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Next: Axiom of Parallels Up: Foundations of geometry Previous: The Axioms of Incidence

Axioms of Order

These axioms were almost ignored by Euclid except the second one below. Their importance was noticed by M. Pasch who saw how they were implicitly being used in many proofs. This is one problem with ``evident truths''; we often forget to state some of the axioms and then the geometry is incomplete without them. The following axioms make clear the notion of a point lying between two other points.
  1. When B is between A and C then, A, B and C are distinct points lying on a line and B is between C and A.
  2. Given a pair of points A and B there is a point C so that B is between A and C.
  3. If B lies between A and C then A does not lie between B and C.
  4. Let A, B and C be three points on a plane $ \alpha$ and a be a line on $ \alpha$ that does not contain any one of these points. If there is a point D on a that is between A and B then either a contains a point between A and C or a contains a point between B and C.

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@font
picture(6309,4845)(2209,-4531) (5551,179)(0,0)[lb]A (2446,-3646)(0,0)[lb]B (6241,-4531)(0,0)[lb]C (3976,-1561)(0,0)[lb]D (2296,-1231)(0,0)[lb]l
The Line l contains a point on one of the other two sides
The first three axioms allow us to introduce the notion of a half-line or ray. Given a pair of points A and B the half ray starting at B and pointing away from A consists of all points C so that B is between A and C. Similarly, the last axiom allows us to introduce the notion of a half plane. Given a point A and a line a the half-plane bounded by a and opposite to A consist of all points B so that a contains a point lying between A and B.

In spite of the axioms of order being ignored for so many hundreds of years they are so important that one can entirely replace the axioms of incidence by giving an extended set of axioms of order. Think of it this way. If a straight line is to be the shortest path from a point to another then we must at least be able to say what are the points ``on the way'' or in-between.

The following theorems can be deduced from the axioms of Incidence and Order.

Theorem 4   Given any two point A and B there is a point C that lies between A and B.

Theorem 5   Given three points A, B and C that lie one a line exactly one point that lies between the other two.

Theorem 6   Given four points on a line they can be labelled A, B, C and D so that B is between A and C and between A and D and C is between B and D and between A and D.

Theorem 7   Given any finite set of points on a line they can be labelled A1, A2, ..., An so that the points are in that order.

An important theorem that can be deduced from the order axioms was first discovered by G. Desargues:

Theorem 8   Given points A, B, C, A', B' and C' so that the lines AA', BB' and CC' all pass through a point O. Further, let AB and A'B' meet in a point C'', AC and A'C' meet in a point B'', BC and B'C' meet in a point A''; moreover, let us assume that the 10 points considered are distinct. Then the points A'', B'' and C'' all lie on a line.

Proof. In case the plane $ \alpha$ containing the points A, B and C does not contain all of the points A', B' and C' then the line containing the points A'', B'' and C'' is just the line of intersection of $ \alpha$ with the plane $ \alpha{^\prime}$ determined by A', B' and C'. Thus, the theorem needs only to be proved under the assumption that all the points lie in a plane. In this case we shall show how to construct A''', B''' and C''' that do not lie in the plane and so that the points A, B, C, A''', B''' and C''' also satisfy the hypothesis of the theorem. Moreover, A''', B''' and C'' are collinear and so on cyclically. Thus, the planar version will then follow from the non-planar version.

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picture(8379,7452)(949,-6688) (4366,-3556)(0,0)[lb]A (4201,-6571)(0,0)[lb]A' (3076,-4066)(0,0)[lb]B' (3526,-2716)(0,0)[lb]B (5581,-3136)(0,0)[lb]C (5716,-3826)(0,0)[lb]C' (6331,-2326)(0,0)[lb]B'' (2041,-826)(0,0)[lb]C'' (8251,-3076)(0,0)[lb]A'' (4936,629)(0,0)[lb]O
The triangles ABC and A'B'C' lie in different planes.
In the remaining cases we examine all the possibilities for between-ness for the triples (A, A', O), (B, B', O) and (C, C', O). By interchanging the $\scriptstyle \prime$ and permuting the letters (A, B, C) we can reduce to the following two cases.
  1. A does not lie between A' and O, B does not lie between B' and O, C does not lie between C' and O.
  2. A' lies between A and O, B' does not lie between B and O, C' does not lie between C and O.

