# Does every circle in $\mathbb{R^2}$ contain a point with rational coordinates? [closed]

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

• Yes, $Q$ is dense un $R$ – Fede Poncio Apr 9 '18 at 1:32
• @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! – nra Apr 9 '18 at 1:33
• Do you require circles to have positive radius? – Eric Towers Apr 9 '18 at 1:33
• 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. – edm Apr 9 '18 at 1:35
• @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? – Carl Mummert Apr 9 '18 at 1:35

No, consider $$x^2+y^2=r^2$$ there are continuum many $r$ giving disjoint circles, but only countably many rational points.
• 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? – nra Apr 9 '18 at 1:38
• $r$ is a real number, it is the bijection with positive reals. – Rene Schipperus Apr 9 '18 at 1:40
• 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 $(0,\infty)$. – Xander Henderson Apr 9 '18 at 1:40
• @nra: this is a simple cardinality argument. There is no surjection from $Q^2$ to R – Carl Mummert Apr 9 '18 at 1:41
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.