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Show that the relation R in the set `A={1,2,3,4,5}` given by ` R={(a,b):|a-b|` is even }, is an equivalence relation. |
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Answer» `A= {1,2,3,4,5}` `R={(a,b):|a-b|` is even } It is clear that for any clement `a in A,` we have `|a - a| = 0` (which is even). Therefore, R is reflexive. Let `(a,b) in R`. `implies |a-b|` is even, ` implies |-(a-b)|=|b-a|` is also even `implies (b,a) in R` Therefore, R is symmetric. Now, let `(a,b) in R ` and `(b,c) in R.` ` implies |a-b|` is even and `|b-c|` is even `implies (a-b)` is even and `(b-c)` is even ` " " ` (assuming that `a gt b gt c`) `implies (a-c)=(a-b)+(b-c)` is even ` " " ` [Sum of two even integers is even] ` implies |a-c|` is even `implies (a,c) in R` Therefore, R is transitive. Hence, R is an equivalence relation. |
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