I've found that there is a definition error in core mathematics, and
I've given the math necessary to understand the problem; however, I
*still* see posters arguing, and it occurs to me that I need to show
you why what they're saying is ludicrous, if any progress is to be
made.

First I have a somewhat forbidding expression:

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)

Rather than fiddle with it, I instead focus on what does NOT have m as
a factor, which is easily done by setting m=0, which gives

P(0) = u^2 f^2 ( 3x + uf)

which is much nicer.

Several posters have, however, continually tried to raise the
possibility that P(0) is in fact STILL a function of m, in a rather
odd way, as they've focused on factors of P(m), as I use the factors
g_1, g_2, and g_3, where

P(m) = g_1 g_2 g_3

and split up P(0) with them, so that I also have that

when m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.

Now a neat opportunity is revealed by the use of these terms
*independent* of m, as I can easily see that f^2 divides off of P(m)
and P(0) in a straightforward way.

I've isolated terms independent of m, so without regard to m, I know
that two and only two of the g's have a factor that is f, when f is
coprime to 3, x and u.

It's that conclusion which posters have argued against, and the
purpose of this post is to allow you to see that what they're posting
is illogical.

That is, focus on objectors trying to claim that the g's have factors
of f that vary as m varies, which would force a dependency back on to
those terms independent of m.

You see that's why I focus on terms *independent* of m, as they have
greater weight than those dependent on m, because they CANNOT be
affected by m's value, because if they were, they'd be dependent on m.

See the tautology?

When objectors try to convince you that the tautology can be broken
remember what's key:

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)

P(0) = u^2 f^2 ( 3x + uf)

P(m) = g_1 g_2 g_3

when m=0,

g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.

Focus on uf, and 3x + uf being independent of m, and the argument at

http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782

is rather straightforward.

So why am I facing posters willing to argue so continually against
rather basic math?

Well it turns out that over a hundred years ago algebraic integers
were defined as roots of monic polynomials with integer coefficients.

It was a good idea for the time, but mathematicians failed to realize
that it was slightly flawed in that some roots of non-monic
polynomials with integer coefficients need to be included or you can
create supposed proofs of two different and opposite conclusions in
the ring of algebraic integers.

That makes it an error in core mathematics, which is kind of big news.

Or as Isaac Asimov might have said, rather funny.


James Harris

James Harris wrote:
I've found that there is a definition error in core mathematics, and
I've given the math necessary to understand the problem; however, I
*still* see posters arguing, and it occurs to me that I need to show
you why what they're saying is ludicrous, if any progress is to be
made.

First I have a somewhat forbidding expression:

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)

Rather than fiddle with it, I instead focus on what does NOT have m as
a factor, which is easily done by setting m=0, which gives

P(0) = u^2 f^2 ( 3x + uf)

which is much nicer.

Agreed. Would you also agree that P(m) P(0)?

Several posters have, however, continually tried to raise the
possibility that P(0) is in fact STILL a function of m, in a rather
odd way, as they've focused on factors of P(m), as I use the factors
g_1, g_2, and g_3, where

P(m) = g_1 g_2 g_3

Would you agree that at least one, if not all, of the g's must vary as m
varies? If so, it may be clearer to write them as g_1(m), g_2(m), g_3(m).


and split up P(0) with them, so that I also have that

when m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.

The notation could be cleaned up, but ok.


Now a neat opportunity is revealed by the use of these terms
*independent* of m, as I can easily see that f^2 divides off of P(m)
and P(0) in a straightforward way.

No. You can see that P(0) is divisible by f^2 in a straitforward way.
To claim that it does so with P(m) is different, unless you are claiming
that P(m) is effectively the same as P(0). This is easily seen by
noting that g_1(m) and g_2(m) have x terms, but g_1(0) and g_2(0) do
not. There is a fundamental difference between the two.


I've isolated terms independent of m, so without regard to m, I know
that two and only two of the g's have a factor that is f, when f is
coprime to 3, x and u.

Only when m=0. This says nothing about when m0.


It's that conclusion which posters have argued against, and the
purpose of this post is to allow you to see that what they're posting
is illogical.

That is, focus on objectors trying to claim that the g's have factors
of f that vary as m varies, which would force a dependency back on to
those terms independent of m.

