Principle of Strong Mathematical Induction

1. Principle of Strong Mathematical Induction • Let P(n) be a statement defined for integers n; a and b be fixed integers with a≤b. • Suppose the following statements are true: 1. P(a), P(a+1), … , P(b) are all true (basis step) 2. For any integer k>b, if P(i) is true for all integers i with a≤i<k, then P(k) is true. (inductive step) • Then P(n) is true for all integers n≥a.

2. Example: Divisibility by a Prime • Theorem: For any integer n≥2, n is divisible by a prime. P(n) • Proof (by strong mathematical induction): 1) Basis step: The statement is truefor n=2 P(2) because2 | 2and2 is a prime number. 2) Inductive step: Assume the statement is true for all i with 2≤i<k P(i) (inductive hypothesis) ; show that it is true for k . P(k)

3. Example: Divisibility by a Prime • Proof (cont.): We have that for all iZ with 2≤i<k, P(i) i is divisible by a prime number. (1) We must show: P(k) k is also divisible by a prime. (2) Consider 2 cases: a) k is prime. Then k is divisible by itself. b) k is composite. Then k=a·b where2≤a<k and 2≤b<k. Based on (1),p|a for some prime p. p|a and a|k imply that p|k(by transitivity). Thus, P(n) is true by strong induction. ■

4. Proving a Property of a Sequence • Proposition: Suppose a0, a1, a2, … is defined as follows: a0=1, a1=2, a2=3, ak = ak-1+ak-2+ak-3for all integers k≥3. Then an ≤ 2n for all integers n≥0. P(n) • Proof (by strong induction): 1) Basis step: The statement is truefor n=0: a0=1≤1=20 P(0) for n=1: a1=2≤2=21 P(1) for n=2: a2=3≤4=22 P(2)

5. Proving a Property of a Sequence • Proof (cont.): 2) Inductive step: For any k>2, Assume P(i) is true for all i with 0≤i<k: ai≤ 2i for all0≤i<k . (1) Show that P(k) is true: ak≤ 2k(2) ak= ak-1+ak-2+ak-3 ≤ 2k-1+2k-2+2k-3 (based on (1)) ≤ 20+21+…+2k-3+2k-2+2k-1 = 2k-1 (as a sum of geometric sequence) ≤ 2k Thus, P(n) is true by strong induction. ■