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Ex 1.1, 3 Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive. R = {(a, b):b = a + 1} where a, b ∈ {1, 2, 3, 4, 5, 6} Check reflexive If the relation is reflexive, then (a, a) ∈ R i.e. a = a + 1 Since a = a + 1 cannot be possible for any value of a, the given relation is not reflexive. Check symmetric To check whether symmetric or not, If (a , b) ∈ R, then (b , a) ∈ R i.e., if b = a + 1, then a = b + 1 Since a = b + 1 is not true for all values of a & b. Hence, the given relation is not symmetric Check transitive To check whether transitive or not, If (a,b) ∈ R & (b,c) ∈ R , then (a,c) ∈ R i.e., if b = a + 1, & c = b + 1 then c = a + 1 Since, b = a + 1 & c = b + 1 Putting value of b in (2) c = (a + 1) + 1 c = a + 2 Hence, c ≠ a + 1 Hence, the given relation it is not transitive

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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