Dan, your proof looks to be Apolstol’s from 2000, replacing a right isosceles triangle with side lengths n,n,m by the similar triangle with side lengths m-n,m-n,2n-m. The same construction, without the geometry, assumes sqrt(2)=m/n, so m=n*sqrt(2). Since 1>sqrt(2)-1>0, n>n*(sqrt(2)-1)>0 so n>m-n>0. Then sqrt(2)=m/n = [m(sqrt(2)-1)]/[n(sqrt(2)-1)] = [m*sqrt(2) - m]/[m-n] = [2n-m][m-n], a fraction in lower terms, use infinite descent. Cut-the-Knot has 29 proofs that sqrt(2) is irrational. Pick your favorite! Steve On May 29, 2018, at 3:56 PM, Dan Asimov <dasimov@earthlink.net<mailto:dasimov@earthlink.net>> wrote: My favorite irrationality proof is this which is almost instant if you allow pictures: Suppose there exist integers p, q with (p/q)^2 = 2. Then p^2 = 2 q^2 implies there is a (p, p, q) right triangle by Pythagoras. Draw this (p, p, q) triangle with the hypotenuse as base: Call it ABC with A = right angle. Draw the circle with center at B and radius = p. This cuts the hypotenuse at the point D into lengths p and q-p. Extend a perpendicular from the hypotenuse at D up to point E on side AC. This cuts side AC into pieces of sides q - p and p - (q - p) = 2p - q. Then ADE is a 45-45-90 right triangle with integer sides that are smaller than the original. Which implies an infinite regress of smaller positive integers, contradiction. —Dan Mike Stay wrote: ----- My favorite irrationality proof is one I heard from John Baez. Suppose the cube root of two were not irrational; then there would be two positive integers p, q such that p/q = ∛2. Multiplying both sides by q and cubing, we get p³ = 2q³ = q³ + q³, which has no solutions in the positive integers by Fermat's Last Theorem! ----- _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com<mailto:math-fun@mailman.xmission.com> https://urldefense.proofpoint.com/v2/url?u=https-3A__mailman.xmission.com_cg...