### Sunday, February 17, 2002

## Newsgroup poll: Do you really all agree about my work?

If you hadn't noticed there are several posters who are posting a

Their position basically is that I don't have any valid mathematical work of any worth, though they seem to be obsessive in their need to reply to my posts.

They seem to think the entire sci.math newsgroup agrees with them.

They also seem to agree with David Ullrich that my saying he'd acted as my lapdog in an instance was an egregious insult worthy of a racial slur (as I'm black) in reply, and they keep stating that the newsgroup agrees with them.

They also seem to be extremely angry that I contacted Oklahoma State University where Ullrich is actually a math professor to express my outrage at his statement about a racial slur being an "appropriate" response, in this case to my saying that he'd acted as my lapdog in an instance, and his continually posting obnoxiously in reply to me.

I like Google because it gives any of you who are curious about the facts an opportunity to check some of the statements being made.

For instance, Ullrich claims that he was trying to educate me on the proper way to talk to people. So why don't you do a search in Google, in the newsgroup sci.math, with David Ullrich as the author, with the key words "James", "Harris", and "idiot".

http://groups.google.com/groups?as_q=James%20Harris%20idiot&as_ugroup=sci.math&as_uauthors=David%20Ullrich&hl=en

http://groups.google.com/advanced_group_search?num=25&hl=en&group=sci.math

Now, it seems to me that with things so obvious Ullrich and his minions believe that you really

Well, I'm thinking that you can contact Oklahoma State University, as Ullrich claims the university supports him, and that despite my emails, he hasn't been told anything is wrong with his behavior, and

Oh yeah, Ullrich's boss, who he claims supports him, who I talked with on the phone, is Benny Evans, the Professor and Department Head, Mathematics. The phone number for the math department is (405) 744–5688.

**lot**in reply to me, and I notice them repeating over and over again that they speak for the newsgroup.Their position basically is that I don't have any valid mathematical work of any worth, though they seem to be obsessive in their need to reply to my posts.

They seem to think the entire sci.math newsgroup agrees with them.

They also seem to agree with David Ullrich that my saying he'd acted as my lapdog in an instance was an egregious insult worthy of a racial slur (as I'm black) in reply, and they keep stating that the newsgroup agrees with them.

They also seem to be extremely angry that I contacted Oklahoma State University where Ullrich is actually a math professor to express my outrage at his statement about a racial slur being an "appropriate" response, in this case to my saying that he'd acted as my lapdog in an instance, and his continually posting obnoxiously in reply to me.

I like Google because it gives any of you who are curious about the facts an opportunity to check some of the statements being made.

For instance, Ullrich claims that he was trying to educate me on the proper way to talk to people. So why don't you do a search in Google, in the newsgroup sci.math, with David Ullrich as the author, with the key words "James", "Harris", and "idiot".

http://groups.google.com/groups?as_q=James%20Harris%20idiot&as_ugroup=sci.math&as_uauthors=David%20Ullrich&hl=en

http://groups.google.com/advanced_group_search?num=25&hl=en&group=sci.math

Now, it seems to me that with things so obvious Ullrich and his minions believe that you really

**do**agree with him, or doesn't believe that any of you can do anything about him.Well, I'm thinking that you can contact Oklahoma State University, as Ullrich claims the university supports him, and that despite my emails, he hasn't been told anything is wrong with his behavior, and

**tell**them what you think.Oh yeah, Ullrich's boss, who he claims supports him, who I talked with on the phone, is Benny Evans, the Professor and Department Head, Mathematics. The phone number for the math department is (405) 744–5688.

### Wednesday, February 13, 2002

## JSH: Enough distractions, back to the math.

Some of you may believe that there's a lot in contest when it comes to my proof of Fermat's Last Theorem but there isn't. What bothers me is an apparent refusal by people who argue with me to admit the truth. They just seem completely irrational to me.

