Sum of Even Integers is Even/Proof 1

Proof
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.

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 is:
 * $x + y = 2 a + 2 b = 2 \left({a + b}\right)$

Thus $x + y$ has $2$ as a divisor and is even by definition.

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.

By the basis for the induction, this is 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.