I've heard that the ancient Greeks had proofs of irrationality for sqrt(2), sqrt(3), sqrt(5), sqrt(6), ... up to some modest bound, and I think they used this type of reduction. So it's possible that this proof is several millenia old. (Does Apostol's article have any info?) For what it's worth, my favorite proof that sqrt(2) is irrational is: Consider the set S = { a + b sqrt(2) | a, b are integers } , and note that S is closed under multiplication. If sqrt(2) were rational, say p/q , then S would be contained in (1/q)Z . However, for large n , we have 0 < (2 - sqrt(2))^n < 1/q , a contradiction, since (2 - sqrt(2))^n is in S . Michael Reid

Does anyone know who first thought of the following proof that sqrt(2) is irrational? It's definitely my favorite.

Suppose that the isosceles right triangle whose legs are 1 has hypotenuse = p/q with p,q in Z+.

By magnifying it by a factor of q, we get the right triangle with sides q, q, p.

Call this triangle ABC, with A at the right angle.

Now draw arc of the circle O with center at B and radius q, from side AB to the hypotenuse. Assume it hits the hypotenuse at point X. Draw a perpendicular to the hypotenuse at X, extended to hit side AC, say at point Y.

Then the segments CX and XY are of equal length p-q. And since the lines AY and XY are tangents to the circle O from an external point (namely Y), they are equal length, namely p-q, an integer.

This means that the segment CY must be of length q-(p-q) = 2q-p, also an integer.

Hence the triangle CXY is another, smaller, isosceles right triangle with integer sides.

Repeating this process leads to an infinite regress of integer right triangles each with smaller sides than the last one: contradiction.

--Dan