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Question

# Let O be the origin. We define a relation between two points P and Q in a plane if OP = OQ. Show that the relation, so defined is an equivalence relation.

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Solution

## Let A be the set of all points in a plane such that $A=\left\{P:P\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{in}\mathrm{the}\mathrm{plane}\right\}\phantom{\rule{0ex}{0ex}}\mathrm{Let}R\mathrm{be}\mathrm{the}\mathrm{relation}\mathrm{such}\mathrm{that}R=\left\{\left(P,Q\right):P,Q\in A\mathrm{and}OP=OQ,\mathrm{where}O\mathrm{is}\mathrm{the}\mathrm{origin}\right\}$ We observe the following properties of R. Reflexivity: Let P be an arbitrary element of R. The distance of a point P will remain the same from the origin. So, OP = OP $⇒\left(P,P\right)\in R\phantom{\rule{0ex}{0ex}}\mathrm{So},R\mathrm{is}\mathrm{reflexive}\mathrm{on}A.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Symmetry}:\mathrm{Let}\left(P,Q\right)\in R\phantom{\rule{0ex}{0ex}}⇒OP=OQ\phantom{\rule{0ex}{0ex}}⇒OQ=OP\phantom{\rule{0ex}{0ex}}⇒\left(Q,P\right)\in R\phantom{\rule{0ex}{0ex}}\mathrm{So},R\mathrm{is}\mathrm{symmetric}\mathrm{on}A.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Transitivity}:\mathrm{Let}\left(P,Q\right),\left(Q,R\right)\in R\phantom{\rule{0ex}{0ex}}⇒OP=OQ\mathrm{and}OQ\mathit{=}OR\phantom{\rule{0ex}{0ex}}⇒OP=OQ=OR\phantom{\rule{0ex}{0ex}}⇒OP=OR\phantom{\rule{0ex}{0ex}}⇒\left(P,R\right)\in R\phantom{\rule{0ex}{0ex}}\mathrm{So},R\mathrm{is}\mathrm{transitive}\mathrm{on}A.$ Hence, R is an equivalence relation on A.

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