Question -
Answer -
(i) Given R1 on Q0 definedby (a, b) ∈ R1 ⇔ a = 1/b.
Reflexivity:
Let a be an arbitrary element of R1.
Then, a ∈ R1
⇒ a ≠1/a for all a ∈ Q0
So, R1 is not reflexive.
Symmetry:
Let (a, b) ∈ R1
Then,(a, b) ∈ R1
Therefore we can write‘a’ as a =1/b
⇒ b = 1/a
⇒ (b, a) ∈ R1
So, R1 is symmetric.
Transitivity:
Here, (a, b) ∈R1 and (b, c) ∈R2
⇒ a = 1/b and b = 1/c
⇒ a = 1/ (1/c) = c
⇒ a ≠ 1/c
⇒ (a, c) ∉ R1
So, R1 is not transitive.
(ii) Given R2 on Z definedby (a, b) ∈ R2 ⇔ |a – b| ≤ 5
Now we have checkwhether R2 is reflexive, symmetric and transitive.
Reflexivity:
Let a be anarbitrary element of R2.
Then, a ∈ R2
On applying the givencondition we get,
⇒ | a−a | = 0 ≤ 5
So, R1 is reflexive.
Symmetry:
Let (a, b) ∈ R2
⇒ |a−b| ≤ 5 [Since, |a−b| = |b−a|]
⇒ |b−a| ≤ 5
⇒ (b, a) ∈ R2
So, R2 is symmetric.
Transitivity:
Let (1, 3) ∈ R2 and (3, 7) ∈R2
⇒|1−3|≤5 and |3−7|≤5
But |1−7| ≰5
⇒ (1, 7) ∉ R2
So, R2 is not transitive.
(iii) Given R3 on R definedby (a, b) ∈ R3 ⇔ a2 –4ab + 3b2 = 0.
Now we have checkwhether R2 is reflexive, symmetric and transitive.
Reflexivity:
Let a be anarbitrary element of R3.
Then, a ∈ R3
⇒ a2 − 4a × a+ 3a2= 0
So, R3 is reflexive
Symmetry:
Let (a, b) ∈ R3
⇒ a2−4ab+3b2=0
But b2−4ba+3a2≠0 for all a, b ∈ R
So, R3 is not symmetric.
Transitivity:
Let (1, 2) ∈ R3 and (2, 3) ∈ R3
⇒ 1 − 8 + 6 =0 and 4 – 24 + 27 = 0
But 1 – 12+ 9 ≠ 0
So, R3 is not transitive.