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0|0|0|||||Desargues flaw|acco||15:04:49|08/24/2010|Dear Alex%0D%0A%0D%0AThere seems to be a flaw in the second  part of the proof%3A%0D%0A%22The proof above then applies to the triangles A1B1C1%27 and A2B2C2%27...%22%0D%0A%0D%0A%22The proof above%22 don%27t apply%2C if the triangles are in the same plane%2C %0D%0Awhich is possible in certain exceptional cases.%0D%0A%0D%0AAnyway this possibility is not commented.%0D%0A%0D%0AWith kind regards%0D%0AAllan
1|1|0|||||RE%3A Desargues flaw|alexb||05:13:36|08/25/2010|Allan%2C would you mind mentioning the exceptional cases. I do not immediately see any.
2|2|1|||||RE%3A Desargues flaw|acco||15:57:56|08/25/2010|One example is the case where%0D%0AA1%2CB1%2CA2%2CB2 are collinear and C1%3DC2%0D%0AThen we get C2%27%3DC1%27%2C and the new triangles are%0D%0Ain the same plane.%0D%0A%0D%0AI find it easier to adjust the proof than%0D%0Ato find all such cases. %0D%0A%0D%0Aacco%0D%0A
4|3|2|||||RE%3A Desargues flaw|Hubert Shutrick|1|05:43:28|08/27/2010|The case where A1%2CB1%2CA2%2CB2 are collinear does indeed cause that the triangles A1B1C%271 and A2B2C%272 are in the same plane because O is on that line%2C but the theorem is then obviously true because that line is also the perspective line. The method of proof still works if we lift the points A1 and A2 out of the plane instead of C1 and C2.%0D%0A%0D%0AI sought this proof because it only uses the axioms of incidence which are the first step when one tries to build projective spaces from axioms. The incidence laws for a plane projective space are that any two different points lie on a just one line and that any two different lines meet in just one point. As far as I know%2C Desargue%27s theorem cannot be proved using those axioms only. However%2C the incidence axioms for 3-dimensional projective geometry are all that is required for Desargue%27s theorem even in the case when the triangles lie in a plane.%0D%0A%0D%0AA 2-dimensional projective space with a pair of triangles that didn%27t satisfy Desargue%27s theorem couldn%27t be embedded as a plane in a higher dimensional projective space.%0D%0A %0D%0AAll the best%2C%0D%0A       Hubert
5|4|4|||||RE%3A Desargues flaw|acco||14:38:26|08/27/2010|Hi Hubert%2C%0D%0AI agree completely with your response.%0D%0AThe existing formulation is fine%2C if none of the corresponding vertices and sides are equal. %0D%0AThen it should be stated as assumption for the theorem. %0D%0A%0D%0ABut if it is decided%2C that the version of Desargues theorem and the proof  should be more general%2C then it would not be a problem%2C %0D%0Aif  A1 may be equal to B1 or  line A1B1 may be equal to A2B2.%0D%0A%0D%0AIndeed this will require the page rewritten starting with%0D%0A%0D%0ADesargues Theorem%0D%0ALet A1B1C1 and A2B2C2 be two triangles. Consider two conditions%3A%0D%0A   1. Lines A1A2%2C B2B2%2C C1C2  through the corresponding vertices can be chosen concurrent.%0D%0A   2. Collinear points ab%2C bc%2C ca can be chosen on the %28extended%29 sides A1B1 and A2B2%2C B1C1 and B2C2%2C C1A1 and C2A2%2C respectively.%0D%0A%0D%0AIf you and Alex are satisfied with the current restrictions %0D%0Athe suggestion is to write near the top of the web page%3A%0D%0A%0D%0AIt is assumed%2C that none of the corresponding vertices and sides are equal. %0D%0A%0D%0AAll the best to you and Alex%2C%0D%0Aacco
6|5|5|||||RE%3A Desargues flaw|alexb||00:38:03|08/28/2010|Thank you both for the discussion.%0D%0A%0D%0AMy feeling is that an appearance of two exclusive conditions needs to be explained. Without an example%2C the reader will be left wondering why the conditions have been introduced and where they have been used.%0D%0A%0D%0AI would much prefer a statement proved of course from incidence axioms as formulated by Hubert%3A if the pair of C-vertices leads to coplanar triangles%2C then either the A- or B-pair will work.
