The Method of Mathematical Induction

Citation preview

I. S. Sominskii

1

+1

m

i + i

POPULAR LECTURES IN MATHEMATICS SERIES E ditors : I. N. S neddon and M. Stark

Volume 1 THE METHOD OF MATHEMATICAL INDUCTION

TITLES IN THE POPULAR LECTURES IN MATHEMATICS SERIES

Vol. 1 The Method of Mathematical Induction By I. S. Sominskii

Vol. 2 Fibonacci Numbers By N. N. Vorob' ev Vol. 3 Some Applications of Mechanics to Mathematics By V. A. Uspenskii Vol. 4 Geometrical Constructions using Compasses Only By A. N. K ostovskii Vol. 5 The Ruler in Geometrical Constructions By A. S. Smogorzhevskii Vol. 6 Inequalities By P. P. K orovkin

THE METHOD OF MATHEMATICAL INDUCTION by

I. S. SOMINSKII

Translated from the Russian by

HALINA MOSS, B.Sc. Translation Editor IAN N. SNEDDON Simson Professor of Mathematics in the University of Glasgow

BLAISDELL PUBLISHING

COMPANY

NEW Y O R K • L O N D O N

A D I VI S I O N OF R A N D O M HOU S E

SOLE D I S T R I B U T O R S IN T HE U N I T E D STATES AND C A N A D A

Blaisdell Publishing Company 22 East 51st Street, New York 22, N. Y.

Copyright © 1961 P ergamon P ress Ltd .

A translation of the original volume Metod matematicheskoi induktsii (Moscow, Fizmatgiz, 1959)

Library of Congress Card Number: 61-11532

Printed in Great Britain by Pergamon Printing and Art Services Limited, London

CONTENTS Page Forew ord

...............................................................................................v i i

INTRODUCTION................................................................................................. 1 CHAPTER I

The Method o f M a th e m a tic a l I n d u c tio n

CHAPTER I I

Exam ples and E x e r c i s e s .........................................12

CHAPTER I I I

The P ro o f by I n d u c tio n o f Some Theorems o f E le m en ta ry A l g e b r a ........................................ 39

CHAPTER IV

S o l u t i o n s ......................................................................... 45

y

. .

3

FOREWORD The method o f m a th e m a tic a l in d u c tio n , w hich i s th e s u b je c t o f t h i s book, i s w id e ly a p p l ic a b le i n a l l d e p a rtm e n ts o f m ath em atics, from th e e le m e n ta ry sc h o o l c o u rs e up to b ran ch es o f h ig h e r m ath em atics o n ly l a t e l y i n v e s t i g a t e d . I t i s c l e a r , th e r e f o r e , t h a t even a sc h o o l c o u rs e o f m a th em atics can n o t be s tu d ie d s e r i o u s l y w ith o u t m a s te rin g t h i s m ethod. Id e a s o f m a th e m a tic a l in d u c tio n , m oreover, have a w ide g en e­ r a l s ig n i f ic a n c e and a c q u a in ta n c e w ith them a l s o h as an im p o rtance f o r th o s e whose i n t e r e s t s a r e f a r removed from m ath em atics and i t s a p p l i c a t i o n s . The e s s e n t i a l s o f th e method and some sim p le exam ples o f i t s u se a r e g iv e n i n C h a p te r I and i n th e f i r s t s e c ti o n o f C h ap ter I I . To s tu d y th e s e i t i s s u f f i c i e n t f o r th e r e a d e r to be f a m i l i a r w ith th e c o u rs e o f m a th em atics i n th e sev en y e a r sc h o o l p e r io d . The re m a in in g s e c ti o n s o f t h i s book a r e f u l l y a c c e s s ib l e to th e r e a d e r who h as m a ste re d th e mathema­ t i c s c o u rse o f a f u l l se c o n d a ry s c h o o l. T h is book i s m eant f o r p u p ils in th e h ig h e r form s o f se co n d ary s c h o o ls , f i r s t y e a r s tu d e n ts i n u n i v e r s i t i e s , te a c h e r t r a i n i n g c o lle g e s and t e c h n ic a l c o l l e g e s . I t w ould a ls o be u s e f u l f o r d is c u s s io n i n a sc h o o l m a th e m a tic a l s o c ie t y .

v ii

I N T R O D U C T I O N

P ro p o s itio n s can be d iv id e d in to g e n e r a l and p a r t i c u l a r . The fo llo w in g a r e exam ples o f g e n e r a l p r o p o s itio n s : A ll c i t i z e n s o f th e U .S .S .R . have th e r i g h t to e d u c a tio n . In e v e ry p a r a lle l lo g r a m th e d ia g o n a ls a r e b is e c t e d a t t h e i r p o in t o f i n t e r s e c t i o n . A ll numbers en d in g w ith a z e ro a r e d i v i s i b l e by 5* The c o rre sp o n d in g exam ples o f p a r t i c u l a r p r o p o s itio n s a r e a s fo llo w s : P e tr o v h as th e r i g h t to e d u c a tio n . I n th e p a r a lle l lo g r a m ABCD th e d ia g o n a ls a r e b is e c t e d a t t h e i r p o in t o f i n t e r s e c t i o n . 140 i s d i v i s i b l e by 5 . The t r a n s i t i o n from g e n e r a l p r o p o s itio n s to p a r t i c u l a r ones i s c a l l e d d e d u c tio n . L e t us c o n s id e r an exam ple: (1 ) A ll c i t i z e n s o f th e U .S .S .R . have th e r i g h t to ed u ca­ tio n . (2 ) P e tro v i s a S o v ie t c i t i z e n . ( 3) P e tro v h as th e r i g h t to e d u c a tio n . We o b ta in th e p a r t i c u l a r p r o p o s itio n ( 3) from th e g e n e r a l p r o p o s itio n ( l ) , w ith th e h e lp o f th e p r o p o s itio n ( 2 ) . P ro g r e s s in g from p a r t i c u l a r p r o p o s itio n s to g e n e r a l ones i s c a lle d in d u c t i o n . In d u c tio n can le a d to c o r r e c t a s w e ll as to i n c o r r e c t c o n c lu s io n s . We s h a l l make t h a t c l e a r by means o f two exam ples: ( 1) 140 i s d i v i s i b l e by 5 .

1

2

THE METHOD OP MATHEMATICAL INDUCTION (2 ) A ll numbers en d in g i n z e ro a r e d i v i s i b l e by r)#

Prom th e p a r t i c u l a r p r o p o s itio n ( l ) we o b ta in e d th e g e n e r a l p r o p o s itio n (2 ) and t h i s p r o p o s itio n i s c o r r e c t . On th e o th e r hand: ( 1 ) 140 i s d i v i s i b l e by 5 . (2 ) A ll t h r e e f i g u r e numbers a r e d i v i s i b l e by 5 . Prom th e p a r t i c u l a r p r o p o s itio n ( l ) we o b ta in e d th e g e n e r a l p r o p o s itio n ( 2 ) and t h i s p r o p o s itio n i s w rong. The q u e s tio n w hich th e n a r i s e s i s how can we u se in d u c tio n i n m a th em atics so a s to o b ta in o n ly c o r r e c t c o n c lu s io n s ? The answ er to t h i s q u e s tio n i s g iv e n i n t h i s book.

CHAPTER

I

THE METHOD OF MATHEMATICAL INDUCTION 1 F i r s t o f a l l we exam ine two exam ples o f in d u c tio n a r e in a d m is s ib le i n m a th e m a tic s. E x a m p l e

1.

w hich

L et 1 n (/i+ l) *

I t i s ea sy to v e r i f y t h a t : c _ _ J _____ 1

“

1-2— T ’

c ____ !___. J ____ 2

1.2 • 2*3

3•

c _ J ___|__ \___|__ L_ 8

1-2 « 2-3 T* 3.4

c —

!.

I

1

““ 1-2 ' 2-3

|

r“ 3. 4 t

3_ 4* *1 ^

4

4 . 5 — -5 •

On th e b a s is o f th e above r e s u l t s we s t a t e , t h a t f o r e v e ry n a tu ra l n n n+ 1 *

*n

E x a m p l e 2 . L et us exam ine th e q u a d r a tic e x p r e s s io n x* + x + 4 1 , f i r s t c o n s id e r e d by th e w e ll known m a th e m a tic ia n L eonard E u le r,o n e o f th e f i r s t o f th e S t . P e te r s b u r g A cade­ m ic ia n s • I f we s u b s t i t u t e z e ro f o r x i n t h i s e x p r e s s io n , we o b ta in th e prim e number 4 1 . A gain, i f we s u b s t i t u t e 1 f o r x i n th e e x p r e s s io n , we o b ta in 43*w hich a g a in i s a prim e num ber. C o n tin u in g to s u b s t i t u t e s u c c e s s iv e ly th e v a lu e s 2, 3 , 4, 5, 6, 7, 8 , 9* 10 f o r x i n th e e x p r e s s io n we o b ta in each tim e a prim e number: 47, 53* 6 l , 7 1 * 8 3 , 97* 113, 131* 151.

