### Thursday, September 30, 2010

## JSH: Puzzling counting prime numbers

Rather than give directly a mathematical find I've found it interesting to see if others can figure out something on their own, thus giving them an intellectual challenge. In this thread the challenge is to find a certain way to count prime numbers.

There are LOTS of resources available with a lot of mathematical literature on the subject, and all of the established literature is available for this challenge. Of course I also have the answer, which is not in established literature, for those who give up.

Challenge:

Find a prime counting function P that counts primes by first counting composites and subtracting that count and 1 from the total count to get primes, which only counts composites at each prime that have not already been counted.

That is, Legendre's method rather inefficiently counts ALL composites at each prime that have that prime as a factor, which is dumb!!! It's so dumb you end up just subtracting those back out which is a naive way to count, don't you think? Why, for instance, if you're counting composites divisible by 3, should you also count evens? Of if you're counting composites with 5 as a factor, count the ones with 3 and 2, and then just subtract them back out? It's WACKY!

Can you find a more efficient prime counting method than Legendre's which more smartly counts ONLY those composites at each prime that have not ALREADY been counted?

If you think the problem is too easy, great! Give the function in reply.

If you have no clue, here's a clue.

Hint: The requirements force a P(x,y) function, so you end up with a multi-dimensional prime counting function instead of a pi(x), single variable one.

Can anyone solve the puzzle on their own? If you get completely lost, of course, the answer is on my math blog. Given lots of places as I've explained, and explained and explained.

But if I can figure it out, can you?

There are LOTS of resources available with a lot of mathematical literature on the subject, and all of the established literature is available for this challenge. Of course I also have the answer, which is not in established literature, for those who give up.

Challenge:

Find a prime counting function P that counts primes by first counting composites and subtracting that count and 1 from the total count to get primes, which only counts composites at each prime that have not already been counted.

That is, Legendre's method rather inefficiently counts ALL composites at each prime that have that prime as a factor, which is dumb!!! It's so dumb you end up just subtracting those back out which is a naive way to count, don't you think? Why, for instance, if you're counting composites divisible by 3, should you also count evens? Of if you're counting composites with 5 as a factor, count the ones with 3 and 2, and then just subtract them back out? It's WACKY!

Can you find a more efficient prime counting method than Legendre's which more smartly counts ONLY those composites at each prime that have not ALREADY been counted?

If you think the problem is too easy, great! Give the function in reply.

If you have no clue, here's a clue.

Hint: The requirements force a P(x,y) function, so you end up with a multi-dimensional prime counting function instead of a pi(x), single variable one.

Can anyone solve the puzzle on their own? If you get completely lost, of course, the answer is on my math blog. Given lots of places as I've explained, and explained and explained.

But if I can figure it out, can you?

### Wednesday, September 29, 2010

## JSH: Puzzling through number theory?

With the success of the approach of presenting a major number theory issue as a puzzle I've began to consider that the problem with my prior approach has been that I've not given people the opportunity for intellectual challenge. Simply explaining mathematics may not be nearly as enticing as puzzling through.

So might that help in other areas? Conceivably here's another area for those who wish to match wits with me.

Find a prime counting function P that counts functions by first counting composites and subtracting that count and 1 from the total count to get primes, which only counts composites not already counted.

That is, Legendre's method rather inefficiently counts ALL composites at each prime that have that prime as a factor, which is dumb!!! Why, for instance, if you're counting composites divisible by 3, should you also count evens? It's WACKY!

Can you find a more efficient prime counting method than Legendre's which more smartly counts ONLY those composites at each prime that have not ALREADY been counted?

As a check, how hard is that as a puzzle? Should a smart, say, undergrad be able to succeed? How about a grad student?

If you think the problem is easy, great! Give the function in reply. If you have no clue, here's a clue.

Hint: The requirements force a P(x,y) function, so you end up with a multi-dimensional prime counting function instead of a pi(x), single variable one.

Can anyone solve the puzzle on their own? If you get completely lost, of course, the answer is on my math blog. Given lots of places as I've explained, and explained and explained.

But if I can figure it out, can you?

So might that help in other areas? Conceivably here's another area for those who wish to match wits with me.

Find a prime counting function P that counts functions by first counting composites and subtracting that count and 1 from the total count to get primes, which only counts composites not already counted.

That is, Legendre's method rather inefficiently counts ALL composites at each prime that have that prime as a factor, which is dumb!!! Why, for instance, if you're counting composites divisible by 3, should you also count evens? It's WACKY!

Can you find a more efficient prime counting method than Legendre's which more smartly counts ONLY those composites at each prime that have not ALREADY been counted?

As a check, how hard is that as a puzzle? Should a smart, say, undergrad be able to succeed? How about a grad student?

If you think the problem is easy, great! Give the function in reply. If you have no clue, here's a clue.

Hint: The requirements force a P(x,y) function, so you end up with a multi-dimensional prime counting function instead of a pi(x), single variable one.

Can anyone solve the puzzle on their own? If you get completely lost, of course, the answer is on my math blog. Given lots of places as I've explained, and explained and explained.

But if I can figure it out, can you?

### Monday, September 27, 2010

## JSH: Attacking the puzzle

Rather than do yet ANOTHER explanation of an issue in number theory, I presented the issue as a puzzle. Some posters have replied in that thread, and those who tried to solve the puzzle failed massively, but protested when I noted that they'd failed.

So here is an attack on the puzzle using the knee-jerk responses that are WRONG. So it's easily to massively fail against this thing.

Here again is the puzzle, and I'll then attack it just a bit.

Try in the ring of algebraic integers to use:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0, as well, while P(x) is a quadratic with integer coefficients.

You can find cases in the ring of algebraic integers where NEITHER of the f's can have 9 as a factor, for certain values of x, so you wander off the to the complex plane—because you've a VERY SMART math student—so you can see exactly what is happening.

There you note that:

(9g_1(x) + 9)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

follows trivially by the distributive property so FOR THAT CASE the ring of algebraic integers presumably would have 9 as a factor for all x, but can be shown to NOT have 9 as a factor as noted above.

Being an EXTREMELY INTELLIGENT math student you find this puzzling, so you consider it until it makes sense.

First off, let's attack the idea of going to the complex plane as a divisibility issue as it DOES NOT MATTER on the complex plane.

Well, note that the result HOLDS for polynomial functions for the f's and g's. So with all integers—valid on the complex plane—you do indeed find that one of the f's has 9 as a factor for all x.

Notice that following that on the complex plane does not remove the result.

The complex plane INCLUDES the ring of integers, so it can look at something happening in that ring easily.

If divisibility were the issue, it would emerge with integers as well, and everything would be divisible by 9. But one of the f's is not in general divisible by 9, so no, divisibility as the issue is a massive fail for this puzzle.

One thing that I think is fascinating as a bizarre issue is that some posters seem lost on the concept that you can use the complex field to simply look at something happening in a ring, like the ring of integers. May seem like overkill, but you can do it!

So making that mistake actually also says something about your sophistication as a math student, as presumably a fairly astute math student would realize that trivially.

Another poster actually said, there was no puzzle!!! Clearly that one is a massive fail as my easy counter is to explain it then. But his answer was to try and claim divisibility, so another massive fail. Claiming there is no puzzle but not being able to answer correctly is not an answer.

Can you now figure out the puzzle?

What others issue need I address? This thread can be for attacking the puzzle, the validity of it as a puzzle, or if you want, try to give an answer. I assure you it's not a trivial task to answer it correctly.

So here is an attack on the puzzle using the knee-jerk responses that are WRONG. So it's easily to massively fail against this thing.

Here again is the puzzle, and I'll then attack it just a bit.

Try in the ring of algebraic integers to use:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0, as well, while P(x) is a quadratic with integer coefficients.

You can find cases in the ring of algebraic integers where NEITHER of the f's can have 9 as a factor, for certain values of x, so you wander off the to the complex plane—because you've a VERY SMART math student—so you can see exactly what is happening.

There you note that:

(9g_1(x) + 9)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

follows trivially by the distributive property so FOR THAT CASE the ring of algebraic integers presumably would have 9 as a factor for all x, but can be shown to NOT have 9 as a factor as noted above.

Being an EXTREMELY INTELLIGENT math student you find this puzzling, so you consider it until it makes sense.

First off, let's attack the idea of going to the complex plane as a divisibility issue as it DOES NOT MATTER on the complex plane.

Well, note that the result HOLDS for polynomial functions for the f's and g's. So with all integers—valid on the complex plane—you do indeed find that one of the f's has 9 as a factor for all x.

Notice that following that on the complex plane does not remove the result.

The complex plane INCLUDES the ring of integers, so it can look at something happening in that ring easily.

If divisibility were the issue, it would emerge with integers as well, and everything would be divisible by 9. But one of the f's is not in general divisible by 9, so no, divisibility as the issue is a massive fail for this puzzle.

One thing that I think is fascinating as a bizarre issue is that some posters seem lost on the concept that you can use the complex field to simply look at something happening in a ring, like the ring of integers. May seem like overkill, but you can do it!

So making that mistake actually also says something about your sophistication as a math student, as presumably a fairly astute math student would realize that trivially.

Another poster actually said, there was no puzzle!!! Clearly that one is a massive fail as my easy counter is to explain it then. But his answer was to try and claim divisibility, so another massive fail. Claiming there is no puzzle but not being able to answer correctly is not an answer.

Can you now figure out the puzzle?

What others issue need I address? This thread can be for attacking the puzzle, the validity of it as a puzzle, or if you want, try to give an answer. I assure you it's not a trivial task to answer it correctly.

## JSH: Better as a puzzle?

I'm kind of excited at the idea of presenting what I claim is a major number theory issue as a puzzle.

And it kind of makes sense to do it that way versus trying to explain it in detail—which I've been doing for years—as it could be more fun that way for those who wish a massive intellectual challenge. After all, the problem has stood for over a hundred years!

Back-story as well is that I faced it as a puzzle years ago. As back in 2002 I had what I thought was a proof of Fermat's Last Theorem using equations that followed from what I call tautological spaces. YEARS of effort but I kept being called on the ring. Posters would bug me about the ring, and for a while I'd even talked about a "flat ring" as I needed a ring without fractions.

Well one poster says that what I really wanted was the ring of algebraic integers, and I said, ok, as they sounded fine, and went on about my way, except with this latest arguments posters noted my result contradicted with the ring of algebraic integers.

I was floored! It was like, uh oh, not good. And I began to puzzle it out. After some time I figured out the puzzle!

And I discovered the object ring.

It IS a puzzle.

Presumably the smartest math people can just figure it out! Which can keep me from having to keep trying to explain, but a major benefit is emerging already in the thread I created presenting the puzzle as some very obsessive posters just fail outright, and badly.

I'm re-thinking how I look at various posters based on responses in that thread. Surprisingly to me, I may have given some of them too much benefit of the doubt as to their, um, mathematical abilities.

That puzzle is a breaker. It will break anyone but the best. And responses to it, are crystal clear as to basic math ability.

Either you have it, or you don't.

And it kind of makes sense to do it that way versus trying to explain it in detail—which I've been doing for years—as it could be more fun that way for those who wish a massive intellectual challenge. After all, the problem has stood for over a hundred years!

Back-story as well is that I faced it as a puzzle years ago. As back in 2002 I had what I thought was a proof of Fermat's Last Theorem using equations that followed from what I call tautological spaces. YEARS of effort but I kept being called on the ring. Posters would bug me about the ring, and for a while I'd even talked about a "flat ring" as I needed a ring without fractions.

Well one poster says that what I really wanted was the ring of algebraic integers, and I said, ok, as they sounded fine, and went on about my way, except with this latest arguments posters noted my result contradicted with the ring of algebraic integers.

I was floored! It was like, uh oh, not good. And I began to puzzle it out. After some time I figured out the puzzle!

And I discovered the object ring.

It IS a puzzle.

Presumably the smartest math people can just figure it out! Which can keep me from having to keep trying to explain, but a major benefit is emerging already in the thread I created presenting the puzzle as some very obsessive posters just fail outright, and badly.

I'm re-thinking how I look at various posters based on responses in that thread. Surprisingly to me, I may have given some of them too much benefit of the doubt as to their, um, mathematical abilities.

That puzzle is a breaker. It will break anyone but the best. And responses to it, are crystal clear as to basic math ability.

Either you have it, or you don't.

### Sunday, September 26, 2010

## JSH: Who is Erik Max Francis?

Oh hey, maybe a better test of my influence is to ask a question that has been bugging me a bit, as yes, there still is that Crank.net webpage slamming me as a crank and crackpot, which still comes up quite high in Google searches when I search on my name.

Near as I can tell "Erik Max Francis" not only has no intentions of ever taking it down, he doesn't even seem to care about updating it any more, so I guess he thinks I'm done and buried—by him.

So I need some info: is that his real name? What is he doing lately? Is he still actively slamming people adding them to his site?

Oh yeah, from what I've gathered from his site he's a computer developer in San Jose. Is that true? What's his community connections?

Why is he such an angry person?

Nothing mean here!!! Just curious about this guy, and if giving info about him seems tasteless, how about answering this question: how compelling is his site?

Do readers find it important as to how they think of me?

Do they think it fair for someone to slam another person like that and just walk away, leaving it up indefinitely?

Or, is it completely fair? Hey if it is, why can't I do it someday? Maybe to YOU?

Near as I can tell "Erik Max Francis" not only has no intentions of ever taking it down, he doesn't even seem to care about updating it any more, so I guess he thinks I'm done and buried—by him.

So I need some info: is that his real name? What is he doing lately? Is he still actively slamming people adding them to his site?

Oh yeah, from what I've gathered from his site he's a computer developer in San Jose. Is that true? What's his community connections?

Why is he such an angry person?

Nothing mean here!!! Just curious about this guy, and if giving info about him seems tasteless, how about answering this question: how compelling is his site?

Do readers find it important as to how they think of me?

Do they think it fair for someone to slam another person like that and just walk away, leaving it up indefinitely?

Or, is it completely fair? Hey if it is, why can't I do it someday? Maybe to YOU?

## JSH: How complicated can it be?

I'm getting replies from posters questioning what I see as a trivial argument, so how hard can it be? Like, once again, as I try to use the simplest expression I've found to explain what I think is trivial, try in the ring of algebraic integers to use:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0, as well, while P(x) is a quadratic with integer coefficients.

You can find cases in the ring of algebraic integers where NEITHER of the f's can have 9 as a factor, for certain values of x, so you wander off the to the complex plane—because you've a VERY SMART math student—so you can see exactly what is happening.

There you note that:

(9g_1(x) + 9)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

follows trivially by the distributive property so FOR THAT CASE the ring of algebraic integers presumably would have 9 as a factor for all x, but can be shown to NOT have 9 as a factor as noted above.

Being an EXTREMELY INTELLIGENT math student you find this puzzling, so you consider it until it makes sense.

Now then, how can you multiply by 9 in the field, but not be allowed to have 9 as a factor in the ring of algebraic integers?

I think the answer is easy, but am I wrong? Can any of you solve the puzzle?

If you get completely lost, just go to Google.

Search on: algebraic integers vs complex numbers

I put it all in a paper, which is on Scribd.

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0, as well, while P(x) is a quadratic with integer coefficients.

You can find cases in the ring of algebraic integers where NEITHER of the f's can have 9 as a factor, for certain values of x, so you wander off the to the complex plane—because you've a VERY SMART math student—so you can see exactly what is happening.

There you note that:

(9g_1(x) + 9)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

follows trivially by the distributive property so FOR THAT CASE the ring of algebraic integers presumably would have 9 as a factor for all x, but can be shown to NOT have 9 as a factor as noted above.

Being an EXTREMELY INTELLIGENT math student you find this puzzling, so you consider it until it makes sense.

Now then, how can you multiply by 9 in the field, but not be allowed to have 9 as a factor in the ring of algebraic integers?

I think the answer is easy, but am I wrong? Can any of you solve the puzzle?

If you get completely lost, just go to Google.

Search on: algebraic integers vs complex numbers

I put it all in a paper, which is on Scribd.

### Thursday, September 23, 2010

## JSH: But is it helpful?

On my side there has been a very strong need to understand and I've had a lot on my plate for quite a few years trying to work through any number of mathematical questions generated in a burst you might say, back in 2002. The arguing has helped me tremendously, as I've developed a full understanding of what I call the coverage issue, as well as been able to get a better grasp of my other result, on the prime counting function.

Over the years I've tried to be very clear about how I use Usenet as I want to be fair. As heated as arguments have gone over the years there is just no way that I think that Usenet is a control area of mathematical society. It is to me, a fringe. It is a place where mostly cast-offs from mainstream math society, can still talk mathematics, though a price they often pay is a very insulting atmosphere, where often your voice can be lost, or seem lost.

To me I've brought quite a few benefits though to posters who may not have known what is still possible, no matter what posters may tell you, or how lost you may feel. As much as some decry the search realities I have now, I think for others it should be comforting to know that if you are right, and argue your positions carefully, it is possible that the world will in some way listen to you.

And technology has so greatly advanced today that yeah, your ideas can be found, no matter how much of a needle in a haystack you may think they are. One of my favorites (warning leads to one of my blogs) is: prime number compression

I only come up about #6 when I do that search in Google, but there is still this kind of solid feeling that, hey, here's a little idea of mine that somehow, someway is found by Google for whatever reason, and I'll take that versus not having it.

What I've shown then I think is that you don't have to believe the nasty people!!!

Posters may say anything to you. You may get the most vicious accusations against you, especially of failure, if it appears that you are right and have something important. There is no rule that it should be otherwise.

Over the years the primary complaint I get is that I keep arguing to people who don't believe me, and won't believe me, so why bother?

