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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. |
Answer» (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. |
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