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.