# Parallel Postulate proven? **URL:** \ **Category:** Factual Questions\ **Created:** [May 4, 2012, 3:42am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834 "2012-05-04T03:42:05Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\ **Post date:** [May 5, 2012, 4:56am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/21 "2012-05-05T04:56:16Z")
> [@supery00n](#): > You’re the author, right? And using this forum to refine your proof. I should have realized it from the May 2012 join date and the ‘y00n’ username vs the J. Yoon in the paper. I feel like an idiot now. > [@supery00n](#): > > According to the Wikipedia article on the parallel postulate, it says that both of the following are equivalent statements to Euclid’s fifth postulate: > [http://en.wikipedia.org/wiki/Parallel\_postulate](http://en.wikipedia.org/wiki/Parallel_postulate) Are you really going to trust Wiki to that extent? Look up “absolute geometry” (it may also be called Neutral geometry). It’s geometry that can be done using only the first 4 postulates. One of the conclusions you can reach is that the angle sum of a triangle is at most 180 degrees. But to conclude from a single 180 degree triangle that all triangles are 180 degrees is equivalent to assuming the Parallel Postulate. > [@supery00n](#): > > With regard to the second point, are you saying that the construction used for that proof is actually a “special case” because of the assumption that a transversal was used? I was thinking that it might be “enough” that the author showed that the transversal does exist rigorously..? The problem is that the author tries to extend the logic without showing the logic can be extended in such a fashion. He starts with a pair of equal and parallel lines and a transversal and shows how that second pair of equal parallel lines (which intersect the original parallel lines) can be constructed. That’s fine. But then he assumes that additional equal/parallel lines can be constructed “next to” (for lack of a better way of putting it) the constructed parallels without showing that they exist or how to do it. Incidentally, this is similar to the mistake Proclus made. Proclus’ statement, for what it’s worth, is that if a straight line intersects one of a pair of parallel lines, it must intersect the other. Even though it’s not exactly the same error being made here, you can see how it’s similar. ---
**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\ **Post date:** [May 5, 2012, 5:01am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/22 "2012-05-05T05:01:35Z")
Missed the edit window, but about the 180 degree triangle thing again. The part about constructing parallels is important here, as well. That part is only true in Euclidean geometry. And since the proof already follows a Euclidean framework, it naturally follows it will preferentially show triangles with 180 degrees. That’s what I get for letting my logic get sloppy by simply letting earlier parts of the proof stand and not going through step by step. Naturally if an earlier step is shown to be false, you can get some wonky conclusions. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 5, 2012, 5:04am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/23 "2012-05-05T05:04:26Z")
> [@supery00n](#): > > It is quite likely that many people, in my view not entirely without justification, felt that arXiv had a lot of nonsense in it when it first came out too. Peer review and endorsement have their purpose. It’s a question of the signal vs. noise ratio and quality control. If there were a process in which all contributions were “signal” and none of them were “noise,” I would support that process wholeheartedly. But I believe that there is a trend in the opposite direction; namely, that people who choose to submit to viXra, arXiv, or any peer reviewed journal for that matter, consist of self-selected groups. > > That’s the key issue IMO. People say peer review is necessary, important, and I agree. I look at peer review as “artificial selection,” and the way viXra does things as “natural selection,” in that the former attempts to solve the problem of quality control with peer review whereas the latter attempts to do so by the principle that those who wish to contribute will “naturally select themselves.” This doesn’t make any sense. Anybody can convince themselves that their work is worth publishing. The only people whose work I’m going to bother reading are the people who can convince somebody else that their work is worth publishing. It may sound somewhat elitist or snobbish, but I have finite time and I don’t want to spend it chasing down crap. ---
