### Sunday, October 26, 2003

## Explaining math definition problem

I'm an independent researcher, which means that I use my

Getting important research findings is one thing, and getting them noticed, is another.

At least here on Usenet I can talk freely to people around the world.

What I'd like to explain is my disturbing and to me fascinating finding of a problem with a math definition that's over a hundred years old. In looking over various replies to my previous posts on this subject, I've seen assertions that definitions can't cause problems, which is something that I can address quickly at the start.

Over a hundred years ago, the great German mathematicians Karl Gauss played with numbers of the form a+bi, where 'a' and 'b' are integers. In his honor they were later called gaussian integers, though a number like 1+2i is not an integer. The "gaussian" up-front is important. Later mathematicians came up with other numbers they called algebraic integers, which include gaussian integers.

They thought they'd found THE set, or superset you might call it, which includes all numbers with certain special properties of integrality.

The most important property to point out is the ability to have primeness between numbers.

For instance, with integers, 2 and 3 are coprime, that is, they don't share non-unit factors, that is, factors other than 1 or -1, with each other.

Just be clear here, factors of 1, are called units or unit factors.

But notice that with rationals, you have 2(3/2) = 3, so 2 and 3

What Gauss had started considering, which other mathematicians extended, was the idea of sets of numbers where you kept interesting properties of the set of integers, like being able to say two numbers were coprime.

What I've found is a problem with their set of algebraic integers, as unfortunately, despite what many mathematicians think, it's too small.

That's it. The definition they use is too small to do what they think it does, which is include all these interesting numbers with special properties.

But because they

It's like when the Greeks with their word "atom" thought they had the smallest thing, and later our civilization used it, and broke atoms apart, though part of the definition is that they are

Now my research finding isn't hard to show quickly in broad strokes.

On of my important analysis tools is a simple technique to factor polynomials into non-polynomial factors.

For instance, with the polynomial

P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078

that technique gives you

P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3

so I can factor to get

P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).

where the a's are the roots of the cubic

a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).

Now despite the complexity, you can rely on

P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)

as the cubic defining the a's with x=0 is

a^3 - 3a^2, which has roots, 0, 0 and 3.

You may not realize it, but what you just saw is revolutionary, both in the special techniques, and most importantly with the consequences that quickly follow.

That's because P(x) has another special feature as

P(x) = 49(300125 x^3 - 18375 x^2 + 360 x + 22)

where that 49 is just begging to be divided off, which gives, of course,

P(x)/49 = 300125 x^3 - 18375 x^2 + 360 x + 22.

But remember, my three factors with the a's from before had

Here the principle is like if you have

S(x) = 7x^2 + 14x + 7 = (7x+7)(x+1)

in that setting x=0 gives you

S(0) = (7(0) + 7)(0 + 1) = 7(1).

The point is that the 7 is constant, so x's value means nothing to it.

So from before with

P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)

I know that dividing through by 49, it must go like

P(0)/49 = (5(0)/7 + 1)(5(0)/7 + 1)(5(3) + 7) = 1(1)(22)

and as the constant terms are

P(x)/49 = (5a_1/7 + 1)(5a_2/7 + 1)(5a_3 + 7).

The problem now though is that conclusion can be used to show that unequivocally beyond any reasonable doubt the definition of algebraic numbers is TOO SMALL, as at times 5a_1/7 and 5a_2/7 are not included.

You see, they get left out, which is a problem because from the

Some of you may find yourselves fearful of using your own mathematical understanding, if you realize I'm right, and then realize that mathematicians are disputing the result, especially if you see posters tossing out far more complicated math in reply to my post, but remember, math isn't magic.

Logic rules mathematics, so look for what makes sense. And remember that you can't assume that posters are on your side. I don't want you to assume that I'm on your side either.

You see, I don't need you to assume anything, as what I need you to do is check.

While some mathematicians may erroneously believe now that it's in their interest to hide the problem I've revealed, that mistake in thinking does not help the rest of the world. After all, what good does it do everyone else for mathematicians to hide their definition problem?

What's in it for you?

**own**funding, and my**own**direction to go out and see what knowledge I can obtain. Some of my research has been in the area of mathematics.Getting important research findings is one thing, and getting them noticed, is another.

At least here on Usenet I can talk freely to people around the world.

What I'd like to explain is my disturbing and to me fascinating finding of a problem with a math definition that's over a hundred years old. In looking over various replies to my previous posts on this subject, I've seen assertions that definitions can't cause problems, which is something that I can address quickly at the start.

Over a hundred years ago, the great German mathematicians Karl Gauss played with numbers of the form a+bi, where 'a' and 'b' are integers. In his honor they were later called gaussian integers, though a number like 1+2i is not an integer. The "gaussian" up-front is important. Later mathematicians came up with other numbers they called algebraic integers, which include gaussian integers.

