Not understanding the concept of equivalence class

Question

Let $U$ be a set defined: $U=\{(x,y)\in \Bbb R^2\mid x^2+y^2=1; xy\neq 0\}$, and let $R$ be relation defined: $(x_1,y_1)R(x_2,y_2) \iff (x_1 \cdot x_2>0∧y_1\cdot y_2>0)$.

I was to prove it's an equivalence relation - which I did. Then I was asked to show its equivalence classes. So, I don't really understand what it is, and I don't understand how to write it with proper notation.

Thanks for any help!

edit: Sorry for all the typos!

Type { to get ${$. The other case is similar and I leave it as an exercise to you. — Git Gud, Apr 21 '13 at 15:06
$R$ is not an equivalence class. But you can find equivalence classes w.r.t. $R$. — Yury, Apr 21 '13 at 15:08
I'm nearly certain you've misread the question. As stated it doesn't make sense. I think maybe you are supposed to write down all of the equivalence classes? — , Apr 21 '13 at 15:08
@ohad I believe there's a typo in the way you define $R$: $(x_1,y_1)R(x_2,y_2) \iff (\color{red}{x_2} \cdot x_2>0∧y_1\cdot y_2>0)$ — Git Gud, Apr 21 '13 at 15:09

score 12 · Answer 1 · answered Apr 21 '13 at 15:17

12

${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

Equivalence classes

answered Apr 21 '13 at 15:17

Git Gud

31,356

I like this. It answers the question, without providing the answer (that could just be copied). In the process of translating this image into a written solution, the student will solidify their understanding of the material. In theory :) – The Chaz 2.0 Apr 21 '13 at 15:33
@TheChaz2.0 Thank you. – Git Gud Apr 21 '13 at 15:34

score 3 · Answer 2 · answered Apr 21 '13 at 15:18

The equivalence classes of $R$ are the sets $C_i$ such that $a,b\in C_i \implies aRb$. Another way of thinking about this is to take a single 'representative' member $c_i$ of each class, and then $C_i \equiv [c_i] = \{a: c_iRa\}$, the set of all things related to $c_i$.

Let's start by picking $c_1=(1/\sqrt 2,1/\sqrt 2)$ to the top right of the circle. Assuming the condition is the product of the $x$ coordinates with each other is positive and similarly the $y$ coordinates: $c_1R(x,y) \iff x>0 \text{ and } y>0$. Therefore $C_1 = \text{part of circle in top right quadrant} $. Can you find the other classes? Just pick points you haven't yet put into a class.

score 2 · Answer 3 · answered Apr 21 '13 at 15:12

Hint: Put another way, assuming that Git Gud is correct about the typo, $(x_1,y_1)\:R\:(x_2,y_2)$ if and only if $x_1,x_2$ have the same sign and $y_1,y_2$ have the same sign. This should readily split $U$ (the points on the unit circle that aren't on the coordinate axes) into $4$ equivalency classes.

score 2 · Answer 4 · answered Apr 21 '13 at 15:15

Think of equivalence classes as subsets of $U$ that consist of elements that are all equivalent to one another. The fact that $R$ is an equivalence relation guarantees that the equivalence classes will partition $U$, that is, each element of $U$ will belong to exactly one equivalence class.

$U$ is the unit circle with its intersections with the coordinate axes removed. Let's consider a point of $U$ that lies in the first quadrant, $(x,y)$. By the equivalence relation, you should be able to show that another point in $U$, say $(x',y')$, is equivalent to $(x,y)$ if and only if $(x',y')$ is also in the first quadrant. This shows that the points of $U$ in the first quadrant form one equivalence class.

See if you can follow a similar train of thought for the other quadrants.

Not understanding the concept of equivalence class

4 Answers4