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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. |
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Answer» Here, `R = {(a,b):a,b in R and (a-b)` is divisible by `5}` 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 `5`. `=> (-(b-a))` is divisible by `5`. `=> (b-a)` is also divisble by `5`. `:. (b,a) in R`. `:. R` is symmetric. Since `(a,b) in R and (b,c) in R` `=> (a-b)` is divisible by `5` & `(b-c)` is divisible by `5`. `=> (a-b+(b-c))` is divisible by `5`. `=> (a-c)` is divisible by `5`. `:. (a,c) in R`. `:. R` is transitive. As `R` is reflexive, symmetric and transitive, `R` is an equivalence relation. |
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