# Square Root of 2 - Proofs of Irrationality - Geometric Proof

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 = mn implies BF = mn. By symmetry, DF = mn, and FDC is also a right isosceles triangle. It also follows that FC = n − (mn) = 2nm.

Hence we have an even smaller right isosceles triangle, with hypotenuse length 2nm and legs mn. 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.

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