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Mathematical Induction

Mathematical Induction. n. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. F(n). F(1) = 1; F(n+1) = F(n) + (2n+1) for n≥1. 1. 4. 9. 16. 25. 36. 49. 64. 81. 100. F(n) =. n 2 for all n ≥ 1. Prove it!. Prove statements of the form: "p(n) for n ≥ 1". Verify p(1). "Base case"

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Mathematical Induction

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  1. Mathematical Induction

  2. n 1 2 3 4 5 6 7 8 9 10 F(n) F(1) = 1; F(n+1) = F(n) + (2n+1) for n≥1 1 4 9 16 25 36 49 64 81 100 F(n) = n2 for all n ≥ 1 Prove it!

  3. 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

  4. Prove: F(n) = n2 for n ≥ 1. • Recall F(1) = 1; F(n+1) = F(n) + (2n+1) for n>1 • Proof: F(1) = 1 = 12 • Assume F(n) = n2 for some n ≥ 1. • Then F(n+1) = F(n) + (2n+1), since n+1 > 1. = n2 + 2n + 1 by hypothesis = (n+1)2 • Therefore F(n) = n2 for all n ≥ 1. Induction hypothesis Base Case Algebra Not just for some!

  5. 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.

  6. What does our proof have to do with S? Let S = { n ≥ 1 | F(n) = n2 } F(n) = 1 = 12 Base case shows 1 is in S F(n) = n2 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) = n2 for all n ≥ 1.

  7. 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.

  8. 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.

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

  10. 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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