Existence of an infinite set included in a circle with rational coordinates.

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 : $$2\vert \sin( \frac{\widehat{AOB}}2) \vert$$

• 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 $\mathcal{D} \subsetneq \mathcal{C} \subsetneq \mathbb{Q}^2 \subsetneq \mathbb{R}^2$ ... – user119228 Feb 1 '14 at 11:25
• Julien, that's something to add to your question. – Asaf Karagila Feb 1 '14 at 11:29
• @AsafKaragila why ? – user119228 Feb 1 '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 1 '14 at 11:32

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.)