Re: Playing with infinities
From: KeithK (me_at_nomail.com)
Date: 01/31/05
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Date: Sun, 30 Jan 2005 20:23:47 -0700
<jstevh@msn.com> wrote in message
news:1107135346.930520.226480@f14g2000cwb.googlegroups.com...
> I'm going to try and explain one more time, what a simple and elegant
> idea surrogate factoring is, and how exactly it relies on traversal of
> the entire set of rationals.
>
> Given
>
> yx^2 + Ax - M^2
>
> you trivially have
>
> x(yx + A) = M^2
>
> and there are an *infinite* number of rational solutions for x and y,
> given integers A and M.
>
> That doesn't help you much though, so I went on to
>
> yz^2 + Az - j^2 = 0
>
> where j^2 + T = M^2, and j is just some number you pick in order to
> factor M.
>
> Now, yz^2 + Az - j^2 = 0, also has an infinite number of rational
> solutions for y and z, given integers A and j.
>
> So you have two sets of infinity, and I look at their intersection.
>
> That intersection is given by the non-trivial set of solutions for y,
> such that both requirments are held. That is, such that
>
> yx^2 + Ax - M^2 = 0
>
> and
>
> yz^2 + yz - j^2 = 0
>
should be:
yz^2 + Az - j^2 = 0 ?
KeithK
> with rational y and x, and integers A, M and j.
>
> It just so happens that the mathematics works out that the non-trivial
> solutions for y are a finite set determined by the factorization of
>
> A^4 T^4.
>
> It is possible to prove--and rather easy to prove--that the set is
> complete in that it must give a factorization of M.
>
> It must give you a factorization of M.
>
> That set of rational solutions for y represents the intersection of two
> infinities, and that is what allows it to always factor M.
>
> Proving that is easy. It's so easy that there's just no excuse for
> mathematicians not believing me, and forcing some implementation to
> take place before they'll acknowledge the truth.
>
> If you people force this to go on because you will not accept
> mathematical proof, then you are complete frauds, and must be, as
> supposedly mathematical proof is what matters to you.
>
> It's not a hard theory.
>
> You can look at the details explained out at
>
> http://groups.yahoo.com/group/sufactor/
>
> where I'm also trying to figure out the implementation as I fear that
> only when there's a full demonstration will anyone really listen, as
> you people are not showing you actually care about mathematical proof.
>
> And the reality of what you think of mathematical proof may hurt a
> tremendous number of people worldwide, as this is not research I can
> control.
>
> Many of you have for years maybe played at being mathematicians
> thinking there was no responsibility attached, but if our economic
> society as we know it goes bye bye because you are really frauds,
> telling the world the factoring problem didn't have such a solution and
> then sitting around when the mathematical proof was thrust in your
> faces, then it will be your fault.
>
> Playtime is over.
>
> Either you are mathematicians or you're not, and if none of you are,
> when everything changes, then you'll understand that mathematics is
> about the truth, and not about what you wish to believe is true.
>
> I fear you are frauds who know next to nothing about mathematics, but
> instead play silly games with each other with abstruse works you don't
> even really understand, and you just got caught by forces you can't
> comprehend, and the consequences are far greater than you can, even
> now, imagine.
>
> And it's completely fair.
>
>
> James Harris
>
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