Chapter 4 Properties of the integers: mathematical induction

1. Chapter 4 Properties of the integers: mathematical induction Yen-Liang Chen Dept of IM NCU

2. 4.1. The well-ordered principle: mathematical induction • Z+={xZx1} but Q+={xQx>0} and R+={xRx>0}. • Z+ is different from Q+ and R+ in the following property . • The well-ordering principle: Every nonempty subset of Z+ contains a smallest element.

3. The principle of mathematical induction • (a) If S(1) is true and (b) If S(k) is true, then S(k+1) is true, (c) then S(n) is true for all n Z+. • The condition (a) is referred to as basis step, while that in part (b) is called the inductive step. The basis step can begin with any integer number, even negative.

4. Ex 4.1 • For all nZ+, S(n)=1+2+…+n=n(n+1)/2 • Basis: it is true for n=1 • Hypothesis: it is true for S(n) • Induction: Adding n+1 to S(n), we find S(n+1) is also true.

5. Ex 4.4 • For all nZ+, S(n)=12+22+…+n2=n (n+1) (2n+1)/6 • Basis: it is true for n=1 • Hypothesis: it is true for S(n)

6. Ex 4.6 • S(n)=1+2+…+n=(n2+n+2)/2 • Obviously, this formula is wrong. • However, we find that S(k)S(k+1) . • Although S(k)S(k+1) but S(1) is not true, S(n) is not true either. This example indicates the need to establish the basis step.

7. Ex 4.8 • For all n6, 4n<(n2-7) • Basis: it is true for n=6 • Hypothesis: it is true for n=k • Induction: it is also true n=k+1.

8. Ex 4.9 • Harmonic number Hn=1+(1/2)+…+(1/n) • For all n Z+, • Basis: it is true for n=1 • Hypothesis: it is true for n=k

9. Ex 4.10 • binary search principle • For all AnR with An=2n and the elements of An are listed in ascending order. To determine if r exist in An or not, we must compare r with no more than n+1 elements in An. • Basis: it is true for A0 and A1. • Hypothesis: it is true for An. • Induction: it is also true An+1.

10. Ex 4.12 • For all k1, S(k): k has 2k-1 compositions. • Basis: S(1) is true. • Given the composition of k, we can produce the composition of k+1 by: • the last summand of k is added by 1. • We append a new summand “1” after the last summand of k. • Therefore, the number of composition of k+1 is twice the number of k.

12. Ex 4.13 • for all n14, S(n): n can be written as a sum of 3’s and/or 8’s • Basis: 14=3+3+8 • Hypothesis: S(k) is true. • Induction: Let us consider k. • If there is one 8 in the sum that equals k, then we can replace this as 3+3+3. • If there is no 8, then since k14, there are at least five 3’s as summands, which can be replaced by two 8’s.

13. Alternative form of mathematical induction (a) If S(n0), S(n0+1), S(n0+2),…, S(n1-1), S(n1) are true, and (b) If S(n0), S(n0+1), …, S(k-1), S(k) are true for some kn1, then S(k+1) is also true, (c) then S(n) is true for all n n0.

14. Ex 4.15 • a0=1, a1=2, a2=3, an=an-1+an-2+an-3 for n3. • S(n): an3n for all nN. • Basis: it is true for n=0, n=1 and n=2. • Hypothesis: S(n) is true for nk.

15. 4.2. Recursive definition • In example 4.15, we have a recursive definition

16. Ex 4.16 • Given the definition that (1) the conjunction of p1, p2 by p1p2, (2) p1p2…pn pn+1 (p1p2… pn)  pn+1. • Show that (p1p2… pr)  (pr+1pr+2… pk) = (p1p2… pr+1pr+2… pk) .

17. Ex 1.47 • Given the definition that (1) the union of A1, A2 by A1A2, (2) A1A2…AnAn+1 (A1A2…An)An+1. • Show (A1A2… Ar)  (Ar+1Ar+2… Ak) = (A1A2… Ar+1Ar+2… Ak) .

18. Ex 4.18 • (A1A2…Ak+1)c=A1c A2c… Ak+1c • Basis: it is true for n=2 • Hypothesis: it is true for n=k (A1A2…Ak)c=A1c A2c… Akc

19. Ex 4.19 • The Fibonacci numbers are defined as • F0=0, F1=1 and • Fn=Fn-1+Fn-2 for n2. • Show that

20. Ex 4.20 • Lucas numbers: (1) L0=2, L1=1; and (2) Ln=Ln-1+Ln-2 for n2. • Then Ln=Fn-1+Fn+1.

21. Ex 4.21 • Eulerian numbers am, k are defined: (1) am, k = (m-k)am-1, k-1 +(k+1)am-1, k for 0km-1. (2) a0,0=1, am, k=0 for km, am, k=0 for k<0.

