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Show that the relation R defined in the set A of all triangles as `R={(T_(1),T_(2)):T_(1)` is similar to `T_(2)`}, is equivalence relation. |
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Answer» `R={(T_(1),T_(2)):T_(1)` is similar to `T_(2)`} R is reflexive since every triangle is similar to itself. Further, if `(T_(1),T_(2)) in R,` then `T_(1)` is similar to `T_(2)`. `implies T_(2)` is similar to `T_(2)` `implies (T_(2),T_(1)) in R`. Therefore, R is symmetric. Now, let `(T_(1),T_(2)),(T_(2),T_(3)) in R`. `implies T_(1)` is similar to `T_(2)` and `T_(2)` is similar to `T_(3)` `implies T_(1)` is similar to `T_(3)` `implies (T_(1), T_(3)) in R` Therefore, R is transitive. Thus, R is an equivalence relation. |
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