Tom Apostol's geometric proof of the irrationality of $\sqrt{2}$

Tom M. Apostol made another geometric reductio ad absurdum argument showing that $\sqrt{2}$ is irrational. It is also an example of proof by infinite descent. 

It makes use of classic compass and straightedge construction, proving the theorem by a method similar to that employed by ancient Greek geometers. It is essentially the same algebraic proof as in the previous paragraph, viewed geometrically in another way.

Let $△ ABC$ be a right isosceles triangle with hypotenuse length $m$ and legs $n$ as shown in Figure 2. 

By the Pythagorean theorem, . Suppose $m$ and $n$ are integers. 

Let $m:n$ be a ratio given in its lowest terms.

Draw the arcs $BD$ and $CE$ with centre $A$. Join DE. It follows that $AB = AD$, $AC = AE$ and $∠BAC$ and $∠DAE$ coincide. Therefore, the triangles $ABC$ and $ADE$ are congruent by SAS.

Because $∠EBF$ is a right angle and $∠BEF$ is half a right angle, $△ BEF$ is also a right isosceles triangle. Hence $BE = m − n$ implies $BF = m − n$. By symmetry, $DF = m − n$, and $△ FDC$ is also a right isosceles triangle. 

It also follows that 

$FC = n − (m − n) = 2n − m$.

Hence, there is an even smaller right isosceles triangle, with hypotenuse length $2n − m$ and legs $m − n$. These values are integers even smaller than $m$ and $n$ and in the same ratio, contradicting the hypothesis that $m:n$ is in lowest terms. 

Therefore, $m$ and $n$ cannot be both integers; hence, is irrational.

From Wikipedia