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Misc 7 Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is (A) 1 (B) 2 (C) 3 (D) 4 Total possible pairs = { (1, 1) , (1, 2), (1, 3), (2, 1) , (2, 2), (2, 3), (3, 1) , (3, 2), (3, 3) } Reflexive means (a, a) should be in relation . So, (1, 1) , (2, 2) , (3, 3) should be in a relation Symmetric means if (a, b) is in relation, then (b, a) should be in relation . So, since (1, 2) is in relation, (2, 1) should also be in relation Transitive means if (a, b) is in relation, & (b, c) is in relation, then (a, c) is in relation So, if (1, 2) is in relation, & (2, 1) is in relation, then (1, 1) should be in relation Relation R1 = Total possible pairs = { (1, 1) , (1, 2), (1, 3), (2, 1) , (2, 2), (2, 3), (3, 1) , (3, 2), (3, 3) } So, smallest relation is R1 = { (1, 2), (2, 1), (1, 1), (2, 2), (3, 3) } If we add (2, 3), then we have to add (3, 2) also , as it is symmetric but, as (1 , 2) & (2, 3) are there, we need to add (1, 3) also , as it is transitive As we are adding (1, 3), we should add (3, 1) also, as it is symmetric Relation R2 = Hence, only 2 possible relations are there which are equivalence ∴ B is the correct answer

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo