferent values of x be constant, between certain limits,
then, to that degree of accuracy at least, the increment of the function
will be proportional to the increment of the argument fcrr all values
within those limits.
§ 1 4.] Let us now state the conclusion of last article under
Fig. 2.
526 RULE OF PROPORTIONAL PARTS chap.
an analytical form, all the limitations before mentioned as to
constancy of tabular (or first) difference being supposed fulfilled.
Let h be the difference of the arguments as they are entered
in the table, D the tabular difference f(a + h) /(«), a + h' a value
of the argument, which does not occur in the table, but which
lies between the values a and a + h, which do occur, so that h'<h.
Then, by last paragraph,
f(a + h') f (a) _ (a + h')  a
f(a + h) f(a) (a + h)  a
Hence f{a + h')f(a) K
D h
Since in (1) f(a), D, h are all known, it gives us a relation
between h' and f(a + h'). When, therefore, one or other of these
is given, we can calculate the other. We have in fact
f(a + h')=f(a) + ^J) (2),
and a + h^a + f(a + h ^ f(a) h (3).
From (2) we find a value of the function corresponding to a
given intermediate value of the argument. From (3) we find an
intermediate value of the argument corresponding to a given
intermediate value of the function.
The equations (2) and (3) are sometimes called the Rule of
Proportional Parts.
Example 1.
To find by means of first differences the value of ^/(2303 , 45) as accurately
as Barlow's Table will allow.
By the rule of proportional parts, we have
^'(2303 45) = ^(2303 00) + T <&( 00191),
= 1320580+ 00086,
= 1320666,
which will be found correct down to the last figure.
The only labour in the above calculation consists in working
out the fraction 45/100 of the tabular difference. In tables of
XXI
EXAMPLES 527
logarithms even this labour is spared the calculator ; for under
each difference there is a small table of proportional parts, giving
the values of 1/10, 2/10, 3/10, 4/10, 5/10, 6/10, 7/10, 8/10,
9/10 of the difference in question (see the last column on p. 520).
It will be observed that, if we strike the last figure off each of
the proportional parts (increasing the last of those left if the
one removed exceeds 5), we have a table of the various hun
dredths, and so on. Hence Ave can use the table twice over (in
some cases it might be oftener), as in the following example : —
Example 2.
To find log 30 81 345.
We may arrange the corresponding contributions as follows ' —
3081300 14887340
40 56
5 7
log 30 81345 = 14887403
Example 3.
To find the number whose logarithm is 14871763.
14871763
14871637 3070200
96
85 60
11 8
antilog 1 4871763 = 3070268
Here we set down under the given logarithm the next lowest in the table,
and opposite to it the corresponding number 30702.
Next, we write down 0000096, the difference of these two logarithms, and
look for the greatest number in the table of proportional parts which does not
exceed 96— this is 85. We set down 85, and opposite to it the corresponding
figure 6.
Lastly, we subtract 85 from 96, the result being 0000011. We then
imagine a figure struck off every number in the table of proportional parts,
look for the remaining one which stands nearest to 11, and set down the
figure, namely 8, corresponding to it, as the last digit of the number we are
seeking.
Exercises XLIII.
(1.) Find the characteristics of log 10 36983, logi 5 8 , logi 5 3 , log 10 '00068
(2.) Find the characteristics of log 5 1067, log 5 0138, log^, logvAjl/8
(3.) Find log 2 8 ^2.
(4.) Calculate log 2 36'432 to two places of decimals.
528 EXERCISES XLIII
CHAP.
Calculate out the values of tlio following as accurately as your tables will
allow: —
(5.) 4163x7835. (6.) *3068 x 0015^0579.
(7.) (50063745) 5 . (8.) ^/(5 0063745).
(9.) (01369) 12 . (10.) (001369)*.
(11.) {15(*318)fyl6}^. (12.) {(1035) 7  1}/ {1035  1}.
(13.) The population of a country increases each year by *13 % 0I " its
amount at the beginning of the year. By how much °/ will it have increased
altogether after 250 years ?
(14.) If the number of births and deaths per annum be 3*5 and 1*2 °/
respectively of the population at the beginning of each year, after how many
years will the population be trebled ?
(15.) Calculate the value of ^(32 6  8 +55 3  6 ).
(16.) Calculate the value of 1 + c + c + . . . +e 19 , where e= 2 71828.
(17.) Find a mean proportional between 3*17934 and 3*987636.
(18.) Insert three mean proportionals between 65"342 and 88*63.
(19.) The 1st and 13th terms of a geometric progression are 3 and 65
respectively : find the common ratio.
