

InterviewSolution
Saved Bookmarks
1. |
For the relation R1 defined on R by the rule (a, b) R1 1 + ab > 0. Prove that : (a, b) R1 and (b,c) R1 (a, c) R1 is not true for all a, b, c R. |
Answer» To prove : (a, b) R1 and (b,c) R1 (a, c) R1 is not true for all a, b, c R. Given, R1 = {(a, b): 1 + ab > 0} Let a = 1, b = - 0.5, c = - 4 Here, (1, -0.5) R1 [∵ 1+(1×-0.5) = 0.5 > 0] And, (-0.5, -4) R1 [∵ 1+(-0.5×-4) = 3 > 0] But, (1, -4) ∉ R1 [∵ 1+(1×-4) = - 3 < 0] ∴ (a, b) R1 and (b,c) R1 (a, c) R1 is not true for all a, b, c R Hence Proved. NOTE : Here R1 is a relation whereas R denotes a real number. |
|