3

4

THE METHOD OP MATHEMATICAL INDUCTION

On th e b a s i s o f r e s u l t s o b ta in e d we s t a t e t h a t th e s u b s t i ­ t u t i o n o f any p o s i t i v e i n t e g e r f o r x i n th e above q u a d r a tic alw ay s r e s u l t s i n a prim e num ber. Why i s th e r e a s o n in g u sed i n th e s e exam ples n o t a c c e p ta b le i n m a th em atics? Where i s th e flaw i n o u r c o n c lu s io n s ? The t r o u b le i s t h a t i n b o th c a s e s we made a g e n e r a l s t a t e ­ ment c o n c e rn in g any n ( i n th e second exam ple, any x) s o le ly on th e g rounds t h a t t h i s p r o p o s itio n tu r n e d o u t to be tr u e f o r some v a lu e s o f n ( o r x ) . I n d u c tio n i s w id e ly used i n m a th e m a tic s, b u t one must u se i t w ith s k i l l . Any u se o f in d u c tio n w hich i s n o t c a r e f u l m ig h t le a d to f a l s e c o n c lu s io n s . T hus, a lth o u g h i n Example 1 o u r g e n e r a l p r o p o s itio n does i n f a c t t u r n o u t to be t r u e ( a s i s p roved below i n Example 5 ) , th e g e n e r a l p r o p o s itio n s t a t e d i n Example 2 tu r n s o u t to be f a l s e . In d e e d , when th e q u a d r a tic + 41 i s s tu d ie d more c l o s e l y , i t i s se en t h a t i t i s e q u a l to a prim e number f o r x = 0 , 1 , 2, . . . , 39, b u t f o r x = 40 i t e q u a ls 41 » t h a t i s , i t i s a co m p o site num ber. 2 I n Example 2 we e n c o u n te re d a p r o p o s itio n v a l i d i n 40 p a r ­ t i c u l a r c a s e s , b u t s t i l l w rong i n g e n e r a l . We s h a l l g iv e two f u r t h e r exam ples o f p r o p o s itio n s w hich a r e t r u e i n s e v e r a l p a r t i c u l a r c a s e s and f a l s e i n g e n e r a l. E x a m p l e 3* The b in o m ia l xn— 1 , w here n i s a p o s i t i v e i n t e g e r , i s o f g r e a t i n t e r e s t to m a th e m a tic ia n s . I t i s s u f f i c i e n t to sa y t h a t i t i s c l o s e l y c o n n e c te d w ith a g e o m e tr ic a l problem ab o u t d iv id i n g a c irc u m fe re n c e i n t o n equal p a r ts . I t i s n o t s u r p r i s i n g , t h e r e f o r e , t h a t t h i s e x p r e s s io n i s s tu d i e d e x t e n s i v e ly i n m a th e m a tic s. I n p a r t i c u l a r , th e p ro ­ blem o f f a c t o r i z i n g t h i s e x p r e s s io n i n t o f a c t o r s w ith i n t e ­ g ra l c o e ffic ie n ts is o f g re a t i n te r e s t. When th e f a c t o r s

xn— 1 w ere w r i t t e n o u t f o r numerous

THE METHOD

5

p a r t i c u l a r v a lu e s o f n , i t was o b se rv e d t h a t none o f th e c o e f f i c i e n t s was n u m e ric a lly l a r g e r th a n u n i t y . I n f a c t X— 1 *s—JC— 1, = ( * - ! ) ( * + 1). x4— 1 =» (x — 1) (x + 1) (*» - f- 1), JC4— 1 = (JC — l ) ( ^ + x» + ^ + x + l ) , **— l = , ( * _ ! ) ( * » + * + 1) (*»— x + 1),

T ab les w ere com piled w ith in w hich a l l c o e f f i c i e n t p o s­ se sse d t h i s p r o p e r ty . A ttem p ts to pro v e th e p r o p e r ty f o r ev e ry n f a i l e d . In 19389 i n a n o te i n th e jo u r n a l U spekhi m a te m a tic h e sk ik h nauk ^ S u c c e s s e s o f M a th em atic al S c ie n c e s 1) th e S o v ie t m ath em a tic ia n , N.G. C h e b o ta re v , c h a lle n g e d m a th e m a tic ia n s to s e t t l e t h i s q u e s tio n . The s o lu t io n was g iv e n by V. Iv an o v * . I t tu r n e d o u t t h a t th e above p r o p e r ty i s p o s s e s s e d by a l l p o ly n o m ia ls o f th e ty p e x*— 1 , whose d e g re e n i s l e s s th a n IO5 . One o f th e f a c t o r s o f xw — 1 , how ever, i s th e p o ly n o m ial x ^ + x 4* — x*s — X4* — 2x41 — X40— x ^ + ^ + —

-f- x19— x9— ac®— — 2*7— x « - -x * + x * + x + l.

w hich no lo n g e r p o s s e s s e s t h i s p r o p e r ty . E x a m p l e 4* I n to how many p a r t s i s sp a ce d iv id e d by n p la n e s a l l p a s s in g th ro u g h one p o i n t , i f no th r e e o f them p a ss th ro u g h th e same s t r a i g h t l i n e ? L et us examine th e s im p le s t c a s e s o f t h i s p ro b lem . One p la n e d iv id e s space in t o 2 p a r t s . Two p la n e s p a s s in g th ro u g h one p o in t d iv id e sp ace i n t o 4 p a r t s . T hree p la n e s p a s s in g th ro u g h one p o i n t , b u t n o t p a s s in g th ro u g h one s t r a i g h t l i n e , d iv id e sp ace in t o 8 p a r t s . At f i r s t s i g h t i t may seem t h a t when th e number o f p la n e s i s in c re a s e d by 1 , th e number o f p a r t s i n t o w hich sp a ce i s d iv id e d i s d oubled and, th u s , 4 p la n e s w i l l d iv id e sp a ce *Y» IVANOVT U sp. m a t, n au k . 4 . 513-317 ( l 9 4 l ) .

6

THE METHOD OP MATHEMATICAL INDUCTION

i n t o 16 p a r t s , f i v e p la n e s i n t o p a rts , e tc . p la n e s w ould s p l i t up sp ace in t o 2n p a r t s .

In g e n e r a l n

I n r e a l i t y t h i s does n o t o ccurs 4 p la n e s s p l i t sp ace up i n t o 14 p a r t s and 5 p la n e s i n t o 22 p a r t s . I n g e n e r a l, n p la n e s s p l i t sp a ce up i n t o

n(n — 1) + 2 p a r t s

Prom th e exam ples d is c u s s e d above we draw a sim p le b u t im p o rta n t c o n c lu sio n s A p r o p o s itio n can be c o r r e c t f o r a g r e a t number o f p a r t i ­ c u l a r c a s e s and a t th e same tim e may be f a l s e i n g e n e r a l.

3 Now, th e f o llo w in g q u e s tio n a r i s e s . We have a p r o p o s itio n j u s t i f i e d i n s e v e r a l p a r t i c u l a r c a s e s . I t i s im p o s s ib le to c o n s id e r a l l p a r t i c u l a r c a s e s . How a r e we to know w h eth er th e p r o p o s itio n i s c o r r e c t i n g e n e r a l? I t i s som etim es p o s s i b le to answ er t h i s q u e s tio n by a p p ly ­ in g a s p e c ia l method o f r e a s o n in g , c a l l e d *the method o f m a th e m a tic a l in d u c tio n * ( f u l l in d u c tio n , com plete in d u c tio n ) At th e b a s is o f t h i s method l i e s th e *p r in c ip le * o f m athe­ m a tic a l in d u c tio n , w hich may be s t a t e d i n th e fo llo w in g ways A p r o p o s itio n i s t r u e f o r ev e ry p o s i t i v e i n t e g e r n i f : (1 ) i t i s t r u e f o r

n = 1 , and

(2 ) i t fo llo w s from th e t r u t h o f th e p r o p o s itio n f o r any p o s i t i v e i n t e g e r n = k t h a t th e p r o p o s itio n i s a ls o tr u e fo r n = k -f-1 . To p ro v e t h i s p r i n c i p l e l e t us suppose th e o p p o s ite , i . e . l e t us suppose t h a t th e p r o p o s itio n i s n o t tr u e f o r e v e ry p o s i t i v e i n t e g e r n . Then t h e r e e x i s t s a n a t u r a l m , such th a t (1 ) th e p r o p o s itio n i s n o t tr u e f o r /* = /* , (2 ) f o r any n s m a lle r th a n m th e p r o p o s itio n i s tr u e ( i n o th e r w o rd s, m i s th e f i r s t i n t e g e r number f o r w hich th e

THE METHOD

7

p r o p o s itio n i s f a l s e ) . O b v io u sly , m > 1 , s in c e f o r n — 1 th e p r o p o s itio n i s t r u e ( c o n d itio n l ) . I t fo llo w s t h a t m — 1 i s a p o s i t i v e i n t e g e r . We th e r e f o r e have th e s i t u a t i o n , t h a t th e p r o p o s itio n i s tr u e f o r th e p o s i t i v e i n t e g e r m — 1 , b u t i s f a l s e f o r th e n e x t in te g e r m . T h is c o n t r a d i c t s c o n d itio n ( 2 ) . N o t e . W hile p ro v in g th e p r i n c i p l e o f m a th e m a tic a l i n ­ d u c tio n we made u se o f th e f a c t t h a t any s e t o f p o s i t i v e in te g e r s c o n ta in s a s m a lle s t num ber. I t *is e a s i l y se en t h a t t h i s p r o p e r ty i n i t s t u r n c o u ld be deduced as a c o r o l l a r y o f th e p r i n c i p l e o f m a th e m a tic a l in d u c tio n . T h e r e f o r e , th e two p r o p o s itio n s a r e e q u i v a le n t. E i t h e r o f them can be ta k e n a s an axiom d e f in i n g th e n a t u r a l s e r i e s ; th e o th e r w i l l th e n be a th e o re m . I t i s i n f a c t u s u a l to ta k e th e p r i n c i p l e o f in d u c tio n a s th e axiom *.

