# please solve with explanationlet 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!

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

Cheers!

• 25

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• -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
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