Let R = {(a, b) : a = b²} for all a, b ∈ N.

Show that R satisfies none of reflexivity, symmetry and transitivity.

We have, R = {(a, b) : a = b²} relation defined on N.

Now,

We observe that, any element a ∈ N cannot be equal to its square except 1.

⇒ (a,a) ∉ R ∀ a ∈ N

For e.g. (2,2) ∉ R ∵ 2 ≠ 2²

⇒ R is not reflexive.

Let (a,b) ∈ R ∀ a, b ∈ N

⇒ a = b²

But b cannot be equal to square of a if a is equal to square of b.

⇒ (b,a) ∉ R

For e.g., we observe that (4,2) ∈ R i.e 4 = 2² but 2 ≠ 4²⇒ (2,4) ∉ R

⇒ R is not symmetric

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ N

⇒ a = b² and b = c²

⇒ a ≠ c²

⇒ (a,c) ∉ R

For e.g., we observe that

(16,4) ∈ R ⇒ 16 = 4² and (4,2) ∈ R ⇒ 4 = 2²

But 16 ≠ 2²

⇒ (16,2) ∉ R

⇒ R is not transitive.

Thus, R is neither reflexive nor symmetric nor transitive.