Principle of Strong Mathematical Induction Let Pn be
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.
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 true for n=2 P(2) because 2 | 2 and 2 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) 2
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. We must show: P(k) k is also divisible by a prime. Consider 2 cases: a) k is prime. Then k is divisible by itself. b) k is composite. Then k=a·b where 2≤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. (1) (2) ■
Proving a Property of a Sequence Ø Proposition: Suppose a 0, a 1, a 2, … is defined as follows: a 0=1, a 1=2, a 2=3, ak = ak-1+ak-2+ak-3 for all integers k≥ 3. Then an ≤ 2 n for all integers n≥ 0. P(n) Ø Proof (by strong induction): 1) Basis step: The statement is true for n=0: a 0=1 ≤ 1=20 for n=1: a 1=2 ≤ 2=21 for n=2: a 2=3 ≤ 4=22 P(0) P(1) P(2)
Proving a Property of a Sequence Ø Proof (cont. ): 2) Inductive step: For any k>2, Assume P(i) is true for all i with 0≤i<k: ai ≤ 2 i for all 0≤i<k. Show that P(k) is true: ak ≤ 2 k ak= ak-1+ak-2+ak-3 ≤ 2 k-1+2 k-2+2 k-3 (based on (1)) ≤ 20+21+…+2 k-3+2 k-2+2 k-1 = 2 k-1 (as a sum of geometric sequence) ≤ 2 k Thus, P(n) is true by strong induction. (1) (2) ■
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