# Sum of Odd Number of Odd Numbers is Odd

## Theorem

In the words of Euclid:

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

## Proof

Let $S = \set {r_1, r_2, \ldots, r_n}$ be a set of $n$ odd numbers, where $n = 2 m + 1$.

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:

 $\ds \sum_{k \mathop = 1}^n r_k$ $=$ $\ds \sum_{k \mathop = 1}^{2 m + 1} \paren {2 s_k + 1}$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^{2 m} \paren {2 s_k + 1} + 2 s_n + 1$ extracting the $n$th element from the summation $\ds$ $=$ $\ds 2 q + 2 s_n + 1$ where $q \in \N$: Sum of Even Number of Odd Numbers is Even $\ds$ $=$ $\ds 2 \paren {q + s_n} + 1$

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

$\blacksquare$

## Historical Note

This proof is Proposition $23$ of Book $\text{IX}$ of Euclid's The Elements.