For all integers n 1, 3n 2 is even. Proof (by mathematical induction): Suppose the

Chapter 5, Problem 31

(choose chapter or problem)

In order for a proof by mathematical induction to be valid, the basis statement must be true for n = a and the argument of the inductive step must be correct for every integer k ≥ a. In 30 and 31 find the mistakes in the “proofs” by mathematical induction.

“Theorem:” For all integers n ≥ 1, 3n − 2 is even. “Proof (by mathematical induction): Suppose the theorem is true for an integer k, where k ≥ 1. That is, suppose that \(3^{k} − 2\) is even. We must show that \(3^{k+1} − 2\) is even. But

\(3^{k+1} − 2 = 3^{k} \cdot 3 − 2 = 3^{k} (1 + 2) − 2\)

\(= (3^{k} − 2) + 3^{k} \cdot 2\).

Now \(3^{k} − 2\) is even by inductive hypothesis and \(3^{k} \cdot 2\) is even by inspection. Hence the sum of the two quantities is even (by Theorem 4.1.1). It follows that \(3^{k+1} − 2\) is even, which is what we needed to show.”

Text Transcription:

3^k − 2

3^k+1 − 2

3^k+1 − 2 = 3^k cdot 3 − 2 = 3^k (1 + 2) − 2

= (3^k − 2) + 3^k cdot 2

3^k − 2

3^k cdot 2