Q172 / 270 · NDA 2017 (Phase 2) · Mathematics

If we define a relation R on the set $\mathrm{N} \times \mathrm{N}$ as $(a, b) R(c, d) \Leftrightarrow a+d=b+c$ for all $(a, b),(c, d) \in N \times N$, then the relation is:

  • A
    symmetric only
  • B
    symmetric and transitive only
  • Correct answer
    C
    equivalence relation
  • D
    reflexive only

Solution

$(a, b) R(c, d) \Leftrightarrow a+d=b+c$ (i) $a+a=a+a$. $\therefore \quad(\mathrm{a}, \mathrm{a}) \mathrm{R}(\mathrm{a}, \mathrm{a}) \Rightarrow \mathrm{R}$ is reflexive. (ii) $(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d}) \Rightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}$ $(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \Rightarrow \mathrm{c}+\mathrm{b}=\mathrm{d}+\mathrm{a}$ $\therefore \mathrm{R}$ is symmetric. $\begin{aligned} & \text { (iii) } \operatorname{Let}(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d}) \text { and }(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{e}, \mathrm{f}) \\ & \Rightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c} \text { and } \mathrm{c}+\mathrm{f}=\mathrm{d}+\mathrm{e}\end{aligned}$ $\begin{aligned} & \Rightarrow \mathrm{a}+\mathrm{d}+\mathrm{c}+\mathrm{f}=\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e} \\ & \Rightarrow \mathrm{a}+\mathrm{f}=\mathrm{b}+\mathrm{e} \\ & \Rightarrow(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{e}, \mathrm{f})\end{aligned}$ $\therefore \mathrm{R}$ is transitive. from (i), (ii), (iii) R is an equivalence relation.

Hence, the correct answer is option: 'C'.

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