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picture(9549,6694)(139,-7790) (601,-2611)(0,0)[lb]A'' (1201,-7336)(0,0)[lb]A (2551,-5236)(0,0)[lb]B (5026,-7261)(0,0)[lb]C (7501,-4186)(0,0)[lb]O (4351,-6211)(0,0)[lb]A' (3751,-5236)(0,0)[lb]B' (5776,-6436)(0,0)[lb]C' (5251,-3211)(0,0)[lb]A''' (4501,-3436)(0,0)[lb]B''' (3526,-3811)(0,0)[lb]C'' (5926,-3736)(0,0)[lb]C''' (9151,-6211)(0,0)[lb]B'' (6076,-2011)(0,0)[lb]O'' (5476,-1261)(0,0)[lb]O'
Lifting A'B'C' in the first case.
Examining the first case, let O' be a point not in the plane of A, B and C. Let A''' be a point between A' and O'. The line joining A and A''' then must contain a point O'' lying between O and O' since its intersection with the line joining A' and O is A which does not lie between these points by hypothesis. Now the line joining O'' and B contains a point O'' between O and O' and thus must contain a point B''' between O' and B' since the point B of intersection of this line with the line joining B' and O is not between these two points. Similarly, we obtain C''' lying on the line joining C and O'' and lying between O' and C'.

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picture(5850,7924)(1351,-8461) (4651,-8461)(0,0)[lb]B (7201,-5761)(0,0)[lb]C' (5326,-3661)(0,0)[lb]B' (5476,-4786)(0,0)[lb]O (3526,-6436)(0,0)[lb]A' (1426,-961)(0,0)[lb]A''' (3226,-5011)(0,0)[lb]O' (4276,-4036)(0,0)[lb]C (4051,-1936)(0,0)[lb]A'' (2551,-4786)(0,0)[lb]C''' (2776,-6436)(0,0)[lb]B'' (1651,-5761)(0,0)[lb]B''' (1351,-5311)(0,0)[lb]O'' (1531,-7636)(0,0)[lb]A (2341,-7786)(0,0)[lb]C''
Lifting A'B'C' in the second case.
In the second case we choose a point O'' which does not line in the plane of A, B and C. Let A''' be a point so that O'' lies between A and A'''. Now the line joining A' and A''' contains the point A' which lies between A and O; moreover its intersection with the line joining A and O'' is A''' which does not lie between these points. Thus the there is a point O' on the line joining A' and A''' which lies between O'' and O'. Now, consider the line joining O' and B' and the triangle of points O'', B and O. As before we find a point B''' which lies between B and O'' and on the line joining O' and B'. Similarly, we find C'''.

In both these cases the line joining A''' and B''' meets the plane $ \alpha$ within the intersection of the plane determined by A', B' and O' and the plane $ \alpha$; this is the line joining A' and B'. Similarly, the intersection of the line joining A''' and B''' with $ \alpha$ lies within the intersection of the plane determined by A, B and O'' with the plane $ \alpha$; this is the line joining A and B. In other words, the line joining A''' and B''' contains the point C''. We prove the other containments cyclically. $ \qedsymbol$

This theorem and the ideas behind its proof allow us to make constructions like:

Exercise 3   Given a pair of non-parallel lines on a piece of paper (in a bounded region of a plane) and a point not on these lines; construct a line through this point that passes through the point of intersection of the two lines (which may not be within the sheet of paper!).


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\end{picture}
picture(5454,6039)(1339,-7138) (1396,-3166)(0,0)[lb]l (2191,-4951)(0,0)[lb]m (2911,-5056)(0,0)[lb]A'' (2866,-2506)(0,0)[lb]B (3526,-3691)(0,0)[lb]B' (4651,-5521)(0,0)[lb]O (2281,-6571)(0,0)[lb]C' (3421,-6301)(0,0)[lb]C (4786,-1741)(0,0)[lb]A (4606,-3046)(0,0)[lb]A' (4081,-4471)(0,0)[lb]B''
The line joining A'' and B'' passes through the intersection of l and m.
Another aspect of Desargues' theorem is that it's proof makes use of the non-planar axioms of incidence. We shall see that

Theorem 9   If we have a plane which satisfies all the planar axioms of Incidence and the axioms of order, then it can be embedded in a geometry that satisfies the axioms of space if and only if Desargues' theorem is valid in this plane.


next up previous
Next: Axiom of Parallels Up: Foundations of geometry Previous: The Axioms of Incidence
Kapil H. Paranjape 2001-01-20