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Published byJeffry Mathews Modified over 5 years ago

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Mathematical Induction

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F(1) = 1; F(n+1) = F(n) + (2n+1) for n≥1 100816449362516941 10987654321 F(n) n F(n) =n 2 for all n ≥ 1 Prove it!

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Prove statements of the form: "p(n) for n ≥ 1" Verify p(1). "Base case" Assume p(n) for some n ≥ 1 "Induction hypothesis" Show p(n+1) must be true for this value of n. "Induction step" Conclude p(n) is true for all n ≥ 1

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Prove: F(n) = n 2 for n ≥ 1. Recall F(1) = 1; F(n+1) = F(n) + (2n+1) for n>1 Proof: F(1) = 1 = 1 2 Assume F(n) = n 2 for some n ≥ 1. Then F(n+1) = F(n) + (2n+1), since n+1 > 1. = n 2 + 2n + 1 by hypothesis = (n+1) 2 Therefore F(n) = n 2 for all n ≥ 1. Induction hypothesis Base Case Algebra Not just for some!

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First principle of mathematical induction Let S be a set of integers containing a. Suppose S has the property that whenever some integer n ≥ a belongs to S, then the integer n + 1 also belongs to S. Then S contains every integer greater than or equal to a.

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What does our proof have to do with S? Let S = { n ≥ 1 | F(n) = n 2 } F(n) = 1 = 1 2 Base case shows 1 is in S F(n) = n 2 for some n ≥ 1 => F(n+1) = (n+1) 2 Induction step shows that if n is in S, then n+1 is in S. By the first principle of induction, S contains all integers ≥ 1. That is, F(n) = n 2 for all n ≥ 1.

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Why does the first principle of induction work? Proof: Given S = { n ≥ a | p(n)} has the property that a is in S and n in S implies n+1 is in S. It remains to show that S contains all integers ≥ a. Let T = { n ≥ a | n is not in S} Assume, towards a contradiction, T is not empty. By the WOP, T has a smallest element, t Now t > a, and n = t – 1 belongs to S. But then n+1 = (t–1)+1 = t belongs to S. This contradiction shows that T is empty.

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Second Principle of Mathematical Induction Let S be the set of integers containing a. Suppose S has the property that n belongs to S whenever every integer less than n and greater than or equal to a belongs to S. Then S contains all integers greater than or equal to a.

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Proof of the 2nd Principle of Mathematical Induction Almost identical to the proof of the first. Try it!

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Proof of the Fundamental Theorem of Arithmetic. Let S be the set of integers greater than 1 which are primes or products of primes. 2 belongs to S Assume that for some integer n ≥ 2, S contains all integers k with 2 ≤ k < n. We must show that n is in S. If n is prime, then n is in S. If n is composite, n = ab where a,b are in S. By hypothesis, a and b are primes or products of primes. So n is a product of primes.

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