### Tuesday, September 28, 2004

## Would NASA impress?

I get a lot of flak from all these posters and people seem to believe them!

I notice that Dr. Beckwith didn't seem to matter much to any of you.

What if I have someone from NASA?

Would that make a difference if a contact who works at NASA as a scientist came forward to post support for any of my work?

Or would you all just attack him as well?

I notice that Dr. Beckwith didn't seem to matter much to any of you.

What if I have someone from NASA?

Would that make a difference if a contact who works at NASA as a scientist came forward to post support for any of my work?

Or would you all just attack him as well?

### Friday, September 17, 2004

## Some math, algebraic integers

Luckily for me the math that explains my work is actually quite simple in many ways. All that is required is that you accept algebra. That might sound weird, but pay attention and you'll understand by the end.

Now with rationals you can have a simple case like

2x^2 + 5x - 3 = 0

where the roots are x=1/2 and x=-3, if I did my algebra correctly.

Now notice you have an integer paired with a fraction, which, of course, is NOT an integer.

Now consider some numbers q_1 and q_2, which I won't say a lot about now, though in a bit I'll put them in a particular ring, where you get a polynomial with

x = q_1/2 and x=-3q_2 as the roots.

That polynomial looks like

2x^2 + (6q_2 - q_1)x - 3q_1 q_2 = 0

and let q_1 q_2 = 1, and 6q_2 - q_1 be an integer.

It turns out that q_2 can't be an algebraic integer.

Mathematicians at this point make an assumption, which is that q_1 MUST have non-unit algebraic integer factors in common with 2.

Well, yeah, but the implication is wrong.

That might sound complicated or weird, but think about the rational example, where you have roots 1/2 and -3, and notice that -3 does not have any non-unit factors in common with 2, but somehow, someway if you have irrational factors, for some reason 3q_2 MUST have non-unit factors in common with 2?

It begs the imagination that despite the infinity of possibility that 3q_2 is so constrained, and in fact, it's not.

Mathematicians made a mistake.

It turns out that q_1 and q_2 CAN be in a ring where 1 and -1 are the only integers that are units, and to be more specific just in case some are still confused, where 1 is the only NATURAL that is a unit.

I figured out a rather basic way to prove that using some very simple algebra.

To date, no one had shown an error in my proof, though the paper in which I go through the proof did have at least on minor error that has been pointed out by a sci.math poster.

What I need is fairness. I can go through the details of that paper, point-by-point, step-by-step and show how it follows from basic algebra.

But some posters cheat, and they are the same posters who succeeded with that email campaign against my paper.

These people don't play. They are very good at manipulating the bulk of you, and they have an agenda at hiding the mathematical truth.

My challenge to you is to stand up for yourselves and actually THINK.

Time will tell if any of you have what it takes to be a real mathematician.

Here it will require that you focus on the math, and accept algebra, no matter what tricks posters throw at you.

So far, most of you have failed…but I have a little bit of faith left.

There may be hope yet.

Now with rationals you can have a simple case like

2x^2 + 5x - 3 = 0

where the roots are x=1/2 and x=-3, if I did my algebra correctly.

Now notice you have an integer paired with a fraction, which, of course, is NOT an integer.

Now consider some numbers q_1 and q_2, which I won't say a lot about now, though in a bit I'll put them in a particular ring, where you get a polynomial with

x = q_1/2 and x=-3q_2 as the roots.

That polynomial looks like

2x^2 + (6q_2 - q_1)x - 3q_1 q_2 = 0

and let q_1 q_2 = 1, and 6q_2 - q_1 be an integer.

It turns out that q_2 can't be an algebraic integer.

Mathematicians at this point make an assumption, which is that q_1 MUST have non-unit algebraic integer factors in common with 2.

Well, yeah, but the implication is wrong.

That might sound complicated or weird, but think about the rational example, where you have roots 1/2 and -3, and notice that -3 does not have any non-unit factors in common with 2, but somehow, someway if you have irrational factors, for some reason 3q_2 MUST have non-unit factors in common with 2?

It begs the imagination that despite the infinity of possibility that 3q_2 is so constrained, and in fact, it's not.

Mathematicians made a mistake.

It turns out that q_1 and q_2 CAN be in a ring where 1 and -1 are the only integers that are units, and to be more specific just in case some are still confused, where 1 is the only NATURAL that is a unit.

I figured out a rather basic way to prove that using some very simple algebra.

To date, no one had shown an error in my proof, though the paper in which I go through the proof did have at least on minor error that has been pointed out by a sci.math poster.

What I need is fairness. I can go through the details of that paper, point-by-point, step-by-step and show how it follows from basic algebra.

