# 4. A geometric proof: sqrt(2) is irrational.

We’ll use the Wikipedia picture of isosceles triangle ABC with hypotenuse ${n}$ and legs ${m}$. Pythagoras tells us that for such a triangle, the ratio ${\dfrac{m}{n}=\sqrt{2}}$. For ${\sqrt{2}}$ to be rational, ${m}$ and ${n}$ must be integers, which we assume.

Inside ABC, construct a smaller triangle CDF with smaller integer hypotenuse ${2n-m}$ and legs ${m-n}$. Repeating in an infinite descent, we must end in a contradiction regarding the integer sides. (Say at the ith iteration, ${(m(i),n(i))=1}$ and the next iteration will be in contradiction to that.)
${\hfill \mbox{\raggedright \rule{.07in}{.1in}}}$

Aside: This ‘proof without words’ was published by Tom Apostol in the American Mathematical Monthly in November 2000. It was shown to me by my supervisor at work, himself a mathematician. One wonders how the ancient Greeks missed this proof. Perhaps they knew of it, but the evidence vanished in Julius Caesar’s Alexandria fire.

Aside: If needed, details of the geometric construction are on Wikipedia, Square root of 2.

Aside: Conway and Guy in their Book of Numbers offer a paper-folding argument along the lines of this proof.

Aside: This result can be derived by algebraic reasoning as well. Assuming ${m}$, ${n}$ integers, ${(m,n)=1}$, then ${\sqrt{2}=\dfrac{m}{n}=\dfrac{m(\sqrt{2}-1)}{n(\sqrt{2}-1)}=\dfrac{2n-m}{m-n}}$, a contradiction since ${1> \sqrt{2}-1> 0}$ implies ${n>m-n}$, but by hypothesis ${n}$ is the least value.