Sets And Relations Question 4
Question 4 - 2024 (27 Jan Shift 1)
Let $S={1,2,3, \ldots, 10}$. Suppose $M$ is the set of all the subsets of $S$, then the relation $\mathrm{R}={(\mathrm{A}, \mathrm{B}): \mathrm{A} \cap \mathrm{B} \neq \phi ; \mathrm{A}, \mathrm{B} \in \mathrm{M}}$ is $:$
(1) symmetric and reflexive only
(2) reflexive only
(3) symmetric and transitive only
(4) symmetric only
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Answer (4)
Solution
Let $S={1,2,3, \ldots, 10}$
$\mathrm{R}={(\mathrm{A}, \mathrm{B}): \mathrm{A} \cap \mathrm{B} \neq \phi ; \mathrm{A}, \mathrm{B} \in \mathrm{M}}$
For Reflexive,
$M$ is subset of ’ $S$ '
So $\phi \in \mathrm{M}$
for $\phi \cap \phi=\phi$
$\Rightarrow$ but relation is $\mathrm{A} \cap \mathrm{B} \neq \phi$
So it is not reflexive.
For symmetric,
$\mathrm{ARB} \quad \mathrm{A} \cap \mathrm{B} \neq \phi$,
$\Rightarrow \mathrm{BRA} \Rightarrow \mathrm{B} \cap \mathrm{A} \neq \phi$,
So it is symmetric.
For transitive,
If $A={(1,2),(2,3)}$
$B={(2,3),(3,4)}$
$C={(3,4),(5,6)}$
$\mathrm{ARB} \backslash & \mathrm{BRC}$ but $\mathrm{A}$ does not relate to $\mathrm{C}$
So it not transitive
