Square Root of Prime is Irrational/Proof 1

Theorem

The square root of a prime number is irrational.

Proof

Let $p$ be prime.

Aiming for a contradiction, suppose that $\sqrt p$ is rational.

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

\(\ds \sqrt p\)

\(=\)

\(\ds \frac m n\)

\(\ds \leadsto \ \ \)

\(\ds p\)

\(=\)

\(\ds \frac {m^2} {n^2}\)

\(\ds \leadsto \ \ \)

\(\ds n^2 p\)

\(=\)

\(\ds m^2\)

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, which is a contradiction.

So $\sqrt p$ must be irrational.

$\blacksquare$