Re: JSH Paper proven totally wrong 5 years ago by many, so why is JSH still whining now ?


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W. Dale Hall View profile
More options Jul 21 2003, 2:44 pm

Newsgroups: sci.math, sci.skeptic, alt.writing
From: "W. Dale Hall" <mailtodh...@xxxxxxxxx>
Date: Mon, 21 Jul 2003 19:50:47 GMT
Local: Mon, Jul 21 2003 2:50 pm
Subject: Re: Reply from German editor about my paper
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James Harris wrote:
Will Twentyman <wtwenty...@xxxxxxxxxxx> wrote in message
<news:3f1aa917$1_2@newsfeed>...

James Harris wrote:


... stuff deleted ...

I noticed a poster DID reply to this classic example, but didn't point
out the obvious, possibly because the assumption is that everyone
knows why the argument is flawed. Now someone might call it a proof,
but it's clearly not, as a proof is correct. That's easy to get
confused because you may hear people talking about proof, when they
have a *claim* of proof.

Most importantly, notice that a short, flawed argument can be handled
by showing a break in the logical chain at a single point.

Then why do you fail to address breaks in your logical chain that others
have pointed out? Why do you fail to deal with the counter-examples and
counter-proofs?

How am I supposed to answer those questions? If you're right then I'm
someone who has deluded himself to the point I'm no longer rational.
Having lost trust in my own mind, what would I have left? Yet if
you're wrong, then what can I say in response?


You can state that the other person is wrong, and point out the error,
as you're demanding of others [yet continue to evade the responses].

I, for one, have pointed out [and will do so again in this article]
a point where your proof breaks down. I do not point out a statement
that fails to follow from its predecessors in the argument, because
your argument is so foggily written that virtually *every* statement
fails to follow from those that precede it.

Rather than slog through a pile of eminently unreadable, ill-formulated
pidgin-mathematical blathering, I point out where you make an incorrect
statement. As you yourself will attest, a correct argument cannot (given
consistency, which you are not addressing, nor do you have the knowledge
to address) produce an incorrect result. Your result is demonstrably
incorrect (as I point out), therefore your argument fails.

What I will say is that if you point to a single logical break in the
chain, that is all that's necessary to sink a proof claim. So your
questions are automatcially suspicious with the claims of
"counter-examples" and "counter-proofs".


That is one way to rebut an incorrect argument, but it is also
sufficient to demonstrate that the argument produces an erroneous
conclusion. You are doing yourself no service by evading that point,
especially when the erroneous conclusion is one you promoted in
your paper. That is important, since you cannot claim that someone
else somehow mangled your argument to produce the error. If anyone
understands that argument, it must be you, and if you produce
errors in it, then no one else can be blamed for being somehow
mis-interpreting what you say, or particularly inept at doing
what you call "simple math".

To claim that my demonstration (that you have a false result) is
somehow suspicious, or that anyone else's questions to you, based
on that result, are suspicious, is unconscionable.

If you can't support your argument in a *particular* case, how then
can you dare to claim that it holds *in general*?

For you to repeatedly state, "It's a correct argument, because no one
has found the statement where it goes awry," is NOT a legitimate defense
against a demonstration that your argument produces an erroneous result.

Now let's suppose that you believe what you're saying, and you're
facing me, whom you must think is a deluded person, would you use your
*best* most succinct evidence i.e. the "smoking gun" or would you try
to build a case?


Stop the posturing. This is not about a legal case, this is not about
a commitment proceedings, this is about your continuing parading of
a thoroughly discredited argument, with claims of its being a new
result about algebraic integers. Given that you don't even know what
a ring is, or what a polynomial is, or a variable, for that matter,
why do you feel that your pitiful efforts have exposed the sham
that you imagine mathematics to be? Maybe you've exposed the sham that
mathematicians imagine you are?

If you're trying to convince others, you might build a case, but why
bother if you have a smoking gun? If you have videotape of a witness
shooting the victim, why bring in character witnesses?


Whom is it that you imagine you're referring to?

Do you imagine that someone other than you needs any convincing? From my
perspective, all this effort is directed towards getting you to realize
that you're playing in a game where you refuse to learn the rules, and
where you have no particular skills, aside from doing prodigious amounts
of trivial calculations, and constructing text that only you can
decipher. Why anyone would care to convince your sorry self of the
error of your ways is at times a puzzle to me, but a large part of this
is to stand up in the face of your incorrigible arrogance and stupidity,
and to refuse to let the newsgroup be taken over by such punks as
yourself.

If you have a victim giving a positive identification of the
perpetrator, why worry about fingerprint evidence?


Why all the flailing about? Why not just say what you mean?

In mathematics, the logical break in a "proof" is THE smoking gun.


Idiot. Your proof is *all* logical break, punctuated only by, well,
punctuation. There is no connected argument in your paper, not one. The
closest you come (which is still laughably distant from a good argument)
is the "Factorization Lemma", and that is either trivially true:

Let g divide P. Then g = r + c, where c is a constant
and r may or may not not be a constant.

Or absolutely irrelevant; others have pointed out how it is a crock,
and I won't be repeating their arguments. You've seen them, but true
to your own no-class upbringing, consistently ignore them.

It is irrefutable, and cannot be bridged or fixed. No witness now
born, nor that will ever be born can stand against it. And no
rational jury will ever rule against you once you present it.


Back to pseudo-argument. Get off that soapbox before you hurt yourself.

A flawed argument cannot stand on its own. That's why it's flawed.


