Q. 115.0( 4 Votes )

# On the set S of all real numbers, define a relation R = {(a, b) : a ≤ b}.

Show that R is (i) reflexive (ii) transitive (iii) not symmetric.

Answer :

Let R = {(a, b) : a ≤ b} be a relation defined on S.

Now,

We observe that any element x ∈ S is less than or equal to itself.

⇒ (x,x) ∈ R ∀ x ∈ S

⇒ R is reflexive.

Let (x,y) ∈ R ∀ x, y ∈ S

⇒ x is less than or equal to y

But y cannot be less than or equal to x if x is less than or equal to y.

⇒ (y,x) ∉ R

For e.g. , we observe that (2,5) ∈ R i.e. 2 < 5 but 5 is not less than or equal to 2 ⇒ (5,2) ∉ R

⇒ R is not symmetric

Let (x,y) ∈ R and (y,z) ∈ R ∀ x, y ,z ∈ S

⇒ x ≤ y and y ≤ z

⇒ x ≤ z

⇒ (x,z) ∈ R

For e.g. , we observe that

(4,5) ∈ R ⇒ 4 ≤ 5 and (5,6) ∈ R ⇒ 5 ≤ 6

And we know that 4 ≤ 6 ∴ (4,6) ∈ R

⇒ R is transitive.

Thus, R is reflexive and transitive but not symmetric.

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