Q52 / 110 · VITEEE 2022 · Mathematics
n the set \(\mathbf{N} \times \mathbf{N}\), the relation \(\mathrm{R}\) is defined by (a, b) \(R(c, d) \Leftrightarrow a d=b c\). Then \(R\) is
- Apartial order relation
- Correct answerBequivalence relation
- Creflexive and transitive but not symmetric
- Dsymmetric and transitive but not reflexive
Solution
The relation is defined on \(\mathbf{N} \times \mathbf{N}\). since the elements of \(\mathbf{N} \times \mathbf{N}\) are ordered pairs hence the relation on \(\mathbf{N} \times \mathbf{N}\) relates two ordered pairs. Now,
(1) \(\forall \mathrm{a}, \mathrm{b} \in \mathbf{N} \quad \mathrm{ab}=\mathrm{ba}\)
\(\therefore(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \forall(\mathrm{a}, \mathrm{b}) \in \mathbf{N} \times \mathbf{N}\). Thus \(\mathrm{R}\) is refiexive
(ii) Let
\(\begin{aligned}
& (a, b) R(c, d) \Leftrightarrow a d=b c \forall(a, b) \text { and }(c, d) \in N \times N \\
& \Rightarrow b c=a d \Rightarrow c b=d a \Rightarrow(c, d) R(a, b) .
\end{aligned}\)
Thus \(R\) is symmetric
(iii) Again let (a,b) R (c,d) and (c,d) R (e,f)
\(\Rightarrow \mathrm{ad}=\mathrm{bc}\) and \(\mathrm{cf}=\mathrm{de}\)
\(\Rightarrow\) adcf \(=\) bcde \(\Rightarrow\) af \(=\) de \(\Rightarrow(a, b) R(e, f)\)
Thus \(\mathrm{R}\) is transitive.
Thus \(\mathrm{R}\) is equivalence relation.
Hence, the correct answer is option: 'B'.