Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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.

$$\:\mathrm{R}_{\left(\mathrm{x},\mathrm{y}\right)} \Rightarrow\:\:\mathrm{y}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \:\Rightarrow\boldsymbol{\mathrm{symmetric}} \\ $$$$\mathrm{v}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{u}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \Leftrightarrow\mathrm{y}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} −\mathrm{u}^{\mathrm{2}} \:\Rightarrow_{\left(\mathrm{x},\mathrm{y}\right)} \mathrm{R}_{\left(\mathrm{u},\mathrm{v}\right)} \Rightarrow_{\left(\mathrm{u},\mathrm{v}\right)} \mathrm{R}_{\left(\mathrm{x},\mathrm{y}\right)} \\ $$$$\mathrm{i}.\mathrm{e}\:\boldsymbol{\mathrm{reflexive}} \\ $$$$\mathrm{v}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} =\:\mathrm{u}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \mathrm{and}\:\mathrm{t}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} \:=\:\mathrm{s}^{\mathrm{2}} −\mathrm{u}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=\:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} \right)−\left(\mathrm{y}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} \right)\:=\:\left(\mathrm{s}^{\mathrm{2}} −\mathrm{u}^{\mathrm{2}} \right)−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{u}^{\mathrm{2}} \right) \\ $$$$=\:\mathrm{s}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \:\mathrm{i}.\mathrm{e}\:_{\left(\mathrm{x},\mathrm{y}\right)} \mathrm{R}_{\left(\mathrm{u},\mathrm{v}\right)} \mathrm{and}_{\left(\mathrm{u},\mathrm{v}\right)} \mathrm{R}_{\left(\mathrm{s},\mathrm{t}\right)} \Rightarrow_{\left(\mathrm{x},\mathrm{y}\right)} \mathrm{R}_{\left(\mathrm{s},\mathrm{t}\right)} \\ $$$$\mathrm{i}.\mathrm{e}\:\boldsymbol{\mathrm{transitive}} \\ $$$$\therefore\:\mathrm{The}\:\mathrm{relation}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equivalance}\:\mathrm{relation}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com