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.
A critical step in the proof is the claim that if 1 is subracted from an odd number, then the remainder is even. This was mentioned in VII.Def.7, but never proved. See the Guide for that definition for details. Unless that gap is filled, this proposition, along with many that depend upon it, are unjustified.
Use of this proposition
This proposition is used in the next one.