**Author:** ![supery00n](https://avatars.discourse-cdn.com/v4/letter/s/5f9b8f/32.png) [@supery00n](https://boards.straightdope.com/u/supery00n)\ **Post date:** [May 5, 2012, 5:33am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/24 "2012-05-05T05:33:47Z")
> [@Great\_Antibob](#): > > You’re the author, right? And using this forum to refine your proof. > > I should have realized it from the May 2012 join date and the ‘y00n’ username vs the J. Yoon in the paper. > > I feel like an idiot now. > > Are you really going to trust Wiki to that extent? > > Look up “absolute geometry” (it may also be called Neutral geometry). It’s geometry that can be done using only the first 4 postulates. One of the conclusions you can reach is that the angle sum of a triangle is at most 180 degrees. > > But to conclude from a single 180 degree triangle that all triangles are 180 degrees is equivalent to assuming the Parallel Postulate. > > The problem is that the author tries to extend the logic without showing the logic can be extended in such a fashion. > > He starts with a pair of equal and parallel lines and a transversal and shows how that second pair of equal parallel lines (which intersect the original parallel lines) can be constructed. > > That’s fine. But then he assumes that additional equal/parallel lines can be constructed “next to” (for lack of a better way of putting it) the constructed parallels without showing that they exist or how to do it. > > Incidentally, this is similar to the mistake Proclus made. > > Proclus’ statement, for what it’s worth, is that if a straight line intersects one of a pair of parallel lines, it must intersect the other. Even though it’s not exactly the same error being made here, you can see how it’s similar. Point _J_ is a specific point on line _CD_; namely that one which is at an equal distance from the point _H_ as the point _E_ is from the point _G_ **(I.3).** Since we know that the lines _AB_ and _CD_ are parallel, they never meet in both directions if produced indefinitely **(I.Def.23).** Therefore, the lines _AB_ and _CD_ do not intersect at the point _J_ on line _CD_. So we know that no point on line _AB_ can pass through the point _J_ since if it did, that would mean that the line _AB_ does intersect the point _J_, which in turn implies that the line _AB_ does intersect the line _CD_ at the point _J_; hence, by contradiction, it has been proven that indeed, no point on line _AB_ can pass through the point _J_. So the point _E_, being a point on line _AB_, does not pass through/coincide with the point _J_. So point _E_ and point _J_ are distinct points. But any two distinct points can be joined by a finite straight line **(I.Post.1)**. So the finite straight line _EJ_, which is a transversal of _AB_ and _CD_, is constructible. I’m not sure whether this argument is sound or not, but it seems that the existence of the transversal (which is just a straight line, after all) is guaranteed by the first postulate (namely the existence of a straight line between any two points). I presented a drawn out argument in order to show that these points E and J are actually distinct…I think the only way that a transversal (straight line) couldn’t be constructed between E and J is if E and J were actually one and the same points, in which the “line” would be “degenerate.” ---
**Author:** ![supery00n](https://avatars.discourse-cdn.com/v4/letter/s/5f9b8f/32.png) [@supery00n](https://boards.straightdope.com/u/supery00n)\ **Post date:** [May 5, 2012, 10:38am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/25 "2012-05-05T10:38:00Z")
> [@](#): > > His set up is designed to prove this. But what he actually showed was the existence of a specific construction for two such new equal/parallel lines which joined the original parallel lines. The construction relied on a transversal. I think I might have responded in my last post to the wrong thing…It seemed more likely after rereading your post that you meant the very first statement of the proof. > [@](#): > > Let the straight line _EF_ falling on the two straight lines AB and CD make the alternate angles AEF and EFD equal to one another. I _think_ that one could start with line CD, and then take some arbitrary point F on the line CD and an arbitrary point E not on the line CD, and then join the points E and F to form a finite straight line EF that stands on CD at the point F. (basic stuff). Then you can construct an angle equal to a given angle on a given straight line with the vertex at a given point on that line. Looking at the diagram, it seems like the following line would be sufficient: Construct an angle, namely AEF, that is equal to angle EFD with vertex E and extend it in both directions, to the left to point A and to the right to point B. So the transversal EF has been constructed to AB and CD. (My argument doesn’t necessarily say that any two parallel lines can be cut by a transversal, it only says that one can construct at least one set of parallel lines that are cut by a transversal). The equality of alternate interior angles (I don’t know why the author just said “alternate angles” but I guess there are no exterior angles here…so that’s okay) then proves, by I.27, that AB and CD are parallel (meaning they never intersect in both directions). Is what I said about the transversal in this post what you meant? ---
