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James Harris October 17th 03 07:14 PM
Core error argument objection refuted, short
 
I've noted a problem in algebraic number theory with the inclusiveness
of the definition of algebraic integers as roots of monic polynomials
with integer coefficients.

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

First I have

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and the factorization

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

My finding that only two of the a's have f as a factor without regard
to the value of m has been vigorously disputed.

However, consider w_1(m), a factor of a_1 that is a factor of f, as
well as a function that varies with m, then it follows that

a_1 x + uf has w_1(m) as a factor,

so dividing through by w_1(m) gives

a_1 x/w_1(m) + uf/w_1(m)

but then uf/w_1(m) cannot in general be an algebraic integer as it's
not representable as a polynomial with a finite number of terms if
w_1(m) varies with m.

My guess is that some may be assuming that f is replaceable by some
function of m, but in fact, its independent of the value of m, so it's
like 1/(x+1) which is also not representable by a polynomial if x is
an algebraic integer not equal to 0 or -2.

So the objection is refuted by the impossibility of uf/w_1(m) being an
algebraic integer, for all algebraic integers m, if w_1(m) varies with
m.

My hope is that posters who have been so successful in convincing
others that my argument is flawed will post concessions.


James Harris

fishfry October 17th 03 09:28 PM
Core error argument objection refuted, short
 
In article ,
(James Harris) wrote:

I've noted a problem in algebraic number theory with the inclusiveness
of the definition of algebraic integers as roots of monic polynomials
with integer coefficients.

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

First I have

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

Can't you articulate your object in English? Such as, "If you define
algebraic integers as roots of monic polys with integer coefficients,
then the following odd thing happens: etc. etc. Just say exactly what
you think the contradiction or problem is.

Virgil October 17th 03 09:34 PM
Core error argument objection refuted, short
 
In article ,
(James Harris) wrote:



where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).


The above cubic non-equation doesn't "give" anything.

James Harris October 17th 03 11:33 PM
Core error argument objection refuted, short
 
(Peter van Rossum) wrote in message ...
In article ,
James Harris wrote:
I've noted a problem in algebraic number theory with the inclusiveness
of the definition of algebraic integers as roots of monic polynomials
with integer coefficients.

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

First I have

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and the factorization

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

If the a_i's depend on m, why don't you write it properly?

P(m) = ( a_1(m) x + u f ) ( a_2(m) x + u f ) ( a_3(m) x + u f )

That's a style issue. I see it as a gesture of futility and anguish
at the reality of this easy refutation.

Mathematicians are such babies.

My finding that only two of the a's have f as a factor without regard
to the value of m has been vigorously disputed.

However, consider w_1(m), a factor of a_1 that is a factor of f, as
well as a function that varies with m, then it follows that

a_1 x + uf has w_1(m) as a factor,

i.e. w_1(m) | a_1(m) x + u f for all algebraic numbers m

Mathematicians in trying to deny the reality here have been hoping on
some w_1(m) that can give a variable factor of f for a_1. I'm just
following that idea through to the necessary conclusion.

so dividing through by w_1(m) gives

a_1 x/w_1(m) + uf/w_1(m)

but then uf/w_1(m) cannot in general be an algebraic integer as it's
not representable as a polynomial with a finite number of terms if
w_1(m) varies with m.

I hope you realize that the functions a_i are also not given by polynomials
in m (i.e., there does not exists a polynomial A_i(M) in A[M], the

That's a rather stupid lie given that I put the polynomial in this
post.

Again it is

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)

and its roots are a_1, a_2 and a_3.

But mathematicians have been persistent in showing a rather pathetic
and stupid refusal to accept rather basic mathematics.

I find them disgusting.

polynomial ring in one variable over the algebraic numbers, such that
A_i(m) = A_i(m) for all algebraic numbers m).
For the function w_1, well, the only thing you've mentioned about it is
that w_1(m) | a_1(m) and w_1(m) | f (for all algebraic numbers m,
I think).
So how could one expect a_1(m) x / w_1(m) + u f / w_1(m), or u f / w_1(m)
to have any particular form at all? Of course, for u f / w_1(m)
to be an algebraic integer for all algebraic integers m, it doesn't
have to be expressible as a polynomial in m over the algebraic integers.

