Question 13 (OR 2 nd question)

Check whether the relation R in the set R of real numbers, defined by R = {(a, b) : 1 + ab > 0}, is reflexive, symmetric or transitive.

Slide43.JPG

Slide44.JPG
Slide45.JPG Slide46.JPG

  1. Class 12
  2. Sample Papers, Previous Year Papers and Other Questions
Ask Download

Transcript

Question 13 (OR 2nd question) Check whether the relation R in the set R of real numbers, defined by R = {(a, b) : 1 + ab > 0}, is reflexive, symmetric or transitive. R = {(a, b) : 1 + ab > 0}, Checking for reflexive If the relation is reflexive, then (a ,a) ∈ R i.e. 1 + a2 > 0 Since square numbers are always positive Hence, 1 + a2 > 0 is true for all values of a. So, the given relation it is reflexive. R = {(a, b) : 1 + ab > 0}, Checking for symmetric, To check whether symmetric or not, If (a, b) ∈ R, then (b,a) ∈ R i.e., if 1+ ab < 0, then 1 + ba > 0 Since if 1 + ab > 0, then 1 + ba > 0 is always true for all value of a & b Hence, the given relation is symmetric R = {(a, b) : 1 + ab > 0}, Checking transitive To check whether transitive or not, If (a, b) ∈ R & (b, c) ∈ R , then (a, c) ∈ R i.e., if 1 + ab > 0, & 1 + bc > 0, then 1 + ac > 0 Let’s take an example a = –8, b = –2, c = 1/4 1 + ab = 1 + (–8) × (–2) = 1 + 16 = 17 > 0 1 + bc = 1 + (–2) × 1/4 = 1 – 1/2 = 1 – 1/2 = 1 – 1/2 1 + ac = 1 + (–8) × 1/4 = 1 – 2 = –1 ≯ 0 Since 1 + ac ≯ 0 when 1 + ab > 0 and 1 + bc > 0 ∴ The condition is not true for all values of a, b, c. Hence, the given relation it is not transitive Therefore, the given relation is reflexive and symmetric, but not transitive 1 + bc 1 + ac

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.