proof writing - Prove that there is no rational number whose square is 12.

Let's assume that $12 = (\frac{p}{q})$, where p,q $\in$ $\mathbb{R}$ and $p$ and $q$ are coprime.

Then we have,

$(\frac{p^2}{q^2})= 12^2 = 144.$

So,

$p^2 = 144*q^2$

and $p^2 = 2*(72)*q^2.$

This implies that p is even.

Then,

$p=(2k)$

So,

$(2k)^2=2*(72)*q^2$

$4k^2 = 2*(72)*q^2$

$2k^2 = 72q^2$

Thus,

$k^2 = 36q^2$

so,

$k=6q$

then $(\frac{p}{q}) = (\frac{12q}{q})$,

which contradicts p and q being coprime. Therefore 12 is irrational. QED

I was wondering if the proof I've provided is sound.

Answer

Assume otherwise. then there exists $p,q\in \Bbb N$ such that $mcd\{p,q\}=1$ and $\frac{p}{q}=\sqrt{12}.$ From here $\frac{p^2}{q^2}=12$ which is equivalent to $p^2=12q^2=3(2q)^2.$ From here we deduce that $3|p,$ so $p=3k$ for some natural number $k.$ The equation is now $9k^2=3(2q)^2,$ or after simplify, $3k^2=(2q)^2.$ Now we deduce that $3|2q,$ and hence $3|q.$ But this is a contradiction because $3|p,$ so that $mcd\{p,q\}\geq 3>1.$ Hence $\sqrt{12}$ is irrational.