JEE Mathematics 2025 Question with Solution

Given: $R = \{(1, 2), (2, 3), (3, 3)\}$ on the set $\{1, 2, 3, 4\}$.

Find: The minimum number of ordered pairs to be added so that $R$ becomes an equivalence relation.

For an equivalence relation, $R$ must be reflexive, symmetric, and transitive.

For reflexivity, all pairs $(1,1), (2,2), (3,3), (4,4)$ must be present. Since $(3,3)$ is already present, we need to add $(1,1), (2,2), (4,4)$. This contributes 3 new elements.

For symmetry:

• from $(1,2)$, we must add $(2,1)$
• from $(2,3)$, we must add $(3,2)$
• $(3,3)$ is already symmetric

This contributes 2 more new elements.

For transitivity:

• from $(1,2)$ and $(2,3)$, we must add $(1,3)$
• using the new pair $(3,2)$ and $(2,1)$, we must add $(3,1)$

This contributes 2 more new elements.

Thus the total added pairs are

So the number of elements added is

Therefore, the minimum number of elements to be added is $7$. Hence, the correct option is D.

Given: $R = \{(1, 2), (2, 3), (3, 3)\}$ on $\{1,2,3,4\}$.

Find: The least number of pairs needed to make $R$ an equivalence relation.

The pairs $(1,2)$ and $(2,3)$ show that the elements $1,2,3$ must lie in the same equivalence class. Therefore, among $1,2,3$, every ordered pair must be present in the final relation.

Hence, on the class $\{1,2,3\}$, we need all $3 \times 3 = 9$ ordered pairs:

Among these, the given relation already contains $(1,2), (2,3), (3,3)$.

So for the class $\{1,2,3\}$, the missing pairs are

which are 6 pairs.

The element $4$ is unrelated to any other element, so to satisfy reflexivity it must at least have $(4,4)$. This adds 1 more pair.

Therefore, the minimum number of pairs to be added is

So the correct option is D.