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Find the mistake in the following proof by induction. Claim: Every nonnegative integer is even. Proof: Let P(n) be the statement that `n is even". We will use strong induction to prove that P(n) is true for all nonnegative integers n. The base case is n = 0, which is clearly an even number. Thus P(0) is true.

Find the mistake in the following proof by induction. Claim: Every nonnegative integer is even. Proof: Let P(n) be the statement that `n is even". We will use strong induction to prove that P(n) is true for all nonnegative integers n. The base case is n = 0, which is clearly an even number. Thus P(0) is true.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Find the mistake

Find the mistake in the following proof by induction. Claim: Every nonnegative integer is even. Proof: Let P(n) be the statement that ``n is even'. We will use strong induction to prove that P(n) is true for all nonnegative integers n. The base case is n 0, which is clearly an even number. Thus P(0) is true. Assume that P(m) has been proven for all integers 0 ≤ m ≤ k where k is a nonnegative integer. We must prove that k + 1 is even. By P(1) and P(k), we know that both 1 and k are even, hence, so is their sum k + 1. Thus P(k + 1) is true. Therefore, we showed that ʼn is even for all nonnegative integers ʼn, as required.
Transcribed Image Text:Find the mistake in the following proof by induction. Claim: Every nonnegative integer is even. Proof: Let P(n) be the statement that ``n is even'. We will use strong induction to prove that P(n) is true for all nonnegative integers n. The base case is n 0, which is clearly an even number. Thus P(0) is true. Assume that P(m) has been proven for all integers 0 ≤ m ≤ k where k is a nonnegative integer. We must prove that k + 1 is even. By P(1) and P(k), we know that both 1 and k are even, hence, so is their sum k + 1. Thus P(k + 1) is true. Therefore, we showed that ʼn is even for all nonnegative integers ʼn, as required.
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