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 = 1/2> 0
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

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.