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`

Explanation

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`.

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