# The Sum of Two Even Numbers

In Middle School and High School Mathematics, the discussions about mathematical proofs often focus on Geometry. In this post, I will discuss a very basic theorem in Algebra and its proof. The theorem states that the sum of two even numbers is even.  Note that in this particular post, I mean integers when I say numbers.

As I have discussed before, I will assume that some readers of this blog are not mathematically inclined, so the proofs in the early part of this blog are going to be quite detailed.

The conjecture that the sum of two even numbers is even came from observations. For example, $latex 2$ and $latex 8$ are even, and $latex 2 + 8 = 10$ is even.  Other examples such as $latex -16 + 4 = -12$ and $latex -22 + -4 = -26$ also support the observation. Listing more example would convince us that this the conjecture is really true. However, no matter how many pairs of numbers we add, it is not yet enough to say that the sum of two numbers is even because we have not listed all pairs. Of course, it is impossible to list all pairs. We need a proof.

Theorem

The sum of two even numbers is even.

Proof

Let $latex m$ and $latex n$ be even numbers.

Then $latex m$ and $latex n$ are both divisible by $latex 2$. That is, if we divide $latex m$ and $latex n$ by $latex 2$, we can find quotients $latex p$ and $latex q$. Writing the two equations, we have

$latex \frac{m}{2} = p$ and $latex \frac{n}{2} = q$.

Clearly $latex q$ and $latex r$ are also integers (Why?). Multiplying both sides of the two equations by $latex 2$, we have

$latex m = 2p$ and $latex n = 2q$.

If we add $latex m$ and $latex n$, we have

$latex m + n = 2p + 2q = 2(p+q)$.

Now, $latex 2(p + q)$ is even since it is divisible by $latex 2$. But $latex 2(p+q) = m + n$ which is the sum of two even numbers. Therefore, the sum of two even numbers is even.

You would have realized by reading the proof above that all even numbers can be represented by $latex 2k$ for any integral $latex k$. This is because in any even number, we can always factor out $latex 2$ and determine the integer $latex k$ as the other factor. As a consequence, if $latex 2k$ is even, the $latex 2k + 1$ or $latex 2k – 1$.