properties of the integers: mathematical induction 1. mathematical induction 2. harmonic, fibonacci,...

Properties of the Integers: Mathematical Induction

1. Mathematical Induction

2. Harmonic, Fibonacci, Lucas Numbers

3. Prime Numbers

Chapter 4

Chapter 4 Properties of the Integers: Mathematical Induction

4.1 The Well-Ordering Principle: Mathematical Induction

The Well-Ordering Principle Any nonempty subset of Z+

contains a smallest element. (We often express this by sayingthat Z+ is well ordered.)

This principle serves to distinguish Z+ from Q+ and R+.

Theorem 4.1 Finite Induction Principle or Principle of Mathematical Induction

S(n): an open statement, n a positive integer

(a) If S(1) is true; and /* basis, not necessarily from 1 */(b) If whenever S(k) is true (for some ), then S(k+1) is true; (inductive step)then S(n) is true for all

k Z

n Z

If



Proof: Let is false}. We wish to provethat so to obtain a contradiction we assume that

Then by the well - ordering principle, has aleast element . Since (1) is true, So > 1, and

consequently -1 Z With -1 (since is the smallest element in ), is true. By condition(b), is also true. A contradiction.

+

F t Z S tF

F Fs S s s

s s F sF S s

S s S s

{ | ( ),

..

.( )

(( ) ) ( )

1

11 1

[ ( ) [ [ ( ) ( )]]], ( ).

S n k n S k S kn n S n

0 00

1



Ex. 4.1. For any =

Proof: =Induction basis: is true.Induction hypothesis: = is true.Establish the truth of

Note the importance of induction basis. For example, in

Ex. 4.4, if = ( , we also have

n Z i n n

S n i n nS

S k k k kS k

S n i n nS k S k

in

in

ik

in

, ( ) /

( ): ( ) /( )

( ): ( ) /( ).

(

( ): ) /( ) ( ). )

1 2

1 21

1 21

2 21

1

1

1

21



Ex. 4.5 Find a formula for ( )2 11

ii

n

Ans: Observe and conjecture.

n=1 1 1=12

n=2 1+3 4=22

n=3 1+3+5 9=32

n=4 1+3+5+7 16=42

conjecture: ( )2 11

ii

n

=n2

then prove by induction



Ex. 4.7 the Harmonic Numbers

H H H

Hn

n H n

n H n H n

n

jj

nn

n

1 2 3

2

1

1 11

21

1

2

1

3

11

2

1

3

1

1

1

, , ,

.

, ( ) .

In general,

Prove N, (a)

(b) If Z then +




For (a), induction hypothesis

Since for all 1 j 2 it follows thatk

S k H k

H

H

j

H H H k k

k

k

k

k k k

k k k k

k k k

k k

kk

( ):

, ,

( ) ( )

2

2

2

2 2 2

1

11

2

1

2

1

2 1

1

2 21

2 1

1

2 21

2

1

2

21

21 1 1 1 1

1

1




For (b), induction hypothesis

.S k H k H k

H H H k H k H

k Hk

k H

k H k

jj

kk

jj

kj

j

kk k k

k k

k

( ): ( )

( )

( )[ ]

( ) ( )

1

1

1

11 1

1 1

1

1

1

11

12 1



Ex. 4.8 For let R, where |An|=2n and the elements ofAn are listed in ascending order. If R, prove that in order todetermine whether An, we must compare r with no more thann+1 elements in An.

n0 An r

r

Proof: When n=0, A0={a} and only one comparison is needed.

Assume the result is true for some and consider the casefor Ak+1.

k 0

kkkkkk |C||B| where,CBA 21 Let and Bk < Ck

(1) Compare once to determine whether r B r Ck k or

(2) Then compare in Bk or in Ck

1

k+1k+2 Total comparisons



Ex. 4.10 Prove S(n): n can be written as a sum of 3's and/or 8's(with no regard to order) for any n14.

basis: n=14=3+3+8 OKinduction hypothesis: k can be added up by 3's and/or 8's.

If k=...+8+..., then k+1=...+3+3+3+...Otherwise, there are at least 5 3's in k's summands.k+1=...3+3+3+3+3+1+...=...8+8+...