This has been explicitly demonstrated using f=2, u=1, m=0 and f=2, u=1,
m=1. The g's are observably different, and there is a change that
occurs where you drop from two g's divisble by 2 to one g divisible by
2. You are claiming otherwise, yet the numbers are there for you to
inspect.


You see that's why I focus on terms *independent* of m, as they have
greater weight than those dependent on m, because they CANNOT be
affected by m's value, because if they were, they'd be dependent on m.

The problem lies in the terms that are dependent on m. They change, and
they are also part of all three g's.

Consider f(m) = mx+2. f(0)=2 is divisible by 2 in the algebraic
integers. f(1)=x+2 is not. f(0) has no x term. f(1) does. This is
the same change that is occurring with your g's.


See the tautology?

No.


When objectors try to convince you that the tautology can be broken
remember what's key:

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)

P(0) = u^2 f^2 ( 3x + uf)

P(m) = g_1 g_2 g_3

when m=0,

g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.

Focus on uf, and 3x + uf being independent of m, and the argument at

http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782

is rather straightforward.

Except for the fact that it ignores two terms, and misrepresents one as
independent of m.


So why am I facing posters willing to argue so continually against
rather basic math?

Because there is a difference between what happens when m=0 and when
m=1. There is a trivial counter-example to your claims. You are
ignoring it rather than realizing that your "proof" *cannot* be true.


Well it turns out that over a hundred years ago algebraic integers
were defined as roots of monic polynomials with integer coefficients.

It was a good idea for the time, but mathematicians failed to realize
that it was slightly flawed in that some roots of non-monic
polynomials with integer coefficients need to be included or you can
create supposed proofs of two different and opposite conclusions in
the ring of algebraic integers.

You misunderstand. The definition came first. Then, people showed that
they are closed under addition and multiplication, as well has holding
the other ring properties. The best you can do is show that the
algebraic integers are not a ring. The definition cannot be flawed,
only our belief that they form a ring. If you believe they are not a
ring, then you must show which ring property they violate.

Please understand me: a clear definition cannot be wrong, only useless.
The algebraic integers *are* the complex roots of monic polynomials.
If they do not have properties that you like, that is an issue with what
you like, not with the definition. If there is a mistake about the
properties they are claimed to have, please clearly demonstrate it.
This is unlikely to occur, however, as the proofs are fairly
straight-forward and well-documented.


That makes it an error in core mathematics, which is kind of big news.

The biggest news you can hope for is that the algebraic integers do not
form a ring. I'm not sure what the repercussions of that would be.



--
Will Twentyman
email: wtwentyman at copper dot net

That is, focus on objectors trying to claim that the g's have factors
of f that vary as m varies, which would force a dependency back on to
those terms independent of m.

How can you say that a dependency is laid upon independent terms? Apart from
that, dependency to what? No ideas of dependency are configured by
mathematics.

JJ


"James Harris" wrote in message
m...
I've found that there is a definition error in core mathematics, and
I've given the math necessary to understand the problem; however, I
*still* see posters arguing, and it occurs to me that I need to show
you why what they're saying is ludicrous, if any progress is to be
made.

First I have a somewhat forbidding expression:

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)

Rather than fiddle with it, I instead focus on what does NOT have m as
a factor, which is easily done by setting m=0, which gives

P(0) = u^2 f^2 ( 3x + uf)

which is much nicer.

Several posters have, however, continually tried to raise the
possibility that P(0) is in fact STILL a function of m, in a rather
odd way, as they've focused on factors of P(m), as I use the factors
g_1, g_2, and g_3, where

P(m) = g_1 g_2 g_3

and split up P(0) with them, so that I also have that

when m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.

Now a neat opportunity is revealed by the use of these terms
*independent* of m, as I can easily see that f^2 divides off of P(m)
and P(0) in a straightforward way.

I've isolated terms independent of m, so without regard to m, I know
that two and only two of the g's have a factor that is f, when f is
coprime to 3, x and u.

It's that conclusion which posters have argued against, and the
purpose of this post is to allow you to see that what they're posting
is illogical.

That is, focus on objectors trying to claim that the g's have factors
of f that vary as m varies, which would force a dependency back on to
those terms independent of m.

You see that's why I focus on terms *independent* of m, as they have
greater weight than those dependent on m, because they CANNOT be
affected by m's value, because if they were, they'd be dependent on m.

