Question

R is a relation on the set Z of integers and it is given by
(x, y) ∈ R ⇔ | x − y | ≤ 1. Then, R is
(a) reflexive and transitive
(b) reflexive and symmetric
(c) symmetric and transitive
(d) an equivalence relation

Answer 1

(b) reflexive and symmetric

$$\begin{gathered} \mathrm{Reflexivity}:\mathrm{Let}x\in R.\mathrm{Then}, \\ x-x=0<1 \\ \Rightarrow \left|x-x\right|\le 1 \\ \Rightarrow \left(x,x\right)\in R\mathrm{for}\mathrm{all}x\in Z \\ \mathrm{So},R\mathrm{is}\mathrm{reflexive}\mathrm{on}Z. \\ \mathrm{Symmetry}:\mathrm{Let}\left(x,y\right)\in R.\mathrm{Then}, \\ \left|x-y\right|\le 0 \\ \Rightarrow \left|-(y-x)\right|\le 1 \\ \Rightarrow \left|y-x\right|\le 1\left[\mathrm{Since}\left|x-y\right|=\left|y-x\right|\right] \\ \Rightarrow \left(y,x\right)\in R\mathrm{for}\mathrm{all}x,y\in Z \\ \mathrm{So},R\mathit{}\mathrm{is}\mathrm{symmetri}c\mathrm{on}Z\mathit{.} \\ \mathrm{Transitivity}:\mathrm{Let}\left(x,y\right)\in R\mathrm{and}\left(y,z\right)\in R.\mathrm{Then}, \\ \left|x-y\right|\le 1\mathrm{and}\left|y-z\right|\le 1 \\ \Rightarrow \text{It is not always true that}\left|x-y\right|\le 1. \\ \Rightarrow \left(x,z\right)\notin R \\ \mathrm{So},R\mathrm{is}\mathrm{not}\mathrm{transitive}\mathrm{on}Z\mathit{.} \end{gathered}$$