Why is $\sqrt {12} = 2 \sqrt 3$? Why $\sqrt {12} = 2 \sqrt 3$? It is obvious? If we considered the function $f(s) = s^2 $ it is injective on positive numbers so we obtain the conclusion. But in  the same time it is an equality between irrational numbers. Suppose that we know just to compute the square roots.

Let's assume (unless that's the part you want proven) that there is only 1 positive number $x$ with the property $x^2=12$.  We'll denote that number by $\sqrt{12}$.  Likewise let $\sqrt{3}$ be the positive $y$ such that $y^2=3$.
We know that $\Bbb R$ is a field and thus multiplication is commutative.  We also know that because $2\gt 0$ and $\sqrt{3}\gt 0$ that $2\sqrt{3}\gt 0$. Then $$(2\sqrt{3})^2 = (2\sqrt{3})(2\sqrt{3}) = (2^2)(\sqrt{3})^2 = 4\cdot 3=12=\sqrt{12}^2 \\ \implies 2\sqrt{3}=\sqrt{12}$$
