Question

Prove that if m and n are integers and mn is even, then m is even or n is even.

Short answer

If\(mn\)is given to be even, then either\(m\)has to be even or\(n\)has to be even.

Step-by-step solution

Introduction

Consider\(m\)and \(n\) are integers and\(mn\)is even.

The purpose is to show that either\(m\)is even or\(n\)is even.

Prove using contraposition

This can be proved by using contraposition.

Assume it is not true that either\(m\)is even or\(n\)is even.

Then\(m\)is odd and\(n\)is odd.

So, by definition of odd numbers\(m = 2x + 1\)and\(n = 2y + 1\), for any integers\(x\)and\(y\).

Now,

\(mn = \left( {2x + 1} \right)\left( {2y + 1} \right)\)

\(\begin{aligned}&= 4xy + 2x + 2y + 1\\ &= 2\left( {2xy + x + y} \right) + 1\\ &= 2z + 1\end{aligned}\)

For some integer\(z = 2xy + x + y\)

Thus,\(mn\)is odd, which is negative of statement\(mn\)is even.

Therefore, if\(mn\)is given to be even, then either\(m\) has to be even or\(n\) has to be even.