4 A p ro o f b ased on th e p r i n c i p l e o f in d u c tio n i s c a l l e d a 1p ro o f by th e method o f m a th e m a tic a l in d u c tio n * o r sim p ly , a *p ro o f by in d u c tio n * . Such a p r o o f m ust n e c e s s a r i l y con­ s i s t o f p ro v in g two in d e p e n d e n t th e o re m s. T h e o r e m

1.

The p r o p o s itio n i s t r u e f o r

n= \ •

T h e o r e m 2. The p r o p o s itio n i s tr u e f o r n = k -f- 1 i f i t i s tr u e f o r n = k * w here k i s any p o s i t i v e i n t e g e r . I f b o th th e s e theorem s a r e p ro v e d , th e n , on th e b a s i s o f th e p r i n c i p l e o f in d u c tio n th e p r o p o s itio n i s t r u e f o r any n. E x a m p l e s ®=

5.

F in d th e sum + -T3 + 3k + • • • +7T7ST!) •

(C f. Example l ) . *For in s ta n c e th e p r i n c i p l e o f in d u c tio n i s th e f i f t h axiom i n Peano*s scheme f o r th e n a t u r a l numbers [ c f . E . L andau, *F oundations o f A n aly sis* (C h e ls e a P ub. C o ., 1 9 5 1 ), p . 2 ] E d ito r .

8

THE METHOD OF MATHEMATICAL INDUCTION We know t h a t

Now we s h a l l n o t r e p e a t o u r m is ta k e , made i n Example 1 , and we s h a l l n o t r u s h to s t a t e t h a t f o r any n a t u r a l n

We s h a l l he c a r e f u l and we s h a l l s a y , t h a t th e ex a m in a tio n o f Sv S9, SB, SA a llo w s us to p u t fo rw ard a h y p o th e s is t h a t Sn = -

^ f o r any p o s i t i v e i n t e g e r n .

We know i n a d d i tio n

t h a t t h i s h y p o th e s is i s t r u e f o r i m l , 2, 3, 4 . To t e s t t h i s h y p o th e s is we s h a l l u se th e method o f in d u c tio n . T h e o r e m s'= j

^

1.

r n

The h y p o th e s is i s t r u e f o r

/ i = l , s in c e

= i •

T h e o r e m 2 . L et us suppose t h a t th e h y p o th e s is i s tr u e fo r # = £ , th a t i s

w here k i s some p o s i t i v e i n t e g e r . We s h a l l p ro v e t h a t , i n t h a t c a s e , th e h y p o th e s is i s hound to he t r u e a l s o f o r n = * + i , th a t is

Sh+ ' - k + 2In d e e d , ■S*+l"='S* +

+ 2)*

i t f o llo w s , a c c o rd in g to th e c o n d itio n o f th e th eo rem , t h a t c

__ *+*

^

i_______1_____

ft + 1 T

a 3 = 5,

m4 =

7, . . .

L et us s e t o u r s e lv e s th e f o llo w in g t a s k : to d e v is e a fo rm u la f o r th e odd number un9 e x p r e s s in g i t i n term s o f n . S o l u t i o n . th u s :

The f i r s t odd number ux can be w r i t t e n “i = 2 . 1 — 1.

(1 )

The second odd number uQ can be w r i t t e n th u s : w2 = 2 • 2 — 1.

(2 )

The t h i r d odd number us can be w r i t t e n th u s : «3=* 2 • 3 — 1.

I f we lo o k c a r e f u l l y a t e q u a tio n s ( l ) ,

12

( 3) (2 ) and ( 3) we a r e

EXAMPLES AND EXERCISES

13

le d to th e h y p o th e s is t h a t to o b ta in th e flth odd number un i t i s s u f f i c i e n t to d o u b le i t s s u f f i x n and th e n s u b t r a c t 1; t h a t i s , f o r th e /ith odd number we have th e fo rm u la

«a = 2/i — l.

(4)

L et us p rove t h a t t h i s fo rm u la i s j u s t i f i e d . T h e o r e m 1. v a l id f o r n = 1 .

E q u a tio n ( l ) shows t h a t fo rm u la ( 4 ) i s

T h e o r e m 2. L et us suppose t h a t th e fo rm u la i s v a l id f o r n * * » A ,i.e . t h a t th e fcth odd number i s o f th e form : 0* «3B2k — 1, We must now t r y to pro v e t h a t fo rm u la ( 4 ) i s bound to be v a l i d a ls o f o r th e (&-(■-1) th odd num ber, i . e . t h a t th e (fe -|-l)th odd number i s o f th e form : «*+i “ 2 ( A + l ) — 1, o r, th a t 0*+l =* 2A-J- 1. To o b ta in th e (A -j-l)th odd p o s i t i v e i n t e g e r i t i s s u f f i c i e n t to add 2 to th e A th odd i n t e g e r , i . e . «*+i =*0*4-2 . Our c o n d itio n i s ak = 2k — 1 . T hat means t h a t 0*+1 = (26 — 1)4" 2 *= 2A4~ 1»

w hich was w hat we s e t o u t to p ro v e . A n s w e r . th e fo rm u la

The « t h odd p o s i t i v e i n t e g e r i s g iv e n by “» = 2« — 1

Problem 2 : To c a l c u l a t e th e sum o f th e f i r s t « odd w hole num bers. L et us c a l l th e r e q u ir e d sum Sn , i . e . S „ = l + 3 + 5 + . . . + (2«— 1).

14

THE METHOD OP MATHEMATICAL INDUCTION

In m ath em atics t h e r e a r e ready-m ade fo rm u lae f o r s o lv in g su ch p ro b le m s. We a r e i n t e r e s t e d i n s o lv in g t h i s problem w ith o u t r e c o u r s e to such a fo rm u la , b u t by u s in g th e method o f in d u c tio n . F or t h i s we m ust f i r s t fo rm u la te a h y p o th e s is , i . e . sim p ly t r y and g u e ss th e an sw e r. We g iv e n p a r t i c u l a r v a lu e s 1, 2, 5 . . . u n t i l we have c o l­ l e c t e d enough m a te r ia l f o r th e b a s is o f a more o r l e s s r e l i a b l e h y p o th e s is . A f te r t h i s a l l t h a t rem ain s i s to t e s t th e h y p o th e s is by th e method o f m a th e m a tic a l in d u c tio n . We have Sj = 1,

S2 — 4,

S3 = 9,

S 4 — 16,

S6 — 25,

S6 = 36.

Now i t a l l depends on how o b s e rv a n t th e p e rso n ta c k li n g th e problem i s , w h eth e r he i s a b le to g u ess a g e n e r a l r e s u l t from th e p a r t i c u l a r o n e s . We ta k e i t

t h a t i n t h i s c a se i t i s e a sy to n o tic e t h a t : Sj — l 2,

5 a= 22,

S8 = 32,

S4 =

On th e s e g rounds we can suppose t h a t i n g e n e r a l = We s h a l l now p ro v e t h a t t h i s h y p o th e s is i s th e c o r r e c t on e. T h e o r e m 1 . F or n — 1 th e sum i s r e p r e s e n te d by one term e q u a l to 1 . F o r /i = l th e e x p r e s s io n n2 a ls o e q u a ls 1 . T h at means t h a t , f o r n — \ , th e h y p o th e s is i s c o r r e c t . T h e o r e m 2 . L et us suppose t h a t th e h y p o th e s is i s c o r r e c t f o r n = k , i . e . S k = k2, and th e n p rove t h a t i n t h a t c a s e th e h y p o th e s is i s a l s o c o r r e c t f o r n = k -{-1 , i . e . S*+i = ( * + l ) 2. Now $k+1— S* + (2^ + 0 -

15

EXAMPLES AND EXERCISES and.

