proof that $\sqrt{2}$ is irrational

Assume that the square root of $2$ (http://planetmath.org/SquareRootOf2) is rational. Then we can write

$$\sqrt{2}=\frac{a}{b},$$

where $a,b\in \mathbb{N}$ and $a$ and $b$ are relatively prime. Then $2={(\sqrt{2})}^2={\left({\displaystyle \frac{a}{b}}\right)}^2={\displaystyle \frac{a^2}{b^2}}$. Thus, $a^2=2b^2$. Therefore, $2\mid a^2$. Since $2$ is prime, it must divide $a$. Then $a=2c$ for some $c\in \mathbb{N}$. 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 $k,m\in \mathbb{N}$ such that $n=p^{k}m$, where $p\nmid m$ and $k$ is odd. Assume that $\sqrt{n}=a/b$, where $a,b\in \mathbb{N}$ and are relatively prime. Then $p^{k}m=n={(\sqrt{n})}^2={\left({\displaystyle \frac{a}{b}}\right)}^2={\displaystyle \frac{a^2}{b^2}}$. Thus, $a^2=p^{k}m{b}^2$. 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 $p^{k}m{b}^2$ 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.