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#1
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Parallel Postulate proven?
What do you think about this proof that was submitted to viXra (something similar to arXiv)? It's not supposed to be possible...what would be the implications if it were true (in mathematics and physics, in particular)? Do you think the proof is valid?
http://vixra.org/abs/1205.0003 The pdf of the supposed "proof" is on the page. |
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#2
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It's not possible to prove the parallel postulate from the rest of Euclidean geometry. We know this because we can point to examples that satisfy all the other axioms but not the parallel postulate.
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#3
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It can't be proven because there are consistent geometrical systems (e.g., the geometry of the surface of a sphere where the great circles are straight lines) embedded in Euclidean geometry, and satisfying the other axioms of Euclidean geometry. If a triangle on the surface of a sphere must have angles summing to 180 degrees, then three-dimensional Euclidean geometry is inconsistent -- indeed, ordinary arithmetic is inconsistent, because you can build 3-D Euclidean space from that -- and very little mathematics is left to be consistent.
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#4
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That's what I was thinking too.
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#5
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Me too.
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#6
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I'm not going to say that everything posted on viXra is nonsense and bullshit, but that's what I'll bet on every single time. There's a reason that these papers aren't posted on arXiv.
Edit: I should've put this on my previous post, but what a lot of people don't understand is exactly what a mathematician means when they say that something is impossible. It's not that we don't know to do it; it's that we know exactly why it can't be done. Unfortunately, the explanations are not usually lay-friendly, so it's a little difficult to impress this on people. Last edited by ultrafilter; 05-03-2012 at 11:05 PM. |
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#7
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Locally speaking, all surfaces with a local tangent plane approximation have are homeomorphic to a plane. A sphere is not a plane globally; it is only a surface that is locally homeomorphic to a plane. Euclid's fifth postulate, I think, is only meant to apply for surfaces that are planes globally, and not merely locally (i.e. "true" planes). I am not comfortable with extending Euclid's fifth postulate to surfaces for which it does not apply. Every conditional statement has its range of validity, and to say that it doesn't apply outside that range does nothing to either bolster or discredit the conditional statement. |
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#8
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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." Last edited by supery00n; 05-03-2012 at 11:32 PM. |
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#9
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I just read it, and my conclusion is it's too good to be true. It probably assumes something equivalent to the parallel postulate/fifth axiom. I didn't see any blatant errors in logic, but it didn't even address, let alone disprove, the proof of the impossibility of any proof of Euclid's fifth postulate.
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The proof was written up by an undergraduate student from UCLA apparently. And his last name is the same as mine... ![]() All that being said, however, I must admit I couldn't find a logical or factual error in the proof, and I knew this stuff pretty well since middle school (olympiads, competitions, etc.)... |
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#10
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*ignore this post*
Last edited by Indistinguishable; 05-04-2012 at 03:13 AM. |
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#11
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The very first line fails to hold in non-Euclidean geometry... It invokes proposition I.27 of Euclid's Elements, whose proof is in turn based on proposition I.16, the first in Euclid's Elements which fails in elliptic geometry.
(I thought, momentarily, that I had perhaps misread it and the first line was simply stating what was about to be proven, but that's not so: The third paragraph invokes the "Because AB and CD are parallel to one another" provided by the first line) Last edited by Indistinguishable; 05-04-2012 at 03:21 AM. |
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#12
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There must also be some other part which fails even in hyperbolic geometry (which does validate I.27), and it should be fairly easy to spot by just going through line by line with a particular hyperbolic counterexample to the conclusion in mind, but I'm lazy right now.
Last edited by Indistinguishable; 05-04-2012 at 04:04 AM. |
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#13
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#14
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#15
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#16
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No, the parallel postulate can't be proven even if you tacitly assume geometry on a plane. The standard notion of a 'plane' is already Euclidean, as Chronos has already noted.
You can have a sort of 'hyperbolic' plane, but lines won't behave on it the way you would expect (which is why the usual assumptions of behavior on a plane are implicitly Euclidean). I've gone through the proof line by line. It's actually rather cleverly written, but the key flaw comes in the 2nd paragraph. The crucial conclusion of the 2nd paragraph is only true for geometry on a Euclidean plane. It's a fairly major assumption, as most of the rest of the conclusions follow from it. If the "world" is hyperbolic, the assumption isn't necessarily true even on a plane. He assumes that if two particular line segments were extended indefinitely, they would be equal to each other. While that would certainly be true in a Euclidean plane, it's not universally true in a general geometric setting relying on just the first 4 postulates. Since that conclusion doesn't necessarily hold in general, the rest of the proof falls apart. Other notes I'd include if I were actually reviewing it: 1) "Euclids" in the title should be Euclid's. Overlooked it, I suppose. 2) Author needs to work on his language. His choice of words and sentence structure only serve to obfuscate the actual proof. There's no need to affect 19th century writing to sound impressive. 3) Author also needs to establish explicit lemmas. He uses lemmas proven within the text in later steps, and it makes the proof less readable when those lemmas aren't clearly noted. |
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#17
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The problem is not that Euclidean geometry (i.e., geometry with the parallel postulate) is inconsistent: it works fine. The problem is that it's not the only geometrical system possible: you can define consistent geometrical systems where the parallel postulate is not true. And that includes systems embedded in Euclidean geometry, where the "lines" defined there are well-defined objects that satisfy the other axioms of Euclidean geometry, but are not the "lines" of Euclidean geometry. So you need to take the parallel postulate (or an equivalent axiom) to define your Euclidean geometry. If you don't, you might have a system where the angles of a triangle add up to less than 180 degrees, or more than 180 degrees. |
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#18
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Oops. I made an error in my previous analysis.
