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.

 

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