=

and th e r e f o r e S*+i =^k2-\- (2k + 1) = (fc + l)a,

w hich was w hat we s e t o u t to p ro v e . A n s w e r .

The sum 6f th e f i r s t n odd i n t e g e r s i s

Problem 5 . F in d un , i f i t i s known t h a t u1 = 1 and t h a t f o r any n a t u r a l k > 1 uk

H i n t; Problem 4 .

a, = 3 • 1 — 2,

= 3 -2

— 2.

F in d th e sum

Sn = l + H i n t :

— u k -1 + 3.

(l)

2 + 2H

2»

+ . . . + 2» - ‘.

S1= 2 — 1; S2 = 22 — 1; *S8 = 2s — 1 ; o r

(2 ) exam ine 2Sn— Sn» Problem b . P rove t h a t th e sum o f th e f i r s t n p o s i t i v e i n te g e r s i s eq u a l to .? ^ . S o l u t i o n . T h is problem d i f f e r s from th e p r e v io u s ones i n t h a t i t i s u n n e c e s sa ry to make up a h y p o th e s is s in c e one i s a lre a d y g iv e n . I t i s o n ly n e c e s s a r y to p ro v e t h a t i t i s c o rre c t. L et th e sum r e q u ir e d be

Sn , i . e . l e t

5n = l + 2 + 3 + T h e o r e m

1.

F or n — 1 , th e h y p o th e s is i s c o r r e c t .

T h e o r e m

2.

Let

s k= 1 4 - 2 + • • • + 6

k(k + l) 2

16

THE METHOD OP MATHEMATICAL INDUCTION

We s h a l l show, t h a t ^Ar+i

(A + l)(A + 2) 2

T h is f o llo w s from S>+1 = S „ + {k + 1 ) - *, P roblem 6 . P ro v e t h a t th e sum o f th e sq u a re s o f th e f i r s t n p o s i t i v e i n t e g e r s i s e q u a l to w(” + l)(2n + 1) . 6

P roblem 7 .

P ro v e t h a t s

Sn = l — 22+ 32 — 42+ . . . + ( — 1)”- ^ = ===( _ n(n + \ ) ' S o l u t i o n . T h e o r e m 1 . F or n = 1 , th e h y p o th e s is i s o b v io u sly c o r r e c t s in c e ( - 1 )° = 1 . T h e o r e m

2.

L et

S* = 1 — 2* + 3=— . . . + ( - 1) * 'V = ( _ 1)*-» *.(* + '). To p ro v e t h a t S*+1 = 1 — 2 2 + 3 * — . . . + ( - l ) * - 1fe2-f-(— 1)*( f e + l ) * =

(

i)*(* + Q(* + 2>y

we n o te t h a t s * + 1 = s * + ( - i ) * ( A + i ) !! =

= (— l ) * - 1 --

+ (— 1)* (A -|-1)* =

= ( - D * [(A + 1 ) - - ] ( A + 1) = ( - 1 ) * and i f f o r e v e ry i n t e g e r k > 2 “* = (a+ P ) “*-i — «Po*.9. P roblem 2 0 . The p ro d u c t 1 .2 .5 . . . n i s d en o ted by n 1 (w hich i s re a d f a c t o r i a l /!*)• I t i s u s e f u l to remember t h a t 11= 1, 21= 2, 31 = 6, 41 = 24, 51= 120 . C a l c u l a t e Sn « 1 • ll -J- 2*21 “J- 3*31 -J- • • •

n • nl

EXAMPLES ARP EXERCISES

19

S, = l . 11= 1, 5 S= 1 . 1 1 + 2 . 2t = 5, S, = 1 • 11+ 2 • 21 + 3 • 31 = 23, «Si = 1 • 11 + 2 • 21 + 3 • 31 + 4 • 41 = 119.

Looking c l o s e l y a t th e s e r e s u l t s we o b se rv e t h a t St = 21— 1,

S, = 3I— 1, 5 , = 41— 1, S, = 51— 1.

T h is s u g g e s ts th e h y p o th e s is : •S,, = (n + 1)1 — 1. L et us check t h i s h y p o th e s is . T h e o r e m

1.

For

« = 1 th e h y p o th e s is i s c o r r e c t , as

5, = 1 . 11 = 21 — 1. T h e o r e m

2.

Let

S * = 1 • 11+ 2 - 21+ . . . + £ • £ ! = (ft+ 1)1 — l. To show t h a t •S*+I = l • H + 2 - 21+ ----- ^ f t.* ! + (*_|_i).(fc + i) !== - ( * + 2)1we o b se rv e t h a t •S*+i = 5 t + (ft + 1) • (ft + 1 ) 1 =

= (I*+ ')! — l) + ( f t +l ) - ( f t + l)! = = (ft+ 1) 1(1 + (ft+ 1)] — 1 = = (ft+ ! ) ! ( * + 2) — 1 = (ft+ 2)1 — 1.

1,

20

THE METHOD OF MATHEMATICAL INDUCTION Problem 2 1 .

P ro v e th e i d e n t i t y

__1 1+ *

- 4

-1 — \ + xi

1 -f* 8

T

1

2- - =

1 f.r» "

___ ■

at— 1 '

P roblem 2 2 .

I t i s g iv e n t h a t a “h P = m* aP =

A, = m ---------a— — ,

i42* =3 m -----n—~I m—

/!« = ,„ -------------2- j --m- - - - - - rn — 1

and so on, t h a t i s , f o r

6>1

Ak+l — m —

---

{mzjz\\

a :£ p ) .

P ro v e t h a t

(i)

S o l u t i o n T h e o r e m l . To b e g in w ith , we s h a l l p ro v e t h a t form ul a ( 1 ) i s t r u e f o r /i = 2 . From th e g iv e n c o n d itio n i t fo llo w s th a t Aq = m------ ( a + B ) ------------------- ?£__ —a*+ P2+ «P —«—P (« + P) — 1

m - \

a + p - l

A cc o rd in g to fo rm u la ( l ) __(«8 — ft3)-—(a2 — ft2)

^

(®*

P2)

(a —P) *

D iv id in g th e to p and bottom o f th e l a t t e r f r a c t i o n by a — p , we have «*+ p3+ ap - q - S «+ P - l . w hich was w hat we s e t o u t to p ro v e .

EXAMPLES AND EXERCISES T h e o r e m 2. n= , i . e . th a t

21

Suppose t h a t fo rm u la ( l ) i s v a l i d f o r

*

(a* — P*) — (a*-1 — p* *) *

(2)

We s h a l l p ro v e , t h a t i n t h a t c a s e i t i s v a l i d a ls o f o r « = A-(- 1 , i . e . (a*+a —

*+1

— (a*+1 —

(a*+l_p*+i) _ (ok _ p*) •

Now =

or

= (« + ? )-£ .

Making u se o f e q u a tio n (2 ) we f in d t h a t t(«* — P*) — («*-1 —p*~')l —(a* “ P*) (.*+*_ P»+a)_ (.* + ‘ _P*+>)

A*+i = ( * + P )

so t h a t th e theorem i s p ro v e d . Problem 2 5 .

S im p lify th e poly n o m ial

1 x I x ( x —\) 1 "“ TM 2!

I / _ , in x { x —\)...(x —n-t-1) ni

A n s w e r . ( i\» I l)

(x—\)(x —2)...(x —n) n\

Problem 2 4 . P rove t h a t i t i s p o s s i b le to p ay , w ith o u t r e ­ q u ir in g change, any w hole number o f r o u b le s , g r e a t e r th a n 7 , w ith b an k n o tes o f v a lu e 5 r o u b le s and 5 r o u b le s . S o l u t i o n . F or 8 r o u b le s th e s ta te m e n t i s t r u e . L et th e sta te m e n t be tr u e f o r k r o u b le s w here ft i s an i n t e g e r g r e a t e r th a n o r e q u a l to 8 . Two c a s e s a r e p o s s ib le s

22

THE METHOD OF MATHEMATICAL INDUCTION

( a ) k r o u b le s b e in g p a id w ith th r e e - r o u b le b an k n o tes o n ly and (b ) k r o u b le s b e in g p a id w ith b an k n o tes i s a t l e a s t one o f v a lu e 5 r o u b le s .

among w hich th e r e

In c a s e (a ) th e r e m ust be n o t l e s s th a n th r e e 3“ro u b le n o te s , a s i n t h i s c a se fc > 8 . To pay fe-f-l ro u b le s we sh o u ld exchange th r e e 3 - ro u b le n o te s f o r two 3 - ro u b le n o te s . I n th e second c a s e , to pay k - \ - \ r o u b le s , we sh o u ld ex­ change one 5 - ro u b le n o te f o r two 3 -ro u b le o n e s . P roblem 2 5 . P ro v e t h a t th e sum o f th e cubes o f th r e e s u c c e s s iv e p o s i t i v e i n t e g e r s i s d i v i s i b l e by 9* S o l u t i o n . The sum l8-{-23- |-3 3 i s d i v i s i b l e by 9* T h is m eans, t h a t th e s ta te m e n t i s c o r r e c t when th e f i r s t o f th e th r e e s u c c e s s iv e n a t u r a l numbers i s 1 . L et th e sum /j3_|_(£_|_i)3-f-(/j-U2)3 , w here k i s some n a t u r a l num ber, be d i v i s i b l e by 9« The sum ( * + i r + ( k + 2)3+ i * + 3 ) 3= = [ft8 4 -(ft4-l)9+ (*-4-2)s]

+ 3*4-3)

can be w r i t t e n as th e sum o f two term s each o f w hich i s d i v i s i b l e by 9« I t i s t h e r e f o r e a ls o d i v i s i b l e by 9 and th e r e s u lt i s e s ta b lis h e d . P roblem 2 6 .