For instance, consider the ring business which is the PRIMARY objection they raise over and over again, which I can explain easily with a simple polynomial-like expression:

(v^3+1)W^3 - 3vW - 2 = (sqrt(v+1)W+b1)(sqrt(v+1)W+b2)((v^2-v+1)W+b3),

where these people have apparently convinced many of you that for some odd reason it's impossible, or at least extraordinarily difficult to determine whether in any sense whether b1, b2, b3, b1 b2, b2 b3, or b1 b3 have a factor of 1/sqrt(v+1).

The argument goes something like this:

To talk about 1/sqrt(v+1) as a factor, it must be in the ring. But if it's in the ring, it's a factor of all members of the ring; therefore, talking of 1/sqrt(v+1) as a factor means that it must be a factor of the numbers.

And notice I've been talking about b1 b2, b2 b3, or b1 b3 because these people berated me for saying that b1, b3 and b3 didn't have a factor of 1/sqrt(v+1), and then there were people who claimed that even that didn't cover it, since they seemed to think that multiplying or adding to another member of the ring formed by adjoining the square roots of the integers to the integers might introduce 1/sqrt(v+1).

Yet, through this all, for months I've done a simple demonstration:

Let me remind you that the expression is

(v^3+1)W^3 - 3vW - 2 = (sqrt(v+1)W+b1)(sqrt(v+1)W+b2)((v^2-v+1)W+b3),

and now let me set v=-1, to get

3W - 2 = ((0)W + b1)((0)W+b2)((3)W+b3) = b1 b2 (3W + b3).

So, how many of you

Well, there's actually an interesting thing you might notice before you jump to any conclusions, and that is b1 b2. Maybe b1 has a factor of sqrt(v+1) and b2 has a factor of 1/sqrt(v+1), and there is an error at v=-1, so that you have to approach it.

Do you all believe that issue cannot be resolved, with all that "modern math" has to offer?

I guess you do.

The other objection to the proof depends on the assertion that somehow two expressions in two unknowns when neither expression is a multiple of the other in any way limits the values for the unknowns.

The equivalent assertion would be something like, given

x+ y = 1, and x+2y = 3, x and y can have an infinite number of integer values.

Well, it just so happens that in the proof the expressions are not so simple, so despite the fact that there are two unknowns with two expressions, many of you apparently have accepted an illogical position. I find that extraordinary.

How could you all apparently be convinced of something like that?

And that's it folks.

And I know that you pride yourselves on thinking I'm lying but you can verify that these are THE objections to my proof of Fermat's Last Theorem.

When people ask if

Well, if you had a proof of Fermat's Last Theorem, that people were getting credit for shooting down using illogic and insults, wouldn't that make you a bit nutty?

And don't you think you'd be trying to get some attention for your proof?

And given the behavior that I've seen for

Especially when maybe, eventually,

http://www.msnusers.com/ProofofFermatsLastTheorem.

I certainly hope it's soon though as I'm getting tired of all these people getting their jollies from insulting me. I think some of them may be doing it

For instance, consider the ring business which is the PRIMARY objection they raise over and over again, which I can explain easily with a simple polynomial-like expression:

(v^3+1)W^3 - 3vW - 2 = (sqrt(v+1)W+b1)(sqrt(v+1)W+b2)((v^2-v+1)W+b3),

where these people have apparently convinced many of you that for some odd reason it's impossible, or at least extraordinarily difficult to determine whether in any sense whether b1, b2, b3, b1 b2, b2 b3, or b1 b3 have a factor of 1/sqrt(v+1).

The argument goes something like this:

To talk about 1/sqrt(v+1) as a factor, it must be in the ring. But if it's in the ring, it's a factor of all members of the ring; therefore, talking of 1/sqrt(v+1) as a factor means that it must be a factor of the numbers.

