Let T be the set of all triangles in a plane with R as relation in T given by $$R = {(T_1,T_2): T_1 \cong T_2}$$. Show that R is an equivalence relation.

Toolbox:
• A relation R in a set A is a equivalance relation if it is symmetric, reflexive and transitive.
• A relation R in a set A is called reflexive if $(a,a) \in R\;$ for every $\; a\in\;A$
• A relation R in a set A is called symmetric, if $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$
• A relation R in a set A is called transitive, if $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$
• If a triangle $T_1$ is similar to $T_2$ then $T_2$ will be similar to $T_1$
Step 1:
Reflexitivity :
We know every triangle is congruent to itself.
$\therefore (T,T)\in R$ all $T\in A$
$\Rightarrow R$ is reflexive.
Step 2:
Symmetry :
Let $(T_1,T_2)\in R$ then
$(T_1,R_2)\in R$
$T_1$ is congruent to $T_2$
$\Rightarrow T_2$ is congruent to $T_1$
$\Rightarrow (T_2,T_1)\in R$
So $R$ is symmetric.
Step 3:
Transitivity :
Let $T_1,T_2,T_3\in A$ .such that $(T_1,T_2)\in R$ and $(T_2,T_3)\in R$
Then $(T_1,T_2)\in R$ and $(T_2,T_3)\in R$
$\Rightarrow T_1$ is congruent to $T_2$ and $T_2$ is congruent to $T_3$
$\Rightarrow T_1$ is congruent to $T_3$
$\therefore (T_1,T_3)\in R$
So $R$ is transitive.
Hence $R$ is an equivalence relation.
answered Sep 24, 2013