Subset of $\mathbb{Q}$ Let $S= \{x_0,\dots,x_n\}$ be a finite subset of $[0,1]$ , $x_0=0$ and $x_1=1$ such that every distance between pair of elements of $S$ occurs at least twice, except for the distance $1$, then we are to show that $S$ is a subset of $\mathbb{Q}$.
 A: Let $V$ be the $\mathbb{Q}$-vector space generated by $S$ and let $\leq$ be any total order on $V$ compatible with addition (forall $x,y,z \in V$, if $x \leq y$ then $x+z \leq y+z$, and more generally, any reasoning you are used to involving $+$ and $\leq$ is valid).
Then, for this particular total order, there is a unique pair $(x_i,x_j) \in S^2$ such that $x_i - x_j$ is the greatest distance between pair of elements of $S$ :
Since $S$ is finite and $\leq$ is total, the set $\{x-y, (x,y)\in S^2\}$ is finite so it has a maximal element $x_i - x_j$ for some pair $(x_i,x_j) \in S^2$.
Then we show it is unique : suppose there is $(x_k,x_l) \in V^2$ such that $(x_i - x_j) = (x_k - x_l)$.
Then $(x_i - x_j) + (x_i - x_j) = (x_k - x_l) + (x_i - x_j) = (x_k - x_j) + (x_i - x_l)$.
Since $(x_i - x_j)$ is maximal, $(x_k - x_j) \leq  (x_i - x_j)$ and $(x_i - x_l) \leq  (x_i - x_j)$. If any of those two inequalities were strict, we would get the contradiction $(x_i - x_j) + (x_i - x_j) &lt (x_i - x_j) + (x_i - x_j)$, so they have to be equalities, which implies that $(x_k,x_l) = (x_i,x_j)$.

Now, the hypothesis on $S$ says that the only pairs $(x_i,x_j)$ such that $x_i-x_j$ is unique are the pairs $(0,1)$ and $(1,0)$. So it implies that for any total order on $V$ compatible with addition, this greatest distance is always $1$ or $-1$.
So we only need to show that if $V \neq \mathbb{Q}$, there must exist total orders $\leq$ on $V$ such that the greatest distance for $\leq$ is not $1$ or $-1$ :
Since $S$ generates $V$, and $(x_1 = 1)$ is free, we can add elements of $S$ to $(x_1)$ to form  a basis $(e_1, \ldots, e_m) = (x_{i_1}, \ldots, x_{i_{m-1}},x_1)$ of $V$.


*

*Pick the lexicographical order induced by this basis. It is caracterised by the property that $0 &lt \sum a_i e_i$ if and only if the first nonzero coefficient is positive.
In particular, $(x_{i_1} - x_0) = x_{i_1} > \pm x_1 = \pm (x_1 - x_0) = \pm 1$, so neither $1$ nor $-1$ can be the greatest distance for $\leq$.

*Define a linear application $f : V \to \mathbb{R}$ with $f(1) = 1$ and $f(x_{i_k}) = \pi^k$. Since $\pi$ is transcendental, this map $f$ is injective. Pick the order defined by $x \leq y \Leftrightarrow f(x) \leq_\mathbb{R} f(y)$, where $\leq_\mathbb{R}$ is the usual order on $\mathbb{R}$. But then, since $f(x_{i_1}) >_\mathbb{R} f(x_1)$, we have again $(x_{i_1} - x_0) = x_{i_1} > \pm x_1 = \pm (x_1 - x_0) = \pm 1$.
