# Parity of a sum of odd numbers [duplicate]

Is there a theorem wherein the sum of the elements of a set odd numbers with even cardinality be even?

• This fact is easily proven by induction. Yes, in general the sum of an even number of odd numbers will be even and the sum of an odd number of odd numbers will be odd. On the other hand, the sum of even numbers will be even regardless how many of them there are. Commented May 29, 2017 at 15:35

Not a theorem per se but a statement exceedingly easy to prove: $\sum_{i=1}^{2n} (2p_i + 1) = 2\sum_{i=1}^{2n} p_i + \sum_{i=1}^{2n} 1= 2\sum_{i=1}^{2n} p_i + 2n = 2(\sum_{i=1}^{2n} p_i + n)$.

Isn't really necessary to give such a basic result the status of "Theorem".

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Maybe more intutive and less notation is:

$even + even = even$ because $2n + 2m = 2(n+m)$

$odd + odd = even$ because $(2n + 1) + (2m + 1) = (2n + 2m) + 2 = even + even = even$.

So $odd$ added an even number of times is $(odd + odd) + (odd + odd) + ..... + (odd + odd) = even + even + .... + even = even$.

Choose two of the numbers; their sum is even because they're both odd.

If the set had two elements, you are done. Otherwise you can take two more elements. Their sum is even, so it's even also the sum with the previous two.

Continue until you finish.

This is actually the sketch of a proof by induction.

If the set has two elements, the sum is even.

Suppose this holds for sets with $2n$ elements and suppose you have a set with $2(n+1)$ elements, say $\{a_1,a_2,\dots,a_{2n},a_{2n+1},a_{2n+2}\}$.

Then the sum $S=a_1+a_2+\dots+a_{2n}$ is even by the induction hypothesis. Therefore also $$S+(a_{2n+1}+a_{2n+2})$$ is even.

Any odd number $q \equiv 1 \bmod 2$

Thus a collection of odd numbers $\{q_i\}$ with $Q$ members has sum $\sum_1^Q q_i \equiv \sum_1^Q 1 \equiv Q \bmod 2$

Thus the parity of the sum depends on the parity of the membership count.