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Mathematical Induction 2 Principle of Mathematical Induction Induction Mathematical induction proves a predicate is true for all integers greater than some number. Every program has at least one bug and can be shortened by at least one instruction—from which, by induction, it is evident that every program can be reduced to one instruction that does not work. (Ken Arnold) The Principle of Mathematical Induction: To prove n ≥ b implies P (n), do the following steps: 1. Basis Step: Prove P (b) is true. 2. Induction Step: Prove k ≥ b ∧ P (k) implies P (k + 1). In combination, the basis and induction steps prove ∀n(n ≥ b → P (n)), or in other words, n ≥ b implies P (n). Mathematical Induction Principle of Mathematical Induction Template for Proofs . . . . . . . . . . . Example 1. . . . . . . . . . . . . . . . . . Example 1, Slide 2 . . . . . . . . . . . . Example 2. . . . . . . . . . . . . . . . . . Example 3. . . . . . . . . . . . . . . . . . Example 4. . . . . . . . . . . . . . . . . . Strong Induction Strong Induction . . . Example 1. . . . . . . . Example 2. . . . . . . . Example 3. . . . . . . . Example 3 Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 3 4 5 6 7 8 . . . . . . . . . . . . . . . . . . . 9 . 9 10 11 12 13 Throughout, the domain is the integers. CS 2233 Discrete Mathematical Structures Mathematical Induction – 2 Template for Proofs 1. Write statement to prove: n ≥ b implies P (n) substituting for b and P (n). 2. Show basis step. Prove that P (b) is true. 3. Show induction step by direct proof. (a) Write inductive hypothesis. Assume k ≥ b and P (k). (b) Write intended conclusion P (k + 1). Substitute n with (k + 1). (c) Prove P (k + 1). A key part of the proof is using P (k). You may also need to use k ≥ b. CS 2233 Discrete Mathematical Structures 1 Mathematical Induction – 3 2 Example 1 Example 2 n n ≥ 1 implies n < 2n n ≥ 1 implies Σ (2i − 1) = n2 i=1 Basis (n = 1): 1 < 21 = 2 Basis (n = 1): Σ1i=1(2i − 1) = 1 = 12 Induction (n > 1): Induction (n > 1): Assume k ≥ 1 and k < 2k Assume k ≥ 1 and Σki=1(2i − 1) = k 2 Want to show k + 1 < 2k+1 k+1 (2i − 1) = (k + 1)2 Want to show Σi=1 Proof: We know (k + 1)2 = k 2 + 2k + 1 and Σk+1 i=1 (2i − 1) = last term + rest of terms = 2(k + 1) − 1 + Σki=1(2i − 1) CS 2233 Discrete Mathematical Structures Mathematical Induction – 4 Proof: We know 2k+1 = 2k + 2k . k < 2k implies 2k+1 = 2k + 2k > k + k k ≥ 1 implies k + k ≥ k + 1 k + 1 ≤ k + k and k + k < 2k+1 imply k + 1 < 2k+1 CS 2233 Discrete Mathematical Structures Example 1, Slide 2 Mathematical Induction – 6 Example 3 Σki=1(2i − 1) = k 2 implies 2(k + 1) − 1 + Σki=1(2i − 1) = 2k + 1 + k 2 n n ≥ 1 implies Σ i > n2/2 i=1 Basis (n = 1): Σ1i=1 i = 1 > 12/2 = 1/2 k Σk+1 i=1 (2i − 1) = 2(k + 1) − 1 + Σi=1(2i − 1) and 2(k + 1) − 1 + Σki=1(2i − 1) = k 2 + 2k + 1 and k 2 + 2k + 1 = (k + 1)2 imply 2 Σk+1 i=1 (2i − 1) = (k + 1) Induction (n > 1): Assume k ≥ 1 and Σki=1 i > k 2 /2 2 Want to show Σk+1 i=1 i > (k + 1) /2 Short Proof: k ≥ 1 ∧ Σki=1(2i − 1) = k 2 implies k Σk+1 i=1 (2i − 1) = 2(k + 1) − 1 + Σi=1(2i − 1) = 2(k + 1) − 1 + k 2 = (k + 1)2 CS 2233 Discrete Mathematical Structures Mathematical Induction – 5 3 Proof Parts: (k + 1)2/2 = (k 2 + 2k + 1)/2 k and Σk+1 i=1 i = k + 1 + Σi=1 i How to use inductive hypothesis? CS 2233 Discrete Mathematical Structures Mathematical Induction – 7 4 Example 4 Example 1 2 n ≥ 5 implies n < 2 n Let Fi be the ith Fibonacci number. Why 5? Basis (n = 5): 52 = 25 < 25 = 32 n ≥ 0 implies Fn < 2n Induction (n > 5): Basis (n ∈ {0, 1}): F0 = 0 < 20 = 1 and F1 = 1 < 21 = 2 Assume k ≥ 5 and k 2 < 2k Induction (n > 1): Want to show (k + 1)2 < 2k+1 Assume k ≥ 1 and F0 < 20, . . . , Fk < 2k Proof Parts: (k + 1)2 = k 2 + 2k + 1 and 2k+1 = 2k + 2k How to use inductive hypothesis? Want to show Fk+1 < 2k+1 CS 2233 Discrete Mathematical Structures Prove this using Fk+1 = Fk + Fk−1 and 2k+1 = 2k + 2k and Fk < 2k and Fk−1 < 2k−1. Mathematical Induction – 8 CS 2233 Discrete Mathematical Structures Mathematical Induction – 10 Example 2 Strong Induction 9 n ≥ 12 implies there exists i ∈ N and j ∈ N such that 4i + 5j = n Strong Induction Basis (n ∈ {12, 13, 14, 15}): How? Strong induction is a more general form of mathematical induction. It also proves a predicate is true for all integers greater than some number. Induction (n > 15): Strong Induction has two steps. Want to show there is a combination for k + 1 1. Basis: Prove P (b) is true. 2. Induction: Prove k ≥ b ∧ P (b) ∧ . . . ∧ P (k) imply P (k + 1). Proof Sketch: Add one more 4 to k − 3 combo. Because k ≥ 15 then k − 3 ≥ 12 has a combination by the inductive hypothesis. As with mathematical induction, the basis and induction steps prove n ≥ b implies P (n). Assume k ≥ 15 and there are combinations for 12, 13, 14, 15, . . . , k CS 2233 Discrete Mathematical Structures Mathematical Induction – 11 The basis often includes more than one base case. CS 2233 Discrete Mathematical Structures Mathematical Induction – 9 5 6 Example 3 Example 3 Continued n Proof Parts: Harmonic Numbers: Let Hn = Σ 1/i i=1 Also, let lg n = log2 n The first half (or more) of the terms of Hk+1 is greater than lg k+1 lg(k + 1) 1 2 = − . 3 3 3 n ≥ 1 implies Hn > (lg n)/3 Basis (n = 1): H1 = 1 > (lg 1)/3 = 0 Induction (n > 1): The last third (or more) of the terms of Hk+1 is greater than or equal to 1/3. (k + 1)/3 terms are each ≥ 1/(k + 1). Assume k ≥ 1 and H1 > (lg 1)/3, . . . , Hk > (lg k)/3 k+1 k+1 + ≤k+1 2 3 CS 2233 Discrete Mathematical Structures Mathematical Induction – 13 Want to show Hk+1 > (lg(k + 1))/3 CS 2233 Discrete Mathematical Structures Mathematical Induction – 12 7 8