Question

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric, or transitive.

Answer 1

Let A = {1, 2, 3, 4, 5, 6}.

A relation R is defined on set A as:

R = {(a, b): b = a + 1}

∴R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

We can find (a, a) ∉ R, where a ∈ A.

For instance,

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R

∴R is not reflexive.

It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Now, (1, 2), (2, 3) ∈ R

But,

(1, 3) ∉ R

∴R is not transitive

Hence, R is neither reflexive, nor symmetric, nor transitive.

Answer 2

(i) Reflexivity:

Letabeanarbitraryelementof R.Then,

a = a + 1 cannot be true for all a ∈ A.

⇒ (a, a) ∉ R 

So, R is not reflexive on A.

(ii) Symmetric:

Let (a, b) ∈ R

⇒ b = a + 1

⇒ −a = −b + 1

⇒ a = b − 1

Thus, (b, a) ∉ R

So, R is not symmetric on A.

(iii) Transitive: 

Let (1, 2) and (2, 3) ∈ R

⇒ (a, b) ∈ R and (b, c) ∈ R

⇒ b = a + 1 and c = b + 1

⇒ c = a+ 2

⇒ (a, c) ∉ R

So, R is not transitive on A.

Hence R is not reflexive, not symmetric and not transitive.