# Simple principle in core error proof

Discussion in 'Undergraduate Math' started by James Harris, Oct 20, 2003.

1. ### James HarrisGuest

What makes my situation especially frustrating is how simple the
argument is that proves there's this problem with algebraic integers,
yet still I have to keep explaining.

The basic principle is like with

P(m) = 2(m^2 + 2m + 1) = (a_1 m + 2)(a_2 m + 1)

where most of you can probably guess what factor a_1 must have!!!

Now the polynomial I use is more complicated such that I need to set
m=0 to figure things out, but notice here what happens:

P(0) = 2, and dividing off 2 from P(m) gives

P(m)/2 = m^2 + 2m + 1

and notice that P(0)/2 = 1, which tell you that the independent term
changed.

Given

(a_1 m + 2)(a_2 m + 1)

it's clear that a_1 has a factor that is 2.

It's a simple idea that I use with a more complicated polynomial, and
I think that the reason so many math people go out of their way to
make it seem like it's wrong is that they're embarrassed by the over
hundred year old error that I found.

I've been surprised at how dedicated they can be at trying to hide the
truth.

After all, I've communicated with top mathematicians like Barry Mazur,
Andrew Granville, and Ralph McKenzie, who may be people many of you
haven't heard of, but in certain math circles they're well-known.

In McKenzie's case I explained it all to him ***in-person*** and he
basically blew me off.

These mathematicians are hellbent on trying to hide that their
discipline could actually have a flaw like this for as long as they
can get away with it.

It's wrong, and it can't help world society.

Meanwhile it's not doing me a bit of good either, and I think that
part of their motivation is making me miserable, knowing that what I

Yup, call me crazy, but I think the bastards are out to get me!!!

James Harris

James Harris, Oct 20, 2003

2. ### VirgilGuest

There is a meaning of "simple" which applies to JSH's explanation,
meaning authored by a simpleton, which JSH is mathematically at
least.

Virgil, Oct 20, 2003

3. ### Arturo MagidinGuest

This is a red herring. Here, the polynomial has a coefficients which
are constant with respect to the variable of factorization, and what
you are varying is the polynomial variable m. In your actual
situation, you are factoring with respect to a DIFFERENT variable, X,
and the coefficients are not constant, but are instead functions of
m.

This example is thus worthless as a guide.
What happens is you engage in a red herring. In your actual
application, your factorization is not with respect to m, but with
respect to x. And you do not set x equal to 0, you set m equal to
0. What you are doing here is in no way parallel to what you do in
your actual application, and thus conclusions you draw from this
example are not necessarily applicable to your actual application.

[.red herring left in place so James cannot complain that I
'removed the math'.]
Because a1 and a2 are constant. They do not vary. So whatever happens
at one point is necessarily what happens at all points. This is not what
the coefficients change as you change m, and it does not follow that
what happens at one value of m is what happens at all values of m. In
fact, it does not, has been amply demonstrated by explicit
calculation.
Then you got confused. here, you have a polynomial in m being factored
as linear terms in m, where the coefficients with respect to m are
constant. In your application, you are factoring a polynomial which, as
m varies, is reducible or irreducible over Q depending on the value of
m; here, you have a polynomial which is always reducible over Q
regardless of the value of the (non-existent) non-polynomial variable.

In your application, you have a polynomial in m being factored into
linear terms WITH RESPECT TO A DIFFERENT VARIABLE x, and the
coefficients with respect to m are ->functions of x<-, and the
coefficients of x are functions of m. Here you set the polynomial
variable equal to 0, but in your application you set m, not the
polynomial variable at issue, equal to 0. Here you KNOW the
coefficients are constant and therefore whatever happens at one point
will happen at all points, but in your application the coefficients
are NOT constant and therefore it does not follow that what happens at
one point will happen at all points. The situations are NOT similar,
and therefore the situations are not comparable.

[.snip.]

======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A great
many people are staggered to this extent, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================

Arturo Magidin

Arturo Magidin, Oct 20, 2003
4. ### fishfryGuest

Suppose I define a set B = {all objects in the universe that are
left-handed giraffes} and then I look at every object in the universe
and ask, is it a left-handed giraffe? If it is, it's in B. If not,
it's not in B. Perhaps there are no left-handed giraffes, in which case
B is still well-defined, it just happens to be empty.

If I define the algebraic integers as real numbers that are the root of
a polynomial with monic leading coefficient, then I can look at each
real number and say it's either in this set or isn't.

Can you explain what you mean when you say you are getting a
contradiction from the definition? For example, have you found a real
number that both is and isn't an algebraic integer? What contradiction,
exactly, have you found?

fishfry, Oct 20, 2003
5. ### James HarrisGuest

Yet readers can see the reality here:

P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7)

The form of the polynomial allows me to factor P(m) into
non-polynomial factors, and the factorization with those factors is

P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)

where ONLY m is a variable.

