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1 Strong Mathematical Induction

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Principle of Strong Mathematical Induction Let P(n) be a predicate 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.

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3 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)

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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 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.■

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Proving a Property of a Sequence Proposition: Suppose a 0, a 1, a 2, … is defined as follows: 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. 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=2: a 2 =3 ≤4=2 2 P(2)

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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: a i ≤ 2 i for all 0≤i<k.(1) Show that P(k) is true: a k ≤ 2 k (2) a k = a k-1 +a k-2 +a k-3 ≤ 2 k-1 +2 k-2 +2 k-3 (based on (1)) ≤ 2 0 +2 1 +…+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.■

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