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જો $\mathrm{T}$ એ સમતલમાં આવેલા બધા જ ત્રિકોણનો ગણ હોય અને $\mathrm{R}$ એ $\mathrm{T}$ પરનો સંબંધ $\mathrm{R} =\left\{\left( \mathrm{T} _{1}, \mathrm{T} _{2}\right): \mathrm{T} _{1}\right.$ એ ${{T_2}}$ ને એકરૂપ છે $\}$ દ્વારા વ્યાખ્યાયિત હોય, તો સાબિત કરો કે $\mathrm{R}$ એ સામ્ય સંબંધ છે.
Solution
$\mathrm{R}$ is reflexive, since every triangle is congruent to it self.
Further, $\left( \mathrm{T} _{1}, \,\mathrm{T}_{2}\right) \in \mathrm{R} \Rightarrow \mathrm{T} _{1}$ is congruent to $\mathrm{T} _{2} \Rightarrow \mathrm{T} _{2}$ is congruent to $\mathrm{T} _{1} \Rightarrow\left( \mathrm{T} _{2}, \mathrm{T} _{1}\right) \in \mathrm{R} .$
Hence, $\mathrm{R}$ is symmetric.
Moreover, $\left( \mathrm{T} _{1},\, \mathrm{T} _{2}\right),\left( \mathrm{T} _{2}, \,\mathrm{T} _{3}\right) \in \mathrm{R} \Rightarrow \mathrm{T} _{1}$ is congruent to $\mathrm{T} _{2}$ and $\mathrm{T} _{2}$ is congruent to $\mathrm{T} _{3} \Rightarrow \mathrm{T} _{1}$ is congruent to $\mathrm{T} _{3} \Rightarrow\left( \mathrm{T} _{1}, \,\mathrm{T} _{3}\right) \in \mathrm{R}$.
Therefore, $\mathrm{R}$ is an equivalence relation.
