In order for a proof by mathematical induction to be valid, the basis statement must be true for the initial value of n and the argument of 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.)

3k + 1 − 2 ≠ (3k − 2) + 3k · 2

(3k − 2) + 3k · 2 ≠ 3k(1 + 2) − 2

3k − 2 is odd by the inductive hypothesis.

The inductive hypothesis is assumed to be true.

The basis step is false. It says that 31 − 2 is even.