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Is it true that any circle in $\mathbb{R^2}$ contains a point with rational coordinates? what about any simple closed curve?

If it is, could you please help me with the proof?

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    $\begingroup$ Yes, $Q$ is dense un $R$ $\endgroup$ – Fede Poncio Apr 9 '18 at 1:32
  • $\begingroup$ @FedePoncio I know that one. I also know that Q^2 is dense in R^2, but still don't know why that implies what I asked! $\endgroup$ – nra Apr 9 '18 at 1:33
  • $\begingroup$ Do you require circles to have positive radius? $\endgroup$ – Eric Towers Apr 9 '18 at 1:33
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    $\begingroup$ For arbitrary closed curves, no. You can easily make a rectangle frame such that for each straight line part, say a horizontal line, the $y$-coordinate is irrational. $\endgroup$ – edm Apr 9 '18 at 1:35
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    $\begingroup$ @Fede Poncio: could you expand on how that answers the question? Is it easy to see that $ (x-\pi)^2 + (y-\pi)^2= e$ contains a rational point? $\endgroup$ – Carl Mummert Apr 9 '18 at 1:35
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No, consider $$x^2+y^2=r^2$$ there are continuum many $r$ giving disjoint circles, but only countably many rational points.

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  • $\begingroup$ How would you go about proving that? How can I find a bijection between disjoint circles in $R^2$ and say, R or a set with cardinality continuum? $\endgroup$ – nra Apr 9 '18 at 1:38
  • $\begingroup$ $r$ is a real number, it is the bijection with positive reals. $\endgroup$ – Rene Schipperus Apr 9 '18 at 1:40
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    $\begingroup$ The bijection is given to you already! Each $r \in (0,\infty)$ defines a distinct circle. There are continuum many points in any interval, including $(0,\infty)$. $\endgroup$ – Xander Henderson Apr 9 '18 at 1:40
  • $\begingroup$ @nra: this is a simple cardinality argument. There is no surjection from $Q^2$ to R $\endgroup$ – Carl Mummert Apr 9 '18 at 1:41
  • $\begingroup$ yep, I see... :-D thanks everyone... $\endgroup$ – nra Apr 9 '18 at 1:42
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For circles, no. Pick any number $r$ with $r^2$ irrational. Then the circle $$x^2+y^2=r^2$$ does not have any rational solution, or else $r^2$ is rational.

For arbitrary closed curves, even more counterexamples exist. Pick a rectangle such that for each straight line part, say a horizontal line of the rectangle, the $y$-coordinate is irrational, while for vertical line, the $x$-coordinate is irrational.

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