JSH: Nearly done
jstevh_at_msn.com
Date: 01/24/05
- Next message: Student911: "Re: Help needed understanding dx,dy terms in integrals"
- Previous message: tomstdenis_at_gmail.com: "Re: Surrogate factoring, theory versus implementation"
- Next in thread: bryant_j_j_at_yahoo.com: "Re: JSH: Nearly done"
- Reply: bryant_j_j_at_yahoo.com: "Re: JSH: Nearly done"
- Reply: David Kastrup: "Re: JSH: Nearly done"
- Reply: ošin: "Re: Nearly done"
- Reply: Christian Bau: "Re: JSH: Nearly done"
- Reply: cbrown_at_cbrownsystems.com: "Re: JSH: Nearly done"
- Reply: Larry Lard: "Re: JSH: Nearly done"
- Reply: Gib Bogle: "Re: JSH: Nearly done"
- Reply: Nora Baron: "Re: JSH: Nearly done"
- Reply: litsohate_at_hotmail.com: "Re: JSH: Nearly done"
- Reply: David Rosoff: "Re: JSH: Nearly done"
- Messages sorted by: [ date ] [ thread ]
Date: 23 Jan 2005 16:34:31 -0800
There has been a lot of verbiage flying back and forth related to my
theory and method for factoring that I call surrogate factoring.
The mathematics though is surprisingly simple, while checking it
thoroughly can test you to your limits. It's some of the best kind of
mathematics to consider as elementary methods are shown to show
surprising results never conceived of before.
At the heart of the theory are two simple quadratics:
yx^2 + Ax - j^2 = T
and
yz^2 + Az - j^2 = 0,
where all are to be rational, while A, j and T are also integers.
The primary question is, do rational non-zero x, y, z exist?
So what does that have to do with factoring?
Well j and T are chosen such that M^2 = j^2 + T, where M is the number
you're trying to factor, and that means that from the first equation
you have
x(yx + A) = M^2
so x is a factor of M, but it can be a fraction, so you concentrate on
its numerator to see if that gives a prime factor of M.
So why M^2 and not M? Well that's where the theory starts pushing you,
as I tried M, and found out that mathematically it didn't give me
something I could work with easily, but M^2 did.
If you let b_1 b_2 = -j^2, and f_1 f_2 = T, it is easy to prove that
x = (b_1 f_2 + b_2 f_1 - 2j^2)/A
meaning that x is defined by the factors of j^2 and T, though it is a
factor of M^2, which is why this is surrogate factoring. The
surrogates are factored to try and factor M.
It's a brilliant idea. No matter what any poster says in reply, what
I've already shown is simply brilliant, as I have x, a factor of M,
defined by factors of numbers other than M, and that is the start.
The first problem you face is that b_1, b_2, f_1 and f_2 may not be
integers, but are in the field of rationals.
My paper goes over methods that put f_1 and f_2 into integers, while
allowing the mathematics to *choose* b_1 and b_2 so that they are still
in the field of rationals, but you get the mathematics selecting them
out of infinity.
Yup, that's why I call it a super sieve, as in picking b_1 and b_2 for
you, the algebra checks against all rationals, the entire set, and
that's an infinite set!
Now we are beyond brilliant into the arena of almost impossible to
imagine, with a technique for factoring, which loops through the entire
field of rationals in searching for a solution.
So must x reveal a non-trivial factor of M?
The answer, amazingly enough, depends on quadratic residues!!!
The mathematics requires only elementary methods, but in the space of a
few paragraphs I've talked about looking for a prime factor of some
target M, where the answer depends on the factorization of numbers
other than M, and you pick one of those numbers, while letting the
algebra factor the other, going through the entire field of rationals
to pick a solution!
All of that is easy to prove, and not only did I prove it in a paper, I
wrote a program that does it.
See http://groups.yahoo.com/group/sufactor/
I'm a person not just making claims--I have the full demonstration of
everything I say here, and I've worked it out, and put it out for
people to consider.
So, how can that be possible? How can what I say here be true, and the
mathematical establishment not pay attention?
There's something wrong with them. I have other big results and
they've tried to ignore those, and I've talked about some of them here
where people have usually lied about my work.
It's weird. But it's the reality.
With "pure math" people can lie. That's just a fact. If some group of
mathematicians write gibberish, and declare it to be a proof of
something, then basically that goes over, if they are considered to be
mathematicians of note.
Sure, mathematicians say that's not the way it is, but that's the way
it is.
And if an amateur mathematician makes major discoveries, it goes the
other way.
Mainstram mathematicians just band together and ignore their work, like
they will try to do with my latest discovery, which they also try to
do because they're very stupid.
The factoring problem is quite important in the real world.
It's not "pure math", so some of those people you respect and admire
may soon be in jail, hated worldwide, and villified by people who will
not be able to comprehend their behavior.
I find it hard to understand myself.
You do not believe that will happen.
It does not matter.
It will happen.
There are probably only a few days before many of your heroes are cast
down, humiliated, ripped from their positions, and put up on public
display as enemies of humanity itself, as people too blind to act in
the best interests of society, and too dumb to realize they wouldn't
get away with it.
You do not believe but it will happen, and the foundations of your
society will not just be shaken--they will be shattered--as I promised
years ago.
The end was never in doubt. You cannot betry the truth, and you cannot
block mathematics.
You betrayed mathematics itself, so you will be destroyed by it.
It is the just solution, the most logical, the most rational one, and
one of absolute perfection.
Those who betrayed the field of mathematics, who sullied that field
with their lies, and their belief that social rules could win that they
could make up "truth" and get away with it are about to learn just how
powerful of a field it truly is as a lesson that humanity will never
forget.
No one will be able to forget.
You will learn this time.
You will learn, never to make these mistakes again, and the lesson will
be complete.
James Harris
- Next message: Student911: "Re: Help needed understanding dx,dy terms in integrals"
- Previous message: tomstdenis_at_gmail.com: "Re: Surrogate factoring, theory versus implementation"
- Next in thread: bryant_j_j_at_yahoo.com: "Re: JSH: Nearly done"
- Reply: bryant_j_j_at_yahoo.com: "Re: JSH: Nearly done"
- Reply: David Kastrup: "Re: JSH: Nearly done"
- Reply: ošin: "Re: Nearly done"
- Reply: Christian Bau: "Re: JSH: Nearly done"
- Reply: cbrown_at_cbrownsystems.com: "Re: JSH: Nearly done"
- Reply: Larry Lard: "Re: JSH: Nearly done"
- Reply: Gib Bogle: "Re: JSH: Nearly done"
- Reply: Nora Baron: "Re: JSH: Nearly done"
- Reply: litsohate_at_hotmail.com: "Re: JSH: Nearly done"
- Reply: David Rosoff: "Re: JSH: Nearly done"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|