23. 4.3. The division algorithm: prime number • Definition 4.1. ba if and only if a=bn • Theorem 4.3.

24. Ex 4.23 • Do there exists an integer x, y, z such that 6x+9y+15z=107 • Since 36, 39 and 315, 3 is a divisor of 107. • But since 3 is not a divisor of 107, there does not exist such x, y, z.

25. Ex 4.24 • If 172a+3b, then 179a+5b • 172a+3b17(-4)(2a+3b). • 17171717a+17b • 17(-4)(2a+3b) +17a+17b179a+5b

26. Prime • Prime: integers that have exactly two divisors. • Composite number: integers that are not primes. • for any composite n, there is a prime p such that pn. (Why?) • There are infinitely many primes. (Theorem 4.4)

27. Proof of Theorem 4.4 • Let p1, p2,…,pk be the set of all primes. • Let B= p1 p2…pk+1 and B is a composite. • There exists pj such that pjB and pj p1 p2…pk. • By Theorem 4.3(e), pj1 ; so, this is a contradiction.

28. Theorem 4.5. • The division algorithm in Fig 4.10 • If a, bZ, then there exist unique q, rZ with a=qb+r, 0r<b. • Here q is called quotient, r remainder, b divisor and a dividend.

29. 4.4. The greatest common divisor: the Euclidean algorithm • Common divisor , Definition 4.2 • Greatest common divisor c, Definition 4.3 (a) ca and cb, (b) For any common divisor d of a and b, dc • Theorem 4.6. gcd(a, b) is unique (Why?) • Let S={as+bts,tZ, as+bt>0}. The least element in S is the greatest common divisor of a , b.

30. Proof of Theorem 4.6 • Since c=ax+by, if da and db, then dc=ax+by. Def 4.3(b) • if c can not divide a, then a=qc+r where 0<r<c. So, we have r=a-qc=a-q(ax+by)=(1-qx)a+(-qy)b , meaning rS. This is a contradiction to the fact that c is the smallest one in S. Hence, ca and cb. Def 4.3(a) • If c1 and c2 both satisfy the above two conditions, let us designate c1 as the gcd(a, b), then we have c2c1. By reversing the role, we have c1c2. So, we conclude that c1=c2.

31. Properties • gcd(a, b) is the smallest positive integer we can write as a linear combination of a and b. • gcd(a, b)= gcd(b, a); gcd(a, b)= gcd(a, -b)= gcd(-a, b)= gcd(-a, -b); gcd(a, 0) =a • gcd(0, 0) not defined. • We call a and b relatively prime if gcd(a, b)=1.

32. Theorem 4.7. Euclidean algorithm.

33. If dr0 and dr1dr2 dr1 and dr2dr3 drn-2 and drn-1drn This satisfies Def 4.3(b) From the last equation, rnrn-1. So, rnrn-2 because rn-2=qn-1rn-1+rn rnrn-1 and rnrn-2 rnrn-3 rnrn-2 and rnrn-3 rnrn-4 rnr2 and rnr1 rnr0 This satisfies Def 4.3(a) Proof

34. Examples • Ex 4.34. Find the gcd of 250 and 111. • Ex 4.35. Show that 8n+3 and 5n+2 are relatively prime. • Ex 4.36. How can we measure exactly one ounce by two containers with capacities 17 ounces and 55 ounces respectively.

35. Least common multiple • Definition 4.4. lcm(a, b) the least common multiple • lcm(1, n)=lcm(n, 1)=n • lcm(c, na)=n  a • If n m, lcm(am, an)=an. • If c=lcm(a,b) and d is a common multiple of a, b, then cd. • a b=lcm(a,b)  gcd(a, b)

36. 4.5. The fundamental theorem of arithmetic • Lemma 4.2. If p is prime and pab, then pa or pb. • If pa, we are finished. • If not, then gcd(p, a)=1. • px+ay=1 • b=p(bx)+(ab)y • Since pab and pp, we have pb. • Lemma 4.3. If p is prime and pa1a2…an, then pai for some i. (Why)

37. is irrational • If not, then we have =a/b, where a and b are relatively prime, gcd(a, b)=1. • 2=a2/b22b2=a22a22a • 2aa=2c • 2b2=a2 =(2c)2=4c2 b2=2c2 • b2=2c22b22b • Since 2a and 2b, gcd(a, b)2

38. Theorem 4.11. • Every integer can be written as a product of primes uniquely. (1) the existence of a prime factorization. (2)The factorization is uniqueness. • The first part is proved by contradiction. Consider the smallest integer m that can not be factorized . • The second part is proved by induction. It is true for 2 and for n-1. Now we prove it holds for n.

39. Examples • Ex 4.42. The prime factorization for 980220. • Ex 4.44. How to count the number of divisors of n? • Ex 4.45. Determine the gcd(a, b) and lcm(a, b) by factorization.