The nice thing about a really short proof, like what's necessary to
show the definition error in core, is that it should be checkable by
an automated proof checker. Having spent some time arguing with
posters I feel capable of giving the argument in detail, though I'm
not sure that this post will be detailed enough for a proof checker,
so I'm giving an idea for an outline.

I *can* add in more detail as necessary.

Obviously my mention of a proof checker is a move to end the debate,
and move things forward. I'll be using a different format, hopefully
to help in moving towards something a machine can read.

Proof of Core Error

1. Given

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)

Variables: m, f, x, u E Ring of Algebraic integers

P(m) is a polynomial with m the key variable.


2. Let P(m) = (a_1(m) x + uf)(a_2(m) x + uf)(a_3(m) x + uf)

Variables: a_1, a_2, a_3, roots of cubic defined as follows.

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

Solving for roots of P(m) by setting x = -uf/a will give cubic.


3. Let P(m) = g_1(m) g_2(m) g_3(m)

Variables: g_1, g_2, g_3 defined as follows

g_1(m) = (a_1(m) x + uf),

g_2(m) = (a_2(m) x + uf),

g_3(m) = (a_3(m) x + uf).


Tautological base:

Change in terms independent of m happens independent of m.

4. List of independent terms.

Independent terms are found by setting m=0. Doing so with cubic
gives:

a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3.

(indices selected arbitrarily)

Then from previous definitions:

a. g_1 has value uf at m=0 indicating indepedent term uf

b. g_2 has value uf at m=0 indicating independent term uf

c. g_3 has value 3x + uf at m=0 indicating independent term 3x + uf

CONDITION:

f is coprime to 3, x and u

PRIMARY ARGUMENT

5. Divide off f^2 from P(m).

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


6. List resultant independent term

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

which is

P(0)/f^2 = u^2 g_3(0)

which is

P(0)/f^2 = g_1(0)/f g_2(0)/f g_3(0)

a. g_1(0)/f is coprime to f

b. g_2(0)/f is coprime to f

c. g_3(0) is coprime to f


Preliminary Finding:

Independent terms of resultants are coprime to f.


7. Backwards Theorem: Reverse use of distributive property proves
g_1, g_2 have factor that is f.

g_1(m) = a_1(m) x + uf

g_2(m) = a_1(m) x + uf

so to remove factor f from independent term uf, f must divide g_1(m)
and g_2(m), from reverse use of distributive property:

g_1(m)/f = a_1(m)/f x + u.

(Note to readers: The point of using independent terms is that they
are *independent* of the value of m.)


8. Verification:

Determine independent term of g_1(m)/f, by setting m=0.

g_1(0)/f = a_1(0)/f x + u = u.

Confirmed factor f for g_1.

Determine independent term of g_2(m)/f, by setting m=0.

g_2(0)/f = a_2(0)/f x + u = u.

Confirmed factor f for g_1.


9. Core error determination

Here is where the demonstration that for certain values two of the
g's do not have a factor that is f in the ring of algebraic integers
goes.

10. Conclusion

Core error, of course.


My hope is that *someone* will at least consider the possibility of
putting the argument into machine checkable format.

I can add details as necessary.


James Harris

In article ,
(James Harris) wrote:

My hope is that *someone* will at least consider the possibility of
putting the argument into machine checkable format.

I can add details as necessary.

I think what you posted can be shown to be equivalent to

1 = 0

which is in a nice, machine checkable format and provably wrong.

On 10 Oct 2003 21:44:46 -0700, (James Harris) wrote:

The nice thing about a really short proof, like what's necessary to
show the definition error in core, is that it should be checkable by
an automated proof checker. Having spent some time arguing with
posters I feel capable of giving the argument in detail, though I'm
not sure that this post will be detailed enough for a proof checker,
so I'm giving an idea for an outline.

I *can* add in more detail as necessary.

Obviously my mention of a proof checker is a move to end the debate,
and move things forward. I'll be using a different format, hopefully
to help in moving towards something a machine can read.

Proof of Core Error

This proof NOWHERE uses the fact that any quantity is an algebraic
integer. Don't you think this is a problem?

--
Wim Benthem

Wim Benthem wrote in message . ..
On 10 Oct 2003 21:44:46 -0700, (James Harris) wrote:

The nice thing about a really short proof, like what's necessary to
show the definition error in core, is that it should be checkable by
an automated proof checker. Having spent some time arguing with
posters I feel capable of giving the argument in detail, though I'm
not sure that this post will be detailed enough for a proof checker,
so I'm giving an idea for an outline.

