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 ?

Commented by Yozzii last updated on 23/Dec/15

- 2 > - 4 \Rightarrow {(- 4)}^{2} > {(- 2)}^{2}

S u c h (x, y) = (- 2, - 4) f o r e x a m p l e .

S u p p o s e x > y a n d x, y \in R^{-} .

T h e n, 0 > x > y . B u t, f o r a \in R = R^{+} \cup R^{-}

\Rightarrow a^{2} ⩾ 0 a n d ∣ a ∣⩾ 0 . S i n c e y < x < 0

w e h a v e t h a t t h e m o d u l u s o f t h e

i n e q u a l i t y r e v e r s e s t h e s i g n s .

\Rightarrow∣ y ∣>∣ x ∣> 0 \Rightarrow∣ y ∣^{2} >∣ x ∣^{2} > 0 . T h i s i s

e q u i v a l e n t t o y^{2} > x^{2} > 0 .

∴ \exists (x, y) \in R^{2} ∣ x > y \land y^{2} > x^{2} .

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