Correct Answer - A::C::D
`aRb iff a - b` is an even integer, `a, b in I`
a - a = 0 (even integer)
`therefore (a,a) in R, AAa inI`
`therefore` R is reflexive relation.
Let `(a, b) in R implies (a-b)` is an even integer.
implies - (b-a) is an even integer
implies (b - a) is an even integer
implies `(b,a) in R`
`therefore R` is symmetric relation.
Now, let `(a, b) in R` and `(b, c) in R`
Then, (a-b) is an even integer and (b-c) is an even integer.
So, let `a-b=2x_(1),x_(1)inI`
and `b-c=2x_(2),x_(2)inI`
`therefore (a-b)+(b-c)=2(x_(1)+x_(2))`
`implies (a-c)=2(x_(1)+x_(2))impliesa-c=2x_(3)`
`therefore` (a-c) is an even integer.
`therefore` aRb and bRc implies aRc. So, R is transitive relation,
Hence, R is an equivalence relation.
