# Example 48 - Chapter 1 Class 12 Relation and Functions

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 48 Show that the number of equivalence relation in the set {1, 2, 3} containing (1, 2) and (2, 1) is two. 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 two possible relations are there which are equivalence

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Chapter 1 Class 12 Relation and Functions

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.