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I am trying the following exercise:

Let $\mathcal {C}$ the set of points of a circle with center $O$ and radius $1$ with rational coordinates. Show that there exists a infinite set $\mathcal{D} \subsetneq \mathcal{C}$ ($\subsetneq \mathbb{Q}^2 \subsetneq \mathbb{R}^2$...) such that for any pair of points $\mathcal{D}$ the distance between these two points is irrational.

However I don't really have ideas to start.

Edit: Let $(\frac{p}{q},\frac{u}{v})\in \mathcal{C}$ such that $(pv)^2+(uq)^2=(qv)^2$, then $(pv,uq,qv)$ is a Pythagorean triple. and reciprocally if we have a Pythagorean triple $(a,b,c)$, then $(\frac{a}{c},\frac{b}{c}) \in \mathcal{C}$.

Therefore, $$\mathcal{C} = \{(\frac{2uv}{u^2+v^2},\frac{u^2-v^2}{u^2+v^2}), (u,v) \in \mathbb{Z}^2 \} \bigcup \{(\frac{u^2-v^2}{u^2+v^2},\frac{2uv}{u^2+v^2}), (u,v) \in \mathbb{Z}^2 \}$$

Edit 2

  • Between two rational numbers there is always an irrational number

Proof. Let $(r,r')\in \mathbb{Q^2}$ with $r<r'$ and $x=r+\frac{\sqrt{2}}2(r'-r)$; Then $x\in ]r,r'[$ (because $0<\frac{\sqrt{2}}2<1$). T herefore $\sqrt{2}(\frac{r-r'}2)\notin \mathbb{Q}$ so that $x\notin \mathbb{Q}$.

  • We consider the Euclidean plane. Let $n\geq 3$ an integer, there exist n points such that the distance between any two of these points is irrational.

Proof. Let $A_n$ a point with coordinates $(n,n^2)$ the distance between $A_n$ et $A_m$ is $$\sqrt{(n-m)^2+(n^2-m^2)^2}=\vert n-m \vert\sqrt{(n+m)^2+1}$$ if $n \neq m$ then $\vert n-m \vert \neq 0$ and $n+m\geq 1$ wich implies $(n+m)^2+1$ is not a square number . Therefore $\vert n-m \vert\sqrt{(n+m)^2+1}$ is an irrational number.

  • Now I know that the distance between two points in the unit circle is : $$2\vert \sin( \frac{\widehat{AOB}}2) \vert$$

Thank you in advance,

Yiorgos S. Smyrlis
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  • Set of points in $\Bbb R^2$ or in $\Bbb Q^2$? Or in some other metric space? Is that space complete or not? – Asaf Karagila Jan 31 '14 at 23:21
  • @AsafKaragila $\mathcal{D} \subsetneq \mathcal{C} \subsetneq \mathbb{Q}^2 \subsetneq \mathbb{R}^2$ ... –  Feb 01 '14 at 11:25
  • Julien, that's something to add to your question. – Asaf Karagila Feb 01 '14 at 11:29
  • @AsafKaragila why ? –  Feb 01 '14 at 11:31
  • Uhh... because it's an information people trying to solve your problem might be interested in? And having them work on a solution, only to see the comments afterward is nothing less than rude? – Asaf Karagila Feb 01 '14 at 11:32
  • @AsafKaragila Uhh..I didn't understand your sentence it is why I said 'why?' –  Feb 01 '14 at 11:35
  • @AsafKaragila Good evening, Do you have an idea for this exercise ? Thanks in advance. –  Feb 01 '14 at 22:09
  • No, sorry. But I do recall seeing it before. You should construct this by induction, proving a lemma that given finitely many points, there is always another one which has an irrational distance from all of them. Then use that to have the induction go through every finite step, and at the end you have an infinite set as wanted. I'm not sure, off hand, how to prove this sort of lemma, though. – Asaf Karagila Feb 01 '14 at 22:11
  • @AsafKaragila Hum okay I will search, Thank you for your time! –  Feb 01 '14 at 22:22
  • (Seeing it before might not have been on this site! But it may have been... I don't know...) – Asaf Karagila Feb 01 '14 at 22:24
  • Note about your edit, you want the points to be in $\cal C$, not just points in $\Bbb Q^2$. – Asaf Karagila Feb 03 '14 at 22:56
  • @AsafKaragila Sorry, I don't understand where is my mistake ? –  Feb 03 '14 at 23:01
  • I'm not saying that there is a mistake. I'm saying that since you wish to prove a lemma "Given $n$ points, there is another with irrational distance from all of them" and then by induction define a subset of $\cal C$ (suppose $x_i$ were defined for $i<n$, let $x_n$ be a point guaranteed from the lemma). That's the suggestion I gave you anyway. But in that case the lemma doesn't ensure that the point found is from $\cal C$, but just from $\Bbb Q^2$. So you need to prove that given $n$ points in $\cal C$ there is a point in $\cal C$ whose distance from each of the $n$ points is irrational. – Asaf Karagila Feb 03 '14 at 23:03
  • @AsafKaragila okay thanks, yes for: "Given n points, there is another with irrational distance from all of them" my edit prove the lemma I think.. And yes I am working on your suggestion. –  Feb 03 '14 at 23:17

1 Answers1

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For integers $u,v$ not both zero let $$ P(u,v) := \Bigl( \frac{2uv}{u^2+v^2}, \frac{u^2-v^2}{u^2+v^2} \Bigr) \in \cal C. $$ Now calculate that the distance between $P(u,v)$ and $P(u',v')$ is $$ \frac{2\left|uv'-u'v\right|}{\sqrt{(u^2+v^2)({u}^2+{v'}^2)}}. $$ Therefore one construction of $\cal D$ is to use all $P(u,v)$ with $0<u<v$ and $u^2+v^2$ prime: by a classical theorem of Fermat, every prime congruent to $1 \bmod 4$ appears exactly once; and the product of any two distinct primes has an irrational square root, so every two points of $\cal D$ are separated by an irrational distance.

(If instead you wanted all distances to be rational, just choose all coprime $u,v$ such that $u^2+v^2$ is itself a square.)

Noam D. Elkies
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