# Sum of Even Integers is Even/Proof 2

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## Theorem

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 $21$)

## Proof

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, $\ds \sum_{k \mathop = 1}^n r_k$ is even.

$\blacksquare$

## Historical Note

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

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{IX}$. Propositions