And notice I've been talking about b1 b2, b2 b3, or b1 b3 because these people berated me for saying that b1, b3 and b3 didn't have a factor of 1/sqrt(v+1), and then there were people who claimed that even that didn't cover it, since they seemed to think that multiplying or adding to another member of the ring formed by adjoining the square roots of the integers to the integers might introduce 1/sqrt(v+1).

Yet, through this all, for months I've done a simple demonstration:

Let me remind you that the expression is

(v^3+1)W^3 - 3vW - 2 = (sqrt(v+1)W+b1)(sqrt(v+1)W+b2)((v^2-v+1)W+b3),

and now let me set v=-1, to get

3W - 2 = ((0)W + b1)((0)W+b2)((3)W+b3) = b1 b2 (3W + b3).

So, how many of you

**really**believe that 1/sqrt(v+1) is in any way a factor of the b's?Well, there's actually an interesting thing you might notice before you jump to any conclusions, and that is b1 b2. Maybe b1 has a factor of sqrt(v+1) and b2 has a factor of 1/sqrt(v+1), and there is an error at v=-1, so that you have to approach it.

Do you all believe that issue cannot be resolved, with all that "modern math" has to offer?

I guess you do.

The other objection to the proof depends on the assertion that somehow two expressions in two unknowns when neither expression is a multiple of the other in any way limits the values for the unknowns.

The equivalent assertion would be something like, given

x+ y = 1, and x+2y = 3, x and y can have an infinite number of integer values.

Well, it just so happens that in the proof the expressions are not so simple, so despite the fact that there are two unknowns with two expressions, many of you apparently have accepted an illogical position. I find that extraordinary.

How could you all apparently be convinced of something like that?

And that's it folks.

And I know that you pride yourselves on thinking I'm lying but you can verify that these are THE objections to my proof of Fermat's Last Theorem.

When people ask if

**anyone**supports my claims they are basically saying that you all believe what they claim. I know that's silly as I think most of you haven't really paid attention as you've been convinced that I'm just some nut or attention seeker.Well, if you had a proof of Fermat's Last Theorem, that people were getting credit for shooting down using illogic and insults, wouldn't that make you a bit nutty?

And don't you think you'd be trying to get some attention for your proof?

And given the behavior that I've seen for

**years**from mathematicians on this newsgroup—what you've seen lately isn't that strange from my experience—why should I trust some unknown editor at a journal to not be of the same type?Especially when maybe, eventually,

**one**of you will see the truth athttp://www.msnusers.com/ProofofFermatsLastTheorem.

I certainly hope it's soon though as I'm getting tired of all these people getting their jollies from insulting me. I think some of them may be doing it

**because**they think the proof is correct, and they won't be able to so easily get away with attacking me in the future. Or, maybe they know they can still verbally assault me, but that then they won't get the kind of applause they now do on sci.math. Then again, maybe they**would**. I think many of you would still applaud them even knowing that I solved Fermat's Last Theorem just because you hate me that much.### Tuesday, February 12, 2002

## JSH: Ullrich and the racial comments.

I've decided that I'd rather not continue in an open ended discusion with Ullrich and the people apparently replying in support of his statement that racial slurs can be appropriate, so I'm pursuing other routes, including sending Ullrich's statements to OSU.

From this point further I'm not interested in replying to Ullrich or the people who are posting in support of his position. I WILL see what his university thinks of his statements, and if they agree with Ullrich.

From this point further I'm not interested in replying to Ullrich or the people who are posting in support of his position. I WILL see what his university thinks of his statements, and if they agree with Ullrich.

### Friday, February 08, 2002

## Credibility check: Alternate approach to polynomial

Yesterday I talked about a way to analyze the roots of the polynomial

W^3 -3vW/(v^3+1) - 2/(v^3+1) = 0,

to show that two of them have the "operator" of 1/sqrt(v+1), while one does not.

I realized that many of you find it easier to dismiss a long post with new ideas than to check to see if it actually works, so here's a short post with an alternate approach using more traditional methods.

