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.
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$\begingroup$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 $$(2\sqrt{3})^2 = (2\sqrt{3})(2\sqrt{3}) = (2^2)(\sqrt{3})^2 = 4\cdot 3=12=\sqrt{12}^2 \\ \implies 2\sqrt{3}=\sqrt{12}$$
$\endgroup$ 2 $\begingroup$By definition, $\sqrt3$ is the positive number $s$ such that $s^2=3.$ Likewise, $\sqrt{12}$ is the positive number $t$ such that $t^2=12.$ The claim, then, is that $t=2s,$ but is it true? Well, $2s$ is a positive number, since it is a product of two positive numbers. Moreover, $$(2s)^2=2^2s^2=4s^2=4\cdot 3=12.$$ Hence, $2s=t,$ as desired.
$\endgroup$ 4 $\begingroup$Since,$$12 = 4\times 3 = 2^2 \times 3$$ $$\sqrt{12} = \sqrt{2^2\times 3} = \sqrt{2^2}\times\sqrt{3}=2\sqrt{3}$$
$\endgroup$ 3 $\begingroup$Let's consider
\begin{aligned} f : \mathbf{R_+} &\to \mathbf{R_+}\\ x &\to x^2 \end{aligned}
then $f$ is bijection, for it is strictly increasing.
$f(\sqrt{12}) = f(2\sqrt{3}) = 12$, so $\sqrt{12} = 2\sqrt{3}$.
Note that $\sqrt{12}$ is just a notation, there is no problem to denote one number with different notations. For example, we can also use $[3, \overline{[2,6]}]$ which means to be a continued fraction to denote $\sqrt{12}$.
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