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Topic: JSH: End of an error
Replies: 59   Last Post: Dec 13, 2006 9:24 PM

 Messages: [ Previous | Next ]
 Rupert Posts: 3,810 Registered: 12/6/04
Re: JSH: End of an error
Posted: Dec 8, 2006 10:44 PM

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.
> >
> >
> > 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
> >
> > 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.
> > where now
> >
> > 7g'(x) = 7*f(x)/2 = 5a_2(x)
> >
> > and f(x) would have to divide out factors from 7, so there would be a
> >
> > Therefore the claim that g(x) can simply divide off factors from 7, so
> > that 7*g(x) does not have 7 as a factor is refuted.
> >
> > Note this claim is an objection raised to protect the current role of
> > the ring of algebraic integers as provably 7*g(x) does NOT have 7 as a
> > factor in that ring, with non-zero integer x as noted above.
> >
> > So what is the resolution to this weirdness?
> >
> > Well, consider evens and notice that 2 is not a factor of 6 with only
> > evens because 3 is not even.
> >
> > So it's not very complicated since the problem with the ring of
> > algebraic integers is about exclusion.
> >
> > Some numbers are excluded because they are not root of monic
> > polynomials with integer coefficients, so you can get weird stuff, and
> > an analogy is with evens where odds are excluded, so 2 is not a factor
> > of 6 with only evens.
> >
> > Proving the odd behavior of the ring of algebraic integer is easy. I
> > just did it yet again above.
> >
> > Explaining it simply is easy as well, as I just did, yet again with the
> > 2 and 6 analogy.
> >
> > But there are a lot of people around the world who learned mathematical
> > ideas that have been shown to be flawed and so far that group has
> > resisted publication, bizarre retraction of these ideas, and the death
> > of the math journal--to keep using the wrong ideas, and teach them to
> > others.
> >
> > While posters have routinely lied about the details of the argument and
> > objected for years in a steady attempt to deny what is mathematically
> > correct.
> >
> > And meanwhile undergraduates have been taught mathematical ideas proven
> > wrong years ago.
> >
> > I do not know what else to do here. If math people will not follow
> > mathematical proof, and even second proof, then what can anyone do?
> >
> > If the will for error is too strong for rational argument to work, then
> > it seems there are few options in this situation except to grieve over
> > the fall of humanity and the failure of civilization in this area.
> >
> >
> > James Harris

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 Michael Press
12/9/06 mensanator
12/9/06 Tim Peters
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 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 Larry Hammick
12/9/06 Ryugyong Hotel
12/9/06 William Hughes
12/9/06 Michael Press