But I argue to understand. And I know that others can follow these arguments from anywhere in the world. And I know that the marginally few people who may reply to you are not necessarily representative of anything at all, except the few people who bother to reply to you on Usenet.

But is it helpful? I ask that question as the subject as I think lots of people can get really confused about the discussions as when they read me arguing with various people those people can seem to be important simply by virtue of them getting a reply or because they're tacking something into one of my threads when more than likely nothing they say matters, especially on the big picture scale.

These are people who will reply with anger if I talk about just my math blog getting hits from 121 countries/territories as Google Analytics puts it because they themselves are not at that level. But they post in reply to me as if I'm inferior to them. Which I think is an unnatural enough thing for many people that it is very disconcerting.

If I'm at that level why do I tolerate people who are not, who are so disrespectful?

Because I need to understand myself, and arguing these things out is a way to understand. There are not a lot of places where I can argue math, and worse, people who think I'm great are useless to me for that purpose.

To the extent that deluded people exist who will trump themselves up over me, then I can do extreme mathematics and talk out ideas to better understand them, which is good in the long-term for an entire world.

So, I think, yes. It IS helpful. And I worry a bit as reality looks to catch up, and one day there will be no people who will dare to argue with me about math, who are at all intelligible, as with a position of authority in people's minds I'll lose one of the greatest things that Usenet has given me—questioning.

Over the years I've tried to be very clear about how I use Usenet as I want to be fair. As heated as arguments have gone over the years there is just no way that I think that Usenet is a control area of mathematical society. It is to me, a fringe. It is a place where mostly cast-offs from mainstream math society, can still talk mathematics, though a price they often pay is a very insulting atmosphere, where often your voice can be lost, or seem lost.

To me I've brought quite a few benefits though to posters who may not have known what is still possible, no matter what posters may tell you, or how lost you may feel. As much as some decry the search realities I have now, I think for others it should be comforting to know that if you are right, and argue your positions carefully, it is possible that the world will in some way listen to you.

And technology has so greatly advanced today that yeah, your ideas can be found, no matter how much of a needle in a haystack you may think they are. One of my favorites (warning leads to one of my blogs) is: prime number compression

I only come up about #6 when I do that search in Google, but there is still this kind of solid feeling that, hey, here's a little idea of mine that somehow, someway is found by Google for whatever reason, and I'll take that versus not having it.

What I've shown then I think is that you don't have to believe the nasty people!!!

Posters may say anything to you. You may get the most vicious accusations against you, especially of failure, if it appears that you are right and have something important. There is no rule that it should be otherwise.

Over the years the primary complaint I get is that I keep arguing to people who don't believe me, and won't believe me, so why bother?

But I argue to understand. And I know that others can follow these arguments from anywhere in the world. And I know that the marginally few people who may reply to you are not necessarily representative of anything at all, except the few people who bother to reply to you on Usenet.

But is it helpful? I ask that question as the subject as I think lots of people can get really confused about the discussions as when they read me arguing with various people those people can seem to be important simply by virtue of them getting a reply or because they're tacking something into one of my threads when more than likely nothing they say matters, especially on the big picture scale.

These are people who will reply with anger if I talk about just my math blog getting hits from 121 countries/territories as Google Analytics puts it because they themselves are not at that level. But they post in reply to me as if I'm inferior to them. Which I think is an unnatural enough thing for many people that it is very disconcerting.

If I'm at that level why do I tolerate people who are not, who are so disrespectful?

Because I need to understand myself, and arguing these things out is a way to understand. There are not a lot of places where I can argue math, and worse, people who think I'm great are useless to me for that purpose.

To the extent that deluded people exist who will trump themselves up over me, then I can do extreme mathematics and talk out ideas to better understand them, which is good in the long-term for an entire world.

So, I think, yes. It IS helpful. And I worry a bit as reality looks to catch up, and one day there will be no people who will dare to argue with me about math, who are at all intelligible, as with a position of authority in people's minds I'll lose one of the greatest things that Usenet has given me—questioning.

### Wednesday, September 22, 2010

## JSH: When error feels natural

The problem I face is not proof. Even a moderately skeptical person would find my ability to use the latest technologies of our world, web search, to give my own personal definition of mathematical proof to be at least of interest.

Yet supposedly intelligent posters on sci.math dismiss it outright.

And dismiss Google along with it! Claiming that search results are meaningless.

Even when I've defined mathematical proof. As if anyone could do it and take over the world of search results with their own definition.

Error on this scale is outside of any of the boundaries of known response.

Quite simply a rather remarkable error took over the mathematical field in the late 18th century. To protect living in error, math people today will deny any and all evidence because the error is their normal.

The error removes the usefulness of Galois Theory showing that to be what I call a display theory only.

Proving the error is trivial. Doing dramatic things is trivial as well.

Taunted about publication I GOT PUBLISHED and the journal editors caved to social pressure, pulled my paper after publication, and then the journal quietly died. Its hosting university scrubbed all mention of it. About ten years of math papers were tossed as trash by the United States—saved by EMIS.

But that's nothing, right?

I'm more important than any one of you by that measure. Than even dozens of you. How many mathematicians were published in that journal? Did it matter?

Do you think any of you would have fared any better?

None of you are important enough. None of your results are important enough. Any journal where I had managed to get published its likely the result would have been the same. Any journal. I've pondered how long the Annals might have survived.

The error is your life now. It has taken over the mathematical world. It feeds on your minds. It needs them.

It is the one true power of the "pure math" world. It may have created the idea of "pure mathematics" in order to survive.

It is a virus of the human intellect. And it is more powerful than any of you.

And more important than all of you, as it demonstrated by crushing that mathematical journal without effort.

It can do the same to any of your research efforts at will. You live at its mercy.

This war is a fight for the soul of the human race. And it is one of the greatest battles in the history of this species.

If it is lost, the future of humanity is lost with it, as mathematics is lost.

And without correct mathematics, the human species has no future.

Well, at least, not a future worth living.

Yet supposedly intelligent posters on sci.math dismiss it outright.

And dismiss Google along with it! Claiming that search results are meaningless.

Even when I've defined mathematical proof. As if anyone could do it and take over the world of search results with their own definition.

Error on this scale is outside of any of the boundaries of known response.

Quite simply a rather remarkable error took over the mathematical field in the late 18th century. To protect living in error, math people today will deny any and all evidence because the error is their normal.

The error removes the usefulness of Galois Theory showing that to be what I call a display theory only.

Proving the error is trivial. Doing dramatic things is trivial as well.

Taunted about publication I GOT PUBLISHED and the journal editors caved to social pressure, pulled my paper after publication, and then the journal quietly died. Its hosting university scrubbed all mention of it. About ten years of math papers were tossed as trash by the United States—saved by EMIS.

But that's nothing, right?

I'm more important than any one of you by that measure. Than even dozens of you. How many mathematicians were published in that journal? Did it matter?

Do you think any of you would have fared any better?

None of you are important enough. None of your results are important enough. Any journal where I had managed to get published its likely the result would have been the same. Any journal. I've pondered how long the Annals might have survived.

The error is your life now. It has taken over the mathematical world. It feeds on your minds. It needs them.

It is the one true power of the "pure math" world. It may have created the idea of "pure mathematics" in order to survive.

It is a virus of the human intellect. And it is more powerful than any of you.

And more important than all of you, as it demonstrated by crushing that mathematical journal without effort.

It can do the same to any of your research efforts at will. You live at its mercy.

This war is a fight for the soul of the human race. And it is one of the greatest battles in the history of this species.

If it is lost, the future of humanity is lost with it, as mathematics is lost.

And without correct mathematics, the human species has no future.

Well, at least, not a future worth living.

### Tuesday, September 21, 2010

## JSH: Leverage in the construction

The secret to the power of the seemingly simple construction I use is as easy as 1, 2, 3:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients.

The leverage I use here is that multiplying times 1 or 2 is trivial. So it actually IS trivial to see how the 9 multiplies, as 9(1) = 9, but 9(2) = 18.

Years ago I called it a balance I think. I'm chaining the functions to easy constants—1 and 2—so that I can force them to behave in a particular way, by the distributive property.

The result is so powerful that you may notice that no actual mathematician has stepped up, even in Usenet posts, to dispute it, and posters who argue with me, have not been established mathematicians!!!

There is no way to dispute the result mathematically. It just is a mathematical fact that what you multiply times an expression will show up, if you multiply 1 or 2. So if it's x, you get 2x, if you multiply times 2. If it's pi, you get 2*pi.

And by the distributive property:

a*(f(x) + b) = a*f(x) + a*b

The triviality of the proof has been there for over seven years, as in, it's easy. I've yet to find a mathematician, an actual professional mathematician, who has disputed it. As how can they? And I welcome any to step up in reply to this post to tell me they are ready to jump in line and dispute the result.

So it's easy to prove 9, or previously I used 7, as a factor in those cases I give all the time in various threads. And the idea of unit factors not units in the ring of algebraic integers is not rocket science, so mathematicians would trivially comprehend it.

The secret is in the construction, forcing the functions across from integers, which is why forcing values at x=0 is so important. In trying to dispute the distributive property, various posters have been very creative in coming up with bizarre "functions", but those are just attempts at disputing the distributive property, as:

a*(f(x) + b) = a*f(x) + a*b

does not distinguish between f(0) and other values. So their "functions" are attempts at refuting the distributive property, which of course sounds ludicrous, so they claim they aren't!!!

Why then would mathematicians not just acknowledge my research?

Good question. Best guess is that they prefer the error!!!

Most of them were brought up in it. It was around before they were born.

It's like a religion. It's the religion they learned.

They live in error because error is what is now natural to them.

It's correct mathematics which may seem strange to a professional mathematician who has known only error his entire life—but believed otherwise, until a powerful, but simple mathematical construction was like a physicist telling a devout Fundamentalist Christian that Jesus didn't walk on water.

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients.

The leverage I use here is that multiplying times 1 or 2 is trivial. So it actually IS trivial to see how the 9 multiplies, as 9(1) = 9, but 9(2) = 18.

Years ago I called it a balance I think. I'm chaining the functions to easy constants—1 and 2—so that I can force them to behave in a particular way, by the distributive property.

The result is so powerful that you may notice that no actual mathematician has stepped up, even in Usenet posts, to dispute it, and posters who argue with me, have not been established mathematicians!!!

There is no way to dispute the result mathematically. It just is a mathematical fact that what you multiply times an expression will show up, if you multiply 1 or 2. So if it's x, you get 2x, if you multiply times 2. If it's pi, you get 2*pi.

And by the distributive property:

a*(f(x) + b) = a*f(x) + a*b

The triviality of the proof has been there for over seven years, as in, it's easy. I've yet to find a mathematician, an actual professional mathematician, who has disputed it. As how can they? And I welcome any to step up in reply to this post to tell me they are ready to jump in line and dispute the result.

So it's easy to prove 9, or previously I used 7, as a factor in those cases I give all the time in various threads. And the idea of unit factors not units in the ring of algebraic integers is not rocket science, so mathematicians would trivially comprehend it.

The secret is in the construction, forcing the functions across from integers, which is why forcing values at x=0 is so important. In trying to dispute the distributive property, various posters have been very creative in coming up with bizarre "functions", but those are just attempts at disputing the distributive property, as:

a*(f(x) + b) = a*f(x) + a*b

does not distinguish between f(0) and other values. So their "functions" are attempts at refuting the distributive property, which of course sounds ludicrous, so they claim they aren't!!!

Why then would mathematicians not just acknowledge my research?

Good question. Best guess is that they prefer the error!!!

Most of them were brought up in it. It was around before they were born.

It's like a religion. It's the religion they learned.

They live in error because error is what is now natural to them.

It's correct mathematics which may seem strange to a professional mathematician who has known only error his entire life—but believed otherwise, until a powerful, but simple mathematical construction was like a physicist telling a devout Fundamentalist Christian that Jesus didn't walk on water.

### Monday, September 20, 2010

## JSH: The real hard part

I find it hard to believe that Usenet matters in terms of what I see as the politics of this saga. I DO use Usenet as I describe to work through problems where a lot of the time for me it's mostly just stating the problem, and proposed solutions, where I don't necessarily need a lot of feedback, but don't mind if someone shoots down my ideas.

All that's easy. What's hard is adjusting to the role. I should put that as The Role. And it is maybe a little comforting to get that disdain from Usenet posters who see me as just some oddball, crackpot and I find their confidence so amazingly bizarre. For such people the world is a simple place and some guy ranting on Usenet about his mathematical discoveries can't really be important.

But I have a world that seems to increasingly listen to me, and a role that requires extraordinary responsibility as I try to grow into it, so I'm posting more about that lately than anything else, if you read between the lines.

But it is so much to adjust to accepting. My world has become simpler in terms of what I think I understand and surreal in terms of what I find difficult to comprehend as I shift from where I came and how I grew up into something I never knew, but only imagined as best I could growing up—yet still, had no clue.

I know things I don't know how I know them. The future sometimes seems like this book already written where I can scan the pages and then seems like this dark and hidden scary thing, to be feared. Possibilities run through my mind on every angle, and I can't seem to nail them down. But when I think I know something, and wish to test it, too often it just seems to happen and I don't really know why.

There is no way if I am who I say I am that there aren't quite a few nations where their leaders know exactly who I am.

For some of them I'm a dream come true, a way that the US is humbled in ways it doesn't yet comprehend. To others I'm a scary potential nightmare, a great unknown whose real powers are a mystery. And maybe to some of them I'm just a guy trying to figure out how he got in this really strange situation.

So no, I don't need Usenet to be known. I don't need the United States. If my research is what I say it is, then there are plenty of people in nations around the world who have known for years. If it's not, then I'm just some crackpot with one more rant, bound to be more down the line. Doesn't matter.

Adjusting to the role means learning to be more than I ever thought I could be, and believing it. Believing first in the mathematics and then that there must be some reason I found it. That it was given to me. Given to me for a purpose bigger than I am for a world that has far greater places to reach, ahead of it.

The role for which I have to be prepared is one that is iconic for an iconoclast. Mainstream for an outlier. Safe for a dreamer. But as concrete as a world requires.

If you've ever been afraid of public speaking, imagine what I face. My audience will be in the billions.

And if I stand in front of that audience, there will be no time then, for stage fright.

So everything before, is practice.

All that's easy. What's hard is adjusting to the role. I should put that as The Role. And it is maybe a little comforting to get that disdain from Usenet posters who see me as just some oddball, crackpot and I find their confidence so amazingly bizarre. For such people the world is a simple place and some guy ranting on Usenet about his mathematical discoveries can't really be important.

But I have a world that seems to increasingly listen to me, and a role that requires extraordinary responsibility as I try to grow into it, so I'm posting more about that lately than anything else, if you read between the lines.

But it is so much to adjust to accepting. My world has become simpler in terms of what I think I understand and surreal in terms of what I find difficult to comprehend as I shift from where I came and how I grew up into something I never knew, but only imagined as best I could growing up—yet still, had no clue.

I know things I don't know how I know them. The future sometimes seems like this book already written where I can scan the pages and then seems like this dark and hidden scary thing, to be feared. Possibilities run through my mind on every angle, and I can't seem to nail them down. But when I think I know something, and wish to test it, too often it just seems to happen and I don't really know why.

There is no way if I am who I say I am that there aren't quite a few nations where their leaders know exactly who I am.

For some of them I'm a dream come true, a way that the US is humbled in ways it doesn't yet comprehend. To others I'm a scary potential nightmare, a great unknown whose real powers are a mystery. And maybe to some of them I'm just a guy trying to figure out how he got in this really strange situation.

So no, I don't need Usenet to be known. I don't need the United States. If my research is what I say it is, then there are plenty of people in nations around the world who have known for years. If it's not, then I'm just some crackpot with one more rant, bound to be more down the line. Doesn't matter.

Adjusting to the role means learning to be more than I ever thought I could be, and believing it. Believing first in the mathematics and then that there must be some reason I found it. That it was given to me. Given to me for a purpose bigger than I am for a world that has far greater places to reach, ahead of it.

The role for which I have to be prepared is one that is iconic for an iconoclast. Mainstream for an outlier. Safe for a dreamer. But as concrete as a world requires.

If you've ever been afraid of public speaking, imagine what I face. My audience will be in the billions.

And if I stand in front of that audience, there will be no time then, for stage fright.

So everything before, is practice.

### Sunday, September 19, 2010

## JSH: But are they serious?

It occurs to me that maybe some of the posters that argue with me have not realized that I've had various tools available for years now to handle discussions on most of my work, where it surprises me that slight changes now may have made an odd difference, like using 9 instead of 7. Why would that matter? I'm not sure, but I THINK it does force you to realize with integers that the result does hold, so with:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and also one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients, it doesn't take much to find that you can't split up that 9. AT least not with polynomial functions. I left it more arbitrary to integers, and of course, some posters came up with convoluted "functions".

But I think that feels more like a dodge. With polynomial functions that 9 will not split into 3. You can't figure out a way to force the f's to have 3 as a factor versus just one having 9, and trying to understand why may make the result fairly easy.

But does it matter really? Well Usenet arguments were never really all that important. And while I think sci.math was instrumental in triggering the death of the journal SWJPAM, I think it also more important that mathematicians behind the scenes have always been in control.

If so, Usenet for them was just a joke, a place where I was stuck as they could close the doors to mainstream, and for years I was convincing myself anyway. Trying to grasp that I had found this thing, and believe it.

I've let that out, as something I'm highlighting because while it is speculation, if true, you were just their cover, but their math students were their sacrifices.

Societal structures are quite powerful. Human beings rely on them for so many things. And people can abuse them when they have the public trust.

If I'm right, then, no, it doesn't change much. All along Usenet could agree with me, and posters could agree with me, and the doors would still be closed.