But some posters cheat, and they are the same posters who succeeded with that email campaign against my paper.

These people don't play. They are very good at manipulating the bulk of you, and they have an agenda at hiding the mathematical truth.

My challenge to you is to stand up for yourselves and actually THINK.

Time will tell if any of you have what it takes to be a real mathematician.

Here it will require that you focus on the math, and accept algebra, no matter what tricks posters throw at you.

So far, most of you have failed…but I have a little bit of faith left.

There may be hope yet.

### Friday, September 10, 2004

## Amateur math, neat relation

I'm going to give a derivation again, and talk about what I face with today's mathematicians so you can see how they operate and how they make certain that amateur mathematicians are cut out.

The simplest way to understand what I did is to consider the count of odd composites up to and including N that are divisible by 3, as

3C <= N

where C is the count of all composites, so C <= N/3, but every odd natural besides one can be written as 2k+1, where k is a natural, so you can replace C with 2k+1, to get

3(2k+1) <= N,

and solving you get

k <= (N-3)/6, and using floor(x) = [x], you have

k = [(N-3)/6] as the count of composites.

When I did my original research over two years ago, knowing that N was to be even, I only cared about odds so I used the equivalent of

3(2k+1) <= N-1

which gives k = [(N-4)/6], which does work as long as N is even.

Here the brackets mean to just take the integer part, like [1.3333...] = 1.

Now, notice that instead of 3 I can use any odd prime, so I have

p(2k+1)<=N, so k = [(N-p)/2p]

is the count of

BUT, you can also get that count by using [N/p] - [N/2p] - 1, so I have as a fundamental relationship that

[N/p] - [N/2p] - 1 = [(N-p)/2p]

and in general

[N/j] - [N/2j] - 1 = [(N-j)/2j]

where j is a natural greater than 1.

You can of course tweak it a bit so that you have

j(2k-1)<=N, so k = [(N+j)/2j]

and then the formula becomes

[N/j] - [N/2j] = [(N+j)/2j]

which was actually guessed by an anonymous poster a while back when I was talking about my "tidbits".

Some poster on sci.math made a post like that was a big deal, but in mathematics guessing is one thing, proving is another.

I've seen a lot of talk about these simple relations already, as if…as if mathematicians knew them before my work, and the discussions I started.

But that's how today's mathematicians operate.

They don't care about the math, only about what's said about the math.

They are all about prestige and holding on to their social position and the math be damned.

Go out, look in number theory texts or other math books and find that relation.

I hope you do find it, as if this is new then mathematicians are worse than pathetic, because if the people of sci.math who are still trying to push the idea that my work is useless or old never learned that relation but STILL are trying to downgrade it, then there's just no doubt that they hate mathematics.

Math is not a social game. People who truly love mathematics don't stop to first decide if they like someone before they accept a result.

That's the beauty of mathematics, but there are these people who call themselves mathematicians who other people call mathematicians who get paid, and they probably figure that if they don't play these social games then maybe they can't get their wife that car, or maybe they can't put their kid through school.

But they are willing to sacrifice you for their needs.

To them amateur mathematicians are worse than scum, and scarier than nuclear bombs.

But the math doesn't care about their mortgages. It doesn't care about their political needs. And it doesn't care about them.

Today's mathematicians have to hate mathematics because mathematics doesn't look out for them. It doesn't pay attention to their needs. It doesn't worry about their bills.

Mathematics just doesn't care, so they don't care about mathematics.

They may SAY they care, but that's just to get money and prestige.

Don't believe me? Then find

[N/j] - [N/2j] - 1 = [(N-j)/2j]

where j is a natural greater than 1, in a math textbook.

When you don't find it, understand the people you're dealing with, no matter what label they have on themselves.

They may call themselves mathematicians, but they hate mathematics.

The simplest way to understand what I did is to consider the count of odd composites up to and including N that are divisible by 3, as

3C <= N

where C is the count of all composites, so C <= N/3, but every odd natural besides one can be written as 2k+1, where k is a natural, so you can replace C with 2k+1, to get

3(2k+1) <= N,

and solving you get

k <= (N-3)/6, and using floor(x) = [x], you have

k = [(N-3)/6] as the count of composites.

When I did my original research over two years ago, knowing that N was to be even, I only cared about odds so I used the equivalent of

3(2k+1) <= N-1

which gives k = [(N-4)/6], which does work as long as N is even.

Here the brackets mean to just take the integer part, like [1.3333...] = 1.