See the tautology?

When objectors try to convince you that the tautology can be broken
remember what's key:

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)

P(0) = u^2 f^2 ( 3x + uf)

P(m) = g_1 g_2 g_3

when m=0,

g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.

Focus on uf, and 3x + uf being independent of m, and the argument at

http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782

is rather straightforward.

So why am I facing posters willing to argue so continually against
rather basic math?

Well it turns out that over a hundred years ago algebraic integers
were defined as roots of monic polynomials with integer coefficients.

It was a good idea for the time, but mathematicians failed to realize
that it was slightly flawed in that some roots of non-monic
polynomials with integer coefficients need to be included or you can
create supposed proofs of two different and opposite conclusions in
the ring of algebraic integers.

That makes it an error in core mathematics, which is kind of big news.

Or as Isaac Asimov might have said, rather funny.


James Harris

(James Harris) wrote in message om...
I've found that there is a definition error in core mathematics, and
I've given the math necessary to understand the problem; however, I
*still* see posters arguing, and it occurs to me that I need to show
you why what they're saying is ludicrous, if any progress is to be
made.

First I have a somewhat forbidding expression:

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)

Rather than fiddle with it, I instead focus on what does NOT have m as
a factor, which is easily done by setting m=0, which gives

P(0) = u^2 f^2 ( 3x + uf)

which is much nicer.

Several posters have, however, continually tried to raise the
possibility that P(0) is in fact STILL a function of m,


This statement is so incorrect and so stupid that it MUST be
an example of deliberately missing the point. We have NOT
said that P(0) is "STILL a function of m". What we HAVE said
is that P(m) for m 0 is a function of m. I doubt you disagree
with that.


in a rather
odd way, as they've focused on factors of P(m), as I use the factors
g_1, g_2, and g_3, where

P(m) = g_1 g_2 g_3

and split up P(0) with them, so that I also have that

when m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.

Now a neat opportunity is revealed by the use of these terms
*independent* of m, as I can easily see that f^2 divides off of P(m)
and P(0) in a straightforward way.


P(0), yes. P(m), no. You have shown nothing about P(m)
for m 0.

Your argument boils down to the following dimwit statement:

Let a(m) be a function of m such that a(0) = 0. Thus a(0)
is divisible by f for any positive integer f. a(0) is
independent of m. Therefore a(m) is also divisible by f for
any positive integer f and m 0.

I daresay even you can think of counterexamples to your under-
lying principle here.


I've isolated terms independent of m, so without regard to m, I know
that two and only two of the g's have a factor that is f, when f is
coprime to 3, x and u.


You have done this only when m = 0. That part is OK. It is
when you claim that because P(0) and g1(0), a1(0), etc., are
"independent" of m that you overreach. Granted, these constants
are "independent" of m. And SO WHAT ? You still need to consider
the stuff that is NOT independent of m: P(m), g1(m), a1(m), etc.,
when m 0. What on earth makes you think that it is sufficient
to consider only m = 0 ???


It's that conclusion which posters have argued against, and the
purpose of this post is to allow you to see that what they're posting
is illogical.

That is, focus on objectors trying to claim that the g's have factors
of f that vary as m varies, which would force a dependency back on to
those terms independent of m.


Here g1 = a1*x + u*f. Agreed?

Also a1 is dependent on m. Agreed?

Therefore g1 is also dependent on m. Right?

Therefore something about g1 that is true when m = 0 need not
necessarily be true when m 0. Right? Or are you saying
that everything that is true about g1 when m = 0 *is* also true
when m 0 ? Is g1 a CONSTANT FUNCTION ? Is that what you
think ?


You see that's why I focus on terms *independent* of m, as they have
greater weight than those dependent on m, because they CANNOT be
affected by m's value, because if they were, they'd be dependent on m.

See the tautology?


The implied tautology is: the constant term is constant.

Unfortunately you must deal with the NONCONSTANT terms
also. Continuing to blather on about how the constant term
P(0) is independent of m, and how the constant term is
constant, etc., is just so much noise. You have confused
yourself to the point that you can no longer think clearly
on this topic. You have lost all perspective.


When objectors try to convince you that the tautology can be broken
remember what's key:

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)

P(0) = u^2 f^2 ( 3x + uf)

P(m) = g_1 g_2 g_3

when m=0,

g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.


