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A-relation-R-defined-by-x-y-R-u-v-v-2-y-2-u-2-x-2-show-that-R-is-an-equivalent-Relation-

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Question Number 67688 by Rio Michael last updated on 30/Aug/19
A relation R defined by _((x,y)) R_((u,v)) ⇔ v^2 −y^2 = u^2 −x^2 show that R is an equivalent Relation.
$${A}\:{relation}\:\mathbb{R}\:{defined}\:{by}\:\:\:_{\left({x},{y}\right)} {R}_{\left({u},{v}\right)} \:\Leftrightarrow\:\:{v}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:{u}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$${show}\:{that}\:{R}\:{is}\:{an}\:{equivalent}\:{Relation}. \\ $$
Commented by Prithwish sen last updated on 30/Aug/19
R_((x,y)) ⇒ y^2 −y^2 = x^2 −x^2 ⇒symmetric v^2 −y^2 = u^2 −x^2 ⇔y^2 −v^2 =x^2 −u^2 ⇒_((x,y)) R_((u,v)) ⇒_((u,v)) R_((x,y)) i.e reflexive v^2 −y^2 = u^2 −x^2 and t^2 −v^2 = s^2 −u^2 ⇒t^2 −y^2 = (t^2 −v^2 )−(y^2 −v^2 ) = (s^2 −u^2 )−(x^2 −u^2 ) = s^2 −x^2 i.e _((x,y)) R_((u,v)) and_((u,v)) R_((s,t)) ⇒_((x,y)) R_((s,t)) i.e transitive ∴ The relation is an equivalance relation.
$$\:\mathrm{R}_{\left(\mathrm{x},\mathrm{y}\right)} \Rightarrow\:\:\mathrm{y}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \:\Rightarrow\boldsymbol{\mathrm{symmetric}} \\ $$$$\mathrm{v}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{u}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \Leftrightarrow\mathrm{y}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} −\mathrm{u}^{\mathrm{2}} \:\Rightarrow_{\left(\mathrm{x},\mathrm{y}\right)} \mathrm{R}_{\left(\mathrm{u},\mathrm{v}\right)} \Rightarrow_{\left(\mathrm{u},\mathrm{v}\right)} \mathrm{R}_{\left(\mathrm{x},\mathrm{y}\right)} \\ $$$$\mathrm{i}.\mathrm{e}\:\boldsymbol{\mathrm{reflexive}} \\ $$$$\mathrm{v}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} =\:\mathrm{u}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \mathrm{and}\:\mathrm{t}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} \:=\:\mathrm{s}^{\mathrm{2}} −\mathrm{u}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=\:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} \right)−\left(\mathrm{y}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} \right)\:=\:\left(\mathrm{s}^{\mathrm{2}} −\mathrm{u}^{\mathrm{2}} \right)−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{u}^{\mathrm{2}} \right) \\ $$$$=\:\mathrm{s}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \:\mathrm{i}.\mathrm{e}\:_{\left(\mathrm{x},\mathrm{y}\right)} \mathrm{R}_{\left(\mathrm{u},\mathrm{v}\right)} \mathrm{and}_{\left(\mathrm{u},\mathrm{v}\right)} \mathrm{R}_{\left(\mathrm{s},\mathrm{t}\right)} \Rightarrow_{\left(\mathrm{x},\mathrm{y}\right)} \mathrm{R}_{\left(\mathrm{s},\mathrm{t}\right)} \\ $$$$\mathrm{i}.\mathrm{e}\:\boldsymbol{\mathrm{transitive}} \\ $$$$\therefore\:\mathrm{The}\:\mathrm{relation}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equivalance}\:\mathrm{relation}. \\ $$

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