Square Root of Prime is Irrational

Theorem

The square root of any prime number is irrational.

Proof

Let $p$ be prime.

Suppose that $\sqrt p$ is rational.

Then there exist natural numbers $m$ and $n$ such that:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sqrt p\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac m n\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle p\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {m^2} {n^2}\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle n^2 p\)	\(=\)	\(\displaystyle \)	\(\displaystyle m^2\)	\(\displaystyle \)	\(\displaystyle \)

Any prime in the prime factorizations of $n^2$ and $m^2$ must occur an even number of times because they are squares.

Thus, $p$ must occur in the prime factorization of $n^2 p$ an odd number of times.

Therefore, $p$ occurs as a factor of $m^2$ an odd number of times, a contradiction.

So $\sqrt p$ must be irrational.

$\blacksquare$

Also see

The special case of $p = 2$ is a well-known mathematical proof.