Let S be the set of all real numbers. Then , the relation `R = {(a, b) : 1+ ab gt 0}` on S is

Options

A. Reflexive and symmetric but not transitive
B. Reflexive and transitive but not symmetric
C. Symmetric and transitive but not reflexive
D. None of the above is true

Answer

A

Explanation

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.