Show that the relation R = {(a, b) : a > b} on N is transitive but nei

1.	Show that the relation R = {(a, b) : a > b} on N is transitive but neither reflexive nor symmetric.
Answer» 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.

Show that the relation R = {(a, b) : a > b} on N is transitive but neither reflexive nor symmetric.

Answer»

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.