Question Number 3874 by Filup last updated on 23/Dec/15
$${x}>{y} \\ $$ $${y}^{\mathrm{2}} >{x}^{\mathrm{2}} \\ $$ $$ \\ $$ $$\mathrm{Does}\:\mathrm{such}\:\mathrm{a}\:\mathrm{pairing}\:\mathrm{exist}? \\ $$ $$\mathrm{How}\:\mathrm{can}\:\mathrm{you}\:\mathrm{prove}\:\mathrm{it}? \\ $$
Commented byYozzii last updated on 23/Dec/15
$$−\mathrm{2}>−\mathrm{4}\Rightarrow\left(−\mathrm{4}\right)^{\mathrm{2}} >\left(−\mathrm{2}\right)^{\mathrm{2}} \\ $$ $${Such}\:\left({x},{y}\right)=\left(−\mathrm{2},−\mathrm{4}\right)\:{for}\:{example}. \\ $$ $${Suppose}\:{x}>{y}\:{and}\:{x},{y}\in\mathbb{R}^{−} . \\ $$ $${Then},\:\mathrm{0}>{x}>{y}.\:{But},\:{for}\:{a}\in\mathbb{R}=\mathbb{R}^{+} \cup\mathbb{R}^{−} \\ $$ $$\Rightarrow{a}^{\mathrm{2}} \geqslant\mathrm{0}\:{and}\:\mid{a}\mid\geqslant\mathrm{0}.\:{Since}\:{y}<{x}<\mathrm{0} \\ $$ $${we}\:{have}\:{that}\:{the}\:{modulus}\:{of}\:{the}\: \\ $$ $${inequality}\:{reverses}\:{the}\:{signs}. \\ $$ $$\Rightarrow\mid{y}\mid>\mid{x}\mid>\mathrm{0}\Rightarrow\mid{y}\mid^{\mathrm{2}} >\mid{x}\mid^{\mathrm{2}} >\mathrm{0}.\:{This}\:{is} \\ $$ $${equivalent}\:{to}\:{y}^{\mathrm{2}} >{x}^{\mathrm{2}} >\mathrm{0}. \\ $$ $$\therefore\:\exists\left({x},{y}\right)\in\mathbb{R}^{\mathrm{2}} \mid{x}>{y}\:\wedge\:{y}^{\mathrm{2}} >{x}^{\mathrm{2}} . \\ $$