Proof by Descent:

Suppose that a prime number does have a rational square root: that is, p being a prime, we can write sqrt(p) = a / b, for natural numbers a and b. Squaring both sides yields p = a2 / b2, so that a2 = p*b2. We now see that p divides a2.

But when a prime number divides the product of two numbers, it must divide at least one of them. Fortunately, a2 = a*a, so we can simply submit that p divides a. Which means, by definition, that we can write a = p*a1. So, we have a^2 = (p*a1)a2 = p*ba2. Dividing through by p, p*a1a2 = ba2.

Clearly, we have the same situation we just dealt with... p divides ba2, so p divides b, so b = p*b1, so we have p*a1a2 = (p*b1)a2. And dividing through, we get a1a2 = p*b1a2.

But this is a solution of the form we had earlier, except that a1 < a, and b1 < b. The solution's existence implies an infinitude of smaller and smaller solutions, and since the natural numbers are bounded at 1, we will eventually imply a nonsense solution with aka2 < 1, a contradiction. So the premise that there are numbers a and b such that sqrt(p) = a / b is incorrect, and the square root of a prime is irrational.