Re: Surrogate Factoring Theorem

From: Proginoskes (proginoskes_at_email.msn.com)
Date: 04/06/05 

Date: 6 Apr 2005 12:22:28 -0700

jstevh@msn.com wrote:
> Bryan Olson wrote:
> > Proginoskes wrote:
> > > When you're in the field of rational numbers, any number
> > > is a factor of
> > > another.
> >
> > Well, all but one of them.

Of course.

> Looking at the surrogate factoring theorem, you have
>
> b_1 b_2 = M^2

This is only given in the proof, not in the statement of the theorem.
(And before you say no, it's in the theorem, take a look at YOUR post
at
http://groups-beta.google.com/group/Surrogate-Factoring/browse_frm/thread/0ac6e9d61827a013/bb2241cca1c6969d#bb2241cca1c6969d
.

> and yes, trivially, you can say that any rational other than
> 0 will work for b_1 or b_2.
>
> But, actually, the theorem *directly* gives you b_2 f_1, where
>
> (b_2 f_1)(b_1 f_2) = M^2 T
>
> so, yes, b_1 and b_2 are rationals, but they can be integers,
> which, you might remember, are rationals as well.

Yes, but they don't have to be integers, and that's the point, which
you seem to be continually missing.* The following problem:

Given: An integer N
Find: Two rationals p and q (neither of which is +/-1 or +/- N) such
that p * q = N

is MUCH easier than the following:

Given: An integer N
Find: Two integers p and q (neither of which is +/-1 or +/-N) such that
p * q = N.

And THAT'S what all the fuss about factoring is; any 6-year old who
knows how to manipulate fractions can solve the second quickly (choose
p = 2, q = N/2, and it'll always work), but NO ONE has been able to
solve the first quickly.

> The theorem doesn't tell you one way or the other.

Which is why it's worthless in relation to factoring. You've just
proved my point.

     --- Christopher Heckman

* "You've managed to kill just about everyone, but like a poor
marksman, you keep missing the target." -- William Shatner, Star Trek
II: The Wrath of Khan (paraphrased).