JEE Main 2025 Jan 23 Shift 1, Mathematics Q11: Relations

The existing pairs $(1,2)$ and $(2,3)$ force $1,2,3$ to be in the same equivalence class. The element $4$ is unrelated to anything, so it forms its own singleton class.

The equivalence relation must therefore contain every pair $(a,b)$ with $a,b\in \{1,2,3\}$, plus $(4,4)$.

That is the $3\times 3=9$ pairs over $\{1,2,3\}$ together with $(4,4)$, totalling $10$ pairs. Two of these, $(1,2)$ and $(2,3)$, are already present (and $(3,3)$ too). So we need to add $10-3=7$ pairs:

$(1,1),(2,2),(4,4),(2,1),(3,2),(3,1),(1,3)$.

$\therefore 7$ elements.