# $x\cdot y = \sqrt2$. What can be said about $x$ and $y$ [closed]

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

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

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• 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. – DonAntonio Nov 5 '18 at 11:37
• The set of all such pairs $(x,y)$ forms an equilateral hyperbola in the usual Cartesian coordinate plane. – Dave L. Renfro Nov 5 '18 at 11:39

Here is a geometric point of view of your problem.

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

• Sujit.Very nice!(The blue curve is a hyperbola:))+ – Peter Szilas Nov 5 '18 at 13:08

Yes, of course, but $$x=1$$ and $$y=\sqrt2$$ are also valid.

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

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

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

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