Sunday, July 22, 2001

 

Proof that the math world is sick

I've been posting for a while that I have a simple proof of Fermat's Last Theorm that's only about a page, which is quite simple.

Lots of other people have been posting that it's not really math, and they say NO ONE in the world believes I have anything at all.

I want you to remember all of that when you get to the end of this post.

Just because I feel a little guilty, remember I posted what my intent has been: I want to show the world that the math community is being rotted out by the strange attitude that mathematicians can be "pure" by producing useless information, and refusing to have the discipline address as a priority the mathematical issues at the frontiers of physics.

My proof of FLT starts with x^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2)

which is exactly the same as

x^2 + y^2 + vz^2 = x^2 + y^2 + vz^2

(which didn't keep one poster from arguing that I was dividing by zero in the first).

The above is a tautology. (My use of that phrase in this context brought objections and derision.)

Of course, I use x^p + y^p = z^p, in there, and eventually come to

(v^p+1)z^{2p}- px^2 y^2 (vz^2)Q'{z^2,x^2 y^2} - 2x^p y^p = 0(mod (x^2+y^2+vz^2))

which is a rather neat expression that says a lot.

(In case you'd like to look at the main expression for a specific case, with p=5

(v^5+ 1)z^10 - 5x^2 y^2 (vz^2)(v^2 z^4 - x^2 y^2) - 2x^5 y^5)).

What it says is that as long as x^p + y^p = z^p, that expression has

x^2 + y^2 + vz^2 as a factor.

Because I'm proving FLT, I look at the case where v is an integer, and quickly noticed that setting v=-1 gave me an interesting result—

px^2 y^2 z^2 Q'{z^2,x^2 y^2} - 2x^p y^p = 0(mod (x^2 + y^2 - z^2)).

So I consider an integer factor of x^2 + y^2 - z^2 that is NOT shared with x or y, which is odd, and I called it F.

So now, I decide to consider v's, where v = -1 + mF, where m is a counting number, which shouldn't be a surprise.

Obviously then, my primary expression—

(v^p+1)z^{2p}- px^2 y^2 (vz^2)Q'{z^2,x^2 y^2} - 2x^p y^p

is divisible by F.

Now I factor that expression into (a1 z^2 + b1 xy)…(ap z^2 + bp xy).

Here it should be obvious enough that

a1…ap have F as a factor.

So at this point, we now have

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2)

(Some of you may be wondering why I don't substitute in -1 + mF for v. Well, it just seems like effort for nothing to me.)

Now I'll pause a bit because it finally became clear that many on the newsgroup weren't sure you could even do that, and then even if they figured such an expression MIGHT exist, they kept bugging me about the ring.

You see, their professors have been teaching them for a long time that you have to have a particular ring, and they are having problems with that programming.

In short, it is confusing them.

Once I finally understood the problem I brought them back to the quadratic case, which I figured would be a good place to clear everything up, so I gave

fx^2 + gx + h = (a1 x + b1)(a2 x + b2), and talked about it a LOT.

Didn't help much.

Today I saw a post from a guy who believes that if the main expression has a degree greater than 4, then you're stuck. I kindly noted that he was trying to refute Gauss. Why?

a1 a2 = f, a1 b2 + a2 b1 = g, and b1 b2 = h, so

you have four unknowns with f,g and h which are the knowns.

In general, you'll have 2n unknowns with n+1 knowns, which allows you to always solve for each 'a' using all of the b's.

That gives you a monic polynomial, which Gauss proved ALWAYS has a solution.

BUT all of the math community on sci.math has had problems with what I just presented, and there's still a big issue that you might not have noticed I glossed over—I didn't give the ring for a1,a2,b1, and b2.

Uh oh!!! According to the math community that means everything is INVALID!!!

Ohmigod, we just wasted our time, oh well.

Folks, this is the math that they're telling you is junk.

Translation: They don't understand it because most of them are parrots, so they call it junk rather than accept mathematical truth.

So, why not specify the ring?