I post it deliberately to show you why many of the people posting on the newsgroup should no longer be considered credible by you. In fact, I think the people still making negative posts are doing so to preserve their perception of a social order, without regard for the truth. Or maybe they believe their social order IS the truth, as if titles and academic positions determine that.

Ok, I assume you know that you can solve the polynomial given, and you also should know that solution has no validity at v=-1 because you have divide by zero errors all over the place.

However, you

In that case, two of your roots start going to infinity, while one approaches 2/3.

That's easy to see by multiplying through by v^3+1 to get

(v^3+1)W^3 - 3vW - 2 = 0, where in the limit as v approaches -1,

the polynomial approaches 3W - 2 = 0.

Oh, and using this approach it's also possible to consider factors of sqrt(v+1).

For instance, in the limit as v approaches -1, sqrt((sqrt(v)-i))/sqrt(v+i) equals 1/sqrt(sqrt(v)+i) = 1/sqrt(2i), which is clearly nowhere to be seen in the result 3W - 2 = 0.

So, it's

That is, while I can say, given 1/3 + 2/3 - 1 = 0, that two of the numbers are fractions while one is not, you've seen a lot of posts—attacking me and trying to use this to cast doubt on my proof of Fermat's Last Theorem—claiming that there's no way to distinguish the roots of the polynomial I've given in the same way.

When I talked of factors of the roots of 1/sqrt(v+1), with one root not having 1/sqrt(v+1), I was attacked.

When I tried to say some of the roots of the polynomial have a denominator of sqrt(v+1) while one does not, I was attacked.

So, yes, you can continue to doubt your own judgement against that of the mathematical establishment, and you can continue to believe that just because mathematicians say so, there's no way to talk of two of the roots being different from the third with regard to 1/sqrt(v+1), but that will say something about you as a person, and nothing about the truth.

I know that most of you have been trained all of your lives that there are certain experts who cannot be challenged. While I was trained to be a scientist while at Vanderbilt University for four years earning my bachelors degree in Physics, which means I was trained to challenge the experts.

Are there any of you who will go by the truth versus your need to believe in the infallibility of a group of human beings?

What? Do you think that they are just too "smart" to be wrong?

W^3 -3vW/(v^3+1) - 2/(v^3+1) = 0,

to show that two of them have the "operator" of 1/sqrt(v+1), while one does not.

I realized that many of you find it easier to dismiss a long post with new ideas than to check to see if it actually works, so here's a short post with an alternate approach using more traditional methods.

I post it deliberately to show you why many of the people posting on the newsgroup should no longer be considered credible by you. In fact, I think the people still making negative posts are doing so to preserve their perception of a social order, without regard for the truth. Or maybe they believe their social order IS the truth, as if titles and academic positions determine that.

Ok, I assume you know that you can solve the polynomial given, and you also should know that solution has no validity at v=-1 because you have divide by zero errors all over the place.

However, you

**can**consider the roots as v approaches -1.In that case, two of your roots start going to infinity, while one approaches 2/3.

That's easy to see by multiplying through by v^3+1 to get

(v^3+1)W^3 - 3vW - 2 = 0, where in the limit as v approaches -1,

the polynomial approaches 3W - 2 = 0.

Oh, and using this approach it's also possible to consider factors of sqrt(v+1).

For instance, in the limit as v approaches -1, sqrt((sqrt(v)-i))/sqrt(v+i) equals 1/sqrt(sqrt(v)+i) = 1/sqrt(2i), which is clearly nowhere to be seen in the result 3W - 2 = 0.

So, it's

**clear**that the current system, which posters have been heatedly claiming can't differentiate between the two roots that go to infinity and the one that goes to 2/3 without something indirect like this, is flawed.That is, while I can say, given 1/3 + 2/3 - 1 = 0, that two of the numbers are fractions while one is not, you've seen a lot of posts—attacking me and trying to use this to cast doubt on my proof of Fermat's Last Theorem—claiming that there's no way to distinguish the roots of the polynomial I've given in the same way.