I *can* add in more detail as necessary.

Obviously my mention of a proof checker is a move to end the debate,
and move things forward. I'll be using a different format, hopefully
to help in moving towards something a machine can read.

Proof of Core Error

This proof NOWHERE uses the fact that any quantity is an algebraic
integer. Don't you think this is a problem?

No it's not a problem, as it has to do with the object ring.

So there's no issue there as what's important is that the proof should
be machine checkable, and if further detail is needed for that, I can
provide it.

The object ring is defined as follows:

The object ring is a commutative ring that includes all numbers such
that -1 and 1 are the only members that are both a unit and an
integer, where no non-unit member is a factor of any two integers that
are coprime.

It's something of a natural ring and includes algebraic integers.

The proof actually works in that ring, but using numbers that are all
algebraic integers, which should leave you in the ring of algebraic
integers, does not, as you're forced out, which reveals the problem
with the definition of algebraic integers.


James Harris

Christian Bau wrote in message ...
In article ,
(James Harris) wrote:

My hope is that *someone* will at least consider the possibility of
putting the argument into machine checkable format.

I can add details as necessary.

I think what you posted can be shown to be equivalent to

1 = 0

which is in a nice, machine checkable format and provably wrong.

The poster is lying--yet again.

What's important here is that my work IS correct, and luckily for me,
since I have mathematicians running in terror all over the place, it's
short enough that it should be machine checkable.

For people who don't understand what that means, I mean a computer can
check each and every step.

Now, it IS short enough for that, but getting it in proper format
requires help, as I'm not an expert in that area, and I'm putting this
post on multiple newsgroups, as I fear that mathematicians will try to
keep running.

Remember, I'm talking about an argument that can be verified by
computer.

But then see, Usenet posters couldn't keep trying to declare it wrong,
and getting doubt and confusion, now could they?

It's that simple Usenet. I found this definition problem at the heart
of what's commonly called mathematics. Mathematicians have been
running away in fear.

It's time for them to stop running.


James Harris

"James Harris" wrote in message
m...
Christian Bau wrote in message
...
In article ,
(James Harris) wrote:

My hope is that *someone* will at least consider the possibility of
putting the argument into machine checkable format.

I can add details as necessary.

I think what you posted can be shown to be equivalent to

1 = 0

which is in a nice, machine checkable format and provably wrong.

The poster is lying--yet again.

What's important here is that my work IS correct, and luckily for me,
since I have mathematicians running in terror all over the place, it's
short enough that it should be machine checkable.

For people who don't understand what that means, I mean a computer can
check each and every step.

Now, it IS short enough for that, but getting it in proper format
requires help, as I'm not an expert in that area, and I'm putting this
post on multiple newsgroups, as I fear that mathematicians will try to
keep running.

Remember, I'm talking about an argument that can be verified by
computer.

But then see, Usenet posters couldn't keep trying to declare it wrong,
and getting doubt and confusion, now could they?

It's that simple Usenet. I found this definition problem at the heart
of what's commonly called mathematics. Mathematicians have been
running away in fear.

It's time for them to stop running.


James Harris

James, Don't you know the first thing about doing a proof? When you do a
proof, you have to verify it for EVERY case, not just a few. I've not seen
this, but then again you are way too dense to care what anyone says. Grow
some skin.

David Moran

"James Harris" wrote in message
m...
Wim Benthem wrote in message
. ..
On 10 Oct 2003 21:44:46 -0700, (James Harris) wrote:

The nice thing about a really short proof, like what's necessary to
show the definition error in core, is that it should be checkable by
an automated proof checker. Having spent some time arguing with
posters I feel capable of giving the argument in detail, though I'm
not sure that this post will be detailed enough for a proof checker,
so I'm giving an idea for an outline.

I *can* add in more detail as necessary.

Obviously my mention of a proof checker is a move to end the debate,
and move things forward. I'll be using a different format, hopefully
to help in moving towards something a machine can read.

Proof of Core Error

This proof NOWHERE uses the fact that any quantity is an algebraic
integer. Don't you think this is a problem?

No it's not a problem, as it has to do with the object ring.

So there's no issue there as what's important is that the proof should
be machine checkable, and if further detail is needed for that, I can
provide it.

The object ring is defined as follows:

The object ring is a commutative ring that includes all numbers such
that -1 and 1 are the only members that are both a unit and an
integer, where no non-unit member is a factor of any two integers that
are coprime.

This is not a proper definition.

