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Let "T" be the set of all triangles in...

Let `"T"` be the set of all triangles in a plane with `"R"` as relation in `"T"` given by
`"R"={("T"_1,"T"_2):("T")_1 "congruent to" "T"_2}`
. Show that `"R"` is an equivalence relation.

Text Solution

Verified by Experts

`R=[(T_1,T_2):T_1` IS Congruent to` T_2`]
Check reflexive
since every triangle is congruent to itself
Triangle T is congruent to Triangle T
`(T,T)inR`
So R is reflexive
check symmetric
If `T_1`is congruent to `T_2`
then `T_2`is congruent to `T_1`
So, if `(T_1,T_2)inR`,Then `(T_2,T_1)inR`
Hence ,R is symmetric
Check transivity
If `T_1` is congruent to `T_2`,&`T_2` is congruent to `T_3`
then `T_1`is congruent to `T_3`
So ,If `(T_1,T_2),(T_2,T_3)inR`
then `(T_1,T_3)inR`
SO,R is Transitive
Since `R` is reflexive,symmetric & transitive.
Therefore ,R is an equivalence relation
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