Re: SF: Some basic facts
From: Rick Decker (rdecker_at_hamilton.edu)
Date: 04/20/05
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Date: Tue, 19 Apr 2005 20:20:02 -0400
jstevh@msn.com wrote:
> First off, I have the surrogate factoring theorem, also known as the
> SFT, which is a theorem.
>
> So, as a theorem, it can't be refuted.
So far, so good. I agree that SFT is correct (but see below).
>
> That's a fact, which makes it nice for me because it's an irrefutable
> point from which to handle posters.
>
> That's a nice thing about mathematics: absolutes.
>
> Now posters have gone on and on about why they say it can't be made
> practical, but, so what?
>
> It's still a theorem, still an absolute, still absolutely true.
>
> I can rest on the SFT itself, without concern about practicality,
> though I've looked and it has NOT been proven that it doesn't lead to
> some practical factoring method from what I've seen posted.
Quite right, but there's empirical evidence that from a practical
point of view it's no better than trying randomly-chosen integers
for factors.
>
> Any such proof would be of interest, but the SFT would still be a
> theorem, regardless.
>
> Yes, I can rely on "pure math" here, as the concept is just amazingly
> clever, even if it turns out it doesn't work.
Here we part company. Far from being clever, the concept is trivial.
Think of what follows as a psychological experiment--I have a hypothesis
about your response that I'll reveal later.
Stripped to essentials, your theorem states:
Given integers M and j and if T is defined to be M^2 - j^2,
then there are functions g_i(p, q, r, s) such that
if k_1 and k_2 are rationals satisfying k_1 k_2 = -Tj^2
then we may express g_i(k_1, k_2, M, T) as the product of two
rational numbers b_2 f_1 where b_1 b_2 = M^2, f_1 f_2 = T.
You see that this is trivial, I hope, since b_2 and f_1
can obviously be chosen to be *any* nonzero rationals
at all and still satisfy the hypothesis of the theorem.
Furthermore, this theorem would still hold NO MATTER
WHAT FUNCTION YOU CHOOSE, as long as g(k_1, k_2, M, T)
is a nonzero rational.
In even simpler terms, you're asserting that a specific
rational may be expressed as the product of two rationals.
What Gertrude Stein said about Philadelphia is true about
your theorem: "There's no there there."
I await your response.
Regards,
Rick
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