Why are prime square roots irrational?

Answer 1

Let's assume that p is prime and the square root of p is rational. This means there are (positive) integers a, b such that sqrt(p) = a/b. Therefore: p = (a/b)*(a/b) = (a^2)/(b^2) This shows that a/b already has to be a (positive) integer, so that (a^2)/(b^2) is also one. (If a/b is not an integer, multiplying it by itself wouldn't create one, since no elements would come in that you could cancel the numerator and the denominator with.) So we have shown that (a^2)/(b^2) = (a/b)^2 = p. But this means that p isn't prime, because it has a/b (an integer) as a divisor, so we have a contradiction of the given fact that p is prime. This makes our assumption that sqrt(p) is rational false, and therefore proves that if p is prime, sqrt(p) is irrational.