Example 48 - Chapter 1 Class 12 Relation and Functions (Term 1)
Last updated at Jan. 28, 2020 by Teachoo
Last updated at Jan. 28, 2020 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) } Each relation should have (1, 2) and (2, 1) in it For other pairs, Let’s check which pairs will be in relation, and which won’t be 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 (1, 1) , (2, 2) , (3, 3)
Finding number of relations
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