If as many odd numbers as we please are added together, and their multitude is even, then the sum is even.

Let as many odd numbers as we please, *AB, BC, CD,* and *DE,* even in multitude, be added together.

I say that the sum *AE* is even.

Since each of the numbers *AB, BC, CD,* and *DE* is odd, if a unit is subtracted from each, then each of the remainders is even, so that the sum of them is even. But the multitude of the units is also even. Therefore the sum *AE* is also even.

Therefore, *if as many odd numbers as we please are added together, and their multitude is even, then the sum is even.*

Q.E.D.