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. 
 A: 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. 
A: Well, we know $\sqrt{12}  = 2\sqrt 3$, and we know that $2$ is rational while $\sqrt 3$ is irrational (proof here). We also know that the product of an irrational number and a non-zero rational number is irrational (proof here).
Thus $\sqrt{12}$ is irrational.
A: 
Let's assume that $12=\frac {p}{q}$, where $p,q ∈ R$ and $p$ and $q$
  are coprime.

Okay, that would have to be either $p = 12;q=1$ or $p=-12; q=-1$.

Then we have, $(p^2q^2)=12^2=144.$

Yep $(\pm 12)^2(\pm 1)^2 = 144$.
That is true.

So,$p^2=144∗q^2$ and $p^2=2∗(72)∗q^2$. This implies that $p$ is even.

Yes, that is all true.  $p=\pm 12$ is even and $(\pm 12)^2 = 2*(72)*(\pm)^2$.

Then, $p=(2k)$

Yes, $k = \pm 6$ and $p = \pm 12 = 2*(\pm 6)$.

So $(2k)^2=2∗(72)∗q^2$
$4k^22=2∗(72)∗q2
> 
> $2k^2=72q^2$
> 
> Thus,
> 
> $k^2=36q^2$
> 
> so,
> 
> $k=6q$

Yes,  $k =\pm 6$ is equal to $6q$ as $q = \pm 1$.

then $\frac pq= \frac {12q}{q}$, which contradicts $p$ and $q$ being coprime.

No, it doesn't  $\frac {12q}q$ may not  be in lowest terms.  Even if you did not know that $q =\pm 1$ this wouldn't conclude anything. 
What you want to say is $q|p = 12q$ so $p$ and $q$ are not relatively prime.  But you would be wrong as $q = \pm 1$.  
What you have proven is that $p = 12q$ (which you knew at the very beginning. so $q|p$ and $\gcd(p,q)= q$.  So $p$ and $q$ are relatively prime if and only if $q = \pm 1$.  Which is not only possible, but inevetible.

Therefore $12$ is irrational. QED

You do realize that $12$ is rational (it's an integer and all integers are rational) and that that was never in question.

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

Since it proved that $12$ was irrational, it obviously is  not.
Not you need to prove that if $r^2 =(\frac pq)^2 = 12$ then $r$ is impossible to be rational.  Not if $r = \frac pq =12$ that $r$ is impossible to be rational.
You attempted to prove the entirely wrong thing.  Proved the trivial result that $12$ was rational, and declared it to be irrational.
