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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: 3,797 Registered: 12/6/04
Re: JSH: End of an error
Posted: Dec 9, 2006 12:04 AM

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.
> > > >
> > > >
> > > > 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...
>

>
> 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".

I have given one interpretation of what this can mean.

We could set

f'(x)=5b_1(x)+5
7g'(x)=5b_2(x)

and then the claim might be that for all algebraic integers x, b_1(x)
and b_2(x) are roots of the same equation. As I pointed out, this is
easily proved to be wrong.

I suspect this is what you do mean, and it's wrong. If you mean
something else, you'll have to explain what.

> But the first way
>
> f(x) = 5a_1(x) + 5
>
> and
>
> 7g(x) =5a_2(x)
>
> so with the second way
>
> f'(x) = 5a_1(x) + 5
>
> and
>
> 7g'(x) =5a_2(x)
>
> where the a's are the same.
>

No, they are *not* the same. You should call them b_1(x) and b_2(x). I
gave you the equations of which they are roots.

Date Subject Author
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