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Let `n`be a fixedpositive integer. Define a relation `R`on Z asfollows:`(a , b)R a-b`is divisibleby `ndot`Show that `R`is anequivalence relation on `Zdot` |
Answer» 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. |
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