Which pair is not present in the smallest equivalence relation on A={1,2,3,4} containing R={(1,2),(2,3),(3,1)}?

Correct answer: (1,4)

The correct answer is D, (1,4). The three given pairs already connect 1, 2, and 3. In the smallest equivalence relation, reflexivity adds (1,1), (2,2), (3,3), and (4,4); symmetry adds the reverse pairs; and transitivity fills in every ordered pair among 1, 2, and 3. Thus {1,2,3} becomes one equivalence class, while 4 remains alone. Since no given pair connects 4 to the other elements, the equivalence closure does not add (1,4), (4,1), or any other pair between 4 and the first class. Options A, B, and C are explicitly present in the original relation, so they certainly remain present in every relation containing it. Option D is therefore the only pair that is absent. Common confusion: an equivalence closure adds what the three properties force, not arbitrary pairs between unrelated classes.