JEE Mathematics 2024 Question with Solution

Given: $A = \{1, 2, 3, 4\}$ and $R = \{(1, 2), (2, 3), (1, 4)\}$ with $R \subset S$, where $S$ is an equivalence relation on $A$.

Find: The minimum number of elements $n$ in $S$.

An equivalence relation must be reflexive, symmetric, and transitive.

For reflexivity, add all pairs of the form

For symmetry, from

we must also add

Now use transitivity:

and by symmetry also

Again, using transitivity with the obtained relations,

and by symmetry also

Similarly, the remaining required pairs are

Thus, the smallest equivalence relation is

Hence, the total number of elements is $16$. Therefore, the correct option is A.

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

Find: The minimum size of an equivalence relation $S$ containing $R$.

In an equivalence relation, if $1 \sim 2$, $2 \sim 3$, and $1 \sim 4$, then by transitivity all four elements become equivalent. So there is only one equivalence class:

An equivalence relation on one class of size $4$ contains every ordered pair from that class, so the number of pairs is

Therefore, the minimum value of $n$ is $16$ and the correct option is A.