Am I a crank?
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From: Virgil on 8 Sep 2006 22:55 In article <d2g3g2d0s2l1u3spbjf6t3p1mg93mubc1v(a)4ax.com>, Lester Zick <dontbother(a)nowhere.net> wrote: > On Fri, 8 Sep 2006 01:00:39 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> > wrote: > >the problem is that it is in general impossible to prove (within > >the theory) that it is consistent. > > Is it also impossible to prove that it is not inconsistent? How does Zick allege that being "not inconsistent" is any different from being consistent?
From: Virgil on 8 Sep 2006 23:00 In article <ouk3g2lbnnmdlgreijtfss88d45cutpu4c(a)4ax.com>, Lester Zick <dontbother(a)nowhere.net> wrote: > On Fri, 08 Sep 2006 12:33:40 -0600, Virgil <virgil(a)comcast.net> wrote: > > >In article <03a3g2p6s0o7jc14jt3b2pcp5remsieb8n(a)4ax.com>, > > Lester Zick <dontbother(a)nowhere.net> wrote: > > > >> On Thu, 07 Sep 2006 17:35:36 -0600, Virgil <virgil(a)comcast.net> wrote: > >> > >> >In article <ij61g2dls6044ds806e87t95r8h4tf1ogv(a)4ax.com>, > >> > Lester Zick <dontbother(a)nowhere.net> wrote: > >> > > >> >> On Thu, 07 Sep 2006 13:26:12 -0600, Virgil <virgil(a)comcast.net> wrote: > >> >> > >> >> >In article <mah0g29jhf7u65h4um3k1jebid22us331o(a)4ax.com>, > >> >> > Lester Zick <dontbother(a)nowhere.net> wrote: > >> >> > > >> >> >> On Wed, 06 Sep 2006 17:13:32 -0600, Virgil <virgil(a)comcast.net> > >> >> >> wrote: > >> >> > > >> >> >> >Zick is the one whose trivia is founded in the trivium. Math is a > >> >> >> >part > >> >> >> >of the quadrivium. > >> >> >> > >> >> >> And modern math is founded, whatever that means, in the trivium and > >> >> >> not in the quadrivium. > >> >> > > >> >> >And how does someone so self-decaredly ignorant of mathematics know > >> >> >this? > >> >> > >> >> Because you self declaredly proclaim assumptions of truth in lieu of > >> >> demonstrations. > >> > > >> >Zick frequently does this, but why does he deceive himself that > >> >mathematicians emulate his idiocies? > >> > >> Because they don't and probably can't demonstrate their trivial > >> assumptions of truth. > >> > >But, unlike Zick, they are careful to point out just what unproven > >assumptions they are making. > > Hell that's easy enough: all of them. Just like Zick, who can't demonstrate any of his trivial assumptions of truth, but carefully hides all his trivial assumptions instead of honestly revealing them. > > > Thus no one need be deceived by them, though one cannot say the > > same about Zick's claims. > > ~v~~
From: Proginoskes on 9 Sep 2006 02:40 icahit(a)gmail.com wrote: > 1. Detailed proof on the spiral chain coloring will be given soon > (nothing will be changed in the structure of the proof) which is an > addition of the use (S,NS)-Kempe chain switching in case of undecided > case on the current spiral segment, where S is a "safe color" and NS is > a non-safe color. > > 2. On my proof of the Steinberg's conjecture, I have just added the > following justification: In the final step of the spiral chain coloring > if the planar graph without cycles of length 4 and 5 is four colorable > then the color four has to be assigned to the last node of the spiral > chain. But then edges incident to the last node has to be in a cycle of > length 4 or 5 which is an contradiction. What if there are other vertices colored with the fourth color? --- Christopher Heckman > Well after my presentation of my 4CT proof at the ICM someone asked me > then I will submit to a journal? I made the following joke, if someone > has the non-computer proof of the four color theorem then historically > (due to Kempe's proof) he has the right to wait at least 10 years. But > I am almost sure that no one will be able to submit an counter-example > or a serious critics. > > Cahit > > > Proginoskes wrote: > > Jesse F. Hughes wrote: > > > John Schutkeker <jschutkeker(a)sbcglobal.net.nospam> writes: > > > > > > > Isn't Arxiv peer-reviewed? > > > > > > No. > > > > There are a couple of Four Color Theorem papers which have obvious > > errors in them. (I found them in < 5 minutes.) One is by Friess, and > > the other is by an Asian mathematician. I haven't yet decided whether > > Cahit's result works, mainly because of the lack of a proof that is > > detailed enough. > > > > You might want to check Cahit's "proof" of Steinberg's Conjecture (any > > planar graph without 4- or 5-cycles is 3-colorable). There's a sentence > > in the proof which says (paraphrased), "We only need to consider the > > one following graph." He offers this statement without any kind of > > justification, and the rest of the paper is unnecessary fluff! > > > > --- Christopher Heckman
