Proof that the square root of any nonperfect square positive integer is irrational
PROOFS
PROOF 1:Let us assume that is rational where
is a nonperfect square positive integer. Then it can be written as
. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore
is irrational.
Proof 2:Suppose to the contrary that for integers a,b, and that this representation is fully reduced, so that
. 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
, since
.
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 a length equal to a-b. Finally, we may write the hypotenuse of the smaller triangle as
, which is also an integer. Should the lengths of the sides of the smaller triangle are integers, but by triangle similarity, the hypotenuse to side-length ratios are equal:
, and obviously from the picture the latter numerator and denominator are smaller numbers. Hence,
was not in lowest terms, a contradiction. This implies that
cannot be rational.