**Geometric Proof**

Another reductio ad absurdum showing that is irrational is less well-known. 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 previous proof viewed geometrically.

Let *ABC* be a right isosceles triangle with hypotenuse length *m* and legs *n*. 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 the ∠*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*) = 2*n* − *m*.

Hence we have an even smaller right isosceles triangle, with hypotenuse length 2*n* − *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.

Read more about this topic: Square Root Of 2, Proofs of Irrationality

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