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Question
The following relation is defined on the set of real numbers.
aRb if 1 + ab > 0
Find whether relation is reflexive, symmetric or transitive.
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Solution
Reflexivity:
Let a be an arbitrary element of R. Then,
a ∈ R
⇒1 + a × a > 0
i.e. 1 + a2 > 0 [Since, square of any number is positive]
So, the given relation is reflexive.
Symmetry :
Let (a, b) ∈ R
⇒ 1+ ab > 0
⇒ 1+ba > 0
⇒ (b, a) ∈ R
So, the given relation is symmetric.
Transitivity :
Let (a, b)∈R and (b, c)∈R
⇒1+ ab > 0 and 1+ bc >0
But 1+ ac ≯ 0
⇒ (a, c) ∉ R
So, the given relation is not transitive.
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