The poster is babbling, possibly in shock. Such weakness from the
math world is telling. Mathematicians don't know how to really think,
but have gotten away with faking it.

With all you assumptions, it is of course trivial that u f / w_1(m)
is an algebraic integer (since f is divisible by w_1(m) in the algebraic
integers).

It's not possible. I'll put in values to help those readers confused
by symbols (though algebra is BASED on symbols) by letting u=2, f=13,
then you have

26/w_1(m)

and you don't have to be a rocket scientist to know that no function
w_1(m) that actually varies with m can exist such that 26/w_1(m) is an
algebraic integer for *all* integer m.

It's a show of how broken math society is that Peter van Rossum would
dare to make that stupid assertion.

My guess is that some may be assuming that f is replaceable by some
function of m, but in fact, its independent of the value of m,

Everybody understands that, I'm sure.

so it's
like 1/(x+1) which is also not representable by a polynomial if x is
an algebraic integer not equal to 0 or -2.

But I wonder who understands this. I definitely don't.

(By the way, if x *is* an algebraic integer unequal to -1, then
1/(x+1) *is* expressible as a polynomial over the algebraic integers -
a constant polynomial. But you probably mean that there is no
polynomial F(X) over the algebraic integers such that F(x) = 1/(x+1)
for all algebraic integers unequal to -1. I still wonder what 0 and
-2 have to do with it - maybe just a mistake.)

I was thinking about x being an integer, though in algebraic integers,
any x such that x=u_1 - 1, where u_1 is a unit in algebraic integers
will work.

So the objection is refuted by the impossibility of uf/w_1(m) being an
algebraic integer, for all algebraic integers m, if w_1(m) varies with
m.

Can you repeat the definition of algebraic integer again for
the newsgroup and tell us how you conclude that u f / w_1(m)
is not an algebraic integer?

Mathematicians are pathetic liars.

I'll use u=2, f=13 again, now then, NO function in algebraic integers
exists such taht 26/w_1(m) is an algebraic integer for all integers m,
if w_1(m) varies with m.

It's just not possible, but it takes a mathematician to lie about it.


James Harris

Arturo Magidin October 17th 03 11:37 PM
Core error argument objection refuted, short
 
In article ,
James Harris wrote:

[.snip.]

It's not possible. I'll put in values to help those readers confused
by symbols (though algebra is BASED on symbols) by letting u=2, f=13,
then you have

26/w_1(m)

and you don't have to be a rocket scientist to know that no function
w_1(m) that actually varies with m can exist such that 26/w_1(m) is an
algebraic integer for *all* integer m.


w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)}

I'm not saying that's the function in question (it is not), but there
is a function that "actually varies with m such that 26/w_1(m) is an
algebraic integer for *all* integer m." So the argument that no such
function can exist is simply bogus.

================================================== ====================
"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

James Harris October 17th 03 11:47 PM
Core error argument objection refuted, short
 
(Arturo Magidin) wrote in message ...
Cc:

In article ,
James Harris wrote:

[.snip.]

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

First I have

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and the factorization

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

My finding that only two of the a's have f as a factor without regard
to the value of m has been vigorously disputed.

However, consider w_1(m), a factor of a_1 that is a factor of f, as
well as a function that varies with m, then it follows that

"Factor" where?

I assume you mean: w_1(m) is, for each value of m, an algebraic
integer which is a factor of a_1(m), and also a factor of f, both in
the ring of algebraic integers.

Mathematicians are disgustingly bad liars when caught, and this one is
no exception.

The issue is whether or not a_1 can have a varying factor of f, which
varies with m; however, my post shows how easy it is to show that
leads to a stupid contradiction.

