Show that the relation R = {(a, b) : a > b} on N is transitive but neither reflexive nor symmetric.
We have, R = {(a, b) : a > b} relation defined on N.
Now,
We observe that, any element a ∈ N cannot be greater than itself.
⇒ (a,a) ∉ R ∀ a ∈ N
⇒ R is not reflexive.
Let (a,b) ∈ R ∀ a, b ∈ N
⇒ a is greater than b
But b cannot be greater than a if a is greater than b.
⇒ (b,a) ∉ R
For e.g., we observe that (5,2) ∈ R i.e 5 > 2 but 2 ≯ 5 ⇒ (2,5) ∉ R
⇒ R is not symmetric
Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ N
⇒ a > b and b > c
⇒ a > c
⇒ (a,c) ∈ R
For e.g., we observe that
(5,4) ∈ R ⇒ 5 > 4 and (4,3) ∈ R ⇒ 4 > 3
And we know that 5 > 3 ∴ (5,3) ∈ R
⇒ R is transitive.
Thus, R is transitive but not reflexive not symmetric.