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) = 2n − m.
Hence we have 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.
Other articles related to "geometric proof, proof, geometric":
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... To complete the proof, Archimedes shows that The formula above is a geometric series—each successive term is one fourth of the previous term ... mathematics, that formula is a special case of the sum formula for a geometric series ... Archimedes evaluates the sum using an entirely geometric method, illustrated in the picture to the right ...
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