Proving $\sqrt 3$ is irrational. There is a very simple proof by means of divisibility that $\sqrt 2$ is irrational. I have to prove that $\sqrt 3$ is irrational too, as a homework. I have done it as follows, ad absurdum:
Suppose
$$\sqrt 3= \frac p q$$
with $p/q$ irreducible, then
$$\begin{align}
& 3q^2=p^2 \\
& 2q^2=p^2-q^2 \\
&2q^2=(p+q)(p-q) \\
\end{align}$$
Now I exploit the fact that $p$ and $q$ can't be both even, so it is the case that they are either both odd, or have different parity. Suppose then that $p=2n+1$ and $q=2m+1$
Then it is the case that
$$\begin{align}
&p-q=2(n-m) \\
&p+q=2(m+n+1) \\
\end{align}$$
Which means that
$$\begin{align}
&2q^2=4(m-n)(m+n+1) \\
&q^2=2(m-n)(m+n+1) \\
\end{align}$$
Then $q^2$ is even, and so is then $q$, which is absurd. Similarly, suppose
$q=2n$ and $p=2m+1$.
Then $p+q=2(m+n)+1$ and $p-q=2(m-n)+1$. So it is the case that
$$\begin{align}
&2q^2=(2(m-n)+1)(2(m+n)+1)\\
&2q^2=4(m^2+m-n^2)+1 \\
\end{align}$$
So $2q^2$ is odd, which is then absurd.

Is this valid?

Here are some proofs I've found (link at bottom):
If $\sqrt 3 = m/n$:
$$ \frac{m}{n} = \sqrt 3 \frac{\sqrt 3 - 1}{\sqrt 3 - 1} = \frac{3-\sqrt 3}{\sqrt 3 - 1}
= \frac{3-m/n}{m/n-1} = \frac{3 n - m}{m-n}$$
and the right side has a smaller denominator, since $m < 2n$ (i.e., $\sqrt 3 < 2$).

$x = \sqrt{3} - 1$ is a root of the equation $x^2 + 2x - 2 = 0$, thus:
$$x(3+x) = 2+x$$
$$x = \frac{2+x}{3+x} = \cfrac{1}{1 + \cfrac{1}{2 + x}}$$
And thus
$$x = [1,2,1,2,\dots]$$
So $\sqrt{3}$ os irrational.

These proofs and others: How to prove that $\sqrt 3$ is an irrational number?
