please solve with explanation

let R be a relation in Rth set of real numbers, defined as (a,b)E R<=>1+AB>0 for all a,b E R. show that R is reflexive, symmetric but not transitive.

Hi!
Here is the answer to your question.
 
R = {a, b}: 1 + ab > 0, a, b ε R}
 
Reflexive:
For any a ε R, a2 ≥ 0
⇒ 1 + a2 ≥ 1
But 1 > 0
∴ 1 + a2 > 0
⇒1 + a. a > 0
⇒ (a, a) ε R
Thus, R is reflexive
 
Symmetric:
Let (a, b) ε R, where a, b ε R
∴ 1 + ab > 0
⇒ 1 + ba > 0
⇒ (b, a) ε R
Thus, R is symmetric
 
Transitive:
(–2, 0) and (0, 1) ε R since 1 + (– 2). 0 = 1 > 0 and 1 + 0.1 = 1 > 0
But 1 + (–2) .1 = 1 – 2 = –1 which is not greater than equal to 0.
Thus, (–2, 1) is not an element in the relation R
This shows that R is not transitive
 
Hope! You got the answer.
 
Cheers! 

  • 25

 yayayayayayayayayayayayya

  • -5

 R = {a, b}: 1 + ab > 0, a, b ε R}

 
Reflexive:
For any a ε Ra 2 ≥ 0
⇒ 1 + a 2 ≥ 1
But 1 > 0
∴ 1 + a 2 > 0
⇒1 + a. a > 0
⇒ (a, a) ε R
Thus, R is reflexive
 
Symmetric:
Let (a, b) ε R, where a, b ε R
∴ 1 + ab > 0
⇒ 1 + ba > 0
⇒ (b, a) ε R
Thus, R is symmetric
 
Transitive:
(–2, 0) and (0, 1) ε R since 1 + (– 2). 0 = 1 > 0 and 1 + 0.1 = 1 > 0
But 1 + (–2) .1 = 1 – 2 = –1 which is not greater than equal to 0.
Thus, (–2, 1) is not an element in the relation R
This shows that R is not transitive
 

  • -4
What are you looking for?