**Author:** ![supery00n](https://avatars.discourse-cdn.com/v4/letter/s/5f9b8f/32.png) [@supery00n](https://boards.straightdope.com/u/supery00n)\ **Post date:** [May 5, 2012, 10:46am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/26 "2012-05-05T10:46:29Z")
I just ran out of the post edit limit too… > [@](#): > > So the transversal EF has been constructed to AB and CD. (My argument doesn’t necessarily say that any two parallel lines can be cut by a transversal, it only says that one can construct at least one set of parallel lines that are cut by a transversal). (quoting my own post above). Continued: But that seems to be ok since the author is just trying to show that there’s one triangle that obeys the 180 degree angle sum law. ---
**Author:** ![supery00n](https://avatars.discourse-cdn.com/v4/letter/s/5f9b8f/32.png) [@supery00n](https://boards.straightdope.com/u/supery00n)\ **Post date:** [May 5, 2012, 10:53am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/27 "2012-05-05T10:53:19Z")
I just ran out of the post edit limit too… > [@](#): > > So the transversal EF has been constructed to AB and CD. (My argument doesn’t necessarily say that any two parallel lines can be cut by a transversal, it only says that one can construct at least one set of parallel lines that are cut by a transversal). (quoting my own post above). Continued: But that seems to be ok since the author is just trying to show that there’s one triangle that obeys the 180 degree angle sum law. I’m not too sure of Wikipedia’s validity either, but I skimmed the talk page and someone recently added a source that contains the proof that the existence of one triangle with angle sum 180 is equivalent to (biconditional) the result that all triangles have an angle sum of 180 degrees; since the latter is equivalent to Euclid’s fifth postulate, by the transitive property, the existence of one triangle with angle sum 180 degrees is equivalent to Euclid’s fifth postulate. A guy in the talk page said that this source, unfortunately with no preview, has the proof of the first equivalence (namely between one triangle 180 and all triangles 180). It’s listed in the further reading section at the bottom of the page on the parallel postulate. Faber, Richard L. (1983), Foundations of Euclidean and Non-Euclidean Geometry, New York: Marcel Dekker Inc., ISBN 0-8247-1748-1 ---
**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\ **Post date:** [May 5, 2012, 11:20am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/28 "2012-05-05T11:20:41Z")
> [@ultrafilter](#): > > It’s not that we don’t know to do it; it’s that we know exactly why it can’t be done. Isn’t that overstating the case a bit? Some proofs are nonconstructive, and while they show that something can (or can’t) be done, they don’t show _exactly_ why… one might show that a counterexample must exist without being able to actually point to one. Just a quibble. Obviously I agree with the point that most people don’t understand the strength of “impossible” when it comes to math. ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [May 5, 2012, 12:10pm UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/29 "2012-05-05T12:10:49Z")
Let me add a point about the sum of the angles in any triangle. In euclidean geometry, the angle sum is exactly 180 (= pi radians) for all triangles. For a hyperbolic space, the angle sum is always \< 180, while for a spherical space it is always \> 180. In fact, on a sphere of radius, the excess over pi is exactly the area of the triangle. For example, imagine the triangle whose edges are the longitudes through 0 and pi/2 and the equator. All three angles are right, so the excess is pi/2. This triangle is 1/8 of the sphere whose area is 4pi. (Alternatively, you could think of the triangle as the outside of that and then every angle is 3/2 pi, the excess is 7/2 pi, which is the area outside.) But here is the main point I wanted to make. Think of the earth. Spherical in the large, but pocked by mountains, valleys, and passes. The geometry in the vicinity of a pass is hyperbolic. So if you smoothly move a small triangle from a pass to a more spherical region its angle sum will slowly change from \< pi to \> pi and there will, by continuity, have to be a point in between where it is exactly pi. (If you wish to know what I mean by smooth, it means having a continuous derivative, which will imply that the angles change continuously.) In such a situation, finding one triangle with angle sum pi doesn’t prove a thing. It would if the geometry had the same curvature everywhere. But the earth, for example, doesn’t. And neither does space since the local geometry depends on the masses lying around. ---