(20.) The 4th and 7th terms of a geometric progression are 31 and 52
respectively : find the 5 th term.
(21.) How many terms of the series s + p+p+ ■ • • must be taken in
order that its sum may differ from unity by less than a millionth ?
(22.) Given log 10 5= *69897, find the number of digits in (\/5) 9a .
(23.) Given log 2673 = 3*427, and log 3267 = 3*51415, find log 11.
(24.) Find the first four significant figures and the number of digits in
1.2.3.4. . . 20.
(25.) How many terms of 3 1 , 3, 3 3 , . . . must I take in order that the
product may just exceed 100000 ?
(26.) Given log x 36 = 1*3678, find a:.
(27.) Given log a: 6£ = 2, find x.
Solve the following equations : —
(28.) 21* = 20. (29.) 2* 15*= 5. (30.) 2* 2 = 5 x 2*.
(31.) 6* = 5y, 7*=3y. (32.) 3*3*=5.
(33.) 2 3 *+ 2 »=5, 4 2 *=2 2 <'+ ;! . (34.) &*&** >;»\ipx m
(35. ) (a + b)^(a 4  2au + b*)* 1 = {a bf*.
(36.) x x +y = y ia , y x +« = x a .
(37.) Find by means of a table of common logarithms log e 16.345, where
« = 2*71828.
(38.) Show that x—a ; and that x = a
(39.) Show that
log„(log„N) _ h^(k>gj^
V(l0ga"i) V(log*'0 : V(l<>go&) \/(l0g»rt)
(40.) Show that the logarithm of any number to base a n is a mean pro
g
portional between its logarithms to the bases a and a" .
xxr HISTORICAL NOTE 529
(41.) If P, Q, R be the pth, qth, rth terms of a geometric progression,
show that 2(y  r) log P = 0.
(42.) If ABC be in harmonic progression, show that log(A + C)
+ log(A + C2B) = 2log(AC).
(43.) If a, b, c be in G.P., show that S{log (6/c)} 1 =3{log 4 (c/a)} 1 .
(44.) If a, b, c be in G.P., and log c «, log 6 c, log a b in A. P., then the
common difference of the latter is 3/2.
(45.) If a 2 + b 2 = c, then log 6+c « + log c _ & a = 2iog 6+c relog c _ 6 «.
/ a /» \ re x (v +a — x) y(z + x  y) z(x + y  z) . , • „ „ . „ .
(46.) If 5 ^. =^r =■ , , then y*0'=f? ! x*=x*y*.
log a; logy logs ' J a
(47.) If :c 2 = \ogx n Xi, x z =\og Xl x 2) K 4 =logx 2 a! s , . . ., av,=logr B _ 2 a3„i, x 1 =
l°S*n\ x »> then XlX ' ■ ■ ' T » = 1 
Historical Note. — The honour of devising the use of logarithms as a means of
abbreviating arithmetical calculations, and of publishing the first logarithmic
table, belongs to John Napier (15501617) of Merchiston (in Napier's day near,
iu our day in, Edinburgh). This invention was not the result of a casual inspira
tion, for we learn from Napier's Rabdolorjla (1617), in which he describes three
other methods for facilitating arithmetical calculations, among them his calculat
ing rods, which, uuder the name of " Napier's Bones," were for long nearly as
famous as his logarithms, that he had devoted a great part of his life to the con
sideration of methods for increasing the power and diminishing the labour of
arithmetical calculation. Napier published his invention in a treatise entitled
"Mirifici Logarithmorum Cauonis Descriptio, ejusque usus, in utraque Trigo
nometria ut etiam in omni Logistica Mathematica, Amplissimi, facillimi, et
expeditissimi explicatio. Authore ac Inventore Ioanne Nepero 5 Barone Merchis
tonii, &c, Scoto, Edinburgi (1614)." In this work he explains the use of
logarithms ; and gives a table of logarithmic sines to 7 figures for every minute
of the quadrant. In the Canon Mirificus the base used was neither 10 nor what
is now called Napier's base (see the chapter on logarithmic series in the second
part of this work). Napier himself appears to have been aware of the advantages
of 10 as a base, and to have projected the calculation of tables on the improved
plan ; but his infirm health prevented him from carrying out this idea ; and his
death three years after the publication of the Canon Mirificus prevented him from
even publishing a description of his methods for calculating logarithms. This
work, entitled Mirifici Logarithmorum Canonis Constructio, &c. , was edited by
one of Napier's sons, assisted by Henry Briggs.
