Is R = {(a, b) | a, b ∈ Z, ∃k ∈ Z s.t. (a − b) = 3k} an equivalence relation? I'm not quite sure what ∃k ∈ Z s.t. (a − b) = 3k describes in the relation. Perhaps if I could figure out what that means, I could prove R is reflexive, symmetric, and transitive.
The question also states that if R is an equivalence relation, I should describe its equivalence classes. Could someone explain more clearly to me what equivalence classes are and how to "describe" them?
Thank you in advance.
 A: Hints: 
$(8,5)\in R$ because $8-5=3=3\times 1~~~~$  (and $1\in \mathbb{Z}$).  $(8,2)\in R$ because $8-2=6=3\times 2$.  $(2,8)\in R$ because $2-8=-6=3\times -2$. $(2,2)\in R$ because $2-2=0=3\times 0$.
  However $(8,3)\notin R$ because $8-3=5=3\times (5/3)$, and $5/3\notin \mathbb{Z}$.
Of the three properties you need to prove, the trickiest one is transitivity.  If $(a,b)\in R$ and $(b,c)\in R$, then $a-b=3m$ and $b-c=3n$.  Sum these to get $a-c=3m+3n=3(m+n)$.
A: Once you are working with an equivalence relation, then you end up with lots of things that are considered the same by that relation. Let $S$ be an equivalence relation. Consider $1Sx$, which is the set of $x$ such that $x$ is equivalent to $1$. This would is the equivalence class for $1$. So if $1S5$, then $5$ is in the equivalence class of $1$ and, while using $S$, the two can be used interchangeably.
For your particular $R$, you just have to work through the definitions to show reflexive, symmetric and transitive.
A: Hint:
$\exists k \in \Bbb Z$ such that $(a - b) = 3k \iff 3 \ | \ (a - b)$ 
Think about the properties of divisibility. $a \ | \  0 \;\; \forall a \in \Bbb Z, \;\; a \ |  \ b \iff a \ | \ (-b)$ and $a\ | \ b,  \ a\ | \ c \implies a$ divides any linear combination of $a$ and $b$.
Say $R$ is an equivalence relation. Then the set $[a]=\{b \ | \ (a,b) \in R \}$ is the equivalence class of $a$.  That is the set of all elements related to $a$ in $R$. 
Shouldn't be hard to finish this off. 
