Understanding Functions: Puzzled by f(x) = sqrt(x)?

[EvLer]

I am going on my own through a chapter on functions and here is something that puzzled me in the definition:

Each element in the domain is paired with just one element in the range.

I guess my calculus knowledge interferes, but what about function like f(x) = sqrt(x). It has two roots: + and -. How does set theory account for that? Or is sqrt(x) not a function in set-theoretic terms
Although f(x) = x^2 fits the definition of the function.
Could someone please explain?

Thanks in advance.

 

[master_coda]

$f(x)=\pm\sqrt{x}$ is not a function, since f(x) has more than one value for all x > 0. However, $f(x)=\sqrt{x}$ is a function; $\sqrt{x}$ always refers to the positive square root.

 

[honestrosewater]

Yes, just look at the definition of function. If something doesn't meet the requirements of the definition, it just simply isn't a function.
But how can you tell whether something is a function until you specify its domain and the superset of its range? $\sqrt{x}$ isn't a function from N to N.

 

[EvLer]

Thanks