To Henry Briggs (15561630), Professor of Geometry at Gresham College, and
afterwards Savilian Professor at Oxford, belongs the place of honour next to
Napier in the development of logarithms. He recognised at once the merit and
seized the spirit of Napier's invention. The idea of the superior advantages
of a decimal base occurred to him independently ; and he visited Napier in
Scotland in order to consult with him regarding a scheme for the calculation
of a logarithmic table of ordinary numbers on the improved plan. Finding
Napier in possession of the same idea in a slightly better form, he adopted his
suggestions, and the result of the visit was that Briggs undertook the work which
Napier's declining health had interrupted. Briggs published the first thousand
of his logarithms in 1617 ; and, in his Arithmetica Logarithmica, gave to 14
places of decimals the logarithms of all integers from 1 to 20,000, and from
90,000 to 100,000. In the preface to the lastmentioned work he explains the
methods used for calculating the logarithms themselves, and the rules for using
them in arithmetical calculation.
VOL. I 2 M
530 HISTOKICAL NOTE chap, xxi
While Briggs was engaged in filling up the gap left in his table, the work of
calculating logarithms was taken up in Holland by Adrian Vlacq, a bookseller of
Gouda. He calculated the 70,000 logarithms which were wanting in Briggs'
Table ; and published, in 1628, a table containing the logarithms to 10 places of
decimals of all numbers from 1 to 100,000. The work of Briggs and Vlacq has
been the basis of all the tables published since their day (with the exception of
the tables of Sang, 1871) ; so that it forms for its authors a monument both
lasting and great.
In order fully to appreciate the brilliancy of Napier's invention and the merit
of the work of Briggs and Vlacq, the reader must bear in mind that even the
exponential notation and the idea of an exponential function, not to speak of the
inverse exponential function, did not form a part of the stockintrade of mathe
maticians till long afterwards. The fundamental idea of the correspondence of
two series of numbers, one in arithmetic, the other in geometric progression, which
is so easily represented by means of indices, was explained by Napier through
the conception of two points moving on sejjarate straight lines, the one with
uniform, the other with accelerated velocity. If the reader, with all his acquired
modern knowledge of the results to be arrived at, will attempt to obtain for him
self in this way a demonstration of the fundamental rules of logarithmic calcula
tion, he will rise from the exercise with an adequate conception of the penetrating
genius of the inventor of logarithms.
For the full details of this interesting part of mathematical history, and
in particular for a statement of the claims of Jost Biirgi, a Swiss contemporary
of Napier's, to credit as an independent inventor of logarithms, we refer the
student to the admirable articles "Logarithms" and "Napier," by J. W. L. Glaisher,
in the Encyclopaedia Britannica (9th ed.). An English translation of the
Constnictio, with valuable bibliographical notes, has been published by Mr. W.
R. Macdonald, F.F.A. (Edinb. 1889).
CHAPTEE XXII.
Theory of Interest and Annuities Certain.
§ 1.] Since the mathematical theory of interest and annuities
affords the best illustration of the principles we have been dis
cussing in the last two chapters, we devote the present chapter
to a few of the more elementary propositions of this important
practical subject. What we shall give will be sufficient to enable
the reader to form a general idea of the principles involved.
Those whose business requires a detailed knowledge of the
matter must consult special textbooks, such as the TextBook of
the Institute of Actuaries, Part I., by Sutton.*
SIMPLE AND COMPOUND INTEREST.
§ 2.] When a sum of money is lent for a time, the borrower
pays to the lender a certain sum for the use of it. The sum
lent is spoken of as the capital or principal ; the payment for
the privilege of using it as interest. There are various ways of
arranging such a transaction ; one of the commonest is that the
borrower repays after a certain time the capital lent, and pays
also at regular intervals during the time a stated sum by way of
interest. This is called paying simple interest on the borrowed
capital. The amount to be paid by way of interest is usually
stated as so much per cent per annum. Thus 5 per cent (5 %)
per annum means £5 to be paid on every £100 of capital, for
* Full references to the various sources of information will be found in
the article "Annuities" (by Sprague), Encyclopaedia Britamica, 9th edition,
vol. ii.
532 AMOUNT PRESENT VALUE DISCOUNT chap.
every year that the capital is lent. In the case of simple
interest, the interest payable is sometimes reckoned strictly in
proportion to the time ; that is to say, allowance is made not
only for whole years or other periods, but also for fractions
of a period. Sometimes interest is allowed only for integral
multiples of a period mutually agreed on. We shall suppose
that the former is the understanding. If then r denote the
interest on £1 for one year, that is to say, onehundredth of
the named rate per cent, n the time reckoned in years and
fractions of a year, P the principal, I the whole interest paid, A
the amount, that is, the sum of the principal and interest, both
reckoned in pounds, we have
I = nrP (1); A = I + P = P(1 + nr) (2).
