Last updated at Dec. 8, 2016 by Teachoo

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Ex 1.1, 2 Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b2} is neither reflexive nor symmetric nor transitive R = {(a, b) : a ≤ b2} Checking for reflexive, If the relation is reflexive, then (a, a) ∈ R i.e. a ≤ a2 Let us check Hence, a ≤ a2 is not true for all values of a. So, the given relation it is not reflexive. Checking for symmetric, To check whether symmetric or not, If (a, b) ∈ R, then (b, a) ∈ R i.e., if a ≤ b2, then b ≤ a2 Since b ≤ a2 is not true for all values of a & b. Hence, the given relation is not symmetric Checking transitive To check whether transitive or not, If (a, b) ∈ R & (b, c) ∈ R , then (a, c) ∈ R i.e., if a ≤ b2, & b ≤ c2 then a ≤ c2 Since if a ≤ b2, & b ≤ c2 then a ≤ c2 is not true for all values of a, b, c. Hence, the given relation it is not transitive Therefore, the given relation is neither reflexive, symmetric or transitive

Chapter 1 Class 12 Relation and Functions

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .