Algebra of segments

It seems obvious that the addition defined in this way is commutative, but to prove it one has to use Desargues' theorem twice.

We construct the point \(K_1\) as above as the intersection of the line through \(A_1\) parallel to \(\ell_2\) with the line through \(B_2\) parallel to \(\ell_1\). Similarly, we construct \(K_2\) as the intersection of the line through \(B_1\) parallel to \(\ell_2\) with the line through \(A_2\) parallel to \(\ell_1\). If the line through \(K_1\) and \(K_2\) is parallel to \(e\) then the addition is commutative.

Consider first the triangles \(\left[A_1, A_2, F_1\right]\) and \(\left[B_1, B_2, F_2\right]\) shown in yellow below.² Their sides are parallel by construction and the three intersection points of the parallel sides lie on the line at infinity, so the two triangles are in axial perspective. By Desargues' theorem they are also in central perspective, so the points \(O\), \(F_1\) and \(F_2\) lie on the same line.

Now consider the blue triangles \(\left[O, A_1, A_2\right]\) and \(\left[K_1, K_2, F_2\right]\). We just concluded that \(O\), \(F_1\) and \(F_2\) lie on the same line. The lines \(\langle A_1, K_1\rangle\) and \(\langle A_2, K_2\rangle\) connecting the other vertices of the triangles also pass through the point \(F_1\). Moreover, two of the sides the triangles are already parallel, so their intersection points lie on the line at infinity. Using again Desargues' theorem (in the other direction compared to before) we conclude that also the third intersection point lies on the infinity line and that the line \(\langle K_1, K_2\rangle\) is parallel to \(\langle A_1, A_2\rangle\) and therefore also to \(e\).

The proof of associativity uses the same trick. One constructs two points \(L_1\) and \(L_2\) corresponding to \((a + b) + c\) and \(a + (b + c)\) respectively. If the line through \(L_1\) and \(L_2\) is parallel to \(e\) then the addition is commutative.

One again uses Desargues' theorem twice, first for the yellow triangles to show that the points \(F_1\), \(F_2\) and \(B_1\) lie on the same line. Then, as above, the theorem is used for the blue triangles to conclude that the line \(\langle L_1, L_2\rangle\) is indeed parallel to \(e\).

The commutativity of the multiplication is interesting because one needs Pascal's theorem to prove it. However, recall that one way to prove Desargues' theorem was to assume Hilbert's axioms I, II and III (including the space axioms). It turns out that these axioms are not strong enough to prove Pascal's theorem and that one needs at least the axioms of Hilbert's group IV or V. If this is the case, we can prove the commutativity as follows.

Pascal's theorem states the following:

If six points points on a conic are connected in such a way that the line segments between them form a hexagon, then the three pairs of lines corresponding to opposite sides of the hexagon intersect in three points that lie on the same line.

We will need this theorem in a degenerate version, where the conic consists of two lines and one of those lines is the infinity line of the projective plane.

We first construct \(a b\) by drawing a line from \(A_1\) to \(E_2\) and another line parallel to it through \(B_2\). This line intersects \(\ell_1\) in a point \(C_1^{ab}\). Similarly, to compute \(b a\) we draw a line from \(B_1\) to \(E_2\) and another line parallel to it through \(A_2\). This line also intersects \(\ell_1\) in a point \(C_1^{ba}\). We have to show that the points \(C_1^{ab}\) and \(C_1^{ba}\) coincide.

By construction the line \(\langle A_1, A_2\rangle\) is parallel to \(\langle B_1, B_2\rangle\), the line \(\langle A_1, E_2\rangle\) is parallel to \(\langle B_2, C_1^{ab}\rangle\) and the line \(\langle B_1, E_2\rangle\) is parallel \(\langle A_2, C_1^{ba}\rangle\). The three pairs of lines intersect in three points that lie on the line at infinity. These three infinite points together with the points \(A_2\), \(E_2\) and \(B_2\) form a hexagon. By Pascal's theorem the intersections points of the opposite sides of the hexagon lie on the same line. These are the points \(A_1\), \(B_1\) and the intersection of \(\langle B_2, C_1^{ab}\rangle\) with \(\langle A_2, C_1^{ba}\rangle\). However, both \(C_1^{ab}\) and \(C_1^{ba}\) already lie on the line \(\langle A_1, B_1\rangle\) so \(C_1^{ab}\) and \(C_1^{ba}\) are the same point.

The proof that \(a (b c) = (a b) c\) uses again Desargues' theorem.

We first compute the segment \(bc\) by multiplying \(b\) and \(c\) as above. This gives us a point \(BC_2\) on \(\ell_2\). We then "multiply by \(a\)" by drawing a line through \(BC_2\) parallel to the line \(A_1 E_2\). The intersection of this line with \(\ell_1\) gives the point \(A(BC)_1\) corresponding to \(a (b c)\).

We can also compute the segment \(a b\) which gives us a point \(AB_1\) on \(\ell_1\). As we already know that the multiplication is commutative, we can "multiply \(c\) by \(ab\)" to obtain \((ab) c\). This is done by drawing a line through \(C_2\) parallel to the line \(\langle AB_1, E_2\rangle\) which as above intersects \(\ell_1\) in a point that we call \((AB)C_1\).

Define the point \(F_1\) as the intersection of the line \(\langle B_1, B_2\rangle\) with the line \(\langle AB_1, E_2\rangle\). Similarly, define \(F_2\) as the intersection point of the lines \(\langle (AB)C_1, C_2\rangle\) and \(\langle A(BC)_1, BC_2\rangle\). The two yellow triangles \(\left[E_2, F_1, B_1\right]\) and \(\left[C_2, F_2, BC_1\right]\) have their three corresponding sides parallel to each other. Threfore, by Desargues' theorem, the points \(F_1\), \(F_2\) and the origin \(O\) lie on the same line.

Now consider the blue triangles \(\left[B_2, AB_1, F_1\right]\) and \(\left[BC_2, (AB)C_1, F_2\right]\). Two of their sides are already parallel and since they are in central perspective in the origin it follows from Desargues' theorem that their third sides are also parallel. In particular, this means that \(\langle AB_1, B_2\rangle\) is parallel to \(\langle (AB)C_1, BC_2\rangle\).

By construction \(\langle AB_1, B_2\rangle\) is however also parallel to \(\langle A(BC)_1, BC_2\rangle\), so the lines \(\langle (AB)C_1, BC_2\rangle\) and \(\langle A(BC)_1, BC_2\rangle\) are the same line and the points \(A(BC)_1\) and \((AB)C_1\) are also the same (because they are both found by intersecting with \(\ell_1\)).