Let A be the set of all triangles in a plane. Show that the relation

Question:

Let $A$ be the set of all triangles in a plane. Show that the relation

$\mathrm{R}=\left\{\left(\Delta_{1}, \Delta_{2}\right): \Delta_{1} \sim \Delta_{2}\right\}$ is an equivalence relation on $\mathrm{A} .$

 

Solution:

Let $R=\left\{\left(\Delta_{1}, \Delta_{2}\right): \Delta_{1} \sim \Delta_{2}\right\}$ be a relation defined on $A$.

Now

$\underline{R}$ is Reflexive if $(\Delta, \Delta) \in \underline{R} \underline{\forall} \Delta \underline{A}$

We observe that for each $\Delta \in$ A we have,

$\Delta \sim \Delta$ since, every triangle is similar to itself.

$\Rightarrow(\Delta, \Delta) \in \mathrm{R} \forall \Delta \in \mathrm{A}$

$\Rightarrow \mathrm{R}$ is reflexive.

$\underline{R}$ is Symmetric if $\left(\Delta_{1}, \Delta_{2}\right) \in R \Rightarrow\left(\Delta_{2}, \Delta_{1}\right) \in \underline{B} \forall \Delta_{1}, \Delta_{2} \in A$

Let $\left(\Delta_{1}, \Delta_{2}\right) \in \mathrm{R} \forall \Delta_{1}, \Delta_{2} \in \mathrm{A}$

$\Rightarrow \Delta_{1} \sim \Delta_{2}$

$\Rightarrow \Delta_{2} \sim \Delta_{1}$

$\Rightarrow\left(\Delta_{2}, \Delta_{1}\right) \in \mathrm{R}$

$\Rightarrow \mathrm{R}$ is symmetric

$\underline{R}$ is Transitive if $\left(\Delta_{1}, \Delta_{2}\right) \in R$ and $\left(\Delta_{2}, \Delta_{3}\right) \in \underline{R} \Rightarrow\left(\Delta_{1}, \Delta_{3}\right) \in \underline{R} \forall \Delta_{1}, \Delta_{2}, \Delta_{3} \in A$

Let $\left(\Delta_{1}, \Delta_{2}\right) \in \mathrm{R}$ and $\left(\left(\Delta_{2}, \Delta_{3}\right) \in \mathrm{R} \forall \Delta_{1}, \Delta_{2}, \Delta_{3} \in \mathrm{A}\right.$

$\Rightarrow \Delta_{1} \sim \Delta_{2}$ and $\Delta_{2} \sim \Delta_{3}$

$\Rightarrow \Delta_{1} \sim \Delta_{3}$

$\Rightarrow\left(\Delta_{1}, \Delta_{3}\right) \in \mathrm{R}$

$\Rightarrow \mathrm{R}$ is transitive.

Since $R$ is reflexive, symmetric and transitive, it is an equivalence relation on $A$.

 

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