# This axiomatic system seems contradictory. Any thoughts?

1. Dec 3, 2011

### jdinatale

1. The problem statement, all variables and given/known data

edit: WHOOPS. Almost forgot to

3. The attempt at a solution

Ok. So I began with the line l_1 with exactly 6 points on it, A, B, C, D, E, F. (Axiom 2). Now, by axiom 1, there must exist 2 additional points not on this line. So I formed those, G and H.

No by Axiom 3, there exists a line between each of the points on L_1 and G and H, as well as a line between G and H. I drew those. Notice, however, that the line HF is parallel to GA, GB, GC, GD, and GE.

This contradicts axiom 4 which states that through G, only 1 line is parallel to HF!

Last edited: Dec 3, 2011
2. Dec 4, 2011

### jdinatale

Geeze, this forum doesn't seem to like geometry very much.

3. Dec 4, 2011

### SammyS

Staff Emeritus
Only three of your lines show 6 points. Those are l1, p and, u. So, those lines you claim are parallel to line q (also known as HF) might actually intersect line q.

4. Dec 4, 2011

### jdinatale

Just because two lines "intersect" in the diagram, doesn't mean that they actually intersect within the axiom system. This model wouldn't work if that was the case because there would exist 7 points on AH (counting all of the intersections).

I have to PROVE that each line is incident with 6 points.