Assume that the square root of $2$ is rational. Then we can write

\begin{equation*} \sqrt{2} = \frac{a}{b}, \end{equation*}where $a,b\in\N$ and $a$ and $b$ are relatively prime. Then $\displaystyle 2=(\sqrt{2})^2=\left(\frac{a}{b}\right)^2=\frac{a^2}{b^2}$ . Thus, $a^2=2b^2$ . Therefore, $2\mid a^2$ . Since $2$ is prime, it must divide $a$ . Then $a=2c$ for some $c\in\N$ . Thus, $2b^2=a^2=(2c)^2=4c^2$ , yielding that $b^2=2c^2$ . Therefore, $2\mid b^2$ . Since $2$ is prime, it must divide $b$ .

Since $2\mid a$ and $2\mid b$ , we have that $a$ and $b$ are not relatively prime, which contradicts the hypothesis. Hence, the initial assumption is false. It follows $\sqrt{2}$ is irrational.

With a little bit of work, this argument can be generalized to any positive integer that is not a square. Let $n$ be such an integer. Then there must exist a prime $p$ and $k,m\in\N$ such that $n=p^km$ , where $p\nmid m$ and $k$ is odd. Assume that $\sqrt{n}=a/b$ , where $a,b\in\N$ and are relatively prime. Then $\displaystyle p^km=n=(\sqrt{n})^2=\left(\frac{a}{b}\right)^2=\frac{a^2}{b^2}$ . Thus, $a^2=p^kmb^2$ . From the fundamental theorem of arithmetic, it is clear that the maximum powers of $p$ that divides $a^2$ and $b^2$ are even. Since $k$ is odd and $p$ does not divide $m$ , the maximum power of $p$ that divides $p^kmb^2$ is also odd. Thus, the same should be true for $a^2$ . Hence, we have reached a contradiction and $\sqrt{n}$ must be irrational.

The same argument can be generalized even more, for example to the case of nonsquare irreducible fractions and to higher order roots.