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

Let as many odd numbers as we please, *AB, BC,* and *CD,* the multitude of which is odd, be added together.

I say that the sum *AD* is also odd.

Subtract the unit *DE* from *CD,* therefore the remainder *CE* is even.

But *CA* is also even, therefore the sum *AE* is also even.

And *DE* is a unit. Therefore *AD* is odd.

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

Q.E.D.

Like the previous proposition, the proof of this one relies on an unproved statement in VII.Def.7.