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

 Messages: [ Previous | Next ]
 William Hughes Posts: 4,277 Registered: 12/6/04
Re: JSH: End of an error
Posted: Dec 9, 2006 12:42 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)
>

No, you have already defined f' and g'. You cannot define them
again.

- William Hughes

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
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12/8/06 David Bernier
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12/8/06 Rupert
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12/8/06 JAMES HARRIS
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12/9/06 William Hughes
12/9/06 Rupert
12/9/06 JAMES HARRIS
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12/10/06 Proginoskes
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12/9/06 William Hughes
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12/10/06 gjedwards@gmail.com
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12/11/06 Bob Marlow
12/10/06 Tim Peters
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12/9/06 jshsucks@yahoo.com
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12/9/06 William Hughes
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