can someone let me know what this question means and how do i go about solving it?
Let Z be the set of integers and R be the relation defined in Z such that aRb if a – b is divisible by 3. Then R partitions the set Z into how many pairwise disjoint sets?
all i know is that this relation would be an equivalence relation.Please help.
Here are the equivalence classes: $$[0]=\{3k\mid k\in\mathbb{Z}\}\\ [1]=\{3k+1\mid k\in\mathbb{Z}\}\\ [2]=\{3k+2\mid k\in\mathbb{Z}\}. $$ and these are the only equivalence classes. To see this, let $[a]$ be any equivalence class, then by division algorithm, we have $a=3q+r$ for some $q\in\mathbb{Z}$ and $r=0,1,2$. Thus $[a]=[r]$, where $r=0,1,2$.
You are correct in pointing out that $R$ is an equivalence relation.
Consider the set $S=\{n| n=a-b\quad \forall a,b \in \mathbb{Z} \}$.
$S$ is clearly the same as $\mathbb{Z}$ since the difference of any two integers is an integer, and any integer can be written as the difference of two integers.
When you think of the equivalence classes of $R$, $aRb$ means $a$ and $b$ belong to the same equivalence class of $R$. Thus, each class will contain all the integers such that the difference of any two of them is divisible by $3$.
Now to answer the question at hand, what are the possible remainders when you divide an integer by $3$? Either $0,1$ or $2$. Certainly $0,1,2$ each belong to different equivalence classes since none of $(1-0), (2-1), (2-0)$ are divisible by $3$.
From the claim in bold above, it follows that you have $3$ equivalence classes:
Hopefully you'll have no trouble filling in the last two.
