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?

I'll just add to this mayhem with this bit of logic.
Assume $\sqrt{3}=\frac{a}{b}\implies3=\frac{a^2}{b^2}\implies3b^2=a^2$. Now, this cannot be true! When you square something, you double all of its prime factors. Then we know $a^2$ and $b^2$ have an even number of factors, therefor $3b^2$ has an odd number of factors because we have added a factor of $3$. So $3b^2 \neq a^2$. 