The math professors at the top of the hierarchy could have ended the debate back in 2003.

Instead, I fear—though it is speculation—they killed a math journal.

None of you ever really mattered.

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and also one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients, it doesn't take much to find that you can't split up that 9. AT least not with polynomial functions. I left it more arbitrary to integers, and of course, some posters came up with convoluted "functions".

But I think that feels more like a dodge. With polynomial functions that 9 will not split into 3. You can't figure out a way to force the f's to have 3 as a factor versus just one having 9, and trying to understand why may make the result fairly easy.

But does it matter really? Well Usenet arguments were never really all that important. And while I think sci.math was instrumental in triggering the death of the journal SWJPAM, I think it also more important that mathematicians behind the scenes have always been in control.

If so, Usenet for them was just a joke, a place where I was stuck as they could close the doors to mainstream, and for years I was convincing myself anyway. Trying to grasp that I had found this thing, and believe it.

I've let that out, as something I'm highlighting because while it is speculation, if true, you were just their cover, but their math students were their sacrifices.

Societal structures are quite powerful. Human beings rely on them for so many things. And people can abuse them when they have the public trust.

If I'm right, then, no, it doesn't change much. All along Usenet could agree with me, and posters could agree with me, and the doors would still be closed.

The math professors at the top of the hierarchy could have ended the debate back in 2003.

Instead, I fear—though it is speculation—they killed a math journal.

None of you ever really mattered.

### Friday, September 17, 2010

## JSH: Begs incredulity

So yeah I had a paper published in 2003, and while that paper used cubics instead of quadratics, and I hadn't figured out all the detail I have today, it still got the job done of showing an issue with the ring of algebraic integers and was appropriately published.

And some sci.math'ers went after it with emails to the editors, who caved, yanked my paper, managed one more edition and then shut down, the hosting university SCRUBBED all mention of the journal—which had about 10 years of existence—and math people went on about their lives. World kept on turning….

(Google: SWJPAM to get the EMIS archives of the dead, dead, dead math journal.)

And you know with a problem of this sort, which is a foundation level problem, and a dead math journal back in 2003, yeah, it begs the limits of incredulity to imagine there are not top mathematicians who know exactly what the problem is and even know how it destroys their own research, or worse, know it invalidates research they are currently doing!!! (Can you imagine? I can. Knowingly doing crap research that is mathematically false. KNOWINGLY.)

But that's speculation.

And I think a MAJOR ISSUE for readers who say, but that's impossible, is that they don't realize maybe that if they are expected to not believe it, then if it is what is happening, then certain people know them well enough to know that's what they'd do.

So it becomes a Catch-22. I can prove there is an issue. I can show all kinds of craziness around the issue, like a dead, dead, dead math journal. But if math society say, oh, that's not possible, you get 7 years plus of people living in error.

So why is it a big deal?

x^2 + bx + 2 = (x + 2u_1)(x + u_2)

is a great expression I think to understand the error and figure things out. Use a non-zero integer b, and ponder how u_1 got left out when mathematicians were working to extend Gauss. Gauss did fine with gaussian integers which don't have a coverage issue.

Coverage was KNOWN as an issue back when algebraic integers were defined, but the tests those mathematicians did weren't it would seem up to the task of finding this problem, and it wipes out about a hundred plus years of number theory.

Just freaking wipes it out. It is an unbelievably devastating error. Unbelievably devastating.

It re-writes history books. Turns math heroes into zeros. And utterly destroys the image of the mathematical community which it holds to most of the world today.

So yeah, some of these professors may believe in some rationalization that they're doing a "good thing for the world", possibly deciding that living in error—ending mathematics as a real discipline in a lot of number theory areas—is better than dealing with the consequences and the damage to prestige and academia.

AFTER all, they would know that the world got along kind of ok for over a hundred years with the error!!!

Oh, so how? Well, "pure mathematics" arose with the error!!! It may have been some kind of organic thing as the math doesn't work! So if that hadn't happened some people might be noting that this stuff doesn't work for anything practical so maybe it's wrong.

But that's speculation. But the math does not work. That is NOT speculation. It just does not work.

So it really begs incredulity that top mathematicians don't know about the error. But hey, have to note that's speculation.

But think about it, all any math student has to do is just throw:

x^2 + bx + 2 = (x + 2u_1)(x + u_2)

up on the board, talk about coverage, with b a non-zero integer, at any university around the world on any given day, and they can just about blow up the classroom in discussion, so does that happen?

Oddly enough, in authoritarian environments, it's probably damn unlikely. So yeah, they can still I'd guess claim ignorance.

And some sci.math'ers went after it with emails to the editors, who caved, yanked my paper, managed one more edition and then shut down, the hosting university SCRUBBED all mention of the journal—which had about 10 years of existence—and math people went on about their lives. World kept on turning….

(Google: SWJPAM to get the EMIS archives of the dead, dead, dead math journal.)

And you know with a problem of this sort, which is a foundation level problem, and a dead math journal back in 2003, yeah, it begs the limits of incredulity to imagine there are not top mathematicians who know exactly what the problem is and even know how it destroys their own research, or worse, know it invalidates research they are currently doing!!! (Can you imagine? I can. Knowingly doing crap research that is mathematically false. KNOWINGLY.)

But that's speculation.

And I think a MAJOR ISSUE for readers who say, but that's impossible, is that they don't realize maybe that if they are expected to not believe it, then if it is what is happening, then certain people know them well enough to know that's what they'd do.

So it becomes a Catch-22. I can prove there is an issue. I can show all kinds of craziness around the issue, like a dead, dead, dead math journal. But if math society say, oh, that's not possible, you get 7 years plus of people living in error.

So why is it a big deal?

x^2 + bx + 2 = (x + 2u_1)(x + u_2)

is a great expression I think to understand the error and figure things out. Use a non-zero integer b, and ponder how u_1 got left out when mathematicians were working to extend Gauss. Gauss did fine with gaussian integers which don't have a coverage issue.

Coverage was KNOWN as an issue back when algebraic integers were defined, but the tests those mathematicians did weren't it would seem up to the task of finding this problem, and it wipes out about a hundred plus years of number theory.

Just freaking wipes it out. It is an unbelievably devastating error. Unbelievably devastating.

It re-writes history books. Turns math heroes into zeros. And utterly destroys the image of the mathematical community which it holds to most of the world today.

So yeah, some of these professors may believe in some rationalization that they're doing a "good thing for the world", possibly deciding that living in error—ending mathematics as a real discipline in a lot of number theory areas—is better than dealing with the consequences and the damage to prestige and academia.

AFTER all, they would know that the world got along kind of ok for over a hundred years with the error!!!

Oh, so how? Well, "pure mathematics" arose with the error!!! It may have been some kind of organic thing as the math doesn't work! So if that hadn't happened some people might be noting that this stuff doesn't work for anything practical so maybe it's wrong.

But that's speculation. But the math does not work. That is NOT speculation. It just does not work.

So it really begs incredulity that top mathematicians don't know about the error. But hey, have to note that's speculation.

But think about it, all any math student has to do is just throw:

x^2 + bx + 2 = (x + 2u_1)(x + u_2)

up on the board, talk about coverage, with b a non-zero integer, at any university around the world on any given day, and they can just about blow up the classroom in discussion, so does that happen?

Oddly enough, in authoritarian environments, it's probably damn unlikely. So yeah, they can still I'd guess claim ignorance.

### Thursday, September 16, 2010

## JSH: More units

Though it took YEARS for me to figure out, for the big picture people trying to get a handle on what could be wrong with the ring of algebraic integers, you can try focusing on one key expression:

x^2 + bx + 2 = (x + 2u_1)(x + u_2)

where 'b' is a non-zero integer chosen such that the roots of the quadratic are non-rational, as, guess what?

Then u_1 cannot exist in the ring of algebraic integers, and u_2 cannot be a unit within that ring.

If you understand the above and figure out why then you should have a good handle on the problem and can begin surveying the scope of destruction this result puts across number theory over the last hundred years.

Sorry, had to toss in some hyperbole there—I'm so good at it!

Clue: Major issue is with defining algebraic integers as roots of

It DID take me years to figure it out, and I've been working at explaining it for more years. I've also fixed the problem—closing what I call now a coverage gap—with a ring I call the object ring.

(A perk of being the discoverer is getting to name things. Really cool perk.)

Google: object ring

Think about that fix and how I solved the problem and you'll see the rational behind that definition which includes the ring of algebraic integers, and the units that ring leaves out.

Try that clue above if you're puzzling over it greatly. The problem has existed for over a hundred years. Seems reasonable to suppose that you may take a little while to really figure it out.

x^2 + bx + 2 = (x + 2u_1)(x + u_2)

where 'b' is a non-zero integer chosen such that the roots of the quadratic are non-rational, as, guess what?

Then u_1 cannot exist in the ring of algebraic integers, and u_2 cannot be a unit within that ring.

If you understand the above and figure out why then you should have a good handle on the problem and can begin surveying the scope of destruction this result puts across number theory over the last hundred years.

Sorry, had to toss in some hyperbole there—I'm so good at it!

Clue: Major issue is with defining algebraic integers as roots of

**monic**polynomials with integer coefficients. With the expression above, consider solving for u_1. You will get a non-monic!!!It DID take me years to figure it out, and I've been working at explaining it for more years. I've also fixed the problem—closing what I call now a coverage gap—with a ring I call the object ring.

(A perk of being the discoverer is getting to name things. Really cool perk.)

Google: object ring

Think about that fix and how I solved the problem and you'll see the rational behind that definition which includes the ring of algebraic integers, and the units that ring leaves out.

Try that clue above if you're puzzling over it greatly. The problem has existed for over a hundred years. Seems reasonable to suppose that you may take a little while to really figure it out.

## JSH: Coverage issue of algebraic integers

Unique issues arise with the ring of algebraic integers as a result of the very definition that gives the set, where the rule that algebraic integers be roots of monic polynomials with integer coefficients can be shown to force a dramatic coverage issue, where key numbers are left out.

The problem is similar to what happens if you consider only evens, so 2 and 6 do NOT then have a factor in common because 3 is odd, so it has been left out.

Seeing the more complex problem with the ring of algebraic integers requires only elementary methods using some basic algebra.

I'll need only one proof:

The equation (g_1(x) + 1)(g_2(x) + 2) = P(x), when P(x) is a quadratic with integer coefficients, and g_1(0) = g_2(0) = 0, where the g's are arbitrary functions of x, with a specific case not required for the proof itself, may for certain highly specific cases not exist in the ring of algebraic integers.

Proof:

Proving that statement requires use of a constant factor, so I'll multiply by 9, and now consider the following expression in the ring of algebraic integers:

9(g_1(x) + 1)(g_2(x) + 2) = 9*P(x)

where g_1(0) = g_2(0) = 0, and P(x) is a quadratic with integer coefficients; therefore, trivially, the last coefficient of P(x) is 2.

Then by the distributive property, I have:

(9g_1(x) + 9)(g_2(x) + 2) = 9*P(x)

which must always exist then if the original expression exists. But now consider functions f_1(x), and f_2(x), such that

f_1(x) = 9g_1(x), and f_2(x) = g_2(x) - 7, so:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where again g_1(0) = g_2(0) = 0, and notice one of the f's equals 0, at x=0, and P(x).

Let f_1(x) and f_2(x) be non-rational roots of the same monic quadratic with integer coefficients—so they are being required to be algebraic integers—then f_1(x) = 9g_1(x), requires one of the f's to have 9 as a factor, while the other must be coprime to 9, since

f_2(x) = g_2(x) - 7

and g_1(0) = g_2(0) = 0.

But it is well-known that in the ring of algebraic integers that just one root of a monic polynomial with integer coefficients irreducible over Q may not have a non-unit rational as a factor if ALL of the roots do not have it as a factor.

Therefore 9 is a factor of NEITHER of the f's in the ring of algebraic integers, and there is a contradiction.

Proof complete.

The result has important implications for the coverage of algebraic integers, as consider the following carefully peculiar way to generate functions.

With P(x) = 441x^2 - 35x + 2, now moving to the ring of algebraic integers, multiplying by 9, and re-grouping terms gives:

9*P(x) = (81x^2 - 18x)7^2 + (9x-1)(9)(7) + 9^2

which allows me now to get a non-polynomial factorization with

9*P(x) = (7a_1(x) + 9)(7a_2(x)+ 9)

where the a's are roots of

a^2 - (9x-1)a + (81x^2 - 18x) = 0

and with f_1(x) = 7a_1(x) and f_2(x) = 7a_2(x), I have functions that fulfill the conditions above.

The very purpose of the odd construction here is to get functions where the f's are roots of the same monic polynomial with integer coefficients, where just one of the f's equals 0 at x=0, and that has been achieved.

Now notice that at x=1, I have the f's are roots of:

a^2 - 8a + 63 = 0

where it is not allowed in the ring of algebraic integers for either of the roots to have 7 as a factor.

Now consider again the expression:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

by the proof above the part 9(g_1(x) + 1)(g_2(x) + 2) is blocked from the ring of algebraic integers at x=1, while:

(f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

is STILL in the ring! And notice that you are not able to in general divide off the 9, in the ring of algebraic integers.

But that is not a problem for the field of complex numbers!

The sudden move to a field may be a surprise, but notice I've hit an odd limitation: the ring of algebraic integers provably will not allow me to generally divide off the 9, but I'm curious! I want to do it anyway! So how? Go to the field of complex numbers and there is no block. Dividing off the 9 there gives:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9)/9 = P(x)

where I haven't guessed at how the 9 divides through the middle but of course there are an

And notice that (g_1(x) + 1)(g_2(x) + 2) = P(x) exists in the field of complex numbers!!!

So then, what was the ring of algebraic integers hiding? One might presume that the g's are forced in some way to be fractions but it's not clear why that would be the case! And worse, what might you presume to be in the denominator for any type of fraction?

Oddly enough, if you check against the solution for the f's—you can solve for the f's using the quadratic formula—you would be forced to say, factors in common with 9.

But there is no 9. And I could just as soon use 7, if I wished, and did so for years with such demonstrations.

So the ring of algebraic integers is hiding, what? And how?

The obvious answer is, the ring of algebraic integers is using unit factors, and holding the 9 hostage, one might say:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/9 + u_1)(f_2(x)*u_2 + 9*u_2) = P(x)

where I'm dividing through as I did because f_1(x) = 9*g_1(x), from above.

Some might ask, why unit factors?

The answer is, only unit factors can be generated spontaneously, like 9 isn't a unit factor and had to multiply times P(x) to be a factor, but once I divide 9 off, P(x) only has 1 itself as a factor, so only factors of 1 are available i.e. unit factors.

But now there is a truly odd conclusion that follows: if mathematicians try to divide off 9, from say, the roots of

a^2 - 8a + 63 = 0

what they are getting are the unit factor u_1, which is an algebraic integer but not a unit in that ring, and also 9*u_2, where u_2 is hidden as it's not a member of the ring! Like with 2 and 6 when you only consider evens, you find that 3 must remain wrapped.

So u_2 is a hidden unit, and the coverage problem is revealed.

Why does it have to be a unit again? Because only units can generate spontaneously.

But what is the ring of algebraic integers DOING??!!! Why is it doing this thing here?

The 'why' of the problem is actually fairly simple.

Answering the mystery is as simple as noting a corollary from the ring of integers that is not allowed in the ring of algebraic integers, as note that in the ring of integers, any member of that ring can be paired with a unit as roots of a quadratic, for example:

x^2 + 3x + 2 = (x + 2)(x + 1)

While that is allowed in the ring of algebraic integers, the more general:

x^2 + bx + 2 = (x + 2u_1)(x + u_2)

is not allowed at all for integer b, if the u's are non-rational. That gives numbers that are properly units which are blocked from being units in that ring.

So why wasn't that noticed over a hundred years ago when mathematicians were testing out the ring of algebraic integers and LOOKING for coverage problems?

Good question.

With a hundred years of naive usage of the ring of algebraic integers a serious amount of resistance has emerged today so that the simple mathematics above is for the most part, years old. I've explained often, trying different ways to explain, as I persist in trying to get acknowledgement of a serious problem.

Some may wonder why it's such a big deal. Well consider one of the tricks mathematicians in number theory often use, which is to use the result that something is not a unit in the ring of algebraic integers to "prove" it has factors like above such a faux proof would involve showing the factors in common with 9 remain for both f's. Any conclusion they get from that wrong argument is not mathematically proven and can actually be false!

Imagine if people DID only use evens, and if you tried to explain to people that 2 can divide from 6 to give 3, and they said that 3 didn't exist, you were crazy and that 2 and 6 were coprime? How might that change our world today?

While mathematicians resist acknowledging this error they can do flawed mathematical arguments, and possibly appear to prove just about anything, which is an odd thing about serious math errors.

That could be enticing for some.

After all, mathematics is VERY hard as a discipline. Can you imagine being a "pure math" mathematician and starting over?

This time to do mathematics that is actually correct?

What if instead you could just play pretend? After all, your entire career has been smoke and mirrors anyway, if it is "pure math" only in areas where this error holds sway.

This post is yet another attempt to help chip away at the resistance. And you know it does amaze me the newer math students who must not wish to achieve greatness in the field, as knowing there is this error, which they can work through themselves as its elementary methods and mostly rather simple algebra, they must know that there is no great proof to be found in actuality if you rely on the error!

So they will never achieve true greatness while under its shadow.

To the extent that they learn in error anyway, they are accepting that they will never make a great discovery—or maybe any real discovery at all—and that could be part of the problem! Possibly deep down many of today's math students don't feel capable of ever making a great discovery anyway.

At least with the error, they can get paid, raise a family maybe. Pay a mortgage.

The problem is similar to what happens if you consider only evens, so 2 and 6 do NOT then have a factor in common because 3 is odd, so it has been left out.

