question_answer

Let T be the set of all triangles in a plane. Let us define a relation \[R=\{({{T}_{1}},\,\,{{T}_{2}}):{{T}_{1}}\,\,is\,\,similar\,\,to\,\,{{T}_{2}};{{T}_{1}},\,{{T}_{2}}\in T\}.\] Show that R is an equivalence relation.

Answer:

Given relation is             \[R=\{({{T}_{1}},\,\,{{T}_{2}}):{{T}_{1}}\,\,is\,\,similar\,\,to\,\,{{T}_{2}};{{T}_{1}},\,{{T}_{2}}\in T\},\] where T is the set of all triangles in a plane. We know that, two triangles are said to be similar, if they have same shape.                           Reflexive Let \[{{T}_{1}}\in T\] be any arbitrary element. As we know that every triangle is similar to itself.           So, \[({{T}_{1}},\,{{T}_{2}})\in R\] \[\therefore \] R is reflexive. Symmetric Let \[{{T}_{1}},\,{{T}_{2}}\in T\] such that \[({{T}_{1}},\,\,{{T}_{2}})\in R\] \[\Rightarrow \] \[{{T}_{1}}\] is similar to \[{{T}_{2}}.\] \[\Rightarrow \] \[{{T}_{2}}\] is similar to \[{{T}_{1}}\]                                         [\[\therefore \] two triangles are similar to each other] \[\Rightarrow \,\,\,\,({{T}_{2}},\,{{T}_{1}})\in R\] \[\therefore \] R is symmetric.            . Transitive Let \[\,\,\,\,{{T}_{1}},\,\,{{T}_{2}},\,\,{{T}_{3}}\in T\] such that             \[({{T}_{1}},\,\,{{T}_{2}})\in R\] and \[({{T}_{2}},\,\,{{T}_{3}})\in R\] \[\Rightarrow \] \[{{T}_{1}}\] is similar to \[{{T}_{2}}\] and is similar to \[{{T}_{3}}\]. \[\Rightarrow \] \[{{T}_{1}}\] is similar to \[{{T}_{3}}.\,\,\,\,\Rightarrow \,\,\,\,({{T}_{1}},\,\,{{T}_{3}})\in R\] \[\therefore \] R is transitive. Thus relation R is reflexive, symmetric and transitive, so R is an equivalence relation.