Re: JSH: Final exam

From: James Harris (jstevh_at_msn.com)
Date: 11/30/04 

Date: 30 Nov 2004 04:04:49 -0800

"*** T. Winter" <***.Winter@cwi.nl> wrote in message news:<I7z4Lz.L88@cwi.nl>...
> In article <3c65f87.0411291820.33ce6c13@posting.google.com> jstevh@msn.com (James Harris) writes:
> > In arguing with me posters continually push that x=0 is a special
> > case.
>
> Yup, I will argue.
>

Here readers can get a good dose of pseudo-math from one the most
obnoxious posters on sci.math as this guy not only lies about basic
mathematics, as you can see in his post, but he copied from my Usenet
posts without my permission on to his own webpage, where he added in
negative commentary!

And refused to quit using my Usenet posts in violation of both the
spirit and letter of international copyright law.

And he lies about math.

> > Well, here's yet another answer and I want you to pay careful
> > attention to what the sci.math'ers do now.
> >
> > Remember
> > 49(300125x^3 - 18375 x^2 - 360 x + 22) =
> > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7)
> > where the a's are the roots of
> >
> > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> >
> > and I find out what the constant terms are by using x=0, but now I'll
> > consider x = 7.
> >
> > Then I have
> > a^3 + 3(342)a^2 - 49(816361) = 0
> > which is irreducible over Q, for the a's and
> > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7)
> > where you'll notice that 102040002 is coprime to 7.
>
> Yup.
>
> > So 7 divides TWO and only TWO of the factors. How do you know?
>
> The question is *why*?
>
> > Because when x=0, two of the a's equal 0, that's why.
>

Notice that you have

49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7)

with 49 on the left side, and it has to divide through in some way.

For two of the a's you know they result from functions that equal 0,
when

x = 0 mod 7 as they equal 0, when x = 0

and you can't get any better than that!

The third equals 3 when x = 0, so it is blocked from havinng factors
in common with 7, when x has 7 itself as a factor, like when x = 7.

It's basic. To believe otherwise you need to challenge algebra.

> Oh, not very convincing. When x = 0, sqrt(x) = 0, so sqrt(x) is divisible
> by 7? It would be true if the a's were polynomials. They are not.
>

Now notice, the poster can't deny the result when the a's are
polynomials as then you can physically SEE how it works, so he handles
that right off the bat by asserting that whether or not they are
polynomials matters.

Also he tossed in a weird assertion that doesn't follow as there's no
reason to suggest that sqrt(7) is divisible by 7. He's just trying to
throw up smoke as one other tactic is to confuse readers to the point
that they just give up and trust that if someone is disagreeing with
me, then I must be wrong.

> > x = 0, is equivalent to x = 0 mod 7.
>
> Using equivalence classes mod 7 is only useful when you are doing
> additions and multiplications. With other functions it just does

So now congruences only matter with additions and multiplications,
according to the sci.math'er, so isn't it interesting what math this
person is teaching you?

But, I didn't necessarily phrase my own sentence well, as my point is
that

x=0 is equivalent to x = 0 mod 7, with respect to having factors of 7

so my using x = 0 mod 7, is a neat way to show that the result at x=0
is NOT a special case at all, as the constant terms work as I've
explained repeatedly, while the claims of people arguing with me fall
flat.

If this weren't such a HUGE issue with massive repercussions for
mathematicians worldwide, then maybe these sci.math-ers might finally
give up and admit the truth, but wait, I'm giving them too much
credit.

It could be a minor issue with few repercussions and these people,
from what I've seen over the years would still lie to you.

Lying is in their nature.

> not work. You are actually saying that sqrt(7) should be 0 mod 7.

No I'm not.

> Which means that sqrt(7) is divisible by 7, which means that 1/7
> is an element of the ring you are in, and so 7 is a unit in that ring.
>

Desperation.

> > Notice though that the cubic defining the a's is still irreducible
> > over Q, and irreducibility has no impact.
>
> Yup, and so the a's are not polynomials and that has a terrific impact.
>

This poster hardly even tries, but why should he?

These TACTICS WORK on sci.math, as I've noted for years now.

The sci.math readership is like some kind of religious cult, willing
to believe just about anything as long as posters disagree with me.

They are true believers on sci.math, and it's freaky.

> > However, in the ring of algebraic integers NONE of the a's have 7 as a
> > factor, and each has a non-unit factor in common with 7.
>
> And the non-unit factors multiply together to give a multiple of 49. So,
> what is the problem?

Oh, hey, it's all ok if it works out in the end, right?

If you want the full picture you need to read my original post, and my
other reply to the other sci.math-er.

The story here is truly sad as the problem I found *was* just a
historical one, where math professors today could simply point out
that they followed what they were taught.

But now as the days turn into months and the months move toward years,
there's little reason to believe that such a huge result hasn't
traveled through the math community, so what was a fascinating
historical error, is now a case of modern fraud.

Math professors who go out today to teach flawed algebraic number
theory to their students may be criminals and it may be possible to
prosecute them under anti-fraud laws, but that sounds so removed from
reality that they will probably continue without real fear.

But, for those of you listening to them, the impact is very real.

You are being taught false information, and being forced to do
homework, and take tests to regurgitate that false information and God
help you if you come back saying that it's false based on what you
learned on newsgroups from a guy called James Harris.

I feel sorry for you, but if you look around you can see how often
society does things that inside you feel are wrong, but people just go
on about their business as that's what they've learned is necessary.

And many of you, unfortunately, will be like them, today, and the next
day.

So, in a great field, where truth is paramount, you will knowingly
learn false things, and maybe even pride yourself on the grades you
get from professors who are compromised.

But you're living in a mirage...

James Harris

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