Yes. All just fine up to this point.


Focus on uf, and 3x + uf being independent of m,


Yes. These are "independent" of m, because these are
the values of the factors when m = 0. But they are not what
you must deal with when m 0. In that case you must
consider

a1*x + u*f, a2*x + u*f, and a3*x + u*f.

None of these are "independent of m". You yourself have
admitted that a1 and a2 are dependent on m.

It's not enough to focus on the constant terms. The
function you are considering, P(m), is not comprised
solely of its own constant term. You are factoring it
for all m, not just m = 0. Its factors when m = 0 do
not determine its factors when m 0. You should know
better by now.


and the argument at

http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782

is rather straightforward.


And incorrect.


So why am I facing posters willing to argue so continually against
rather basic math?


You are NOT facing them. You started two new threads on this
topic today alone, ostensibly because things were getting too
hot for you in "old" threads. You come here and try to
grandstand as if you were some poor victim of evil lying
mathematicians. In fact you are running away and hoping we will
not follow.



Well it turns out that over a hundred years ago algebraic integers
were defined as roots of monic polynomials with integer coefficients.

It was a good idea for the time, but mathematicians failed to realize
that it was slightly flawed in that some roots of non-monic
polynomials with integer coefficients need to be included or you can
create supposed proofs of two different and opposite conclusions in
the ring of algebraic integers.


Your previous posts on this indicate that you think there
are algebraic integers A and B such that A * B is NOT an
algebraic integer. This does not imply that the definition of
algebraic integer is flawed. If it were true, it would imply
that the SET of algebraic integers is not closed under
multiplication. There are several problems with this:

1) You have never specified what A and B might be.

2) There is a very old theorem which says: the set of
algebraic integers forms a ring. That implies that
the algebraic integers are closed under multiplication.

3) If what you claimed were correct, it would not indicate
a problem with "core" (whatever that is). It would
indicate that mathematics is inconsistent. You have
drawn back from making that claim because you know that
it would imply that your FLT "proof", and almost all
the rest of mathematics, is meaningless. Yet if you
were right, this would be the inescapable conclusion.
You have therefore tried to pass it off as an error
in a definition, which really makes no sense at all.


You could clear this up a bit by specifying A and B. It
is certainly a mystery why you have not done this before.


That makes it an error in core mathematics, which is kind of big news.

Or as Isaac Asimov might have said, rather funny.


Bizarre. Why is this worth quoting?


Nora B.


James Harris

You're right again James.
Give up on them.
Surely, _they_ (the "evil ones" that is), are not worth your precious time.
I mean, just think of all the great contributions you could have made in the
name of mathematics if you did not have to argue with those submental
types.Sure they pointed errors out from time to time,but you would have seen
them anyhow, right?
In fact, it could be made a rule that nobody replies to James' posts, so he
can get more work done....
(Guess we tried that one before, eh? Oh well, but it was fun trying ;))

M

JSH writes:
I've found that there is a definition error in core mathematics, and
I've given the math necessary to understand the problem; however, I
*still* see posters arguing, and it occurs to me that I need to show
you why what they're saying is ludicrous, if any progress is to be
made.....

snip (some awesome math. 'Twas a pity to have to snip it.)....


So why am I facing posters willing to argue so continually against
rather basic math?

(Nora Baron) wrote in message om...
(James Harris) wrote in message om...
I've found that there is a definition error in core mathematics, and
I've given the math necessary to understand the problem; however, I
*still* see posters arguing, and it occurs to me that I need to show
you why what they're saying is ludicrous, if any progress is to be
made.

First I have a somewhat forbidding expression:

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)

Rather than fiddle with it, I instead focus on what does NOT have m as
a factor, which is easily done by setting m=0, which gives

P(0) = u^2 f^2 ( 3x + uf)

which is much nicer.

Several posters have, however, continually tried to raise the
possibility that P(0) is in fact STILL a function of m,


This statement is so incorrect and so stupid that it MUST be
an example of deliberately missing the point. We have NOT
said that P(0) is "STILL a function of m". What we HAVE said
is that P(m) for m 0 is a function of m. I doubt you disagree
with that.

The problem with a poster arguing against mathematical logic is that
when it's carefully explained *they* look stupid, so this poster
decides to angrily deny reality.

