Is inversion logically consistent in neutral geometry?

Question

A description of inversion can be found here.

Since the process uses points, circles, lines, extending lines, right angles, and similar triangles, it seems to me that this process could be proven to be consistent in neutral geometry.

If it is helpful, Mohr-Mascheroni proved that any straightedge and compass construction can be performed using only a compass. Therefore, it seems you could perform this process using only the concept of points and distance/circles.

It seems like the largest issue might be the finite nature of reflected straight lines. From what I have seen, the popular approach is that when a straight line is reflected, it becomes a circular arc: a circle with a single point excluded (the origin of the circle of inversion).

Regarding "consistency":

From this source: "An axiomatic system is consistent if there is no statement such that both the statement and its negation are axioms or theorems of the axiomatic system."

"In an axiomatic system, an axiom is independent if it is not a theorem that follows from the other axioms."

Since a major aspect of neutral geometry is the exclusion of the parallel postulate, the question is intended to ask if the process of circle inversion (when considered as a statement or theorem) does not force independence from neutral geometry, but instead, establishes logic consistent with it.

Considering these factors, is it sensible to attempt to make or find a proof of consistency?

Answer 1

It is a theorem of neutral geometry that any point interior to a circle has at most one inverse with respect to that circle. But you need the parallel postulate to show that this inverse actually exists.

To see this, suppose $P'$ is a point inside a circle centered at $O$ with $P' \ne O$, and you want to construct the inverse $P$ of $P'$. Explicitly, the construction runs as follows:

1. Construct the line $OP'$.
2. Draw a perpendicular to that line through $P'$. Let $Q$ be the intersection of this perpendicular with the circle.
3. Construct the line $OQ$.
4. Draw a perpendicular to $OQ$ through $Q$; call this line $\ell$. Let $P$ be the intersection of $\ell$ with the line $OP'$.

In order for this construction to work, we need to know that $\ell$ is not parallel to $OP'$. The Euclidean proof of this fact runs something like:

• Draw a perpendicular line $m$ to $P'Q$ through $Q$. Then $m$ is parallel to $OP'$, because both $m$ and $OP'$ are perpendicular to $P'Q$.
• Because $P'$ and $O$ are distinct points, $m$ and $\ell$ are distinct lines.
• Because $m \ne \ell$ and $m$ and $\ell$ both pass through the point $Q$, it follows by the uniqueness of parallels that $\ell$ is not parallel to $OP'$.

But the last step is an application of the parallel postulate, so this is not valid neutrally. Indeed, the construction can fail in the hyperbolic plane.

For an explicit example of this in the Poincaré disk model, let $O=(0,0)$ and $P'=\left(\frac{1}{2},0\right)$, and take the circle $x^2+y^2=\frac{1}{2}$ centered at $O$. The hyperbolic line $OP'$ is the a segment of the $x$-axis; the perpendicular to that line through $P'$ is a segment of the circle $x^2-\frac{5}{2}x+y^2+1=0$. This perpendicular intersects the given circle $x^2+y^2=\frac{1}{2}$ when $x=\frac{3}{5}$, so we can take $Q=\left(\frac{3}{5},\sqrt{\frac{7}{50}}\right)$. But then the hyperbolic line $\ell$ ends up being a segment of the circle $\frac{6}{5}x+\frac{\sqrt{14}}{5}y=\frac{2}{3}(x^2+y^2+1)$. This circle does not intersect the $x$-axis, so the line $\ell$ does not intersect the line $OP'$.

That is, points need not have inverses. On the other hand, the point $P$ is defined as an intersection of two lines, which is unique whenever it exists. And the first three steps in the construction are all legitimate in neutral geometry. So this construction always produces at most one point, which completes the proof of the claim in my first paragraph.