Let S = 1, 2, 3, ......, 10. Suppose M is the set of all the subsets of S, then the relation R = (A, B) : A ∩ B ≠ φ; A, B ∈ M is :

Step-by-step solution

1. 1

   Given Relation

   Set $S = \{1, 2, \dots, 10\}$, $M = P(S)$, $R = \{(A, B) : A \cap B \neq \phi\}$.
2. 2

   Relation Properties

   Recall definitions of reflexive, symmetric, and transitive relations.
3. 3

   Check Reflexivity

   For $R$ to be reflexive, $(A, A) \in R$ for all $A \in M$. This means $A \cap A \neq \phi$.
4. 4

   Test Reflexivity

   If $A = \phi$, then $A \cap A = \phi \cap \phi = \phi$. So $(\phi, \phi) \notin R$.
5. 5

   Check Symmetry

   For $R$ to be symmetric, if $(A, B) \in R$, then $(B, A) \in R$.
6. 6

   Test Symmetry

   If $A \cap B \neq \phi$, then $B \cap A \neq \phi$ is always true.
7. 7

   Check Transitivity

   For $R$ to be transitive, if $(A, B) \in R$ and $(B, C) \in R$, then $(A, C) \in R$.
8. 8

   Test Transitivity

   Let $A=\{1\}$, $B=\{1,2\}$, $C=\{2\}$. $A \cap B \neq \phi$ and $B \cap C \neq \phi$.
9. 9

   Transitivity Result

   But $A \cap C = \{1\} \cap \{2\} = \phi$. So $(A, C) \notin R$. Not transitive.