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x-gt-y-y-2-gt-x-2-Does-such-a-pairing-exist-How-can-you-prove-it-




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

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