Let $T$ be the set of all triangles in a plane with $R$ a relation in $T$ given by $R = {(T_{1}, T_{2})} : T_{1}$ congruent to $T_{2}$ }. Show that $R$ is an equivalence relation.

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Solution

Since a relation $R$ in $T$ is said to be an equivalence relation if $R$ is reflexive, symmetric and transitive.
(i) Since every triangle is congruent itself
$∴$ $R$ is reflexive
$(T_{1}, T_{1}) ε R \Rightarrow T_{1}$ is congruent to itself $∴$ R reflexive
(ii) $(T_{1}, T_{2}) ϵ R \Rightarrow T_{1}$ is congruent to $T_{2}$
$\Rightarrow T_{2}$ is congruent to $T_{1}$
$\Rightarrow (T_{2}, T_{1}) ϵ R$
Hence $R$ is symmetric
(iii) Let $(T_{1}, T_{2}) ϵ R$ and $(T_{2}, T_{3}) ϵ R$
Then $T_{1}$ is congruent ot $T_{2}$ and $(T_{2})$ is congruent to $(T_{3})$
$\Rightarrow T_{1}$ is congruent to $T_{3}$
$\Rightarrow (T_{1}, T_{3}) ϵ R$
$∴$ $R$ is transitive
Hence $R$ is an equivalence relation.

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