Rather than concede this Arturo Magidin is back to try and keep the
lie going.

a_1 x + uf has w_1(m) as a factor,

Should be a_1(m)x + uf, but yes.

so dividing through by w_1(m) gives

a_1 x/w_1(m) + uf/w_1(m)

Should be a_1(m)x/w_1(m) + uf/w_1(m), so this is g_1(m)/w_1(m), yes.

but then uf/w_1(m) cannot in general be an algebraic integer


(Yes, I'm interrupting mid-sentence). I'll repeat the full sentence in
a second)

Why not? You said that w_1(m) was an algebraic integer which is both a
factor of a_1 and a factor of f.

That's the assertion that is shown to be impossible if w_1(m) varies
as m varies.

It's impossible because uf is not a function of m.

Basic, and easy stuff, but it takes a mathematician to try and lie
anyway, even when beaten.

Since w_1(m) is a factor of f, f/w_1(m) is an algebraic integer. And
if u is an algebraic integer, then u*[f/w_1(m)] is an algebraic integer.

So this is an algebraic integer for every value of m for which the
function is defined.

Unless you are saying that the product of two algebraic integers is
not necessarily an algebraic integer? (That's a theorem of Eisenstein,
Gauss, and Dedekind)?

Rather than tell the truth this mathematician is trying to distract.

But it doesn't take a rocket scientist to know that given an integer
f, you can't have a function w_1(m) in algebraic integers such that
for *any* integer m, f/w_1(m) is an integer.

Some people may be confused by the symbol "f", so as it's independent
and I can give it a value, let f=13.

Now then, Arturo Magidin needs you to believe that maybe there exists
a function in algebraic integers such that 13/w_1(m) is an algebraic
integer for *all* integer m.

but then uf/w_1(m) cannot in general be an algebraic integer
as it's
not representable as a polynomial with a finite number of terms if
w_1(m) varies with m.

No, this is wrong. There does not have to be a single polynomial that
you can plug m into and that gives you a polynomial for w_1(m). So
you just need an expression with the property that for -each- value
of m, what you get is a monic polynomial with integer coefficients
which has w_1(m) as a root.

You want the VALUES uf/w_1(m) each to be an algebraic integer. That
means that for EACH value of m there must be a polynomial, monic with
integer coefficients, that has uf/w_1(m) as a root. There does not
have to be a single polynomial, nor a single family of polynomials
with easy formulas depending on m. All there has to be is for -each-
value of m, -some- polynomial with integer coefficients that works
for -that- value. The polynomials don't have to be related to one
another by any one finite formula.


There is no finite expression with a finite number of terms with the
property that for every value of m, you get a monic polynomial with
integer coefficients that has all of

-|m|-1, -|m|, -|m|+1,....,0,....,|m|-2, |m|-1

as roots. But there are expressions that for each value of m, give you
a monic polynomial with integer coefficients that has all of them as roots:

For example, let g(n,m) = { 1 if |n|=|m|
{ 0 if |n||m|.

Then take

f(m,x) = \prod_{i=-infinity}^{infinity} (g(i,m)x - [g(i,m)i - 1])

where the index i ranges over all integers.

Or, there is no single expression with a finite number of terms which,
for each nonzero integer value of m gives you a polynomial that has
the largest prime p that divides m, 1 if m=1 or -1, and 0 if m=0, as a
root. But there is a single (infinite) expression that, for each value
of m, gives you a polynomial that has such a root. For instance,

x*(Sum (from i=0 to infinity) [m/i]\prod_{j=1 to [m/i]}(x-j))

where [m/i] is the floor of m/i.


There is no reason to demand that there be a single finite expression
that works; such an expression would be -sufficient-, but not
necessary, for the w_1(m) to be, each, an algebraic integer. In fact,
there is no need to demand a single expression either, although it is
trivial to construct one: for each value of m, let g_m(x) be a
monic polynomial with integer coefficients that has w_1(m) as a root,
and simply take a sum over all m's of xi_{n}*g_n(x), where xi_{n} is
the characteristic function of the singleton {n}: equal to 1 when the
input is n, and equal to zero otherwise.

Readers on sci.physics and sci.logic note that the mathematicians have
been caught in a base lie trying to hide an over one hundred year old
error.

