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$ xRy$ iff $x^2 -y^2 = x-y$ is relation we have defined over $\mathbb{R} $

I have shown this is equivalence relation.

Now we have asked to find equivalence class of 3 which can be found out to be -2 and 3 .

Also next we need to find equivalence class of general x which turn out out to be 1-x. (Check this also)

So next question is given like $ xRy$ iff $f(x) = f(y)$ ; now I have to find all such function from $\mathbb{R} $ to $\mathbb{R} $. I stuck there any hint or help appreciated.

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  • $\begingroup$ $x^2-y^2=x-y$ is the same thing as $x^2-x=y^2-y$. So one of them is $f(t)=t^2-t$. But it is not the only one, because you can compose $f$ with any other bijection of the reals to get a new one. So definitely, there are infinitely many of them. $\endgroup$ – Crostul Nov 21 '19 at 7:51
  • $\begingroup$ @Crostul It doesn't have to be a bijection. It just has to be injective on $[-1/4,\infty)$, and then we don't care what it does to any number below $-\frac14$. It doesn't even have to be defined elsewhere. $\endgroup$ – Arthur Nov 21 '19 at 7:56
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As noted in the comments, $f(t)=t^2-t$ is over such function. However, looking at the graph, what does it mean that $f(x)=f(y)$? It means exactly that $x$ and $y$ are equally far away from $\frac12$.

There are many other functions that do this. A full characterisation might be phrased along the lines of

$f:\Bbb R\to\Bbb R$ characterizes this equivalence relation iff it is injective on $[\frac12,\infty)$ and symmetric about $\frac12$.

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