Because it interferes with the proof, getting back to it:

Then for a given factor a1 of F

a1 z^2 + b1 xy = = 0(mod a1), and in general

an z^2 + bn xy = = 0(mod an), 0<n<p+1, so b1…bp = 0(mod F),

but

b1…bp=-2,

and there is a contradiction if F is a counting number greater than 2 (if F = 1, then I'd just use x^2 + z^2 - y^2).

Q.E.D.

Ooh that was fast! I bet most of you are wondering what all of that meant, which is a pity because it's basic math.

Ok, enough rhetoric, let's go back through it line by line.

"for a given factor a1 of F

a1 z^2 + b1 xy = = 0(mod a1)"

Ok, remember we have that a1…ap = F (for p=5, a1 a2 a3 a4 a5 = F), so obviously a1 is a factor of F.

Oh yeah, they've been telling you that I'm using some funky definition of factor.

Translation: They're confused, so I must be doing something wrong.

No, they're just confused.

I'm using factor the same way everybody does. You know 2(3)=6, so 2 is a FACTOR of 6.

Now remember, we have that

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2)

which means that

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod F)

(Why didn't I just write that above? It seems SO obvious that I just figured it'd be a waste.)

So,

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod a1…ap).

Wait! You may holler. How do I know that F factors that way on the right?

Because I didn't specify a1…ap, that's why.

Understand why I don't want to specify a ring at the start now?

(In case you missed it, the ring is specified when I set my a's. For instance, with x^2 + 4x + 4, a1=a2=1, forces b1=b2=2, but you could have a1 = 1+sqrt(3)i, and a2 = 1-sqrt(3)i, the math doesn't care which you use but some of you get confused.)

Now I'm going to give you an example with integers and see where we can go from there,

(3x + 2)(5x+a) = 0(mod 3(5)), and let me specify that must be true for ALL x's.

But, it's not, now is it?

'a' must be divisible by 5, but 2 isn't divisible by 3, so I can stick in an x where 5x+a isn't divisible by 3, and the expression is false.

The math is basically just a use of the distributive property, which should be basic enough.

In the proof I talk about sharing of factors but it might just confuse people here at this point.

Besides, it occurs to me that some of you are saying, well that's integers, other types of numbers can be really different.

Well that's why I've been trying to educate the community about discrete sets and the continuous field. That is, I hate to break it to you, but mathematicians got things a bit wrong. Just a little bit wrong, so don't get too excited. In any event, they ARE supposed to be taught about the difference ring operations make, so it shouldn't matter, they should still be able to get on the same page with me.

(Any of you use p-adics? Then you have NO excuse for not getting it.)


"a1 z^2 + b1 xy = = 0(mod a1), and in general

an z^2 + bn xy = = 0(mod an), 0<n<p+1, so b1…bp = 0(mod F)"

Well, a1 z^2 + b1 xy = 0(mod a1) means b1 xy = 0(mod a1), and I said that F is coprime to x and y, so it's a no-brainer that b1 = 0(mod a1).

Go down the line and b1…bp = 0(mod F).

Simple.

"but

b1…bp=-2,

and there is a contradiction if F is a counting number greater than 2 (if F = 1, then I'd just use x^2 + z^2 - y^2)."

That is, I arbitrarily picked x^2 + y^2 + vz^2 at the start of the proof, so if necessary I can switch things around a bit and start over, which would force F to be something other than 1.

So why are people still running around telling you there's nothing to my proof?

You'd think they'd want to be in on something like this, wouldn't you?

Well, I put a lot of it on bad training, and the rest is on the human need for consistency.

Think about it, if you see a bum on the street begging for change, you don't want to deal with him being an oil tycoon the next day.

Sure, we can handle it if he won the lottery or something but there MUST be an explanation like that or people get VERY unsettled.

I'm very unsettling. Ok. Get over it.

Your feelings are not a reason but an excuse, and the consequences will be the same because I believe bad training is the PRIMARY reason, and incompetence is not a good reason.





<< Home

This page is powered by Blogger. Isn't yours?