Let `Z` be the set of all integers and `R` be the relation on `Z` defined as `R={(a , b); a ,\ b\ in Z ,` and `(a-b)` is divisible by `5.}` . Prove that `R` is an equivalence relation.

Explanation

The given relation is ` R={(a,b):a,b in Z " and "a-b " is divisible by " 5}.`
We shall prove that R is reflexive, symmetric and transitive.
(i) R is reflexive as as for any ` x in Z`, we have `x-x=0` and 0 is divisible by 5
`implies x-x` is divisible by 5
`implies (x,x) in R, AA x in Z`
` :. `R is reflexive.
(ii) R is symmetric As `(x,y) in R, " where " x,y in Z`
`implies x-y ` is divisible by 5 `" " ` [By defnition of R]
`implies x-y=5lambda " for some " lambda in Z`
`implies y-x=5(-lambda)`
`implies y-x` is also divisible by 5
` :. R` is symmetric.
(iii) R is transitive AS `(x,y) in R,` where `x,y in Z`
`implies x-y` is divisible by 5
`implies x-y=5 lambda_(1)` for some `lambda_(1) in Z`
Again, for `(y,z) in R` where, `y,z in Z`
`implies y-z` is divisible by 5
`implies y-z=5lambda_(2)` for some `lambda_(2) in Z`
Now, `(x-y) +(y-z)=5lambda_(1)+5lambda_(2)`
`impliesx-z=5(lambda_(1)+lambda_(2))`
`implies x-z` is divisible by 5 for some `(lambda_(1)+lambda_(2)) in Z`
` :. ` R is transitive.
Since, R is reflexive, symmetric and transitive.
Therefore, it is an equivalence relation.