Seeing the more complex problem with the ring of algebraic integers requires only elementary methods using some basic algebra.

I'll need only one proof:

The equation (g_1(x) + 1)(g_2(x) + 2) = P(x), when P(x) is a quadratic with integer coefficients, and g_1(0) = g_2(0) = 0, where the g's are arbitrary functions of x, with a specific case not required for the proof itself, may for certain highly specific cases not exist in the ring of algebraic integers.

Proof:

Proving that statement requires use of a constant factor, so I'll multiply by 9, and now consider the following expression in the ring of algebraic integers:

9(g_1(x) + 1)(g_2(x) + 2) = 9*P(x)

where g_1(0) = g_2(0) = 0, and P(x) is a quadratic with integer coefficients; therefore, trivially, the last coefficient of P(x) is 2.

Then by the distributive property, I have:

(9g_1(x) + 9)(g_2(x) + 2) = 9*P(x)

which must always exist then if the original expression exists. But now consider functions f_1(x), and f_2(x), such that

f_1(x) = 9g_1(x), and f_2(x) = g_2(x) - 7, so:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where again g_1(0) = g_2(0) = 0, and notice one of the f's equals 0, at x=0, and P(x).

Let f_1(x) and f_2(x) be non-rational roots of the same monic quadratic with integer coefficients—so they are being required to be algebraic integers—then f_1(x) = 9g_1(x), requires one of the f's to have 9 as a factor, while the other must be coprime to 9, since

f_2(x) = g_2(x) - 7

and g_1(0) = g_2(0) = 0.

But it is well-known that in the ring of algebraic integers that just one root of a monic polynomial with integer coefficients irreducible over Q may not have a non-unit rational as a factor if ALL of the roots do not have it as a factor.

Therefore 9 is a factor of NEITHER of the f's in the ring of algebraic integers, and there is a contradiction.

Proof complete.

The result has important implications for the coverage of algebraic integers, as consider the following carefully peculiar way to generate functions.

With P(x) = 441x^2 - 35x + 2, now moving to the ring of algebraic integers, multiplying by 9, and re-grouping terms gives:

9*P(x) = (81x^2 - 18x)7^2 + (9x-1)(9)(7) + 9^2

which allows me now to get a non-polynomial factorization with

9*P(x) = (7a_1(x) + 9)(7a_2(x)+ 9)

where the a's are roots of

a^2 - (9x-1)a + (81x^2 - 18x) = 0

and with f_1(x) = 7a_1(x) and f_2(x) = 7a_2(x), I have functions that fulfill the conditions above.

The very purpose of the odd construction here is to get functions where the f's are roots of the same monic polynomial with integer coefficients, where just one of the f's equals 0 at x=0, and that has been achieved.

Now notice that at x=1, I have the f's are roots of:

a^2 - 8a + 63 = 0

where it is not allowed in the ring of algebraic integers for either of the roots to have 7 as a factor.

Now consider again the expression:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

by the proof above the part 9(g_1(x) + 1)(g_2(x) + 2) is blocked from the ring of algebraic integers at x=1, while:

(f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

is STILL in the ring! And notice that you are not able to in general divide off the 9, in the ring of algebraic integers.

But that is not a problem for the field of complex numbers!

The sudden move to a field may be a surprise, but notice I've hit an odd limitation: the ring of algebraic integers provably will not allow me to generally divide off the 9, but I'm curious! I want to do it anyway! So how? Go to the field of complex numbers and there is no block. Dividing off the 9 there gives:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9)/9 = P(x)

where I haven't guessed at how the 9 divides through the middle but of course there are an

**infinity**of ways to do it in the field.And notice that (g_1(x) + 1)(g_2(x) + 2) = P(x) exists in the field of complex numbers!!!

So then, what was the ring of algebraic integers hiding? One might presume that the g's are forced in some way to be fractions but it's not clear why that would be the case! And worse, what might you presume to be in the denominator for any type of fraction?

Oddly enough, if you check against the solution for the f's—you can solve for the f's using the quadratic formula—you would be forced to say, factors in common with 9.

But there is no 9. And I could just as soon use 7, if I wished, and did so for years with such demonstrations.

So the ring of algebraic integers is hiding, what? And how?

The obvious answer is, the ring of algebraic integers is using unit factors, and holding the 9 hostage, one might say:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/9 + u_1)(f_2(x)*u_2 + 9*u_2) = P(x)

where I'm dividing through as I did because f_1(x) = 9*g_1(x), from above.

Some might ask, why unit factors?

The answer is, only unit factors can be generated spontaneously, like 9 isn't a unit factor and had to multiply times P(x) to be a factor, but once I divide 9 off, P(x) only has 1 itself as a factor, so only factors of 1 are available i.e. unit factors.

But now there is a truly odd conclusion that follows: if mathematicians try to divide off 9, from say, the roots of

a^2 - 8a + 63 = 0

what they are getting are the unit factor u_1, which is an algebraic integer but not a unit in that ring, and also 9*u_2, where u_2 is hidden as it's not a member of the ring! Like with 2 and 6 when you only consider evens, you find that 3 must remain wrapped.

So u_2 is a hidden unit, and the coverage problem is revealed.

Why does it have to be a unit again? Because only units can generate spontaneously.

But what is the ring of algebraic integers DOING??!!! Why is it doing this thing here?

The 'why' of the problem is actually fairly simple.

Answering the mystery is as simple as noting a corollary from the ring of integers that is not allowed in the ring of algebraic integers, as note that in the ring of integers, any member of that ring can be paired with a unit as roots of a quadratic, for example:

x^2 + 3x + 2 = (x + 2)(x + 1)

While that is allowed in the ring of algebraic integers, the more general:

x^2 + bx + 2 = (x + 2u_1)(x + u_2)

is not allowed at all for integer b, if the u's are non-rational. That gives numbers that are properly units which are blocked from being units in that ring.

So why wasn't that noticed over a hundred years ago when mathematicians were testing out the ring of algebraic integers and LOOKING for coverage problems?

Good question.

With a hundred years of naive usage of the ring of algebraic integers a serious amount of resistance has emerged today so that the simple mathematics above is for the most part, years old. I've explained often, trying different ways to explain, as I persist in trying to get acknowledgement of a serious problem.

Some may wonder why it's such a big deal. Well consider one of the tricks mathematicians in number theory often use, which is to use the result that something is not a unit in the ring of algebraic integers to "prove" it has factors like above such a faux proof would involve showing the factors in common with 9 remain for both f's. Any conclusion they get from that wrong argument is not mathematically proven and can actually be false!

Imagine if people DID only use evens, and if you tried to explain to people that 2 can divide from 6 to give 3, and they said that 3 didn't exist, you were crazy and that 2 and 6 were coprime? How might that change our world today?

While mathematicians resist acknowledging this error they can do flawed mathematical arguments, and possibly appear to prove just about anything, which is an odd thing about serious math errors.

That could be enticing for some.

After all, mathematics is VERY hard as a discipline. Can you imagine being a "pure math" mathematician and starting over?

This time to do mathematics that is actually correct?

What if instead you could just play pretend? After all, your entire career has been smoke and mirrors anyway, if it is "pure math" only in areas where this error holds sway.

This post is yet another attempt to help chip away at the resistance. And you know it does amaze me the newer math students who must not wish to achieve greatness in the field, as knowing there is this error, which they can work through themselves as its elementary methods and mostly rather simple algebra, they must know that there is no great proof to be found in actuality if you rely on the error!

So they will never achieve true greatness while under its shadow.

To the extent that they learn in error anyway, they are accepting that they will never make a great discovery—or maybe any real discovery at all—and that could be part of the problem! Possibly deep down many of today's math students don't feel capable of ever making a great discovery anyway.

At least with the error, they can get paid, raise a family maybe. Pay a mortgage.

### Wednesday, September 15, 2010

## JSH: Ran into a wall

Nothing has really worked. Publication just ended up with the journal dead. I've used tautological spaces against quadratic Diophantines, so what. Worked out a prime counting function that actually leads directly to continuous functions. Nothing.

I branched out. Moved into pop culture and general problem solving. Gave a healthcare plan in a single 140 character "perfect" tweet. Nothing.

Gave ideas for YouTube to Google. Ideas to Yahoo. Gave political ideas. Economic ideas. Nothing.

Gave a freaking idea to the post office. Trivial little thing, but still. NOTHING.

My analysis is indicating that the world as it claims to exist does not. There seems to be a single global governing body made up of hereditary members primarily located in Great Britain. A hidden royal family.

I have indications that world history is mostly fictional.

That this world is mostly a well-told lie. Ruled by a small group of people. Who have near absolute power.

I branched out. Moved into pop culture and general problem solving. Gave a healthcare plan in a single 140 character "perfect" tweet. Nothing.

Gave ideas for YouTube to Google. Ideas to Yahoo. Gave political ideas. Economic ideas. Nothing.

Gave a freaking idea to the post office. Trivial little thing, but still. NOTHING.

My analysis is indicating that the world as it claims to exist does not. There seems to be a single global governing body made up of hereditary members primarily located in Great Britain. A hidden royal family.

I have indications that world history is mostly fictional.

That this world is mostly a well-told lie. Ruled by a small group of people. Who have near absolute power.

### Tuesday, September 14, 2010

## JSH: How do we resolve this situation?

The "we" in the subject line is not royal, as I'm kind of perplexed, and I'm looking for feedback. We need to figure out this situation. Even these latest explanations on the subtle behavior of the ring of algebraic integers are just re-hashes of the same thing repeated. I've known it all for years and talked about it for years.

Latest major change was using 9 instead of 7, so I'm going to just ask the newsgroup, how do we resolve this situation?

The mathematics will not change.

Over the years I've watched posters go through any number of gyrations in trying to obfuscate this result, but the big question in my mind is, why bother?

WHAT are they trying to protect? The ring of algebraic integers has this behavior. It's an objective thing. People didn't realize it had this behavior before I started howling about it (we should hope), and now it's known, so why would mathematical society pretend it does not exist?

MY guess has always been its a social thing where a society is protecting itself.

But what about any love of mathematical truth, or even just curiosity?

By now I have a lot more than just the discovery of the core error. I've wondered about curiosity—lack of it—with quite a few of my results, especially some of the later ones.

Can a need to hold on to a particular social order be that great? What do any of you actually believe in?

I've noted before that by now I can walk into any math office of any supposedly great "pure math" mathematician and shred his research, at will. At will.

His protection, however, remains the willingness of not only supposedly smart people worldwide to ignore a fascinating issue with algebraic integers, but for new students to willingly give up any hope of mathematical greatness for themselves!

It's almost as if, none of you wish to be great.

I find that puzzling.

Not a single math student over all the world with a desire for greatness?

When did that happen?

Where is the human species going then? If none of the people who pursue mathematics today care if they're actually right, or care if history will ever remember them?

If there is a human future, I will be in it, as one of the great discoverers. But it appears that none of you wish to be in it at all.

Why not?

Latest major change was using 9 instead of 7, so I'm going to just ask the newsgroup, how do we resolve this situation?

The mathematics will not change.

Over the years I've watched posters go through any number of gyrations in trying to obfuscate this result, but the big question in my mind is, why bother?

WHAT are they trying to protect? The ring of algebraic integers has this behavior. It's an objective thing. People didn't realize it had this behavior before I started howling about it (we should hope), and now it's known, so why would mathematical society pretend it does not exist?

MY guess has always been its a social thing where a society is protecting itself.

But what about any love of mathematical truth, or even just curiosity?

By now I have a lot more than just the discovery of the core error. I've wondered about curiosity—lack of it—with quite a few of my results, especially some of the later ones.

Can a need to hold on to a particular social order be that great? What do any of you actually believe in?

I've noted before that by now I can walk into any math office of any supposedly great "pure math" mathematician and shred his research, at will. At will.

His protection, however, remains the willingness of not only supposedly smart people worldwide to ignore a fascinating issue with algebraic integers, but for new students to willingly give up any hope of mathematical greatness for themselves!

It's almost as if, none of you wish to be great.

I find that puzzling.

Not a single math student over all the world with a desire for greatness?

When did that happen?

Where is the human species going then? If none of the people who pursue mathematics today care if they're actually right, or care if history will ever remember them?

If there is a human future, I will be in it, as one of the great discoverers. But it appears that none of you wish to be in it at all.

Why not?

## JSH: Didn't occur to you?

I think Gauss hid things too. It occurs to me that he allowed the "core error" itself. I also figure Einstein hid things. And maybe all of them did.

The world kind of treats you like scum, and the turns the story around later.

This world will never really know this story. Never.

At least I'll let you in on what the others may not have, but it probably won't matter.

You'll just lie. History will claim to know what happened. But no one ever will. Not the full story.

Not the real one.

The world kind of treats you like scum, and the turns the story around later.

This world will never really know this story. Never.

At least I'll let you in on what the others may not have, but it probably won't matter.

You'll just lie. History will claim to know what happened. But no one ever will. Not the full story.

Not the real one.

## JSH: Understanding the proof forward thread

If people would help in trying to understand the error versus working desperately to always obfuscate it, things would be easier, but as I have an active gallery of people trying to embrace error I need to explain, explain, explain, so here's a thread to explain my new proof forward thread, where I just go ahead and use the distributive property to shut-up the people claiming that the result doesn't follow by the distributive property.

Key to everything is:

9(g_1(x) + 1)(g_2(x) + 2) = 9*P(x)

which more simply of course is:

(g_1(x) + 1)(g_2(x) + 2) = P(x)

and the proof forward thread is about showing that the ring of algebraic integers will deliberately exclude that expression in

Now that specificity is important, as going the OTHER WAY posters would work desperately to try and claim that whatever example I gave didn't apply! Obfuscation maximum.

Going THIS WAY all I have to do is show the existence of cases which is easy!

The other way also there was this dispute that started about functions, but now I can ignore that entirely by simply giving the functions themselves, or proving they exist.

So why should anyone care if the ring of algebraic integers will in certain

(g_1(x) + 1)(g_2(x) + 2) = P(x) ?

Because it reveals a coverage problem.

In the case that I use, the fuller expression:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9P(x)

is important as: (f_1(x) + 9)(f_2(x) + 9) = 9P(x) does exist.

So I focus on the case where the 9 is generally divided off, which in certain particular cases does not exist!

But that's kind of nutty! What happens if you move to some other ring where you CAN divide the 9 off?

Well you get: (g_1(x) + 1)(g_2(x) + 2) = P(x)

Expecting something else?

So what happens in the ring of algebraic integers when it refuses to allow you to have that? It uses units:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 9*u_1)(f_2(x)*u_2/9 + u_2) = P(x)

And mathematicians when they perform calculations suppose the unit factor u_2 still has non-unit factors in common with 9, as u_2 is not a unit in the ring of algebraic integers, and u_1 is not in the ring of algebraic integers at all.

They are units in my object ring.

Ok, so I figured all of this out YEARS AGO.

I can explain exactly what the ring of algebraic integers has to do. Can tell why it has to do it. And can prove the issue forwards or backwards. Proving it forward allows me to definitely use the distributive property—removing claims that the distributive property is not relevant.

Why doesn't proof matter?

Because the people arguing with me don't give a damn about the truth, and the mathematicians who know better refuse to speak up. Usenet posters don't matter. I'm sure you're all aware of THAT at least. So it's not like it's a big deal if some obsessive people reply to me objecting no matter what and refusing the explanations, unless there are mathematicians around the world looking closely at them and supposing that the continuing objections means they can safely ignore the error.

So what can you do?

I don't know, maybe a petition? If some math people who know of the error send something to some prestigious department, like Berkeley, noting that they are aware of the error, and don't care about knuckleheads on Usenet objecting and arguing over points well explained then they may disabuse mathematicians—if they ARE looking to Usenet for guidance—of the notion that somehow they are acting in secret.

If some "leading" whatever realizes that people all over the world know that he is a fraud, and is stunned to find out that a half a dozen obsessive Usenet posters don't mean squat about world knowledge of his fraudulent behavior then he may get off his ass and do something.

I sometimes imagine these supposedly top mathematicians are routinely reading sci.math to get a feel for how safe they are!!!

Until some of you let them realize that it does not matter what certain posters are saying, they may continue indefinitely.

Key to everything is:

9(g_1(x) + 1)(g_2(x) + 2) = 9*P(x)

which more simply of course is:

(g_1(x) + 1)(g_2(x) + 2) = P(x)

and the proof forward thread is about showing that the ring of algebraic integers will deliberately exclude that expression in

**certain**cases!Now that specificity is important, as going the OTHER WAY posters would work desperately to try and claim that whatever example I gave didn't apply! Obfuscation maximum.

Going THIS WAY all I have to do is show the existence of cases which is easy!

The other way also there was this dispute that started about functions, but now I can ignore that entirely by simply giving the functions themselves, or proving they exist.

So why should anyone care if the ring of algebraic integers will in certain

**particular cases**refuse to allow:(g_1(x) + 1)(g_2(x) + 2) = P(x) ?

Because it reveals a coverage problem.

In the case that I use, the fuller expression:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9P(x)

is important as: (f_1(x) + 9)(f_2(x) + 9) = 9P(x) does exist.

So I focus on the case where the 9 is generally divided off, which in certain particular cases does not exist!

But that's kind of nutty! What happens if you move to some other ring where you CAN divide the 9 off?

Well you get: (g_1(x) + 1)(g_2(x) + 2) = P(x)

Expecting something else?

So what happens in the ring of algebraic integers when it refuses to allow you to have that? It uses units:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 9*u_1)(f_2(x)*u_2/9 + u_2) = P(x)

And mathematicians when they perform calculations suppose the unit factor u_2 still has non-unit factors in common with 9, as u_2 is not a unit in the ring of algebraic integers, and u_1 is not in the ring of algebraic integers at all.

