The Parallel Postulate
Background Information
The parallel postulate is a geometrical mathematical approach which states that in a two dimensional geometry, if a line segment intersects two straight lines forming interior angles whose sum are less than 1800 on the same side, then the two lines if extrapolated meet on that side where the sum of the two angles are less than 1800 .
This postulate is also called the Euclid’s fifth postulate because of the fifth postulate in the Euclid elements which is a distinctive element in the Euclidean geometry. Euclidean geometry is the study of geometry that fulfills the axioms of Euclid including the parallel postulate. In the geometries where the parallel postulate does not hold is considered non-Euclidean geometry. For close to two thousand years now, the history of the parallel postulate has been under several proofs using the Euclid’s first four postulates. The parallel attracts several proofs because it is not self-evident. If the order of the Euclid’s postulates were listed in order of their significance, then the parallel postulate would be of no much significance but Euclid only included it when he realized he could not proceed without it. The fifth postulate was proved from the other four and many of the proofs have been accepted as proofs until a mistake was found. The mistake was assuming some obvious property which became an equivalent to the Playfair’s axiom. This was named after a written commentary by John Playfair in 1795 where he proposed the replacement of the Euclid’s fifth postulate by this axiom. Nevertheless, Euclid did not postulate the converse of this parallel postulate. This distinguished his fifth postulate from the elliptical geometry. The converse of the parallel postulate states that if the sum of the two interior angles equals to 1800, the lines are parallel and will never intersect. The Hilbert’s definitions and axioms were used as a modified version of the Euclid’s definition (Dodgson, 209). As mentioned before, these attempts to prove the Euclidean parallel postulate were done by outstanding mathematicians of their time (210). Don't use plagiarised sources.Get your custom essay just from $11/page
Proclus
Proclus was a well-known Greek mathematician who founded some of the basic arguments about the geometry. He was the head of the Napoleonic school in Athens after Euclid. He gave a short sketch of the early geometry which were founded in much older books Eudemus. It also determines the approximate dates which would remain unidentified over a very long period of time. Proclus was mainly a philosopher not a mathematician but his contributions on the first book of Euclid element which gives a rich source to the Greek Geometry (210). Proclus criticized Euclid’s fifth postulate that some of its sections ought to be struck out of the postulate, for the statement was involving many difficulties. The statement about the meeting of two extrapolated lines is rather plausible but not necessary according to Proclus. He offered an example of hyperbola that approaches asymptotes as closely you like without meeting them. This is an example that the Euclid’s conclusion can at least be imagined. He added that “it is clear that we must seek the proof of the present theorem and how it is alienated to the present postulates.” (210)
Proclus’s Proof
Given two parallel lines l and m, suppose n cuts m at any given point let’s say P. we wish to show that this line n also cuts m. Now let Q be the foot of the perpendicular bisector from P to l. suppose n coincides with line PQ, then it intersects l at Q. Otherwise, one ray PY of n emerging from P lies between the line PQ and ray PX of m. Take X to be the foot of the perpendicular line from Y to m. Proclus the presented an argument that as point Y extends endlessly from P on n, segment XY increases without boundary according to Aristotle’s axiom. Finally, XY becomes greater than the fixed segment PQ. Here, Y must be on the other side of l thus that position and its starting position, Y must have hit l, which means that line n intersects l. This argument is however sophisticated involving motion and between. Additionally, every step in the argument can be assumed to be correct on condition that Aristotle’s axiom is assumed. To justify the last step of this prof, drop a perpendicular from Y to meet l. this may also test the collinearity for X, Y, and Z and XZ is equivalent to PQ. So when XY becomes greater than PQ, XY must be larger than the XZ and so that Y must be on the other side of l. lest have a look at the diagram below.
P X m
S Y
n
Q Z l
Let S be the foot of the perpendicular from Y to PQ. S is on the same side of m as Y and Q since SY is parallel to m. we form rectangle SPXY, hence its opposite sides are equal and parallel hence congruent. Applying Proclus argument and Aristotle’s axiom: A point Y exists on the given ray of n so that XY>PQ. Then PS, which is congruent to XY, is also >PQ.
Clairaut’s Axiom and Proclus Theorem
Clairaut was a leading French mathematician who made important contributions to modern geometry. He lived between 1713 and 1765 when he did his work. According to Clairaut axiom, triangles exists. He did not try to prove Euclid V in neutral geometry but assumed instead in his work in 1741. He claims that drawing the physical rectangles is impossible since if one believes so, then it’s also believed then that the world is also spherical. Even though this axiom sought to replace the Euclid’s fifth postulate, it was weaker and logically equivalent to it and so nothing serious was gained logically by the replacement.
Clavius Axiom
Clavius’ axiom asserts that all points equidistant from a given straight line, on a given side of it, there is a straight line. For any line l and any point P not on l, the equidistant locus to l through P is the set of all points on the line through P which is also parallel to l. other mathematicians like Ibn tried to justify this argument through the kinematics argument (213). To illuminate the status of the Clavius axiom in neutral geometry, Clavius considered the following theorem. One, the plane is semi-Euclidean and second, for any line l and any point P not on l, the equidistant locus to l, through P is the set of all points on the parallel to l, through P obtained by a standard construction on the line through P orthogonal to PQ, where Q is the foot of the perpendicular from P to l and lastly on this axiom itself.
Proof