Question

Let T be the set of all triangles in a plane with R as relation in T given by R = {(T1 , T2 ) :T1 ≅T2 } . Show that R is an equivalence relation.

Answer 1

For the sake of convenient , i use ' ~ ' as congruent symbol...
( 1 ) Reflexivity
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Clearly T ~ T ( any triangle is congruent to itself )
=> ( T , T ) € R ( € => belongs to )
=> R is reflextive

( 2 ) Symmetry
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Let ( T1 , T2 ) € R be any arbitrary element
To show : ( T2 , T1 ) € R
As ( T1 , T2 ) € R
=> T1 ~ T2 , by definition of R
=> T2 ~ T1
=> ( T1 , T2 ) € R
=> R is symmetric

( 3 ) Transitivity
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Let ( T1 , T2 ) € R and ( T2 , T3 ) € R be any arbitrary elements ( to show : ( T1 , T3) € R )
As ( T1 , T2) € R and ( T2 , T3 ) € R , then by definition of R we must have
T1 ~ T2 and T2 ~ T3
=> T1 ~ T3
=> ( T1 , T3 ) € R
=> R is transitive

Since R is reflexive , symmetric and transitive. Hence R is an equivalence relation.