TITLE: Proving $\sqrt 3$ is irrational.
QUESTION [30 upvotes]: 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?

REPLY [4 votes]: One can show a more general result, that is for any prime number $p$ we must have $\sqrt{p}$ is irrational. see my answer here.