Theorem 4.2 Finite Induction Principle--Alternate Form

Let with (a) If are true and(b) If whenever are

true for some where then is also true.then is true for all

n n Z n nS n S n S n

S n S n S n S k

k Z k n S k

S n n n

0 1 0 10 0 1

0 0 1

1

0

11

1

, .( ), ( ), , ( )

( ), ( ), , ( ), , ( )

, , ( )

( ) .



Ex. 4.11 Prove S(n): n can be written as a sum of 3's and/or 8's(with no regard to order) for any n14.

Assume the truth of the statements:

S S S k S k k Zk n k k

k k S kS k

( ), ( ), , ( ), ( ) ,. ,

( ) . ( )( ).

14 15 116 1 17

1 2 3 21

for some where And now if then and The truth of implies the truth of



12

12222323

12212323

11

1012

122

2373

3

222

1211

2

2

2110

k

kkk

kkkkk

nn

nn

nnn

)()(

)()(ppp

.kn Consider true. all are )k(S),k(S

,),(S),(S Assume .p :)n(S :Proof

.p that Show.n whereZn for

,ppp and ,p,p Let

(recurrence relation)


4.2 Recursive Definitions

0,2,4,6,8,10,12,...: bn=2n

1,2,3,6,11,20,37,68,125,...:an=?

explicit definition

a a a an n n n 1 2 3

implicit recursivedefinition

Examples:

factorial: (n+1)!=(n+1)(n!)

Harmonic Number: H Hnn n 11

1



Ex. 4.16 The Fibonacci Numbers

(1) and

(2) for with

F F

F F F n Z nn n n

0 1

1 2

0 1

2

, ;

,

Prove that (a)

for all N.

n Z F F F

b F n

ii

nn n

n

n

2

01

5

3( )



Ex. 4.16 The Fibonacci Numbers

( ,

( )

(

a) Induction hypothesis:

b)

F F F

F F F F F F

F F F F F

F F F

ii

kk k

ii

ki

i

kk k k k

k k k k k

k k k

k k

k k k

2

01

2

0

1 2

01

21 1

2

1 1 1 2

1 1

1

1 1 1

5

3

5

3

5

3

8

3

5

3

24

9

5

3

25

9

5

3

1k

Please check out the induction basis for yourselves first.



Ex. 4.17 The Lucas Number

( ) , ;

.

.

1 2 1

2

1

0 1

1 2

02

1 1

and

(2) for Z with

Prove

(a) For N,

(b) Z

+

+

L L

L L L n n

n L L

n L F F

n n n

ii

nn

n n n



Ex. 4.17 The Lucas Number

(

, .

.

a) Assume

Then =

(b) Assume for the integers= 1,2,3, . . . , where

Then

L L

L L L L

L F Fn k k k

L L L F FF F F F

ii

kk

ii

kk k k

n n n

k k k k kk k k k

02

0

12 1 3

1 1

1 1 1 12 2

1

1 1

1 2

Please check out the induction basis for yourselves first.


4.3 The Division Algorithm: Prime Numbers

Def. 4.1 For integers a,b, we say b divides a, and we write b|a,if there is an integer n, such that a=bn. We say b is a divisorof a or a is a multiple of b.

Theorem 4.3. For any Z (a) 1| (b) (c) (d) Z(e) If , for some Z, and divides two ofthe three integers, then divides the remaining integer.(f) Z.(g) For let Z. If divides each then

where Z for

a b ca a a b b a a ba b b c a c a b a bx xx y z x y z a

aa b a c a bx cy x y

i n c a ca c x c x c x x i

i i

n n i

, ,| [( | ) ( | )]

[( | ) ( | )] | | | ,, ,

[( | ) ( | )] |( ), ,, ,

| ( ), all

0

111 1 2 2 n.



proof of (f):If and then for some

Z. So If follows that

a b a c b am c anm n bx cy am x an y a mx ny

a bx cy

| | , ,, ( ) ( ) ( ).

|( ).

Ex. 4.21 Let a,b in Z so that 2a+3b is a multiple of 17. Prove that17 divides 9a+5b.