And yes I used more symbols before but found that posters like Arturo
Magidin could too easily confuse people about what was going on, so
I've put in values where once there were symbols.

Unfortunately, Arturo Magidin is a rather evil person who lies about
the math as if the core error should stay hidden.

He's bad, he's evil, and I'm sick of his crap.

James Harris

James Harris, Oct 20, 2003
6. ### Arturo MagidinGuest

And can you PROVE that a_1, a_2, and a_3 will satisfy the properties
you want under these conditions? What are those properties?

You can ALWAYS get such a factorization, if you allow a_1, a_2, and
a_3 to be algebraic numbers; but then your congruence arguments bite
the dust.
Unfortunately, James Harris is a rather dishonest person who lies about me,
having been unable to find a single instance of me "lying about the
math", and who, each time he accuses me of such behavior, ends up
retracting his claim but not his accusation.

======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A great
many people are staggered to this extent, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================

Arturo Magidin

Arturo Magidin, Oct 21, 2003
7. ### Arturo MagidinGuest

Yet, when I offered to stop posting if you would only tell me to do
so, you ->refused<-. Were you too scared to tell someone to stop
posting, or are you just lying here about being "sick of [my] crap"?
(No, I won't count as a lie the fact that you have given no evidence
whatsoever of anything I have posted being false, let along 'crap';
that's just your usual lying to yourself).

Why, if as you claim I am "bad", "evil", "rather evil", someone who
"lies about the math", and if you are "sick of [my] crap", did you
refuse my offer to 'step out of your way', as it were, by no longer
posting to anything you might say?

======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A great
many people are staggered to this extent, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================

Arturo Magidin

Arturo Magidin, Oct 21, 2003
8. ### flipGuest

.... Worthless dribble deleted! ***

I'll go for calling you crazy.

Does anyone second the motion.

You are like a broken record, replaying the same old nonsense and you have
lied to yourself for so long, you believe your own incompetence.

Please go get some psychotropic drugs or jump off a bridge, you raving
lunatic.

How pathetic you have proven with your feeble attempts for fame, fortune and
recognition. You are so sad and an example of schooling gone horribly wrong
in the US (as much as it hurts to admit that).

Perhaps you should renounce your degree and go back to grade school and see
if people still think you are "smart".

Get a life!

flip, Oct 21, 2003
9. ### Wayne BrownGuest

Wayne Brown, Oct 21, 2003
10. ### Brian Quincy HutchingsGuest

isn't a "term," just a monomial, 6abm^2 e.g.?... so,
what do you mean by "independent term?..." so,
i followed that example, but what is it supposed to prove?... I mean,
it's a good, artificial construction to prove ... what?
you could clear this up, by using an even simpler equation, and
applying your alleged method to it. as far as I can see,
you simply refuse to assert your definitions --
whether or not they are from "over-100-year-old math."
what is it in "substitution of variables,"
that exposes the wrongness of the definition
of algebraic integers (and, of course,
the evil of Arturo) ??

<deletives impleted>

--les ducs d'Enron!
http://larouchepub.com/other/2003/3041shultz_arnie.html
http://larouchein2004.net/pages/speeches/2003/031022webcast.htm

Brian Quincy Hutchings, Oct 22, 2003
11. ### Andrzej KolowskiGuest

Tell me what happens when

P(m) = 5*(m^2 + 7*m + 5)

and you assume

P(m) = (a_1*m + 5)*(a_2*m + 5).

Andrzej

Andrzej Kolowski, Oct 22, 2003
12. ### James HarrisGuest

Readers should note the challenge to the *reality* that setting m=0
gives the independent, or constant term.

But trivially with the poster's example the constant term is 25.

The independent term is 25.
There's nothing to assume as what I do is focus on constant terms.

Again, trivially, the constant term is 25.

The independent term is 25.

I want readers to pay *careful* attention as in fact several posters
have gotten away with misrepresenting what I do when in fact I focus
on constant or independent terms.

The math is basic. These posters, however, need to get you to
question basic algebra which many of you learned while still children.

And they've apparently been VERY successful up to now.

I focus on the independent terms.

That's how the argument works.

To win the war of confidence, posters like Andrzej Kolowski have to
get you to question your own mathematical knowledge, to get you to
lack confidence in your own mathematical understanding, so that you
rely on them as the experts, and then they lie to you.

It's not complicated people, and it's an old story in human history as
I'm challenging a powerful group that'd rather not let the truth get
in its way.

James Harris

James Harris, Oct 22, 2003
13. ### C. BondGuest

James Harris wrote:

[snip]
Your use of the qualifier 'independent' is incorrect. The function, P(m), is a polynomial in 'm'. Presumably *all*
coefficients of the individual terms involving products with powers of 'm' are 'independent' of 'm'. Each of the
coefficients in the polynomial is a constant, including the coefficient of the term corresponding to m^0 ( the standard
constant term). If you *evaluate* the polynomial at m=0, only the constant term remains, but not because it is
'independent' of 'm' -- is is neither more nor less dependent on 'm' than any other coefficient -- rather that you have
simply replaced 'm' with 0 to zero out all other coefficients and evaluated P(0).

