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Given, $x,y\in\Bbb R$ and $x\cdot y=\sqrt{2}$ .Can $x$ and $y$ be taken as $\sqrt[4]{2}$?

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closed as unclear what you're asking by Arnaud D., Namaste, Batominovski, GNUSupporter 8964民主女神 地下教會, Connor Harris Nov 5 '18 at 14:33

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  • $\begingroup$ Yes, they can...but also you can take $\;x=\sqrt2,\,y=1\;$ , or $\;y=-4,\,x=-\frac1{2\sqrt2}\;$ , etc. The only things you can say for sure is that they both are non-zero and both are positive or both are negative. $\endgroup$ – DonAntonio Nov 5 '18 at 11:37
  • $\begingroup$ The set of all such pairs $(x,y)$ forms an equilateral hyperbola in the usual Cartesian coordinate plane. $\endgroup$ – Dave L. Renfro Nov 5 '18 at 11:39
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Here is a geometric point of view of your problem.

All the points lying on the blue curve is the solution of your equation $xy=\sqrt{2}$

enter image description here

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  • $\begingroup$ Sujit.Very nice!(The blue curve is a hyperbola:))+ $\endgroup$ – Peter Szilas Nov 5 '18 at 13:08
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Yes, of course, but $x=1$ and $y=\sqrt2$ are also valid.

You can take $y=\frac{\sqrt2}{x}$ for all $x\neq0.$

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Yes, $x=y= \sqrt[4]{2}$ satisfy the equation $xy= \sqrt{2}$. But there are infinitely many other solutions of the equation $xy= \sqrt{2}$.

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$|x+y| \ge 2\sqrt[4]{2}$

and one or both of $x$ and $y$ are irrational.

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