When I talked of factors of the roots of 1/sqrt(v+1), with one root not having 1/sqrt(v+1), I was attacked.

When I tried to say some of the roots of the polynomial have a denominator of sqrt(v+1) while one does not, I was attacked.

So, yes, you can continue to doubt your own judgement against that of the mathematical establishment, and you can continue to believe that just because mathematicians say so, there's no way to talk of two of the roots being different from the third with regard to 1/sqrt(v+1), but that will say something about you as a person, and nothing about the truth.

I know that most of you have been trained all of your lives that there are certain experts who cannot be challenged. While I was trained to be a scientist while at Vanderbilt University for four years earning my bachelors degree in Physics, which means I was trained to challenge the experts.

Are there any of you who will go by the truth versus your need to believe in the infallibility of a group of human beings?

What? Do you think that they are just too "smart" to be wrong?

### Saturday, February 02, 2002

## JSH: Set theory, some thoughts

A while back I talked a bit about why I don't believe fractions are objects. Not surprisingly, that was met with a great deal of derision on the newsgroup. I don't know how many of you can appreciate what it feels to be an object of ridicule for presenting such thoughts, but I do.

Now, there has been a lot of discussion about a ring I use in my proof of Fermat's Last Theorem.

That ring is formed by adjoining the square roots of the integers to the integers and also adjoining the numbers b1,…,bp to the resulting ring.

An assertion was made that the ring

While dealing with this objection, I realized something interesting, since people kept arguing with me

It looked like an impasse until I realized I could produce a contradiction with their argument using a simple example:

Consider the ring formed by adjoining the number x to the ring of integers.

Now, prove that 1/3 is not in the ring.

I realized, as I also realized that none of these people had the skills to answer the challenge, that a property of being a fraction or being fractional, as I've called it, is what I'm abstracting, but current mathematics doesn't seem to have it.

I've tried to generalize the definition of fraction myself, and consistently failed.

However, that comes from thinking fractions are objects. As, I've talked about fractions like 1/2 not being objects but being

It basically has to do with the label of object.

That is, a fundamental postulate is that something is either an object or an object operator.

I don't like that "something", but I seem to be stuck with it.

Then 1/2 is not in and of itself and object because you can't have 1/2 of an object unless the object is a set with two objects.

Then 1/2 means take 1 of the 2 objects within the set the operator is being applied to, which is the way it's

For instance, you can have a pie, but that pie can be cut into two pieces. So you now have a set of two pieces and to take 1/2 the pie, you take one of the two pieces.

There are no exceptions to the use of 1/2 as an operator in the real world. That is, 1/2 is

(Consider, what's 1/2 of an electron?)

So, I've used the term set without defining it, when I think it's clear that a set is a group of one or more objects.

So what is a group by my definition, why it's a set, of course!

Circularity among definitions is not a surprise when you're dealing with things at the fundamental level. If you look at established mathematics at the level of traditional set theory, you will find axioms, and if you push the axioms, you will see they are circular.

Or, you can just get a dictionary, look up a word, and then look up the words that make up the definition, and continue with this, and you will find yourself going in a circle.

To get all of mathematics I need only a few operations for my objects. I need addition, multiplication, and extraction.

I've used extraction by considering 1 piece of pie out of two that make up the entire pie.

I don't think I need to explain the addition and subtraction operators.

Now, what's neat about this is that gives me two types of numbers as objects. There are the integers, and the numbers of the continuous field. For instance, sqrt(2) (note we define it with an operator) is an object that multiplies times sqrt(2) to give the object 2 and is a member of the continuous field.

Now here's where things get fascinating (at least to me).

While i is an object, e and pi are not, as they are both operators.

That's fun to explain, and it goes back to the real world.

