Equidistance is Independent of Betweenness

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Theorem

Let $\mathcal G$ be a formal systematic treatment of geometry containing only:

The language and axioms of first-order logic, and the disciplines preceding it
The undefined terms of Tarski's Geometry (excluding equidistance)
Some or all of Tarski's Axioms of Geometry.


In $\mathcal G$, equidistance $\equiv$ is necessarily an undefined term with respect to betweenness $\mathsf B$.


Proof

Our assertion is that $\equiv$ cannot be defined in terms of $\mathsf B$.

Seeking a contradiction, assume that it can.

Call this assumption $\left({A}\right)$.

If $\left({A}\right)$ holds, it must hold in all systems.

Let one such system be $\left({\R^2, \mathsf B_1, \equiv_1}\right)$ where:

$\R^2$ is the cartesian product of the set of real numbers with itself
$\mathsf B_1$ is a ternary relation of betweenness
$\equiv_1$ is a quaternary relation of equidistance


Let $\mathcal G$ be the discipline preceding the given discipline, where $\mathcal G$ is as defined above (excluding both $\equiv$ and $\mathsf B$).


Define $\mathsf B_1$ as follows:

Define the following coordinates in the $xy$-plane:

\(\displaystyle a\) \(=\) \(\displaystyle \left({x_1, x_2}\right)\) $\quad$ $\quad$
\(\displaystyle b\) \(=\) \(\displaystyle \left({y_1, y_2}\right)\) $\quad$ $\quad$
\(\displaystyle c\) \(=\) \(\displaystyle \left({z_1, z_2}\right)\) $\quad$ $\quad$

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

Let:

\(\displaystyle \Delta x_1\) \(=\) \(\displaystyle x_3 - x_2\) $\quad$ $\quad$
\(\displaystyle \Delta x_2\) \(=\) \(\displaystyle x_2 - x_1\) $\quad$ $\quad$
\(\displaystyle \Delta y_1\) \(=\) \(\displaystyle y_2 - y_1\) $\quad$ $\quad$
\(\displaystyle \Delta y_2\) \(=\) \(\displaystyle y_3 - y_2\) $\quad$ $\quad$

Then:

Betweenness(Analytic Def'n).png
$\mathsf{B}abc \dashv \vdash \left({\Delta x_1 \Delta y_1 = \Delta x_2 \Delta y_2}\right) \land$
$\left({0 \le \Delta x_1 \Delta y_1 \land 0 \le \Delta x_2 \Delta y_2}\right)$


Define $\equiv_1$ as follows:

Define the following coordinates in the $xy$-plane:

$a = \left({x_1,x_2}\right)$
$b = \left({y_1,y_2}\right)$
$c = \left({z_1,z_2}\right)$
$d = \left({u_1,u_2}\right)$

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

Equidistance(Analytic Def'n).png

$ab \equiv cd \dashv \vdash \left[{\left({x_1-y_1}\right)^2 + \left({x_2 - y_2}\right)^2 = \left({z_1-u_1}\right)^2 + \left({z_2-u_2}\right)^2}\right]$


Now, define the isomorphism $\phi$ on $\left({\R^2, \mathsf B_2, \equiv_2}\right)$ as:

$\phi: \R^2 \to \R^2$ on $\left({\R^2, \mathsf B_1, \equiv_1}\right), \left({x_1, x_2}\right) \mapsto \left({x_1, 2 x_2}\right)$



Now consider the system:

$\left({\R^2, \mathsf B_2, \equiv_2}\right)$

where $\mathsf B_2$ and $\equiv_2$ are the relations defined as above, but on the elements in the images of $\mathsf B_1$ and $\equiv_1$, respectively.

Observe that $\mathsf B_1$ and $\mathsf B_2$ coincide, because in:

$\left({x_1 - y_1}\right) \cdot \left({2 y_2 - 2 z_2}\right) = \left({2 x_2 - 2 y_2}\right) \cdot \left({y_1 - z_1}\right) \land$
$\left({0 \le \left({x_1 - y_1}\right) \cdot \left({y_1 - z_1}\right)}\right) \land \left({0 \le \left({2 x_2 - 2 y_2}\right) \cdot \left({2 y_2 - 2 z_2}\right)}\right)$

we can simply factor out the $2$ and divide both sides of the equality of inequality by $2$.


But consider the elements:

$p_1 = \left({0, 0}\right)$
$p_2 = \left({0, 1}\right)$
$p_3 = \left({1, 0}\right)$


Observe that $p_1 p_2 \equiv_1 p_1 p_3$:

$\left({0 - 0}\right)^2 + \left({0 - 1}\right)^2 = \left({0 - 1}\right)^2 + \left({0 - 0}\right)^2$


But $\neg \left({p_1 p_2 \equiv_2 p_1 p_3}\right)$:

$\left({0 - 0}\right)^2 + \left({0 - 2}\right)^2 \ne \left({0 - 1}\right)^2 + \left({0 - 0}\right)^2$

But both $\left({\R^2, \mathsf B_1, \equiv_1}\right)$ and $\left({\R^2, \mathsf B_2, \equiv_2}\right)$ are both models of $\mathcal G$.



Recall that if $\left({A}\right)$ holds, it must hold in all systems.

But it does not.

Hence $\left({A}\right)$ is false, from Proof by Contradiction.

$\blacksquare$


Also see


Sources