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Over the years, mathematicians had been bothered by the unstated assumptions of Euclidean geometry, which has been the foundation of geometry for about 2000 years. One of these unstated assumptions is the concept of 'betweenness'. In the early 20th century, David Hilbert (1862-1943) reworked the axioms of Euclid's work. He expanded Euclid's original five axioms to 21 axioms in five groups. This article is based largely on David Hilbert's Foundations of Geometry1, 2nd English edition.
David Hilbert identified three basic sets of objects for geometry: points, lines, and planes. The element of line geometry is the point. The elements of plane geometry are the point and line. The elements of space geometry are points, lines, and planes.
Points, lines and planes have specific relationships. These relationship are described by the axioms of geometry:
The word 'incidence' means one geometric object may contain another (see incidence, Dictionary.com). For example, a line may contain a point and, in fact, must be made up of at least two points. The axioms of incidence are:
For every two points A and B, there exists a line that contains each of these points.
Axiom I,1 means if one picks any two points, there must be a line that connects those two points. To see the logic of this, try to think of two points in space that can not be connected by a line. In figure 1, click on the yellow points and drag them. Where ever the points are placed (except on top of each other), there is a line that passes through them.
For every two points A and B, there exists no more than one line that contains each of these points.
Axiom I.2 states that there is exactly one line that passes through two given points. In figure 2, click on one of the yellow points and move it right on top of the other yellow point. Do you see one line or two?
|I,3.||There exists at least two
points on a line. There exists at least three points that do not lie on a line.
Try to imagine a line that has only one point. It is not a line, but a point. So each line must have at least two points. In figure 3, click on one of the yellow points and drag it on top of the other. Does the line still exist?
The second statement implies the existence of a plane. If there are at least three points that do not lie on a line, there must exist some mathematical construct to contain these three points. This construct is the plane. This implies that lines exist within planes.
For any three points A, B, and C, there exists a plane α that contains the points A, B, and C. For every plane there exists a point that lies in the plane.
For any three points A, B, and C that do not lie on one and the same line, there exists no more than one plane that contains each of the three points.
If two points A and B of line a lie in a plane α, then every point of a lies in α.
If two planes, α and β have a point A in common, then the have at least one point B in common.
There exist at least four points that do not lie in a plane.
|II,1.||If a point B lies between a point A and a point C then the points A, B, and C are three distinct points and B also lies between C and A.||
|II,2.||For two points A and C, there always exists at least one point B on the line AC such that C lies between A and B.||
|II,3.||Of any three points on a line there exists no more than one that lies between the other two.||
|II,4.||Let A, B, and C be three points that do not lie on a line and let a be a line in the plane ABC which does not meet any of the points A, B, or C. If the line a passes through a point of the segment AB, it also passes through a point of segment AC or through a point of the segment BC.||
|III,1.||If A and B are two points on line a, and A' is a point on the same or on another line a', then it is always possible to find another point B' on a given side of the line a' through A' such that segment AB is congruent or equal to the segment A'B'. In symbols, AB ≡ a'b'.||
|III,2.||If a segment A'B' and a segment A''B'' are congruent to the same segment AB, then the segment A'B' is also congruent to the segment A''B''. Briefly, if two segments are congruent to a third one the are congruent to each other.||
|III,3.||On a line let AB and BC be two segments which, except for B have no point in common. Furthermore, on the same or another line a', let A'B' and B'C' be two segments which, except for B' also have not points in common. In that case, if AB ≡ A'B' and BC ≡ B'C', then AC ≡ A'C'.||
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