From: Chip Eastham on 9 Sep 2006 09:30 Tony Orlow wrote: [snipping much, jumping in after Moeblee cites transfinite induction] > Yes, but is that what I'm talking about? If it were allowable to prove > that 2x>x for all x>0, and omega or aleph_0 were considered greater > than 0, then 2*aleph_0>aleph_0 would be a fact. However, it's not in > the standard construction. Hi, Tony: Naturally you are the expert about what you are trying to talk about. In the interest of better communication, let me try my hand at what Moeblee has been trying to point out. Transfinite induction involves a case (limit ordinals) which do not arise in the case of finite induction (a limit ordinal is necessarily infinite). The usual (finite) induction step only deals with the successor ordinals. Here one can prove that if 2*x > x for some ordinal x, and y is the successor ordinal to x, then 2*y > y. So the case involving successor ordinals is okay for this hypothesis. What breaks down is the limit ordinal case, and the instance you cite of omega (an ordinal) or aleph_0 (a cardinal) is such a case. omega is _not_ a successor ordinal. It lacks an immediate predecessor (all predecessors are finite). While there are inductive hypotheses for which the argument can be made in the case of a limit ordinal, that the truth of the hypothesis for all preceding ordinals implies that truth of the hypothesis for the limit ordinal, the hypothesis 2*x > x is not such an example. So, while you are certainly free to claim that "transfinite induction", whatever it might be, is not what you wanted to talk about, the example you give above turns on the distinction between the case of a successor ordinal and the case of a limit ordinal. I hope it is not unreasonable to try to bring this distinction to your attention. regards, chip
From: icahit on 9 Sep 2006 13:06
Again the same argument applies that is there must be at least two triangles with a common edge which contains the vertex colored with the fourth color. But of course if we assume that G, planar graph without cycles of length four and five is chromatic critical then this situation occured at the last vertex of the spiral chain. Cahit Proginoskes wrote: > icahit(a)gmail.com wrote: > > 1. Detailed proof on the spiral chain coloring will be given soon > > (nothing will be changed in the structure of the proof) which is an > > addition of the use (S,NS)-Kempe chain switching in case of undecided > > case on the current spiral segment, where S is a "safe color" and NS is > > a non-safe color. > > > > 2. On my proof of the Steinberg's conjecture, I have just added the > > following justification: In the final step of the spiral chain coloring > > if the planar graph without cycles of length 4 and 5 is four colorable > > then the color four has to be assigned to the last node of the spiral > > chain. But then edges incident to the last node has to be in a cycle of > > length 4 or 5 which is an contradiction. > > What if there are other vertices colored with the fourth color? > > --- Christopher Heckman > > > Well after my presentation of my 4CT proof at the ICM someone asked me > > then I will submit to a journal? I made the following joke, if someone > > has the non-computer proof of the four color theorem then historically > > (due to Kempe's proof) he has the right to wait at least 10 years. But > > I am almost sure that no one will be able to submit an counter-example > > or a serious critics. > > > > Cahit > > > > > > Proginoskes wrote: > > > Jesse F. Hughes wrote: > > > > John Schutkeker <jschutkeker(a)sbcglobal.net.nospam> writes: > > > > > > > > > Isn't Arxiv peer-reviewed? > > > > > > > > No. > > > > > > There are a couple of Four Color Theorem papers which have obvious > > > errors in them. (I found them in < 5 minutes.) One is by Friess, and > > > the other is by an Asian mathematician. I haven't yet decided whether > > > Cahit's result works, mainly because of the lack of a proof that is > > > detailed enough. > > > > > > You might want to check Cahit's "proof" of Steinberg's Conjecture (any > > > planar graph without 4- or 5-cycles is 3-colorable). There's a sentence > > > in the proof which says (paraphrased), "We only need to consider the > > > one following graph." He offers this statement without any kind of > > > justification, and the rest of the paper is unnecessary fluff! > > > > > > --- Christopher Heckman |