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Graphically, it passes the vertical line test. However, I am confused because for $x=0$, there are infinite $y$ values that satisfy $xy=0$. Can someone help me understand this?

user
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Dan H.
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    It is not. Let $x=0$. What $y$ value does this "function" output? – Rushabh Mehta Nov 07 '18 at 22:27
  • @RushabhMehta thank you! – Dan H. Nov 07 '18 at 22:30
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    "for x=0, there are infinite y values that satisfy xy=0": so it does not pass the vertical line test, does it? –  Nov 07 '18 at 22:36
  • @RushabhMehta It is not a function as a general statement is not completely correct. We need to specify which domain we are referring to. For $x\neq 0$ it defines indeed the finction $y=0$. – user Nov 07 '18 at 22:38
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    Maybe I am being pedantic, but I don't think that the question is well-posed. A vital part of the definition of a function is a specification of the domain and codomain of that function. In this case, are we to assume that $x$ and $y$ are real numbers? integers? complex numbers? something else? If they are real numbers, do we allow any real numbers, or some subset of real numbers? Moreover, a pedantic and snarky response to your question is "No, it is an equation, not a function." Perhaps the question is "Does the equation $xy=1$ define a function $y$ in terms of $x$?" – Xander Henderson Nov 07 '18 at 22:40
  • If that is the question, the domain becomes important. For example, the answer could be "Yes. The equation defines a function $y(x) = 1/x$ on the domain $(0,\infty)$." – Xander Henderson Nov 07 '18 at 22:41
  • @RushabhMehta What about my consideration? Do you agree with that? – user Nov 09 '18 at 08:50
  • @gimusi yes, I agree – Rushabh Mehta Nov 09 '18 at 15:00
  • @RushabhMehta That's nice! Thanks – user Nov 10 '18 at 11:03

2 Answers2

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As a function of one variable it is not (in general) a function, indeed recall that by definition

$$f:A\subseteq\mathbb{R}\to \mathbb{R}\quad\forall x\in A \quad \exists! y \quad y=f(x)$$

and, as noticed by Rushabh Mehta, for $x=0$ we have that $y$ can assume any value.

It would be a function, in implicit form, if we assume for example

$$f:\mathbb{R}\setminus\{0\}\to \mathbb{R}$$

which correspons to the function $y=0$ with domain $x\neq 0$.

Refer also to the related

user
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It is not a function (as it was pointed out in a comment), but a relation, like a circle (a set of ordered pairs): $R=\{(x,y)|xy=0\}$.

Botond
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