Re: SF: Areas of confusion, infinity

From: Nora Baron (norabaron_at_hotmail.com)
Date: 04/15/05 

Date: 14 Apr 2005 15:47:34 -0700

From: "Nora Baron" <norabaron@hotmail.com>
Newsgroups: sci.crypt,alt.math,sci.math
Subject: Re: SF: Areas of confusion, infinity
Date: Thu, 14 Apr 2005 15:13:01 -0700

JSH said:

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One of the problems with surrogate factoring in terms of understanding
it, is that it utilizes infinite sets, notably the set of rationals.

The naive view is that since in the set of rationals every number
except 0 is a factor of every other number that you can't use it in
factoring, which may be why I discovered the surrogate factoring
theorem, while others did not, as it necessarily has to use rationals
because of equations like

yz^2 - Az + j^2 = 0

and I say the naive view as I've already proven that is is naive with
the surrogate factoring theorem.

How?

Well, the surrogate factoring theorem links rational factorizations,
but in so doing, it shows no inclination for trivial versus non-trivial
factorizations.

It simply doesn't indicate a preference either way.

So you can pick rational factors as you see fit, and you'll get
rational factors in exchange, where there's no reason I've yet seen or
heard for the math to be choosy.

That means that I can use rationals and rely on human choice: a person
will pick factors.

The mathematics then gives you factors in exchange.

Now here's where it's not really tricky, but I've seen posters working
to try and force the issue that the mathematics IS picky, and they rely
on pseudo-mathematics.

If you in picking factors choose to pick trivial and non-trivial
factors, why, in response to your choices, should the theorem only
provide trivial factors?

There's no mathematical reason why.

But posters confuse on this issue by pointing out that in the set of
rationals every rational except 0 is a factor of every other rational,
so they argue that in exchange for your trivial and non-trivial factors
you will get trivial factors because if you randomly pick a rational
from the set of rationals it will tend to have a numerator coprime to
any integer you might want to factor.

Basically, they argue as if there isn't a link.

However, the surrogate factoring theorem is not playing social games.
It's a mathematical theorem which allows you to choose factors and in
exchange you get factors.

For the theorem to pick trivial factors only in exchange would be a
choice.

In mathematics there has to be a reason for a theorem to make a choice.

Posters don't give any reason.

Some of you rely on others to try and learn mathematics. I've seen
posts where people reply at how happy they are to be learning this or
from posters replying to me, when the sad reality is that posters here
who know mathematics are usually replying to me to confuse a particular
issue.

So they're teaching you bogus stuff.

Ok, so naturally, you may tell yourself, I'd say that posters are
teaching your wrong information, when you may feel confident that I'm
the person who is wrong.

But, notice, the surrogate factoring theorem simply links factors.

You have factors of j^2 T, where j is some number you choose, and T is
given by

T = M^2 - j^2

where M is presumably the number you're trying to factor.

If you get factors of j^2 T, then using the theorem you get factors of

M^2 T

and if you are doing the factoring of j^2 T, you can use trivial and
non-trivial factors, right?

For those confused on trivial versus non-trivial, consider 15.

15 is a factor of 15, since 15(1) = 15, but it's a trivial, as in easy,
factor, as it doesn't take any work for people to figure that out.

However 3 is a non-trivial factor, as 3(5) = 15, and you need to know a
little more to realize that it's a factor.

Human beings distinguish between hard and easy factors.

Now, if the surrogate factoring theorem, like human beings,
distinguishes between hard and trivial factors then it must have a
reason, as mathematics is that way.

There is a reason for everything in mathematics.

If you choose trivial and non-trivial factors to use with the theorem,
why should it link your factor to only or mostly trivial factors?

If it does there has to be a mathematical reason. Understand?

However, if it does not, then guess what?

--------------------------------------------------------------------

NB: Sure, guess what: the burden of proof is on your side.
A rational factor of M is trivial if its numerator has no factors
in common with M or if its numerator is divisible by M. The numerator
is an integer,not a non-integer rational. Say M = p * q, where p
and q are primes. Let n be the numerator of a rational which
divides M. The n either is, or is not, divisible by p. In real
applications, p is a VERY large prime. MOST integers are coprime
to it. If p > 10^50, then the chance that n is divisible by p is
less than 1/10^50 - unless you can prove otherwise. True, there
is a link between the divisors of T and the divisors of M. What
you must show, and have NOT shown, is that that link produces
numerators of the dividing rational which beat the 1/10^50 odds.

  The easiest way to find out is to write a little program.

  Why don't you do that?

--------------------------------------------------------------------
JSH:

If it does not then it means that people could develop ideas from that
theorem and build algorithms that could factor very efficiently, and
very quickly.

If they do so, then while we're arguing out these points on Usenet,
there may be some people who are factoring rather large numbers.

All it takes is for them to be less gullible than those of you who
listen to people who lie to you about basic mathematics.
----------------------------------------------------------------------

NB: I don't think anybody is listening to you any more. You
keep saying your method is going to work and change the world,
etc. But you are afraid to even write a little program and try
it.

--------------------------------------------------------------------
JSH:

Remember the link aspect of the surrogate factoring theorem, and ask
yourself, why should the theorem give only trivial factors in exchange
for trivial and non-trivial factors?

--------------------------------------------------------------------

NB: Again, the burden of proof is on you. If you think it has a
high probability of giving nontrivial factors, you should either
(1) provide a proof, or (2) write a little program and test it.
If all your "theorem" does is produce numerators which are no
better than random integers, then you will get only trivial
factors 99.99999999999999999999% of the time (for large M).

