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 = a^{2} / b^{2}, so that a^{2} = p*b^{2}. We now see that p divides a^{2}.

But when a prime number divides the product of two numbers, it must divide at least one of them. Fortunately, a^{2} = a*a, so we can simply submit that p divides a. Which means, by definition, that we can write a = p*a_{1}. So, we have a^2 = (p*a_{1})a^{2} = p*ba^{2}. Dividing through by p, p*a_{1}a^{2} = ba^{2}.

Clearly, we have the same situation we just dealt with... p divides ba^{2}, so p divides b, so b = p*b_{1}, so we have p*a_{1}a^{2} = (p*b_{1})a^{2}. And dividing through, we get a_{1}a^{2} = p*b_{1}a^{2}.

But this is a solution of the form we had earlier, except that a_{1} < a, and b_{1} < 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 a_{k}a^{2} < 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.