Let S=1,2,3, , 10.

Question

Let $S=\{1,2,3, \ldots, 10\}$. Suppose $M$ is the set of all the subsets of $S$, then the relation $\mathrm{R}=\{(\mathrm{A}, \mathrm{B}): \mathrm{A} \cap \mathrm{B} 
eq \phi ; \mathrm{A}, \mathrm{B} \in \mathrm{M}\}$ is :

Accepted Answer

Let $$S=\{1,2,3, \ldots, 10\}$$

$$R=\{(A, B): A \cap B \neq \phi ; A, B \in M\}$$

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,

$$\[\begin{array}{ll} \text { ARB } & \mathrm{A} \cap \mathrm{B} \neq \phi, \ \Rightarrow \mathrm{BRA} & \Rightarrow \mathrm{B} \cap \mathrm{A} \neq \phi, \end{array}$$\]

So it is symmetric.

For transitive,

$$\begin{aligned}
\text { If } A & =\{(1,2),(2,3)\} \
B & =\{(2,3),(3,4)\} \
C & =\{(3,4),(5,6)\}
\end{aligned}$$

$$\mathrm{ARB}$$ \& $$\mathrm{BRC}$$ but $$\mathrm{A}$$ does not relate to $$\mathrm{C}$$ So it not transitive

Solutions · 1