Thursday, October 14, 2004


Long threads, tactics

If you look over the thread where I've been working on another paper you'll see a good reason for creating new threads.

That one is now coming up on a 100 posts, with a lot of them coming from people who spend a lot of their time on Usenet trying to obscure what I'm saying.

They do this deliberately, and it's an effective tactic.

I've checked in various ways to watch interest in my work as it works its way around the world, as I move Google search results. As I post here Google and also Yahoo search results shift, so I watch them, and I can sort of map how effective some of these posters are.

Real mathematicians would want to get to the bottom and find out what's true, but these people are practiced at trying to control what you think on the subject.

If you didn't know it's worth mentioning that Erik Max Francis, a person who put up an insulting but popular webpage—according to Google which ranks it now about 9 for the name "James Harris"—used to post a lot on sci.math, and by now you should know about the sci.math'ers who got together to send emails and censor my paper out of an electronic math journal.

These people are serious, and they have been practicing for years, as have I.

As of now, I do more than move Google search results, as what I say gets read by quite a few people worldwide and it travels.

Kids read it, adults read it. People who think read what I say. They're learning more about how even the math world is corrupted in ways they're seeing all over the world in many places, from Catholic priests, to businesses, to politics.

They're a new generation, and they're learning rapidly.

And what people arguing with me say, travels.

The words weave their way out from here, and I can watch the war as it progresses around the globe, in waves, as attacks and battles play out.

It's a fascinating struggle with me on one side and a dedicated group of people on the other, and they're not stupid people.

By myself I have had to slowly slog through in what is basically a war of attrition, and part of that war is moving on when a thread gets clogged up by these people.

Tomorrow I'll be looking at Google search results again to see how they've moved.

Now, as I've put out this latest flurry of postings, they're moving on a day by day basis.

It's that big. The impact on the math field is like nothing that has ever come before, and even at the fringes, like replies from obsessive posters, the impact is already worldwide.

It's like a growing hurricane.

Wednesday, October 13, 2004


JSH: Now a change

Working out the details in that new paper thread allows me from now on to refer to sections of it when dealing with the obnoxious posters who refuse to accept mathematics.

For instance, when someone claims that W. Dale Hall has a counterexample, I can point to Section 2, and how it shows that finding algebraic integers does not show a contradiction.

If you look over my current draft you can see from beginning to end a fairly basic argument that handles every issue raised over years in arguments, and shows how mathematically a fairly straightforward conclusion is necessary to avoid contradiction.

My case begins with the simple position that constants are constant, and notice from Section 2 and Section 3 that sci.math'ers arguing with me have been relying on unit factors—just not units in the ring of algebraic integers.

For those who don't understand what I mean about constants and unit factors, consider units in the ring of algebraic integers with

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

as the constant terms are 1 and 2 of the factors, but you can multiply by u_1, and u_2, where u_1 u_2 = 1, and u_1 and u_2 are algebraic integers to get

P(x) = (u_1 x + u_1)(u_2 x + 2u_2)

and someone might jump up and down and now claim that the constants are not constant.

That is essentially what posters like Dik Winter have done, for months now, except the units are provably not in the ring of algebraic integers, but in a ring that includes the ring of algebraic integers, where two conditions rule:
  1. No rational other than 1 or -1 is a unit.

  2. No non-unit member of the ring is a factor of any two rationals that are coprime to each other.
Those two requirements determine rings of a special type, like the ring of integers or the ring of algebraic integers.

They are the key requirements.

So, when posters have put forward what they claim are counterexamples, they have simply put forward algebraic integers multiplied by units that act as conversion factors.

They are units that convert non-algebraic integers in a ring defined by the conditions 1. and 2. into an algebraic integer, and it's not hard to prove that such will always exist, as that is basically Section 2. of my draft paper.

I've covered all the bases, and shown what must be true mathematically.

That now leaves the issue of your egos and other psychological issues you may have, but some of you must know that I'm not surprised, as you're not mathematicians.

And make no mistake, I'm emphasizing that point for a purpose, as mathematicians necessarily would care more for what is mathematically true.

Mathematics is a hard discipline. Behaving in one way proves your not a mathematician, without any room for argument, just like a math proof leaves no room for argument.

So later, when you fail as I expect you will, then I will say that you have proven you are not mathematicians, so that other necessary things can be done.

After all, what's mathematically true was always true, and those wrong now, were always wrong.

No one who understand mathematics would hold on to a false position, especially one so easily proven to be false, if they were a true mathematician.

Friday, October 01, 2004


Latest fuss, my apologies

Unfortunately I was just wrong recently as I fiddled with

x^2 + (as_1 + bs_2)x + ab s_1 s_2 = (x + as_1)(x + bs_2)

and tried to generalize beyond the result with

x^2 + (2s_1 + 3s_2)x + 6 s_1 s_2 = (x + 2s_1)(x + 3s_2)

where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer,

where it's easy to show that s_1 and s_2 cannot be irrational algebraic integers, to show that the unit case is blocked in the ring of algebraic integers, for irrationals.

I hoped that if I could show that in general 'a' and b can't be algebraic integers without the same thing happening, then I could show that my other work is quite correct and there is a problem with the ring.

However, the result does not travel beyond 'a' and b being integers, and I was just wrong.

I did make a leap.

My apologies for jumping at that one and pushing people to toss easy examples on math professors like with that x^2 + x + 6, thingy.

I continually get frustrated in what I call conventional attempts to prove my advanced polynomial factorization techniques are correct.

The proper and final coverage of this result is shown with

x^2 + (2s_1 + 3s_2)x + 6 s_1 s_2 = (x + 2s_1)(x + 3s_2)

where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer,

as s_1 and s_2 cannot be irrational algebraic integers, showing that the unit case is blocked for irrationals, though s_1 = s_2 = 1 or s_1 = s_2 = -1, work just fine.

The traditional teaching is that the irrationals are just different, and that in fact the unit case IS blocked, while I can show, using advanced techniques, that the unit case is not blocked, as it's actually just not available in the ring of algebraic integers.

I apologize again for the screw-ups, and only hope that maybe, some people will bother to acknowledge the advanced techniques, as I can't find a conventional approach that does more than just hint at the problem, like with s_1 and s_2 not
being able to be algebraic integers if irrational.

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