However, P(0) = u^2 f^2( 3x + uf) which should be VERY familiar, and
this poster and others have claimed an m dependency on how f^2 divides
off, as P(m) also has a factor that is f^2, which is why P(0) does.

By claiming a dependency on m for how f^2 divides off, this poster and
others have been doing exactly what I said.

Now it turns out that P(m) = g_1 g_2 g_3, so their assertions have to
do with the value of the g's at m=0 as then g_1 = uf, g_2 = uf, g_3 =
3x + uf, which are values independent of m. Bbut in claiming a
dependency on m for how the f^2 divides off from the g's, when m does
not equal 0, when I've proven that two of them must have f as a factor
as revealed by terms INDEPENDENT of m, they are ignoring rather basic
algebra and logic.

Anger doesn't change that reality.

in a rather
odd way, as they've focused on factors of P(m), as I use the factors
g_1, g_2, and g_3, where

P(m) = g_1 g_2 g_3

and split up P(0) with them, so that I also have that

when m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.

Now a neat opportunity is revealed by the use of these terms
*independent* of m, as I can easily see that f^2 divides off of P(m)
and P(0) in a straightforward way.


P(0), yes. P(m), no. You have shown nothing about P(m)
for m 0.

The irrationality here is almost scary.

I focus on terms *independent* of m, so LOGICALLY, there wouldn't be
an m dependency on the results.

Yet here you have this poster crying about m not equal to 0, as if it
matters what value m is!!!

The behavior is irrational.

Your argument boils down to the following dimwit statement:

At least this poster seems to have turned to anger versus the fear
that seems to grip so many others.

That's interesting.

Let a(m) be a function of m such that a(0) = 0. Thus a(0)
is divisible by f for any positive integer f. a(0) is
independent of m. Therefore a(m) is also divisible by f for
any positive integer f and m 0.

Notice though that this poster IS still acting from fear as that's not
the actual argument.

I focus on terms independent of m for a reason, but in replies,
posters like this one repeatedly make up something, and claim that's
the argument.

It's kind of sad, actually.

But remember, you can't eat a math proof.

Human beings are wired to act more for social reasons than for
abstract things like mathematical truth.

This poster is clearly well aware of the *social* consequences of
telling the truth, and is, like human beings so often are, willing to
toss out the truth, for this poster's society.

The poster is protecting mathematicians and the protective instinct
may be strong, but is actually *against* world society, which gains no
benefit from mathematicians hiding the definition error in core.

I daresay even you can think of counterexamples to your under-
lying principle here.

The underlying principle is that terms independent of m are
independent of m.

So changes in those terms occur, yup, independent of m.

That simple principle gives the result, and the result is, independent
of m.

I've isolated terms independent of m, so without regard to m, I know
that two and only two of the g's have a factor that is f, when f is
coprime to 3, x and u.


You have done this only when m = 0. That part is OK. It is
when you claim that because P(0) and g1(0), a1(0), etc., are
"independent" of m that you overreach. Granted, these constants
are "independent" of m. And SO WHAT ? You still need to consider
the stuff that is NOT independent of m: P(m), g1(m), a1(m), etc.,
when m 0. What on earth makes you think that it is sufficient
to consider only m = 0 ???

Readers should note that the poster is behaving irrationally as in
acknowledging that I've found terms *independent* of m using m=0, the
poster *rationally* should then accept that the value of m is
irrelevant to those terms.

The technique of setting m=0 is just a way to find them.

It's a neat technique that works quite well.

Now then, if the poster can recognize that they are found, why then
does the poster bother with mentioning m not equal 0?

After all, for terms independent of m, m's value doesn't matter.

The best explanation for this poster is fear and anger.

Many of you probably know about the effects of those in other
contexts, but here you can see it with mathematics.

It's that conclusion which posters have argued against, and the
purpose of this post is to allow you to see that what they're posting
is illogical.

That is, focus on objectors trying to claim that the g's have factors
of f that vary as m varies, which would force a dependency back on to
those terms independent of m.


Here g1 = a1*x + u*f. Agreed?

Yup.

Also a1 is dependent on m. Agreed?

Yup.

Therefore g1 is also dependent on m. Right?

Yup.

Therefore something about g1 that is true when m = 0 need not
necessarily be true when m 0. Right? Or are you saying
that everything that is true about g1 when m = 0 *is* also true
when m 0 ? Is g1 a CONSTANT FUNCTION ? Is that what you
think ?

