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.


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