No, a flawed argument is flawed because it contains errors. It cannot
stand on its own because those errors lead to more errors.

Just like the error that I pointed out. Error. Yours.

You don't need to find another proof to duel with it. You don't need
to attack the person finding the argument. You don't need to argue
persuasively.


Can somebody say Amen, brother! Can I get a Hallelujah? Glory be!

You present the logical break--the smoking gun.

That's mathematics.


Do you still deny that your argument is producing an error?

If you have the evidence, present THE logical break.


The evidence,

From the paper:

Therefore, with the factorization

65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)

one of the a's is coprime to 5, [...].

This does not follow from the preceding text [the whole earlier part of
the paper].

Why not? Because it's false:

I have found common factors between the a's in the following
factorization:

65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)

and the number 5. Those factors are algebraic integers and not units.

You claim, by virtue of your own apparently unassailable method of
argument, that the a's are coprime to 5. Does your definition of
coprime numbers allow for common factors between the numbers?
I have proposed that these polynomials

q(x) = 8 x^2 - 76 x -185
r(x) = 8 x^2 - 4 x - 45
s(x) = 4 x^2 - 37 x - 104

have the property that, for any root z of the polynomial

p(x)= x^3 - 12 x^2 + 65,

we have

q(z)*r(z) = 5
r(z)*s(z) = z.

I've shown how this can be verified, by doing the following
multiplications (courtesy of DOE Macsyma):

First, here are the products that I'm making claims about:

q(x)*r(x) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8325
r(x)*s(x) = 32 x^4 - 312 x^3 - 864 x^2 + 2081 x + 4680

Next, a couple of products of p(x) = x^3 - 12 x^2 + 65 with
polynomials of degree 1:

(64 x + 128)*(x^3 - 12 x^2 + 65)
= 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320

(32 x + 72)*(x^3 - 12 x^2 + 65)
= 32 x^4 - 312 x^3 - 864 x^2 + 2080 x + 4680

Finally, we compare the results and see this:

q(x)*r(x) = (64 x + 128)*p(x) + 5
r(x)*s(x) = (32 x + 72)*p(x) + x,

Note that, for any value xo that makes p(xo) = 0,
that same value xo will make q(xo)r(xo) = 5, so
r(xo) is a factor of 5.

That value of xo also makes r(xo)*s(xo) = xo, so
r(xo) is a factor of xo.

In short, r(xo) becomes a factor of *both* xo and 5.

Since r(x) is a polynomial with integral coefficients,
r(xo) is an algebraic integer whenever xo is.

It is similarly simple to demonstrate that, whenever
xo is a root of p(x), then r(xo) is a root of the
polynomial mpr(x) = x^3 - 969 x^2 + 315 x + 5:

First, expand the polynomial mpr(r(x)):

mpr(r(x)) = (r(x))^3 - 969 (r(x))^2 + 315 (r(x)) + 5

= (8 x^2 - 4 x - 45)^3 - 969 (8 x^2 - 4 x - 45)^2
+ 315 (8 x^2 - 4 x - 45) + 5

= 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3
+ 731136 x^2 - 374400 x - 2067520

Next, multiply these two polynomials:

p(x) = x^3 - 12 x^2 + 65,

and

w(x) = 512 x^3 + 5376 x^2 - 5760 x - 31808

to get this:

p(x)*w(x) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3
+ 731136 x^2 - 374400 x - 2067520

Notice the equality

mpr(r(x)) = p(x)*w(x).

That means for every value of x, the polynomial you get by computing
r(x), then evaluating mpr(x) at that value, is equal to the product of
p(x) and w(x).

If xo is a root of p(x), you have p(xo) = 0, so p(xo)*w(xo) = 0, and
therefore r(xo) is a root of mpr(x).

Note that there are three such roots of mpr(x), and (taking the three
roots x1,x2,x3 of p(x)), three values

r1 = r(x1) ~ 968.67481
r2 = r(x2) ~ -0.01517
r3 = r(x3) ~ 0.34036

correspond to the three real roots of mpr(x). As such, their product
must be -5.

Your earlier claim that the a's must be coprime to 5 implies that the
r's must be units (since they're common factors of ai with 5, in the
ring of algebraic integers).

If you multiply units, even you must realize that the product is again
a unit.

Therefore, -5 (and equivalently, 5 itself) must be a unit in the ring
of algebraic integers.

Do you agree or not?

If so, then say so. I'll make it easy; here's a form for you
to fill out and post to your full array of newsgroups:

I, James S. Harris, affirm my belief that, in
the ring of algebraic integers, the (rational)
integer 5 [five] is a unit.

James S. Harris.

If not, then show me (hey, don't worry about me, show your public!)
where my error is. I've done all the multiplication; you can verify
or refute all this very easily, given the ability to multiply
or expand polynomial expressions.

Show how highly you value algebra: Do some.

It's easy: ordinary polynomials, ordinary polynomial multiplication,
nothing up my sleeves, no salesman will call. You don't even need
to solve any equations. Just multiply, combine terms, show me wrong!

Prove me wrong. Your mama says you can't.

Show us all how much you got that power, how powerful your powers
are, and how you aren't afraid to use your powers for good instead
of evil!

Crush that damnable Evil Mathematics Cabal, once and for all!

The proof is out there.

James Harris


So is your published error.

Unfortunately, error trumps proof.

You lose.

Dale.


So the paper JSH says he had published in a Journal is full of mistakes.
And it should have and was rejected because of flawed algebra, and it was
wrong anyway.

Since JSH is still wining about that, it proves JSH 100% troll.


.