P ro v e t h a t f o r a w hole //>. 0 An— 1l n+2—J—122n+1

i s d i v i s i b l e by 1 5 3 • Problem 2 7 . // + 1 numbers a r e p ic k e d a t random from th e 2/i i n t e g e r s 1 , 2, ..........2n . P rove t h a t among th e numbers p ic k e d we can f in d a t l e a s t tw o, one o f w hich i s d i v i s i b l e by th e o t h e r . S o l u t i o n . L et us su p p o se , t h a t o u t o f th e numbers 1 , 2 , ............ 2//, w here / / > 2 , we managed to p ic k //-f-i num­ b e r s i n such a way t h a t none o f them i s d i v i s i b l e by any o t h e r . We s h a l l d e n o te th e s e t o f a l l th e s e numbers by /Wn+i . We s h a l l p ro v e t h a t i n t h a t c a se i t i s p o s s i b le to

EXAMPLES M L EXERCISES

23

p ic k n numbers o u t o f th e 2n — 2 num bers: 1 , 2 , 2n—2 * such t h a t a g a in none i s d i v i s i b l e by any o t h e r . F our c a s e s can a r i s e . (a ) iWn+1 c o n ta in s n e i t h e r 2n — 1 n o r 2n. (b) Afn+, c o n ta in s 2rt— 1 b u t n o t 2n • (c ) Mn+1 c o n ta in s 2n b u t n o t 2n — 1. (d ) Mn+i c o n ta in s b o th 2n— 1 and 2n. C a s e ( a ) : L et us e x c lu d e any one number from Mn+j • n numbers w i l l rem a in , none o f w hich i s g r e a t e r th a n 2n — 2. None o f th e s e numbers i s d i v i s i b l e by any o t h e r . C a s e ( b ) : L e t us e x c lu d e th e number 2rt — 1 from Mn+1. n numbers re m a in , none o f w hich i s g r e a t e r th a n 2n — 2 • None o f th e s e numbers i s d i v i s i b l e by any o t h e r . C a s e ( c ) : L et us e x c lu d e 2rt from Mn+1 and a g a in we o b ta in th e same r e s u l t . C a s e ( d ) : F i r s t o f a l l l e t us n o te t h a t Mn+1 ca n n o t c o n ta in th e number n , o th e rw is e th e r e w ould have been two numbers ( 2n and n) one o f w hich i s d i v i s i b l e by th e o t h e r . L et us ex c lu d e 2n— 1 and 2n . We s h a l l d e n o te th e re m a in in g s e t o f n — 1 numbers by iWn-1 . L et us now j o i n th e number n to Mn_t . We o b ta in n num bers, none o f w hich i s g r e a t e r th a n 2/i — 2 . I t rem a in s to show t h a t among th e s e numbers none i s d i v i s i b l e by any o t h e r . In Afn+1 th e r e w ere no o th e r . T h is m eans, t h a t e i t h e r . I t rem ain s o n ly ap p eared when n jo in e d

numbers d i v i s i b l e one by th e th e r e w ere no such numbers i n Mn_1 to make s u re t h a t no such numbers Afn-1 .

F or t h i s p u rp o se i t i s s u f f i c i e n t to make s u re t h a t : (1 ) None o f th e numbers and in

m aking up Mn_i i s d i v i s i b l e by n

( 2 ) th e number n i s n o t d i v i s i b l e by any o f th e numbers .

The fo rm er fo llo w s from th e f a c t t h a t none o f th e numbers in Mn_t i s g r e a t e r th a n 2/i — 2 . The l a t t e r fo llo w s from

24

THE METHOD OF MATHEMATICAL INDUCTION

th e f a c t t h a t 2n i s i n d i v i s i b l e by any o f th e numbers making up iWn_ j. T hus, i f i t supposed t h a t th e p r o p o s itio n i s wrong f o r th e 2n num bers 1 , 2 , ..........2 n , i t i s a l s o wrong f o r th e 2(n — 1) numbers 1 , 2, ..........2n — 2 . T hat m eans, i f th e p ro ­ p o s i t i o n i s c o r r e c t f o r th e 2[n — 1) num bers 1 , 2 , ..........2n — 2 , t h a t i t i s c o r r e c t a ls o f o r th e 2n numbers 1 , 2 , ......... 2/i. F or th e two numbers 1 , 2 th e p r o p o s itio n i s t r u e . T hat m eans, t h a t th e p r o p o s itio n i s t r u e f o r th e 2n numbers 1 , 2 , ..........2/i, w here /i i s any n a t u r a l num ber. We sh o u ld n o te t h a t t h i s problem h as th e f o llo w in g sim p le s o l u t i o n . L et us p ic k any n + l num bers o u t o f th e 2/i num­ b e r s 1 , 2, ...........2/i. L et us d e n o te t h i s s e t o f numbers by ^M+l • We s h a l l d iv id e each even number i n Mn+, by such a power o f 2 t h a t th e q u o tie n t i s odd. Out o f th e s e q u o tie n ts and odd num bers we s h a l l make up th e s e t Afn+1. I n Afn+1 th e r e a r e « + 1 odd num bers, each o f w hich i s l e s s th a n 2n . As th e r e a r e , i n a l l , n p o s i t i v e odd numbers l e s s th a n 2rt Afn+1 m ust c o n ta in a t l e a s t two e q u a l num bers. L et us d en o te each o f th e s e e q u a l num bers by k . T h is r e s u l t means t h a t i n Mn+1 t h e r e w ere a t l e a s t two num­ b e rs 2*k and 2*k . One i s d i v i s i b l e by th e o t h e r . P roblem 2 8 . P rove t h a t n v a r io u s s t r a i g h t l i n e s in a p la n e , p a s s in g th ro u g h one p o i n t , d iv id e th e p la n e in to 2n p a rts . P roblem 2 9 . P ro v e t h a t n s t r a i g h t l i n e s ly in g i n a p la n e b re a k up th e p la n e in t o p a r t s , w hich i t i s p o s s i b le to p a i n t w ith b la c k and w h ite p a i n t i n such a way, t h a t a l l a d ja c e n t p a r t s ( i . e . p a r t s w hich have a segm ent o f a s t r a i g h t l i n e i n common) a r e p a i n te d i n d i f f e r e n t c o l o u r s . 2 P roblem 3 0 « P ro v e t h a t n p la n e ^ p a s s in g th ro u g h one p o in t i n su ch a way t h a t no t h r e e o f them p a s s th ro u g h th e same s t r a i g h t l i n e d iv id e sp a ce i n t o An = n(n — 1) + 2 p a r t s .

EXAMPLES AND EXERCISES

25

S o l u t i o n , ( l ) One p la n e d iv id e s sp a ce in to 2 p a r t s and A , = 2 . F or n = i , th e p r o p o s itio n i s th e r e f o r e c o r r e c t . (2) Lee us suppose t h a t th e p r o p o s itio n i s j u s t i f i e d f o r n = k f i . e . k p la n e s d iv id e sp a ce i n t o k ( k — 1) + 2 p a r t s . We shall then prove that fc-f-l planes divide space into k(k-\- l ) - f 2 Parts. L et P be th e ( f c + l ) t h p la n e . The p la n e P i s i n t e r s e c t e d by each o f th e f i r s t k p la n e s a lo n g some s t r a i g h t l i n e . In t h i s way th e p la n e P i s d iv id e d i n t o p a r t s by k d i f f e r e n t s t r a i g h t l i n e s , w hich p a s s th ro u g h one p o i n t . T ak in g P ro ­ blem 28 a s th e b a s is o f o u r argum ent we s t a t e t h a t th e p la n e P i s broken up in t o 2k p a r t s , each o f w hich r e p r e s e n ts a p la n e a n g le w ith i t s apex a t th e g iv e n p o i n t . The f i r s t k p la n e s d iv id e sp ace i n t o c e r t a i n p o ly h e d ra l a n g le s. Some o f th e s e p o ly h e d ra l a n g le s a r e d iv id e d by p la n e P in to two p a r t s . The common f a c e o f two such p a r t s i s a p a r t o f th e p la n e lim ite d by th e two r a y s a lo n g w hich P i n t e r s e c t s th e f a c e s o f th e p a r t i c u l a r p o ly h e d ra l a n g le , i . e . i t i s one o f th e 2k p la n e a n g le s in to w hich th e p la n e P was bro k en u p . T h is means t h a t th e number o f p o ly h e d ra l a n g le s w hich a r e b roken in t o two p a r t s by th e p la n e P ca n n o t be g r e a t e r th a n 2k. On th e o th e r hand, each o f th e 2k p a r t s i n t o w hich th e p la n e P i s d iv id e d a s a r e s u l t o f i t s i n t e r s e c t i o n w ith th e f i r s t k p la n e s i s a common f a c e o f two p o ly h e d ra l a n g le s and so i t d iv id e s th e p o ly h e d ra l a n g le form ed by th e f i r s t k p la n e s in to two p a r t s . T h is means t h a t th e number o f p o ly h e d ra l a n g le s w hich a r e bro k en in t o two p a r t s by th e p la n e P ca n n o t be l e s s th a n 2k. Thus, th e p la n e P b re a k s up in t o two p a r t s e x a c t ly 2k p a r t s o f sp ace form ed by th e f i r s t k p la n e s . I f k p la n e s , th e r e f o r e , b re a k sp a ce up i n t o 6 (6 — l) - ( - 2 p a r t s , th e n 6 - |- l p la n e s b re a k sp ace up in t o \k {k — l)-}-2]-|-26 = 6 (6 -{~ l)4 - 2

26 p a rts .

