Re: JSH: Explaining the huge math error

On Sep 2, 8:40 pm, JSH <jst...@xxxxxxxxx> wrote:
On Sep 2, 4:50 pm, Raphanus <lester.we...@xxxxxxxxx> wrote:





On Sep 1, 5:01 pm, JSH <jst...@xxxxxxxxx> wrote:

For years now I've tried to raise the alarm about a huge problem in an
esoteric branch of number theory where there may be now a huge case of
academic fraud, where the problem for me is that the mathematical
community itself is so impacted by the size of the error that I
haven't found a way to get mathematical proof of it accepted.

Luckily it's easy to explain and crucially depends on the distributive
property so my hope is that if I can convince mathematically
experienced members of the physics community of the existence of the
error they can help with the daunting task of handling the non-
response from the math community and the issue of possible widespread
academic fraud.

It's simple to explain the error.

Trivially if I have a polynomial factorization like x^2 + 3x + 2 = (x
+1)*(x+2), I can multiply both sides by 7, like

7*(x^2 + 3x + 2) = (7x+7)*(x+2)

and divide that 7 off, just as easily, without a trace, as I could
multiply by anything.

But over six years ago I came up with a technique where you have a
polynomial multiplied by a constant but factored into non-linear
functions:

P(x) = 175x^2 - 15x + 2

and

7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0,

So now you have something more complicated, where I've enmeshed the
quadratic P(x) with a quadratic generator, and it turns out that now
in an areas where mathematicians routinely operate to try and prove
things, you cannot divide the 7 off, when with an integer x:

a^2 - (7x-1)a + (49x^2 - 14x) = 0

has non-rational roots.

For instance, with x=1, you have the a's are the roots of

a^2 - 6a + 35 = 0

and if you try to use the quadratic formula you have a = (6 +/-
sqrt(-104))/2, and if 7 divides just one of those there is a problem
in SEEING it because sqrt(-104) is imaginary, and in fact you can
rigorously prove that in something mathematicians call the ring of
algebraic integers it is IMPOSSIBLE for either of the roots to have 7
as a factor.

So I can prove that one of the roots has 7 as a factor by the
distributive property AND prove that it cannot have 7 as a factor in
the ring of algebraic integers.  So there is an apparent
contradiction.

But the ring of algebraic integers is the ring mathematicians have
used for over a hundred years for arguments they think are proofs in
number theory and it just betrayed a huge problem because think back
to

7*(x^2 + 3x + 2) = (7x+7)*(x+2).

I can use functions f_1(x) = 7x and f_2(x) = x - 5, and have

7*(x^2 + 3x + 2) = (f_1(x) + 7)*(f_2(x) + 7)

and that doesn't change how things work so what's changed with

7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0?

The TYPE of function has changed.  I've gone from linear function to
non-linear ones that are the roots of a quadratic generator.

But if

a*(f(x) + b) = a*f(x) + a*b

by the distributive property, how can the TYPE of function change the
behavior?

a*f(x)

does not say, yeah, only if f(x) is linear, now does it?

Sure, maybe if one of the functions has 1/7 in it, the value can
change but the distributive property remains the same.  But the ring
of algebraic integers cannot have 1/7 in it.  It does not.

So something weirder is happening.

Turns out that no matter how trivial all this sounds the problem is
huge enough to invalidate mathematical arguments thought to be proofs
over the entire 20th century until now, as the ring of algebraic
integers became widely used by mathematicians in the late 1800's.

I have gone to the mathematical community with this problem, and even
got a paper bringing attention to the problem published in the now
dead math journal Southwest Journal of Pure and Applied Mathematics,
or SWJPAM for short.

Members of the sci.math newsgroup mounted an email campaign against
the paper.

The journal pulled it after publication.

It managed one more edition and then quietly shut down.

See:

http://www.emis.de/journals/SWJPAM/

and

http://www.emis.de/journals/SWJPAM/vol2-03.html

If the simple mathematical argument I gave you above is correct, then
it implies that ALL number theorists today who use the ring of
algebraic integers may have flawed results which are not correct
because it has a completeness problem, so the very people who are
tasked with accepting this error can be completely invalidated by it.

It may remove some of the arguments considered great works over a
period of a hundred years, and make useless a tremendous amount of
mathematical knowledge which can be shown to lack value by it.

At this point in time, rather than acknowledge the error,
mathematicians are continuing to TEACH the flawed mathematical ideas,
and have shut all doors--like publication.

The Catch-22 here is that the professors who would be tasked with
acknowledging this error are the very people who would find tremendous
invalidation from it, so much so in fact, that many of them may have
no real mathematical accomplishments at all: from their doctoral
theses to their latest research.

Admitting the truth would remove the very basis for their positions.

So then, what is the solution here?  How do you handle an error of
this magnitude?

James Harris

James,...I'm not a believer in conspiracy theories because they assume
the highly improbable (in any scenario) homogeneity of a large number
of people.  People are too egotistical and self-serving to "maintain a
silence".  Some math professor in Albania or Bulgaria , France or The
University of Podunk, America would have a lot to gain by publishing
in their local journal the refutation of a scared mathematical theorem
for it not to happen.  I don't know enough math to look for your
mistake, but I'm sure it is there.  Paranoia is terrible - especially
when they really are out to get you!

Well, I did get published.  The mathematical journal is now dead.
Google SWJPAM

I can do wacky demonstrations like suggest you Google "definition of
mathematical proof" where you should get my personal definition at my
math blog.

Ok, I can puzzle out ways to try and convince but it might be simpler
to ask, what if anything would convince you that there is an actual
problem with a HUGE over 100 year old error sitting at the core of
accepted number theory corrupting out the modern mathematical system?

Anything?  Or is there no way you could ever be convinced?

James Harris- Hide quoted text -

- Show quoted text -

All scientific experiments require verification before they are
accepted. All mathematical proofs (Fermat's Last Theorem, etc.)
require a vetting by competent peers. If you can get another
competent mathematician to confirm your proof, I'll be convinced - but
that would be of little matter.
.

Relevant Pages

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  • Re: A simple error
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  • Re: Now a change
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