Arrow Space Construction Is there a paper or book that has rigorously constructed the space of "arrow vectors" and shown that it is a vector space?  
By "arrow vectors" I mean oriented line segments in Euclidean n-space.  This space will be over the field of real numbers and the operations of vector addition and scalar multiplication are defined as usual:


*

*Vector Addition is defined via the parallelogram method


*Scalar multiplication is defined by scaling a line segment by the amount of the scalar.  Where multiplication by positive numbers preserves direction and negative numbers reverses it.

I'm just wondering how far anyone has followed the heuristic.
 A: All what follows is purely speculative, since I don't have enough time to check all the details, but I'd be glad to help if someone is stuck in a proof.
Let's first define what the arrow space is. Given an Euclidean space $S$ (a model of  Tarski's axioms ), consider the following equivalence relation $\operatorname{E}$ on $S^2$ :
$(P,Q) \operatorname{E} (P',Q') :\Longleftrightarrow$ the segments $[PQ']$ and $[P'Q]$ have the same middle point (ie if $PQQ'P'$ is a parallelogramm).
A vector is an element of $S^2/\operatorname{E}$ (an equivalence class). For convenience, one writes $\vec{PQ}$ for the equivalence class of $(P,Q)$.
You'll need the following lemma :
lemma For any points $P,Q,R$, there is a (unique) point $S$ such that $\vec{PQ} = \vec{RS}$.
eg, here's a way to define the addition of 2 vectors $\vec{PQ}$ and $\vec{P'Q'}$ :
Let $S$ be the point such that $\vec{QS} = \vec{P'Q'}$, one defines $\vec{PQ} + \vec{P'Q'} := \vec{PS}$. 
proposition Addition is well defined!
As for product of a vector by a scalar, you'll need a suited field. If you work with a model $(S,\operatorname{B},\equiv)$ of Tarski's axiom, then take any line $\mathcal{D} = (PQ)$. We'll work with the following (ordered) field $(F,+,\cdot,\leqslant)$ :


*

*domain : $F := \{\vec{AB} \,|\, A,B \in \mathcal{D}\}$ . Hence, $F = \{\vec{PA} \,|\, A \in \mathcal{D}\}$ .

*addition : vector addition, as above.
$(F,+)$ is a commutative group with neutral element $\vec{PP}$.

*order : the order is induced by the betweenness relation $\operatorname{B}$ and the addition
$$\vec{PA} \leqslant \vec{PB} :\Longleftrightarrow \operatorname{B}(P,C,Q) \vee \operatorname{B}(P,Q,C) \textrm{ where }C \textrm{ is such that} \vec{AB} = \vec{PC}$$

*multiplication : that's the trickiest part, it mimics Thales's theorem. Let $x,y \in F$, say $x = \vec{PA}$ and $y = \vec{PB}$. Here's how one computes $x \cdot y$ :


Up to swaping $x$ and $y$, we can wlog assume that there is $C \notin \mathcal{D}$ such that $CQ \equiv PB$. Let $D$ be the intersection of $(PC)$ and the parallel to $(CQ)$ passing trough $A$. The product $x \cdot y$ will be equal to $\vec{PE}$, where $E \in \mathcal{D}$ satisfies $PE \equiv AD$. There are two possibilities for $E$, $\vec{PE} \leqslant \vec{PP}$ or  $\vec{PE} \geqslant \vec{PP}$:


*

*If both $\vec{PA}$ and $\vec{PB}$ are $\leqslant \vec{PP}$ (resp. $\geqslant \vec{PP}$ ) one takes $E$ such that $\vec{PE} \geqslant \vec{PP}$.

*Otherwise, one takes $E$ such that $\vec{PE} \leqslant \vec{PP}$.


proposition $(F,+,\cdot,\leqslant)$ is a real closed field with zero $\vec{PP}$ and unit $\vec{PQ}$.
remark $F=F_{P,Q}$ seems to depend on the choice of $P$ and $Q$, but for any $P' \neq Q' \in S$ such that $PQ \equiv P'Q'$, one gets $F_{P,Q} \cong F_{P',Q'}$ canonically.