To believe them you need to believe in an impossible function which
would allow 13/w_1(m) to be an algebraic integer for ALL integer m.

Note that I've repeatedly shown that mathematicians have been lying on
this issue but they've escaped by continuing to argue where it has
been difficult for me to show that they MUST be lying, but here your
own mathematical knowledge should let you see the truth.

Anyone who disagrees need just give an algebraic integer function
w_1(m) such that 13/w_1(m) is an algebraic integer for all integer m.

Of course, mathematicians are stepping into the problem that given
some thing like 13/w_1(m) you need, say, r_1(m) as an algebraic
integer function such that

13/w_1(m) = r_1(m), so

w_1(m) r_1(m) - 13 = 0

which necessarily has zeroes. That is, only SOME values where it will
work.

I want you to note just how disgusting mathematicians can be in
trashing mathematics, such that they challenge such a basic reality.

They are rogues who've placed themselves outside of decent society.


James Harris

W. Dale Hall October 18th 03 12:03 AM
Core error argument objection refuted, short
 


James Harris wrote:
(Peter van Rossum) wrote in message ...


... stuff deleted ...



I hope you realize that the functions a_i are also not given by polynomials
in m (i.e., there does not exists a polynomial A_i(M) in A[M], the


That's a rather stupid lie given that I put the polynomial in this
post.

Again it is

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)

and its roots are a_1, a_2 and a_3.


Two points:

1. This is the first time (in the present thread) that you have
stated that the a's are *roots* of that polynomial. All you
said was that the a's are *given* by that polynomial. You
may have read Virgil's article (also this thread), pointing
out that the polynomial doesn't give you anything.

2. You have misread the Mr. van Rossum's remark, preferring to
label it as a lie. Note what he *actually* said, rather than
what you *thought* he said:

I hope you realize that the functions a_i are also not
given by polynomials in m (i.e., there does not exists
a polynomial A_i(M) in A[M], the polynomial ring in one
variable over the algebraic numbers, such that A_i(m) =
A_i(m) for all algebraic numbers m).

It is incontrovertible that the a's, even as roots of the
above polynomial, are not given as the values of a polynomial
in m.


But mathematicians have been persistent in showing a rather pathetic
and stupid refusal to accept rather basic mathematics.

I find them disgusting.


Yes, of course. You are delighted in your constant ability to misread,
to misunderstand, and to operate at the level of a 10-year-old.




The poster is babbling, possibly in shock. Such weakness from the
math world is telling. Mathematicians don't know how to really think,
but have gotten away with faking it.


Ah, yes. Everyone is cowering in fear. No matter that you are running
away from a direct refutation of your position about the coefficients
a1, a2, a3, from the factorization:

65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)


That information has been given to you, and since you couldn't follow
a direct computation showing *each* of the a's to have a (separate)
non-unit factor in common with 5, you ran away.

Brave Sir Robin ran away,
Bravely ran away, away.
When danger reared its ugly head, he bravely turned his tail
and fled.
Yes, brave Sir Robin turned about
And gallantly, he chickened out. Bravely taking to his feet,
He beat a very brave retreat,
Bravest of the brave, Sir Robin.


Can you repeat the definition of algebraic integer again for
the newsgroup and tell us how you conclude that u f / w_1(m)
is not an algebraic integer?


Mathematicians are pathetic liars.

I'll use u=2, f=13 again, now then, NO function in algebraic integers
exists such taht 26/w_1(m) is an algebraic integer for all integers m,
if w_1(m) varies with m.

It's just not possible, but it takes a mathematician to lie about it.



He is packing it in and packing it up
And sneaking away and buggering up
And chickening out and ****ing off home,
Yes, bravely he is throwing in the sponge.


James Harris

Dale

(Lyrics to "Brave Sir Robin", from "Monty Python and the Holy Grail")

Nora Baron October 18th 03 01:28 AM
Core error argument objection refuted, short
 
(James Harris) wrote in message om...
I've noted a problem in algebraic number theory with the inclusiveness
of the definition of algebraic integers as roots of monic polynomials
with integer coefficients.

