### Core Logic S$S$ must be reflexive, symmetric, and transitive, containing (1,2)$(1,2)$, (2,3)$(2,3)$, and (1,4)$(1,4)$. Symmetric property forces (2,1), (3,2), (4,1) in S$(2,1), (3,2), (4,1) \in S$. Transitive property: (1,2)$(1,2)$ and (2,3) implies (1,3) in S$(2,3) \implies (1,3) \in S$. Symmetric implies (3,1) in S$\implies (3,1) \in S$. (4,1)$(4,1)$ and (1,2) implies (4,2) in S$(1,2) \implies (4,2) \in S$. Symmetric implies (2,4) in S$\implies (2,4) \in S$. (4,1)$(4,1)$ and (1,3) implies (4,3) in S$(1,3) \implies (4,3) \in S$. Symmetric implies (3,4) in S$\implies (3,4) \in S$. ### Step 1: Universal Relation Since 1 is related to 2, 3, 4 and the relation is an equivalence relation (which creates partitions), all elements 1, 2, 3, and 4 must fall into the same single equivalence class. Thus, S$S$ must contain all possible ordered pairs in A times A$A \times A$. ### Step 2: Final Count Number of elements in A times A = 4 times 4 = 16$A \times A = 4 \times 4 = 16$. Minimum value of n$n$ is 16$16$. ### Pattern Recognition If a relation connects all elements in a set to each other through a chain, its equivalence closure is the universal relation A times A$A \times A$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Relations and Functions