Consider that the poster is, yet again, trying to find a way to make
independent terms, dependent.

Here what's key is that uf is independent and looking at P(0)/f^2
reveals that when f^2 is divided from P(m), that is, with P(m)/f^2,
the *independent* term must lose the factor f, and then is u.

Now then, as that is INDEPENDENT of m, it has to occur, independent of
m.


You see that's why I focus on terms *independent* of m, as they have
greater weight than those dependent on m, because they CANNOT be
affected by m's value, because if they were, they'd be dependent on m.

See the tautology?


The implied tautology is: the constant term is constant.

The value of m does not affect terms that are independent of m.

Readers should note the recalcitrance. It looks like anger and fear
to me.

Unfortunately you must deal with the NONCONSTANT terms
also. Continuing to blather on about how the constant term
P(0) is independent of m, and how the constant term is
constant, etc., is just so much noise. You have confused
yourself to the point that you can no longer think clearly
on this topic. You have lost all perspective.

Yet I easily explain how while g_1 has uf as an independent term,
P(m)/f^2 doesn't have independent terms with f as a factor AT ALL.

That happens *independent* of the value of m.

When objectors try to convince you that the tautology can be broken
remember what's key:

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)

P(0) = u^2 f^2 ( 3x + uf)

P(m) = g_1 g_2 g_3

when m=0,

g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.


Yes. All just fine up to this point.


Focus on uf, and 3x + uf being independent of m,


Yes. These are "independent" of m, because these are
the values of the factors when m = 0. But they are not what
you must deal with when m 0. In that case you must
consider

a1*x + u*f, a2*x + u*f, and a3*x + u*f.

Notice the odd irrationality here. The poster accepts that the terms
are independent of m, but then turns right around and talks about m
dependency.

To those on sci.physics and sci.logic, I hope you understand now my
predicament.

I've contacted TOP mathematicians and they ran away, and you can see
the behavior here with this poster.

Unfortunately, math society seems to be frozen by the trauma of this
definition error, and it needs help.

None of these are "independent of m". You yourself have
admitted that a1 and a2 are dependent on m.

However, they are paired with with terms INDEPENDENT of m, and those
terms rule because they are.

It's not enough to focus on the constant terms. The
function you are considering, P(m), is not comprised
solely of its own constant term. You are factoring it
for all m, not just m = 0. Its factors when m = 0 do
not determine its factors when m 0. You should know
better by now.


and the argument at

http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782

Yup.


is rather straightforward.


And incorrect.

Stated without proof.

So why am I facing posters willing to argue so continually against
rather basic math?


You are NOT facing them. You started two new threads on this
topic today alone, ostensibly because things were getting too
hot for you in "old" threads. You come here and try to
grandstand as if you were some poor victim of evil lying
mathematicians. In fact you are running away and hoping we will
not follow.


I'm replying in all the threads.

I thought it important to note the FEAR that explains this phenomenom,
so I made a couple of threads, as well as to pull in other newsgroups,
as I need help with this level of group irrationality.

Also I thought it important to point out that the argument is short
enough to be checked by computer, so I made another thread.


Well it turns out that over a hundred years ago algebraic integers
were defined as roots of monic polynomials with integer coefficients.

It was a good idea for the time, but mathematicians failed to realize
that it was slightly flawed in that some roots of non-monic
polynomials with integer coefficients need to be included or you can
create supposed proofs of two different and opposite conclusions in
the ring of algebraic integers.


Your previous posts on this indicate that you think there
are algebraic integers A and B such that A * B is NOT an
algebraic integer. This does not imply that the definition of
algebraic integer is flawed. If it were true, it would imply
that the SET of algebraic integers is not closed under
multiplication. There are several problems with this:

What previous posts?


1) You have never specified what A and B might be.

2) There is a very old theorem which says: the set of
algebraic integers forms a ring. That implies that
the algebraic integers are closed under multiplication.

3) If what you claimed were correct, it would not indicate
a problem with "core" (whatever that is). It would
indicate that mathematics is inconsistent. You have
drawn back from making that claim because you know that
it would imply that your FLT "proof", and almost all
the rest of mathematics, is meaningless. Yet if you
were right, this would be the inescapable conclusion.
You have therefore tried to pass it off as an error
in a definition, which really makes no sense at all.


