Prove that if $n$ is not the square of a natural number, then $\sqrt{n}$ is irrational. [duplicate]

Let $n$ be a positive integer such that there is no $m$ such that $n = m^2$. Suppose $\sqrt{n}$ is rational. Then there exists $p$ and $q$ with no common factor (beside 1) such that

$\sqrt{n} = \frac{p}{q}$

Then

$n = \frac{p^2}{q^2}$.

However, $n$ is an positive integer and $p$ and $q$ have no common factors beside $1$. So $q = 1$. This gives that

$n = p^2$

Contradiction since it was assumed that $n \neq m^2$ for any $m$.

Here’s an explanation that I find clearer, and that uses unique factorization explicitly: If a positive number $n$ is not the square of any integer, then when you write it as a product of primes, at least one prime shows up to an odd power. Let one such prime be $p$, and look at the supposed equation $\sqrt{n}=a/b$, with $a$ and $b$ positive integers. This gives $n=a^2/b^2$, hence $nb^2=a^2$. How many times does $p$ show up in the factorization of the left side and of the right? Oddly many times on the left, evenly many on the right. Contradiction to the unique factorization of $nb^2$.

This can also be done with the rational root test: consider the polynomial equation $$x^2 - n = 0$$

and suppose that it has a rational root. Then, this rational root must be an (integer) factor of $n$. So, if $\sqrt{n}$ is rational, then there exists $t\in \mathbb{N}$ (since $x^2 - n$ is an even function of $x$, we may assume, without loss of generality, that $t>0$) with $t \vert n$ such that $$t^2 - n = 0$$ which is to say $$n = t^2$$ and hence $n$ is the square of a natural number.

In fact, this argument generalizes to showing that if $\sqrt[m]{n}$ is rational, then $n$ is an $m^{th}$ power.