Question

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 1

(i) Reflexivity

Consider a as an arbitrary element on the set S

So we get a ≤ a where (a, a) ∈ R

Hence, R is reflective.

(ii) Transitivity

Consider a, b and c ∈ S where (a, b) and (b, c) ∈ S

We get

(a, b) ∈ R => a ≤ b and (b, c) ∈ R => b ≤ c

Based on the above equation we get

(a, c) ∈ R => a ≤ c

Hence, R is transitive.

(iii) Non symmetry

We know that

(5, 6) ∈ R => 5 ≤ 6

In the same way

(6, 5) ∈ R => 6 ≰ 5

Hence, R is non symmetric.