elementary number theory - 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.

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Is this valid?

Answer

It works, but can be simplified: $\rm\:mod\ 2\!: p\equiv p^2 = 3q^2 \equiv q,\:$ so $\rm\:p,q,\:$ being coprime, are odd. $\rm\:mod\ 4\!:\ odd\equiv \pm 1,\:$ so $\rm\:odd^2\equiv 1,\:$ so $\rm\: 1\equiv p^2 = 3q^2\equiv 3\ \Rightarrow\ 4\:|\:3-1\:\Rightarrow\Leftarrow$