Suppose not. [We take the negation of the theorem and suppose it to be true.] Suppose ∃ a rational number x and an irrational number y such that (x − y) is rational. [We must derive a contradiction.] By definition of rational, we have
x = a/b for some integers a and b with b ≠ 0.
and x − y = c/d for some integers c and d with d ≠ 0.
By substitution, we have
x − y = c/d
a/b − y = c/d
y = a/b − c/d
= (ad − bc)/bd
But (ad − bc) are integers [because a, b, c, d are all integers and products and differences of integers are integers], and bd ≠ 0 [by zero product property]. Therefore, by definition of rational, y is rational. This contradicts the supposition that y is rational. [Hence, the supposition is false and the theorem is true.]
And this completes the proof.
