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.
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.