# equivalence relation over real number

$$xRy$$ 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.

• $x^2-y^2=x-y$ 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. – Crostul Nov 21 '19 at 7:51
• @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. – Arthur Nov 21 '19 at 7:56

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 $$\frac12$$.