**Author:** ![Topologist](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/topologist/32/3208_2.png) [@Topologist](https://boards.straightdope.com/u/Topologist)\ **Post date:** [May 5, 2012, 12:22pm UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/30 "2012-05-05T12:22:58Z")
> [@Dr.Strangelove](#): > > Isn’t that overstating the case a bit? Some proofs are nonconstructive, and while they show that something can (or can’t) be done, they don’t show _exactly_ why… one might show that a counterexample must exist without being able to actually point to one. > > Just a quibble. Obviously I agree with the point that most people don’t understand the strength of “impossible” when it comes to math. It depends on what you mean by _exactly why._ A nonconstructive proof by contradiction may not point you to a counterexample, but will show you exactly why, if one didn’t exist, nonsense would follow. ---
**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\ **Post date:** [May 5, 2012, 3:34pm UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/31 "2012-05-05T15:34:14Z")
> [@supery00n](#): > > I’m not sure whether this argument is sound or not, but it seems that the existence of the transversal (which is just a straight line, after all) is guaranteed by the first postulate (namely the existence of a straight line between any two points). You’re not getting my point. The simple existence of a transversal doesn’t actually matter. As you note, you can always construct one. Actually, you can always construct several, though they may not have exactly the properties you want. But the proof relies on showing that you can always generate additional equal/parallel lines off of two already constructed equal/parallel lines. The author (I still think that’s you, but as it doesn’t actually matter and I have no proof, we can stick with a more general ‘author’) never really showed how that could be performed. The original construction of the equal/parallel lines was done in a very SPECIFIC way (involving a particular transversal that intersected these lines). The fact that a 3rd or additional equal/parallel lines could also be constructed was never proven. We don’t even know if they exist. Additional parallels certainly exist. Will they have the same length? Additional equal lines exist. Will they be parallel to the original 2? These things are never shown by the author - they are merely assumed to be true in general, when the general statement was never proven in the first place - only a specific statement about 2 PARTICULAR equal/parallel lines if given a pair of equal/parallels and a transversal. The construction of additional parallels was hand-waved away. Why is it so trivial as to be hand-waved away? There’s no clear way they actually can be constructed in the way the author thinks they can, unless you assume the Parallel Postulate is already true. I could be wrong about the previous sentence, but the author certainly doesn’t show how. > [@supery00n](#): > > I just ran out of the post edit limit too… > > (quoting my own post above). > > Continued: > But that seems to be ok since the author is just trying to show that there’s one triangle that obeys the 180 degree angle sum law. Actually, it matters more than you think it is. An additional equal/paralllel line is used to construct a 180 degree triangle. Without the additional equal/parallel line, the proof doesn’t work anyway. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 5, 2012, 9:22pm UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/32 "2012-05-05T21:22:01Z")