These formulae enable us to solve all the ordinary problems of
simple interest.
If any three of the four I, n, r, P, or of the four A, n, r, P,
be given, (1) or (2) enables us to find the fourth.
Of the various problems that thus arise, that of finding P
when A, n, r are given is the most interesting. We suppose
that a sum of money A is due n years hence, and it is required
to find what sum paid down at once would be an equitable
equivalent for this debt. If simple interest is allowed, the
answer is, such a sum P as would at simple interest amount in n
years to A. In this case P = A/(l + nr) is called the present
value of A, and the difference AP = A{1 1/(1 + nr)}
= A?w/(1 + nr) is called the discount. Discount is therefore the
deduction allowed for immediate payment of a sum due at some
future time. The discount is less than the simple interest
(namely Anr) on the sum for the period in question. When n
is not large, this difference is slight.
Example.
Find the difference between the interest and the discount on £1525
due nine months hence, reckoning simple interest at 3£ %• The difference
in question is given by
Anr  An?i(l + nr) = Anr/(l +nr).
XXH COMPOUND INTEREST 533
In the present case A *= 1525, n = 9/12 = 3/4, r= 35/100= 035. Hence
Difference = 1525 x ( 02625) 2 f(l 02625),
= £1 :0:5£.
§ 3.] In last paragraph we supposed that the borrower paid
up the interest at the end of each period as it became due. In
many cases that occur in practice this is not done ; but, instead,
the borrower pays at the end of the whole time for which the
money was lent a single sum to cover both principal and
interest. In this case, since the lender loses for a time the use
of the sums accruing as interest, it is clearly equitable that the
borrower should pay interest on the interest ; in other words,
that the interest should be added to the principal as it becomes
due. In this case the principal or interestbearing capital
periodically increases, and the borrower is said to pay compound
interest. It is important to attend carefully here to the under
standing as to the period at which the interest is supposed to
become due, or, as it is put technically, to be convertible (into
principal) ; for it is clear that £100 will mount up more rapidly
at 5 % compound interest convertible halfyearly than it will at
5 % compound interest payable annually. In one year, for in
stance, the amount on the latter hypothesis will be £105, on the
former £105 plus the interest on £2 : 10s. for a halfyear, that
is, £105 : 1 :3.
In what follows we shall suppose that no interest is allowed
for fractions of the interval (conversion period) between the
terms at which the interest is convertible, and we shall take the
conversionperiod as unit of time. Let P denote the principal, A
the accumulated value of P, that is, the principal together with
the compound interest, in n periods ; r the interest on £1 for
one period ; 1 + r = R the amount of £1 at the end of one period.
At the end of the first period P will have accumulated to
P + Pr, that is, to PR. The interestbearing capital or principal
during the second period is PR ; and this at the end of the
second period will have accumulated to PR + PRr, that is, to
PR 2 . The principal during the third period is PR 2 , and the
amount at the end of that period PR 3 , and so on. In short, the
534 EXAMPLES CHA1\
amount increases in a geometrical progression whose common ratio is
R ; and at the end of n periods we shall liave
A = PR» (1).
By means of this equation we can solve all the ordinary
problems of compound interest ; for it enables us, when any three
of the four quantities A, P, R, n are given, to determine the
fourth. In most cases the calculation is greatly facilitated by
the use of logarithms. See the examples worked below.
Cor. I. If I denote the whole compound interest on P during the
n periods, we have
I=AP = P(R"1) (2).
Cor. 2. If P denote the present value of a sum A due n periods
hence, compound interest being allowed, then, since P must in n periods
amount to A, we have
A = PR",
so that P = A/R Jl (3).
The discount on the present understanding is therefore
A(l  1/R' 1 ) (4).
Example 1.
Find the amount in two years of £2350 : 5 : 9 at Z\ % compound interest,
convertible quarterly.
Here P = 23502875, n = S, r = 35/400 =00875, R = l00875.
log A = log P + n log R,
log P = 33711210
?ilogR* = 0302684
34013894
A = £2519936 = £2519 : 18 : 8.
Example 2.
How long will it take £186 : 14 : 9 to amount to £216 : 9 : 7 at 6 % com
pound interest, convertible halfyearly.
