Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 1.1

Ex 1.1, 1 (i)

Ex 1.1, 1 (ii)

Ex 1.1, 1 (iii) Important

Ex 1.1, 1 (iv)

Ex 1.1, 1 (v)

Ex 1.1, 2

Ex 1.1, 3 You are here

Ex 1.1, 4

Ex 1.1, 5 Important

Ex 1.1, 6

Ex 1.1, 7

Ex 1.1, 8

Ex 1.1, 9 (i)

Ex 1.1, 9 (ii)

Ex 1.1, 10 (i)

Ex 1.1, 10 (ii)

Ex 1.1, 10 (iii) Important

Ex 1.1, 10 (iv)

Ex 1.1, 10 (v)

Ex 1.1, 11

Ex 1.1, 12 Important

Ex 1.1, 13

Ex 1.1, 14

Ex 1.1, 15 (MCQ) Important

Ex 1.1, 16 (MCQ)

Last updated at June 5, 2023 by Teachoo

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