Re: SF: Two Az's
From: Proginoskes (proginoskes_at_email.msn.com)
Date: 04/25/05
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Date: 24 Apr 2005 16:30:38 -0700
jstevh@msn.com wrote:
> It turns out that the 50% factoring percentage can be seen
> a different way,
There is ONLY wone way to view the "50% factoring percentage"; that it
is wrong. You claim to have two sets of rational numbers, one where the
factors were trivial, and one where they weren't. Even if the
probabiliy of each of these infinite sets is the same, you STILL don't
get a 50% average, because I demonstrated that there was an infinite
set of rational numbers which are in NEITHER of your sets.
Look it up on Google Groups's Archive if you can't remember.
> which goes to focusing on what I call Az, in the SFT (which
> you can see further down in this post).
You may call it Az, but I call it zA.
> As if you have
>
> b_1 b_2 = M^2
>
> with rational b_1 and b_2, you can express those rationals
> using integers as
>
> b_1 = (g_1 d_2)/d_1
>
> and
>
> b_2 = (g_2 d_1)/d_2
>
> where g_1, g_2, d_1 and d_2 are integers, and
>
> g_1 g_2 = M^2
This is trivial to show: Let d_1 be the denominator of b_1 and d_2 the
denominator of b_2, then define g_1 and g_2 from this. I'm surprised
you haven't turned this derivation into another "SF Theorem".
(Oh, I figured out why! Because you're not sure how to prove
something's an integer!)
> and then you find from an equation in the derivation of the
> SFT that
>
> Az = (-f_1 g_1 d_2^2 - f_2 g_2 d_1^2 + 2M^2 d_1 d_2 )/d_1 d_2
>
> where f_1 f_2 = T.
>
> So if g_1 and g_2 are trivial then you have one value for Az,
> but notice, you can use the same d_1 and d_2, with a
> non-trivial value, which will give you a second value for Az.
>
> For every trivial Az value there is a second non-trivial value.
This is badly-worded at best. What your conclusion should be is: Az
depends on the factors of M^2 that you use.
> What's intriguing here is that mathematically there isn't a lot of
> indication that the mathematics cares either way.
Mathematics doesn't care about triviality? Have you bothered to check
when
(n-1)! divided by n has a remainder of n-1?
Mathematics can (and does) sort numbers into classes.
> What posters who disagree with me are basically arguing
> is that for some reason g_1 and g_2 will be picked
> mathematically as trivial factors, more often than not.
That is not true. The fact is that there is no way to randomly choose a
rational number, where all are equally likely, so you have to say _how_
you're choosing your rational numbers (what's the probability of each).
There IS a probability distribution where you get trivial factors 100%
of the time. There IS a probability distribution where you get
non-trivial factors 100% of the time. But to figure them out, you would
need to factor M^2 ahead of time, which isn't allowed.
The argument over the 50% result is due to the fact that you're not
saying how f_1 and g_1 are to be picked. If you say how to do it, the
50% result will be resolved once and for all.
But, of course, by not explaining how you choose your factors ("at
random" isn't good enough), you're just keeping the fire going.
> There's just no mathematical reason that I can see
Aha. Now the picture comes clear.
If you learned more about mathematics, you might be able to "see"
better.
This is me like saying: "I can't solve the differential equation y' = x
y, so solving any differential equation must be difficult."
> from that equation, and knowing that given d_1 and d_2,
> you can use g_1 and g_2 that are trivial or non-trivial,
> depending on your mood, it hardly makes sense that the
> mathematics would tend to go one way.
d_1 and d_2 are not the issue here. The issue is to choose b_1 so that,
in the end, g_1 and g_2 are non-trivial (i.e., not equal to 1, M, or
M^2).
Once again, you've shown that you're just ignoring the most fruitful
portion of SF. And you definitely should know about it, because it's
been brought up many times this month.
> But from a social perspective, if you accept what is
> mathematically true,
But you don't have to ACCEPT it; for any mathematical result, you can
read through the proof and decide, on the basis of that proof, whether
the result really is true. But since you don't know how to read a
proof, you HAVE to accept it on faith.
In short, it's your problem, not the world's.
> then it's this really big deal.
>
> Part of my point is that math society is corrupt.
Oh no! Next you'll be telling me political society is corrupt!
> It claims that what's mathematically true is all it cares
> about, but here, with something where the truth is
> embarrassing to math society, and potentially dangerous to
> the world's economy, they fudge.
I can only think of one example where the truth was buried because
mathematicians were too horrified by the results: Ancient Greece, when
they found out that sqrt(2) is irrational.
As far as "dangerous to the world's economy", do you really believe
that a system for security was chosen first, found to be easy to crack,
and the only thing keeping criminals from breaking into everyone's
accounts is the "mathematical propaganda"?
Do you really believe that this hypothesis is more likely than the
hypothesis that factoring REALLY IS hard, and that fact was why it was
chosen to be used for security?
It's like saying that all keys are alike, any door will open any key,
and that the only thing keeping burglars out of your apartment/house is
their belief that the key on their keyring will not open the door?
C'mon. Burglars at least jiggle the knob before breaking in.
> Mathematicians lie.
As you've said before, you're a mathematician.
Therefore, you're lying.
> And they will lie stupidly, as here, there really isn't a lot
> of room with the SFT to lie in an intelligent way, except to
> people who are not mathematically sophisticated.
Or to people who haven't checked out what exactly SF is.
> Here's the SFT.
>
> Surrogate Factoring Theorem:
>
> Given M, a target natural number to be factored, and j, an
> integer chosen such that j^2 > M^2, a rational factor b_2
> of M is given by
>
> b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2
>
> where T = M^2 - j^2, and f_1 is a rational factor of T,
> and Az is given by
>
> Az = Ax(Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2Ax - 2M^2)
>
> and Ax is given by
>
> Ax = +/- (k_1 + k_2) + 2j^2
>
> where k_1 k_2 = -Tj^2, and k_1 and k_2 are rationals.
And this "Theorem" doesn't say anything about non-trivial
factorizations; it just talks about rational numbers. (Nowhere in its
statement or "proof" does it even use the word "non-trivial", which is
the important part of factoring. It's easy to show that 131 = 1 * 131.)
--- Christopher Heckman
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