The square root of any prime number is irrational (idea) by srkorn

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 a_ka² < 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.

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