good demonstration, "Randy Poe."

poespam-trap@yahoo.com (Randy Poe) wrote in message news:<df76407e.0310240647.13b76c02@posting.google.com>...

Algebraic integers are defined to be roots of monic polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where
"monic" refers to the leading coefficient.

If we are interested in the set of numbers which are
roots of monic polynomials with integer coefficients,
what does it mean that the set "should" include numbers
that are not such roots?

You've also never answered this one: It is possible to
form a product ab = c with a and c in the set Z but b
not in the set Z. Does that mean the set Z is incomplete
and there is an error in the definition? (Ex: a=3, c=5,
b = 5/3).

brianscsmith@yahoo.com (Brian Smith) wrote in message

jstevh@msn.com (James Harris) wrote in message

Luckily for me the mathematical argument that proves that I've been
correct all along can be further simplified by the use of *numbers*
instead of variables, as while algebra as an idea is well-founded, so
letter symbols should do, when mathematicians are lying about the
mathematics, you need to use what you can, and pray.

1. First the problematic definition:

Algebraic integers are defined to be roots of monic polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where
"monic" refers to the leading coefficient.

If the definition is so problematic, what should the definition be?

The problem has to do with *assuming* that the ring doesn't have
certain interesting properties, so it's not like the definition needs
to be changed.

It's just that its limitations must be noted.

Here the problem is that the emphasis on monic polynomials clips some
numbers.

Not a big deal, if you know that's happening, but mathematicians
didn't realize it for over a hundred years, so now it's a big deal.

My assertion is that the over hundred year old definition excludes
numbers that have to be included to keep from having contradiction
i.e. mathematical inconsistency.

2. The important tool I use is a polynomial:

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

where m varies in the ring of algebraic integers.

Demonstrate your claim using P(m)=m^3-m+6.

[remainder cut]

What? Are you wondering if that polynomial can be factored into
non-polynomial factors? The answer is, yes, it can be.

What exactly do you think I'm claiming that you just toss out some
polynomial?

I'm curious what your reasoning is.

Remember, all I do basically is factor a polynomial into
non-polynomial factors with a neat technique, look at constant terms
of those factors, divide the polynomial by some integer, and consider
the constant terms again to draw a rather direct and obvious
conclusion.


James Harris

brianscsmith@yahoo.com (Brian Smith) wrote in message news:<12f59340.0310271026.5019b55f@posting.google.com>...

Luckily for me the mathematical argument that proves that I've been
correct all along can be further simplified by the use of *numbers*
instead of variables, as while algebra as an idea is well-founded, so
letter symbols should do, when mathematicians are lying about the
mathematics, you need to use what you can, and pray.

1. First the problematic definition:

Algebraic integers are defined to be roots of monic polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where
"monic" refers to the leading coefficient.

If the definition is so problematic, what should the definition be?

The problem has to do with *assuming* that the ring doesn't have
certain interesting properties, so it's not like the definition needs
to be changed.

It's just that its limitations must be noted.

Here the problem is that the emphasis on monic polynomials clips some
numbers.

Not a big deal, if you know that's happening, but mathematicians
didn't realize it for over a hundred years, so now it's a big deal.

My assertion is that the over hundred year old definition excludes
numbers that have to be included to keep from having contradiction
i.e. mathematical inconsistency.

2. The important tool I use is a polynomial:

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

where m varies in the ring of algebraic integers.

Demonstrate your claim using P(m)=m^3-m+6.

[remainder cut]

What? Are you wondering if that polynomial can be factored into
non-polynomial factors? The answer is, yes, it can be.

What exactly do you think I'm claiming that you just toss out some
polynomial?

I'm curious what your reasoning is.

Remember, all I do basically is factor a polynomial into
non-polynomial factors with a neat technique, look at constant terms
of those factors, divide the polynomial by some integer, and consider
the constant terms again to draw a rather direct and obvious
conclusion.


James Harris

brianscsmith@yahoo.com (Brian Smith) wrote in message news:<12f59340.0310271026.5019b55f@posting.google.com>...

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0310201424.39b5d6cd@posting.google.com>...

