For real numbers `x` and `y` , define `x\ R\ y` iff `x-y+sqrt(2)` is an irrational number. Then the relation `R` is (a) reflexive (b) symmetric (c) transitive (d) none of these

Options

A. reflexive
B. symmetric
C. transitive
D. None of these

Answer

A

Explanation

`xRy iff (x - y + sqrt(2))` is an irrational number.
Let `(x, x) in R`.
Then, `x-x+sqrt(2)=sqrt(2)` which is an irrational number,
`therefore xRx, AAx inR`
`therefore` R is an reflexive relation.
`xRyimplies(x-y+sqrt(2))` is an irrational number.
`implies-(y-x-sqrt(2))` is an irrational number.
`yRximplies(y-x+sqrt(2))` is an irrational number.
So, `xRycancelimpliesyRx thereforeR` is not a symmetric relation.
Let (1, 2) `in R`, then `(1-2+sqrt(2))` is an irrational number.
implies `(sqrt(2)-1)` is an irrational number. and (2, 3) `in R`, then `(2-3+sqrt(2))` is an irrational number.
`implies (sqrt(2)-1)` is an irrational number.
`(1, 3)inRimplies(1-3+sqrt(2))` is an irrational number.
implies `(sqrt(2)-2)` is an irrational number.
So, `(1, 2) in R and (2, 3) in R cancelimplies (1, 3) in R` (by any way)
`therefore R` is not transitive relation.