Square Root of Prime is Irrational/Proof 2

Theorem

The square root of a prime number is irrational.

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.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Example $60$