The Square Root of 2 is Irrational (Geometric Proof)

Problem: Show that \sqrt{2} is an irrational number (can’t be expressed as a fraction of integers).

Solution: Suppose to the contrary that \sqrt{2} = a/b for integers a,b, and that this representation is fully reduced, so that \textup{gcd}(a,b) = 1. Consider the isosceles right triangle with side length b and hypotenuse length a, as in the picture on the left. Indeed, by the Pythagorean theorem, the length of the hypotenuse is \sqrt{b^2 + b^2} = b \sqrt{2} = a, since \sqrt{2} = a/b.


Swinging a b-leg to the hypotenuse, as shown, we see that the hypotenuse can be split into parts b, a-b, and hence a-b is an integer. Call the point where the b and a-b parts meet P. If we extend a perpendicular line from P to the other leg, as shown, we get a second, smaller isosceles right triangle. Since the segments PQ and QR are symmetrically aligned (they are tangents to the same circle from the same point), they too have length equal to a-b. Finally, we may write the hypotenuse of the smaller triangle as b-(a-b) = 2b-a, which is also an integer.

So the lengths of the sides of the smaller triangle are integers, but by triangle similarity, the hypotenuse to side-length ratios are equal: \sqrt{2} = a/b = (2b-a)/(a-b), and obviously from the picture the latter numerator and denominator are smaller numbers. Hence, a/b was not in lowest terms, a contradiction. This implies that \sqrt{2} cannot be rational. \square

This proof is a prime example of the cooperation of two different fields of mathematics. We just translated a purely number-theoretical problem into a problem about triangle similarity, and used our result there to solve our original problem. This technique is widely used all over higher-level mathematics, even between things as seemingly unrelated as topological curves and groups. Finally, we leave it as an exercise to the reader to extend this proof to a proof that whenever k is not a perfect square, then \sqrt{k} is irrational. The proof is quite similar, but strays from nice isosceles right triangles

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