They are units in my object ring.

Ok, so I figured all of this out YEARS AGO.

I can explain exactly what the ring of algebraic integers has to do. Can tell why it has to do it. And can prove the issue forwards or backwards. Proving it forward allows me to definitely use the distributive property—removing claims that the distributive property is not relevant.

Why doesn't proof matter?

Because the people arguing with me don't give a damn about the truth, and the mathematicians who know better refuse to speak up. Usenet posters don't matter. I'm sure you're all aware of THAT at least. So it's not like it's a big deal if some obsessive people reply to me objecting no matter what and refusing the explanations, unless there are mathematicians around the world looking closely at them and supposing that the continuing objections means they can safely ignore the error.

So what can you do?

I don't know, maybe a petition? If some math people who know of the error send something to some prestigious department, like Berkeley, noting that they are aware of the error, and don't care about knuckleheads on Usenet objecting and arguing over points well explained then they may disabuse mathematicians—if they ARE looking to Usenet for guidance—of the notion that somehow they are acting in secret.

If some "leading" whatever realizes that people all over the world know that he is a fraud, and is stunned to find out that a half a dozen obsessive Usenet posters don't mean squat about world knowledge of his fraudulent behavior then he may get off his ass and do something.

I sometimes imagine these supposedly top mathematicians are routinely reading sci.math to get a feel for how safe they are!!!

Until some of you let them realize that it does not matter what certain posters are saying, they may continue indefinitely.

## JSH: Proof forward of core error

A remarkable problem in number theory, a "core error" is shown in a forward direction in order to foil obfuscation attempts by posters dedicated to protecting the error.

Consider the following expression in the ring of algebraic integers:

9(g_1(x) + 1)(g_2(x) + 2) = 9*P(x)

where g_1(0) = g_2(0) = 0, and P(x) is a quadratic with integer coefficients; therefore, trivially, the last coefficient of P(x) is 2.

It is trivial to prove that it may not always exist in the ring of algebraic integers.

Proof:

Let P(x) be irreducible over Q, that is, not have rational roots for P(x) = 0. Then by the distributive property, I have trivially:

(9g_1(x) + 9)(g_2(x) + 2) = 9*P(x)

which must always exist then if the original expression exists. But now consider functions f_1(x), and f_2(x), such that

f_1(x) = 9g_1(x), and f_2(x) = g_2(x) - 7, so:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where again g_1(0) = g_2(0) = 0, and notice one of the f's equals 0, at x=0, and P(x) is again a quadratic with integer coefficients.

By symmetry f_1(x) and f_2(x) will be roots of the same monic polynomial, which must itself be irreducible over Q, but then just one of the f's has 9 as a factor, and it is well-known that in the ring of algebraic integers that just one root of a monic polynomial with integer coefficients irreducible over Q may not have a non-unit rational as a factor if ALL of the roots do not have it as a factor.

Therefore 9 is a factor of NEITHER of the f's in the ring of algebraic integers, and there is a contradiction.

Proof complete.

Going forward allows me to get away from arguments by posters claiming that the distributive property does not apply by directly using it in a key step.

Notice then that the ring of algebraic integers can be picky!

It will not allow under certain circumstance:

9(g_1(x) + 1)(g_2(x) + 2) = 9*P(x)

where g_1(0) = g_2(0) = 0, and P(x) is a quadratic with integer coefficients.

But that shows gaps in the numbers covered as, why can there not be numbers that exist that will fulfill those conditions?

Those gaps are what are important, as the lack of full coverage by the ring of algebraic integers is a "core error" with critical implications for number theoretic results over 100+ years.

Having people arguing to hide an error, fighting for years to defend an error, is not surprising given the human condition, but it does show that error in and of itself is not something that people will reject!

Sometimes they embrace it, instead.

Sometimes people love error. Error becomes their true love.

Consider the following expression in the ring of algebraic integers:

9(g_1(x) + 1)(g_2(x) + 2) = 9*P(x)

where g_1(0) = g_2(0) = 0, and P(x) is a quadratic with integer coefficients; therefore, trivially, the last coefficient of P(x) is 2.

It is trivial to prove that it may not always exist in the ring of algebraic integers.

Proof:

Let P(x) be irreducible over Q, that is, not have rational roots for P(x) = 0. Then by the distributive property, I have trivially:

(9g_1(x) + 9)(g_2(x) + 2) = 9*P(x)

which must always exist then if the original expression exists. But now consider functions f_1(x), and f_2(x), such that

f_1(x) = 9g_1(x), and f_2(x) = g_2(x) - 7, so:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where again g_1(0) = g_2(0) = 0, and notice one of the f's equals 0, at x=0, and P(x) is again a quadratic with integer coefficients.

By symmetry f_1(x) and f_2(x) will be roots of the same monic polynomial, which must itself be irreducible over Q, but then just one of the f's has 9 as a factor, and it is well-known that in the ring of algebraic integers that just one root of a monic polynomial with integer coefficients irreducible over Q may not have a non-unit rational as a factor if ALL of the roots do not have it as a factor.

Therefore 9 is a factor of NEITHER of the f's in the ring of algebraic integers, and there is a contradiction.

Proof complete.

Going forward allows me to get away from arguments by posters claiming that the distributive property does not apply by directly using it in a key step.

Notice then that the ring of algebraic integers can be picky!

It will not allow under certain circumstance:

9(g_1(x) + 1)(g_2(x) + 2) = 9*P(x)

where g_1(0) = g_2(0) = 0, and P(x) is a quadratic with integer coefficients.

But that shows gaps in the numbers covered as, why can there not be numbers that exist that will fulfill those conditions?

Those gaps are what are important, as the lack of full coverage by the ring of algebraic integers is a "core error" with critical implications for number theoretic results over 100+ years.

Having people arguing to hide an error, fighting for years to defend an error, is not surprising given the human condition, but it does show that error in and of itself is not something that people will reject!

Sometimes they embrace it, instead.

Sometimes people love error. Error becomes their true love.

### Monday, September 13, 2010

## JSH: Defining function?

Doing foundational work on mathematics is not for the faint-hearted, but it is something I've been forced into doing before and now I'm facing an oddity with "function".

Here's an example that I need shot down, as in, given the current usage of "function", why is this wrong?

Let f(x) = 0, when x=1, and f(x) = 1, when x=1.

Under current usage, is that blocked? If so, how?

Here's an example that I need shot down, as in, given the current usage of "function", why is this wrong?

Let f(x) = 0, when x=1, and f(x) = 1, when x=1.

Under current usage, is that blocked? If so, how?

## JSH: Looking at math resistance now

Back in 2002 I found out that a mathematical analysis technique I'd discovered that I've come to call tautological spaces lead me to a stunning error in "core" mathematics, and arguments have ensued ever since and a math journal was even destroyed (Google: SWJPAM), but the arguments continue and I think one problem has been that people suppose that I must be wrong in that case.

So I've recently used—in integers:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients.

The entire point of the construction is to show that 9 must be a factor of only one of the f's even though theoretically you might think that it can split up! Where of course that is easy enough to imagine I'd think as 9 = 3(3).

Should be a trivial exercise but you can now see arguments erupting disputing it!!! And begin to understand how I could have found a core error back in 2002 and been arguing about it ever since, as the problem is: people just say no.

And when people just say no, it doesn't matter if in other contexts they claim to be mathematicians or claim that mathematical proof matters to them or anything else they claim as it's a groupthink exercise that is about group power: the ability to just refuse to accept an uncomfortable truth.

Why is it a big deal then? Well 9 follows as a factor of one of the f's by the distributive property, as a*(b+c) = a*b+ a*c, means that you have:

9*(b+c) = 9*b + 9*c, and 9*(b+1) = 9*b + 9, and if b is a function g(x), and you have it 0 at x=0:

9(g(0) + 1) = 9*0 + 9 = 0 + 9, and in general 9*(g(x) + 1) = 9*g(x) + 9.

So the result follows by the distributive property.

But if I move beyond integers to algebraic integers I can break modern number theory.

Crush it. Take away Galois Theory. And upend over a hundred years of mathematical efforts.

So you see posters fight it, and they've fought it for years now. Possibly with the complicity of mathematicians smart enough to realize—yeah, it follows by the distributive property!

So why would they?

Well the result is brutal. If you're a "pure math" mathematician with no applied mathematical results then it can simply remove all of your "accomplishments" showing them to just be false beliefs—a human artifact.

And now consider, if you're a "top" mathematician, say at Harvard University, and ALL of your supposed accomplishments that got you there go away, then why are you still there? Why would you expect to stay there if people found out?

And if you are Harvard University, how might such a situation impact how people think of you as an institution?

The emperor has no clothes. Oh, gee, isn't there some story about that somewhere? A parable? Yes, of course there is.

And you're seeing a modern example in all its glory.

They've fought since 2002 and seem to have no desire to ever stop. Not ever.

The emperor prefers to walk around naked.

[A reply to someone who claimed to have proved in another thread that a certain assertion by James was false.]

I defined mathematical proof.

I can define functions, if necessary.

No one will accept your examples down the line as disproof of the problem, as the functions I use to blow away the ring of algebraic integers aren't human choice things, like the crap you tossed out.

To me what you did isn't even close to mathematics. It's a human being saying he can choose.

Sometimes you can't choose.

So I've recently used—in integers:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients.

The entire point of the construction is to show that 9 must be a factor of only one of the f's even though theoretically you might think that it can split up! Where of course that is easy enough to imagine I'd think as 9 = 3(3).

Should be a trivial exercise but you can now see arguments erupting disputing it!!! And begin to understand how I could have found a core error back in 2002 and been arguing about it ever since, as the problem is: people just say no.

And when people just say no, it doesn't matter if in other contexts they claim to be mathematicians or claim that mathematical proof matters to them or anything else they claim as it's a groupthink exercise that is about group power: the ability to just refuse to accept an uncomfortable truth.

Why is it a big deal then? Well 9 follows as a factor of one of the f's by the distributive property, as a*(b+c) = a*b+ a*c, means that you have:

9*(b+c) = 9*b + 9*c, and 9*(b+1) = 9*b + 9, and if b is a function g(x), and you have it 0 at x=0:

9(g(0) + 1) = 9*0 + 9 = 0 + 9, and in general 9*(g(x) + 1) = 9*g(x) + 9.

So the result follows by the distributive property.

But if I move beyond integers to algebraic integers I can break modern number theory.

Crush it. Take away Galois Theory. And upend over a hundred years of mathematical efforts.

So you see posters fight it, and they've fought it for years now. Possibly with the complicity of mathematicians smart enough to realize—yeah, it follows by the distributive property!

So why would they?

Well the result is brutal. If you're a "pure math" mathematician with no applied mathematical results then it can simply remove all of your "accomplishments" showing them to just be false beliefs—a human artifact.

And now consider, if you're a "top" mathematician, say at Harvard University, and ALL of your supposed accomplishments that got you there go away, then why are you still there? Why would you expect to stay there if people found out?

And if you are Harvard University, how might such a situation impact how people think of you as an institution?

The emperor has no clothes. Oh, gee, isn't there some story about that somewhere? A parable? Yes, of course there is.

And you're seeing a modern example in all its glory.

They've fought since 2002 and seem to have no desire to ever stop. Not ever.

The emperor prefers to walk around naked.

[A reply to someone who claimed to have proved in another thread that a certain assertion by James was false.]

I defined mathematical proof.

I can define functions, if necessary.

No one will accept your examples down the line as disproof of the problem, as the functions I use to blow away the ring of algebraic integers aren't human choice things, like the crap you tossed out.

To me what you did isn't even close to mathematics. It's a human being saying he can choose.

Sometimes you can't choose.

### Friday, September 10, 2010

## JSH: Using rationals to understand the error

Mathematics is easy, explaining to people who hate a result, can be damned hard. One bizarre thing that has emerged in the last couple of days is that using the ring of integers and also the field of rationals with 9 instead of 7, can block all the various tricks that posters have used to try and obfuscate a stunning error at the heart of modern number theory.

Damn strange though. Problem seems to be that a lot of people focus on 7 being prime, when that is irrelevant. So I use 9, and it flummoxes posters. Oddity of the human brain. So now I'll explain one of the most far-reaching and tragic errors in human thought.

Consider in integers:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients.

The 9 is forced into one of the f's, though symmetry actually means you do not know which one.

By the distributive property, if you have 9(g_1(x) + 1) = (9*g_1(x) + 9) = (f_1(x) + 9) choosing f_1 to match indices, you know that x=0 is not a special case, and that the result holds for all x.

The issue may seem trivial, but those people who argue with me incessantly want the 9 to be able to split up. So that maybe one of the f's has 3 as a factor, for certain values of x, but that is nonsensical by the distributive property.

One way to FORCE a split is to use something like:

9(g_1(x) + 1)(g_2(x) + 2) = (3f_1(x) + 9)(3f_2(x) + 9) = 9*P(x)

which pushes you out of integers and into rationals.

Notice that FORCING 3 to be a factor of both, splitting up the 9 by human conscious effort, makes the mathematics shift out of the ring of integers, as it has no choice. You've then given it no choice.

So what does any of the above have to do with error?

Well if you understand why 9 is a factor of just one of the f's and realize it's not about factorization, but about the distributive property, well the distributive property is valid all the way up to the field of complex numbers, so the result should hold all the way up to the field of complex numbers, and it does hold in the field of complex numbers.

But it does NOT always hold in the ring of algebraic integers.

Proving that took some clever algebraic manipulations, where I used non-polynomial factors.

And that is the easy explanation.

So the result holds by the distributive property through the complex plane with one exception being the ring of algebraic integers. Explaining exactly what happens there is where things get actually interesting as that ring has to do some clever things to stick with mathematical rules!

And those things are not hard to understand. The 9 can't split up, so the ring of algebraic integers masks it with factors that behave as units, though because of an arbitrary rule they are NOT units in the ring of algebraic integers.

What arbitrary rule?

Well the rule that algebraic integers are roots of monic polynomials with integer coefficients induces some subtle requirements, for instance it prevents any non-rational unit in the ring of algebraic integers from being the root of anything but a monic polynomial with integer coefficients where the last is 1 or -1.

Notice integers do not have that problem as consider:

x^2 + 3x + 2 = (x+2)(x+1)

where a unit is paired with a non-unit in the ring of integers.

The corresponding case with non-rational algebraic integer solutions is NOT allowed by the ring of algebraic integers.

And that is weird. It turns out that having only roots of monic polynomials with integer coefficients leaves out numbers, and is a display property, which leads us to Galois Theory.

Galois Theory is correct in showing you what solutions can be displayed by integers, +, -, and radicals. And is correct in telling you when numbers cannot be so displayed. But that is just a display issue.

To the mathematics it's a non-issue as the math doesn't see sqrt(7). It sees two numbers that give 7 when squared. Human beings cannot get beyond the square root there, but may say +/- sqrt(7) or put up rational approximations like:

sqrt(7) approximately equals: 2.64575131

So to us it's a big deal if we can use radical or not, but to the mathematics it means nothing, so the mathematics doesn't see Galois Theory as doing anything at all.

From today's perspective though it kind of makes sense that over a hundred years ago to the mathematicians then, it may have seemed a big deal whether or not you could actually display a solution, which is why the quintic proof was a big deal, but to us it really doesn't mean much. Like, even with the general solution for cubics or quartics the answers you get are fairly useless and can actually be integers, obscured by radicals.

So why would people argue with me incessantly for years—I first discovered this problem in 2002—and modern mathematicians continue as if the issue does not exist?

Good question! Supposedly they wouldn't. But they have. More remarkably as my body of research has grown the protests by Usenet posters against it have remained steady, but spread to fairly odd things like claiming that Google search results are meaningless.

And those things are scary in a way. We have a world of 6.8 billion people. A lot of them are now on-line. In that world, my research is often coming out as #1. I look at search results where my ideas are at the top out of hundreds of millions of searches, and it's a weird feeling.

So what gives? How come I'm not famous? My best guess is that, um, I actually am? But that social order is more important than individual human beings. For instance, how many of you are worth, say, your country? None of you are, which is why we have soldiers. People go to fight to die for the good of the whole.

The disruptive social impact of this result—like on people's opinions of universities and university professors—seems for the moment to be balanced against the value society sees in keeping the current ideas about same in place!

So it's more important to world society I'm guessing to keep universities valued as they are, rather than have correct mathematics in this area, and the students sacrificed in mathematics aren't nearly important enough in that equation.

And I'm not losing anything myself as I have the freedom to keep researching and live a normal life, when by all rights I should be one of the most famous people on the planet now, and living a really abnormal one.

So the sacrifices here are the college students. The world considers them less valuable than the social systems that are in place, but over time the information is going to be more valuable. But it's up to the world to balance that equation.

Worse though for me is that I get bigger as a historical figure the longer it takes! So I'm less inclined to bother with it all anyway, but am probably stuck at this point.

So yeah, your world is smarter than any and all of you. And it has a world to run. None of you are worth the entire world in the balance so any one or all of you can be potentially sacrificed by your world for its own benefit.

After all, it can always make more people!

The cold calculation is not refutable.

Damn strange though. Problem seems to be that a lot of people focus on 7 being prime, when that is irrelevant. So I use 9, and it flummoxes posters. Oddity of the human brain. So now I'll explain one of the most far-reaching and tragic errors in human thought.

Consider in integers:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients.

The 9 is forced into one of the f's, though symmetry actually means you do not know which one.

By the distributive property, if you have 9(g_1(x) + 1) = (9*g_1(x) + 9) = (f_1(x) + 9) choosing f_1 to match indices, you know that x=0 is not a special case, and that the result holds for all x.

