# Domain of the $f(x) = \sqrt{x}$

Domain of the $$f(x) = \sqrt{x}$$

So typically, when introducing functions to students, teachers will say that the domain of this function is $$[0,\infty)$$.

However, in the curriculum I'm following, one of the first things we did was introduce complex numbers. Thus, shouldn't I be teaching that the domain of this function is all real numbers?

Maybe if we were to specify that the codomain is real numbers then I could understand having the domain be $$[0,\infty)$$, but we never make such a restriction.

• The main arguments for giving a lot of attention to real-valued square roots, and more generally to real-valued non-integer powers, are that: 1) even if you work with a branch cut of the inverse relation of $z^2$, the standard one still has issues on the negative reals; 2) phenomena like that carry on more generally and they deserve an attention in their own right, so that would not be much of a solution; 3) it is generally believed that it is important to exploit differentiability of $(x,y)\mapsto x^y$ on $(0,\infty)\times(0,\infty)$. Commented Oct 3, 2022 at 13:52
• Indeed, once you go into complex numbers, every number $x\in\mathbb C$ (except $0$) has two "candidates" for a square root (if $y^2=x$, so is $(-y)^2=x$) and you can in fact decide arbitrarily for each complex number which of the two to pick as $\sqrt{x}$. The best we can do to preserve at least some fine properties of the function (e.g. continuity, differentiability) is to make those "branch cuts" from complex analysis, but they are also ambiguous. (Where do you cut? The rule $\sqrt{xy}=\sqrt{x}\sqrt{y}$ stops working. $\sqrt{-1}$ may be $i$ or may be $-i$. Etc. etc.)