* When n is very large, the seven figures given in ordinary tables hardly
afford the necessary accuracy in the product n log R. To remedy this defect,
supplementary tables are usually given, which enable the computer to find
very readily to 9 or 10 places the logarithms of numbers (such as R) which
differ little from unity.
xxn NOMINAL AND EFFECTIVE RATE .535
Here P= 186 7375, A = 216"4792, R=1'03.
»=(logAlogP)/logR
•0641847
= 500
•0128372
Hence the required time is five halfyears, that is, 1\ years.
Example 3.
To find the present value of £1000 due 50 years hence, allowing compound
interest at 4 %> convertible halfyearly.
HereA = 1000, ?i = 100, R = l02. We have P=A/R».
log P = log 1000  100 log 102,
= 3  100 x 0086002,
= 21399800.
P = £138032 = £138 : : 8.
§ 4.] In reckoning compound interest it is very usual in
practice to reckon by the year instead of by the conversion
period, as we have done above, the reason being that different
rates of interest are thus more readily comparable. It must be
noticed, however, that when this is done the rate of interest to
be used must not be the nominal rate at which the interest due
at each period is reckoned, but such a rate (commonly called the
effective rate) as would, if convertible annually, be equivalent to
the nominal rate convertible as given.
Let r n denote the effective rate of interest per pound which
is equivalent to the nominal rate r convertible every l/?ith part
of a year ; then, since the amount of £ 1 in one year at the two
rates must be the same, we have
(l+r) n =l+r n ,
that is, r n = (\+r)' l l (1),
and r = (l+r n ) 1 / B l (2).
The equations (1) and (2) enable us to deduce the effective rate
from the nominal rate, and vice versa.
Example.
The nominal rate of interest is 5 %> convertible monthly, to find the
effective rate.
Here r= 05/12= 004166.
Hence ? 12 =(l004166) 12  1,
= 1051141.
r 12 = 05114.
Hence the effective rate is 5 "11 4 °/ .
536 ANNUITIES TERMINABLE OE PERPETUAL CHAP.
ANNUITIES CERTAIN.
§ 5.] When a person has the right to receive every year a
certain sum of money, say £A, he is said to possess an annuity
of £A. This right may continue for a fixed number of years and
then lapse, or it may be vested in the individual and his heirs
for ever ; in the former case the annuity is said to be terminable,
in the latter perpetual. A good example of a terminable annuity
is the not uncommon arrangement in lending money where B
lends C a certain sum, and C repays by a certain number of
equal annual instalments, which are so adjusted as to cover both
principal and interest. The simplest example of a perpetual
annuity is the case of a freehold estate which brings its owner a
fixed income of £A per annum.
Although in valuing annuities it is usual to speak of the
whole sum which is paid yearly, yet, as a matter of practice, the
payment may be by halfyearly, quarterly, &c. instalments ; and
this must be attended to in annuity calculations. Just as in
compound interest, the simplest plan is to take the interval
between two consecutive payments, or the conversionperiod, as
the unit of time, and adjust the rate of interest accordingly.
In many cases an annuity lasts only during the life of a cer
tain named individual, called the nominee, who may or may not
be the annuitant. In this and in similar cases an estimate of
the probable duration of human life enters into the calculations,
and the annuity is said to be contingent. In the second part of
this work we shall discuss this kind of annuities. For the
present we confine ourselves to cases where the annual payment
is certainly due either for a definite succession of years or in
perpetuity.
§ 6.] One very commonly occurring annuity problem is to
find the accumulated value of a FORBORNE annuity. An annuitant
B, who had the right to receive n successive payments at n suc
cessive equidistant terms, has for some reason or other not
received these payments. The question is, What sum should he
receive in compensation ?
XXII
ACCUMULATION OF FORBORNE ANNUITY 537
To make the question general, let us suppose that the last of
the n instalments was due m periods ago.
It is clear that the whole accumulated value of the annuity
is the sum of the accumulated values of the n instalments,
and that compound interest must in equity be allowed on each
instalment.
Now the nth instalment, due for m periods, amounts to AR" ! ,
the n  1th to AE" l+1 , the n  2th to AR m + 2 ; and so on. Hence,
if V denote the whole accumulated value, we have
V = AR m + AR w+1 + . . . +AR m +» 1 (1).
Summing the geometric series, we have
V = AR™(R n l)/(Rl) (2).
Cor. If the last instalment be only just due, m = 0, and the
accumulated value of the forborne annuity is given by
V = A(R»1)/(R1) (3).
Example.
A farmer's rent is £156 per annum, payable halfyearly. He was unable
to pay for five successive years, the last halfyear's rent having been due three
years ago. Find how much he owes his landlord, allowing compound interest
at 3%.