The issue may seem trivial, but those people who argue with me incessantly want the 9 to be able to split up. So that maybe one of the f's has 3 as a factor, for certain values of x, but that is nonsensical by the distributive property.

One way to FORCE a split is to use something like:

9(g_1(x) + 1)(g_2(x) + 2) = (3f_1(x) + 9)(3f_2(x) + 9) = 9*P(x)

which pushes you out of integers and into rationals.

Notice that FORCING 3 to be a factor of both, splitting up the 9 by human conscious effort, makes the mathematics shift out of the ring of integers, as it has no choice. You've then given it no choice.

So what does any of the above have to do with error?

Well if you understand why 9 is a factor of just one of the f's and realize it's not about factorization, but about the distributive property, well the distributive property is valid all the way up to the field of complex numbers, so the result should hold all the way up to the field of complex numbers, and it does hold in the field of complex numbers.

But it does NOT always hold in the ring of algebraic integers.

Proving that took some clever algebraic manipulations, where I used non-polynomial factors.

And that is the easy explanation.

So the result holds by the distributive property through the complex plane with one exception being the ring of algebraic integers. Explaining exactly what happens there is where things get actually interesting as that ring has to do some clever things to stick with mathematical rules!

And those things are not hard to understand. The 9 can't split up, so the ring of algebraic integers masks it with factors that behave as units, though because of an arbitrary rule they are NOT units in the ring of algebraic integers.

What arbitrary rule?

Well the rule that algebraic integers are roots of monic polynomials with integer coefficients induces some subtle requirements, for instance it prevents any non-rational unit in the ring of algebraic integers from being the root of anything but a monic polynomial with integer coefficients where the last is 1 or -1.

Notice integers do not have that problem as consider:

x^2 + 3x + 2 = (x+2)(x+1)

where a unit is paired with a non-unit in the ring of integers.

The corresponding case with non-rational algebraic integer solutions is NOT allowed by the ring of algebraic integers.

And that is weird. It turns out that having only roots of monic polynomials with integer coefficients leaves out numbers, and is a display property, which leads us to Galois Theory.

Galois Theory is correct in showing you what solutions can be displayed by integers, +, -, and radicals. And is correct in telling you when numbers cannot be so displayed. But that is just a display issue.

To the mathematics it's a non-issue as the math doesn't see sqrt(7). It sees two numbers that give 7 when squared. Human beings cannot get beyond the square root there, but may say +/- sqrt(7) or put up rational approximations like:

sqrt(7) approximately equals: 2.64575131

So to us it's a big deal if we can use radical or not, but to the mathematics it means nothing, so the mathematics doesn't see Galois Theory as doing anything at all.

From today's perspective though it kind of makes sense that over a hundred years ago to the mathematicians then, it may have seemed a big deal whether or not you could actually display a solution, which is why the quintic proof was a big deal, but to us it really doesn't mean much. Like, even with the general solution for cubics or quartics the answers you get are fairly useless and can actually be integers, obscured by radicals.

So why would people argue with me incessantly for years—I first discovered this problem in 2002—and modern mathematicians continue as if the issue does not exist?

Good question! Supposedly they wouldn't. But they have. More remarkably as my body of research has grown the protests by Usenet posters against it have remained steady, but spread to fairly odd things like claiming that Google search results are meaningless.

And those things are scary in a way. We have a world of 6.8 billion people. A lot of them are now on-line. In that world, my research is often coming out as #1. I look at search results where my ideas are at the top out of hundreds of millions of searches, and it's a weird feeling.

So what gives? How come I'm not famous? My best guess is that, um, I actually am? But that social order is more important than individual human beings. For instance, how many of you are worth, say, your country? None of you are, which is why we have soldiers. People go to fight to die for the good of the whole.

The disruptive social impact of this result—like on people's opinions of universities and university professors—seems for the moment to be balanced against the value society sees in keeping the current ideas about same in place!

So it's more important to world society I'm guessing to keep universities valued as they are, rather than have correct mathematics in this area, and the students sacrificed in mathematics aren't nearly important enough in that equation.

And I'm not losing anything myself as I have the freedom to keep researching and live a normal life, when by all rights I should be one of the most famous people on the planet now, and living a really abnormal one.

So the sacrifices here are the college students. The world considers them less valuable than the social systems that are in place, but over time the information is going to be more valuable. But it's up to the world to balance that equation.

Worse though for me is that I get bigger as a historical figure the longer it takes! So I'm less inclined to bother with it all anyway, but am probably stuck at this point.

So yeah, your world is smarter than any and all of you. And it has a world to run. None of you are worth the entire world in the balance so any one or all of you can be potentially sacrificed by your world for its own benefit.

After all, it can always make more people!

The cold calculation is not refutable.

## JSH: But does it matter?

I feel a certain responsibility to try and explain how math students are being screwed by the refusal to acknowledge a certain math error, but posters seem to think I personally need the result known, when that's silly. My position as a major discoverer is strengthened by the denial.

Worse, as I know my ideas travel the world, I can get a sense of where that can lead which doesn't require that things move quickly.

But, um, more importantly, I kind of do ok anyway, so it's not like I necessarily need my life turned upside down by fame.

I've had the argument with myself quite a bit, and most of the time I kind of lean towards, ok, I'll do enough as I can see where to go, but thank God for the time to think! The time to plan. The time to watch "celebs" and wonder what to do with the attention.

Fame is not easy. You lose a lot of privacy. You enter a different world.

Once I go through that door, it's not like I can turn around and change my mind!

Worse I don't get celebrity like some movie star, rock star or actor. They quit looking for attention and the world will go away.

With me, the attention will only grow. I won't be able to walk away. Won't be able to say I don't want it.

All these names of famous people that you know, all the celebs of today have an out. I won't.

I think it important that humanity does correct mathematics. But there is a price for me to pay. As I could just wander off and do other things. It IS tempting. Has been tempting for years.

To just walk away now, and let things happen in the fullness of time, like that 2030 date I tossed out a while back. That's kind of like the fantasy. Ok world, how about I take 20 more years, eh? Will it really hurt the human species that much for me to do my own thing, to live a normal life, to be free of fame and attention for 20 more years?

If not, then hey, maybe that can happen! I know that's what I'd like, but then I think about those students whose lives will be thrown away in the process. Dare I sacrifice all those young minds? If it's my choice then, no.

But so far it has been the world's choice, and the world has said they are expendable.

And if the world says they are expendable, then they are.

Worse, as I know my ideas travel the world, I can get a sense of where that can lead which doesn't require that things move quickly.

But, um, more importantly, I kind of do ok anyway, so it's not like I necessarily need my life turned upside down by fame.

I've had the argument with myself quite a bit, and most of the time I kind of lean towards, ok, I'll do enough as I can see where to go, but thank God for the time to think! The time to plan. The time to watch "celebs" and wonder what to do with the attention.

Fame is not easy. You lose a lot of privacy. You enter a different world.

Once I go through that door, it's not like I can turn around and change my mind!

Worse I don't get celebrity like some movie star, rock star or actor. They quit looking for attention and the world will go away.

With me, the attention will only grow. I won't be able to walk away. Won't be able to say I don't want it.

All these names of famous people that you know, all the celebs of today have an out. I won't.

I think it important that humanity does correct mathematics. But there is a price for me to pay. As I could just wander off and do other things. It IS tempting. Has been tempting for years.

To just walk away now, and let things happen in the fullness of time, like that 2030 date I tossed out a while back. That's kind of like the fantasy. Ok world, how about I take 20 more years, eh? Will it really hurt the human species that much for me to do my own thing, to live a normal life, to be free of fame and attention for 20 more years?

If not, then hey, maybe that can happen! I know that's what I'd like, but then I think about those students whose lives will be thrown away in the process. Dare I sacrifice all those young minds? If it's my choice then, no.

But so far it has been the world's choice, and the world has said they are expendable.

And if the world says they are expendable, then they are.

### Wednesday, September 08, 2010

## JSH: Using composites to understand the arguments

For years I've used 7 as my favorite constant with equations that I use to try and show a remarkable problem with established number theory, which has for years left open the door for posters to claim that primeness is critical, so here's an explanation with composites! The 7 will be replaced by 9.

Consider:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients.

Those rules force one of the f's to have 9 as a factor when the f's are polynomials.

Trivially if the f's are linear functions 9 is a factor of only one of them, with those rules.

That follows by the distributive property. But if so, then the f's being linear is immaterial, right? So

Posters replying here should explain:

mathematicians in any way shape or form, shouldn't they be able to answer all 3 questions?

They aren't trick questions! There is no hiding of some obfuscation or some weird psychological trick.

Let's see what actually happens then.

Oh, my point is that 9 is the factor, but provably in the ring of algebraic integers you can have f's where NEITHER of the f's can have 9 as a factor. Some posters pointed out primeness with my 7, so I'm using 9. Curious readers might try to split it up, say, into 3 being a factor of each of the f's just to see if you can—with the rules given about zeros.

If I'm right, then I can collapse much of modern number theory, and show that Galois Theory isn't really useful after all, so a lot at stake.

Ok, enough chatter. Let's see what happens…

Consider:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients.

Those rules force one of the f's to have 9 as a factor when the f's are polynomials.

Trivially if the f's are linear functions 9 is a factor of only one of them, with those rules.

That follows by the distributive property. But if so, then the f's being linear is immaterial, right? So

**whatever**types of functions they are, then 9 should be the factor, by the distributive property.Posters replying here should explain:

- Is 9 the factor as stated above, and if so, why?
- If they believe it can split up, can they demonstrate with integers?
- Why would it matter if the f's are non-polynomials? Shouldn't 9 still be the factor?

mathematicians in any way shape or form, shouldn't they be able to answer all 3 questions?

They aren't trick questions! There is no hiding of some obfuscation or some weird psychological trick.

Let's see what actually happens then.

Oh, my point is that 9 is the factor, but provably in the ring of algebraic integers you can have f's where NEITHER of the f's can have 9 as a factor. Some posters pointed out primeness with my 7, so I'm using 9. Curious readers might try to split it up, say, into 3 being a factor of each of the f's just to see if you can—with the rules given about zeros.

If I'm right, then I can collapse much of modern number theory, and show that Galois Theory isn't really useful after all, so a lot at stake.

Ok, enough chatter. Let's see what happens…

### Tuesday, September 07, 2010

## JSH: Understanding the arguments

Years ago I found a rather fascinating problem with the use of algebraic integers which is so upsetting that despite the ease of its proof, mainstream mathematicians have as far as I can tell almost completely ignored it, except for a brief publication of one of my early papers in the now defunct journal SWJPAM, where the editors yanked my paper after publication.

For those who wonder what the "crackpot" is arguing about with these people over and over again, year after year, I have this thread with a simple explanation using VERY elementary expressions and almost no complicated math, so that you too can understand the arguments.

Consider:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic with integer coefficients, g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0.

That construct is setup deliberately by me to indicate that the 7 only multiplies times g_1(x) + 1, so one of the f's has 7 as a factor, which is trivially true when you have polynomials. For instance, just change the functions to x:

7(x + 1)(x + 2) = (7x + 7)(x + 7) = 7*P(x)

and notice the 7 is FORCED to show up for one of the f's, if you are to stay in the ring of algebraic integers.

What I did though was figure out how to make the f's non-polynomial functions, with the same rules about 0, BUT if you try to put everything in the ring of algebraic integers, you find that there are situations where NEITHER of the f's can have 7 as a factor.

And THAT is the gist of it.

So how can people argue for years on such a thing? Well I say—there must be something wrong with the ring of algebraic integers then! Other posters say I'm wrong—or nastier things—and claim that there is nothing wrong with it.

And round and round we go!

Notice their idea is that somehow the FUNCTIONS can change how the 7 multiplies and often they would say that using x=0 is specious because they'd claim it was a "special case". And they'd go on and on about how the 7 could "split up" how it multiplied with:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

depending on the value of x.

I'd say the "tail does not wag the dog" and that the distributive property doesn't allow what's being multiplied to control how it is multiplied and that what is true for linear functions—where there is no argument—doesn't change because you shift to non-linear, non- polynomial functions, and they'd claim I was wrong!!! Or they'd claim it wasn't the distributive property. Or they'd claim anything rather than agree with me.

With a journal dead already, where I think it was over this issue, and with mainstream mathematicians not doing anything that I can notice, posters on Usenet produce most of the discussion of which I am aware, and there seems to be an established feeling that nothing will change soon!

Some of the posters who reply to me ask then why I bother to repeat the arguments that I say prove I'm right, if the group will never accept them, and of course, if I believe humanity will never accept the mathematical truth then human progress in mathematics ends.

Because the ideas here are trivially easy. There is no way the functions can push back outward to change how that 7 multiplies, so these errors are real, but well established. Worse, "pure math" arose—if you check you history—about the same time that the ring of algebraic integers arose, because I suggest, of the error!

So if this battle is lost, number theory progress for the human species, ends.

Which it may. So those who know the error is real, know that here and now they may be seeing the end of humanity's rise of knowledge in this area. The end of the growth that this species has had for thousands of years in mathematics.

The end of mathematical discovery happening right in front of them.

And how? Human nature is quirky. There is no rule that humanity must always progress necessarily as one day the human species itself will die. It will go extinct. So for those in the know, there may be an odd feeling at seeing one part of the human story end here and now. We at least can mourn its passing.

For those of us then, the discussion is worth having. And for us, hoping that humanity has a little more juice within it, is a hope worth having.

For the others though, the soul of mathematics is dead.

For those who wonder what the "crackpot" is arguing about with these people over and over again, year after year, I have this thread with a simple explanation using VERY elementary expressions and almost no complicated math, so that you too can understand the arguments.

Consider:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic with integer coefficients, g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0.

That construct is setup deliberately by me to indicate that the 7 only multiplies times g_1(x) + 1, so one of the f's has 7 as a factor, which is trivially true when you have polynomials. For instance, just change the functions to x:

7(x + 1)(x + 2) = (7x + 7)(x + 7) = 7*P(x)

and notice the 7 is FORCED to show up for one of the f's, if you are to stay in the ring of algebraic integers.

What I did though was figure out how to make the f's non-polynomial functions, with the same rules about 0, BUT if you try to put everything in the ring of algebraic integers, you find that there are situations where NEITHER of the f's can have 7 as a factor.

And THAT is the gist of it.

So how can people argue for years on such a thing? Well I say—there must be something wrong with the ring of algebraic integers then! Other posters say I'm wrong—or nastier things—and claim that there is nothing wrong with it.

And round and round we go!

Notice their idea is that somehow the FUNCTIONS can change how the 7 multiplies and often they would say that using x=0 is specious because they'd claim it was a "special case". And they'd go on and on about how the 7 could "split up" how it multiplied with:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

depending on the value of x.

I'd say the "tail does not wag the dog" and that the distributive property doesn't allow what's being multiplied to control how it is multiplied and that what is true for linear functions—where there is no argument—doesn't change because you shift to non-linear, non- polynomial functions, and they'd claim I was wrong!!! Or they'd claim it wasn't the distributive property. Or they'd claim anything rather than agree with me.

With a journal dead already, where I think it was over this issue, and with mainstream mathematicians not doing anything that I can notice, posters on Usenet produce most of the discussion of which I am aware, and there seems to be an established feeling that nothing will change soon!

Some of the posters who reply to me ask then why I bother to repeat the arguments that I say prove I'm right, if the group will never accept them, and of course, if I believe humanity will never accept the mathematical truth then human progress in mathematics ends.

Because the ideas here are trivially easy. There is no way the functions can push back outward to change how that 7 multiplies, so these errors are real, but well established. Worse, "pure math" arose—if you check you history—about the same time that the ring of algebraic integers arose, because I suggest, of the error!

So if this battle is lost, number theory progress for the human species, ends.

Which it may. So those who know the error is real, know that here and now they may be seeing the end of humanity's rise of knowledge in this area. The end of the growth that this species has had for thousands of years in mathematics.

The end of mathematical discovery happening right in front of them.

And how? Human nature is quirky. There is no rule that humanity must always progress necessarily as one day the human species itself will die. It will go extinct. So for those in the know, there may be an odd feeling at seeing one part of the human story end here and now. We at least can mourn its passing.

For those of us then, the discussion is worth having. And for us, hoping that humanity has a little more juice within it, is a hope worth having.

For the others though, the soul of mathematics is dead.

### Sunday, September 05, 2010

## JSH: Algebraic integers and Galois Theory

So I've used a simple expression to try and help math students comprehend a VERY important error in "core".

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic with integer coefficients, g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0.

A simple exercise reveals that the 7 cannot split up in any way to give:

(f_1(x) + 7)(f_2(x) + 7)

in the middle, which is a BIG DEAL because the f's can be algebraic integers that are roots of a polynomial irreducible over Q, and the result should force only one of them to have 7 as a factor, but PROVABLY neither can in that situation in the ring of algebraic integers!!!

For years posters insulted me in this situation claiming they could arithmetically show that I was wrong, as they'd trot out results where the 7 had been divided off, they thought—having split up in a way that is commonly believed.

But the correct answer requires a complete ring, which is my ring of objects. I discovered the ring of objects BECAUSE of this problem. Now consider unit factors u_1 and u_2 in the ring of objects which are NOT units in the ring of algebraic integers. If you say, but that's not right, see why it is the only possibility allowed, as now you can have:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

as being unit factors they can spontaneously arise, and NOW divide off the 7 to get:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/7 + u_1)(f_2(x)*u_2 + 7*u_2) = P(x)

and u_1 is an algebraic integer that would be found as a solution, which is NOT a unit in the ring of algebraic integers, while 7u_2 IS an algebraic integer, but u_2 is NOT an algebraic integer, so you have the 7 wrapped up neatly.