Luckily for me the mathematical argument that proves that I've been
correct all along can be further simplified by the use of *numbers*
instead of variables, as while algebra as an idea is well-founded, so
letter symbols should do, when mathematicians are lying about the
mathematics, you need to use what you can, and pray.

1. First the problematic definition:

Algebraic integers are defined to be roots of monic polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where
"monic" refers to the leading coefficient.

If the definition is so problematic, what should the definition be?

The problem has to do with *assuming* that the ring doesn't have
certain interesting properties, so it's not like the definition needs
to be changed.

brianscsmith@yahoo.com (Brian Smith) wrote in message news:<12f59340.0310271026.5019b55f@posting.google.com>...

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0310201424.39b5d6cd@posting.google.com>...

Luckily for me the mathematical argument that proves that I've been
correct all along can be further simplified by the use of *numbers*
instead of variables, as while algebra as an idea is well-founded, so
letter symbols should do, when mathematicians are lying about the
mathematics, you need to use what you can, and pray.

1. First the problematic definition:

Algebraic integers are defined to be roots of monic polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where
"monic" refers to the leading coefficient.

If the definition is so problematic, what should the definition be?

The problem has to do with *assuming* that the ring doesn't have
certain interesting properties, so it's not like the definition needs
to be changed.

It's just that its limitations must be noted.

Here the problem is that the emphasis on monic polynomials clips some
numbers.

Not a big deal, if you know that's happening, but mathematicians
didn't realize it for over a hundred years, so now it's a big deal.

brianscsmith@yahoo.com (Brian Smith) wrote in message news:<12f59340.0310271026.5019b55f@posting.google.com>...

Luckily for me the mathematical argument that proves that I've been
correct all along can be further simplified by the use of *numbers*
instead of variables, as while algebra as an idea is well-founded, so
letter symbols should do, when mathematicians are lying about the
mathematics, you need to use what you can, and pray.

1. First the problematic definition:

Algebraic integers are defined to be roots of monic polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where
"monic" refers to the leading coefficient.

If the definition is so problematic, what should the definition be?

The problem has to do with *assuming* that the ring doesn't have
certain interesting properties, so it's not like the definition needs
to be changed.

It's just that its limitations must be noted.

Here the problem is that the emphasis on monic polynomials clips some
numbers.

Not a big deal, if you know that's happening, but mathematicians
didn't realize it for over a hundred years, so now it's a big deal.

My assertion is that the over hundred year old definition excludes
numbers that have to be included to keep from having contradiction
i.e. mathematical inconsistency.

2. The important tool I use is a polynomial:

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

where m varies in the ring of algebraic integers.

Demonstrate your claim using P(m)=m^3-m+6.

[remainder cut]

What? Are you wondering if that polynomial can be factored into
non-polynomial factors? The answer is, yes, it can be.

What exactly do you think I'm claiming that you just toss out some
polynomial?

I'm curious what your reasoning is.

Remember, all I do basically is factor a polynomial into
non-polynomial factors with a neat technique, look at constant terms
of those factors, divide the polynomial by some integer, and consider
the constant terms again to draw a rather direct and obvious
conclusion.


James Harris

Will Twentyman <wtwentyman@read.my.sig> wrote in message

James Harris wrote:

Luckily for me the mathematical argument that proves that I've been
correct all along can be further simplified by the use of *numbers*
instead of variables, as while algebra as an idea is well-founded, so
letter symbols should do, when mathematicians are lying about the
mathematics, you need to use what you can, and pray.

1. First the problematic definition:

Algebraic integers are defined to be roots of monic polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where
"monic" refers to the leading coefficient.

My assertion is that the over hundred year old definition excludes
numbers that have to be included to keep from having contradiction
i.e. mathematical inconsistency.

No, the definition just doesn't include enough to make things work the
way you should. If the definition had an error of the type you suggest,
the contradiction would lie in one of the theorems *about* algebraic
integers, not in the definition. For example, the theorem that the
algebraic integers form a ring.

I'm merely stating the problem and giving my assertion, so there's no
reason to get into a debate here.

2. The important tool I use is a polynomial:

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

Oh yeah, that's

P(m) = 14706125 m^3 - 900375 m^2 + 899640 m + 22

and I think it's easier to work with as I give it originally.

Notice that what I do is creatively break the polynomial up, so that I
can factor it in a special way.

