The Straight Dope

Go Back   Straight Dope Message Board > Main > General Questions

Reply
 
Thread Tools Display Modes
  #1  
Old 05-03-2012, 10:42 PM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
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.
Reply With Quote
Advertisements  
  #2  
Old 05-03-2012, 10:44 PM
ultrafilter ultrafilter is offline
Guest
 
Join Date: May 2001
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.
Reply With Quote
  #3  
Old 05-03-2012, 10:48 PM
Giles Giles is offline
Charter Member
 
Join Date: Apr 2004
Location: Newcastle NSW
Posts: 11,554
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.
Reply With Quote
  #4  
Old 05-03-2012, 10:48 PM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
Quote:
Originally Posted by ultrafilter View Post
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.
That's what I was thinking too.
Reply With Quote
  #5  
Old 05-03-2012, 10:54 PM
zoid zoid is offline
Charter Member
 
Join Date: Sep 2001
Location: Chicago Il
Posts: 5,494
Me too.
Reply With Quote
  #6  
Old 05-03-2012, 11:03 PM
ultrafilter ultrafilter is offline
Guest
 
Join Date: May 2001
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.
Reply With Quote
  #7  
Old 05-03-2012, 11:06 PM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
Quote:
Originally Posted by Giles View Post
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.
Globally speaking, not all surfaces are planes. But all planes are surfaces.

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.
Reply With Quote
  #8  
Old 05-03-2012, 11:29 PM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
Quote:
Originally Posted by ultrafilter View Post
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.
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."

Last edited by supery00n; 05-03-2012 at 11:32 PM.
Reply With Quote
  #9  
Old 05-04-2012, 01:44 AM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
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.

Quote:
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.
I agree with this point. It's not like a proof of impossibility is like "it's too hard and we're saying it's impossible because it's been 2,000 years and we haven't managed to find a proof." I guess it's like if it's proven that the number of primes is infinite, it's proven that it's impossible to find a largest prime. It's not because we've always found bigger primes than what we thought might have been the largest one that we just gave up and said "there's no largest one..."

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.)...
Reply With Quote
  #10  
Old 05-04-2012, 03:11 AM
Indistinguishable Indistinguishable is offline
Guest
 
Join Date: Apr 2007
*ignore this post*

Last edited by Indistinguishable; 05-04-2012 at 03:13 AM.
Reply With Quote
  #11  
Old 05-04-2012, 03:18 AM
Indistinguishable Indistinguishable is offline
Guest
 
Join Date: Apr 2007
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.
Reply With Quote
  #12  
Old 05-04-2012, 04:03 AM
Indistinguishable Indistinguishable is offline
Guest
 
Join Date: Apr 2007
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.
Reply With Quote
  #13  
Old 05-04-2012, 12:47 PM
Omphaloskeptic Omphaloskeptic is offline
Guest
 
Join Date: Oct 2001
Quote:
Originally Posted by Indistinguishable View Post
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.
The initial assumption is impossible for elliptic geometry, so it can't rule out that non-Euclidean geometry; but it is consistent with hyperbolic geometry, so the result could still have nontrivial content (proving hyperbolic geometry impossible), if it weren't wrong.

Quote:
Originally Posted by Indistinguishable View Post
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.
Right, this is the obvious way to localize the error. I'm too busy to look at it closely now, but my first thought is that there's a confusion between a result holding for "a parallel" and holding for "all parallels," the distinction only becoming relevant in hyperbolic geometry.
Reply With Quote
  #14  
Old 05-04-2012, 12:54 PM
Saint Cad Saint Cad is online now
Guest
 
Join Date: Jul 2005
Quote:
Originally Posted by Giles View Post
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.
But can it be proven if we (as Euclid and apparently the author) tacitly assume geometry on a plane?
Reply With Quote
  #15  
Old 05-04-2012, 01:12 PM
Chronos Chronos is offline
Charter Member
 
Join Date: Jan 2000
Location: The Land of Cleves
Posts: 47,909
Quote:
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).
What this line of reasoning will eventually turn into is a definition, with a "true plane" being defined as one for which Euclid's fifth holds. In which case Euclid's fifth would be true but uninteresting. One might as well have an axiom that "all true triangles have exactly one right angle", where "true triangle" is defined as triangles with exactly one right angle.
Reply With Quote
  #16  
Old 05-04-2012, 01:23 PM
Great Antibob Great Antibob is offline
Guest
 
