Let a and b represent even integers. Then by definition of an even integer, we can say that: a = 2n and b = 2m
Proof by contradiction:
a + b is not an even integer.
Using substitution, 2n + 2m is not an even integer.
Using distibutive property, 2(n + m) is not an even integer.
Consider that (n + m) = k, then we have 2k is not an even integer.
This is a False statement since 2 times any integer is an even integer.
Therefore, a + b is an even integer, so the sum of two even integers is an even integer. QED
