TITLE: First known proof of $\sqrt 2$ is irrational with prime factorization?
QUESTION [6 upvotes]: Do any of you happen to know the history of the standard prime factorization proof of $\sqrt 2$ is irrational?  I know this theorem was known to Aristotle, and that the Fundamental Theorem of Arithmetic, on which the proof rests, is found already in Euclid, but I've not been able to track down the origin of this particular proof.  
These sites I know about:  http://www.cut-the-knot.org/proofs/sq_root.shtml, http://www.math.ufl.edu/~rcrew/texts/pythagoras.html, and of course http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
But any other references, online or in paper form, would be greatly appreciated!

REPLY [3 votes]: Here is a proof based on the well-ordering of the positive integers rather than the FTA:
To begin with, we 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.