Consider that when you see a circle, for instance, drawn on your computer screen, what you're seeing are a lot of little dots. The number of dots can be given by multiplying pi times the diameter of the drawn circle, which defines the pi operator. Note: The result is an

It turns out that if you work things out this way, you can build all of mathematics without the interesting little ambiguities that plague modern mathematics, like the one I pointed out at the top of this post, and no, I'm not thinking of any more off-hand, I just feel they have to be there.

Oh well, time for a lot more ridicule. Since I expect nothing but ridicule, I'm not likely to read any replies to this post.

Now, there has been a lot of discussion about a ring I use in my proof of Fermat's Last Theorem.

That ring is formed by adjoining the square roots of the integers to the integers and also adjoining the numbers b1,…,bp to the resulting ring.

An assertion was made that the ring

**might**have f as a unit, where f is an important variable in my proof that is a prime number (like 3, 5, or 7).While dealing with this objection, I realized something interesting, since people kept arguing with me

**after**I felt I'd clearly shown that f was not a unit in my ring.It looked like an impasse until I realized I could produce a contradiction with their argument using a simple example:

Consider the ring formed by adjoining the number x to the ring of integers.

Now, prove that 1/3 is not in the ring.

I realized, as I also realized that none of these people had the skills to answer the challenge, that a property of being a fraction or being fractional, as I've called it, is what I'm abstracting, but current mathematics doesn't seem to have it.

I've tried to generalize the definition of fraction myself, and consistently failed.

However, that comes from thinking fractions are objects. As, I've talked about fractions like 1/2 not being objects but being

**operators**, I thought it'd be worth it to explain what I mean.It basically has to do with the label of object.

That is, a fundamental postulate is that something is either an object or an object operator.

I don't like that "something", but I seem to be stuck with it.

Then 1/2 is not in and of itself and object because you can't have 1/2 of an object unless the object is a set with two objects.

Then 1/2 means take 1 of the 2 objects within the set the operator is being applied to, which is the way it's

**always**used in the real world.For instance, you can have a pie, but that pie can be cut into two pieces. So you now have a set of two pieces and to take 1/2 the pie, you take one of the two pieces.

There are no exceptions to the use of 1/2 as an operator in the real world. That is, 1/2 is

**never**in and of itself an object but is used on a set with two objects, where the objects can themselves be sets.(Consider, what's 1/2 of an electron?)

So, I've used the term set without defining it, when I think it's clear that a set is a group of one or more objects.

So what is a group by my definition, why it's a set, of course!

Circularity among definitions is not a surprise when you're dealing with things at the fundamental level. If you look at established mathematics at the level of traditional set theory, you will find axioms, and if you push the axioms, you will see they are circular.

Or, you can just get a dictionary, look up a word, and then look up the words that make up the definition, and continue with this, and you will find yourself going in a circle.

To get all of mathematics I need only a few operations for my objects. I need addition, multiplication, and extraction.

I've used extraction by considering 1 piece of pie out of two that make up the entire pie.

I don't think I need to explain the addition and subtraction operators.

Now, what's neat about this is that gives me two types of numbers as objects. There are the integers, and the numbers of the continuous field. For instance, sqrt(2) (note we define it with an operator) is an object that multiplies times sqrt(2) to give the object 2 and is a member of the continuous field.

Now here's where things get fascinating (at least to me).

While i is an object, e and pi are not, as they are both operators.

That's fun to explain, and it goes back to the real world.

Consider that when you see a circle, for instance, drawn on your computer screen, what you're seeing are a lot of little dots. The number of dots can be given by multiplying pi times the diameter of the drawn circle, which defines the pi operator. Note: The result is an

**integer**.It turns out that if you work things out this way, you can build all of mathematics without the interesting little ambiguities that plague modern mathematics, like the one I pointed out at the top of this post, and no, I'm not thinking of any more off-hand, I just feel they have to be there.

Oh well, time for a lot more ridicule. Since I expect nothing but ridicule, I'm not likely to read any replies to this post.