> [@](#): > > Quoth **Great Antibob** : > > Look up “absolute geometry” (it may also be called Neutral geometry). It’s geometry that can be done using only the first 4 postulates. One of the conclusions you can reach is that the angle sum of a triangle is at most 180 degrees. This is not true. There are geometries which obey the first four postulates where the angle sum of a triangle is always greater than 180 degrees. > [@](#): > > Quoth **Hari Seldon** : > > But here is the main point I wanted to make. Think of the earth. Spherical in the large, but pocked by mountains, valleys, and passes. The geometry in the vicinity of a pass is hyperbolic. So if you smoothly move a small triangle from a pass to a more spherical region its angle sum will slowly change from \< pi to \> pi and there will, by continuity, have to be a point in between where it is exactly pi. (If you wish to know what I mean by smooth, it means having a continuous derivative, which will imply that the angles change continuously.) In such a situation, finding one triangle with angle sum pi doesn’t prove a thing. It would if the geometry had the same curvature everywhere. But the earth, for example, doesn’t. And neither does space since the local geometry depends on the masses lying around. But doesn’t a nonuniform curvature violate the Fourth Postulate? ---
**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\ **Post date:** [May 5, 2012, 10:03pm UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/33 "2012-05-05T22:03:57Z")
> [@Chronos](#): > > This is not true. There are geometries which obey the first four postulates where the angle sum of a triangle is always greater than 180 degrees. Really? I’d like to see that sometime. I recall the proof limiting it to 180 from school, but I could be wrong. Of course, it’s greater than 180 for elliptic geometry, but, depending on how you interpret the first two postulates, elliptic geometry can be thought to violate either of the first two postulates. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [May 6, 2012, 12:12am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/34 "2012-05-06T00:12:53Z")
> [@Chronos](#): > > But doesn’t a nonuniform curvature violate the Fourth Postulate? How so? I don’t know what exactly Euclid meant by “All right angles are equal”, but certainly, along the pockmarked Earth, if we take measure and equality of angles in the usual way, all right angles are equal (all equally 90 degrees), yet there is nonuniform curvature. ---
**Author:** ![leahcim](https://avatars.discourse-cdn.com/v4/letter/l/b4bc9f/32.png) [@leahcim](https://boards.straightdope.com/u/leahcim)\ **Post date:** [May 6, 2012, 12:33am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/35 "2012-05-06T00:33:02Z")
> [@Chronos](#): > > This is not true. There are geometries which obey the first four postulates where the angle sum of a triangle is always greater than 180 degrees. The wikipedia article on absolute geometry mentions that absolute geometry can be proved to contain parallel lines so it is inconsistent with elliptic geometry which doesn’t have them. So the theorem is probably, “in any geometry with parallel lines, the maximum angle sum of a triangle is 180 degrees”. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 6, 2012, 12:34am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/36 "2012-05-06T00:34:25Z")
Well, if you define a right angle to be one quarter of a complete revolution, that can vary with curvature. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [May 6, 2012, 12:53am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/37 "2012-05-06T00:53:56Z")
> [@Chronos](#): > > Well, if you define a right angle to be one quarter of a complete revolution, that can vary with curvature. In the sense of “The angle making up one quarter of the circumference of a circle”, I suppose? Yes, you can even find that the arclength-to-circumference ratio of the angle formed by two rays depends on the radius you pick, without varying the rays; that is, the angle between two intersecting lines is not well-defined, only between two intersecting line segments of the same length, with the angle possibly changing if the line segments are extended. But nonetheless, on the interpretation of “angle” as “Arclength-to-circumference ratio in the limit as the radius goes to 0”, the pockmarked Earth serves as a model of non-Euclidean geomtry in which all right angles are equal and yet there is nonconstant curvature, and thus, “All right angles are equal” does not suffice to establish constant curvature. [For what it’s worth, I believe Euclid also defines “right angles”, as an angle such as POL or POR, where O lies on the line segment from L to R and the angles POL and POR are equal. In other words, half the angle between opposite directions. Equality of angles, though, is never defined.] ---
