Question

Let n be a fixed positive integer. Define a relation R on Z as follows:
(a, b) ∈ R ⇔ a − b is divisible by n.
Show that R is an equivalence relation on Z.

Answer 1

We observe the following properties of R. Then,
Reflexivity:
$$\begin{gathered} \mathrm{Let}a\in N \\ \mathrm{Here}, \\ a-a=0=0\times n \\ \Rightarrow a-a\mathrm{is}\mathrm{divisible}\mathrm{by}n \\ \Rightarrow \left(a,a\right)\in R \\ \Rightarrow \left(a,a\right)\in R\mathrm{for}\mathrm{all}a\in Z \\ \mathrm{So},R\mathrm{is}\mathrm{reflexive}\mathrm{on}Z. \end{gathered}$$

Symmetry:
$$\begin{gathered} \mathrm{Let}\left(a,b\right)\in R \\ \mathrm{Here}, \\ a\mathit{-}b\mathit{}\mathrm{is}\mathrm{divisible}\mathrm{by}\mathit{}n \\ \mathit{\Rightarrow }a-b=np\mathrm{for}\mathrm{some}p\in Z \\ \Rightarrow b-a=n\left(-p\right) \\ \Rightarrow b-a\mathrm{is}\mathrm{divisible}\mathrm{by}n[p\in Z\Rightarrow -p\in Z] \\ \Rightarrow \left(b,a\right)\in R \\ \mathrm{So},R\mathrm{is}\mathrm{symmetric}\mathrm{on}Z. \end{gathered}$$

Transitivity:
$$\begin{gathered} \mathrm{Let}\left(a,b\right)\mathrm{and}\left(b,c\right)\in R \\ \mathrm{Here},a-b\mathrm{is}\mathrm{divisible}\mathrm{by}n\mathrm{and}b-c\mathrm{is}\mathrm{divisible}\mathrm{by}n. \\ \Rightarrow a-b=np\mathrm{for}\mathrm{some}p\in Z \\ \mathrm{and}b-c=nq\mathrm{for}\mathrm{some}q\in Z \\ \mathrm{Adding}\mathrm{the}\mathrm{above}\mathrm{two},\mathrm{we}\mathrm{get} \\ a-b+b-c=np+nq \\ \Rightarrow a-c=n(p+q) \\ \mathrm{Here},p+q\in Z \\ \Rightarrow \left(a,c\right)\in R\mathrm{for}\mathrm{all}a,c\in Z \\ \mathrm{So},R\mathrm{is}\mathrm{transitive}\mathrm{on}Z. \end{gathered}$$

Hence, R is an equivalence relation on Z.