Ex 1.1, 5 -  R = {(a, b) : a <= b3} is reflexive, symmetric - Ex 1.1

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  1. Chapter 1 Class 12 Relation and Functions
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Ex 1.1, 5 Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is reflexive, symmetric or transitive. R = {(a, b) : a ≤ b3} Here R is set of real numbers Hence, both a and b are real numbers Check reflexive If the relation is reflexive, then (a, a) ∈ R i.e. a ≤ a3 Let us check Hence, a ≤ a3 is not true for all values of a. So, the given relation it is not reflexive Check symmetric To check whether symmetric or not, If (a,b) ∈ R, then (b,a) ∈ R i.e., if a ≤ b3, then b ≤ a3 Since b ≤ a3 is not true for all values of a & b. Hence, the given relation it 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 a ≤ b3, & b ≤ c3 then a ≤ c3 Since if a ≤ b3, & b ≤ c3 then a ≤ c3 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

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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.
  • Dharmendra Patidar's image
    Dharmendra Patidar
    June 24, 2018, 4 p.m.
    Concept of equivalence class

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  • Payal Verma's image
    Sir  my doubt is from 1.1 exercise from question no.*2   *3...and questionno.*5....

    while checking a relation is reflexive symmetric transitive why we are everytime proving it not reflexive  not symmetric not transitive . I have a doubt why we are always proving that it is not equivalence why we are not proving it as equivalence.
    Sir I've seen in many questions like ... whenever such type of questions come like .. show that or check whether the relation is reflexive..symmetric.or transitive....
    We are always proving that question is neither reflexive nor transitive or symmetric......
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  • Rishabh Sharma's image
    Rishabh Sharma
    Aug. 5, 2017, 5:38 p.m.
    How will we find the values against which to check transitivity or semitricity? They are too complex to derive.
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