Question

The relation 'R' in N × N such that
(a, b) R (c, d) ⇔ a + d = b + c is
(a) reflexive but not symmetric
(b) reflexive and transitive but not symmetric
(c) an equivalence relation
(d) none of the these

Answer 1

(c) an equivalence relation

We observe the following properties of relation R.

$$\begin{gathered} \mathrm{Reflexivity}:\mathrm{Let}(a,b)\in N\times N \\ \Rightarrow a,b\in N \\ \Rightarrow a+b=b+a \\ \Rightarrow \left(a,b\right)\in R \\ \mathrm{So},R\mathrm{is}\mathrm{reflexive}\mathrm{on}N\times N. \\ \mathrm{Symmetry}:\mathrm{Let}\left(a,b\right),\left(c,d\right)\in \mathrm{N}\times \mathrm{N}\mathrm{such}\mathrm{that}\left(a,b\right)R\left(c,d\right) \\ \Rightarrow a+d=b+c \\ \Rightarrow d+a=c+b \\ \Rightarrow \left(d,c\right),\left(b,a\right)\in R \\ \mathrm{So},R\mathrm{is}\mathrm{symmetric}\mathrm{on}\mathrm{N}\times \mathrm{N}. \\ \mathrm{Transitivity}:\mathrm{Let}\left(a,b\right),\left(c,d\right),\left(e,f\right)\in N\times N\mathrm{such}\mathrm{that}\left(a,b\right)R\left(c,d\right)\mathrm{and}\left(c,d\right)R\left(e,f\right) \\ \Rightarrow a+d=b+c\mathrm{and}c+f=d+e \\ \Rightarrow a+d+c+f=b+c+d+e \\ \Rightarrow a+f=b+e \\ \Rightarrow \left(a,b\right)R\left(e,f\right) \\ \mathrm{So},R\mathrm{is}\mathrm{transitive}\mathrm{on}N\times N. \end{gathered}$$

Hence, R is an equivalence relation on N.