Irrationality of $\sqrt{2}$

A cool geometric proof first presented by Tom Apostol. If $\sqrt{2}$ is rational we can draw an isosceles right triangle with integer sides. Consider the smallest such triangle ABC -- that is with shortest hypotenuse. Let the apex be A and hypotenuse AC. Draw a circle centered at A with radius the length of AB and mark point D where it intersects the hypotenuse. Now draw the tangent to the circle at D and mark the point E where it cuts the base of the triangle. Then DEC is a smaller isosceles right triangle with integer sides. This is a contradiction.

Algebraically this leads to the proof that nonperfect squares have irrational roots. Suppose $p/q = \sqrt{n}$. Then $$ \frac{p'}{q'} = \frac{nq - \lfloor \sqrt{n} \rfloor p}{p - \lfloor \sqrt{n} \rfloor q} = \frac{p}{q}$$and $0 < p' < p$ and $0 < q' < q$ (You can follow the lines in the diagram). Exercise: what is the total area of the wedges? $(\sqrt{2}+1)\pi/16 = .47403$

Borwein and Bailey. Mathematics by Experiment. p73



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