Square Root of Prime is Irrational/Proof 2

Proof
Let $p \in \Z$ be a prime number.

Consider the polynomial:
 * $\map P x = x^2 - p$

over the ring of polynomials $\Q \sqbrk X$ over the rational numbers.

From Difference of Two Squares:
 * $x^2 - p = \paren {x + \sqrt p} \paren {x - \sqrt p}$

Because $p$ is prime, $\sqrt p$ is not an integer.

From Polynomial which is Irreducible over Integers is Irreducible over Rationals it follows that $\paren {x + \sqrt p}$ and $\paren {x - \sqrt p}$ do not have rational coefficients.

That is:
 * $\sqrt p$ is not rational.

Hence the result.