There're no $x \in \mathbb{Q}$ which verifies $x^2=12$. Proposition : $\nexists{x}\in\mathbb{Q}\;|\;x^2=12$.
Hypothesis : We can rewrite $x$ like $\left(\frac{p}{q}\right)^2$ where $\forall{p}\in\mathbb{Z}$ and $\forall{q}\in\mathbb{N}$ and $\left(\frac{p}{q}\right)$ can't be reduced ($p$ and $q$ can't be both even).
Demonstration : We have to reject the hypothesis.
$$\left(\frac{p}{q}\right)^2=12 \rightarrow \frac{p^2}{q^2}=12 \rightarrow p^2=12q^2.$$
That means that $p^2$ must be even. That can be possible only if $p$ is even too. Therefore, we can rewrite $p$ like $2k$ where $\forall{k}\in\mathbb{Z}$.
$$p^2=(2k)^2=4k^2=12q^2\\k^2=3q^2$$
And there's where I stuck!
Can your help me?

Now continue,
with 3 instead of two.
Since
$k^2 = 3q^2$,
$k$ must be divisible by $3$.
Writing
$k = 3j$,
$9j^2 = 3q^2$,
or
$3j^2 = q^2$,
so $q$ must
also be divisible by $3$.