**Author:** ![supery00n](https://avatars.discourse-cdn.com/v4/letter/s/5f9b8f/32.png) [@supery00n](https://boards.straightdope.com/u/supery00n)\ **Post date:** [May 6, 2012, 5:16am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/38 "2012-05-06T05:16:05Z")
> [@Hari\_Seldon](#): > > Let me add a point about the sum of the angles in any triangle. In euclidean geometry, the angle sum is exactly 180 (= pi radians) for all triangles. For a hyperbolic space, the angle sum is always \< 180, while for a spherical space it is always \> 180. In fact, on a sphere of radius, the excess over pi is exactly the area of the triangle. For example, imagine the triangle whose edges are the longitudes through 0 and pi/2 and the equator. All three angles are right, so the excess is pi/2. This triangle is 1/8 of the sphere whose area is 4pi. (Alternatively, you could think of the triangle as the outside of that and then every angle is 3/2 pi, the excess is 7/2 pi, which is the area outside.) > > But here is the main point I wanted to make. Think of the earth. Spherical in the large, but pocked by mountains, valleys, and passes. The geometry in the vicinity of a pass is hyperbolic. So if you smoothly move a small triangle from a pass to a more spherical region its angle sum will slowly change from \< pi to \> pi and there will, by continuity, have to be a point in between where it is exactly pi. (If you wish to know what I mean by smooth, it means having a continuous derivative, which will imply that the angles change continuously.) In such a situation, finding one triangle with angle sum pi doesn’t prove a thing. It would if the geometry had the same curvature everywhere. But the earth, for example, doesn’t. And neither does space since the local geometry depends on the masses lying around. This is a really good argument, and I can’t really come up with a counterargument… ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [May 6, 2012, 11:19am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/39 "2012-05-06T11:19:36Z")
> [@Chronos](#): > > This is not true. There are geometries which obey the first four postulates where the angle sum of a triangle is always greater than 180 degrees. > > But doesn’t a nonuniform curvature violate the Fourth Postulate? I stand corrected. It certainly does. So constant curvature is built in. I never noticed that before. More precisely, I knew it, but never realized its implications. But looking at the axioms, I realized something else. Suppose Euclid had simply stated the sum of the angles of any triangle was 180. The objection to the fifth axiom is essentially that it supposes two lines meet at some undefined point, possibly far in the distance. But to assert that the angles of a triangle add up to two right angles is just as limited as the other axioms. Certainly the assumption that any two right angles are congruent is basically the assertion that you can lay one on the other and the sides will continue indefinitely to coincide. In fact, it is less and less clear, the more I think of it what that fourth postulate means. If I move the one right angle on to the other, staying inside the surface, the curvature will bring the one on to the other globally. And if you think of the right angle as living in some kind of platonic plane–well, I am at a loss. I think I’ll stick to my original claim. ---
**Author:** ![supery00n](https://avatars.discourse-cdn.com/v4/letter/s/5f9b8f/32.png) [@supery00n](https://boards.straightdope.com/u/supery00n)\ **Post date:** [May 14, 2012, 12:16am UTC](https://boards.straightdope.com/t/parallel-postulate-proven/620834/40 "2012-05-14T00:16:06Z")
What would you say is the consensus on the proof? My personal opinion is that it, if there are no logical errors, has a chance of falsifying hyperbolic geometry, although not elliptical geometry. I don’t think this has any implications for general relativity since (as far as I know), general relativity is based on a **four-dimensional** hyperbolic space **time** whereas the hyperbolic geometry that would potentially be falsified by this proof would pertain to **three-dimensional** hyperbolic **space**. What do you think about the fact that a parallelogram was constructed (a figure with opposite sides parallel and equal, and opposite angles equal) without invoking Euclid’s fifth postulate or any of its equivalents? In the Elements book I, it seems that propositions for parallelograms and their properties are proven only after I.29, which is the first proposition that requires Euclid’s fifth axiom to hold, is demonstrated. I tend to think that the construction of a parallelogram about its diagonals first, with the sides being constructed only after the diagonals are, is an interesting result in and of itself, and up to there, I see no error. But after that, I’m not completely sure… [Previous page](https://boards.straightdope.com/t/parallel-postulate-proven/620834.md?page=1) [Next page](https://boards.straightdope.com/t/parallel-postulate-proven/620834.md?page=3)