Though Nitin showed an excellent solution (much better than mine), I've already typed the routine solution approach. So I'm posting it anyway:
If it is rational, we can write it in the form of the fraction $m/n$ where $m$ and $n$ are integers and have no common factors – always, otherwise it is not rational.
So we have
$$\frac{\sqrt5+1}{2}=\frac{m}{n}$$
$$\sqrt5=\frac{2m}{n}-1$$
But $\frac{2m}{n}-1$ is rational because $\frac{m}{n}$ is rational.
Therefore $\sqrt5$ is rational. But it is not. We have a contradition, so the whole number must be irrational. q.e.d.
If you want to prove that $\sqrt5$ is irrational. Do the same: if it is rational, we can write it in the form of the fraction $m/n$ where $m$ and $n$ are integers and have no common factors (We can factor out any common factors). So we have
$$\sqrt5=\frac{m}{n}$$
$$5=\frac{m^2}{n^2}$$
$$5n^2=m^2$$
Therefore $m$ is divisible by $5$ and we can re-rewrite it as $m=5k$. So we have
$$5n^2=25k^2$$
$$n^2=5k^2$$
Therefore $n$ is also divisible by 5 and we can re-rewrite it as $n=5p$. Therefore, the fraction $\frac{m}{n}$ can be simplified because the numerator and denominator have a common factor of $5$, which contradicts our assumption that there will be no common factors. This cannot be, so $\sqrt5$ cannot be rational. Therefore it’s irrational. q.e.d.