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.)

3k − 2 is odd by the inductive hypothesis.The property is not true for n = 1.(3k − 2) + 3k · 2 ≠ 3k(1 + 2) − 2The inductive hypothesis is assumed to be true.3k + 1 − 2 ≠ (3k − 2) + 3k · 2