if $x$ is rational and $x^2$ is natural, prove that $x$ is integer If $x$ is rational: There exists $\frac{a}{b}$ such that $a$, $b$ are integers.
If $x^2$ is natural:
$\left(\frac{a}{b}\right)^2$ is natural => $\frac{a^2}{b^2}$ is natural
Then $a^2$ divides $b^2$ => $a$ divides $b$
If $a$ divides $b$ and $a$, $b$ are integers, $\frac{a}{b}$ is integer
so $x$ is integer.

Let $x^2 = n \in \mathbb{N}\,$, then $x$ is a root of $x^2-n=0$. Since $x^2-n$ is a monic polynomial, any rational root must be an integer by the rational root theorem, so $x$ is an integer.
