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