# Define equivalence relation on set of integers with 6 distinct equivalence classes

I know an equivalence relation onto a set of integers, Z, is a relation that is reflexive, symmetric, and transitive. I also know that an equivalence class of "a" in "Z" is the set of elements "x" in "Z" such that (a,x) follow the relation. So I need exactly 6 different equivalence classes. I can't get my brain to just come up with a relation for any set of integers... Please help :)

Tl;dr - TITLE

## 3 Answers

A ~ B <=> A mod 6 = B mod 6 is probably the simplest example

Choose six integers, say $\{0,1,2,3,4,5\}$

Pick some relation such that none of these are in the same equivalence class, and that any other integer is in the same equivalence class as one of these.

For $n,m\in\mathbb Z$, define the equivalence relation $n\sim m$ if, and only if, the remainders of the division of n and m by 6 are equal.

• if dividing n and m by 6 results in n=m then aren't they the same integer from the start? – Rob Sny Nov 7 '16 at 23:09
• @RobSny He says "the rests of the division ... are equal". One could also phrase it as "$m-n$ is divisible by $6$". – Arthur Nov 7 '16 at 23:11
• Yes! I meant the remainder. Forgive my english. – André Porto Nov 7 '16 at 23:12
• Could anyone explain why this results in exactly 6 different equivalence classes? Any suggestions to better understand questions like this? – Rob Sny Nov 7 '16 at 23:17