Question

Prove that if \(n\)is an integer and \(3n + 2\)is even, then \(n\) is even using

a) a proof by contraposition.

b) a proof by contradiction.

Short answer

If\(n\)is an integer and\(3n + 2\) is even then\(n\) has to be even is true using both methods of contraposition and contradiction.

Step-by-step solution

Introduction

Consider\(n\)is an integer and\(3n + 2\) is even.

(a) Using contraposition method to prove that\(3n + 2\)has to be even when\(n\) is an integer.

And then is \(n\)even.

Contraposition of the statement:

\(p \to q\)is \( - q \to - p\).

Here\(p\):\(n\) is an integer and \(3n + 2\)has to be even

\( - p\): \(3n + 2\)is not even

\(q\): \(n\)is even

\( - q\): \(n\)is odd

Prove using contraposition

Assume that \(n\)is odd.

Then by definition of odd numbers, \(n = 2k + 1\), for any integer \(k\).

Now, substitute\(n\) value in \(3n + 2\).

\(3n + 2 = 3\left( {2k + 1} \right) + 2\)

\(\begin{aligned}{l} = 6k + 3 + 2\\ = 6k + 5\\ = 6k + 4 + 1\\ = 2\left( {3k + 2} \right) + 1\end{aligned}\)

,which is odd.

i.e., \(3n + 2\)is odd.

Therefore, if\(n\) is an integer and\(n\)is odd then \(3n + 2\)has to be odd.

Thus, by the contraposition if \(n\) is an integer and \(3n + 2\)is even then \(n\)is even.

Prove using contradiction

(b) Using contradiction method to prove that\(3n + 2\)has to be even when\(n\)is an integer. And then \(n\)is even.

Assume that \(3n + 2\)is even and \(n\)is odd.

Since \(n\)is odd and then multiplication of two odd numbers is odd, it follows that \(3n\)is odd and then that \(3n + 2\)is odd, using properties of odd numbers.

Therefore, the assumption that\(3n + 2\) is even and is\(n\) odd is wrong.

This is a contradiction.

Thus, if\(n\) is an integer and\(3n + 2\) is even then\(n\) has to be even.