Joe Near

If sqrt(2) is rational, then there is some positive integer q such
that q*sqrt(2) is an integer. Assume that q is the smallest number for
which this holds. Since 1 < sqrt(2) < 2, we know that sqrt(2)-1 < 1.
Next, we multiply both sides of that equation by q, resulting in the
new equation q*(sqrt(2) - 1) < q, which becomes q*sqrt(2) - q < q.
Since q*sqrt(2) is an integer, q*sqrt(2) - q is also an
integer. Let us call this new integer p.

We can say that p*sqrt(2) is also an integer, because p*sqrt(2) =
(q*sqrt(2) - q)*sqrt(2) = 2q - q*sqrt(2). Since q*sqrt(2) is an
integer, p*sqrt(2) must be also. But we have seen that p < q, even
though we began with the assumption that q was the smallest integer
such that q*sqrt(2) is an integer. Thus, we have a contradiction, and
there can be no rational q such that q*sqrt(2) is an integer.