7|5|5|||||RE%3A Desargues flaw|Hubert|1|05:07:39|08/28/2010|Thanks Acco%2C%0D%0A%0D%0AYes%2C I was rather careless with the formulation.  As I implied in my first reply%2C my interest in the proof was to illustrate that it could be done and I seem to have left the special cases to the reader.  %0D%0A%0D%0A All the best%2C%0D%0A%0D%0A    Hubert
8|2|1|||||RE%3A Desargues flaw|acco||17:35:38|08/28/2010|Dear Alex%2C%0D%0AThank you and also Hubert for the kind answers.%0D%0AI don%27t think the Desargues page are meant for arbitrary positions of the triangles%2C as there would be%0D%0Amany %28may be ten or more%29 violations of the incident axioms%3A%0D%0A  e.g %0D%0A  - the places where lines intersect or meet%2C it should be different lines%0D%0A  - the places where point  join%2C it should be different points%0D%0A  More concrete %0D%0A  - %22Let O be the common point of A1A2%2C B1B2%2C C1C2%22 in case of A1%3DA2 and B1%3DB2%2C what is O%3F%0D%0A  -%22the plane OA1B1%22 in case of O%3DA1%2C what plane is it%3F%0D%0A  etc.%0D%0A%0D%0ATherefore two suggestions%0D%0A1. State as an assumption%2C that none of the corresponding vertices and sides are equal. %0D%0A OR%0D%0A2. You may use my note%2C formulating a proof in a symmetrical way%2C at%0D%0Ahttp%3A%2F%2Flanco.host22.com%2F%0D%0Aas you like. %0D%0A%0D%0AAll the best%2C%0D%0Aacco
9|3|8|||||RE%3A Desargues flaw|Hubert|1|12:16:42|08/30/2010|acco%2C%0D%0AI have suggested that Alex adds a two sentences saying that the proofs assume that the triangles are in general position and that special cases need special attention but the result is often obvious then.%0D%0A%0D%0AThe fact that lines%2C points or planes that are not uniquely defined usually means that any one satisfying the conditions will serve.%0D%0A%0D%0AI suppose you know that each field F gives a projective space FP%5En for each dimension n.  They all satisfy Desargue%27s although it is rather meaningless for F %3D Z%2F2 since there are only 2 points on each line.%0D%0A%0D%0AThanks again for the comments%2C%0D%0A%0D%0AHubert
10|2|1|||||RE%3A Desargues flaw|acco||18:37:13|08/30/2010|Dear Hubert and Alex%2C%0D%0AAs Hubert hints a general position is not possible for algebraic projective planes over the finite field Z%2F%282%29.%0D%0AIn such a Fano plane each line contain 3 points%2C as illustrated at%0D%0A  http%3A%2F%2Fweb.maths.monash.edu.au%2F%7Ebpolster%2Ffano.html%0D%0A  http%3A%2F%2Fplanetmath.org%2Fencyclopedia%2FFanoPlane.html%0D%0A  %0D%0AI have on the internet shortly tried to find proofs for Desargues theorem covering arbitrary positions of the triangles and found none. %0D%0A%0D%0AThe proof I suggest is found in a version 2 at%0D%0A  http%3A%2F%2Flanco.host22.com%2F%0D%0Aand is also valid for the Fano plane%2C which may be mentioned%2C if adapted. As you will see%2C the extra work is not much.%0D%0A%0D%0AKind regards%0D%0Aacco
11|3|10|||||RE%3A Desargues flaw|Hubert|1|04:31:29|09/02/2010|Dear Acco%0D%0A%0D%0AYes indeed%2C there are three points on each line. Sorry.%0D%0A%0D%0AIs there an example of a two dimensional space where the incidence axioms hold but Desargue%27s theorem is false%3F%0D%0A%0D%0ARegards%2C%0D%0A%0D%0AHubert
12|4|11|||||RE%3A Desargues flaw|acco||11:59:40|09/02/2010|Yes%2C Hubert%2C %0D%0AIn the link%0D%0Ahttp%3A%2F%2Fwapedia.mobi%2Fen%2FNon-Desarguesian_projective_plane%0D%0Aone example is  the Moulton affine plane%2C which seems to be less complicated than the others.%0D%0ABy supplementing an affine plane to a projective plane in the usual way%2C i.e.%0D%0Aletting infinite points be parallel bundles of lines%2C etc.%2C%0D%0Awe get a  non-Desargues projective plane with at least 3 point on each line.%0D%0A%0D%0AMany regards%0D%0AAllan Cortzen
13|4|11|||||RE%3A Desargues flaw|acco||15:45:21|09/02/2010|I better correct the typos in the last message. What is needed is the contra-positive of%3A%0D%0AAn affine sub-plane of a projective Desargues plane is always Desarguian.%0D%0A%0D%0AAllan Cortzen%0D%0A
