Check whether the relation R defined on the set A={1,2,3,4,5,6} as R= {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
We have the set A = {1, 2, 3, 4, 5, 6}
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.
We have
R = {(a, b): b = a + 1}
∵ Every a, b ∈ A.
And A = {1, 2, 3, 4, 5, 6}
The relation R on set A can be defined as:
Put a = 1
⇒ b = a + 1
⇒ b = 1 + 1
⇒ b = 2
⇒ (a, b) ≡ (1, 2)
Put a = 2
⇒ b = 2 + 1
⇒ b = 3
⇒ (a, b) ≡ (2, 3)
Put a = 3
⇒ b = 3 + 1
⇒ b = 4
⇒ (a, b) ≡ (3, 4)
Put a = 4
⇒ b = 4 + 1
⇒ b = 5
⇒ (a, b) ≡ (4, 5)
Put a = 5
⇒ b = 5 + 1
⇒ b = 6
⇒ (a, b) ≡ (5, 6)
Put a = 6
⇒ b = 6 + 1
⇒ b = 7
⇒ (a, b) ≠ (6, 7) [∵ 7 ∉ A]
Hence, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
Check for Reflexivity:
For 1, 2, …, 6 ∈ A [∵ A = {1, 2, 3, 4, 5, 6}]
(1, 1) ∉ R
(2, 2) ∉ R
…
(6, 6) ∉ R
So, ∀ a ∈ A, then (a, a) ∉ R.
⇒ R is not reflexive.
∴ R is not reflexive.
Check for Symmetry:
∀ 1, 2 ∈ A [∵ A = {1, 2, 3, 4, 5, 6}]
If (1, 2) ∈ R
Then, (2, 1) ∉ R
[∵ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}]
So, if (a, b) ∈ R, then (b, a) ∉ R
∀ a, b ∈ A
⇒ R is not symmetric.
∴ R is not symmetric.
Check for Transitivity:
∀ 1, 2, 3 ∈ A
If (1, 2) ∈ R and (2, 3) ∈ R
Then, (1, 3) ∉ R
[∵ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}]
So, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∉ R.
∀ a, b, c ∈ A
⇒ R is not transitive.
∴ R is not transitive.