They thought they'd found THE set, or superset you might call it, which includes all numbers with certain special properties of integrality.

The most important property to point out is the ability to have primeness between numbers.

For instance, with integers, 2 and 3 are coprime, that is, they don't share non-unit factors, that is, factors other than 1 or -1, with each other.

Just be clear here, factors of 1, are called units or unit factors.

But notice that with rationals, you have 2(3/2) = 3, so 2 and 3

**do**share a factor and are not coprime in that ring, which is typically called a field**because**every element except 0, has a multiplicative inverse.What Gauss had started considering, which other mathematicians extended, was the idea of sets of numbers where you kept interesting properties of the set of integers, like being able to say two numbers were coprime.

What I've found is a problem with their set of algebraic integers, as unfortunately, despite what many mathematicians think, it's too small.

That's it. The definition they use is too small to do what they think it does, which is include all these interesting numbers with special properties.

But because they

**think**it's big enough, mathematicians have an error in their discipline based on their false assumption, as they've come up with more arguments based on that assumption, which then aren't actually proven.It's like when the Greeks with their word "atom" thought they had the smallest thing, and later our civilization used it, and broke atoms apart, though part of the definition is that they are

**indivisible**, as people can define things, and later**refine**their definitions.Now my research finding isn't hard to show quickly in broad strokes.

On of my important analysis tools is a simple technique to factor polynomials into non-polynomial factors.

For instance, with the polynomial

P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078

that technique gives you

P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3

so I can factor to get

P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).

where the a's are the roots of the cubic

a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).

Now despite the complexity, you can rely on

**simple**ideas still, by noticing that setting x=0, pulls out constant terms, asP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)

as the cubic defining the a's with x=0 is

a^3 - 3a^2, which has roots, 0, 0 and 3.

You may not realize it, but what you just saw is revolutionary, both in the special techniques, and most importantly with the consequences that quickly follow.

That's because P(x) has another special feature as

P(x) = 49(300125 x^3 - 18375 x^2 + 360 x + 22)

where that 49 is just begging to be divided off, which gives, of course,

P(x)/49 = 300125 x^3 - 18375 x^2 + 360 x + 22.

But remember, my three factors with the a's from before had

**constant**terms of 7, 7 and 22, so dividing by 49 must give constant terms of 1, 1, and 22, which is the result that is so earth shattering.Here the principle is like if you have

S(x) = 7x^2 + 14x + 7 = (7x+7)(x+1)

in that setting x=0 gives you

**constant**terms within the expression, which you can conveniently, also look at to see how it works.S(0) = (7(0) + 7)(0 + 1) = 7(1).

The point is that the 7 is constant, so x's value means nothing to it.

So from before with

P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)

I know that dividing through by 49, it must go like

P(0)/49 = (5(0)/7 + 1)(5(0)/7 + 1)(5(3) + 7) = 1(1)(22)

and as the constant terms are

**independent**of the value of x, it MUST be that in generalP(x)/49 = (5a_1/7 + 1)(5a_2/7 + 1)(5a_3 + 7).

The problem now though is that conclusion can be used to show that unequivocally beyond any reasonable doubt the definition of algebraic numbers is TOO SMALL, as at times 5a_1/7 and 5a_2/7 are not included.

You see, they get left out, which is a problem because from the

**assumption**of mathematicians, they should be included, if the ring of algebraic integers is the ring that mathematicians thought it was.Some of you may find yourselves fearful of using your own mathematical understanding, if you realize I'm right, and then realize that mathematicians are disputing the result, especially if you see posters tossing out far more complicated math in reply to my post, but remember, math isn't magic.

Logic rules mathematics, so look for what makes sense. And remember that you can't assume that posters are on your side. I don't want you to assume that I'm on your side either.

You see, I don't need you to assume anything, as what I need you to do is check.

While some mathematicians may erroneously believe now that it's in their interest to hide the problem I've revealed, that mistake in thinking does not help the rest of the world. After all, what good does it do everyone else for mathematicians to hide their definition problem?

What's in it for you?

### Monday, October 20, 2003

## Simple principle in core error proof

What makes my situation especially frustrating is how simple the argument is that proves there's this problem with algebraic integers, yet still I have to keep explaining.

The basic principle is like with

P(m) = 2(m^2 + 2m + 1) = (a_1 m + 2)(a_2 m + 1)

where most of you can probably guess what factor a_1 must have!!!

Now the polynomial I use is more complicated such that I need to set m=0 to figure things out, but notice here what happens:

P(0) = 2, and dividing off 2 from P(m) gives

P(m)/2 = m^2 + 2m + 1

and notice that P(0)/2 = 1, which tell you that the independent term changed.

Given

(a_1 m + 2)(a_2 m + 1)

it's clear that a_1 has a factor that is 2.

