Proof that $\frac{1 + \sqrt{5}}{2}$ is irrational.

Another completely different approach: One can easily see (by squaring the golden ratio) that it satisfies the quadratic equation $x^2-x-1=0$. Using the Rational Root Theorem we concude that it must be irrational.

Another approach:

Continued fractions are finite for rationals. Now, try to find the continued fraction representation for $\frac{1+\sqrt{5}}{2}\big($you may use the fact that it satisfies the equation $x^2-x-1=0\big)$, it will be $[1;1 ,1,1,\cdots] $, which has infinitely many $1$. Conidering the contrapositive statement of the statement above, if a number has infinitely many terms in it's cont. fraction representation then it is not rational. Hence, $\frac{1+\sqrt{5}}{2} $ is not rational or irrational.

Another one:

Let, $\frac{1+\sqrt{5}}{2}=\frac{p}{q},p,q\in\mathbb{Z}$, then $\sqrt{5}=\frac{2p-q}{q}$. Now, we can prove that $\sqrt{5}$ is irrational using this argument:

$\sqrt{5}$ satisfies the monic polynomial $x^2-5=0$, hence, if it is an rational algebraic number it need to be an integer. But, $2<\sqrt{5}<3$ (as, $4<5<9$). Hence, $\sqrt{5}$ is irrational.

Now come back to the equation $\sqrt{5}=\frac{2p-q}{q}$. As, $p,q\in\mathbb{Z}$, $\frac{2p-q}{q}\in\mathbb{Q}$, as, both $2p-q,q\in\mathbb{Z}$. But, we have shown that $\sqrt{5}$ is irrational. Hence, contradiction!