THE METHOD OF MATHEMATICAL INDUCTION The p r o p o s itio n i s t h e r e f o r e p ro v e d .

P roblem 5 1 .

P rove th e i d e n t i t y cos a cos 2a cos 4a. . . cos 2Wa = sln2W*lg 2n+1sln a *

S o l u t i o n ,

( l ) F or /i=:0 th e i d e n t i t y i s tr u e as cos a

sin 2a 2 sin a *

(2 ) Suppose th e i d e n t i t y to be t r u e f o r n = k , i . e . cos a cos 2a . . . cos 2* a = sin 2k+1a

2k+1sin a *

Then i t i s a l s o t r u e f o r

n = k-{-1 .

F or

cos a cos 2a . . . cos 2* a cos 2*+1 a = _sin 2*+1a cos 2k+1 a “ 2ft+1sina

sin2*+ ,a 2*+2 sin a *

P roblem 5 2 . P ro v e t h a t i4n = cos/i0 , i f i t i s g iv e n t h a t i4J = cosO, i42 = cos20 , and f o r an i n t e g e r 6 > 2 th e r e e x i s t s th e f o llo w in g r e l a t i o n s h i p A k =

S o l u t i o n , and f o r n = 2 • (2 ) L et ^*-9

(l)

2 co s 0i4fc_j — i4k_9. The p r o p o s itio n i s tr u e f o r /i = l

= cos (6 — 2)0,

Ak_l = c o s ( k — 1)0.

Then Ak = 2 cos 6 cos f&— 1)0 — cos (k — 2) 0 = cos /j0.

Problem 5 5 .

P ro v e t h a t

s.n 5 + i , sin jc-f-dn2jc - f - .. .-{-sin«jc = ----------- sin5 ^. *in4

27

EXAMPLES AMD EXERCISES S o l u t i o n , (2 ) Let

( l ) F o r « = 1 th e p r o p o s itio n i s c o r r e c t .

, s*n 2 x kx sin jc— sin 2x-{- . . . -{-sin kx = ------ — sin y , sinY

Then

sin x-|-sin 2.ir-f-. . . + s in ^jc-f-sin (fe-f-1) x = i —j k “|”— 1x . sin - sin y -f-sin (fc-f-1) x = i x 8 n ~2 . *—+1 jc sin ~ — sin y - { - 2 sin sin j

. k+ 2 sln~ Y ~ x k+\ jc = »----- -— sin - - - x t x I sinT

x cos —

becau se n k 4* 1 * X . k -i- 2 . 2 cos —j —x sin y = sin —y - * — sin y • Problem 5 4 .

P rove t h a t tin ^ ± 1

* Problem 55.

cosx -1-cos2 x-f- . . . 4 - cosnx =

*

2 sin~

P rove t h a t

sin x~Y 2 sin 2x + 3 sin Sx -{- . . . + n sin nx ==* (n + l)sin nx —n sin (n -fr-1) x 4 sin2y Problem 5 6 .

P rove t h a t

cos x -J- 2 cos 2x -f- • • • -{- n cos nx =» (/t-f 1) cos n x ~ n cos (n -f-1) * —»1 4sin‘y Problem 5 7 . P rove t h a t l

x . I

x .

.1

x

y ta n y - { - y ta n y + . . . -f-y j-ta n ^ ^

Problem 5 8 .

1- COt 75X—-- COt X 75= 2* 2n P rove t h a t

cot"1 3 - f “

cot*1 5

■=» ta ir 1 2 -{ - ta ir 1

-f- c o r 1 (2/i -f* ta irl

(X zfz mv).

1) =* n tanrl I*

28

THE METHOD OF MATHEMATICAL INDUCTION

P roblem 5 9 .

P ro v e t h a t O + O* = 2^ (cos ^ + /sin ^ ) .

S o l u t i o n ,

( l ) F o r /i = l ,

th e p r o p o s itio n i s t r u e ,

as J

= 2» (c o s ^ + isln -J).

(2 ) L et (l + 0* = 2 ^ (c o s-£ + /s in — ) . Then a + /) * + * =

= 2T(cos—•+ /sin 4p) • & (c o s - + /sln^==i = 2^ ( COs < i ^ + ,s ln i* ± ilii). P roblem 4 0 .

P rove t h a t

( y ^ - « ) " = 2 n( c o s f . - / s l n £ ) . P roblem 4 1 .

P rove th e f o llo w in g th e o re m .

I f , a s a r e s u l t o f a f i n i t e number o f r a t i o n a l o p e r a tio n s ( i . e . a d d i t i o n , s u b t r a c t i o n , m u l t i p l i c a t i o n and d iv is io n ) c a r r i e d o u t w ith th e com plex numbers xlt x2, . xn , we o b ta in th e number u , th e r e s u l t o f th e same o p e r a tio n s c a r r i e d o u t w ith th e c o n ju g a te com plex numbers x.2, . xn w i l l be « , th e c o n ju g a te o f u . S o l u t i o n . We s h a l l show, f i r s t o f a l l , t h a t th e p r o p o s itio n i s c o r r e c t f o r each o f th e f o u r o p e r a tio n s c a r r i e d o u t w ith 2 com plex num bers. L et xi = a +

x2==c-\- di.

Then *i + *2 = (a + c) + (b + d) i = u; x i + *2 = ( , S k. . But s h> — , ♦ w hich p ro v e s t h a t P roblem 4 4 .

13

5 k+1> - l - . .

^

F in d th e e r r o r i n th e f o llo w in g p r o o f.

P r o p o s i t i o n . F or any p o s i t i v e i n t e g e r n th e f o llo w in g i n e q u a l i t y h o ld s : 2n > 2/ i + l . P r o o f :

L et th e i n e q u a l i t y be j u s t i f i e d f o r n = k .

EXAMPLES AND EXERCISES

31

w here k i s some n a t u r a l num ber, i . e . 2* > 2A + 1.

( 1)

We s h a l l p rove t h a t th e n th e i n e q u a l i t y i s j u s t i f i e d a ls o f o r /iaaA+ 1 , i . e .

2**1> 2 ( H - l ) + l .

(2)

And r e a l l y , 2* i s n o t l e s s th a n 2 f o r any n a t u r a l k . L e t u s add 2* to th e le f t - h a n d s id e o f th e i n e q u a l i t y ( l ) , and add 2 to th e r ig h t- h a n d s id e o f t h a t i n e q u a l i t y . We o b ta in th e c o r r e c t i n e q u a l i t y 2*-f- 2* > 2k- \ - 1- ( - 2, or 2*+1 > 2 ( * + l ) - j - l . The p r o p o s itio n i s t h e r e f o r e p ro v e d . Problem 4 5 . F o r w hich n a t u r a l n does th e f o llo w in g in e q u a ­ l i t y h o ld tru e * 2n> 2/1+ 1 ?

Problem 4 6 . F o r w hich p o s i t i v e i n t e g e r s n i s th e f o llo w ­ in g in e q u a l i t y v a l i d : 2n > /i*?

S o l u t i o n . F o r /i = t th e i n e q u a l i t y i s t r u e , a s 21 > 1* For

2 th e i n e q u a l i t y i s f a l s e , a s 22 = 2a.

