4

Parallel postulate: If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

Playfair's axiom: Given a line and a point not on it, at most one parallel to the given line can be drawn through the point.

Prove the Parallel postulate from Playfair's axiom.

user5402
  • 3,006

3 Answers3

8

To show Playfair implies Parallel is the same as to show (not Parallel) implies (not Playfair). So assume (not Parallel), i.e. there is a line $t$ (extension of the segment you refer to) intersecting lines $l,l'$ and such that the sum of the interior angles on one of the sides of $t$ is less than $180^{\circ}$ (two right angles), yet $l,l'$ do not meet on that side of $t$. They cannot meet on the other side either, since on that other side the sum of interior angles exceeds $180^{\circ}$, and even in neutral geometry (i.e. with no parallel postulate) the sum of two angles in a triangle cannot exceed $180^{\circ}$. Therefore $l,l'$ are parallel. We may also construct another line $l''$ through the point $P$ on $l'$ where the transversal $t$ meets it, such that for this line the sum of its internal angles on the same side of $t$ is equal to $180^{\circ}$, and this line $l''$ is parallel to $l$ by neutral geometry only. So we now have two distinct parallels to $l$ through $P$ and have arrived at the negation of the Playfair axiom. ‍

Cameron Williams
  • 29,432
coffeemath
  • 29,884
  • 2
  • 31
  • 52
  • 1
    Where did you get "the sum of two angles in a triangle cannot exceed 180" from? I don't agree with Euclid's proof. – user21820 Jan 25 '15 at 15:36
  • 1
    See http://en.wikipedia.org/wiki/Absolute_geometry about this issue. Absolute geometry is another term for "neutral geometry" and refers to what can be deduced from using all of Euclid's axioms except for any axiom equivalent to the parallel postulate. One of the theorems in absolute geometry is the Saccheri-Legendre Theorem, that the sum of the three angles of a triangle is at most 180 degrees. In particular if one adds two of the angles of a nondegenerate triangle in neutral geometry, the sum is in fact less than 180 degrees. – coffeemath Jan 25 '15 at 18:36
  • I'm not convinced. At least the proof given by Euclid is invalid, so I don't accept it unless you show me a different proof. The wikipedia page says "the first 28 Propositions and Proposition 31 avoid using the parallel postulate, and therefore are valid in absolute geometry." which is similarly invalid. – user21820 Jan 26 '15 at 07:41
  • 1
    I cannot see what your complaint is, @user2180 -- what the topic of neutral geometry (also called absolute geometry) is about is to see what can be deduced from Euclid's axioms without the parallel postulate. Saccheri-Legendre Theorem is really the thing to look at, and see if any of the steps in the proof are objectionable to you. Saccheri at least was one of the mathematicians who was trying to prove the parallel postulate from the other Euclid axioma, so was pretty careful not to use them in his work. – coffeemath Jan 26 '15 at 08:05
  • @user21820 By the way, are you the person who posted the question, maybe having changed your user name? In what I see that OP (original poster) was "metacompactness", and it is only that user who can "accept" the answer or not. – coffeemath Jan 26 '15 at 08:19
  • No I'm not the asker. When you change your user name it changes everywhere accordingly. But even if I were the asker, why would I accept an answer I don't even find valid? I can't accept the validity of a proof unless I can personally see and verify it myself or I am certain that the person making the claim is reliable. Unfortunately, I don't know where I can get a full proof down to all the axioms. It is not enough to simply say "we can ..." or "it must be ..." unless justified by previous statements. I am well versed in geometry and logic, which is precisely why I can see a ton of loopholes. – user21820 Jan 26 '15 at 11:52
  • Let us continue this discussion in chat. – coffeemath Jan 26 '15 at 13:41
2

History is rich of incorrect proofs about the parallel postulate. We have developed very rigorous proofs checked using the Coq proof assistant.

You can find a computer checked proof here: http://geocoq.github.io/GeoCoq/ and the English version of the proofs in the following paper: https://hal.inria.fr/hal-01178236

You can also find the proof in Greenberg's book "Euclidean and non-euclidean geometries".

Julien Narboux
  • 1,000
1

I think that the proof can be given directly, which gives it more clarity, using Propositions I.13, I.17, I.28 and I.31 (which do not depend on the Fifth Postulate), following the demonstration in the book Introduction to non-Euclidean geometry by Harold E. Wolfe (Chapter 2).

First, note that the line parallel to a line given by a point not on the line constructed in Proposition I.31 ("To draw a straight line through a given point parallel to a given straight line") produces internal angles of the same side. whose sum is equal to two right angles, this is deduced by virtue of Proposition I.13 ("If a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles").

Now, to prove that Playfair Axiom implies the Fifth Postulate it is as follows:

Given lines $AB$ and $CD$ cut by the transversal $ST$ in such a way that the sum of angles $BST$ and $DTS$ is less than two right angles. Construct through $S$ the line $QSR$, making the sum of angles $RST$ and $DTS$ equal to two right angles. This line is parallel to $CD$ by Proposition I.28 ("If a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another"). Since lines $QSR$ and $ASB$ are different lines and, by Playfair’s Axiom, only one line can be drawn through $S$ parallel to $CD$, we conclude that $AB$ meets $CD$. These lines meet in the direction of $B$ and $D$, for, if they met in the opposite direction, a triangle would be formed with the sum of two angles greater than two right angles, contrary to Proposition I.17 ("In any triangle the sum of any two angles is less than two right angles").

enter image description here

G al Cubo
  • 153