Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
sci.math.*
»
sci.math
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
My paper, and the cheaters
Replies:
116
Last Post:
Oct 5, 2004 7:46 PM




Re: My paper, and the cheaters
Posted:
Sep 18, 2004 7:47 PM


jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409171443.51d8ff25@posting.google.com>... > norabaron@hotmail.com (Nora Baron) wrote in message news:<36024859.0409170732.566a5e55@posting.google.com>... > > jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409161404.365eaa1e@posting.google.com>... > > > I've made some pointed criticisms of math society but if you actually > > > pay attention you'll realize that the evidence against math society is > > > nastier than I go into detail about usually. > > > > > > Supposedly a correct math result is what's important, but despite my > > > having a quite correct paper, which was to be published, sci.math'ers > > > managed to get it yanked IN ONE DAY with some hostile emails. > > > > > > The paper has been by a lot of editors and mathematicians and to date, > > > no major error has been found, which I say because there was one minor > > > error that actually got pointed out by a sci.math poster. > > > > > > > Please do not ever again claim to be an honest person. > > Your "proof" contradicts known mathematics. At least 4 > > counterproofs have been given. Dale Hall's counterproof > > can be checked with simple arithmetic. The exact spot > > in your "proof" where you make your major error has been pointed > > out repeatedly. You know all this and yet you continue to > > deny it. > > That's not true. > > In fact, I don't disagree with much of what you say. > > And I'd like you to admit that in a post, so that I can move on to > what I actually *DO* say. > > First though, I need you to read what I have below and acnowledge that > you see the agreement on a key point. >
What??? You want people to agree to your own statement that you understand and accept a proof? Why don't you just say you accept it and go on? Why the gameplaying?
> > > > > Below is one of the simplest counterproofs. Feel free > > to point out errors (assuming you are still interested in > > math as opposed to social studies). This is, incidentally, > > for your future reference, what real proof is supposed to > > look like. > > > > ================================================================ > > > > James Harris has written a paper called "Advanced Polynomial > > Factorization" in which he claims the following: if the > > polynomial > > > > 65*x^3  12*x + 1 > > > > is factored in the form > > > > (a1*x + 1)*(a2*x + 1)*(a3*x + 1), > > > > where a1, a2 and a3 are algebraic integers, then exactly > > two of a1, a2, and a3 are divisible in the algebraic integers > > by sqrt(5), and the third one is coprime to 5. > > > > It should be noted that a1, a2, and a3 are roots of > > > > x^3 + 12*x^2  65. > > > > That Harris's claim is false can be seen from the following: > > > > ========================================================================== > > > > Assume m(x) is an irreducible monic polynomial with integer coefficients > > and algebraic integers a and b are two roots of m(x). > > > > Theorem. If p is a nonzero rational integer and a is divisible by > > sqrt(p) in the algebraic integers, then b is also divisible > > by sqrt(p). > > That last should say, in the algebraic integers. > > Theorem accepted. Proof not needed here so I deleted it out to focus > on what's important. It is in fact true that if 'a' is divisible by > sqrt(p) in the algebraic integers then b is also divisible by sqrt(p) > ***in the ring of algebraic integers***. >
Seems obvious that's what she's saying.
> Concede agreement "Nora Baron" and then I'll explain the rest. >
She did I believe.
> > > > > This polynomial is monic and irreducible with integer coefficients. > > By the theorem above, if one of the roots is divisible by sqrt(5), > > then they all are. This is not possible because the product of the > > roots is 65 = 5 * 13. Therefore Harris's claim that ANY of the > > roots are divisible by sqrt(5) is false. > > > The poster has shifted from the true claimthat the result applies to > algebraic integersto a far vaguer and more inclusive claim, > basically that it applies in general. >
I don't think that was her intention. I think she showed that a/sqrt(5) is not an algebraic integer. Look at her proof. She isn't claiming more than that.
> Repeatedly, posters keep making claims true in the ring of algebraic > integers, I accept those claims in that ring, and they then act like I > don't!!! > > I repeat, the result "Nora Baron" gave is correct *in the ring of > algebraic integers* and I now challenge that poster and others in that > camp to finally and for once concede that I am not arguing against > that point. >
You want the people in "that camp" to agree that you agree with them ?
> Then I'd like, with the permission of "Nora Baron" to explain exactly > what my proof actually says. >
You need PERMISSION to do this ???
> But first I want the poster "Nora Baron" to accept that I am not > challenging on this specific point. > > That's the first step. > > It's up to that poster to respond. >
Like I say, she did.
Incidentally, what did they say about APF at the journal to which you submitted it earlier in the summer?
Andrzej
> > James Harris



