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

Here, `R = {(a,b):a,b in R and (a-b)` is divisible by `5n}`
For all `a in R`,
`=> (a-a) =0` and `0` is divisible by `5`.
`:. R` is refexive.
Since in `R` for every `(a,b) in R`
`=> (a-b)` is divisible by `n`.
`=> (-(b-a))` is divisible by `n`.
`=> (b-a)` is also divisble by `n`.
`:. (b,a) in R`.
`:. R` is symmetric.
Since `(a,b) in R and (b,c) in R`
`=> (a-b)` is divisible by `n` &  `(b-c)` is divisible by `n`.
`=> (a-b+(b-c))` is divisible by `n`.
`=> (a-c)` is divisible by `n`.
`:. (a,c) in R`.
`:. R` is transitive.
As `R` is reflexive, symmetric and transitive, `R` is an equivalence relation.