A) \[R\text{ }\cup \text{ }S\] is an equivalence relation on A
B) \[R\text{ }\cap \text{ }S\] is an equivalence relation on A
C) \[R-S\] is an equivalence relation on A
D) None of these
Correct Answer: B
Solution :
Given, R and S are relations on set A. \ \[R\subseteq A\times A\]and \[S\subseteq A\times A\] Þ \[R\cap C\subseteq A\times A\] Þ \[R\cap S\]is also a relation on A. Reflexivity: Let a be an arbitrary element of A. Then, \[a\in A\]Þ \[(a,a)\in R\]and\[(a,a)\in S\], [\[\because \]R and S are reflexive] Þ \[(a,\,a)\in R\cap S\] Thus, \[(a,a)\in R\cap S\]for all\[a\in A\]. So, \[R\cap S\]is a reflexive relation on A. Symmetry: Let \[a,b\in A\]such that\[(a,b)\in R\cap S\]. Then, \[(a,b)\in R\cap S\] Þ \[(a,b)\in R\]and \[(a,b)\in S\] Þ \[(b,a)\in R\] and \[(b,a)\in S\], [\[\because \]R and S are symmetric] Þ \[(b,a)\in R\cap S\] Thus, \[(a,b)\in R\cap S\] Þ \[(b,a)\in R\cap S\] for all \[(a,b)\in R\cap S\]. So, \[R\cap S\]is symmetric on A. Transitivity: Let \[a,b,c\in A\]such that \[(a,b)\in R\cap S\] and \[(b,c)\in R\cap S\]. Then, \[(a,b)\in R\cap S\]and \[(b,c)\in R\cap S\] Þ \[\{((a,b)\in R\,\,\text{and}\,(a,b)\in S\,)\}\] and \[\{((b,c)\in R\,\text{and}\,\text{(}b,c\text{)}\in S\}\] Þ \[\{(a,b)\in R,(b,c)\in R\}\]and \[\{(a,b)\in S,(b,c)\in S\}\] Þ \[(a,c)\in R\]and \[(a,c)\in S\] \[\left[ \begin{align} & \because R\,\text{and}\,S\,\text{are}\,\text{transitive}\,\text{So} \\ & \text{(}a,b\text{)}\in R\,\text{and (}b,c\text{)}\in R\Rightarrow \text{(}a\text{,}c\text{)}\in R \\ & \text{(}a,b\text{)}\in S\text{ and (}b\text{,}c\text{)}\in S\Rightarrow \text{(}a\text{,}c\text{)}\in S \\ \end{align} \right.\] Þ \[(a,c)\in R\cap S\] Thus,\[(a,b)\in R\cap S\] and\[(b,c)\in R\cap S\Rightarrow (a,c)\in R\cap S\]. So, \[R\cap S\]is transitive on A. Hence, R is an equivalence relation on A.
You need to login to perform this action.
You will be redirected in
3 sec
