Proof
Pappus's theorem is equivalent to its dual (in the presence of the basic axioms for a projective plane). The dual of Pappus's theorem can be proved for the projective plane over any field using projective coordinates as follows.
The dual of the theorem states that if all but one of the nine sets of three lines are concurrent, then all of them are. Choose projective coordinates with
- C=(1,0,0), c=(0,1,0), X=(0,0,1), A=(1,1,1).
On the lines AC, Ac, AX, given by x2=x3, x1=x3, x2=x1, take the points B, b, Y to be
- B=(p,1,1), b=(1,q,1), Y=(1,1,r)
for some p, q, r. The three lines XB, cb, CY are x1=px2, x2=qx3, x3=rx1, so all pass through the same point (called a in the diagram) if and only if pqr=1. The condition for the three lines Cb, cB and XY x2=qx1, x1=px3, x3=rx2 to pass through the same point (Z in the diagram) is qpr=1. So this last set of three lines is concurrent if all the other eight sets are because multiplication is commutative, so pqr=qpr, which proves the dual of Pappus's theorem.
The proof above also shows that if Pappus's theorem holds for a projective space over a division ring, then the division ring is a (commutative) field. German mathematician Gerhard Hessenberg proved that Pappus's theorem implies Desargues's theorem. In general, Pappus's theorem holds for some projective space if and only if it is a projective space over a commutative field. The projective spaces in which Pappus's theorem does not hold are projective spaces over noncommutative division rings, and non-Desarguesian planes.
Read more about this topic: Pappus's Hexagon Theorem
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