Sum of Even Integers is Even

Theorem

The sum of any two even integers is itself even.

General Result

The sum of any finite number of even integers is itself even.

Proof

Consider two even integers $x$ and $y$.

Since they are even, they can be written as $x = 2a$ and $y = 2b$ respectively for integers $a$ and $b$.

Therefore, the sum $x + y = 2a + 2b = 2(a + b)$.

From this sum, it can be clearly seen that $x + y$ has $2$ as a factor and therefore is even.

Hence the result: that the sum of any two even integers is even.

$\blacksquare$

Proof of General Result

Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:

The sum of $n$ even integers is an even integer.

$P(1)$ is trivially true, as this just says:

The sum of $1$ even integers is an even integer.

The sum of $0$ even integers is understood, from the definition of a vacuous summation, to be $0$, which is even.

So $P(0)$ is also true.

Basis for the Induction

$P(2)$ is the case:

The sum of any two even integers is itself even

which has been proved above.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

The sum of any $k$ even integers is itself even.

Then we need to show:

The sum of any $k+1$ even integers is itself even.

Induction Step

This is our induction step:

Consider the sum of any $k+1$ even integers.

This is the sum of $k$ even integers (which is even by the induction hypothesis) and another even integer.

That is, it is the sum of two even integers, and therefore, by the basis for the induction, also even.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

That is:

The sum of any finite number of even integers is itself even.

$\blacksquare$