# Devising an equivalence relation with exactly two (or three) equivalence classes

I'm very stuck on a simple problem, asking me to devise an equivalence relation on R that has exactly two equivalence classes.

I've been struggling to think of a relation that has only two classes while maintaining the symmetry, reflexivity, and transitivity necessary for an equivalence relation. I at first thought to define it as aRb if a = 5 or something similar, but none of these are equivalence relations.

I've also tried to think in terms of deriving an equivalence relation from a partition, but for example making the two partitions the positive and negative sides of the real number line seem to resemble a partial ordering rather than an equivalence relation.

Is there anything else I should be thinking about? (I don't really know how to approach the second part of the problem, the same but with three classes, either). Is my idea of how partitions work wrong? It just seems very unintuitive for an equivalence relation to only have two classes when it is on R.

• There are many equivalence relations that can also be used to define partial orderings. An equivalence relation with exactly two or three non-empty equivalence classes will always imply a partial ordering as soon as you list the classes in a sequence. Don't discard a perfectly good relation just because of that. – David K Apr 24 '18 at 4:41

And when you define a partition of a set, you can define an equivalence relation by saying that $aRb$ if and only if $a$ and $b$ are in the same "box" (Prove it!).
So if you wanna describe an equivalence relation on $\mathbb{R}$ that has two equivalence classes, you only need to partition $\mathbb{R}$ in two sets (Let's say $(- \infty, 0]$ and $(0, \infty)$, but you have a LOT of choices). Now your equivalence relation is $aRb$ if and only if $a,b \leq 0$ or $a,b > 0$.
Exactly the same thing happens if you're looking for an equivalence relation with $n$ equivalence classes. You only need to partition $\mathbb{R}$ in $n$ subsets.