1 Inductive Proofs Rosen 6 th ed., §

2 Mathematical Induction A powerful, rigorous technique for proving that a predicate P(n) is true for every natural number n, no matter how large.A powerful, rigorous technique for proving that a predicate P(n) is true for every natural number n, no matter how large. Based on a predicate-logic inference rule:Based on a predicate-logic inference rule: P(0)  n  0 (P(n)  P(n+1))  n  0 P(n)P(0)  n  0 (P(n)  P(n+1))  n  0 P(n)

3 Outline of an Inductive Proof Want to prove  n P(n) …Want to prove  n P(n) … Base case (or basis step): Prove P(0).Base case (or basis step): Prove P(0). Inductive step: Prove  n P(n)  P(n+1).Inductive step: Prove  n P(n)  P(n+1). –e.g. use a direct proof: Let n  N, assume P(n). (inductive hypothesis)Let n  N, assume P(n). (inductive hypothesis) Under this assumption, prove P(n+1).Under this assumption, prove P(n+1). Inductive inference rule then gives  n P(n).Inductive inference rule then gives  n P(n).

4 Induction Example Prove thatProve that

5 Another Induction Example Prove that , n<2 n. Let P(n)=(n<2 n )Prove that , n<2 n. Let P(n)=(n<2 n ) –Base case: P(0)=(0<2 0 )=(0<1)=T. –Inductive step: For prove P(n)  P(n+1). Assuming n<2 n, prove n+1 < 2 n+1.Assuming n<2 n, prove n+1 < 2 n+1. Note n + 1 < 2 n + 1 (by inductive hypothesis) < 2 n + 2 n (because 1<2=2  2   2  2 n-1 = 2 n ) = 2 n+1Note n + 1 < 2 n + 1 (by inductive hypothesis) < 2 n + 2 n (because 1<2=2  2   2  2 n-1 = 2 n ) = 2 n+1 So n + 1 < 2 n+1, and we’re done.So n + 1 < 2 n+1, and we’re done.

6 Validity of Induction Prove: if  n  0 (P(n)  P(n+1)), then  k  0 P(k) Prove: if  n  0 (P(n)  P(n+1)), then  k  0 P(k) (a) Given any k  0,  n  0 (P(n)  P(n+1)) implies (P(0)  P(1))  (P(1)  P(2))  …  (P(k  1)  P(k)) (P(0)  P(1))  (P(1)  P(2))  …  (P(k  1)  P(k)) Using hypothetical syllogism k-1 times we have (b) Using hypothetical syllogism k-1 times we have P(0)  P(k) P(0)  P(k) (c) P(0) and modus ponens gives P(k). Thus  k  0 P(k). Thus  k  0 P(k).

7 The Well-Ordering Property The validity of the inductive inference rule can also be proved using the well-ordering property, which says:The validity of the inductive inference rule can also be proved using the well-ordering property, which says: –Every non-empty set of non-negative integers has a minimum (smallest) element. –  S  N :  m  S :  n  S : m  n

8 Use Well-Ordering Property and Proof by Contradiction Suppose P(0) is true and that P(k)  P(k+1) is true for all positive k.Suppose P(0) is true and that P(k)  P(k+1) is true for all positive k..Assume P(n) is false for some positive n. Implies S={n|  P(n)} is non-empty and has a minimum element m (i.e., P(m)=false)Implies S={n|  P(n)} is non-empty and has a minimum element m (i.e., P(m)=false) But then, P(m-1)  P((m-1)+1)=P(m) which is a contradiction!But then, P(m-1)  P((m-1)+1)=P(m) which is a contradiction!

9 Generalizing Induction Can also be used to prove  n  c P(n) for a given constant c  Z, where maybe c  0, then:Can also be used to prove  n  c P(n) for a given constant c  Z, where maybe c  0, then: –Base case: prove P(c) rather than P(0) –The inductive step is to prove:  n  c (P(n)  P(n+1)).  n  c (P(n)  P(n+1)).

