Two Attempts to Prove the Parallel Postulate

Term Paper in Geometry D by

Patrik Andersson

 

Introduction

This paper presents two historical attempts of proving Euclid's famous Parallel Postulate. The first was made by B. F. Thibaut in 1809 and can be found in Marvin Jay Greenberg's book [2] as exercise 52 of chapter 9, while the second is in a letter from the Danish mathematician Schumacher to Carl Friedrich Gauss originally written in 1831 and published in volume eight of The Collected Works of Gauss [1].

Both attempts are naturally flawed in some way, since the Parallel Postulate is known to be impossible to prove from the rest of Hilbert's postulates – a fact first shown by Eugenio Beltrami in 1868. In this paper I will state both proofs, point out the flaws and give a short discussion.

 

The First Attempt

Here Thibaut is trying to prove that the angle sum of an arbitrary triangle is equal to p, which is known to be equivalent to the parallel postulate.

Given any triangle rABC, extend the lines AB, AC and BC to points D, E and F respectively (see Figure 1.1).

 

Figure 1.1

Now rotate the line AB about A through the angle <DAC to the line AC, then at C rotate AC to BC through <ECB and finally at B, rotate BC to AB though <FBA.

After these three rotations, Thibaut claims that the line AB has returned to itself and hence it has been rotated by an angle of 2p. Addition of the three angles of rotation gives that

(p - <A) + (p - <C) + (p - <B) = 2p Þ <A + <B + <C = p,

and hence Euclid's parallel postulate follows.

The error lies in the fact that to set up the formula in the last line of the proof, Thibaut uses a theorem valid only in Euclidean Geometry, namely the one below (see also [3]).

Theorem: The composition of multiple rotations turns the plane through an angle equal to the sum of the component rotation angles (and gives a translation precisely when this sum is a multiple of 2p).

Proof: With notation as in Figure 1.2, writing Rl for the reflection in the line l and RotP(q) for the rotation about the point P through the angle q.

RotA(a)·RotB(b) = (Rm·Rn)·(Rn·Rp) = Rm·Rp = RotC(-g) = RotC(a+b), since a + b = 2p - g

 

Figure 1.2

Clearly the mistake Thibaut does is to put the sum of the three individual rotation angles equal to the angle of rotation for the total transformation, thus calling on the result above and making a circular proof, using the very fact he wants to prove.

However, he is correct when he claims that the line AB is invariant under the composition of the three rotations - though it is not fixed point-by-point. As shown below, no point on AB is fixed and hence the composition cannot be a rotation at all.

As can be seen in Figure 1.3, points A and B are translated along AB, to A''' and B''' respectively, by the total transformation T = RotA(a)·RotC(g)·RotB(b) and it is easy to see that this transformation is in fact a translation of length a + b + c, with notation as in the figure.

Here A' = RotA(a)(A), A'' = RotC(g)(A'), A''' = RotB(b)(A'') and in the same way for B', B'', B''' so that T(A) = A''' and T(B) = B'''.

Figure 1.3

This suggests that the following proposition can be salvaged from Thibaut's proof:

Proposition: Given any triangle rABC, the transformation T = RotA(a)·RotC(g)·RotB(b) is a translation of length a + b + c along the line AB in the direction from A to B.

A perhaps interesting result, since it is obviously valid even in Neutral Geometry.

 

The Second Attempt

The first letter from Schumacher to Gauss is dated Copenhagen, May the 3rd, 1831. In it he, just as Thibaut above, tries to prove that the angle sum of an arbitrary triangle is equal to p.

Schumacher's reasoning starts with Figure 2.1 below, from which he notes that

2a + 2a = 2b + 2b = 2c + 2g = 2p and hence a + b + g = 3p – (a + b + c).

Figure 2.1

Then he draws a new line FJ, intersecting both EH and DG at the point A and chooses it so that the angle <FAH equals the "old" angle between FJ and EH, i.e. <FAH = b. The result is shown in Figure 2.2 and it is seen that a + b + c = 2p so that a + b + g = p.

Figure 2.2

The answer from Gauss comes in a letter dated Göttingen, May 17th, 1831. It is short and to the point.

"When two intersecting straight lines (1), (2) meet a third straight line (3) at angles of intersection a and a' respectively, and a fourth straight line (4) intersects (1) at an angle equal to a, then it does not necessarily follow that (4) and (2) intersect at an angle equal to a'."

Thus the error in Schumacher's proof lies in the fact that although he chooses the new line FJ so that <FAH = b, it is not true because of this that <DAF = c. Hence the angle marked c in Figure 2.1 and the angle marked c in Figure 2.2 are not known to be equal, which breaks down Schumacher's argument effectively. In fact, Gauss'statement quoted above is equivalent to the Parallel Postulate.

The exchange of letters continue for a few more rounds, where Schumacher tries to alter his arguments slightly and explain things more carefully, only to have Gauss pointing out the errors made. In a final letter, dated Altona, July 19th, 1831, Schumacher finally accepts it all and thanks Gauss for his kind efforts.

 

References

[1] Gauss, "Werke: Achter Band", Königlichen Gesellschaft der Wissenschaften, Göttingen,

1900.

[2] Greenberg, "Euclidean and Non-Euclidean Geometries – Development and History",

W. H. Freeman and Company, New York, 1994.

[3] Hoggar, "Mathematics for Computer Graphics", Cambridge University Press, 1994.


To Geometry D