HereA = 78, R = l015, m = 6, n = 10.
V = 78 x 1015 6 (1015 10 1)/015.
10 log 1015 =0646600,
1015 10 = 116054.
V = 78 x l015 6 x 16054/015.
log 78 =18920946
6 log 1015 = 0387960
log 16054
= 12055833
log 015
11364739
= 21760913
logV
= 29603826
Y = £912814 = £912
16 : 3.
§ 7.] Another fundamental problem is to calculate the purchase
price of a given annuity. Let us suppose that B wishes, by paying
down at once a sum £P, to acquire for himself and his heirs the
right of receiving n periodic payments of £A each, the first pa}*
ment to be made m periods hence. We have to find P.
P is obviously the sum of the present values of the n pay
538 PURCHASE PRICE OF AX ANNUITY chap.
ments. Now the first of these is due m years hence ; its present
value is therefore A/K" 1 . The second is due ra + 1 years hence ;
A's present value is therefore A/K ,n+1 , and so on. Hence
V— A A .
Hence
P = K^( 1 "^)/( 1 "F.
A — l (Rn_ l)/(R _ l) (2) .
K m+r
Cor. 1. The ratio of the purchase price of an annuity to the
annual payment is often spoken of as the number of years' purchase
which the annuity is worth. If the "period " understood in the
above investigation be a year, and p be the number of years' purchase,
then we have from (2)
p = (R«  l)[R m + n  x (T&  1) (3).
If the period be 1/qth of a year, since the annual payment is then
qA., we have
p = (R n  1 )/qR m + n \Rl) (4 ) .
Cor. 2. If the annuity be not " deferred" as it is called, but
begin to run at once, that is to say, if the first payment be due one
period hence* then m 1, and we have
P = A(R»1)/R"(R1),
= A(lR»)/(Rl) (5).
Also
2> = (R"1)/R»(R1),
= (lR»)/(Rl) (6);
or p = (\B n )/q(Rl) (7),
according as the period of conversion is a year or the qth part of a year.
Cor. 3. To obtain the present value of a deferred perpetual
annuity, or, as it is often put, the present value of the reversion of a
perpetual annuity, we have merely to make n infinitely great in the
equation (2). We thus obtain
* This is the usual meaning of "heginning to run at once." In some
cases the tir*t payment is made at once. In that case, of course. ?n. = 0.
xxu NUMBER OF YEARS' PURCHASE OF A FREEHOLD
539
p4/fi 1
= A/R'« 1 (R1) (8).
Hence, for the number of years' purchase, we have
p = l/E^^R  1) (9),
or p=l/ ff R TO  1 (El) (10),
according as the period of conversion is a year or 1/qth of a yea?:
Cor. 4. JFlien the perpetual annuity begins to run at once the
formulce (8), (9), (10) become very simple. Putting m= 1 we have
P = A/(R1),
= A/(l+rl),
= A/r (11).
For the number of years' purchase
p=\jr (12);
or p=l/ qr (13),
according as the period of conversion is a year or 1/qth of a year.
If the period be a year, remembering that, if s be the rate
per cent of interest allowed, then r = sj 100, we see that
p=100/s (14).
Hence the following very simple rule for the value of a perpetual
annuity. To find the number of years' purcfoise, divide 100 by the
rate per cent of interest corresponding to the kind of investment in
question. This rule is much used by practical men. The following
table will illustrate its application : —
Rate % . . .
3
H
4
4 i 5 5i ' 6
1
No. of years' purchase .
33
28
25
22
20
18
17
Example.
A sum of £30,000 is borrowed, to be repaid in 30 equal yearly instalments
which are to cover both principal and interest. To find the yearly payment,
allowing compound interest at 4 %.
Let A be the annual payment, then £30,000 is the present value of an
540 INTEREST AND ANNUITY TABLES CHAf.
annuity of <£A payable yearly, the annuity to begin at once and run for 30
years. Hence, by (5) above,
30,000 = A(l  r045 30 )/045,
A = 1350/(ll045 30 ),
 30 log 1 045 = 1 4265110,
l045~ so = 267000.
A = 1350/733,
= £1841 : 14 : 11.
§ 8.] It would be easy, by assuming the periodic instalments
or the periods of an annuity to vary according to given laws, to
complicate the details of annuity calculations veiy seriously ;
but, as we should in this way illustrate no general principle of
any importance, it will be sufficient to refer the student to one
or two instances of this kind given among the examples at the
end of this chapter.
