Question

Prove that the relation R on Z defined by
(a, b) ∈ R ⇔ a − b is divisible by 5
is an equivalence relation on Z.

Answer 1

We observe the following properties of relation R.

Reflexivity:
$$\begin{gathered} \mathrm{Let}a\mathrm{be}\mathrm{an}\mathrm{arbitrary}\mathrm{element}\mathrm{of}R.\mathrm{Then}, \\ \Rightarrow a-a=0=0\times 5 \\ \Rightarrow a-a\mathrm{is}\mathrm{divisible}\mathrm{by}5 \\ \Rightarrow \left(a,a\right)\in R\mathrm{for}\mathrm{all}a\in Z \\ \mathrm{So},R\mathrm{is}\mathrm{reflexive}\mathrm{on}Z. \end{gathered}$$

Symmetry:
$$\begin{gathered} \mathrm{Let}\left(a,b\right)\in R \\ \Rightarrow a-b\mathrm{is}\mathrm{divisible}\mathrm{by}5 \\ \Rightarrow a-b=5p\mathrm{for}\mathrm{some}p\in Z \\ \Rightarrow b-a=5\left(-p\right) \\ \mathrm{Here},-p\in Z[\mathrm{Since}p\in Z] \\ \Rightarrow b-a\mathrm{is}\mathrm{divisible}\mathrm{by}5 \\ \Rightarrow \left(b,a\right)\in R\mathrm{for}\mathrm{all}a,b\in Z \\ \mathrm{So},R\mathrm{is}\mathrm{symmetric}\mathrm{on}Z. \end{gathered}$$

Transitivity:
$$\begin{gathered} \mathrm{Let}\left(a,b\right)\mathrm{and}\left(b,c\right)\in R \\ \Rightarrow a-b\mathrm{is}\mathrm{divisible}\mathrm{by}5 \\ \Rightarrow a-b=5p\mathrm{for}\mathrm{some}Z \\ \mathrm{Also},b-c\mathrm{is}\mathrm{divisible}\mathrm{by}5 \\ \Rightarrow b-c=5q\mathrm{for}\mathrm{some}Z \\ \mathrm{Adding}\mathrm{the}\mathrm{above}\mathrm{two},\mathrm{we}\mathrm{get} \\ a-b+b-c=5p+5q \\ \Rightarrow a-c=5(p+q) \\ \Rightarrow a-c\mathrm{is}\mathrm{divisible}\mathrm{by}5 \\ \mathrm{Here},p+q\in Z \\ \Rightarrow \left(a,c\right)\in R\mathrm{for}\mathrm{all}a,c\in Z \\ \mathrm{So},R\mathrm{is}\mathrm{transitive}\mathrm{on}Z. \end{gathered}$$

Hence, R is an equivalence relation on Z.