14|2|1|||||RE%3A Desargues flaw|acco||09:56:08|09/07/2010|Dear Alex%0D%0AI don%27t think the change of the Desargues page is in you or the users best interest.%0D%0AYou should just assume general position%2C %0D%0Awhich here means corresponding triangle vertices are different and also%0D%0Acorresponding triangle sides are different.%0D%0AAnd nothing about exceptions%2C as they would require a lot of work to handle.%0D%0A%0D%0AAs I have mentioned before%2C many places %2810 or more%29 in the proof else are misleading.%0D%0A%0D%0AAlone from the first line of the proof%3A%0D%0A1. %22O be the common point of A1A2%2C B1B2%2C C1C2 %22 is without meaning is the triangles are identical.%0D%0A2. Suppose O%3DA1%3DA2 and the triangle are in the same plane. %0D%0AThat %22C1 is not in the plane OA1B1%22 is unclear%2C as many planes contains the line OA1B1%2C and we may or may not think of a plane containing C1.%0D%0A%0D%0AIt continues.%0D%0ALet me just mention in the converse proof dually it may happen%2C that E%3DO%27. Now from E the point O%27 is projected onto the original plane. This is meaningless.%0D%0A%0D%0AKind Regards%0D%0AAllan Cortzen
15|3|14|||||RE%3A Desargues flaw|alexb||15:50:31|09/07/2010|Alan%2C%0D%0A%0D%0A%3EAnd nothing about exceptions%2C %0D%0A%0D%0AWhy%3F I think it is useful to point out that the proof fails in some cases. Especially because it never explicitly makes use of the %22general position%22 assumption.%0D%0A%0D%0A%3Eas they would require a lot of %0D%0A%3Ework to handle. %0D%0A%0D%0AThat%27s is correct. And I have no intention to dig into that. But the lack of motivation to consider the exceptions is no reason not to mention their existence.%0D%0A%0D%0A%3EAs I have mentioned before%2C many places %2810 or more%29 in the %0D%0A%3Eproof else are misleading. %0D%0A%3E%0D%0A%3EAlone from the first line of the proof%3A %0D%0A%3E1. %22O be the common point of A1A2%2C B1B2%2C C1C2 %22 is without %0D%0A%3Emeaning is the triangles are identical. %0D%0A%3E2. Suppose O%3DA1%3DA2 and the triangle are in the same plane. %0D%0A%3EThat %22C1 is not in the plane OA1B1%22 is unclear%2C as many %0D%0A%3Eplanes contains the line OA1B1%2C and we may or may not think %0D%0A%3Eof a plane containing C1. %0D%0A%3E%0D%0A%3EIt continues. %0D%0A%0D%0AI agree with that. These are the examples where the proof fails. So what%27s wrong with alerting the reader up front to the fact that in some circumstances the proof fails%3F%0D%0A%0D%0A%3ELet me just mention in the converse proof dually it may %0D%0A%3Ehappen%2C that E%3DO%27. Now from E the point O%27 is projected onto %0D%0A%3Ethe original plane. This is meaningless. %0D%0A%0D%0AA construct invoked by the proof is meaningless in some circumstances. This in particular means that in these circumstances the proof fails.%0D%0A%0D%0AActually%2C I am not sure I am getting your point.%0D%0A%0D%0AIf we assume the general position the proof goes through. I believe we both agree on that point. We diverge in regard to my mention of the existence of special cases. You appear to claim that under the general position assumption there are no special cases. I agree with that. The special cases violate that assumption and%2C in addition%2C make some steps in the proof meaningless. Why this can%27t be mentioned is beyond me.%0D%0A%0D%0A%0D%0A
16|4|15|||||RE%3A Desargues flaw|acco||18:45:50|09/07/2010|Dear Alex%0D%0AFinally down to business.%0D%0AIn the latest version%3A %0D%0A%22We simply restrict the theorem to two triangles in general position%2C i.e.%2C assuming that no two points coincide and no three are collinear.%22%0D%0AI would prefer %0D%0A%22In the proof we simply restrict the situation to two triangles in general position%2C %0D%0Ai.e.%2C assuming that A1%21%3DA2%2C B1%21%3DB2%2C C1%21%3DC2 and similar for the sides.%22%0D%0A%0D%0AThe reasons are%0D%0A1.  The theorem is OK%2C only the proof is executed under limitations%0D%0A2.  In general position  e.g. A2%3DB1 is allowed%0D%0A3.  In general position  e.g. A1A2 and B1B2 should intersect in one point%2C so A1A2%21%3DB1B2%0D%0A4.  The triangle concept has not yet been discussed%2C and has the whole time been three non-collinear points.%0D%0A%0D%0AAbout the exceptions%3A%0D%0A%22These are simple enough to be treated individually and will not be considered here.%22%0D%0AI agree%2C but it is not so simple to find these cases%2C because of the language in proof. %28Remember your first response%29%0D%0AThe general position is used again and again in the proof without explicitly mentioning it.%0D%0ATwo cases already in the first line has been pointed out. It is not only cases similar to the case where A1%2CB1%2CA2%2CB2 are collinear and C1%3DC2. %0D%0A%0D%0AGreetings%0D%0AAllan 
17|5|16|||||RE%3A Desargues flaw|alexb||18:50:42|09/07/2010|Allan%2C many thanks for your attention and the time you devoted to that page. I hope now it is OK and may safely rest up there on the web.%0D%0A%0D%0AAlex