Is a circle a multivalued function? I don't really understand multi-valued function. I hope one of you can make me understand it. What I've learned from google, I suppose that a multi-valued function is a binary relation that maps the domain more than once to a single point (i can't describe it well).
So, here, can I conveniently claim that a circle (on the real plane) is a multi-valued function since from one point on the domain we can get $2$ points as the range?
Thanks in advance.
 A: The circle $S$ can be viewed as a binary relation on $\Bbb{R}$, simply because it is a subset of $\Bbb{R}^2$. And indeed if $(x,y)\in S$ then also $(x,-y)\in S$, so if $y\neq0$ then this shows that $S$ is multivalued. But $S$ is not a function on $\Bbb{R}$, because it is not defined on all of $\Bbb{R}$; there is no $y\in\Bbb{R}$ such that $(2,y)\in S$, for example. So the circle cannot be viewed as a multivalued function on $\Bbb{R}$.
There are two ways to adjust your example to valid example of a multivalued function:

*

*We can adjust the domain; viewing the circle $S$ as a subset of $[-1,1]\times\Bbb{R}$, or even $[-1,1]^2$, we do get a multivalued function. This of course relates to the fact that every positive real number has precisely $2$ square roots, i.e. the function
$$f(x)=\sqrt{1-x^2},$$
can also be interpreted as a $2$-valued function on the interval $[-1,1]$.

*We can consider a hyperbola instead, for example the set of all $(x,y)\in\Bbb{R}^2$ satisfying
$$x^2-y^2=-1.$$
I leave it to you to verify that this defines a $2$-valued function on all of $\Bbb{R}$.

A: $y=(R^2-x^2)^\frac{1}{2}$ is multivalue function because exponentiation to a negative even is multivalue by disjunction.
Yes, it is correct that any function which graphical representation has multiple y for single x would be multi-value by disjunction at that point.
