Observe that if $\sqrt 2$ is rational, then there is some positive integer q such that q × $\sqrt 2$ is an integer. Since the positive integers are well ordered, we may suppose that q is the smallest such number.
We next observe that since 1 < $\sqrt 2$ < 2, then $\sqrt 2$ – 1 < 1, and consequently q × ($\sqrt 2$ – 1) = (q ×$\sqrt 2$ – q ) is less than q. Let us call this new number r, and observe that it too is a positive integer. But we now have r × $\sqrt 2$ is also an integer, since r × $\sqrt 2$ = (q × $\sqrt 2$– q ) ×$\sqrt 2$ = (2q – q ×$\sqrt 2$ ). In short, r is a positive integer less than q and r ×$\sqrt 2$ is an integer. But we said that q was the smallest positive integer with this property, and so we have a contradiction.
The nice thing about this proof is how easily it generalizes. Let us denote by $|\sqrt n|$ the integer part of $\sqrt n$ . For example, since the square root of 5 is approximately 2.236, the integer part is 2. For any n that is not a perfect square, we may prove that is irrational exactly as above by considering q × ( $\sqrt n$– $|\sqrt n|$). (On the other hand, if n is a perfect square (so that $\sqrt n$ = $|\sqrt n|$) then there is no contradiction.)
More generally still, if x is a rational but not integral zero of a monic integer polynomial of degree d, let q be the least positive integer so that q$x^j$ is integral for all j < d. Then, considering q(x – n) where n is an integer with n < x < n + 1, we get a contradiction. In other words, we have proved that every rational “algebraic integer” is an integer.