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Topic: JSH: End of an error
Replies: 71   Last Post: Dec 13, 2006 9:24 PM

 Messages: [ Previous | Next ]
 jshsucks@yahoo.com Posts: 471 Registered: 3/26/05
Re: JSH: End of an error
Posted: Dec 9, 2006 12:20 PM

jstevh@msn.com wrote:
> Rupert wrote:
> > jst...@msn.com wrote:
> > > Rupert wrote:
> > > > jstevh@msn.com wrote:
> > > > > Rupert wrote:
> > > > > > jstevh@msn.com wrote:
> > > > > > > Rupert wrote:
> > > > > > > > Rupert wrote:
> > > > > > > > > jstevh@msn.com wrote:
> > > > > > > > > > The widespread usage of flawed ideas in number theory has put me in the
> > > > > > > > > > position of not just proving the errors and getting publication of
> > > > > > > > > > proof in a peer reviewed math journal, but necesitated the finding of a
> > > > > > > > > > second proof as well due to extraordinary resistance to such a shocking
> > > > > > > > > > result.
> > > > > > > > > >
> > > > > > > > > > I have finally found the long sought after second proof.
> > > > > > > > > >
> > > > > > > > > > Start with
> > > > > > > > > >
> > > > > > > > > > P(x) = 175x^2 - 15x + 2
> > > > > > > > > >
> > > > > > > > > > and do your analysis on the factorization
> > > > > > > > > >
> > > > > > > > > > P(x) = (f(x) + 2)*(g(x) + 1)
> > > > > > > > > >
> > > > > > > > > > where f(0) = g(0) = 0, where the functions are otherwise to be
> > > > > > > > > > determined.
> > > > > > > > > >
> > > > > > > > > > So I have
> > > > > > > > > >
> > > > > > > > > > 175x^2 - 15x + 2 = (f(x) + 2)*(g(x) + 1)
> > > > > > > > > >
> > > > > > > > > > and next I do some simple algebra starting with multiplying both sides
> > > > > > > > > > by 7:
> > > > > > > > > >
> > > > > > > > > > 7*(175x^2 - 15x + 2) = 7*(f(x) + 2)*(g(x) + 1)
> > > > > > > > > >
> > > > > > > > > > now I re-order in a special way on the left side and pick one way to
> > > > > > > > > > multiply by 7 on the right:
> > > > > > > > > >
> > > > > > > > > > (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2 = (f(x) + 2)*(7g(x) + 7)
> > > > > > > > > >
> > > > > > > > > > That work is done so that you have 5 and 7 on the left like a
> > > > > > > > > > polynomial with variables, except of course 5 and 7 are not variable,
> > > > > > > > > > but the operations are valid to this point.
> > > > > > > > > >
> > > > > > > > > > But now I need to balance the right side so let
> > > > > > > > > >
> > > > > > > > > > f(x) = 5a_1(x) + 5
> > > > > > > > > >
> > > > > > > > > > and
> > > > > > > > > >
> > > > > > > > > > 7g(x) =5a_2(x)
> > > > > > > > > >
> > > > > > > > > > which gives
> > > > > > > > > >
> > > > > > > > > > (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 = (5a_1(x) + 7)*(5a_2(x) + 7)
> > > > > > > > > >
> > > > > > > > > > allowing me to find a solution of the a's where they are roots of
> > > > > > > > > >
> > > > > > > > > > a^2 - (7x-1)a + (49x^2 - 14x) = 0.
> > > > > > > > > >
> > > > > > > > > > The problem though is in getting to this point--using some VERY simple
> > > > > > > > > > algebra--I multiplied g(x) times 7 to get 5a_2(x), so it makes sense
> > > > > > > > > > that a_2(x) has 7 as a factor.
> > > > > > > > > >

> > > > > > > > >
> > > > > > > > > Only if you have some reason to think that g(x) is an algebraic
> > > > > > > > > integer. You have no reason to think that whatsoever.
> > > > > > > > >

