InterviewSolution
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InterviewSolution
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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. |
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Answer» (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. |
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