You could clear this up a bit by specifying A and B. It
is certainly a mystery why you have not done this before.

What?

That makes it an error in core mathematics, which is kind of big news.

Or as Isaac Asimov might have said, rather funny.


Bizarre. Why is this worth quoting?


Nora B.

Inside joke.


James Harris

"John Jones" wrote in message ...
That is, focus on objectors trying to claim that the g's have factors
of f that vary as m varies, which would force a dependency back on to
those terms independent of m.

How can you say that a dependency is laid upon independent terms? Apart from
that, dependency to what? No ideas of dependency are configured by
mathematics.

JJ

Well that looks like my writing above, though I don't see reference.

In any event, it sound like "John Jones" is just reiterating my point,
which shows that posters arguing against me are full of it.

Unfortunately I only recently discovered that use of the word
"independent" works better than "constant term" which is just one of
those things I guess.

People are interesting in terms of what works.

In any event, for those of you STILL trying to deny basic algebra who
are familiar with the argument showing the problem in core, consider
the alternative:

What if g_1 *does* have a factor of f that varies with m?

Well then, since g_1 = a_1 x + uf, if, say, at m=5, the factor is
sqrt(f), then necessarily you have g_1/sqrt(f) =a_1 x + u sqrt(f), but
your *independent* term just changed with a dependency on m.

Why does it *have* to change? Because part of my argument is dividing
P(m) by f^2, and P(m) = g_1 g_2 g_3.

So, yes, posters arguing with me, have been arguing the illogical
position that the independent term is dependent, and getting away with
it, because, I think, FEAR rules the math community at this point.

I'll leave in the remainder from the original post as it looks like
this poster top-posted, changed the subject line, and then copied in
some text from another of my posts. Why? Who knows?


James Harris

"James Harris" wrote in message
m...
I've found that there is a definition error in core mathematics, and
I've given the math necessary to understand the problem; however, I
*still* see posters arguing, and it occurs to me that I need to show
you why what they're saying is ludicrous, if any progress is to be
made.

First I have a somewhat forbidding expression:

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)

Rather than fiddle with it, I instead focus on what does NOT have m as
a factor, which is easily done by setting m=0, which gives

P(0) = u^2 f^2 ( 3x + uf)

which is much nicer.

Several posters have, however, continually tried to raise the
possibility that P(0) is in fact STILL a function of m, in a rather
odd way, as they've focused on factors of P(m), as I use the factors
g_1, g_2, and g_3, where

P(m) = g_1 g_2 g_3

and split up P(0) with them, so that I also have that

when m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.

Now a neat opportunity is revealed by the use of these terms
*independent* of m, as I can easily see that f^2 divides off of P(m)
and P(0) in a straightforward way.

I've isolated terms independent of m, so without regard to m, I know
that two and only two of the g's have a factor that is f, when f is
coprime to 3, x and u.

It's that conclusion which posters have argued against, and the
purpose of this post is to allow you to see that what they're posting
is illogical.

That is, focus on objectors trying to claim that the g's have factors
of f that vary as m varies, which would force a dependency back on to
those terms independent of m.

You see that's why I focus on terms *independent* of m, as they have
greater weight than those dependent on m, because they CANNOT be
affected by m's value, because if they were, they'd be dependent on m.

See the tautology?

When objectors try to convince you that the tautology can be broken
remember what's key:

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)

P(0) = u^2 f^2 ( 3x + uf)

P(m) = g_1 g_2 g_3

when m=0,

g_1 = uf, g_2 = uf, and g_3 = 3x + uf

and I add in the requirement that f be coprime to 3, x and u.

Focus on uf, and 3x + uf being independent of m, and the argument at

http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782

is rather straightforward.

So why am I facing posters willing to argue so continually against
rather basic math?

Well it turns out that over a hundred years ago algebraic integers
were defined as roots of monic polynomials with integer coefficients.

It was a good idea for the time, but mathematicians failed to realize
that it was slightly flawed in that some roots of non-monic
polynomials with integer coefficients need to be included or you can
create supposed proofs of two different and opposite conclusions in
the ring of algebraic integers.

That makes it an error in core mathematics, which is kind of big news.

Or as Isaac Asimov might have said, rather funny.


James Harris