Definition:Between (Geometry)

Definition
Betweenness is one of the undefined terms in Tarski's Geometry.

Intuitively, a point $b$ is between two others $a$ and $c$ if it lies on the line segment $a c$.

However as line segment has not yet been defined, we are not allowed to call upon it at this stage.

We offer an ostensive definition:


 * [[File:Betweenness.png]]

In the picture, point $b$ is between the two points $a, c$, and we write:
 * $\mathsf{B}abc$

However, point $d$ is not between the two points $a$ and $c$, and we write:
 * $\neg \left({\mathsf B a d c}\right)$

In Euclidean 2-Space
Define the following coordinates in the $xy$-plane:

where $a, b, c \in \R^2$.

Let:

Then:


 * [[File:Betweenness(Analytic Def'n).png]]


 * $\mathsf{B}abc \dashv \vdash \left({\Delta x_1 \Delta y_1 = \Delta x_2 \Delta y_2}\right) \land$


 * $\left({0 \le \Delta x_1 \Delta y_1 \land 0 \le \Delta x_2 \Delta y_2}\right)$

As a justification of this definition, consider the case where $\Delta x_1, \Delta x_2 \ne 0$.

Hence, the right triangles with hypotenuses $ab$ and $bc$ are similar.

Furthermore, the hypotenuses are parallel, because they have the same slope.

They are similarly oriented because $\Delta x$ is by construction parallel to the $x$-axis, $\Delta y$ to the $y$-axis.

They are touching because there are only three points under consideration.

Lastly, the inequalities assure that vertex $b$ lies between the two triangles, because otherwise the inequalities wouldn't hold.

In Euclidean $n$-Space
In the Euclidean $n$-dimensional case, consider three points $A$,$B$ and $C$. We say that $B$ is between $A$ and $C$ if $$\vec{AB} \cdot\vec{AC} = \norm{\vec{AB}} * \norm{\vec{AC}} $$

The intuition behind this definition comes from the fact that when we think that $B$ is between $A$ and $C$, we think of two things. The first thing is that $A$, $B$ and $C$ are collinear, so that $\cos{(\angle{BAC})}$ has the absolute value 1. Hence, $|\vec{AB} \cdot \vec{AC}| = \norm{\vec{AB}} * \norm{\vec{AC}}$

Secondly, the vectors $\vec{AB}$ and $\vec{AC}$ have the same direction. Hence, their dot product should be positive.