Now, notice that instead of 3 I can use any odd prime, so I have

p(2k+1)<=N, so k = [(N-p)/2p]

is the count of

**odd**composites up to and including N that have p as a factor, where p is an odd prime.BUT, you can also get that count by using [N/p] - [N/2p] - 1, so I have as a fundamental relationship that

[N/p] - [N/2p] - 1 = [(N-p)/2p]

and in general

[N/j] - [N/2j] - 1 = [(N-j)/2j]

where j is a natural greater than 1.

You can of course tweak it a bit so that you have

j(2k-1)<=N, so k = [(N+j)/2j]

and then the formula becomes

[N/j] - [N/2j] = [(N+j)/2j]

which was actually guessed by an anonymous poster a while back when I was talking about my "tidbits".

Some poster on sci.math made a post like that was a big deal, but in mathematics guessing is one thing, proving is another.

I've seen a lot of talk about these simple relations already, as if…as if mathematicians knew them before my work, and the discussions I started.

But that's how today's mathematicians operate.

They don't care about the math, only about what's said about the math.

They are all about prestige and holding on to their social position and the math be damned.

Go out, look in number theory texts or other math books and find that relation.

I hope you do find it, as if this is new then mathematicians are worse than pathetic, because if the people of sci.math who are still trying to push the idea that my work is useless or old never learned that relation but STILL are trying to downgrade it, then there's just no doubt that they hate mathematics.

Math is not a social game. People who truly love mathematics don't stop to first decide if they like someone before they accept a result.

That's the beauty of mathematics, but there are these people who call themselves mathematicians who other people call mathematicians who get paid, and they probably figure that if they don't play these social games then maybe they can't get their wife that car, or maybe they can't put their kid through school.

But they are willing to sacrifice you for their needs.

To them amateur mathematicians are worse than scum, and scarier than nuclear bombs.

But the math doesn't care about their mortgages. It doesn't care about their political needs. And it doesn't care about them.

Today's mathematicians have to hate mathematics because mathematics doesn't look out for them. It doesn't pay attention to their needs. It doesn't worry about their bills.

Mathematics just doesn't care, so they don't care about mathematics.

They may SAY they care, but that's just to get money and prestige.

Don't believe me? Then find

[N/j] - [N/2j] - 1 = [(N-j)/2j]

where j is a natural greater than 1, in a math textbook.

When you don't find it, understand the people you're dealing with, no matter what label they have on themselves.

They may call themselves mathematicians, but they hate mathematics.

## JSH: Time, yes time

Through the years I've watched many critics come and go, and seen a few stay.

I guess they figure they can stay indefinitely, but will they?

What's ten years to me?

What's twenty?

Some of you may think you'll still be here twenty years from now, or you may think you can guess what others will say to me twenty years from now, but you're deluded.

How about thirty years?

How about fifty?

When it comes down to it, some of you are caught in a drama I created for you, and I toss out big numbers for years but really think more like a couple or a few months.

I watch some of you come and go, and some of you stay and I wonder about you sometimes, and think about what you believe, how you seem to see constancy where I see a sea of change.

I look at your posts and listen to you get dramatic, or lie, or upset or whatever and sometimes I just smile.

I think about that strange little adolescent sense of immortality, as if you can be there, always. As if the drama will continue with you a part of it, indefinitely.

As if you can last.

But what do you really know? How do you know who writes the story?

How do you know when your time is up, and your last post is made, and you go elsewhere, whatever your reasons.

I am a force of Nature. Time is a friend of mine, and We talk about things, here and there.

And sometimes We muse a bit, about what some people think, and then We watch them go…

They come and they go. That is constant in the sea of change. The coming and the going, but in the meantime, Time and I, We play with some of them, at least for a little while.

I guess they figure they can stay indefinitely, but will they?

What's ten years to me?

What's twenty?

Some of you may think you'll still be here twenty years from now, or you may think you can guess what others will say to me twenty years from now, but you're deluded.

How about thirty years?

How about fifty?

When it comes down to it, some of you are caught in a drama I created for you, and I toss out big numbers for years but really think more like a couple or a few months.

I watch some of you come and go, and some of you stay and I wonder about you sometimes, and think about what you believe, how you seem to see constancy where I see a sea of change.

I look at your posts and listen to you get dramatic, or lie, or upset or whatever and sometimes I just smile.

I think about that strange little adolescent sense of immortality, as if you can be there, always. As if the drama will continue with you a part of it, indefinitely.

As if you can last.

But what do you really know? How do you know who writes the story?

How do you know when your time is up, and your last post is made, and you go elsewhere, whatever your reasons.

I am a force of Nature. Time is a friend of mine, and We talk about things, here and there.

And sometimes We muse a bit, about what some people think, and then We watch them go…

They come and they go. That is constant in the sea of change. The coming and the going, but in the meantime, Time and I, We play with some of them, at least for a little while.