It only remains to mention that in practice the calculation
of interest and annuities is much facilitated by the use of tables
(such as those of Turnbull, for example), in which the values of the
functions (1 + r) n , (1 +r)~ n , {(1 + r) n  1 }jr, {1  (1 + r) n }jr,
rj{\ (1 +r)~ n }, &c, are tabulated for various values of lOOr
and ii. For further information on this subject see the TextBook
of the Institute of Actuaries, Part I., p. 151.
Exercises XLIV.
^1.) The difference between the true discount and the interest on £40,400
for a period x is £4, simple interest being allowed at 4 % > fiud x.
(2.) Find the present value of £15,000 due 50 years hence, allowing 4i °/ B
compound interest, convertible yearly.
(3.) Find the amount of £150 at the end of 14 years, allowing 3 °/ com
pound interest, convertible halfyearly, and deducting 6d. per £ for income
tax.
(4.) How long will it take for a sum to double itself at 6 % compound
interest, convertible annually ?
(5.) How long will it take for one penny to amount to £1000 at 5 °/ com
pound interest, convertible annually ?
(6.) On a salary of £100, what difference does it make whether it is paid
quarterly or monthly ? Work out the result both for simple and for compound
interest at the rate of 4  2 °/ .
(7.) A sum £A is laid out at 10 % compound interest, convertible annually,
arid a sum £2.V at 5 % compound interest, convertible halfyearly. After
how many years will the amounts be ecpual ?
XXII
EXERCISES XLIV 541
(8.) Show that the difference between bankers' discount and true discount,
simple interest being supposed, is
A?) V 2 {1  nr + nh 2  iflr 3 + . . . ad oo } .
(9.) If r> 5/100, n ■> 10, find an upper limit for the error in taking
100(1 + n C i r + nC2r 2  + n C 3 r :] ) as the amount of £100 in n years at 100/ % com
pound interest, convertible annually.
(10.) If £I C and £l s denote the whole compound and the whole simple
interest on £P for n years at 100? %> convertible annually, show that
I r I s = P(, 1 C 2 r 2 + „C 3 r ! + . . .+ '/■»).
(11.) A man owes £P, on which he pays lOOr % annually, the principal
to be paid up after n years. What sum must he invest, at 100r' °/ , so as to be
just able to pay the interest annually, and the principal £P when it falls due ?
(12.) B has a debenture bond of £500 on a railway. When the bond lias
still five years to run, the company lower the interest from 5 %> which was
the rate agreed upon, to 4 °/ , and, in compensation, increase the amount of
B's bond by x °/ a . Find x, supposing that B can always invest his interest
at 5 %.
(13.) A person owes £20,122 payable 12 years hence, and offers £10,000
down to liquidate the debt. What rate of compound interest, convertible
annually, does he demand ?
(14.) A testator directed that his trustees, in arranging his affairs, should
set apart such sums for each of his three sons that each might receive the
same amount when he came of age. When he died his estate was worth
£150,000, and the ages of his sons were 8, 12, and 17 respectively. Find
what sum was set apart for each, reckoning 4 % compound interest for
accumulations.
(15.) B owes to C the sums Ai, A 2 , . . ., A r at dates Bi, n 2 , . . ., » r
years hence. Find at what date B may equitably discharge his debt to C
by paying all 'the sums together, supposing that they all bear the same rate
of interest ; and
1st. Allowing interest and interest in lieu of discount where discount is due.
2nd. Allowing compound interest, and true discount at compound interest.
(16.) Required the accumulated value at the end of 15 years of an annuity
of £50, payable in quarterly instalments. Allow compound interest at 5 %•
(17.) A loan of £100 is to be paid off in 10 equal monthly instalments.
Find the monthly payment, reckoning compound interest at 6 %•
(18.) I borrow £1000, and repay £10 at the end of every month for 10
years. Find an equation for the rate of interest I pay. What kind of
interest table would help you in practically solving such a question as this ?
(19.) The reversion after 2 years of a freehold worth £168 : 2s. a year is to
be sold : find its present value, allowing interest at 2 %> convertible annually.
(20.) Find the present value of a freehold of £365 a year, reckoning com
pound interest at 3J %, convertible halfyearly, and deducting 6d. per £ of
incometax.
(21.) If a perpetual annuity be worth 25 years' purchase, what annuity to
542 EXERCISES XLIV chap, xxii
continue for 3 years can be bought for £5000 so as to bring the same rate of
interest ?
(22. ) If 20 years' purchase be paid for an annuity to continue for a certain
number of years, and 24 years' purchase for one to continue twice as long, find
the rate of interest (convertible annually).
