lecture12(complete) - MA1100 Lecture 12 Mathematical...

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1 MA1100 Lecture 12 Mathematical Induction Using PMI on other universal sets Strong PMI Variations of strong PMI Chartrand: section 6.4
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Lecture 12 2 Announcement ± Collection of clickers ( No.1 to 150) ± Mid-term test : ² Oct 2 (12.00-1.30pm) at MPSH1. ² Handouts (please get one) ± Online survey (open soon) ± Collection of HW2 ² Between next Tuesday and Friday, at Math Dept General Office Counter (S14 Level 3) ² During lecture on Sep 29
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Lecture 12 3 Other Universal Sets ( " n œ Z )P(n) 1.P(0) is true 2.For all k ¥ 0, if P(k) is true, then P(k+1) is true 3.For all k § 0, if P(k) is true, then P(k - 1) is true Then P(n) is true for all n œ Z P(0) Ø P(1) Ø P(2) Ø Ø P(n)… …P(-n) P(-2) P(-1) P(0) Possible to apply PMI to prove: What about ( " x œ Q + ) P(x) and ( " x œ R + ) P(x) ? possible impossible
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Lecture 12 4 Other Universal Sets ( " q œ Q + )P(q) 1.P( 1 ) is true 2.For all k œ N ,i fP( k ) is true, then P( k+1 ) is true Then P(m/n) is true for all m/n œ Q + where m, m n q n = N Fix m œ N Then P(n) is true for all n œ N 3.P(m/ 1 ) is true (from 1 and 2 above) 4.For all k œ N fP(m/ k ) is true, then P(m/ (k+1) ) is true
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Lecture 12 5 Other Universal Sets ( " q œ Q + )P(q) + where m, n m q n = Z P(1) Ø P(2) Ø P(3) Ø Ø P(m) … ) n 2 ( P ) 3 2 ( P ) 2 2 ( P # By (1) and (2) ) n 1 ( P ) 3 1 ( P ) 2 1 ( P # By (3) and (4) ) n 3 ( P ) 3 3 ( P ) 2 3 ( P # ) n m ( P ) 3 m ( P ) 2 m ( P #
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Lecture 12 6 Binary Sum Proposition Every positive integer can be written as the sum of distinct powers of 2 . 4 = 2 2 10 = 2 3 + 2 1 45 = 2 5 + 2 3 + 2 2 + 2 0 Example not 2 1 + 2 1 Proof by Mathematical Induction Base case : prove P(1) is true P(n): n is a sum of distinct powers of 2 . Since 1 = 2 0 , So P(1) is true. Binary Sum 2 0 occurs only in binary sum of odd numbers
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Lecture 12 7 Binary Sum Inductive step : P(k) P(k+1) for all k ¥ 1 hypothesis P(k): want to get P(k+1): Is knowing P(k) enough to prove P(k+1) ? P(n): n is a sum of distinct power of 2 . k is a sum of distinct power of 2 k + 1 is a sum of distinct power of 2 k = sum of distinct power of 2 k + 1 = (sum of distinct power of 2) + 2 0 Is this valid? 2 0 may occur here can’t prove P(k) P(k+1) This proves P(k) P(k+1)
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Lecture 12 8 Strong PMI Let P(n) be a predicate such that: 1.P(1) is true 2.For all k œ N , if P(1), P(2), …, P(k) are true, then P(k+1) is true. Then P(n) is true for all n œ N Theorem ( Strong Principle of Mathematical Induction) stronger hypothesis The base case may start from another integer
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Lecture 12 9 Strong PMI To prove ( " n œ N ) P(n) is true Base case: Prove P(1) is true Inductive Step: Prove " k œ N , [P(1) P(2) ... P(k)] P(k+1) is true We can then conclude that P(n) is true for all n œ N k = 1 k = 2 k = 3 P(1) P(2) P(3) P(4) K terms
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lecture12(complete) - MA1100 Lecture 12 Mathematical...

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