This axiomatic system seems contradictory. Any thoughts?

[jdinatale]

Homework Statement

edit: WHOOPS. Almost forgot to

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!

 

[jdinatale]

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

 

[SammyS]

Homework Statement

edit: WHOOPS. Almost forgot to

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!

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.

 

[jdinatale]

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.

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.