Test whether the following relations R1, R2 and R3 are (i) reflexive (ii) symmetric and (iii) transitive :
R3 on R defined by (a, b) ϵ R3⇔ a2 – 4 ab + 3b2 = 0.
Here, R1, R2, R3, and R4 are the binary relations.
So, recall that for any binary relation R on set A. We have,
R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
So, using these results let us start determining given relations.
We have
R3 on R defined by (a, b) ∈ R3⇔ a2 – 4ab + 3b2 = 0
Check for Reflexivity:
∀ a ∈ R,
(a, a) ∈ R3 needs to be proved for reflexivity.
If (a, b) ∈ R3, then we have
a2 – 4ab + 3b2 = 0
Replace b by a, we get
a2 – 4aa + 3a2 = 0
⇒ a2 – 4a2 + 3a2 = 0
⇒ –3a2 + 3a2 = 0
⇒ 0 = 0, which is true.
⇒ (a, a) ∈ R3
So, ∀ a ∈ R, (a, a) ∈ R3
∴ R3 is reflexive.
Check for Symmetry:
∀ a, b ∈ R
If (a, b) ∈ R3, then we have
a2 – 4ab + 3b2 = 0
⇒ a2 – 3ab – ab + 3b2 = 0
⇒ a (a – 3b) – b (a – 3b) = 0
⇒ (a – b) (a – 3b) = 0
⇒ (a – b) = 0 or (a – 3b) = 0
⇒ a = b or a = 3b …(1)
Replace a by b and b by a in equation (1), we get
⇒ b = a or b = 3a …(2)
Results (1) and (2) does not match.
⇒ (b, a) ∉ R3
∴ R3 is not symmetric.
Check for Transitivity:
∀ a, b, c ∈ R
If (a, b) ∈ R3 and (b, c) ∈ R3
⇒ a2 – 4ab + 3b2 = 0 and b2 – 4bc + 3c2 = 0
⇒ a2 – 3ab – ab + 3b2 = 0 and b2 – 3bc – bc + 3c2
⇒ a (a – 3b) – b (a – 3b) = 0 and b (b – 3c) – c (b – 3c) = 0
⇒ (a – b) (a – 3b) = 0 and (b – c) (b – 3c) = 0
⇒ (a – b) = 0 or (a – 3b) = 0
And (b – c) = 0 or (b – 3c) = 0
⇒ a = b or a = 3b And b = c or b = 3c
What we need to prove here is that, a = c or a = 3c
Take a = b and b = c
Clearly implies that a = c.
[∵ if a = b, just substitute a in place of b in b = c. We get, a = c]
Now, take a = 3b and b = 3c
If a = 3b
Substitute in b = 3c. We get
⇒ a = 9c, which is not the desired result.
⇒ (a, c) ∉ R3
∴ R3 is not transitive.