> > > > > > > > > > BUT if you plug in some numbers like let x=1, you have
> > > > > > > > > >
> > > > > > > > > > a^2 - 6a + 35 = 0
> > > > > > > > > >
> > > > > > > > > > which solves as
> > > > > > > > > >
> > > > > > > > > > a = (6 +/- sqrt(-104))/2
> > > > > > > > > >
> > > > > > > > > > and the result is that one of the roots has 7 as a factor.
> > > > > > > > > >
> > > > > > > > > > But provably neither do in the ring of algebraic integers.
> > > > > > > > > >
> > > > > > > > > > In arguing against this result posters have claimed that neither root
> > > > > > > > > > has 7 itself as a factor, but instead each root shares factors in
> > > > > > > > > > common with 7, but from the start with
> > > > > > > > > >
> > > > > > > > > > 7g(x) =5a_2(x)
> > > > > > > > > >
> > > > > > > > > > if a_2(x) does NOT have 7 itself as a factor then g(x) must divide off
> > > > > > > > > > some factors in common with 7.
> > > > > > > > > >
> > > > > > > > > > But going back to
> > > > > > > > > >
> > > > > > > > > > 175x^2 - 15x + 2 = (f(x) + 2)*(g(x) + 1)
> > > > > > > > > >
> > > > > > > > > > note that 175x^2 - 15x + 2 is always even when x is an integer.
> > > > > > > > > >
> > > > > > > > > > But now let
> > > > > > > > > >
> > > > > > > > > > f(x) = 2*g'(x) and g(x) = f'(x)/2
> > > > > > > > > >
> > > > > > > > > > and making the substitutions and simplifying gives
> > > > > > > > > >
> > > > > > > > > > 175x^2 - 15x + 2 = (g'(x) + 1)*(f'(x) + 2)
> > > > > > > > > >
> > > > > > > > > > and the same approach as before gives the same solution for the a's,

> > > > > > > > >
> > > > > > > > > No, it does not.
> > > > > > > > >
> > > > > > > > > We have
> > > > > > > > >
> > > > > > > > > f(x)=5a_1(x)+5
> > > > > > > > > g(x)=5a_2(x)/7
> > > > > > > > >
> > > > > > > > > where the a's are roots of
> > > > > > > > >
> > > > > > > > > a^2-(7x-1)a+(49x^2-14x)=0.
> > > > > > > > >
> > > > > > > > > Now, if we let g'(x)=f(x)/2 and f'(x)=2g(x), we have
> > > > > > > > >
> > > > > > > > > f'(x)=2(5a_2(x)/7)=5b_1(x)+5 where b_1(x)=2(a_2(x)/7)-1
> > > > > > > > > g'(x)=(5a_1(x)+5)/2=5b_2(x)/7 where b_2(x)=7(a_1(x)+1)/2.
> > > > > > > > >
> > > > > > > > > We have a_1(x)+a_2(x)=7x-1, a_1(x)a_2(x)=49x^2-14x.
> > > > > > > > >
> > > > > > > > > By working out b_1(x)+b_2(x) and b_1(x)b_2(x), we can get an equation
> > > > > > > > > of which the b's will be roots. It will not be the same equation.
> > > > > > > > >

> > > > > > > >
> > > > > > > > My apologies, this is not the way to get the minimum polynomial of the
> > > > > > > > b's. The two b's will have different minimum polynomials.
> > > > > > > >
> > > > > > > > b_1(x) will be a root of
> > > > > > > >
> > > > > > > > ((7/2)(b+1))^2-(7x-1)((7/2)(b+1))+(49x^2-14x)=0
> > > > > > > >
> > > > > > > > whereas
> > > > > > > >
> > > > > > > > b_2(x) will be a root of
> > > > > > > >
> > > > > > > > ((2/7)b-1)^2-(7x-1)((2/7)b-1)+(49x^2-14x)=0.
> > > > > > > >
> > > > > > > > As I say, I assume this is what you are trying to do, namely you are
> > > > > > > > letting g'(x)=f(x)/2 and f'(x)=2g(x), and then you are defining b's so
> > > > > > > > that
> > > > > > > >
> > > > > > > > f'(x)=5b_1(x)+5
> > > > > > > > 7g'(x)=5b_2(x).
> > > > > > > >
> > > > > > > > If that is what you are doing, then the b's will not be roots of the
> > > > > > > > same equation as the a's.
> > > > > > > >
> > > > > > > > If you are doing something else, you'd better explain what.

> > > > > > >
> > > > > > > I know, a good bit of shock to handle. Day before yesterday you were
> > > > > > > probably confident about your view of the world and one simple
> > > > > > > mathematical argument later, so much is shattered...
> > > > > > >

> > > > > >
> > > > > > Your delusions are very humorous.

