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CA Foundation · Quantitative Aptitude · Sets, Relations and Functions, Limits and Continuity

Let A = {1, 2, 3, 4} and R = {(1,1), (2,2), (3,3), (4,4), (1,2), (2,1), (3,4), (4,3)}. The number of ordered pairs that must be added to R, at minimum, to make it an equivalence relation with just one class containing all elements is:

Eight pairs must be added. One equivalence class containing all four elements requires the universal relation with 4×4 = 16 pairs. R currently has 8 pairs, so the 8 cross pairs between {1,2} and {3,4}, in both directions, are needed.

  1. A4
  2. B8Correct
  3. C6
  4. D10

Explanation

R is already an equivalence relation with classes {1,2} and {3,4}. A single class needs all 16 pairs of A×A. R has 8 pairs, so 16 − 8 = 8 pairs must be added: (1,3),(1,4),(2,3),(2,4) and their reverses. Adding only 4 would break symmetry, so 4 is wrong.

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