It's a simple idea that I use with a more complicated polynomial, and I think that the reason so many math people go out of their way to make it seem like it's wrong is that they're embarrassed by the over hundred year old error that I found.

I've been surprised at how dedicated they can be at trying to hide the truth.

After all, I've communicated with top mathematicians like Barry Mazur, Andrew Granville, and Ralph McKenzie, who may be people many of you haven't heard of, but in certain math circles they're well-known.

In McKenzie's case I explained it all to him

These mathematicians are hellbent on trying to hide that their discipline could actually have a flaw like this for as long as they can get away with it.

It's wrong, and it can't help world society.

Meanwhile it's not doing me a bit of good either, and I think that part of their motivation is making me miserable, knowing that what I found *should* get me accolades.

Yup, call me crazy, but I think the bastards are out to get me!!!

The basic principle is like with

P(m) = 2(m^2 + 2m + 1) = (a_1 m + 2)(a_2 m + 1)

where most of you can probably guess what factor a_1 must have!!!

Now the polynomial I use is more complicated such that I need to set m=0 to figure things out, but notice here what happens:

P(0) = 2, and dividing off 2 from P(m) gives

P(m)/2 = m^2 + 2m + 1

and notice that P(0)/2 = 1, which tell you that the independent term changed.

Given

(a_1 m + 2)(a_2 m + 1)

it's clear that a_1 has a factor that is 2.

It's a simple idea that I use with a more complicated polynomial, and I think that the reason so many math people go out of their way to make it seem like it's wrong is that they're embarrassed by the over hundred year old error that I found.

I've been surprised at how dedicated they can be at trying to hide the truth.

After all, I've communicated with top mathematicians like Barry Mazur, Andrew Granville, and Ralph McKenzie, who may be people many of you haven't heard of, but in certain math circles they're well-known.

In McKenzie's case I explained it all to him

**in-person**and he basically blew me off.These mathematicians are hellbent on trying to hide that their discipline could actually have a flaw like this for as long as they can get away with it.

It's wrong, and it can't help world society.

Meanwhile it's not doing me a bit of good either, and I think that part of their motivation is making me miserable, knowing that what I found *should* get me accolades.

Yup, call me crazy, but I think the bastards are out to get me!!!

### Sunday, October 12, 2003

## Finishing argument, core error proven

For me there have been two perspectives as I work to figure out how to explain the definition problem in mathematics with LOTS of opposition, and I wonder about mathematicians so dedicated to attacking an argument that is clearly correct.

I remind of that as I present what should finish their ability to distract, as I've seen a strange and dedicated effort to ignore the actual math, and simply toss up just about anything rather than face the truth.

All variables are in the ring of algebraic integers unless otherwise stated.

Let

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

and let

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

so P(m) = f^2 R(m).

Now consider

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

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

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

Then it must be true that the following factorization exists

R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf)

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

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

where a_3 = b_3, and at m=0, b_3 = 3.

PROOF:

From

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

at m=0, R(0) = u^2(3x + uf), and the cubic for the b's

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

gives b^3 - 3b^2 = 0, which gives that two of the b's are 0, while the third is 3.

That gives

R(0) = (0 x + u)(0 x + u)(3 x + uf) = u^2( 3x + uf)

as required.

No other primitive (non-primitive is like 2x+2 which has a factor of 2) cubic has the required roots, so the cubic

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

is the correct one.

Then it follows that two of the a's have a factor that is f.

However, it is possible to show that for certain integer values of f, and integer values of m, the a's do not

Mathematicians can continue to run from the truth, but they must keep attacking algebra itself to do so.

I remind of that as I present what should finish their ability to distract, as I've seen a strange and dedicated effort to ignore the actual math, and simply toss up just about anything rather than face the truth.

All variables are in the ring of algebraic integers unless otherwise stated.

Let

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

and let

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

so P(m) = f^2 R(m).

Now consider

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

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

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

Then it must be true that the following factorization exists

R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf)

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

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

where a_3 = b_3, and at m=0, b_3 = 3.

PROOF:

From

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

at m=0, R(0) = u^2(3x + uf), and the cubic for the b's

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

gives b^3 - 3b^2 = 0, which gives that two of the b's are 0, while the third is 3.

That gives

R(0) = (0 x + u)(0 x + u)(3 x + uf) = u^2( 3x + uf)

as required.

No other primitive (non-primitive is like 2x+2 which has a factor of 2) cubic has the required roots, so the cubic

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

is the correct one.

Then it follows that two of the a's have a factor that is f.

However, it is possible to show that for certain integer values of f, and integer values of m, the a's do not

**in the ring of algebraic integers**have f as a factor, which is a contradiction.Mathematicians can continue to run from the truth, but they must keep attacking algebra itself to do so.