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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.

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    $\begingroup$ 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. $\endgroup$
    – user765629
    Oct 26, 2020 at 22:33
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    $\begingroup$ 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 $x^2$. $\endgroup$
    – Mummy the turkey
    Oct 26, 2020 at 22:35
  • $\begingroup$ @epiliam , please, make clear, why you made that opinion that the circle may not be a function. $\endgroup$
    – Sergei Krivonos
    Oct 27, 2020 at 3:30
  • $\begingroup$ @epiliam , why multivalue may not be a function? This also doesn’t seem a correct statement. $\endgroup$
    – Sergei Krivonos
    Oct 27, 2020 at 3:31
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    $\begingroup$ @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. $\endgroup$
    – user765629
    Oct 27, 2020 at 4:12

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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:

  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 $\Bbb{R}$.
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  • $\begingroup$ 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 $(1,-3) \notin S$ $\endgroup$
    – user516076
    Nov 1, 2020 at 3:55
  • $\begingroup$ 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. $\endgroup$
    – Servaes
    Nov 1, 2020 at 9:24
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$y=(R^2-x^2)^\frac{1}{2}$ 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.

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