Question -
Answer -
(i) ConsideraRb if a – b > 0
Now for this relationwe have to check whether it is reflexive, transitive and symmetric.
Reflexivity:
Let a be anarbitrary element of R.
Then, a ∈ R
But a −a = 0 ≯ 0
So, this relation is not reflexive.
Symmetry:
Let (a, b) ∈ R
⇒ a −b > 0
⇒ − (b −a) >0
⇒ b −a < 0
So, the given relation is not symmetric.
Transitivity:
Let (a, b) ∈R and (b, c) ∈R.
Then, a −b > 0 and b − c > 0
Adding the two, we get
a – b +b − c > 0
⇒ a – c> 0
⇒ (a, c) ∈ R.
So, the given relation is transitive.
(ii) Consider aRb iff1 + a b > 0
Now for this relationwe have to check whether it is reflexive, transitive and symmetric.
Reflexivity:
Let a be anarbitrary 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 + b c >0
But 1+ ac ≯ 0
⇒ (a, c) ∉ R
So, the given relation is not transitive.
(iii) Consider aRb if|a| ≤ b.
Now for this relationwe have to check whether it is reflexive, transitive and symmetric.
Reflexivity:
Let a be anarbitrary element of R.
Then, a ∈ R [Since, |a|=a]
⇒ |a|≮ a
So, R is not reflexive.
Symmetry:
Let (a, b) ∈ R
⇒ |a| ≤ b
⇒ |b| ≰ a for all a, b ∈ R
⇒ (b, a) ∉ R
So, R is not symmetric.
Transitivity:
Let (a, b) ∈ R and (b, c) ∈ R
⇒ |a| ≤ b and |b| ≤ c
Multiplying the corresponding sides, we get
|a| × |b| ≤ bc
⇒ |a| ≤ c
⇒ (a, c) ∈ R
Thus, R is transitive.