Tuesday, September 14, 2010


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 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.

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