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Consider the set S of strings defined recursively in Problem 1. Prove that for any given string in S, the number of a's in it is the same as the number of b's in it. Prove each statement using either weak, strong, or structural induction. Make sure to clearly indicate the different parts of your proof: the basis step, the inductive hypothesis, what you will show in the inductive step, and the inductive step. Make sure to clearly format your proofs and to write in complete, clear sentences. EXAMPLE: Prove that for any nonnegative integer n, Σ i = (n+1) Answer: Proof. (by weak induction) Basis step: n = 1 Σ=1 1(1+1)==1 Therefore, (n+1) when n = 1. = Inductive hypothesis: Assume that Inductive step: We will show that i=1 i=1 i= = (+1) for some integer k > 1. i= (k+1)((k+1)+1) k+1 Σ=Σ+ (κ + 1) i=1 By inductive hypothesis, k+1 Σ IME i=1 k(k+1) = +k+1 2 k(k+1)+2(k+1) = 2 (k+2)(k+1) = 2 (k+1)((k+1)+1) 2 Therefore, by weak induction, we have shown that = (n+1) for all nonnegative integers n.

Consider the set S of strings defined recursively in Problem 1. Prove that for any given string in S, the number of a's in it is the same as the number of b's in it. Prove each statement using either weak, strong, or structural induction. Make sure to clearly indicate the different parts of your proof: the basis step, the inductive hypothesis, what you will show in the inductive step, and the inductive step. Make sure to clearly format your proofs and to write in complete, clear sentences. EXAMPLE: Prove that for any nonnegative integer n, Σ i = (n+1) Answer: Proof. (by weak induction) Basis step: n = 1 Σ=1 1(1+1)==1 Therefore, (n+1) when n = 1. = Inductive hypothesis: Assume that Inductive step: We will show that i=1 i=1 i= = (+1) for some integer k > 1. i= (k+1)((k+1)+1) k+1 Σ=Σ+ (κ + 1) i=1 By inductive hypothesis, k+1 Σ IME i=1 k(k+1) = +k+1 2 k(k+1)+2(k+1) = 2 (k+2)(k+1) = 2 (k+1)((k+1)+1) 2 Therefore, by weak induction, we have shown that = (n+1) for all nonnegative integers n.

Operations Research : Applications and Algorithms
Operations Research : Applications and Algorithms
4th Edition
ISBN:9780534380588
Author:Wayne L. Winston
Publisher:Wayne L. Winston
Chapter2: Basic Linear Algebra
Section: Chapter Questions
Problem 15RP
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Consider the set S of strings defined recursively in Problem 1. Prove that for any given string in S, the number of a's in it is the same as the number of b's in it.
Transcribed Image Text:Consider the set S of strings defined recursively in Problem 1. Prove that for any given string in S, the number of a's in it is the same as the number of b's in it.
Prove each statement using either weak, strong, or structural induction. Make sure to clearly indicate the different parts of your proof: the basis step, the inductive hypothesis, what you will show in the inductive step, and the inductive step. Make sure to clearly format your proofs and to write in complete, clear sentences. EXAMPLE: Prove that for any nonnegative integer n, Σ i = (n+1) Answer: Proof. (by weak induction) Basis step: n = 1 Σ=1 1(1+1)==1 Therefore, (n+1) when n = 1. = Inductive hypothesis: Assume that Inductive step: We will show that i=1 i=1 i= = (+1) for some integer k > 1. i= (k+1)((k+1)+1) k+1 Σ=Σ+ (κ + 1) i=1 By inductive hypothesis, k+1 Σ IME i=1 k(k+1) = +k+1 2 k(k+1)+2(k+1) = 2 (k+2)(k+1) = 2 (k+1)((k+1)+1) 2 Therefore, by weak induction, we have shown that = (n+1) for all nonnegative integers n.
Transcribed Image Text:Prove each statement using either weak, strong, or structural induction. Make sure to clearly indicate the different parts of your proof: the basis step, the inductive hypothesis, what you will show in the inductive step, and the inductive step. Make sure to clearly format your proofs and to write in complete, clear sentences. EXAMPLE: Prove that for any nonnegative integer n, Σ i = (n+1) Answer: Proof. (by weak induction) Basis step: n = 1 Σ=1 1(1+1)==1 Therefore, (n+1) when n = 1. = Inductive hypothesis: Assume that Inductive step: We will show that i=1 i=1 i= = (+1) for some integer k > 1. i= (k+1)((k+1)+1) k+1 Σ=Σ+ (κ + 1) i=1 By inductive hypothesis, k+1 Σ IME i=1 k(k+1) = +k+1 2 k(k+1)+2(k+1) = 2 (k+2)(k+1) = 2 (k+1)((k+1)+1) 2 Therefore, by weak induction, we have shown that = (n+1) for all nonnegative integers n.
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