Proof by contradiction

Question

For all x, y, z ∈ Z. At least one of x−y , x−z and y−z is even.

Answer 1

Suppose that all of them are odd. Then we add two odd numbers (x-y)+(y-z)=x-z and we get an even number, since the sum of two odd numbers is always even. But by assumption x-z is odd and now we showed that x-z is even. This is a contradiction. Hence, at least one of the x-y, x-z and y-z is even.