I am trying to find the equivalence classes of a relation $R$ which is given by:
$A=\mathbb R\times\mathbb R\setminus\{(0,0)\}$ with relation $R$ on $A$ given by $(x_1,y_1)\sim(x_2,y_2)$ if and only if $(x_1,y_1)$ and $(x_2,y_2)$ lie on a line which passes through $(0,0)$}.
I know that $R$ is an equivalence relation since I have proved it before, but I am still confused on how to find equivalence classes. I know have found all of them if the union of all equivalence classes is equal to $A$ and that two equivalence classes are either equal or disjoint. I also know given $(x,y)\in A$, an equivalence class with respect to $(x,y) $ is $[(x,y)]_R=\{(a,b):(a,b)\sim(x,y)\}$.
Here's what I have so far:
We will show the equivalence classes of $R$ in terms of polar coordinates. We can consider that the radius is some integer and the angle is from 0 to $\pi$. With this we know that the angle lies in the first two quadrants, but the possibility of a negative radius allow us to span all of $\mathbb R$. Thus the equivalence classes can be written as:
$x=r\cos(\theta)$, $y=r\sin(\theta)$, where $r\in \mathbb R\setminus \{0\}$ and $0\leq \theta \leq \pi$
Then, $[(x,y)]_R=\{(a,b): a=r\cos(\theta)$, $b=r\sin(\theta)$, $r\in \mathbb R\setminus \{0\}$}$
From these polar coordinates we see that the angle must remain the same as the angle given by our original pair, but the radius can be infinitely small or infinitely large.
I'm not even sure I am taking the right approach to this or if my answer is even correct. Any help or feedback is appreciated.
Best Answer
The equivalence classes are the lines passing through $(0,0)$ (or, to be more precise, the sets of the form $l\setminus\{(0,0)\}$, where $l$ is a line passing through $(0,0)$). It follows from the definition of $R$ that if $l$ is such a lime and if $(x_1,y_1),(x_2,y_2)\in l$, then $(x_1,x_2)\mathrel R(y_1,y_2)$. On the other hand if $(x_1,x_2)\mathrel R(y_1,y_2)$, then there is such a line containig both of them.