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Two points in $\mathbb Q$ are called connected, if their euclidean distance is exactly $1$.
You are now allowed to also jump from any point in $\mathbb Q^n$ to another, if they are connected in $\mathbb Q^n$. The question now is: What is the smallest n that allows you to reach any point in $\mathbb Q$ , if you start in an abitrary point in $\mathbb Q$, if you are also allowed to jump to a point in $\mathbb Q^n$ , if they were connected in $\mathbb Q^n$.

For example it is to see, that you can connect all points in $\mathbb N$ already for n = 1. 1 is connected to 2 and 2 is connected to 3 and so on... Ultimately you know, that you can pick two arbitrary points of $\mathbb N$ and that there is a way to get from one point to the other by just hopping from one point to the other. n = 1 doesn't do the job for Q, because if you start in the point 1 you can actually reach all points $\mathbb N$, however you cannot reach the point 0.5 which is also in $\mathbb Q$

Does someone know the name of that problem? I really want to read up on that topic, especially for generalizations.

// I found the name of information to the problem: It is related to http://en.wikipedia.org/wiki/Beckman%E2%80%93Quarles_theorem

Imago
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    What do you mean "an n in N that allows you to reach all points in Q^n"? What do you mean by "reach"? – Harry Wilson Sep 08 '14 at 13:48
  • Intuitively, the answer for $N > 1$ should be yes. Also you shouldn't reuse $n$ twice. I am assuming $\mathbb Q^N$ and $n\in\mathbb N^N$. – AlexR Sep 08 '14 at 13:49
  • You can "reach" the rational $n$-sphere, so the question boils down to whether you can realize multiplication with $1/q$, $q \in \mathbb Q$. – Daniel Valenzuela Sep 08 '14 at 13:56
  • @DanValenzuela $|q|$ need not be in $\mathbb Q$ for $q\in\mathbb Q^N$, right? – AlexR Sep 08 '14 at 13:58
  • @AlexR (okay I will reformulate) but nevertheless: you can reach any point on the lattice $\mathbb Z^N$. So multplication with $1/q$ will still be sufficient. But thanks! – Daniel Valenzuela Sep 08 '14 at 14:03

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