F o r n = 3 th e i n e q u a l i t y i s f a l s e , a s 2® 5 a. F o r n = 8 th e i n e q u a l i t y i s t r u e , a s 2* > 62. I t w ould seem t h a t th e i n e q u a l i t y i s t r u e f o r /i = l and f o r any n > 4 • L e t us p ro v e i t . (1 ) F o r n = 5 th e i n e q u a l i t y h o ld s . (2 ) L e t 2* > ft',

(l)

w here k i s some p o s i t i v e i n t e g e r g r e a t e r th a n 4 * We s h a l l p ro v e t h a t ( 2) We know t h a t 2* > 2 f t + 1 f o r ft > 4 (P roblem 45)* T h e r e f o r e , i f we add 2* to th e l e f t - h a n d s id e o f th e i n e q u a l i t y a n d 2A-)-l to i t s r ig h t- h a n d s id e , we s h a l l o b ta in th e tr u e i n e q u a l i t y

( 2) . A n s w e r . P roblem 4 7 .

2n > /i8 when /i = 1 and when n > 4 • P ro v e t h a t (l+ o t)n > l + / i « ,

w here a > — 1, a : £ 0, n i s a p o s i t i v e i n t e g e r g r e a t e r th a n 1 . S o l u t i o n , ( s in c e a* > 0 ) •

( l ) F o r /i = 2 , th e i n e q u a l i t y h o ld s

(2 ) L e t th e i n e q u a l i t y h o ld when n = p o s itiv e in te g e r , i . e . (1+ 0.)* > l + f t « .

w here k i s some (1)

We s h a l l show t h a t i n t h a t c a se th e i n e q u a l i t y h o ld s a ls o f o r /i=efe-}-l , i . e . we s h a l l show t h a t ( l + a ) * + 1> l + (ft+ l)o ..

( 2)

EXAMPLES AND EXERCISES

33

In d ee d , i t fo llo w s from th e c o n d itio n s o f th e problem t h a t , so t h a t th e f o llo w in g i n e q u a l i t y h o ld s :

i-H > o

( l + a ) * + 1> ( l + f t « ) ( l + < x ) ,

(3 )

T his in e q u a l i t y i s o b ta in e d from ( l ) by m u ltip ly in g b o th s id e s by l + « . We can r e w r i te th e i n e q u a l i t y ( 3) i n th e form ^ 1—{—(A—f—1) 8—f—

N e g le c tin g th e p o s i t i v e term koP on th e r ig h t- h a n d s id e we f in d t h a t th e i n e q u a l i t y (2 ) i s e s t a b l i s h e d . Problem 4 8 .

P rove t h a t f o r any p o s i t i v e i n t e g e r n > 1

T r + W + ---+ y ^ > v ^ ' Problem 4 9 .

P ro v e t h a t f o r any p o s i t i v e i n t e g e r n > 1 4n „ (2/i)l n + l ^ (n!)> *

Problem SO.

P rove t h a t 2 n~ 1(an+ b n) > { a + b ) n1

(1)

where « + £ > < ) , a^ tb, and n i s a p o s i t i v e i n t e g e r g r e a t e r th a n u n ity . S o l u t i o n , th e form

l)

When n — 2 , th e i n e q u a l i t y ( l ) ta k e s

2{a' + b*)>{a + b)\ As

( 2)

a4^b , th e fo llo w in g i n e q u a l i t y h o ld s (a — «)2>0.

(3)

Adding (0 -M )2 to each s id e o f th e i n e q u a l i t y ( 3) we o b ta in th e in e q u a l i t y ( 2) .

34

THE METHOD OF MATHEMATICAL INDUCTION Thus, i t i s p roved t h a t th e i n e q u a l i t y ( l ) i s tr u e f o r n = 2 .

2) Suppose t h a t th e i n e q u a l i t y ( l ) h o ld s f o r ti — k w here k i s some p o s i t i v e i n t e g e r , i . e . suppose t h a t 2*-> (a* + * * )> (« + *)*,

(4 )

We s h a l l p ro v e t h a t , i n t h a t c a s e , th e i n e q u a l i t y ( l ) h o ld s a l s o f o r * - * + » , i . e . 2* («*+i

(a+ * )» + '.

( 5)

L e t us m u lt ip ly b o th s id e s o f th e i n e q u a l i t y (4 ) by a-f-A . S in c e i t i s g iv e n t h a t O , we o b ta in th e fo llo w in g c o rre c t in e q u a lity : 2 * - i ( a * + A * )(a +

A )> (a+ A )* + » .

f(L\

To p ro v e th e t r u t h o f th e i n e q u a l i t y ( 5) i t i s s u f f i c i e n t to show t h a t 2*(a*+> -I- *’■+')> 2*-» (a* +

+ b),

(7 )

o r , w hat i s th e same th in g , t h a t ( 8)

The i n e q u a l i t y ( 8 ) can be r e w r i t t e n i n th e form (a* — M)(a — />)> 0.

(9 )

Suppose t h a t a > b . As, i n a d d i tio n to t h a t , i t i s g iv e n t h a t a > — b , we have a > | £ | and i t fo llo w s t h a t ak > b k On th e l e f t - h a n d s id e o f th e i n e q u a l i t y (9) we have a p ro d u c t o f two p o s i t i v e n um bers, w hich i s , o f c o u r s e , p o s i t i v e . I f a < b , u s in g th e same arg u m e n ts, we e s t a b l i s h t h a t ak < b k . I n t h i s c a s e we have a p ro d u c t o f two n e g a tiv e num bers on th e l e f t - h a n d s id e o f th e i n e q u a l i t y ( 9 ) , and t h i s to o i s p o s i t i v e .

EXAMPLES AND EXERCISES

35

In b o th c a s e s t h e r e f o r e th e i n e q u a l i t y ( 9 ) i s v a l i d . T h is p r o v e s , t h a t th e t r u t h o f th e i n e q u a l i t y ( l ) = fo llo w s from i t s t r u t h f o r n = k .

fo r

Problem 5 1 . P rove t h a t f o r any v > 0 , and any p o s i t i v e in t e g e r n th e fo llo w in g i n e q u a l i t y h o ld s

S o l u t i o n , th e form

l a ) F or 0 = 1 th e i n e q u a l i t y ( l ) ta k e s

* + T > 2-

( 2)

The i n e q u a l i t y (2 ) stem s from th e o b v io u s i n e q u a l i t y

lb ) F or rte=s2

th e i n e q u a l i t y ( l ) ta k e s th e form

x* + 1+ 4 r > 3-

(3)

The in e q u a l i t y (2 ) h o ld s f o r any * > 0 . T h is means i t a ls o h o ld s when x i s s u b s t i t u t e d f o r x9, i . e .

th a t

* + - h > 2Adding 1 to b o th s id e s o f t h i s i n e q u a l i t y e q u a lit y ( 3) .

we o b ta in th e i n ­

2) Suppose t h a t th e i n e q u a l i t y ( l ) h o ld s f o r ji = A, w here k i s some p o s i t i v e i n t e g e r , i . e .

**+**"’+ • ..+ -J t + -L> a+ 1.

(4)

We s h a l l p ro v e t h a t , i n t h a t c a s e , th e i n e q u a l i t y ( l ) a l s o h o ld s f o r n = k - f>2 , i . e . ^ * + x *+ ^ - + . . . + _si r + ^ + _ ^ > * + 3.

( 5)

36

THE METHOD OP MATHEMATICAL INDUCTION

S u b s titu tin g

xk+2 f o r x i n th e i n e q u a l i t y (2 ) we g e t **+a+ - ^ T > 2-

( 6)

A dding t o g e th e r th e l e f t - h a n d s id e s o f th e i n e q u a l i t i e s ( 4 ) and ( 6 ) and th e n t h e i r r ig h t- h a n d s id e s we o b ta in th e in e q u a l i t y ( 5) . Now, l e t us sum u p . I n th e s u b s e c tio n s l a ) and lb ) we pro v ed t h a t th e in e q u a li ty ( l ) i s v a l i d f o r /i = 1 and /i = 2 # I n s u b s e c tio n 2) we p roved t h a t th e t r u t h o f th e in e q u a l i t y ( l ) f o r /* = £ + 2 i s a consequence o f th e t r u t h o f t h a t i n ­ e q u a l i t y f o r n = k . I n o th e r w ords, s u b s e c tio n 2) g iv e s us th e r i g h t to p a s s from n = k to /*=*=£-(-2 . The r e s u l t s o f s u b s e c tio n l a ) and 2) g iv e s us th e r i g h t to s t a t e t h a t th e i n e q u a l i t y ( l ) i s t r u e f o r any odd i n t e g e r n. E x a c tly i n th e same way th e r e s u l t s o f s u b s e c tio n s lb ) and 2 ) g iv e us th e r i g h t to s t a t e t h a t th e i n e q u a l i t y ( l ) i s t r u e f o r any even i n t e g e r n . On th e w h o le, we can s t a t e t h a t th e i n e q u a l i t y ( l ) i s tr u e f o r any p o s i t i v e i n t e g e r n . P roblem 5 2 .

P ro v e th e f o llo w in g th e o re m .

The g e o m e tric mean o f s e v e r a l p o s i t i v e numbers i s n o t g r e a t e r th a n t h e i r a r i t h m e t i c mean, i . e . i f av a2, . . . , a n a r e p o s i t i v e , th e n (!) S o l u t i o n . th e sim p le form

l ) F o r /i = 2 th e i n e q u a l i t y ( l ) ta k e s

( 2) F o r any p o s i t i v e ax and a2 , we have th e f o llo w in g i n e q u a l i t y (V ^ -V T ^ V o . From w hich i t i s e a sy to o b ta in i n e q u a l i t y ( 2 ) .

