Question

Let `n` be a fixed positive integer. Define a relation `R` on Z as follows: `(a , b)R a-b` is divisible by `ndot` Show that `R` is an equivalence relation on `Zdot`

Answer 1

aRb iff n|(a-b)| i.e. (a-b) is divisible by n.
Reflexivity a - a = 0 which is divisible by n.
So, `(a, a) in R, AA a in I`
`therefore` R is reflexive relation.
Symmetry Let `(a, b) in R`
Then, `(a, b) in R implies (a-b)` is divisible by n.
implies -(b - a) is divisible by n.
implies (b - a) is divisible by n.
implies (b, a) `in R`
`therefore` R is symmetric relation.
Transitivity Let (a, b) `in R`, (b, c) `in R`, then (a - b) and (b - c) are divisible by n.
implies (a - b) + (b - c) = `n(k_(1) + k_(2))`
`implies a - c = n (k_(1) + k_(2))`
implies (a - c) is divisible by n.
implies `(a, c) in R`
`therefore R` is transitive relation.
`therefore` R is an equivalence relation.