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Question Number 67688 by Rio Michael last updated on 30/Aug/19

$$Arelation\mathbb{R}definedby{}_{(x,y)}{R}_{(u,v)}\iff v^2-y^2=u^2-x^2$$$$showthatRisanequivalentRelation.$$

Commented by Prithwish sen last updated on 30/Aug/19

$${\mathrm{R}}_{(\mathrm{x},\mathrm{y})}\Rightarrow {\mathrm{y}}^2-{\mathrm{y}}^2={\mathrm{x}}^2-{\mathrm{x}}^2\Rightarrow \mathbf{symmetric}$$$${\mathrm{v}}^2-{\mathrm{y}}^2={\mathrm{u}}^2-{\mathrm{x}}^2\iff {\mathrm{y}}^2-{\mathrm{v}}^2={\mathrm{x}}^2-{\mathrm{u}}^2{\Rightarrow }_{(\mathrm{x},\mathrm{y})}{\mathrm{R}}_{(\mathrm{u},\mathrm{v})}{\Rightarrow }_{(\mathrm{u},\mathrm{v})}{\mathrm{R}}_{(\mathrm{x},\mathrm{y})}$$$$\mathrm{i}.\mathrm{e}\mathbf{reflexive}$$$${\mathrm{v}}^2-{\mathrm{y}}^2={\mathrm{u}}^2-{\mathrm{x}}^2\mathrm{and}{\mathrm{t}}^2-{\mathrm{v}}^2={\mathrm{s}}^2-{\mathrm{u}}^2$$$$\Rightarrow {\mathrm{t}}^2-{\mathrm{y}}^2=({\mathrm{t}}^2-{\mathrm{v}}^2)-({\mathrm{y}}^2-{\mathrm{v}}^2)=({\mathrm{s}}^2-{\mathrm{u}}^2)-({\mathrm{x}}^2-{\mathrm{u}}^2)$$$$={\mathrm{s}}^2-{\mathrm{x}}^2\mathrm{i}.\mathrm{e}{}_{(\mathrm{x},\mathrm{y})}{\mathrm{R}}_{(\mathrm{u},\mathrm{v})}{\mathrm{and}}_{(\mathrm{u},\mathrm{v})}{\mathrm{R}}_{(\mathrm{s},\mathrm{t})}{\Rightarrow }_{(\mathrm{x},\mathrm{y})}{\mathrm{R}}_{(\mathrm{s},\mathrm{t})}$$$$\mathrm{i}.\mathrm{e}\mathbf{transitive}$$$$\therefore \mathrm{The}\mathrm{relation}\mathrm{is}\mathrm{an}\mathrm{equivalance}\mathrm{relation}.$$

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