EXAMPLES AND EXERCISES

37

The in e q u a l i t y (2 ) h as a sim p le g e o m e tr ic a l m ean in g . On th e s t r a i g h t l i n e AB mark o f f segm ents o f le n g th ax and a2 c o n s e c u tiv e ly . U sing t h e i r sum a s a d ia m e te r d e s c r ib e a c irc le .

Then

2

i s th e d ia m e te r o f t h a t c i r c l e , and

i s h a l f th e ch o rd w hich i s p e r p e n d ic u la r to t h a t d i a ­ m eter a t th e p o in t common to ax and a2 . The i n e q u a l i t y (2 ) s t a t e s t h a t no chord o f a c i r c l e h as a le n g th g r e a t e r th a n th e d ia m e te r. 2) L et us suppose t h a t i n e q u a l i t y ( l ) i s t r u e f o r n = k . We s h a l l prove t h a t i n t h a t c a se T h is fo llo w s from

i t i s tr u e a ls o f o r / i = 2&.

V ala%• • • a2k — y 1'V ala%•••*»• V ak■»l •/ • a2k ^ k ————— k — —— —— ^ a xat ... ak + V**+i“ ‘ *ft ^ ----------------2---------------»■» -Mfc , gfc-n-h »»» — at 4* g»~h »»»4- ak 4-»»• 4* 4 - ••.

* -i

^

+

+ •••

3 m ---------

or

Now, l e t m be any n a t u r a l num ber. I f m =»2*, th e n a c c o rd ­ in g to (2 ) th e i n e q u a l i t y h o ld s f o r i t . I f on th e o th e r hand mzjz2* , we s h a l l f i n d such an $ , t h a t m i s l e s s th a n 2*, and th e n , on th e b a s is o f (2 ) and ( 3)* we s t a t e t h a t th e i n e q u a l i t y i s c o r r e c t f o r n = m»

CHAPTER

III

THE PROOF BY INDUCTION OF SOME THEOREMS OF ELEMENTARY ALGEBRA T h e o r e m 1 . The sq u a re o f a p o ly n o m ial i s e q u a l to th e sum o f th e sq u a re s o f a l l i t s te rm s added to tw ic e a l l p o s s ib le p ro d u c ts o f th e term s ta k e n two a t a tim e , i . e .

( vt - s+—

«;+«;+ •••+«:+ + 2 (a,a,+

+

m

• • • + a „ . ,a„).

1'

F o r n = 2 th e fo rm u la ( l ) can be p ro v ed by d i r e c t m u lt ip lic a tio n . Suppose t h a t th e fo rm u la ( l ) i s t r u e f o r th a t

n= k—1,

i.e .

w here S i s th e sum o f a l l p o s s i b le p r o d u c ts o f alf aQ, . ta k e n two a t a tim e . We s h a l l p ro v e t h a t (*i + a%+ • • • + 0fc-i + ahy = - « ! + « ? + . . . + « i - . + «i + 2S,. w here Sj i s th e sum o f a l l th e p o s s i b le p r o d u c ts ta k e n two a t a tim e o f ...» ak_lt ak, i . e . $1 = S + (al + a2 + • • • + a* -l)afr T h is fo llo w s from th e f a c t t h a t (fli -{-••• +

+ 2a

’

1

jc 1 .r 1 CO' a 2*+> 1 co«2» ~ C0‘* + 2»+r'*n'2*+T ” p t T — ------------ + cot2 *+1 - — cot* =

2*+1 ■

2*+i co‘ 2*+1

2k+l

5 8 . l ) We have tan( tan 1 2 — tan"1 1) = 2 — 1 2 l-f-2 “ 3 *

T h e re fo re t«n"1 2 — tan 1 1 = tan 1

T h is means t h a t f o r

n « l

=

cot"1 3.

th e p r o p o s itio n i s t r u e .

2) To b e g in w ith , we s h a l l show t h a t cot'1 (2£ + 3) =s tan" 1

In d ee d

— »•«'1 L

55

SOLUTIONS T h is means t h a t t 1 _ m 2* + 3 “~ a cot"1

( 2 k -f- 3 ) =

tan"1

j — tan 1 1.

Suppose t h a t th e p r o p o s itio n i s tr u e f o r n = k , i * e .

c o t" 1 3 -j“

c o t" 1 5 - h

- f - a co t” 1

as» ta n 1 2 -j- ta n 1 y

f

t t

I) a tan "1 — y

—

k

tan 1 1.

( 2)

We s h a l l p rove t h a t , i n t h a t c a s e , i t i s t r u e a l s o f o r n*+ l , i . e .

cot 1 3 +

cot”1 5 + . . . -f- cot”1 (26 -f- 3) s=s = tan"1 2 4- . . . -f- tar"1

— {k - f 1) twfl

^

Adding up th e l e f t - h a n d s id e s o f e q u a l i t i e s ( l ) and (2 ) and th e n t h e i r r ig h t- h a n d s id e s we o b ta in th e e q u a l i t y ( 3 ) . 4 0 . l ) F or

1 th e p r o p o s itio n i s c o r r e c t as y j — i =. 2 (cos-jr — i sin

2) L et ( \ r3 — 0* = 2* (cos — — 1sin

Then (V

j

— /)*+1 — 2A^cos y — / sin ~ ^ *2^ cos y — /sin

4 2 . l ) F or /i = l th e p r o p o s itio n i s j u s t i f i e d . 2) L et (cos x -f- / sin *)* =* cos kx 4- i ^in &*•

56

THE METHOD OP MATHEMATICAL INDUCTION

Then (cua x

-f- isin

jr)**1 a (cos kx -f -I sin kx) (cos x -f- / sin x) » cos (k 4- 1) jc 4- i sin (k -f-1) x.

=»

4 4 . The v e r y l a s t p h ra s e fThe p r o p o s itio n i s proved* i s f a l l a c i o u s . What i s r e a l l y pro v ed i s t h a t th e i n e q u a l i t y 2n > 2 /i+ 1 i s tr u e f o r n a t u r a l num ber.

i f i t i s tru e fo r

n=*k9 w here k i s any

I t does n o t n e c e s s a r i l y fo llo w from t h i s t h a t t h i s in e q u a ­ l i t y i s j u s t i f i e d a t l e a s t f o r one v a lu e o f n , and even l e s s f o r any p o s i t i v e i n t e g e r n . I n s h o r t , th e m is ta k e i s t h i s : t h a t o n ly Theorem 2 was p ro v ed and th e Theorem 1 was n o t c o n s id e r e d and no b a s is f o r in d u c tio n was c r e a t e d . The u s u a l way to p ro v e Theorem 1 w ould be to c o n s id e r n — 1 o r /j = 2 ; i n b o th o f th e s e c a s e s th e p r o p o s itio n i s f a l s e ( b u t se e th e n e x t p ro b le m ).

4 5 . I t i s ea sy to se e t h a t 3 i s ih e l e a s t n a t u r a l v a lu e o f ft f o r w hich th e i n e q u a l i t y 2 » > 2 /i-fI i s t r u e . T ak ing i n t o a c c o u n t t h a t th e t r u t h o f th e i n e q u a l i t y f o r n= *+ 1 fo llo w s from i t s t r u t h f o r n*~k (Problem 44) we s t a t e t h a t th e i n e q u a l i t y i s t r u e f o r any n a t u r a l n > 3 .

4 8 . l ) F o r /:« 2 th e i n e q u a l i t y i s tr u e f o r

2) L et

¥T+ 7 f + —+ We s h a l l p ro v e t h a t i

V*

( 1)

( 2)

SOLUTIONS

57

F o r any * > 0 th e f o llo w in g i n e q u a l i t y h o ld s (5 ) In d ee d , th e i n e q u a l i t y ( 3) i s o b ta in a b le from th e i n e q u a l i t y l+ l/s ^ T > ' by m u ltip ly in g b o th s id e s by — Yk . Adding up s i d e w ise i n e q u a l i t i e s ( l ) and ( 3) we o b ta in th e i n e q u a l i t y ( 2 ) .

4 9 « l ) F o r rt —2 th e i n e q u a l i t y ta k e s th e form o f ^ < 6 th e re fo re , i t i s c o rre c t. 2) L et 4*

where

*>2 .

I t i s e a sy to v e r i f y t h a t f o r * > 0 4 ( » + l ) ^ ( 2 k + \ ) ( 2 * + 2)

k+ 2 ^

(*+!)»

*

T h e re fo re 4*

5+1

4 (* + D ^ (2 ^ * + 2 < (*i)s

( 2 * + l ) (2* + 2)

< *+ « *

th a t i s 4*+‘ ^ (2 * + 2 )t

*+ 2 ^ U 5 + W ’

»

an d ,