Relation of equivalence with sgn Test if the relation
$$(x, y)ρ(a, b)\Leftarrow\Rightarrow sgn(y+\pi x) = sgn(b + \pi a)$$
is a relation of equivalence on $R^2$ and if so, determine the quotient set and $C_{(2, \pi)}$.
Also, $sgn(x)=1$ if $x>0$, $sgn(x)=0$ if $x=0$, and $sgn(x)=-1$ if $x<0$.
I've started with the usual definition of equivalence.
Reflexivity:
$sgn(y+\pi x)=sgn(y+\pi x)$
Symmetry and transitivity - I really have no idea how to do this. I'm not sure if I did reflexivity right, because there must be different cases for $x$ because of the $sgn$, am I correct? Because of that, I don't know how to find the quotient set.
Can someone help me in demistifying this problem?
 A: It's much easier if you consider the function $f\colon \mathbb{R}^2\to\{-1,0,1\}$ defined by
$$
f(x,y)=\operatorname{sgn}(y+\pi x)
$$
Then you can rewrite your relation by stating that
$$
(x,y)\mathrel{\rho}(a,b)
\quad\text{if and only if}\quad
f(x,y)=f(a,b)
$$
whereby proving that it's an equivalence relation is very simple.
Moreover, the equivalence classes can be parametrized by the possible values of $f$, which are $-1$, $0$ and $1$. As each value is reached, there are three equivalence classes: for instance, $f(0,1)=1$, $f(0,-1)=-1$ and $f(-1,\pi)=0$.
Since $f(2,\pi)=\operatorname{sgn}(\pi+2\pi)=\operatorname{sgn}(3\pi)=1$, the equivalence class of $(2,\pi)$ is the set of pairs $(x,y)$ such that $y+\pi x>0$ (or, which is the same, $\operatorname{sgn}(y+\pi x)=1$).
A: reflection :  (x,y)~(x,y) ; it is obvious because of "=" sign.
symmetry: if (x,y)~(a,b) than it has to be (a,b)~(x,y) you don't change the position of x,y because you work with ordered pair in R2.
transitivity: if (x,y)~(a,b) and (a,b)~(c,d) than it has to be (x,y)~(c,d).This is obvious again because of "=" sign. Function sgn() gives you 1/0/-1 so it is like comparing integers, for example 1=1 and 1=1 => 1=1
$sgn(y+πx) = sgn(b+πa) $
$sgn(b+πa)=sgn(d+πc)$
$ => sgn(y+πx)=sgn(d+πc)$
quotient set: pairs that are in relation must give the same sgn() result. 1 or 0 or -1 ,so our R2 is divided  in 3 classes.
1. all pairs which sgn() function gives 1.
sgn(y+πx)=1 when $y+πx >0$, $y >-πx$ our represent can be (1,1)
class is[(1,1)]
2. all pairs which sgn() function gives 0.
sgn(y+πx)=0 when $y+πx =0$, $y =-πx$ our represent can be (1,-π)
class is[(1,-π)]
3. all pairs which sgn() function gives -1.
sgn(y+πx)=-1 when $y+πx <0$, $y<-πx$ our represent can be (0,-1)
class is[(0,-1)]
