Sum of Even Number of Odd Numbers is Even

Theorem

In the words of Euclid:

If as many odd numbers as we please be added together, and their multitude is even, the whole is even.

Let $S = \set {r_1, r_2, \ldots, r_n}$ be a set of $n$ odd numbers, where $n = 2 m$.

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

$S = \set {2 s_1 + 1, 2 s_2 + 1, \ldots, 2 s_n + 1}$

where:

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

Then:

$\ds \sum_{k \mathop = 1}^n r_k$

$=$

$\ds \sum_{k \mathop = 1}^{2 m} 2 s_k + 1$

$\ds $

$=$

$\ds \sum_{k \mathop = 1}^{2 m} 2 s_k + \sum_{k \mathop = 1}^{2 m} 1$

$\ds $

$=$

$\ds 2 \times 2 m \sum_{k \mathop = 1}^{2 m} s_k + 2 m$

$\ds $

$=$

$\ds 2 \paren {2 m \sum_{k \mathop = 1}^{2 m} s_k + m}$

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

$\blacksquare$