InterviewSolution
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InterviewSolution
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Let `S` be the set of all real numbers. Then the relation `R= ` `{(a,b):1+abgt0}` on `S` isA. Reflexive and symmetric but not transitiveB. Reflexive and transitive but not symmetricC. Symmetric and transitive but not reflexiveD. None of the above is true |
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Answer» Correct Answer - A We observe the following properties: Reflexivity: Let a be an arbitrary element of R. Then, `a in R` `implies 1+a.a = 1+a^(2)gt0 " "[because a^(2) gt 0 " for all a"inR]` `implies (a,a)inR_(1)" [By def. of "R_(1)]` Thus, `(a,a) in R_(1)` for all `a in R`. So `R_(1)` is reflexive on R. Symmetry: Let `(a,b) in R`. Then, `(a,b) in R_(1)` `implies 1 + ab gt 0` `implies 1+ba gt 0 " "[because ab = ba " for all a, b"in R]` `implies (b,a) in R_(1) " [By def. of "R_(1)]` Thus, (a, b) `in R_(1) implies (b,a) in R_(1)` for all a, b `in R`. So, `R_(1)` is symmetric on R. Transitivity: We observe that `(1, 1//2) in R_(1)` and `(1//2, -1) in R_(1)` but `(1, -1) cancelin R_(1)` because `1 + 1xx(-1) = 0 cancelgt0`. So, `R_(1)` is not transitive on R. |
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