Can one extend the domain of the square root $\sqrt{\cdot}$ to $\mathbb R$?

Question

Is it even allowed to do that? I know that it is defined to only return the positive solutions and is a function as such, but can one extend the domain with it still being called a function?

Answer 1

It's a function from $[0,\infty) = \mathbb R^+$ to $\mathbb R$ (although not onto).

You can not extend a domain to where it is undefined. But you can restrict it to where it is defined.

$\sqrt{}: \mathbb R^+ \to \mathbb R^+ \subset \mathbb R$ is a function.

Another way to put this is to note that part of the declaration of a function is a precise declaration of what the domain is. So $\sqrt{}$ is a function upon the correct domain. But the correct domain is NOT $\mathbb R$.

The declaration of the domain is often neglected, overlooked and misunderstood. But it is a essential part of defining a function. In fact, maybe the most important part.

Answer 2

If $f(x) = \sqrt{x}$ is considered will all real numbers as its domain, how do you define $f(-1)$?

Normally we do that by saying $f(-1) = i \in \mathbb{C}$ but you are asking about $f:\mathbb{R} \to \mathbb{R}$, I don't think there is a way to do that.