Proof that The Sum of Two Odd Integers is Even

Introduction

We have learned that an even integers can be expressed as $2k$, where $k$ is an integer. Now if we add one to an even integer, it becomes odd. Similarly, if we subtract one from an even integer, it also becomes odd. Therefore, odd integers can be expressed as $2k + 1$ or $2k - 1$, where $k$ is an integer. Now, we proceed with the theorem and the proof.

We proved that the sum of two even integers is even.  In this post, we show that the sum of two odd integers is even.

Theorem

The sum of two odd integers is even.

Proof

Let $m$ and $n$ be odd integers. Then, $m$ and $n$ can be expressed as $2r + 1$ and $2s + 1$ respectively, where $r$ and $s$ are integers. This only means that any odd number can be written as the sum of some even integer and one.

Now, substituting we have $m + n = (2r + 1) + 2s + 1 = 2r + 2s + 2$.

Since $r$ and $s$ are integers, $2r + 2s + 2$ is also an integer. It is clear that $2r + 2s + 2$ is an integer is divisible by 2. Therefore,  $2r + 2s + 2 = m + n$ is even. So, the sum of two odd integers is even.