Sum of Even Integers is Even/Proof 2

From ProofWiki
Jump to navigation Jump to search


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

In the words of Euclid:

If as many even numbers as we please be added together, the whole is even.

(The Elements: Book $\text{IX}$: Proposition $21$)


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:

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


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


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

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


Historical Note

This theorem is Proposition $21$ of Book $\text{IX}$ of Euclid's The Elements.