Question

Show that the relation R on the set Z of integers, given by
R = {(a, b) : 2 divides a – b}, is an equivalence relation.

Answer 1

We observe the following properties of relation R.

Reflexivity:
$$\begin{gathered} \mathrm{Let}a\mathrm{be}\mathrm{an}\mathrm{arbitrary}\mathrm{element}\mathrm{of}\mathrm{the}\mathrm{set}Z.\mathrm{Then}, \\ a\in R \\ \Rightarrow a-a=0=0\times 2 \\ \Rightarrow 2\mathrm{divides}a-a \\ \Rightarrow \left(a,a\right)\in R\mathrm{for}\mathrm{all}a\in Z \\ \mathrm{So},R\mathrm{is}\mathrm{reflexive}\mathrm{on}Z. \end{gathered}$$

Symmetry:
$$\begin{gathered} \mathrm{Let}\left(a,b\right)\in R \\ \Rightarrow 2\mathrm{divides}a-b \\ \Rightarrow \frac{a-b}{2}=p\mathrm{for}\mathrm{some}p\in Z \\ \Rightarrow \frac{b-a}{2}=-p \\ \mathrm{Here},-p\in Z \\ \Rightarrow 2\mathrm{divides}b-a \\ \Rightarrow \left(b,a\right)\in R\mathrm{for}\mathrm{all}a,b\in Z \\ \mathrm{So},R\mathrm{is}\mathrm{symmetric}\mathrm{on}Z. \end{gathered}$$

Transitivity:
$$\begin{gathered} \mathrm{Let}\left(a,b\right)\mathrm{and}\left(b,c\right)\in R \\ \Rightarrow 2\mathrm{divides}a-b\mathrm{and}2\mathrm{divides}b-c \\ \Rightarrow \frac{a-b}{2}=p\mathrm{and}\frac{b-c}{2}=q\mathrm{for}\mathrm{some}p,q\in Z \\ \mathrm{Adding}\mathrm{the}\mathrm{above}\mathrm{two},\mathrm{we}\mathrm{get} \\ \frac{a-b}{2}+\frac{b-c}{2}=p+q \\ \Rightarrow \frac{a-c}{2}=p+q \\ \mathrm{Here},p+q\in Z \\ \Rightarrow 2\mathrm{divides}a-c \\ \Rightarrow \left(a,c\right)\in R\mathrm{for}\mathrm{all}a,c\in Z \\ \mathrm{So},R\mathrm{is}\mathrm{transitive}\mathrm{on}Z. \end{gathered}$$

Hence, R is an equivalence relation on Z.