Sum of Even Integers is Even

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Theorem

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


Proof 1

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.

$\Box$


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.

$\blacksquare$


Proof 2

Let $S = \set {r_1, r_2, \ldots, r_n}$ be a set of $n$ even numbers.

By definition of even number, this can be expressed as:

$S = \set {2 s_1, 2 s_2, \ldots, 2 s_n}$

where:

$\forall k \in \closedint 1 n: r_k = 2 s_k$

Then:

\(\ds \sum_{k \mathop = 1}^n r_k\) \(=\) \(\ds \sum_{k \mathop = 1}^n 2 s_k\)
\(\ds \) \(=\) \(\ds 2 \sum_{k \mathop = 1}^n s_k\)

Thus, by definition, $\displaystyle \sum_{k \mathop = 1}^n r_k$ is even.

$\blacksquare$