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Unformatted text preview: 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=ab where 2a<k and 2b<k. Based on (1), pa for some prime p. pa and ak imply that pk (by transitivity). Thus, P(n) is true by strong induction. Proving a Property of a Sequence Proving a Property of a Sequence Proposition: Suppose a , a , a , is defined as follows: Proving a Property of a Sequence Proving a Property of a Sequence Proof ( cont. ): 2) Inductive step: For any k>2 , Assume P(i) is true for all i with 0i<k: a i 2 i for all 0i<k . (1) Show that P(k) is true: a k 2 k (2) a k = a k1 +a k2 +a k3 2 k1 +2 k2 +2 k3 (based on (1)) 2 +2 1 ++2 k3 +2 k2 +2 k1 = 2 k1 (as a sum of geometric sequence) 2 k Thus, P(n) is true by strong induction....
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This note was uploaded on 05/25/2008 for the course MATH 21127 taught by Professor Gheorghiciuc during the Fall '07 term at Carnegie Mellon.
 Fall '07
 GHEORGHICIUC
 Integers, Mathematical Induction

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