Proof: We observe that 17|Also Hence 17|

( ) |( )( ).|( ). [( ) ( )]

|( ).

2 3 17 4 2 317 17 17 17 17 8 12

17 9 5

a b a ba b a b a b

a b



For n in Z+ where n>1, n is a prime number if n has only two divisors, 1 and n. Otherwise n is a composite.

Lemma 4.1. If n in Z+ and n is composite, then there is aprime p such that p|n.Proof: If not, let S be the set of all composite integers that have noprime divisors. If S is not empty, then by the well-orderingprinciple, S has a least element m. But if m is composite,m=m1m2 with 1<m1<m and 1<m2<m. Since m1 is not in S, m1 is a prime or divisible by a prime, which means m is also divisibleby a prime, a contradiction.

lemma, theorem, corollary



Theorem 4.4 (Euclid) There are infinitely many primes.

Proof: If not, let be the finite set of primes,and let Since for all

cannot be a prime. Hence is a composite. So by Lemma 4.1 there is a prime and Since

and it follows that a contradiction.

p p pB p p p B p i k

B Bp p B p B

p p p p p

k

k i

j j j

j k j

1 2

1 2

1 2

1 1

1

, , ,. ,

| . |

| , | ,



Theorem 4.5 (The Division Algorithm)

.0 , with

uniqueexist then 0, with , If

brrqbaZq,r

therebZa, b

a: dividendb: divisorq: quotientr: remainder



Ex. 4.28 If Z and is composite, then there existsa prime such that and Proof: Since is composite, we can write where1 < and 1 < We claim that one of the integers

must be less than or equal to If not, then and give the contradiction

Without loss of generality, assume If is a prime, the result follows. Otherwise, by Lemma4.1, there exists a prime where So and

+n np p n p n

n n n nn n n n

n n nn n n n n n n

n n n n nn

p n p n p np n

| .,

., .

. .

| . |.

1 2

1 2

1 2

1 2 1 2

1

1

1 1


4.4 The Greatest Common Divisor: the Euclidean Algorithm

Def. 4.2 For a, b in Z, a position integer c is said to be a commondivisor of a and b if c|a and c|b.

Def 4.3 Greatest Common Divisor

For any common divisor d of a and b, we have d|c. Then c is a greatest common divisor of a and b.

Theorem 4.6 For any a,b in Z+, there exists a unique c in Z+

that is the greatest common divisor of a and b.



Proof: Given , Z let = { + | , ,+ > }. Since , by the Well - Ordering principle

has a least element . We claim that is a greatestcommon divisor of , . Since , = + for some

, . Consequently, if and | and | , then| + , so | . If | , = + , < < . Then = -

= - ( + ) = ( - ) + (- ) , so , contradictory to is the least in . Consequen

+a b S as bt s t Zas bt SS c c

a b c S c ax byx y Z d Z d a d bd ax by d c c a a qc r r c r a qc

a q ax by qx a qy b r Sc S

,0

01

tly, | and | . If both and satisfy, then and which implies

1

2 1 2 1

c a c b cc c c c c c c| | , .2 1 2

(existence and uniqueness)



From Theorem 4.6, gcd(a,b) is the smallest positive integer wecan write as a linear combination of a and b.

Integers a and b are called relative prime when gcd(a,b)=1.That is, when there exist x,y in Z with ax+by=1.

Ex. 4.30 Since gcd(42,70)=14, we can find x,y in Z with 42x+70y=14, or 3x+5y=1. By inspection, x=2, y=-1 is asolution. But, general solution? x=2-5k, y=-1+3k orx=2+5k, y=-1-3k..



Theorem 4.7 Euclidean Algorithm

If , Z we apply the division algorithm as follows:

Then , the last nonzero remainder, equals gcd(

+a ba q b r r bb q r r r rr q r r r r

r q r r r r

r q r r r rr q r r r rr q r

r

i i i i i i

k k k k k k

k k k k k k

k k k

k

,,,,

,

,,

1 1 1

2 1 2 2 1

1 3 2 3 3 2

2 1 2 2 1

3 1 2 3 1 2

2 1 1

1 1

000

0

00

a b, ).