----------------------------------------------------------------------
JSH:

I've been looking at that question as I want to know the actual answer,

----------------------------------------------------------------------

NB: No you don't! You absolutely do NOT want to know the answer.
If you did, you would have written a little program long ago and
tested it. But you are afraid to do so. You know what will
happen. It will fail, and you will no longer be able to keep up
this false front. Go ahead, prove me wrong. It should be easy.
What's stopping you?

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JSH:

but posters are working to convince others for whatever reasons
motivate these people, so they basically just say it will only give
trivial factors for really big numbers, with only two very large prime
factors.

Why?

----------------------------------------------------------------------

NB: Why not?

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JSH:

Well really big numbers are important for encryption schemes used
to protect the Internet and lots of other stuff.

So they just pick an area where this work would be dramatic, and
say that it's not.

----------------------------------------------------------------------

NB: And you just sit there and cry, because you are afraid to even
test your own hypothesis! WHO ARE YOU TRYING TO FOOL ???

----------------------------------------------------------------------
JSH:

They are acting on social realities, and not mathematical ones.

----------------------------------------------------------------------

NB: Irony, thy name is Harris.

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JSH:

Notice, what happened when I presented the surrogate factoring theorem.
Some posters promptly tried to challenge whether or not it was a
theorem. Then most settled on calling it trivial and challenging me to
factor some number with it.

----------------------------------------------------------------------

NB: And you could have easily answered the challenge. But you
didn't. You were afraid. You know that the result will cause
your fantasies to come crashing down. You prefer to hang onto
them like a drowning man.

----------------------------------------------------------------------
JSH:

However, the theorem is enough to explain the link. The link between
factors is enough to show importance.

----------------------------------------------------------------------

NB: No, it absolutely is not. The fact that there is a link
shows nothing. YOU must show that that link implies that you have
a higher-than-random chance that the numerators yield nontrivial
factors of M. Or, you could write a little program and show it.
That is, you could if you could just scrape up the courage.

  Some men are afraid of terrorists or of bombs or of enemy
weapons, etc.. Most of them make good soldiers anyway and risk
their lives even though they are almost peeing their pants. You
however are afraid to write a little program. What you are doing
here is peeing your pants because you don't have the guts to test
your own ideas. You want to just keep bragging about it. It's
way past time to put up or shut up. Your pants are pretty much
soaked with pee at this point.

----------------------------------------------------------------------
JSH:

Making it work practically is a whole other arena.

But I fear it can be done, and I'm not going to do it.

----------------------------------------------------------------------

NB: Because you are chicken, not because you actually think it
will make a difference. This pretence that you actually have any
feelings for the human race doesn't ring the least bit true - not
coming from you. The real reason you are afraid to test it is,
you know it will fail.

----------------------------------------------------------------------
JSH:

So we're all waiting for those people who will, and we're waiting to
see what they do.

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NB: No one else thinks it is going to work. That is why you
have not seen any results. Too often in the past people have
done your work for you. So far this time it hasn't happened.
Let's say you were right. The sooner the world knows, the
better. Better that you should demonstrate it, so people could
start abandoning RSA, than that Al Qaeda does it surreptitiously
and does a lot of harm before anyone notices. No, your motives
are not altruistic. They are basically just cowardice. You are
afraid to test it because you know it will fail.

---------------------------------------------------------------------
JSH:

We can talk here as much as we want, but the real power has passed to
the people who are checking, and making it work, assuming they're out
there.

----------------------------------------------------------------------

NB: Could be you, if you were not a coward.

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JSH:

And I think they would be out there, as not everyone is gullible and
willing to listen to sci.math'ers with an agenda, who just always say
that whatever I have is wrong or unimportant.

-----------------------------------------------------------------------

NB: The gullible ones would be listening to you. Either that or
giving some Nigerian official's widow their bank account numbers.

----------------------------------------------------------------------

JSH:

Some people are bound to check.

----------------------------------------------------------------------

NB: Somebody might, but evidently not you.

Nora B.

-----------------------------------------------------------------------------------------------------------------

James Harris

Relevant Pages

  • SF: Areas of confusion, infinity
    ... One of the problems with surrogate factoring in terms of understanding ... it, is that it utilizes infinite sets, notably the set of rationals. ... Well, the surrogate factoring theorem links rational factorizations, ... The mathematics then gives you factors in exchange. ...
    (sci.crypt)
  • Re: Now you can see, my "crank" status
    ... >> Now you can see from the surrogate factoring theorem, ... rationals, as there's a mapping. ... The same argument can apply to any mapping if you say that a mapping is ... For people who've learned a lot of advanced mathematics, ...
    (sci.crypt)
  • Re: SF: Areas of confusion, infinity
    ... >> clearly that from the theorem only, but math people are pushing me ... > rationals will produce non-trivial factorisations of M, ... the mathematics does not distinguish between trivial and non-trivial ... The surrogate factoring theorem does in fact connect two infinite sets. ...
    (sci.crypt)
  • JSH: Explaining surrogate factoring, again
    ... Now I figure that many of you don't have a clue about the mathematics ... behind what I call surrogate factoring, as you listen to people who ... The aim is to find a *rational* y such that x and z are rationals. ... shown to depend on quadratic residues, such that there's about a 50% ...
    (sci.math)
  • JSH: Nearly done
    ... theory and method for factoring that I call surrogate factoring. ... The mathematics though is surprisingly simple, ... but are in the field of rationals. ...
    (sci.math)