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

First I have

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and the factorization

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

My finding that only two of the a's have f as a factor without regard
to the value of m has been vigorously disputed.


It's incorrect. See below.



However, consider w_1(m), a factor of a_1 that is a factor of f, as
well as a function that varies with m, then it follows that

a_1 x + uf has w_1(m) as a factor,

so dividing through by w_1(m) gives

a_1 x/w_1(m) + uf/w_1(m)

but then uf/w_1(m) cannot in general be an algebraic integer as it's
not representable as a polynomial with a finite number of terms if
w_1(m) varies with m.


A totally off-the-wall, unjustified statement, and, as it
so happens, incorrect. But for now, if you want to claim
it is true, the shoe is on your foot: try to prove it.



My guess is that some may be assuming that f is replaceable by some
function of m, but in fact, its independent of the value of m,

No, we are not assuming f is "replaceable by some function of m".
We assume f itself is *constant* with respect to m. However, the factorization
of a polynomial P(m) whose coefficients are functions of m, hence
dependent on m, is also in general dependent on m. We do *not* assume
that the corresponding factorization of f is constant with respect
to m.


so it's
like 1/(x+1) which is also not representable by a polynomial if x is
an algebraic integer not equal to 0 or -2.

So the objection is refuted by the impossibility of uf/w_1(m) being an
algebraic integer,


Also incorrect and false. Remember what you said above:

"... consider w_1(m), a factor of a_1 that is a factor of f..."

If w_1(m) is a "factor of f", that can only mean f/w_1(m) is
an algebraic integer, which of course implies that uf/w_1(m)
is an algebraic integer: this is your *assumption* here.


for all algebraic integers m, if w_1(m) varies with
m.

My hope is that posters who have been so successful in convincing
others that my argument is flawed will post concessions.



Absolutely not!

You are inching closer to the truth in this. You are seeing
how this can work. You are correct that w_1(m) is a factor of
f that depends on m. As noted above this implies that
uf/w_1(m) is an A.I., and a1(m)/w_1(m) is also. I think you
are beginning to see how this can happen.



I note a couple of other things. First, the polynomial in "a"
that you mention above:

[#] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

Your claim is that two roots of this polynomial have a factor
that is f. This means that if r is one of those roots, then
r = f*c, where c is an algebraic integer. This implies that

f^3*c^3 + 3(-1 +m*f^2)*f^2*c^2 - f^2*(m^3*f^4 - 3*m*f^2 + 3*m) = 0,

or, factoring out f^2,

f*c^3 + 3*(-1 + m*f^2)*c^2 - (m^3*f^4 - 3*m*f^2 + 3*m) = 0.


The polynomial in c on the left is non-monic and primitive
(if f is not a multiple of 3). If it is irreducible, then
c cannot be an algebraic integer. When f = 5 and m = 1, this
equation is

5*c^3 + 72*c^2 - 553 = 0,

and the polynomial in c is easily shown to be irreducible.
Therefore c cannot be an algebraic integer. This contradicts
your central claim: NONE of the roots of [#] can have a
factor that is 5 = f.


Second note: It may be worthwhile to see how irreducibility
is inextricably tied to factorization of roots. Assume that


Q(m) = x^2 + m*x + 30,

where m is an integer.

The constant term is divisible by 5 (also by 2 and 3, but I
will focus on 5 here).

For certain values of m, this polynomial is reducible, and the
corresponding roots are not both divisible by 5:

m = 11: r1 = 5, r2 = 6

m = 13: r1 = 10, r2 = 3

m = 17: r1 = 15, r2 = 2

m = 31: r1 = 30, r2 = 1.

In all these examples, obviously one root is *divisible* by 5
and the other is *coprime* to 5. And of course in all these
examples, the polynomial is reducible. This is parallel to
Harris's cubic when m = 0: two of the roots are divisible
by f and one is coprime to f. In this particular case his polynomial
[#] is reducible. In general it is not.

