2
$\begingroup$

Elliptic parallel property: Given a line $L$ and a point $a\notin L$, there exist no lines parallel to $L$ passing through $a$.

Euclidean parallel property: Given a line $L$ and a point $a\notin L$, there exists exactly one line parallel to $L$ passing through $a$.

Hyperbolic parallel property: Given a line $L$ and a point $a\notin L$, there exist at least two lines parallel to $L$ passing through $a$.

Note that each of these properties states existence of particular number of parallel lines for every line $L$ and every point $a\notin L$. It means that negation of one of these properties does not imply one of the other two without additional assumptions.

As you know, the whole set of Hilbert's axioms describes Euclidean geometry. If we replace parallel postulate with it's negation we get hyperbolic geometry. In other words, assuming Hilbert's axioms for neutral geometry (i.e. without parallel postulate or its negation) we can prove that euclidean or hyperbolic parallel property holds. However, the proof relies on congruence and continuity axioms. What if we omit congruence and continuity axioms and restrict only to incidence and order? Can we then prove the existence of at least one parallel line to a given line passing through a given point? Or there is a model of ordered geometry in which there exist a line $L$, point $a\notin L$ and no lines parallel to $L$ passing through $a$ ? We can ask even more: Does there exist a model of ordered geometry in which elliptic parallel property holds?

EDIT. Apparently the answer to my question is in Veblen's paper "A system of axioms for geometry". On page 370 he proves the existence of at least one parallel line using only order and continuity axioms (no congruence and parallel axioms). On page 348 he provides a model for ordered geometry (without continuity axioms) in which elliptic property holds.

You can find the paper here: https://archive.org/details/jstor-1986462

$\endgroup$
  • $\begingroup$ To prove that given a point and line there is a parallel line going through this point, in fact you don't need the continuity, nor Playfair axiom. We have a machine checked proof of this in GeoCoq. But I don't know if you can prove it only with incidence and order axioms. $\endgroup$ – Julien Narboux Oct 13 '17 at 5:20
-1
$\begingroup$

Yes

without congruence, the frame is so rough that It has enough room to have such model as you mentioned.

with congruence (even without continuity), such model dose not exist.

In Hilbert plane (Euclidean plane without any form of parallel postulate and continuous), the parallel lines do exit. You can always use double-perpendicula to do so. Namely, drop a perpendicular , say M, from point A to L and then erect a perpendicular ,say N, at point A to M. N is always parallel to L (parallel means they do not intersect)

thanks to Gupta’s construction, to drop a perpendicular (and from that, erect perpendicular s and get midpoint of any segment) can be always done without help from any any form of parallel postulate and continuous. This construction relys on the facts about the base midpoint of isosceles triangles and congruence of triangles so it indeed needs congruence axioms

So, to eliminate the elliptic property, you need either (incidence+order+congruence) or (incidencr+order+continuity)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.