Ex. 4.31 gcd(250,111)=?

250 111222 28 84 27 27 1

*2

*3

*1

gcd(250,111)=1, what is x, and y suchthat 250x+111y=1?

1=28-1(27)=28-1[111-3(28)]=(-1)111+4(28)=(-1)111+4[250-2(111)]=4(250)+(-9)(111)

x=4+111k, y=-9-250k



Ex. 4.32 For any n in Z+, prove that 8n+3 and 5n+2 are relativeprime.

8n+3 5n+25n+23n+1 3n+1 2n+12n+1n 2n 1

gcd(8n+3,5n+2)=1. But we could alsoarrived at this conclusion if we hadnoticed that(8n+3)(-5)+(5n+2)(8)=1



Ex. 4.34. Griffin has two unmarked containers. One containerholds 17 ounces and the other holds 55 ounces. Explain how Griffin can use his two containers to have exactly one ounce.

gcd(17,55)=1, 1=13(17)-4(55), Consequently, Griffin mustfill his smaller container 13 times and empty the contentsinto the larger container.



Ex. 4.35. Diophantine equation

Find nonnegative integer solutions for 6x+10y=104.

Ans: 3x+5y=52, gcd(3,5)=1, 1=3(2)+5(-1), 52=3(104)+5(-52)x=104-5k, y=-52+3k, 104-5k 0 and -52+3k 0. Sok=18,19,20.

Theorem 4.8. If a,b,c in Z+, the Diophantine equcation ax+by=c has an integer solution if and only if gcd(a,b) divides c.



Def 4.4. For a,b,c in Z+, c is called a common multiple of a,b ifc is a multiple of both a and b. Furthermore, c is the least commonmultiple if it is the smallest of all positive integers that are commonmultiple of a,b. We denote c by lcm(a,b).

Theorem 4.10 For a,b in Z+, ab=lcm(a,b)gcd(a,b)


4.5 The Fundamental Theorem of Arithmetic

Lemma 4.2 If , Z and is a prime, then | | or | .

Lemma 4.3 Let Z for all 1 . If is a primeand | then | for some 1 .

+a b pp ab p a p b

a i n pp a a a p a i n

in i

1 2 ,

Ex 4.38 Show that 2 is irrational.

Proof: If not, let 2 where , Z and

gcd( , ) = 1. Then 2 =

which contradicts theearlier claim that gcd( , ) = 1.

+

a b a b

a b a b b a a

a a c a c b c b c

b b a ba b

/ ,

/ |

|

| | gcd( , ) ,

2 2 2 2 2

2 2 2 2 2 2

2

2 2

2 2 4 2 4 2

2 2 2



Theorem 4.11 (The fundamental theorem of arithmetic) Everyinteger n>1 can be written as a product of primes uniquely, upto the order of the primes.

Ex. 4.41 For n in Z+, we want to count the number of positivedivisors of n.

By Theorem 4.11, = where foreach 1 , is a prime and If | , then

= where 0 for all 1 . So by the rule of product, the number of positive divisors of n is ( 1

n p p pi k p e m n

m p p p f ei k

e e e

e eke

i if f

kf

i i

k

k

k

1 2

1 2

2

1 2

1 2

0

1 1 1

,.

)( ) ( ).



Ex. 4.43 Can we find three consecutive positive integers whoseproduct is a perfect square, that is, do there exist m,n in Z+ with(m)(m+1)(m+2)=n2?

Sol: Suppose m,n do exist. Since gcd(m,m+1)=gcd(m+1,m+2)=1,so for any prime p, if p|(m+1), then p|m and p|m+2. Furthermore,if p|(m+1), then p|n2. Since n2 is a perfect square, the exponentson p in the prime factorizations of both m+1 and n2 must be thesame even integer. So m+1 is a perfect square. And m(m+2) mustbe a perfect square too. But m2<m(m+2)<m2+2m+1<(m+1)2. Som(m+2) cannot be a perfect square. m,n do not exist.

properties of the integers: mathematical induction 1. mathematical induction 2. harmonic, fibonacci,...

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