TITLE: Why is $\sqrt {12} = 2 \sqrt 3$?
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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 $\sqrt{12}$.