Join Date: Feb 2003
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.
Reply With Quote
  #17  
Old 05-04-2012, 01:28 PM
Giles Giles is offline
Charter Member
 
Join Date: Apr 2004
Location: Newcastle NSW
Posts: 11,554
Quote:
Originally Posted by Saint Cad View Post
But can it be proven if we (as Euclid and apparently the author) tacitly assume geometry on a plane?
Yes, but that's not the problem. Euclid, and later geometers, tried to prove various theorems based on a set of axioms. He (and they) found that you need the parallel postulate as an axiom to prove other things that they found to be obviously true (e.g., that the sum of the angles of a triangle is 180 degrees) -- or you need some other axiom that is equivalent to it. If your definition of a "plane" includes some equivalent to the parallel postulate, then you haven't solved the problem.

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.
Reply With Quote
  #18  
Old 05-04-2012, 03:30 PM
Great Antibob Great Antibob is offline
Guest
 
Join Date: Feb 2003
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:
Originally Posted by J. Yoon
...the existence of one triangle whose interior angle sum equals two right angles implies that the interior angle sum of all triangles is equal to two right angles (I.32)
(Bolding mine)

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:
Originally Posted by J. Yoon
Therefore, straight lines which join the ends of equal and parallel straight lines in the same direction are themselves equal and parallel
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.

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.
Reply With Quote
  #19  
Old 05-04-2012, 08:55 PM
Hari Seldon Hari Seldon is offline
Guest
 
Join Date: Mar 2002
Quote:
Originally Posted by supery00n View Post
Globally speaking, not all surfaces are planes. But all planes are surfaces.

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.
Euclid's fifth postulate was not a postulate for planes at all. It was a postulate for geometry. Coordinates were unknown. He has five (or nine or ten, since there were other notions we would now call axioms) and wanted to know if this one was a consequence of the remaining ones. It isn't and cannot be. Spherical geometry is one example, but slightly dubious (because one of the postulates that a line can be continued indefinitely and the statement is sufficiently ambiguous that it is not clear that a great circle on a sphere satisfies it), but a hyperbolic space certainly does. Therefore there's no need to look at some quack paper and find the error. Incidentally, one of the earliest discoveries of non-euclidean geometry was by someone trying to find a contradiction if you assume not PP and finding a perfectly consistent geometry (hyperbolic).

I wouldn't read angle trisections or cube duplications either because I have read and understood the impossibility proofs.
Reply With Quote
  #20  
Old 05-04-2012, 09:35 PM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
Quote:
Originally Posted by Great Antibob View Post
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:



(Bolding mine)

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:



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.

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.
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
Quote:
2. The sum of the angles in every triangle is 180° (triangle postulate).
3. There exists a triangle whose angles add up to 180°.
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..?

Thanks.

Last edited by supery00n; 05-04-2012 at 09:37 PM.
Reply With Quote
  #21  
Old 05-04-2012, 11:56 PM
Great Antibob Great Antibob is offline
Guest
 
Join Date: Feb 2003
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:
Originally Posted by supery00n View Post
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
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.

Quote:
Originally Posted by supery00n View Post
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.
Reply With Quote
  #22  
Old 05-05-2012, 12:01 AM
Great Antibob Great Antibob is offline
Guest
 
Join Date: Feb 2003
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.
Reply With Quote
  #23  
Old 05-05-2012, 12:04 AM
ultrafilter ultrafilter is offline
Guest
 
Join Date: May 2001
Quote:
Originally Posted by supery00n View Post
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.
Reply With Quote
  #24  
Old 05-05-2012, 12:33 AM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
Quote:
Originally Posted by Great Antibob View Post
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."
Reply With Quote
  #25  
Old 05-05-2012, 05:38 AM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
Quote:
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.

Quote:
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?

Last edited by supery00n; 05-05-2012 at 05:40 AM.
Reply With Quote
  #26  
Old 05-05-2012, 05:46 AM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
I just ran out of the post edit limit too...

Quote:
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.
Reply With Quote
  #27  
Old 05-05-2012, 05:53 AM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
I just ran out of the post edit limit too...

