Is there any circle in the $\Bbb{R} \times \Bbb{R}$ such that coordinates of each its point are rational numbers?


I know I have asked an elementary question, but because of I had left Mathematics for $14$ years and I have forgotten everything, but I would ask another related question, how can it be proved by concept of measure of a set in the Euclidean plane of course not by sets in the x-axis and y-axis?

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    $\begingroup$ Nope. there are only countablely many points with rational coordinates. Unless you include degenerate circle with zero radius, all other circle has uncountablely many points. $\endgroup$ – achille hui May 21 '17 at 20:15
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    $\begingroup$ No (unless you consider a degenerate circle of radius $0$). By the Intermediate Value Theorem, any curve between two points $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1\ne x_2$ passes, for every $x$ between $x_1$ and $x_2$, through some point $(x,y)$. $\endgroup$ – Hagen von Eitzen May 21 '17 at 20:17
  • $\begingroup$ Is the "Euclidean page" (your title) double sided ? $\endgroup$ – Jean Marie May 21 '17 at 20:19
  • $\begingroup$ It is a joke based on your "Euclidean page" that should be the "Euclidean plane" $\endgroup$ – Jean Marie May 21 '17 at 20:39
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    $\begingroup$ Please don't vandalize your posts. $\endgroup$ – Glorfindel Jun 30 '17 at 8:09

Another way to think about it: $\Bbb Q^2$ (the set of all ordered pairs of rational numbers) is countable, as will any subset of this set. However, a circle with center $\langle x_0,y_0\rangle$ and radius $r>0$ can be obtained as the image of the injective function $[0,2\pi)\to\Bbb R^2$ given by $$t\mapsto\bigl\langle x_0+r\cos t,y_0+r\sin t\bigr\rangle.$$ Since $[0,2\pi)$ is uncountable, so is its image, and so its image is not a subset of $\Bbb Q^2.$

  • $\begingroup$ @Alireza: Actually, continuity was never used, only cardinality (size) of sets. $\endgroup$ – Cameron Buie May 23 '17 at 22:18
  • $\begingroup$ Ah! It looks like you mean to use measure of sets. I'm not sure that will work, unfortunately. I'll have to think about that. $\endgroup$ – Cameron Buie May 24 '17 at 11:09

Consider the unit circle centered at the origin. it contains the points $(1,0), (\frac1{\sqrt{2}},\frac1{\sqrt{2}})$. The first of these points is rational, the second irrational.

Scaling the radius by either a rational or irrational takes the two points to $\mathbb{Q}, \mathbb{R\setminus Q}$ or $\mathbb{R\setminus Q}, \mathbb{Q}$ respectively.

Similarly for translating the centre.

  • $\begingroup$ The first answer is much more precise but I find this one simpler and probably more useful for OP and an idiot like me. Thanks! $\endgroup$ – Jerry Qu May 23 '17 at 9:34

Consider 1) the intermediate value theorem and 2) the density of the irrationals in the reals.

Let $C = $ the circle centered at $(a,b)$ with radius $r$ then $C = \{(x,y)| (x-a)^2 + (x -b)^2 = r^2\}$.

Now let $f:[0,2\pi) \rightarrow C; f(\theta) = (a + r*\cos \theta, b+r*\sin \theta)$. That is $f$ maps the incidental angle of a point to the cooresponding point of the circle.

Let $f_x(\theta) = a+r\cos \theta$ is a function mapping the angle to the $x$ coordinate and $f_y(\theta = a + r\sin \theta$ is a function mapping the angle to the $y$ coordinate of the circle. These are both continuous functions.

$f_x(0) = a + r$ and $f_x(\pi/2) = a$. So by the intermediate value theorem for any $c \in (a, a+ r)$ there is $\theta \in (0, \pi/2)$ so that $f_x(\theta) = c$.

And as the irrationals are dense in $\mathbb R$ we may choose $c$ so that $c$ is irrational.

So that means $(f_x(\theta), f_y(\theta)) = (c, b+ r\sin (\theta)) \in C$ and $c $ is not rational.

So no circle exists with only rational coordinates.

Basically, you should develop an intuitive sense that no continuous curve through $\mathbb R^n$ can have only rational or irrational terms.


In general, to be a "shape" there is some continuous $f:[0,1]\rightarrow shape$ so that $f(t) = (x,y) \in shape$. For any $t_0 < t_1$ with $f(t_0) = (x_0, y_0)$ and $f(t_1) = (x_1, y_1)$ then as $t$ "passes" from $t_0$ to $t_1$, $x$ "passes" from $x_0$ to $x_1$ and $y$ "passes" from $y_0$ to $y_1$... and they pass through irrational points.


But is possible to have a shape so that for all $(x,y)$ in the shape either $x$ is rational of $y$ is rational and they are never both irrational. Example a square with rational corners.


No. Suppose there is a circle whose points only lie at rational coordinates, with center $(\frac{p}{q}, \frac{r}{s})$ and radius $R$ (they must be rational coordinates- why?), then there is obviously also a point on the circle at $(\frac{p}{q}+\frac{R}{\sqrt{2}}, \frac{r}{s}+\frac{R}{\sqrt{2}})$ by trigonometry. If $R$ is rational*, then obviously the coordinates of that point are irrational.

In addition, we can't make a square whose points are only rational, by the same reasoning. If two points have rational coordinates, the differences in coordinates will be rational also, but the distance between the points can be irrational. Can you take it from there?

Edit In the case of the circle, if $R$ is not a rational multiple of $\sqrt{2}$, then the method holds. If $R$ is a rational multiple of $\sqrt{2}$, choose the point on the circle $(\frac{p}{q}+\frac{R \cdot \sqrt{3}}{2}, \frac{r}{s}+\frac{R}{2})$

  • $\begingroup$ I saw that my proof was insufficient- see my edit. $\endgroup$ – Harry May 22 '17 at 14:45
  • $\begingroup$ I'm not sure how one would prove it with sets, but @JonMark Perry gave a more succinct version of my proof. $\endgroup$ – Harry May 22 '17 at 18:06

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