Relation of equivalence with sgn

Question

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?

Answer 1

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)]

Answer 2

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, $\operatorname{sgn}(y+\pi x)=1$).