# 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?

• A circle is a shape, not a function. You can describe a circle in the plane centered at the origin via $x^2+y^2=r^2$. However, if you were to express this as $y=\pm\sqrt{r^2-x^2}$ you see that for every value of $x$, there are two values of $y$. So $y$ is multi-valued, therefore not a function.
– user765629
Oct 26, 2020 at 22:33
• The circle itself is not a function. It is described as the image of the map $\mathbb{R} \to \mathbb{R}^2: t \to (r\sin(t), r\cos(t))$ which is single-valued. If you consider it to be the solution to $y^2 + x^2 = r^2$ then this is a multi-valued function of $x$. The more pedagogical example is the inverse of the function $x^2$. Oct 26, 2020 at 22:35
• @epiliam , please, make clear, why you made that opinion that the circle may not be a function. Oct 27, 2020 at 3:30
• @epiliam , why multivalue may not be a function? This also doesn’t seem a correct statement. Oct 27, 2020 at 3:31
• @Sergei A shape is a geometric object. A function is a mapping $f:X\to Y$. It consists of a rule, and two sets. By its very definition, a function cannot be multi-valued. Note also that the square-root function is a well defined, single-valued function, so your answer is wrong.
– user765629
Oct 27, 2020 at 4:12

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 $$\Bbb{R}$$.
1. 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 $$[-1,1]$$.
2. 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 $$\Bbb{R}$$.
• Can i have a question please? I'm not sure or maybe don't understand about your first statement. I mean, if $(x,y)\in S \Rightarrow (x,-y)\in S$ I mean, for example i have a unit circle at $(1,2)$ as a center that means $(1,3)\in S$. But we know $(1,-3) \notin S$ Nov 1, 2020 at 3:55
• By "the circle" I meant "the unit circle" throughout my answer, which is the circle of radius $1$ centered at the origin. For your circle you get $(x,4-y)\in S$ if $(x,y)\in S$, and I'm sure you can generalize this to any circle. Nov 1, 2020 at 9:24
$$y=(R^2-x^2)^\frac{1}{2}$$ is multivalue function because exponentiation to a negative even is multivalue by disjunction.