You know that is the only possibility because if the 7 cannot split up, and 7 cannot be a factor in the ring of algebraic integers for a particular case then that's the only way mathematically to avoid a contradiction. The 7 gets wrapped up.

And you need unit factors to do it!

So why is u_1 not a unit in the ring of algebraic integers? Because u_1 and 7u_2 are roots of a monic quadratic with integer coefficients! That's why!

Turns out that the ring of algebraic integers will not allow a number to be a unit unless it is a root of a monic polynomial with integer coefficients that has 1 or -1 as the last coefficient, so it throws out u_2 from the ring, and doesn't consider u_1 to be a unit within the ring.

Clever.

All of the above has implications for Galois Theory as then, Galois Theory doesn't tell you when 7 really is or is not a factor of one of the roots! And all the machinery of Galois Theory isn't giving you useful information when it looks like the 7 is "split up" as you consider different algebraic integer factors appearing as you try to divide that 7 off.

What you are seeing then are UNIT FACTORS that are appearing as needed to wrap up the actual factors. So the real factors are being wrapped up and you learn NOTHING from Galois Theory.

When I figured this out several years ago I noted that you can do the same things with integer solutions to understand what Galois Theory is actually telling you—by not taking the square root.

For instance with x^2 +3x + 2 = 0, you'd have

x = (-3 +/- sqrt(1))/2

and if you refuse to take sqrt(1)—I know sounds silly but use your imagination—then you can use all the mechanisms of Galois theory and get a class whatever, and do all those things, but notice if you do the same with:

x^2 + 7x + 10 = 0, you'd have

x = (-7 +/- sqrt(9))/2

and a different class number, but so what? Remember NOT taking the square root.

So that is MEANINGLESS. It's just noticing that you're NOT taking the square root, and ignores the reality that one of the x's is the same in both cases.

With non-rationals where you can't take the square root the mathematics is simply acknowledging that, and giving you no additional information.

So Galois Theory tells you nothing about the underlying factors. Nothing at all.

And supposed solutions where you try to divide off integers like 7, are simply unit factors wrapping around the underlying factors, which remain hidden.

Years of arguing with posters who tried to fight the distributive property. And claimed they didn't. But to try and attack the above— serious math people are welcome and encouraged to try—you need what I call coupling.

So with:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

you need (g_1(x) + 1)(g_2(x) + 2) to somehow push BACK out against the 7 and tell it how to multiply in order to get:

(f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2)

doing so as a function of x, as at x=0, you're forced to have 7(g_1(x) + 1) = 7*g_1(x) + 7, by the distributive property, as at that value g_1(0) = 0, and one of the f's equals 0, which requires that solution!!!

So it should be set, right? Well they'd argue that it was a special case!!! And that the 7 multiplied *differently* for differing values of x.

Which I call coupling. The 7 then becomes a function of x, though some weird mechanism they argued about endlessly which would frustrate me so much that I'd come up with names for it, like voodoo mathematics.

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic with integer coefficients, g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0.

A simple exercise reveals that the 7 cannot split up in any way to give:

(f_1(x) + 7)(f_2(x) + 7)

in the middle, which is a BIG DEAL because the f's can be algebraic integers that are roots of a polynomial irreducible over Q, and the result should force only one of them to have 7 as a factor, but PROVABLY neither can in that situation in the ring of algebraic integers!!!

For years posters insulted me in this situation claiming they could arithmetically show that I was wrong, as they'd trot out results where the 7 had been divided off, they thought—having split up in a way that is commonly believed.

But the correct answer requires a complete ring, which is my ring of objects. I discovered the ring of objects BECAUSE of this problem. Now consider unit factors u_1 and u_2 in the ring of objects which are NOT units in the ring of algebraic integers. If you say, but that's not right, see why it is the only possibility allowed, as now you can have:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

as being unit factors they can spontaneously arise, and NOW divide off the 7 to get:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/7 + u_1)(f_2(x)*u_2 + 7*u_2) = P(x)

and u_1 is an algebraic integer that would be found as a solution, which is NOT a unit in the ring of algebraic integers, while 7u_2 IS an algebraic integer, but u_2 is NOT an algebraic integer, so you have the 7 wrapped up neatly.

You know that is the only possibility because if the 7 cannot split up, and 7 cannot be a factor in the ring of algebraic integers for a particular case then that's the only way mathematically to avoid a contradiction. The 7 gets wrapped up.

And you need unit factors to do it!

So why is u_1 not a unit in the ring of algebraic integers? Because u_1 and 7u_2 are roots of a monic quadratic with integer coefficients! That's why!

Turns out that the ring of algebraic integers will not allow a number to be a unit unless it is a root of a monic polynomial with integer coefficients that has 1 or -1 as the last coefficient, so it throws out u_2 from the ring, and doesn't consider u_1 to be a unit within the ring.

Clever.

All of the above has implications for Galois Theory as then, Galois Theory doesn't tell you when 7 really is or is not a factor of one of the roots! And all the machinery of Galois Theory isn't giving you useful information when it looks like the 7 is "split up" as you consider different algebraic integer factors appearing as you try to divide that 7 off.

What you are seeing then are UNIT FACTORS that are appearing as needed to wrap up the actual factors. So the real factors are being wrapped up and you learn NOTHING from Galois Theory.

When I figured this out several years ago I noted that you can do the same things with integer solutions to understand what Galois Theory is actually telling you—by not taking the square root.

For instance with x^2 +3x + 2 = 0, you'd have

x = (-3 +/- sqrt(1))/2

and if you refuse to take sqrt(1)—I know sounds silly but use your imagination—then you can use all the mechanisms of Galois theory and get a class whatever, and do all those things, but notice if you do the same with:

x^2 + 7x + 10 = 0, you'd have

x = (-7 +/- sqrt(9))/2

and a different class number, but so what? Remember NOT taking the square root.

So that is MEANINGLESS. It's just noticing that you're NOT taking the square root, and ignores the reality that one of the x's is the same in both cases.

With non-rationals where you can't take the square root the mathematics is simply acknowledging that, and giving you no additional information.

So Galois Theory tells you nothing about the underlying factors. Nothing at all.

And supposed solutions where you try to divide off integers like 7, are simply unit factors wrapping around the underlying factors, which remain hidden.

Years of arguing with posters who tried to fight the distributive property. And claimed they didn't. But to try and attack the above— serious math people are welcome and encouraged to try—you need what I call coupling.

So with:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

you need (g_1(x) + 1)(g_2(x) + 2) to somehow push BACK out against the 7 and tell it how to multiply in order to get:

(f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2)

doing so as a function of x, as at x=0, you're forced to have 7(g_1(x) + 1) = 7*g_1(x) + 7, by the distributive property, as at that value g_1(0) = 0, and one of the f's equals 0, which requires that solution!!!

So it should be set, right? Well they'd argue that it was a special case!!! And that the 7 multiplied *differently* for differing values of x.

Which I call coupling. The 7 then becomes a function of x, though some weird mechanism they argued about endlessly which would frustrate me so much that I'd come up with names for it, like voodoo mathematics.

### Saturday, September 04, 2010

## JSH: Maybe I was luckier

As a child I had indoctrinated into me the ideas of fundamentalist Christianity that are peculiar to the sect of Jehovah's Witnesses, which among others things taught that evolution was wrong, and that the earth had been created by God in a few thousand years—a little better than other fundamentalist Christians who say exactly 7 days.

Letting go of what I'd been taught was a necessity for me to move on and grow in this world. Even my act of joining the military was an act of rebellion against a religion that teaches its practitioners to be "separate from the world". (Oh yeah, it also teaches that all nations are ruled literally by Satan the Devil. Ok, one wonders if maybe that one was right—)

Oh, and I got my degree in physics, so the idea of overturning results was part of the science that I learned.

Math students are taught differently and now some can see from my results that they can get taught dogma that is little better than the ideas of any religion, and over the years as I've tried to get them to accept mathematical truth, I've had time to contemplate how hard it can be for people to learn that deeply held beliefs are just wrong.

So maybe I was luckier.

I can walk through the math error easily. Talk about my object ring and unit factors that wrap to give the appearance that is false and even see something fascinating in how the ring of algebraic integers has to follow mathematical rules to force it to do what it does.

But rigid mathematical rules care nothing for human social needs, and human beings for years have just thumbed their noses at them, and played pretend.

I wonder at times how they feel but sort of know. After all there were times as a kid I believed in the religion I was taught. I know what it feels like. And with a controversial religion that would get challenged by others, I know what it feels like to hold on to your beliefs against challenges.

But it's still sad, even understanding why. So yes, I was luckier. And I do understand.

It's so hard to believe, really, truly believe, and then one day, find out that you believed wrong.

Letting go of what I'd been taught was a necessity for me to move on and grow in this world. Even my act of joining the military was an act of rebellion against a religion that teaches its practitioners to be "separate from the world". (Oh yeah, it also teaches that all nations are ruled literally by Satan the Devil. Ok, one wonders if maybe that one was right—)

Oh, and I got my degree in physics, so the idea of overturning results was part of the science that I learned.

Math students are taught differently and now some can see from my results that they can get taught dogma that is little better than the ideas of any religion, and over the years as I've tried to get them to accept mathematical truth, I've had time to contemplate how hard it can be for people to learn that deeply held beliefs are just wrong.

So maybe I was luckier.

I can walk through the math error easily. Talk about my object ring and unit factors that wrap to give the appearance that is false and even see something fascinating in how the ring of algebraic integers has to follow mathematical rules to force it to do what it does.

But rigid mathematical rules care nothing for human social needs, and human beings for years have just thumbed their noses at them, and played pretend.

I wonder at times how they feel but sort of know. After all there were times as a kid I believed in the religion I was taught. I know what it feels like. And with a controversial religion that would get challenged by others, I know what it feels like to hold on to your beliefs against challenges.

But it's still sad, even understanding why. So yes, I was luckier. And I do understand.

It's so hard to believe, really, truly believe, and then one day, find out that you believed wrong.

## JSH: So do they know?

The error with algebraic integers is so easily demonstrated, and so easily explained that it's hard to believe that people with decades of experience in mathematics wouldn't know it does exist and understand exactly how it functions.

And if so they've known for years.

I think you can wonder a lot and puzzle over a lot of things trying to comprehend how they've been operating with the error, and I've come up with various theories.

But make no mistake, if they DO KNOW then these are highly placed and often powerful people, even at top universities like Harvard and Yale who are fully aware they are teaching their students false information.

Fully aware? God help them. How can they? How could human beings do that to other human beings?

And if so they've known for years.

I think you can wonder a lot and puzzle over a lot of things trying to comprehend how they've been operating with the error, and I've come up with various theories.

But make no mistake, if they DO KNOW then these are highly placed and often powerful people, even at top universities like Harvard and Yale who are fully aware they are teaching their students false information.

Fully aware? God help them. How can they? How could human beings do that to other human beings?

## JSH: Decoupling and algebraic integers

Simple expressions can show a problem with the ring of algebraic integers with minimal difficulty. My efforts in this regard are meant to help math students with a difficult concept—a substantial error in "core".

A primary tool is a simple set of equations:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic polynomial with integer coefficients, and g_1(0) = g_2(0) = 0, and at least one of the f's equals 0 as well when x = 0.

As a preparatory exercise one might consider ways to split the 7 up on the left-hand side in order to get it yet again in the middle, given those rules.

For instance imagine 7 splits into r_1 and r_2, where, say y^2 + by + 7 = 0.

It is trivial to prove that r_1 or r_2 must equal 7 itself, which allows demonstration of a direct contradiction, and as I've argued about that for years I'll leave out some details to note how posters have tried to answer.

They've claimed that the r's must be functions of x, so you have r_1(x) and r_2(x), where it is merely a "special case" at x=0 that one of them equals 7.

Remarkably one can actually prove that in the ring of algebraic integers you CAN find cases where neither of the r's can be 7, which posters have claimed bolsters their case with arithmetic. For years they have then claimed that my refusal to accept that argument is simply a refusal to accept the truth.

Puzzling over the situation years ago I realized there was a simple answer as I noted that given:

7*(f_1(x) + 1) = 7*f_1(x) + 7

that the VALUE of f_1(x) didn't matter to the distributive property, so why should it matter above? Yet it is mathematically correct that the ring of algebraic integers will give you cases where 7 is NOT ALLOWED to be a factor!

So I figured out the object ring:

The object ring is defined by two conditions, and includes all numbers such that these conditions are true:

So now consider in the object ring, unit factors u_1 and u_2, and:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

The unit factors can, of course, move at will and now can "wrap" the 7's, as they are not algebraic integers.

So you get the appearance that 7 is not a factor, even though it is in the complete object ring, so the ring of algebraic integers gives you false information.

What's interesting about that explanation is that it is easy. The distributive property does not allow the 7 to be split up, which is trivial to prove. Coupling the 7 to x makes no sense, except as a dodge of the result! 7 is a constant. It's not a function of x.

So why would number theorists ignore such a result if it's so easily explained and represents an error in "core"?

Well, one possibility is that with the error they can appear to prove anything. So yeah, for instance, Andrew Wiles can claim to have proven FLT. If he worked hard enough he can claim to have disproven it—though that should collapse on the issue of counterexample.

With the error you might claim proof of the Riemann Hypothesis. And someone else with it can claim to have disproven it, with the same problem of a counterexample.

Quite simply, with the error you might be able to appear to prove just about anything.

As long as you pick carefully, and have a group that works with you in this effort, you can do what you wish, give prizes as you wish, control people's careers as you wish.

It gives you complete control because with this flawed mathematics—you can make it say whatever you want.

And for certain kinds of people that could be VERY appealing.

They do control your career if you are a math student. While they have the error, they have you as well.

They OWN your career with this error.

A primary tool is a simple set of equations:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic polynomial with integer coefficients, and g_1(0) = g_2(0) = 0, and at least one of the f's equals 0 as well when x = 0.

As a preparatory exercise one might consider ways to split the 7 up on the left-hand side in order to get it yet again in the middle, given those rules.

For instance imagine 7 splits into r_1 and r_2, where, say y^2 + by + 7 = 0.

It is trivial to prove that r_1 or r_2 must equal 7 itself, which allows demonstration of a direct contradiction, and as I've argued about that for years I'll leave out some details to note how posters have tried to answer.

They've claimed that the r's must be functions of x, so you have r_1(x) and r_2(x), where it is merely a "special case" at x=0 that one of them equals 7.

Remarkably one can actually prove that in the ring of algebraic integers you CAN find cases where neither of the r's can be 7, which posters have claimed bolsters their case with arithmetic. For years they have then claimed that my refusal to accept that argument is simply a refusal to accept the truth.

Puzzling over the situation years ago I realized there was a simple answer as I noted that given:

7*(f_1(x) + 1) = 7*f_1(x) + 7

that the VALUE of f_1(x) didn't matter to the distributive property, so why should it matter above? Yet it is mathematically correct that the ring of algebraic integers will give you cases where 7 is NOT ALLOWED to be a factor!

So I figured out the object ring:

The object ring is defined by two conditions, and includes all numbers such that these conditions are true:

- 1 and -1 are the only rationals that are units in the ring.
- Given a member m of the ring there must exist a non-zero member n

such that mn is an integer, and if mn is not a factor of m, then n

cannot be a unit in the ring.

So now consider in the object ring, unit factors u_1 and u_2, and:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

The unit factors can, of course, move at will and now can "wrap" the 7's, as they are not algebraic integers.

So you get the appearance that 7 is not a factor, even though it is in the complete object ring, so the ring of algebraic integers gives you false information.

What's interesting about that explanation is that it is easy. The distributive property does not allow the 7 to be split up, which is trivial to prove. Coupling the 7 to x makes no sense, except as a dodge of the result! 7 is a constant. It's not a function of x.

So why would number theorists ignore such a result if it's so easily explained and represents an error in "core"?

Well, one possibility is that with the error they can appear to prove anything. So yeah, for instance, Andrew Wiles can claim to have proven FLT. If he worked hard enough he can claim to have disproven it—though that should collapse on the issue of counterexample.

With the error you might claim proof of the Riemann Hypothesis. And someone else with it can claim to have disproven it, with the same problem of a counterexample.

Quite simply, with the error you might be able to appear to prove just about anything.

As long as you pick carefully, and have a group that works with you in this effort, you can do what you wish, give prizes as you wish, control people's careers as you wish.

It gives you complete control because with this flawed mathematics—you can make it say whatever you want.

And for certain kinds of people that could be VERY appealing.

They do control your career if you are a math student. While they have the error, they have you as well.

They OWN your career with this error.

## JSH: Algebraic integers and Galois Theory

So I've used a simple expression to try and help math students comprehend a VERY important error in "core".

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic with integer coefficients, g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0.

A simple exercise reveals that the 7 cannot split up in any way to give:

(f_1(x) + 7)(f_2(x) + 7)

in the middle, which is a BIG DEAL because the f's can be algebraic integers that are roots of a polynomial irreducible over Q, and the result should force only one of them to have 7 as a factor, but PROVABLY neither can in that situation in the ring of algebraic integers!!!

For years posters insulted me in this situation claiming they could arithmetically show that I was wrong, as they'd trot out results where the 7 had been divided off, they thought—having split up in a way that is commonly believed.

But the correct answer requires a complete ring, which is my ring of objects. I discovered the ring of objects BECAUSE of this problem. Now consider unit factors u_1 and u_2 in the ring of objects which are NOT units in the ring of algebraic integers. If you say, but that's not right, see why it is the only possibility allowed, as now you can have:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

as being unit factors they can spontaneously arise, and NOW divide off the 7 to get:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/7 + u_1)(f_2(x)*u_2 + 7*u_2) = P(x)

and u_1 is an algebraic integer that would be found as a solution, which is NOT a unit in the ring of algebraic integers, while 7u_2 IS an algebraic integer, but u_2 is NOT an algebraic integer, so you have the 7 wrapped up neatly.