His actual statement in the 2nd paragraph works just fine. But his final paragraph has a MAJOR leap of logic that just isn't true at all. Here it is: Quote:
In absolute geometry (the geometry you can do with just the first 4 postulates), this statement is not supported. In absolute geometry, the interior angle sum is at most two right angles. The existence of a single example of a triangle with interior angle sum of two right angles does not contradict the possibility of the existence of other triangles with interior angle sum less than 180 degrees. It's only if you assume some kind of Parallel postulate (0, exactly 1, > 1 parallels) that the interior angle sum can be used to define a geometry (Elliptic, Euclidean, Hyperbolic, respectively). But, of course, that simply leads to a tautology. While this error is bad, it doesn't necessarily prevent the possibility that Hyperbolic geometry is invalid while leaving Euclidean, Elliptic, and absolute geometry alone. Fortunately, he made an earlier error with a similar generalization that isn't backed up by the work he already did and is not necessarily true, unless one assumes a Euclidean framework. The statement he makes is: Quote:
The generalization is that such equal/parallel lines can be generated even without the conditions used for the construction (though he clearly didn't understand he was doing this). Even though it will work in a Euclidean world, that generalization fails if you don't have the Parallel Postulate. Last edited by Great Antibob; 05-04-2012 at 03:32 PM. |
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#19
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I wouldn't read angle trisections or cube duplications either because I have read and understood the impossibility proofs. |
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#20
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http://en.wikipedia.org/wiki/Parallel_postulate Quote:
Thanks. Last edited by supery00n; 05-04-2012 at 09:37 PM. |
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#21
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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. Quote:
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. Quote:
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. |
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#22
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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. |
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#23
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#24
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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." |
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#25
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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? Last edited by supery00n; 05-05-2012 at 05:40 AM. |
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#26
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I just ran out of the post edit limit too...
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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. |
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#27
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I just ran out of the post edit limit too...
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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 |
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#28
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Just a quibble. Obviously I agree with the point that most people don't understand the strength of "impossible" when it comes to math. |
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#29
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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. |
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#30
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#31
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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. Quote:
Without the additional equal/parallel line, the proof doesn't work anyway. |
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#32
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#33
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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. |
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#34
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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.
Last edited by Indistinguishable; 05-05-2012 at 07:13 PM. |
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#35
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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".
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#36
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Well, if you define a right angle to be one quarter of a complete revolution, that can vary with curvature.
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#37
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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.] Last edited by Indistinguishable; 05-05-2012 at 07:57 PM. |
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#38
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Last edited by supery00n; 05-06-2012 at 12:16 AM. |
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#39
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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. |
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#40
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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 spacetime 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... Last edited by supery00n; 05-13-2012 at 07:17 PM. |
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#41
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It's wrong. We have systems that satisfy the axioms of non-Euclidean geometry, and so we know that the parallel postulate does not follow from the other axioms. Since you're obviously interested in geometry, you should probably spend some time reading about them. There's a really nice textbook that covers various geometries at an undergrad level, and I recommend it highly.
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#42
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I would say you have smuggled in assumptions (about the truth of parallel postulate), which is easy to do given the subject matter, and that you should try and understand where in the proof such assumptions are made. And there is material in this thread which you can use to do so.
We have relative consistency proofs which show that Euclidean geometry is consistent iff hyperbolic geometry is consistent. If the parallel postulate follows from the first 4 axioms, then hyperbolic geometry is inconsistent, and thus Euclidean geometry is inconsistent. As un-intuitive as it might seem, and contra your project, proving that the parallel postulate follows from the first 4 axioms would prove that Euclidean geometry is inconsistent. |
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#43
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As ultrafilter says, it's wrong. A good way to see why it's wrong, as already recommended above, is to try to perform the construction in hyperbolic space, e.g. on the Poincare disk, and see where things go wrong. This is better done as an exercise than just having somebody show it to you, so I won't draw the picture for you.
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