Can I use the following definitions?
let $(a_1,a_2,..,a_{n+1}) \in \Bbb{R}^{n+1}$, $(a_1,a_2,..,a_{n+1})$ is a point of $\Bbb{R}^n$ if $$a_1=1 \wedge \forall i \in \{2,...,n+1\} (a_i \in \Bbb{R})$$
let $(a_1,a_2,..,a_{n+1}) \in \Bbb{R}^{n+1}$, $(a_1,a_2,..,a_{n+1})$ is a vector of $\Bbb{R}^n$ if $$a_1=0 \wedge \forall i \in \{2,...,n+1\} (a_i \in \Bbb{R})$$
I consider:
- $\mbox{Points}_{\Bbb{R}^n}:=\{(a_1,a_2,..,a_{n+1}) \in \Bbb{R}^{n+1}|(a_1,a_2,..,a_{n+1}) \mbox{ is a point of }\Bbb{R}^n\}$
- $\mbox{Vectors}_{\Bbb{R}^n}:=\{(a_1,a_2,..,a_{n+1}) \in \Bbb{R}^{n+1}|(a_1,a_2,..,a_{n+1}) \mbox{ is a vector of }\Bbb{R}^n\}$
- $\mathfrak{f}:\mbox{Points}_{\Bbb{R}^n} \times \mbox{Vectors}_{\Bbb{R}^n} \to \mbox{Points}_{\Bbb{R}^n}, (a,b)\to \mathfrak{f}(a,b)=a+b$
then $$\mbox{Points}_{\Bbb{R}^n} \mbox{ is affine space on }\mbox{Vectors}_{\Bbb{R}^n}$$ Thanks in advance!
Usually one takes the last coordinate or zeroth coordinate to indicate which points belong to the affine space and what points are vectors in the associated vector space of translations. So that the index range $1,...,n$ remains untouched for the $n$ coordinates.
Note that this is a concrete parametrization of the idea of affine space, the abstract definition does not have this 0-1-indicator.
And the usual name for the affine point space is $\mathbb A^n(\mathbb R)$.
Add: Since there seems to be some confusion in the comments of the question: The reason that an additional coordinate is added in this fashion is that general movements, that is, combinations of translations and linear mappings, can be combined into one matrix multiplication, since $$ \begin{bmatrix}A&b\\0&1\end{bmatrix}\, \begin{bmatrix}P&|&v\\1&|&0\end{bmatrix} = \begin{bmatrix}AP+b&|&Av\\1&|&0\end{bmatrix} $$ i.e., the translation only applies to the points while the linear transformation applies both to points and vectors. This idea is generally used in computer graphics, from Postscript to OpenGL.
Another nice use of this definition is if one wants to check if three points in 2D are on the same line, one only has to check $$ 0=\det\begin{bmatrix}P_1&P_2&P_3\\1&1&1\end{bmatrix}=\det\begin{bmatrix}P_1&P_2-P_1&P_3-P_2\\1&0&0\end{bmatrix} $$