Now look at an other example. Say, m = 14. The
polynomial is irreducible, because the discriminant

D^2 = m^ - 4*30 = 76,

which is not a perfect square. The roots of the
polynomial a


r1 = (-14 + sqrt(76))/2 = -7 + sqrt(19) and

r2 = -7 - sqrt(19).

Clearly both r1 and r2 are algebraic integers.

Assume that r1 is a multiple of 5: r1 = 5*s1. Then

25*s1^2 + 70*s1 + 30 = 0.

Factor out 5:
[*] 5*s1^2 + 14*s1 + 6 = 0.

This happens to have discriminant D^2 = 76. [This is
not a coincidence!]. Thus the polynomial in[*] is
*also* irreducible. Therefore s1 cannot be an algebraic
integer.

Therefore r1 cannot be divisible by 5.

The same can similarly be shown for r2.

However, it is now easy to show that both r1 and r2 must both
be *non-coprime* to 5. For, suppose r1 is coprime to 5.
The fact that

r1*r2 = 5*6

would then imply that r2 is DIVISIBLE by 5, which
contradicts the result above that both r1 and r2 are
NOT divisible by 5.

Conclusion: for m = 14, the polynomial is irreducible,
and both roots have a nonunit algebraic integer factor
in common with 5.

This is true more generally. There is nothing special
about 14. Try, for example, m = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12,
14, 15, 16, 18, ... MOST values of m yield an irreducible
polynomial, and both of the roots will both share algebraic
integer factors in common with 5.

In general it is hard to write down exactly what these
factors are, even in the quadratic case. The important
thing to know about them here is that, as suggested above
by the w_1(m) notation, *they will be dependent on m*.


Nora B.




James Harris

Maxim Stepin October 18th 03 03:40 AM
Core error argument objection refuted, short
 

"fishfry" wrote in message
...
In article ,
(James Harris) wrote:

I've noted a problem in algebraic number theory with the inclusiveness
of the definition of algebraic integers as roots of monic polynomials
with integer coefficients.

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

First I have

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

Can't you articulate your object in English? Such as, "If you define
algebraic integers as roots of monic polys with integer coefficients,
then the following odd thing happens: etc. etc. Just say exactly what
you think the contradiction or problem is.

Yes! That's what I would like to hear too.
It seems like he's trying to prove some theorem, but unable to formulate this theorem.

James Harris October 18th 03 05:12 PM
Core error argument objection refuted, short
 
(Arturo Magidin) wrote in message ...
In article ,
James Harris wrote:

[.snip.]

It's not possible. I'll put in values to help those readers confused
by symbols (though algebra is BASED on symbols) by letting u=2, f=13,
then you have

26/w_1(m)

and you don't have to be a rocket scientist to know that no function
w_1(m) that actually varies with m can exist such that 26/w_1(m) is an
algebraic integer for *all* integer m.


w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)}

I'm not saying that's the function in question (it is not), but there
is a function that "actually varies with m such that 26/w_1(m) is an
algebraic integer for *all* integer m." So the argument that no such
function can exist is simply bogus.

Well that example works, so now I'll say algebraic integer m, and it
blows up at m^2+1 = 0.

For those confused by the discussion, I've pointed out that
mathematicians arguing with me have been arguing for a relation
relating the multiplicative inverse of one algebraic integer function
to another over all algebraic integers.

For instance, with u=2, f=7

14/w_1(m) = r(m)

which would relate w_1(m) to r(m) for all algebraic integer m, which
is like claiming that xy=2 for all integer x, with an integer y.

It's that multiplicative inverse that blows away the objection as in
fact, w_1(m) is a constant with regard to m--independent of m--as I've
repeatedly shown.

However, accepting basic algebra would mean that mathematicians accept
that there's an over one hundred year old definition problem in core
mathematics.

Many of you may have believed that I was making something up, but you
should have enough mathematical knowledge to understand that the
relations these people are arguing for, cannot exist.

Unless you're lost on the example of xy=2, in the ring of integers, in
comparison to 14/w_1(m) = r(m) in the ring of algebraic integers.

It's that "/" that makes the difference. Math people are lying here
because the information is so earthshaking, and they are corrupt.


James Harris


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