10 Induction Example Prove that the sum of the first n odd positive integers is n 2. That is, prove:Prove that the sum of the first n odd positive integers is n 2. That is, prove: Proof by induction.Proof by induction. –Base case: Let n=1. The sum of the first 1 odd positive integer is 1 which equals 1 2. (Cont…) P(n)P(n)

11 Example cont. Inductive step: Prove  n  1: P(n)  P(n+1).Inductive step: Prove  n  1: P(n)  P(n+1). –Let n  1, assume P(n), and prove P(n+1). By inductive hypothesis P(n)

12 Strong Induction Characterized by another inference rule: P(0)  n  0: (  0  k  n P(k))  P(n+1)  n  0: P(n)Characterized by another inference rule: P(0)  n  0: (  0  k  n P(k))  P(n+1)  n  0: P(n) Difference with previous version is that the inductive step uses the fact that P(k) is true for all smaller, not just for k=n.Difference with previous version is that the inductive step uses the fact that P(k) is true for all smaller, not just for k=n. P is true in all previous cases

13 Example 1 Show that every n>1 can be written as a product p 1 p 2 …p s of some series of s prime numbers. Let P(n)=“n has that property”Show that every n>1 can be written as a product p 1 p 2 …p s of some series of s prime numbers. Let P(n)=“n has that property” Base case: n=2, let s=1, p 1 =2.Base case: n=2, let s=1, p 1 =2. Inductive step: Let n  2. Assume  2  k  n: P(k). Consider n+1. If prime, let s=1, p 1 =n+1. Else n+1=ab, where 1  a  n and 1  b  n. Then a=p 1 p 2 …p t and b=q 1 q 2 …q u. Then n+1= p 1 p 2 …p t q 1 q 2 …q u, a product of s=t+u primes.Inductive step: Let n  2. Assume  2  k  n: P(k). Consider n+1. If prime, let s=1, p 1 =n+1. Else n+1=ab, where 1  a  n and 1  b  n. Then a=p 1 p 2 …p t and b=q 1 q 2 …q u. Then n+1= p 1 p 2 …p t q 1 q 2 …q u, a product of s=t+u primes.

14 Example 2 Prove that every amount of postage of 12 cents or more can be formed using just 4- cent and 5-cent stamps.Prove that every amount of postage of 12 cents or more can be formed using just 4- cent and 5-cent stamps. Base case: 12=3  4), 13=2  4)+1(5), 14=1(4)+2(5), 15=3(5), so  12  n  15, P(n).Base case: 12=3  4), 13=2  4)+1(5), 14=1(4)+2(5), 15=3(5), so  12  n  15, P(n). Inductive step: Let n  15, assume  12  k  n P(k). Note 12  n  3  n, so P(n  3), so add a 4-cent stamp to get postage for n+1.Inductive step: Let n  15, assume  12  k  n P(k). Note 12  n  3  n, so P(n  3), so add a 4-cent stamp to get postage for n+1.

15 Example 3 Suppose a 0, a 1, a 2, … is defined as follows: a 0 =1, a 1 =2, a 2 =3, a 0 =1, a 1 =2, a 2 =3, a k = a k-1 +a k-2 +a k-3 for all integers k≥3. a k = a k-1 +a k-2 +a k-3 for all integers k≥3. Then a n ≤ 2 n for all integers n≥0.P(n)  Proof (by strong induction): 1) Basis step: The statement is true for n=0: a 0 =1 ≤1=2 0 P(0) for n=1: a 1 =2 ≤2=2 1 P(1) for n=1: a 1 =2 ≤2=2 1 P(1) for n=2: a 2 =3 ≤4=2 2 P(2) for n=2: a 2 =3 ≤4=2 2 P(2)

16 Example 3 (cont’d) 2) Inductive hypothesis: For any k>2, Assume P(i) is true for all i with 0≤i<k: a i ≤ 2 i for all 0≤i<k. a i ≤ 2 i for all 0≤i<k. 3) Inductive step: Show that P(k) is true: a k ≤ 2 k a k = a k-1 +a k-2 +a k-3 ≤ 2 k-1 +2 k-2 +2 k-3 (using inductive hypothesis) ≤ 2 k-1 +2 k-2 +2 k-3 (using inductive hypothesis) ≤ …+2 k-3 +2 k-2 +2 k-1 ≤ …+2 k-3 +2 k-2 +2 k-1 = 2 k -1 (as a sum of geometric sequence) = 2 k -1 (as a sum of geometric sequence) ≤ 2 k ≤ 2 k Thus, P(n) is true by strong induction.