A) R and S transitive\[\Rightarrow \] \[R\cup S\] is transitive
B) R and S transitive\[\Rightarrow R\cap S\] is transitive
C) R and S transitive\[\Rightarrow R\cup S\] is symmetric
D) R and S reflexive\[\Rightarrow R\cap S\] is reflexive
Correct Answer: A
Solution :
| [a] Let \[(a,b),(b,c)\in R\cup S.\] |
| It is possible that \[(a,b)\in R-S\]and \[(b,c)\in S-R.\] |
| In such a case, we cannot say that |
| \[(a,c)\in R\]or \[(a,c)\in S.\] |
| \[\therefore (a,c)\]may not be in \[R\cup S.\] |
| \[\therefore R\cup S\]is not transitive. |
| [b] Let \[(a,b),(b,c)\in R\cap S\] |
| \[\therefore (a,b),(b,c)\in R\] and \[(a,b),(b,c)\in S\] |
| \[\therefore (a,c)\in R\] and \[(a,c)\in S\] |
| \[\therefore (a,c)\in R\cap S\] \[\therefore R\cap S\] is transitive. |
| [c] Let \[(a,b)\in R\cup S\] \[\therefore (a,b)\in Ror(a,b)\in S\] |
| Now, \[(a,b)\in R\Rightarrow (b,a)\in R\] (\[\because \] R is symmetric) |
| \[(a,b)\in S\Rightarrow (b,a)\in S\] (\[\because \] S is symmetric) |
| \[\therefore (b,a)\in R\cup S\therefore R\cup S\] is symmetric. |
| [d] Let \[a\in A.\therefore (a,a)\in R\] and \[(a,a)\in S.\] |
| \[\therefore (a,a)\in R\cap S.\] \[\therefore R\cap S\] is reflexive. |
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