C. Bond, Oct 22, 2003
14. ### Andrzej KolowskiGuest

I appreciate all the comments, but I don't think you
little vague.

What I wanted to know was about the divisibility of a_1 and
a_2 by 5 for various values of m. Is one or the other of
them always divisible by 5 regardless of m, or what? Your
far as I am concerned.

Andrzej

Andrzej Kolowski, Oct 22, 2003
15. ### James HarrisGuest

Here's what you have.
The constant term is 25. The most you can divide through by with m an
algebraic integer m, is 5, which gives you a constant term of 5 for

So there's no relevance to my argument as you still have 5 as a factor
of the constant term.

The simple and basic algebra argument that proves my case, and shows
that there IS an error in core because of the definition of algebraic
integers, depends on coprimeness of the constant term, which is how I
filled what had been called a "gap".

Remember that Andrzej Kolowski? It was almost two years ago, when I
admitted to having a gap, as several posters had claimed, and then
when I filled it, you all just lied, so I know about credibility, and
you have none.

You've been lying for almost two years now, over an issue that I
handled back in November of 2001.

People like you've been torturing me, in an inhumane behavior that's
hard to explain, as I see you as evil incarnate, dedicated to doing
evil until someone stops you.

And I will stop you.

James Harris

James Harris, Oct 22, 2003
16. ### Andrzej KolowskiGuest

Which is somewhat parallel to what happens in your
argument when you divide through by f^2: note that
when you factor f^2 out of the first *two* terms of

(a1*x + uf)*(a2*x + uf)*(a3*x + uf),

you have

f^2*((a1/f)*x + u)*((a2/f)*x + u)*(a3*x + uf),

and you still have "uf" in that last term.

I note that you didn't really answer my question, whether
one, both, or neither of a1 and a2 are divisible
by 5 when m <> 0.

the constant term with respect to m, not the constant
term with respect to x. My example, unlike your P(m),
doesn't have any x's in it.

Let me modify my example a little bit to get closer

Q(m) = f*(m^2*x^2 + 3*x + f).

Then Q(0) = f*3*x + f^2 = f*(3*x + f).

This is the constant term, "independent of m".

Note that Q(0)/f = 3*x + f. This is similar to

P(0)/f^2 = u^2*(3*x + u*f), or, when u = 1,

P(0)/f^2 = 3*x + f.

Now assume that my polynomial Q(m) is factored in the form

Q(m) = (b1*x + f)*(b2*x + f).

As in your P(m), b1 and b2 are going to be functions
of m: b1(m) and b2(m).

The fact that Q(0) = f*(3*x + f) is consistent with
b1(0) = 0 and b2(0) = 3. Agree?

So b1(0) is divisible by f and b2(0) is relatively
prime to f (if f is a prime <> 3).

This too is similar to your polynomial P(m), where
a1(0) = a2(0) = 0 and a3(0) = 3, the latter being
coprime to f (if f is a prime <> 3).

Anyway, for my new polynomial Q, I would like to see
in general, is b1(m) or b2(m) divisible by f? Both,
only one, or neither ?

Yes, well, of course thanks for the reply.

Andrzej

Andrzej Kolowski, Oct 23, 2003
17. ### Brian Quincy HutchingsGuest

ho hum; same ****, different day.

--les ducs d'Enron!
http://larouchepub.com/other/2003/3041cody_on_recall.html

Brian Quincy Hutchings, Oct 23, 2003
18. ### James HarrisGuest

I figured out that posters were having a field day with all the
variables, and that it made it *easy* for them to lie and confuse
people.

But

P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078

takes that away.

But notice how they try to go back to more variables, like the twisted
and despicable people they are.

I'm telling you, these people are EVIL.

Now if they just admit that constant terms are constant, then the rest
of the argument follows.

That's because the constant terms CANNOT change dependent on the value
of m.

That means that focusing exclusively on the constant terms and how
they do change when 49 divides off of

P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078

tells the tale.

Now I'm informing all of you that the people arguing against me are
EVIL, yes they are real, live EVIL people as mathematics is that
important, so it's important enough for Evil itself to send minions
like them.

James Harris

James Harris, Oct 23, 2003
19. ### Brian Quincy HutchingsGuest

other than the Axis of Ones Who Argue with JSH?
Arturo gave a "tutorial on indepeence of terms"
in another item, in reply to you, but
i have a feeling taht that's the last you'll "go"
to it.
I'm not arguing *against* you, just with you!

--les ducs d'Enron!
http://larouchepub.com/other/2003/3041cody_on_recall.html

Brian Quincy Hutchings, Oct 23, 2003
20. ### Dik T. WinterGuest

Q(m) = 6125 m^3 + 6125 m^2 - 6370 m + 1078

tells another tale.

Dik T. Winter, Oct 24, 2003