JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Sets Relations and Functions

For $R$ on $\{1,2,3,4\}$ to be an equivalence relation it must be reflexive, symmetric and transitive.

Reflexive: need $(1,1),(2,2),(3,3),(4,4)$; only $(3,3)$ is present, so add $(1,1),(2,2),(4,4)$ — 3 pairs.

Symmetric: $(1,2)$ needs $(2,1)$, and $(2,3)$ needs $(3,2)$ — 2 pairs.

The relation now contains $(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(2,3),(3,2)$.

Transitive closure: $(1,2)$ and $(2,3)$ force $(1,3)$; $(3,2)$ and $(2,1)$ force $(3,1)$ — 2 more pairs. Checking all remaining compositions (e.g. $(1,3)\&(3,1)\to (1,1)$, $(1,3)\&(3,2)\to (1,2)$) shows no further pair is needed.

Total elements added: $3+2+2=7$.