A relation R is defined on the set of integers as follows :
`a "" R "" b hArr(a - b)` , is divisible by 6 where a, b, `in` I. prove that R is an equivalence relation.

Explanation

(i) For each a `in` I
`a- a = 0 = 0 xx 6`
`rArr a - a` , is divisible by 6
`rArr _(a)R_(a), AA a in I`
`:.` R is reflexive.
(ii) Let a, b `in` I such that `""_(a)R_(b)`
`:. ""_(a)R_(b) rArr a-b`, is divisible by 6
`rArr - (b - a)`, is divisible by 6
`rArr ""_(b)R_(a)`
`:.` R is symmetric.
(iii) Let a, b, c `in` I such that `""_(a)R_(b) and"" _(b)R_(c)`
Now `""_(a)R_(b) and ""_(b)R_(c)`
`rArr a - b `, and b - c, both are divisible by 6
`rArr [(a - b) + (b - c)]`, is divisibe by 6
`rArr(a - c)`, is divisible by 6
`rArr ""_(a)R_(c)`
`:.` R is transitive.
Hence, R is reflexive, symmetric and transitive `rArr` R is an equivalence relation. Hence Proved.