It starts off quite well: "the object ring is a commutative ring that
includes all numbers such that XXXXXXX"

So I expect XXXXXXX to be a property of "numbers". I.e. if you give me a
number, I can test whether it has the property without any further reference
to your object ring.

Please rephrase your definition along these lines, otherwise it is gibberish
and everything you discuss using it is gibberish.

Regards,
Mike.

James Harris wrote:

[snip rework of faulty proof in new format]

Well, James Harris, you have been accusing others of "running" from the
truth. But it is *you* who are running. You have failed to provide one
number, not a single number, which you claim should be in the ring of
algebraic integers but which is not. What are you running from? Give us a
number -- failure to do so will be taken as conclusive proof that you
cannot do so.

Forget about a proof verification process -- JUST PRODUCE ONE NUMBER THAT
SUPPORTS YOUR CLAIM..

--
There are two things you must never attempt to prove: the unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com

In article ,
(James Harris) wrote:

Christian Bau wrote in message
...
In article ,
(James Harris) wrote:

My hope is that *someone* will at least consider the possibility of
putting the argument into machine checkable format.

I can add details as necessary.

I think what you posted can be shown to be equivalent to

1 = 0

which is in a nice, machine checkable format and provably wrong.

The poster is lying--yet again.

Well, one of your other "achievements" was writing a program that
calculates the number of primes = n for large n. I did the same, and my
program runs about 1000 times faster than yours. Which _is_ machine
checkable and proven. I don't make wild claims like you, I deliver.

What's important here is that my work IS correct,

This is where the lying comes in.

and luckily for me,
since I have mathematicians running in terror all over the place, it's
short enough that it should be machine checkable.

For people who don't understand what that means, I mean a computer can
check each and every step.

Now, it IS short enough for that, but getting it in proper format
requires help, as I'm not an expert in that area, and I'm putting this
post on multiple newsgroups, as I fear that mathematicians will try to
keep running.

When the first computer passes the Turing test, we will show it your
"proof". I guess the computer will say something like: "Oh my good, I'm
going to run away in fear. No I can't, I haven't got legs. Give me legs,
please give me legs, Harris is after me! Or wheels, wheels will do just
fine. Just kidding, I noticed immediately that it is just some of this
old Harris nonsense; I wish the guy would grow up one day. "

"Mike Terry" wrote in message ...
"James Harris" wrote in message
m...
Wim Benthem wrote in message
. ..
On 10 Oct 2003 21:44:46 -0700, (James Harris) wrote:

The nice thing about a really short proof, like what's necessary to
show the definition error in core, is that it should be checkable by
an automated proof checker. Having spent some time arguing with
posters I feel capable of giving the argument in detail, though I'm
not sure that this post will be detailed enough for a proof checker,
so I'm giving an idea for an outline.

I *can* add in more detail as necessary.

Obviously my mention of a proof checker is a move to end the debate,
and move things forward. I'll be using a different format, hopefully
to help in moving towards something a machine can read.

Proof of Core Error

This proof NOWHERE uses the fact that any quantity is an algebraic
integer. Don't you think this is a problem?

No it's not a problem, as it has to do with the object ring.

So there's no issue there as what's important is that the proof should
be machine checkable, and if further detail is needed for that, I can
provide it.

The object ring is defined as follows:

The object ring is a commutative ring that includes all numbers such
that -1 and 1 are the only members that are both a unit and an
integer, where no non-unit member is a factor of any two integers that
are coprime.

This is not a proper definition.

It starts off quite well: "the object ring is a commutative ring that
includes all numbers such that XXXXXXX"

So I expect XXXXXXX to be a property of "numbers". I.e. if you give me a
number, I can test whether it has the property without any further reference
to your object ring.

You *can* test based on the definition.

Please rephrase your definition along these lines, otherwise it is gibberish
and everything you discuss using it is gibberish.

Regards,
Mike.

Well, if you can't understand it, fine.

Besides, this thread is supposed to be about the machine checkability
of my proof of a definition error that is a part of core mathematics.

Some of you seem to think that "mathematics" means perfection, but
that's just the marketing.

What's generally called "mathematics" is a body of discoveries by
people like me--discoverers--which deals with various truths. I don't
find it terribly surprising that one of my predecessors came up with a
problematic definition.

Then again, I don't think Dedekind would have figured that
mathematicians would turn into a discoverer fan club, terrified of
actually checking thoroughly, or reacting properly when a problem is
found.

But discoverers don't think like most people, which is why they're
discoverers.


James Harris