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Question Number 3874 by Filup last updated on 23/Dec/15

$$x>y$$ $$y^2>x^2$$ $$$$ $$\mathrm{Does}\mathrm{such}\mathrm{a}\mathrm{pairing}\mathrm{exist}?$$ $$\mathrm{How}\mathrm{can}\mathrm{you}\mathrm{prove}\mathrm{it}?$$

Commented byYozzii last updated on 23/Dec/15

$$-2>-4\Rightarrow {(-4)}^2>{(-2)}^2$$ $$Such(x,y)=(-2,-4)forexample.$$ $$Supposex>yandx,y\in {\mathbb{R}}^{-}.$$ $$Then,0>x>y.But,fora\in \mathbb{R}={\mathbb{R}}^{+}\cup {\mathbb{R}}^{-}$$ $$\Rightarrow a^2⩾0and\mid a\mid ⩾0.Sincey<x<0$$ $$wehavethatthemodulusofthe$$ $$inequalityreversesthesigns.$$ $$\Rightarrow \mid y\mid >\mid x\mid >0\Rightarrow \mid y{\mid }^2>\mid x{\mid }^2>0.Thisis$$ $$equivalentto{y}^2>x^2>0.$$ $$\therefore \mathrm{\exists }(x,y)\in {\mathbb{R}}^2\mid x>y\wedge y^2>x^2.$$

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