(23.) Two proprietors have equal shares in an estate of £500 a year. One
buys the other out by assigning him a terminable annuity to last for 20 years.
Find the annuity, reckoning 3 % compound interest, convertible annually.
(24.) The reversion of an estate of £150 a year is sold for £2000. How
long ought the entry to be deferred if the rate of interest on the investment
is to be 4 %> convertible annually ?
(25.) If a lease of 19 years at a nominal rent be purchased for £1000, what
ought the real rent to be in order that the purchaser may get 4 °/ on his
investment (interest convertible halfyearly) ?
(26.) B and C have equal interests in an annuity of £A for 2n years (pay
able annually). They agree to take the payments alternately, B taking the
first. What ought B to pay to C for the privilege he thus receives ?
(27. ) A farmer bought a lease for 20 years of his farm at a rent of £50,
payable halfyearly. After 10 years had run he determined to buy the free
hold of the farm. "What ought he to pay the landlord if the full rent of the
farm be £100 payable halfyearly, and 3 % be the rate of interest on invest
ments in land ?
(28.) What annuity beginning n years hence and lasting for n years is
equivalent to an annuity of £A, beginning now and lasting for n years ?
(29.) A testator left £100,000 to be shared equally between two institu
tions B and C ; B to enjoy the interest for a certain number of years, C to
have the reversion. How many years ought B to receive the interest if the
rate be 3^ %> convertible annually ?
(30.) If a man live m years, for how many years must he pay an annuity
of £A in order that he may receive an annuity of the same amount for the
rest of his life ? Show that, if the annuity to be acquired is to continue for
ever, then the number of years is that in which a sum of money would doublo
itself at the supposed rate of interest.
(31.) A gentleman's estate was subject to an annual burden of £100. His
expenses in any year varied as the number of years he had lived, and his
income as the square of that number. In his 21st year he spent £10,458,
and his income, before deducting the annual burden, was £4410. Show that
he ran in debt every year till he was 50.
(32.) A feu is sold for £1500, with a feuduty of £18 payable annually,
and a casualty of £100 payable every 50 years. What would have been the
price of the feu if it had been bought outright ? Beckon interest at 4£ %•
(33.) Find the accumulation and also the present value of an annuity
when the annual payments increase in A.r.
(34.) Solve the same problem when the increase is in G.P.
(35.) The rental of an estate is £mA to begin with ; but at the end of
every q years the rental is diminished by £A, owing to the incidence of fresh
taxation. Find the present value of the estate.
APPENDIX
ON THE GENERAL SOLUTION OF CUBIC AND BIQUAD
RATIC EQUATIONS; AND ON THE CASES WHERE
SUCH EQUATIONS CAN BE SOLVED BY MEANS OF
QUADRATIC EQUATIONS.
§ 1.] Since cubic and biquadratic equations are of frequent
occurrence in elementary mathematics, and many interesting
geometric problems can be made to depend on their solution, a
brief account of their leading properties may be useful to readers
of this book. Incidentally, we shall meet with some principles
of importance in the General Theory of Equations.
COMMENSURABLE ROOTS AND REDUCTIBILITY.
§ 2.] We shall suppose in all that follows that the coefficients
Po • • •> Pn °f an y equation,
p x n +p,x n  1 + . . . + p„, = (1),
are all real commensurable numbers. If, as in chap, xv., § 21,
we put x = £/m, we derive from (1) the equivalent equation
Poi m +p l m$' 1  1 + . . . +p n ,m n ' 1 $ +p n m n = (2),
each of whose roots is m times a corresponding root of (1). If
we then choose m so that mp,jp , . . ., m n ~ 1 p n . 1 /p , m n p n fp are
all integral — for example, by taking for m the L.C.M. of the de
nominators of the fractions p x jp m . . ., p n i/Po> Pn/po — we shall
reduce (2) to the form
£■ + },£»! + . . .+q n = (3),
in which all the coefficients are positive or negative integers, and the
544 APPENDIX
coefficients of the highest term unity. We may call this the Special
Integral Form.
§ 3.] If, as in chap, xv., § 22, we put x = £ + a, we transform
the equation (1) into
i>o£" + fl , i£"~ 1 + . •  + <z» = o,
where q 1 = np t) a+p 1 . Hence, if we take a= —pijnp m the trans
formed equation becomes
&?+.?&* + • • . + q n = o (4),
wherein q 2 , . . ., q' n have now determinate values. It follows
that
By a proper linear transformation, we can always deprive an
equation of the nth degree of its hig