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?

Even/odd won't help here. Try division by $3$:
$$p^2=12q^2\implies 3\mid p^2\implies 3\mid p\implies (3k)^2=12q^2\implies 3k^2=4q^2\implies 3\mid 4q^2\implies 3\mid q$$

There's also an alternative way which makes it very easy to see the problem:
If $\frac pq$ is not an integer, then $\bigl(\frac pq\bigr)^2$ also isn't, so the square root must be an integer if it's rational. But
$$12=2^2\cdot 3=n^2$$
is clearly impossible since $3\mid n^2\implies 9\mid n^2$. This method works for any square root.