> > >
> > > Yours are not.
> > >
> > > I have a simple request at the end.
> > >
> > > If any of those people I'm trying to protect die, then you will be held
> > > accountable.
> > >

> > > > > >
> > > > > > > Listen carefully, please.
> > > > > > >
> > > > > > > What I do is equivalent to realizing that with
> > > > > > >
> > > > > > > 175x^2 - 15x + 2 = (f(x) + 2)*(g(x) + 1)
> > > > > > >
> > > > > > > you can have
> > > > > > >
> > > > > > > 175x^2 - 15x + 2 = (g'(x) + 1)*(f'(x) + 2)
> > > > > > >
> > > > > > > where I just moved a factor of 2 around, and now the same argument that
> > > > > > > I used to give the a's now works with that expression to give the same
> > > > > > > equation which is
> > > > > > >
> > > > > > > a^2 - (7x-1)a + (49x^2 - 14x) = 0.
> > > > > > >

> > > > > >
> > > > > > Okay, let's spell this out.
> > > > > >
> > > > > > You have 175x^2-15x+2=(f(x)+2)(g(x)+1) where
> > > > > >
> > > > > > f(x)=5a_1(x)+5
> > > > > > 7g(x)=5a_2(x)
> > > > > >
> > > > > > where, for all, algebraic integers x, a_1(x) and a_2(x) are roots of
> > > > > >
> > > > > > a^2-(7x-1)a+(49x^2-14x)=0.
> > > > > >
> > > > > > Then you say, let g'(x)=f(x)/2 and f'(x)=2g(x).
> > > > > >
> > > > > > What you need to do is clarify your statement "now that same argument
> > > > > > that I used to give the a's now works with that expression to give the
> > > > > > same equation".
> > > > > >

> > > > >
> > > > > Um, dude, I showed you in my reply:
> > > > >
> > > > > 175x^2 - 15x + 2 = (f(x) + 2)*(g(x) + 1)
> > > > >
> > > > > and
> > > > >
> > > > > 175x^2 - 15x + 2 = (g'(x) + 1)*(f'(x) + 2)
> > > > >
> > > > > where I just moved a factor of 2 around, and you notice that one
> > > > > equation looks exactly like the first except with one you have f(x) and
> > > > > g(x), but with the other you have f'(x) and g'(x)?
> > > > >

> > > >
> > > > Yes, I know. And you wrongly think that that means if we set
> > > >
> > > > f(x)=5a_1(x)+5
> > > > 7g(x)=5a_2(x)
> > > >
> > > > and
> > > >
> > > > f'(x)=5b_1(x)+5
> > > > 7g'(x)=5b_2(x)
> > > >
> > > > then both {a_1(x),a_2(x)} and {b_1(x),b_2(x)} will be the solution set
> > > > for the equation
> > > >
> > > > a^2-(7x-1)a+(49x^2-14x)=0.
> > > >
> > > > But it's just not so.
> > > >

> > >
> > > Then give the polynomial.
> > >

> >
> > I did. Here we go again.
> >
> > If we let
> >
> > f'(x)=5b_1(x)+5
> > 7g'(x)=5b_2(x)
> >
> > then
> >
> > b_1(x) will be a root of
> >
> >
> > ((7/2)(b+1))^2-(7x-1)((7/2)(b+1))+(49x^2-14x)=0
> >
> >
> > whereas
> >
> >
> > b_2(x) will be a root of
> >
> >
> > ((2/7)b-1)^2-(7x-1)((2/7)b-1)+(49x^2-14x)=0.
> >

>
> That's not the technique I used in the original post, now is it?
>
> Starting with
>
> 175x^2 - 15x + 2 = (g'(x) + 1)*(f'(x) + 2)
>
> and using
>
> f'(x)=5b_1(x)+5
>
> 7g'(x)=5b_2(x)
>
> gives
>
> b^2 - (7x-1)b + (49x^2 - 14x) = 0
>
> if you follow the method I outlined in the original post.
>
> It's easy math Rupert, why lie about it?
>

> >
> > > > I've given you the minimum polynomials for b_1(x) and b_2(x) over Q
> > > > (assuming x is rational).
> > > >

> > >
> > > Give it in terms of g'(x) and f'(x).
> > >

> >
> > I did. I'm assuming you want
> >
> > f'(x)=5b_1(x)+5
> > 7g'(x)=5b_2(x)
> >
> > If you want something else, you've got to say what.
> >

>
> I demonstrated exactly.
>
> You dodged repeatedly.
>
> With math so trivial it's basic algebra.
>

> > > > > Let me make it even easier for you:
> > > > >
> > > > >
> > > > > 175x^2 - 15x + 2 = (f(x) + 2)*(g(x) + 1)
> > > > >
> > > > > and
> > > > >
> > > > > 175x^2 - 15x + 2 = (f'(x) + 2)*(g'(x) + 1)
> > > > >
> > > > > Now how do you think you can rationally claim that you can say one
> > > > > gives one answer while the other gives a totally different answer for
> > > > > the a's when there is just this minor notational difference?
> > > > >

> > > >
> > > > Because they are different functions.
> > > >

> > >
> > > Then give the equation for the a's or b's if that's what you wish to
> > > call them that results for g'(x) and f'(x).
> > >

> >
> > I did. Twice now.

