relations and equivalent classes

Question

I've been working on this question on relations and finding at least one equivalent class. I'm done with finding the equivalence relation of this set, But i fail to understand the concept of classes, I looked at an example thrown by somebody on quora, where he explains this concept using a family tree where the siblings are related to each other, and instead of writing it all out, you could just write out the elements related in a single set, showing they're relations to each other.

In my question, x~y -> |x+2| = |y+2|, Prove that the relation defined on the set [-5, 5] is an equivalence relation and find a class.

My approach:

(x~x): |x+2| = |x+2| = which is reflexive (a = a)

(y~x) -> (x~y) = |y+2| = |x+2| => |x+2| = |y+2| which is symmetric considering the ( a = b -> b = a)

(x~y) ^ (y ~ z) -> (x ~ z): |x+2| = |y+2| ^ |y+2| = |z+2| - > |x+2| = |z+2| since |y+2| = |z+2|, |x+2| = |z+2| and |x+2| = |z+2| - > |x+2| = |z+2| , and is transitive, and hence this is am equivalent relation. But how do i derive the class?

An explanation would help a lot.

Answer 1

Assuming you are working with integers on $[-5,5]$. You enumerate them within their equivalence class.

E.g.

• $-5$ with that rule $|(-5)+2| = 3 = |1+2|$, so $1$ is in the same class and you have $C_{-5} = \{-5,1\}$
• $|(-4)+2| = 2 = |0 + 2|$ hence $C_{-4} = \{-4,0\}$

Do the rest yourself. You will have some classes with only one element, and if you find some elements in the same class as the one you already covered, just join them.

Answer 2

The class of $x$ is the set of all elements $y$ such that $$|y+2|=|x+2|\iff y+2=x+2\text{ or } y+2=-(x+2)\iff y=x\text{ or } y=-4-x$$ So if $-5\le x\le 1$, the class of $x$ is $\left\{x,-4-x\right\}$, and if $1<x\le 5$, the class of $x$ is $\left\{x\right\}$.

Another way to see this is that $x\sim y$ iff the distances from $x$ and $y$ to $-2$ are equal, which you can solve with a drawing.

So two examples of classes are $\left\{2\right\}$ and $\left\{-3,-1\right\}$.