But it's still just a polynomial.

where m varies in the ring of algebraic integers.

Some may find it looks odd. But the entire point of that form is to
do something a little different than anyone else apparently has done
before, which is to get 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 the a's are roots of the following cubic:

a^3 + 3(-1 + 49m)a^2 - 49(2401 m^3 - 147 m^2 + 3m).

So you can see that the a's are functions of m, and if you really want
headaches you can go ahead and solve the cubic to get a view of them.

However, I can move on without needing their explicit form.

From the factorization I have three factors

g_1 = (5 a_1 + 7), g_2 =(5 a_2 + 7), g_3=(5 a_1 + 7)

and I can find those terms that are free of m by setting m=0,

or by inspection you could simply rewrite a_1 as a_1(m) etc, and realize
that the 7's are the constant terms, and therefor *independent* of m.

That's wrong. Posters in the past have made a similar claim, but
readers should note that the reality is that the term independent of m
with ONLY TWO of the factors is 7, while for the third it is 3(5) + 7,
that is 22.

Now here's a good place for me to emphasize to readers to read
*carefully* as the problem I'm facing is that mathematicians appear to
be trying to protect themselves from an over one hundred year old
problem, so they will just say wrong things to you, hoping you'll just
trust.

just
like with any polynomial, you know like with S(m) = 2(m^2 + 2m + 1),
S(0) = 2, which you can just look at and see here, but with my P(m)
it's just a little harder, so I set m=0, which gives me

P(0) = 49(3(5) + 7),

which fits with the cubic as at m=0 it gives

a^3 -3a^2 = 0, so a_1 = a_2=0, a_3 = 3,

to show that at m=0, the three factors are

g_1 = 7, g_2 = 7, g_3 = 3(5) + 7 = 22.

Now dividing P(m) by 49 gives

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

and the question is what happens to the g's, but look now at P(0)/49
as that is

P(0)/49 = 3(5) + 7

as two factors of 7, each 7, have beeen divided off, which is easy to
see.

So far, so good.

The math is rather basic, and like I said, it's easy to see.

The only way that can happen is if

P(m)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7)

where only two of the a's have 7 as a factor,

Wrong. Factors of 7 will be distributed as determined by the values of
each of the a's for each value of m. The factorization depends on m
because the a's depend on m. This has been demonstrated with other
polynomials through explicit computations. I don't recall if Arturo has
done the computations for this particular one or not.

That's not how it works, but the problem is that for many "a_1", "a_2"
and "a_3" are mysteries, so I'll help out by asking some "what if"
questions.

What if, a_1 has sqrt(m) as a factor? It has some factor of m, since
it equals 0, when m is a factor. So what then?

What if that factor could be m, would you believe "Will Twentyman"
then?

Some of you may be confused by a_3, but what if it were m+3?

Let's look at that possibility with x, y and z algebraic integers
introduced for any remaining factors:

Then you'd have

P'(m) = (5xm + 7)(5ym + 7)(5z(m+3) + 7)

and how many of you NOW would accept that 7 divides off *dependent* on
the value of m?

I hope none.

What works with the "what if" scenario is

P'(m) = (5(7)m + 7)(5(7)m + 7)(5(m+3) + 7)

and hopefully some of you can now see how simple it really is, and how
diabolical mathematicians in arguing against my result have been.
They've convinced many of you that independent, constant terms can
vary based on the value of m.

Here you can see how it can happen that only two of the factor have 7
as a factor while for the last it's blocked. Now the a's aren't quite
that simple in one sense, but they are in another as you can see at
m=0, that

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

P(0) = (7)(7)( 5(3) + 7).

Now finally, can any of you come up with a *rational* way for 49 to
divide off P(m), such that the independent terms vary based on the
value of m?

But, they're independent terms, so they can't--rationally.

I'm telling you that mathematicians are trying to hide an error in
their discipline. As they are mathematicians they KNOW HOW TO LIE TO
YOU, and you can see that they're quite willing to do it.

Their lies are messing up my life, and can't be helping world society.

Meanwhile we, the public, can't know just how bad the problem is,
though the longer mathematicians lie about the over hundred year old
error, the more likely it seems that it is a meltdown.


James Harris