Question

Let R and S be two non – void relation in a set A. which of the following statements is false?

• A. R and S transitive $\Rightarrow$ R $\cup$ S is not transitive
• B. R and S transitive $\Rightarrow$R $\cap$ S is transitive
• C. R and S symmetric $\Rightarrow$ R $\cup$ S is symmetric
• D. R and S reflexive $\Rightarrow$ R $\cap$ S is reflexive

Answer 1

The correct option is A R and S transitive $\Rightarrow$ R $\cup$ S is not transitive

(a) Let (a, b), (b, c) $ϵ$ R $\cup$ S.
It is possible that (a, b) $ϵ$ R - S and (b, c) $ϵ$ S - R
In such a case, we cannot say that
(a, c) $ϵ$ R or (a, c) $ϵ$ S.
$\therefore$ may not be in R $\cup$ S.
$\therefore$ R $\cup$ S is not transitive
(b) Let (a, b), (b, c) $ϵ$R $\cap$S.
$\therefore$ (a,b), (b,c) $ϵ$ and (a,b), (b,c) $ϵ$ S
$\therefore$ (a,c) $ϵ$ R and (a,c) $ϵ$ S
$\therefore$ (a,c) $ϵ$ R $\cap$ $\therefore$ R $\cap$ S is transitive.
(c) Let (a,b) $ϵ$ R $\cup$ S
$\therefore$ (a,b) $ϵ$ R or (a,b) $ϵ$ S
Now, (a,b) $ϵ$ R $\Rightarrow$ (b,a) $ϵ$ R ($\because$ R is symmetric)
(a,b) $ϵ$ $\Rightarrow$ (b, a) $ϵ$ S ($\therefore$ S is symmetric)
$\therefore$ (b,a) $ϵ$ R $\cup$ S $\therefore$ R $\cup$ S is symmetric
(d) Let a $ϵ$ A.
$\therefore$ (a,a) $ϵ$ R and (a,a) $ϵ$ S
$\therefore$ (a,a) $ϵ$ R $\cap$ S is reflexive