$x\cdot y = \sqrt2$. What can be said about $x$ and $y$

Question

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

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 05 '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 05 '18 at 11:39

score 3 · Answer 1 · answered Nov 05 '18 at 11:41

3

Here is a geometric point of view of your problem.

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

answered Nov 05 '18 at 11:41

Sujit.Very nice!(The blue curve is a hyperbola:))+ – Peter Szilas Nov 05 '18 at 13:08

score 1 · Answer 2 · answered Nov 05 '18 at 11:36

1

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

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

answered Nov 05 '18 at 11:36

Michael Rozenberg

score 1 · Answer 3 · edited Nov 05 '18 at 11:40

1

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

edited Nov 05 '18 at 11:40

Arnaud D.

answered Nov 05 '18 at 11:38

Fred

score 0 · Answer 4 · answered Nov 05 '18 at 11:57

0

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

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

answered Nov 05 '18 at 11:57

gandalf61

4 Answers4