>
> No you didn't as the correct answer is
>
> b^2 - (7x-1)b + (49x^2 - 14x) = 0
>
> and you lied about that to avoid acknowledging that the polynomial has
> the same roots as before.
>
> A very deliberate dodge
>

> > > A simple request.
> > >
> > > Mathematics is a beautiful and logical discipline, but human beings
> > > have one annoying ability--the ability to lie about the details.
> > >
> > > If any of those people I am trying to protect die, then you will learn
> > > what brave new world we have entered.
> > >
> > > You will learn it personally and your own people in Australia will be
> > > the ones to teach it to you.
> > >

> >
> > Get a grip.

>
> I should get a grip?
>
> A published result is lied about by sci.math'ers who not only manage to
> convince the bushwhacked editors into pulling the paper, but later the
> entire math journal dies.
>
> Posters keep lying about the details of the proof forcing me to find
> another simpler proof.
>
> And here you are lying about it, such that you are avoiding some of the
> easiest algebra possible.
>
> And I should get a grip?
>
> The only reason you're not being howled off the newsgroup is that
> mathematicians have this over hundred year old error that these methods
> show, and people are trying to sit quietly as if the math will change
> and they can feel brilliant again.
>
> But the math will not change.
>
> The mistake was made, the error came into number theory, and the math
> will not change.
>
> You need to get a grip as does the rest of the math community.
>
>
> James Harris

You are the only one in denial.

How about giving the "math suicide" fantasy a rest now. It is getting
quite disturbing. No mathematician will ever kill themselves over
anything you come up with. The ones you are talking about don't read
your posts and will probably never know what a raving loony you are, so
don't lose any sleep over their mortality.

Date Subject Author
12/8/06 JAMES HARRIS
12/8/06 Lord Protector
12/8/06 William Hughes
12/8/06 JAMES HARRIS
12/9/06 Jan Kristian Haugland
12/9/06 Michael Press
12/9/06 mensanator
12/9/06 Tim Peters
12/8/06 Jan Kristian Haugland
12/8/06 Jesse F. Hughes
12/8/06 David Bernier
12/8/06 hagman
12/8/06 Richard Tobin
12/8/06 Ryugyong Hotel
12/8/06 marcus_b
12/8/06 Gib Bogle
12/8/06 Brian Quincy Hutchings
12/8/06 Jesse F. Hughes
12/12/06 Brian Quincy Hutchings
12/8/06 Rupert
12/8/06 Rupert
12/8/06 JAMES HARRIS
12/9/06 Rupert
12/9/06 JAMES HARRIS
12/9/06 William Hughes
12/9/06 Rupert
12/9/06 JAMES HARRIS
12/9/06 Rupert
12/9/06 Ryugyong Hotel
12/10/06 Proginoskes
12/13/06 Odysseus
12/9/06 JAMES HARRIS
12/9/06 jshsucks@yahoo.com
12/9/06 William Hughes
12/9/06 Rupert
12/9/06 Rupert
12/9/06 JAMES HARRIS
12/9/06 Rupert
12/10/06 JAMES HARRIS
12/10/06 William Hughes
12/10/06 David Moran
12/10/06 Rupert
12/10/06 JAMES HARRIS
12/10/06 Rupert
12/10/06 Rupert
12/10/06 Rupert
12/10/06 JAMES HARRIS
12/10/06 Rupert
12/10/06 JAMES HARRIS
12/10/06 Rupert
12/10/06 gjedwards@gmail.com
12/10/06 JAMES HARRIS
12/11/06 Bob Marlow
12/10/06 Tim Peters
12/10/06 Rupert
12/10/06 Tim Peters
12/10/06 Rupert
12/10/06 William Hughes
12/10/06 Rupert
12/11/06 Tim Peters
12/9/06 jshsucks@yahoo.com
12/9/06 Ryugyong Hotel
12/9/06 Jesse F. Hughes
12/9/06 Tim Peters
12/9/06 Jesse F. Hughes
12/9/06 Ryugyong Hotel
12/9/06 Tim Peters
12/9/06 Jesse F. Hughes
12/9/06 William Hughes
12/9/06 Jesse F. Hughes
12/9/06 Michael Press
12/10/06 Random maxi