Quote:
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
Reply With Quote
  #28  
Old 05-05-2012, 06:20 AM
Dr. Strangelove Dr. Strangelove is offline
Guest
 
Join Date: Dec 2010
Quote:
Originally Posted by ultrafilter View Post
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.
Reply With Quote
  #29  
Old 05-05-2012, 07:10 AM
Hari Seldon Hari Seldon is offline
Guest
 
Join Date: Mar 2002
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.
Reply With Quote
  #30  
Old 05-05-2012, 07:22 AM
Topologist Topologist is offline
Charter Member
 
Join Date: Aug 2001
Location: Long Island, NY
Posts: 228
Quote:
Originally Posted by Dr. Strangelove View Post
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.
Reply With Quote
  #31  
Old 05-05-2012, 10:34 AM
Great Antibob Great Antibob is offline
Guest
 
Join Date: Feb 2003
Quote:
Originally Posted by supery00n View Post
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.


Quote:
Originally Posted by supery00n View Post
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.
Reply With Quote
  #32  
Old 05-05-2012, 04:22 PM
Chronos Chronos is offline
Charter Member
 
Join Date: Jan 2000
Location: The Land of Cleves
Posts: 47,909
Quote:
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.

Quote:
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?
Reply With Quote
  #33  
Old 05-05-2012, 05:03 PM
Great Antibob Great Antibob is offline
Guest
 
Join Date: Feb 2003
Quote:
Originally Posted by Chronos View Post
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.
Reply With Quote
  #34  
Old 05-05-2012, 07:12 PM
Indistinguishable Indistinguishable is offline
Guest
 
Join Date: Apr 2007
Quote:
Originally Posted by Chronos View Post
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.

Last edited by Indistinguishable; 05-05-2012 at 07:13 PM.
Reply With Quote
  #35  
Old 05-05-2012, 07:33 PM
leahcim leahcim is offline
Member
 
Join Date: Dec 2010
Location: New York
Posts: 1,194
Quote:
Originally Posted by Chronos View Post
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".
Reply With Quote
  #36  
Old 05-05-2012, 07:34 PM
Chronos Chronos is offline
Charter Member
 
Join Date: Jan 2000
Location: The Land of Cleves
Posts: 47,909
Well, if you define a right angle to be one quarter of a complete revolution, that can vary with curvature.
Reply With Quote
  #37  
Old 05-05-2012, 07:53 PM
Indistinguishable Indistinguishable is offline
Guest
 
Join Date: Apr 2007
Quote:
Originally Posted by Chronos View Post
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.]

Last edited by Indistinguishable; 05-05-2012 at 07:57 PM.
Reply With Quote
  #38  
Old 05-06-2012, 12:16 AM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
Quote:
Originally Posted by Hari Seldon View Post
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...

Last edited by supery00n; 05-06-2012 at 12:16 AM.
Reply With Quote
  #39  
Old 05-06-2012, 06:19 AM
Hari Seldon Hari Seldon is offline
Guest
 
Join Date: Mar 2002
Quote:
Originally Posted by Chronos View Post
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.
Reply With Quote
  #40  
Old 05-13-2012, 07:16 PM
supery00n supery00n is offline
Guest
 
Join Date: May 2012
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.
Reply With Quote
  #41  
Old 05-13-2012, 08:18 PM
ultrafilter ultrafilter is offline
Guest
 
Join Date: May 2001
Quote:
Originally Posted by supery00n View Post
What would you say is the consensus on the proof?
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.
Reply With Quote
  #42  
Old 05-13-2012, 09:19 PM
TATG TATG is offline
Guest
 
Join Date: Aug 2008
Quote:
Originally Posted by supery00n View Post
What would you say is the consensus on the proof?
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.
Quote:
Originally Posted by supery00n View Post
My personal opinion is that it, if there are no logical errors, has a chance of falsifying hyperbolic geometry, although not elliptical geometry.
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.
Reply With Quote
  #43  
Old 05-13-2012, 09:21 PM
Omphaloskeptic Omphaloskeptic is offline
Guest
 
Join Date: Oct 2001
Quote:
Originally Posted by supery00n View Post
What would you say is the consensus on the proof?
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.
Reply With Quote
Reply

Bookmarks

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is Off
HTML code is Off
Forum Jump


All times are GMT -5. The time now is 12:29 PM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2013, Jelsoft Enterprises Ltd.

Send questions for Cecil Adams to: cecil@chicagoreader.com

Send comments about this website to: webmaster@straightdope.com

Terms of Use / Privacy Policy

Advertise on the Straight Dope!
(Your direct line to thousands of the smartest, hippest people on the planet, plus a few total dipsticks.)

Publishers - interested in subscribing to the Straight Dope?
Write to: sdsubscriptions@chicagoreader.com.

Copyright © 2013 Sun-Times Media, LLC.