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Restrict attention to $\mathbb R^2$ (2D/cartesian plane).

Attempted definition: Given an equation (in variables $x$ and $y$), define the graph of that equation to be the set of points $(x,y) \in \mathbb{R}^2$ for which the equation holds. (Now we can as usual talk about the graphs of $y=x$, $x^2+y^2=1$, etc.)

The above definition seems reasonable but now suppose we're given the equation $\sqrt{1-x}=\sqrt{x-3}$. Observe that $2$ solves this equation but with $\sqrt{1-2}=\sqrt{2-3}=-\mathrm{i}$. Here there's the question of whether or not we want to accept that $2$ is a solution to the given equation. Then how might we correct the above definition depending on whether

  1. We do want to accept that $2$ is a solution--so that the graph of the given equation is the vertical line $x=2$; or
  2. We don't (say because we're in a high-school setting where complex numbers haven't been introduced)--so that the graph of the given equation is the empty set?
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    The given square roots aren't defined at $x=2$, so $2$ isn't even in the domain to begin with. So I would contend that $2$ isn't a solution. Indeed, when graphed over the reals, the two functions given have empty intersection. – Raad Shaikh Oct 03 '22 at 03:34
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    I recommend simplifying your question to "Is $x=2$ a solution to $\sqrt{1-x}=\sqrt{x-3}$?" The graphing context is an unnecessary distraction. (Presumably, the phrase "satisfies the equation" in your definition of graph incorporates whatever policy you establish regarding that particular situation, so the definition wouldn't really need correcting.) – Blue Oct 03 '22 at 03:47
  • @Blue: Thanks for the recommendation. I've made a small edit. –  Oct 03 '22 at 03:56
  • @RaadShaikh (& Blue): The thing is, I want to try to stick to a definition (say because I'm teaching high-school students) of "graph of an equation" that is simple yet rigorous/precise and before we've introduced functions and domains. So, how might we alter my attempted definition so that we can also take care of the above issue? (Perhaps this is simply impossible, but I'm not sure which is why I posted this Q.) –  Oct 03 '22 at 03:56
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    Here's a tricky twist on the question: sketch the graph of $y=\sqrt{1-x}/\sqrt{x-3}$. One could argue that there is no (real) $x$ for which both numerator and denominator are defined, hence, the graph is empty. Or one could argue $y=\sqrt{1-x}/\sqrt{x-3}=\sqrt{(1-x)/(x-3)}$ and now the right side is perfectly well-defined for $1\le x<3$. – Gerry Myerson Oct 03 '22 at 04:01
  • @user24096 first, let's distinguish between 'function' and 'equation'. To not run into any difficulties I think it is best to say that the graph of a function $f(x,y)$ is the set of all points $(x,y)$ that satisfy $f(x,y)=0$. Now we have to be careful of the domain - $(x,y)\in\text{Domain}(f)$, not $\mathbb{R}^2$. Essentially, you can only consider points where the function can be defined in the first place, anything outside of that must be ignored. – Raad Shaikh Oct 04 '22 at 00:10

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