Parity of a sum of odd numbers

Question

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

Answer 1

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 $(odd + odd) + (odd + odd) + ..... + (odd + odd) = even + even + .... + even = even$.

Answer 2

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.

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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.

Answer 3

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.