Question

A proof by mathematical induction is supposed to show that a given property is true for every integer greater than or equal to an initial value. In order for it to be valid, the property must be true for the initial value, and the argument in the inductive step must be correct for every integer greater than or equal to the initial value. Consider the following statement. For every integer n ≥ 1, 3n − 2 is even. The following is a proposed proof by mathematical induction for the statement. Since the property is true for n = 1, the basis step is true. Suppose the property is true for an integer k, where k ≥ 1. That is, suppose that 3k − 2 is even. We must show that 3k + 1 − 2 is even. Observe that 3k + 1 − 2 = 3k · 3 − 2 = 3k(1 + 2) − 2 = (3k − 2) + 3k · 2. Now 3k − 2 is even by inductive hypothesis, and 3k · 2 is even by inspection. Hence the sum of the two quantities is even by (Theorem 4.1.1). It follows that 3k + 1 − 2 is even, which is what we needed to show. Identify the error(s) in the proof. (Select all that apply.) The inductive hypothesis is assumed to be true. (3k − 2) + 3k · 2 ≠ 3k(1 + 2) − 2 The property is not true for n = 1. 3k + 1 − 2 ≠ (3k − 2) + 3k · 2 3k − 2 is odd by the inductive hypothesis.

Answer 1

Answer:

The property is not true for n = 1.

Step-by-step explanation:

Given

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Required

Identify the error in the proof

From the attachment, the conclusion is that:

$3^{k} - 2$ is even, by inductive

However, there is no indication as to what the base case is.

To prove using mathematical induction implies that we prove the base case. (i.e. n = 1)

So, in order to prove $P(k + 1)$ is true, then $P(k)$ must be true.

Since there is no proof that $P(k)$ is true, then the proof is invalid.