You know that is the only possibility because if the 7 cannot split up, and 7 cannot be a factor in the ring of algebraic integers for a particular case then that's the only way mathematically to avoid a contradiction. The 7 gets wrapped up.

And you need unit factors to do it!

So why is u_1 not a unit in the ring of algebraic integers? Because u_1 and 7u_2 are roots of a monic quadratic with integer coefficients! That's why!

Turns out that the ring of algebraic integers will not allow a number to be a unit unless it is a root of a monic polynomial with integer coefficients that has 1 or -1 as the last coefficient, so it throws out u_2 from the ring, and doesn't consider u_1 to be a unit within the ring.

Clever.

All of the above has implications for Galois Theory as then, Galois Theory doesn't tell you when 7 really is or is not a factor of one of the roots! And all the machinery of Galois Theory isn't giving you useful information when it looks like the 7 is "split up" as you consider different algebraic integer factors appearing as you try to divide that 7 off.

What you are seeing then are UNIT FACTORS that are appearing as needed to wrap up the actual factors. So the real factors are being wrapped up and you learn NOTHING from Galois Theory.

When I figured this out several years ago I noted that you can do the same things with integer solutions to understand what Galois Theory is actually telling you—by not taking the square root.

For instance with x^2 +3x + 2 = 0, you'd have

x = (-3 +/- sqrt(1))/2

and if you refuse to take sqrt(1)—I know sounds silly but use your imagination—then you can use all the mechanisms of Galois theory and get a class whatever, and do all those things, but notice if you do the same with:

x^2 + 7x + 10 = 0, you'd have

x = (-7 +/- sqrt(9))/2

and a different class number, but so what? Remember NOT taking the square root.

So that is MEANINGLESS. It's just noticing that you're NOT taking the square root, and ignores the reality that one of the x's is the same in both cases.

With non-rationals where you can't take the square root the mathematics is simply acknowledging that, and giving you no additional information.

So Galois Theory tells you nothing about the underlying factors. Nothing at all.

And supposed solutions where you try to divide off integers like 7, are simply unit factors wrapping around the underlying factors, which remain hidden.

Years of arguing with posters who tried to fight the distributive property. And claimed they didn't. But to try and attack the above— serious math people are welcome and encouraged to try—you need what I call coupling.

So with:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

you need (g_1(x) + 1)(g_2(x) + 2) to somehow push BACK out against the 7 and tell it how to multiply in order to get:

(f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2)

doing so as a function of x, as at x=0, you're forced to have 7(g_1(x) + 1) = 7*g_1(x) + 7, by the distributive property, as at that value g_1(0) = 0, and one of the f's equals 0, which requires that solution!!!

So it should be set, right? Well they'd argue that it was a special case!!! And that the 7 multiplied *differently* for differing values of x.

Which I call coupling. The 7 then becomes a function of x, though some weird mechanism they argued about endlessly which would frustrate me so much that I'd come up with names for it, like voodoo mathematics.

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic with integer coefficients, g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0.

A simple exercise reveals that the 7 cannot split up in any way to give:

(f_1(x) + 7)(f_2(x) + 7)

in the middle, which is a BIG DEAL because the f's can be algebraic integers that are roots of a polynomial irreducible over Q, and the result should force only one of them to have 7 as a factor, but PROVABLY neither can in that situation in the ring of algebraic integers!!!

For years posters insulted me in this situation claiming they could arithmetically show that I was wrong, as they'd trot out results where the 7 had been divided off, they thought—having split up in a way that is commonly believed.

But the correct answer requires a complete ring, which is my ring of objects. I discovered the ring of objects BECAUSE of this problem. Now consider unit factors u_1 and u_2 in the ring of objects which are NOT units in the ring of algebraic integers. If you say, but that's not right, see why it is the only possibility allowed, as now you can have:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

as being unit factors they can spontaneously arise, and NOW divide off the 7 to get:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/7 + u_1)(f_2(x)*u_2 + 7*u_2) = P(x)

and u_1 is an algebraic integer that would be found as a solution, which is NOT a unit in the ring of algebraic integers, while 7u_2 IS an algebraic integer, but u_2 is NOT an algebraic integer, so you have the 7 wrapped up neatly.

You know that is the only possibility because if the 7 cannot split up, and 7 cannot be a factor in the ring of algebraic integers for a particular case then that's the only way mathematically to avoid a contradiction. The 7 gets wrapped up.

And you need unit factors to do it!

So why is u_1 not a unit in the ring of algebraic integers? Because u_1 and 7u_2 are roots of a monic quadratic with integer coefficients! That's why!

Turns out that the ring of algebraic integers will not allow a number to be a unit unless it is a root of a monic polynomial with integer coefficients that has 1 or -1 as the last coefficient, so it throws out u_2 from the ring, and doesn't consider u_1 to be a unit within the ring.

Clever.

All of the above has implications for Galois Theory as then, Galois Theory doesn't tell you when 7 really is or is not a factor of one of the roots! And all the machinery of Galois Theory isn't giving you useful information when it looks like the 7 is "split up" as you consider different algebraic integer factors appearing as you try to divide that 7 off.

What you are seeing then are UNIT FACTORS that are appearing as needed to wrap up the actual factors. So the real factors are being wrapped up and you learn NOTHING from Galois Theory.

When I figured this out several years ago I noted that you can do the same things with integer solutions to understand what Galois Theory is actually telling you—by not taking the square root.

For instance with x^2 +3x + 2 = 0, you'd have

x = (-3 +/- sqrt(1))/2

and if you refuse to take sqrt(1)—I know sounds silly but use your imagination—then you can use all the mechanisms of Galois theory and get a class whatever, and do all those things, but notice if you do the same with:

x^2 + 7x + 10 = 0, you'd have

x = (-7 +/- sqrt(9))/2

and a different class number, but so what? Remember NOT taking the square root.

So that is MEANINGLESS. It's just noticing that you're NOT taking the square root, and ignores the reality that one of the x's is the same in both cases.

With non-rationals where you can't take the square root the mathematics is simply acknowledging that, and giving you no additional information.

So Galois Theory tells you nothing about the underlying factors. Nothing at all.

And supposed solutions where you try to divide off integers like 7, are simply unit factors wrapping around the underlying factors, which remain hidden.

Years of arguing with posters who tried to fight the distributive property. And claimed they didn't. But to try and attack the above— serious math people are welcome and encouraged to try—you need what I call coupling.

So with:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

you need (g_1(x) + 1)(g_2(x) + 2) to somehow push BACK out against the 7 and tell it how to multiply in order to get:

(f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2)

doing so as a function of x, as at x=0, you're forced to have 7(g_1(x) + 1) = 7*g_1(x) + 7, by the distributive property, as at that value g_1(0) = 0, and one of the f's equals 0, which requires that solution!!!

So it should be set, right? Well they'd argue that it was a special case!!! And that the 7 multiplied *differently* for differing values of x.

Which I call coupling. The 7 then becomes a function of x, though some weird mechanism they argued about endlessly which would frustrate me so much that I'd come up with names for it, like voodoo mathematics.

## JSH: Algebraic integers and Galois Theory

So I've used a simple expression to try and help math students comprehend a VERY important error in "core".

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic with integer coefficients, g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0.

A simple exercise reveals that the 7 cannot split up in any way to give:

(f_1(x) + 7)(f_2(x) + 7)

in the middle, which is a BIG DEAL because the f's can be algebraic integers that are roots of a polynomial irreducible over Q, and the result should force only one of them to have 7 as a factor, but PROVABLY neither can in that situation in the ring of algebraic integers!!!

For years posters insulted me in this situation claiming they could arithmetically show that I was wrong, as they'd trot out results where the 7 had been divided off, they thought—having split up in a way that is commonly believed.

But the correct answer requires a complete ring, which is my ring of objects. I discovered the ring of objects BECAUSE of this problem. Now consider unit factors u_1 and u_2 in the ring of objects which are NOT units in the ring of algebraic integers. If you say, but that's not right, see why it is the only possibility allowed, as now you can have:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

as being unit factors they can spontaneously arise, and NOW divide off the 7 to get:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/7 + u_1)(f_2(x)*u_2 + 7*u_2) = P(x)

and u_1 is an algebraic integer that would be found as a solution, which is NOT a unit in the ring of algebraic integers, while 7u_2 IS an algebraic integer, but u_2 is NOT an algebraic integer, so you have the 7 wrapped up neatly.

You know that is the only possibility because if the 7 cannot split up, and 7 cannot be a factor in the ring of algebraic integers for a particular case then that's the only way mathematically to avoid a contradiction. The 7 gets wrapped up.

And you need unit factors to do it!

So why is u_1 not a unit in the ring of algebraic integers? Because u_1 and 7u_2 are roots of a monic quadratic with integer coefficients! That's why!

Turns out that the ring of algebraic integers will not allow a number to be a unit unless it is a root of a monic polynomial with integer coefficients that has 1 or -1 as the last coefficient, so it throws out u_2 from the ring, and doesn't consider u_1 to be a unit within the ring.

Clever.

All of the above has implications for Galois Theory as then, Galois Theory doesn't tell you when 7 really is or is not a factor of one of the roots! And all the machinery of Galois Theory isn't giving you useful information when it looks like the 7 is "split up" as you consider different algebraic integer factors appearing as you try to divide that 7 off.

What you are seeing then are UNIT FACTORS that are appearing as needed to wrap up the actual factors. So the real factors are being wrapped up and you learn NOTHING from Galois Theory.

When I figured this out several years ago I noted that you can do the same things with integer solutions to understand what Galois Theory is actually telling you—by not taking the square root.

For instance with x^2 +3x + 2 = 0, you'd have

x = (-3 +/- sqrt(1))/2

and if you refuse to take sqrt(1)—I know sounds silly but use your imagination—then you can use all the mechanisms of Galois theory and get a class whatever, and do all those things, but notice if you do the same with:

x^2 + 7x + 10 = 0, you'd have

x = (-7 +/- sqrt(9))/2

and a different class number, but so what? Remember NOT taking the square root.

So that is MEANINGLESS. It's just noticing that you're NOT taking the square root, and ignores the reality that one of the x's is the same in both cases.

With non-rationals where you can't take the square root the mathematics is simply acknowledging that, and giving you no additional information.

So Galois Theory tells you nothing about the underlying factors. Nothing at all.

And supposed solutions where you try to divide off integers like 7, are simply unit factors wrapping around the underlying factors, which remain hidden.

Years of arguing with posters who tried to fight the distributive property. And claimed they didn't. But to try and attack the above— serious math people are welcome and encouraged to try—you need what I call coupling.

So with:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

you need (g_1(x) + 1)(g_2(x) + 2) to somehow push BACK out against the 7 and tell it how to multiply in order to get:

(f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2)

doing so as a function of x, as at x=0, you're forced to have 7(g_1(x) + 1) = 7*g_1(x) + 7, by the distributive property, as at that value g_1(0) = 0, and one of the f's equals 0, which requires that solution!!!

So it should be set, right? Well they'd argue that it was a special case!!! And that the 7 multiplied *differently* for differing values of x.

Which I call coupling. The 7 then becomes a function of x, though some weird mechanism they argued about endlessly which would frustrate me so much that I'd come up with names for it, like voodoo mathematics.

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic with integer coefficients, g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0.

A simple exercise reveals that the 7 cannot split up in any way to give:

(f_1(x) + 7)(f_2(x) + 7)

in the middle, which is a BIG DEAL because the f's can be algebraic integers that are roots of a polynomial irreducible over Q, and the result should force only one of them to have 7 as a factor, but PROVABLY neither can in that situation in the ring of algebraic integers!!!

For years posters insulted me in this situation claiming they could arithmetically show that I was wrong, as they'd trot out results where the 7 had been divided off, they thought—having split up in a way that is commonly believed.

But the correct answer requires a complete ring, which is my ring of objects. I discovered the ring of objects BECAUSE of this problem. Now consider unit factors u_1 and u_2 in the ring of objects which are NOT units in the ring of algebraic integers. If you say, but that's not right, see why it is the only possibility allowed, as now you can have:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

as being unit factors they can spontaneously arise, and NOW divide off the 7 to get:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/7 + u_1)(f_2(x)*u_2 + 7*u_2) = P(x)

and u_1 is an algebraic integer that would be found as a solution, which is NOT a unit in the ring of algebraic integers, while 7u_2 IS an algebraic integer, but u_2 is NOT an algebraic integer, so you have the 7 wrapped up neatly.

You know that is the only possibility because if the 7 cannot split up, and 7 cannot be a factor in the ring of algebraic integers for a particular case then that's the only way mathematically to avoid a contradiction. The 7 gets wrapped up.

And you need unit factors to do it!

So why is u_1 not a unit in the ring of algebraic integers? Because u_1 and 7u_2 are roots of a monic quadratic with integer coefficients! That's why!

Turns out that the ring of algebraic integers will not allow a number to be a unit unless it is a root of a monic polynomial with integer coefficients that has 1 or -1 as the last coefficient, so it throws out u_2 from the ring, and doesn't consider u_1 to be a unit within the ring.

Clever.

All of the above has implications for Galois Theory as then, Galois Theory doesn't tell you when 7 really is or is not a factor of one of the roots! And all the machinery of Galois Theory isn't giving you useful information when it looks like the 7 is "split up" as you consider different algebraic integer factors appearing as you try to divide that 7 off.

What you are seeing then are UNIT FACTORS that are appearing as needed to wrap up the actual factors. So the real factors are being wrapped up and you learn NOTHING from Galois Theory.

When I figured this out several years ago I noted that you can do the same things with integer solutions to understand what Galois Theory is actually telling you—by not taking the square root.

For instance with x^2 +3x + 2 = 0, you'd have

x = (-3 +/- sqrt(1))/2

and if you refuse to take sqrt(1)—I know sounds silly but use your imagination—then you can use all the mechanisms of Galois theory and get a class whatever, and do all those things, but notice if you do the same with:

x^2 + 7x + 10 = 0, you'd have

x = (-7 +/- sqrt(9))/2

and a different class number, but so what? Remember NOT taking the square root.

So that is MEANINGLESS. It's just noticing that you're NOT taking the square root, and ignores the reality that one of the x's is the same in both cases.

With non-rationals where you can't take the square root the mathematics is simply acknowledging that, and giving you no additional information.

So Galois Theory tells you nothing about the underlying factors. Nothing at all.

And supposed solutions where you try to divide off integers like 7, are simply unit factors wrapping around the underlying factors, which remain hidden.

Years of arguing with posters who tried to fight the distributive property. And claimed they didn't. But to try and attack the above— serious math people are welcome and encouraged to try—you need what I call coupling.

So with:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

you need (g_1(x) + 1)(g_2(x) + 2) to somehow push BACK out against the 7 and tell it how to multiply in order to get:

(f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2)

doing so as a function of x, as at x=0, you're forced to have 7(g_1(x) + 1) = 7*g_1(x) + 7, by the distributive property, as at that value g_1(0) = 0, and one of the f's equals 0, which requires that solution!!!

So it should be set, right? Well they'd argue that it was a special case!!! And that the 7 multiplied *differently* for differing values of x.

Which I call coupling. The 7 then becomes a function of x, though some weird mechanism they argued about endlessly which would frustrate me so much that I'd come up with names for it, like voodoo mathematics.

### Friday, September 03, 2010

## JSH: Routinely beating mathematicians

One thing worth mentioning—again—is that now I'm routinely beating even supposedly top mathematicians in their own fields when it comes to web search results. (Great fun of course for a supposed "crank" or "crackpot". I can beat them head-to-head, toe-to-toe. My research crushing theirs. THAT is sweet victory.)

One of my favorite examples.

Search: binary quadratic Diophantine equations

I like that esoteric example as posters ROUTINELY—people who nobody really knows of course as who else would try?—claim that I'm simply giving highly specific searches that link back to me for that reason. Or they claim that Google is biased towards me for various often bizarre and I'd say nonsensical reasons, for the searches that I note primarily seem to come up highly in Google.

In case after case you can compare me to any elite researcher in the world and note his or her position.

Of course web searches are not scholarly journals or rankings within the math industry itself. Big deal. If I'm the guy that most people are directed to then in time I will BE the establishment.

It is just about time.

That's how it has always worked historically. Always.

The Old Guard fights it, of course, and they always lose.

More search results with which to consider:

definition of mathematical proof

prime residue axiom

solving residues

object ring

REALITY for many a math student as she or he goes to class today is that when they do searches on the web on various math topics—if they do I don't know if they do—then my research may often be appearing there as well.

So why aren't mathematicians howling to the search engines asking them to correct their search rankings?

One of my favorite examples.

Search: binary quadratic Diophantine equations

I like that esoteric example as posters ROUTINELY—people who nobody really knows of course as who else would try?—claim that I'm simply giving highly specific searches that link back to me for that reason. Or they claim that Google is biased towards me for various often bizarre and I'd say nonsensical reasons, for the searches that I note primarily seem to come up highly in Google.

In case after case you can compare me to any elite researcher in the world and note his or her position.

Of course web searches are not scholarly journals or rankings within the math industry itself. Big deal. If I'm the guy that most people are directed to then in time I will BE the establishment.

It is just about time.

That's how it has always worked historically. Always.

The Old Guard fights it, of course, and they always lose.

More search results with which to consider:

definition of mathematical proof

prime residue axiom

solving residues

object ring

REALITY for many a math student as she or he goes to class today is that when they do searches on the web on various math topics—if they do I don't know if they do—then my research may often be appearing there as well.

So why aren't mathematicians howling to the search engines asking them to correct their search rankings?