Prove the following theorem Each point in Pappuss geometry l

Prove the following theorem: Each point in Pappuss geometry lies on exactly three lines.

Solution

AXIOMS:

1. There exists at least one line.

2. Every line has exactly three points.

3. Not all lines are on the same point.

4. If a point is not on a given line, then there exists exactly one line on the point that is parallel to the given line.

5. If P is a point not on a line, there exists exactly one point P\' on the line such that no line joins P and P\'.

6. With the exception in Axiom 5, if P and Q are distinct points, then exactly one line contains both of them.

PROOF: Let X be any point. By corrected axiom 3, there is a line not containing X. This line contains points A,B,C [Axiom 2]. X lies on lines meeting two of these points, say B and C [Axiom 5]. There is exactly one line through X parallel to BC [Axiom 4].

There can be no other line through X since by Axiom 4 it would have to meet BC at a point other than A, B or C [Axioms 6 and 5], and this would contradict Axiom 2.