JSH: Ranting aside, a remarkable error
JSH
can be amazingly hard to get your point across where with this
particular error, a hundred years plus of research built on the error
gives some people (unfortunately) a lot of motivation to just fail as
mathematicians in the face of it.
In a lot of ways it's a trivial demonstration, given
7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
the mathematics does not allow that 7 to be controlled by the
factorization of what it is multiplying, so you have with a simple
example:
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
a CHOICE.
That's because it could be: 7(x^2 + 3x + 2) = (x + 1)(7x + 14).
Now I've pointed this out many times over the years as I've repeatedly
explained this error!
No person thinking rationally would suppose that 7 is being told how
to multiply by mysterious forces. I used to say, years ago, that the
tail does not wag the dog.
So, ok, a non-standard factorization reveals a bizarre problem with
the ring of algebraic integers. The construct creatively FORCES the
roots of monic polynomial with integer coefficients when x is an
integer i.e.
a^2 - (7x-1)a + (49x^2 - 14x) = 0
to be part of a factorization:
7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
so yes you can solve for the a's with the quadratic formula:
a_1(x) = ((7x-1)+/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2,
a_2(x) = ((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2
And you can make the substitutions and see that it all works. If
Dedekind had been given this demonstration over a hundred years ago,
you would not be learning the ring of algebraic integers today, except
maybe as an oddity: a fairly useless ring which has a fatal flaw.
But a hundred years of error built up around this error.
It takes away the usefulness of Galois Theory. It removes the ring of
algebraic integers as a useful ring.
Ok, since I found this error students have continued to be
indoctrinated into it. That is a lot of human waste.
Professors who keep feeding it to students may in some cases be
functionally insane due to the enormity of the problem.
Or they may just be cynically determined not to face consequences of
it. Who knows.
But if you learn it, you are training your brain useless knowledge
which does not work.
The math done with it is pure--purely wrong.
So yeah, they'll teach you style. It's style WITHOUT substance.
They force style because the math itself is wrong. Style crap also
helps to keep out outsiders like me!
Rigor in error is still error. You are just rigorously wrong.
The problem is a remarkable error. It has taken over a hundred years
to be revealed. Puzzle over it. Take your time.
Ask yourself: why can't I divide that 7 off?
There is only one answer because it's mathematics.
Remember that word? "Mathematics"
There is an answer. The right one.
James Harris
Mark Murray
> When people refuse to acknowledge even the most basic mathematics it
> can be amazingly hard to get your point across where with this
> particular error, a hundred years plus of research built on the error
> gives some people (unfortunately) a lot of motivation to just fail as
> mathematicians in the face of it.
You have a verly clear non-understanding of your own mathematical
limitation, let alone the maths involved here.
Save what little is left of your reputation and drop this.
> In a lot of ways it's a trivial demonstration, ...
Yes. But you don't understand it.
M
Mark Murray
> Now I've pointed this out many times over the years as I've repeatedly
> explained this error!
What you haven't figured out is that there are an infinite number of ways
of "hiding" the 7.
Example:
7 = (1 - 2i)(7/2 + 14/5i)
Now see if you can divide the 7 out of either of the factors of the right
hand side. I dare you.
M
David C. Ullrich
wrote:
>When people refuse to acknowledge even the most basic mathematics it
>can be amazingly hard to get your point across
You really have no idea how funny it is when _you_ say this?
>where with this
>particular error, a hundred years plus of research built on the error
>gives some people (unfortunately) a lot of motivation to just fail as
>mathematicians in the face of it.
>
>In a lot of ways it's a trivial demonstration, given
>
>7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
>
>where the a's are roots of
>
>a^2 - (7x-1)a + (49x^2 - 14x) = 0
>
>the mathematics does not allow that 7 to be controlled by the
>factorization of what it is multiplying, so you have with a simple
>example:
>
>7(x^2 + 3x + 2) = (7x + 7)(x + 2)
>
>a CHOICE.
>
>That's because it could be: 7(x^2 + 3x + 2) = (x + 1)(7x + 14).
>
>Now I've pointed this out many times over the years as I've repeatedly
>explained this error!
>
>No person thinking rationally would suppose that 7 is being told how
>to multiply by mysterious forces.
Amd if anyone who's pointed out that there's simply no problem here
had _said_ that the 7 is being told how to multiply by mysterious
forces you might have a point.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Ki Song
I think you make a good point, but I don't think this is particularly
a good example, in the sense that you could write
7 = (1 - 2i)(7/2 + 14/5i)
7*1 = 7*(1-2i)(1/2 + 2/5i)
So you could still "divide off the 7."
Perhaps, a better example would be something like:
7 = (( -1-3i\sqrt[3])/2)((-1+3i\sqrt[3])/2)
Here, you don't see the 7 on the right hand side at all.
Or something even simpler:
7 = \sqrt[7] * \sqrt[7]
Of course, I'm just using roots of monic binomials with integer
coefficients whose constant term is 7 to produce these factorizations.
Mark Murray
> I think you make a good point, but I don't think this is particularly
> a good example, in the sense that you could write
>
> 7 = (1 - 2i)(7/2 + 14/5i)
>
> 7*1 = 7*(1-2i)(1/2 + 2/5i)
>
> So you could still "divide off the 7."
Yeah - I realised that as I headed out of the door :-]
> Perhaps, a better example would be something like:
>
> 7 = (( -1-3i\sqrt[3])/2)((-1+3i\sqrt[3])/2)
>
> Here, you don't see the 7 on the right hand side at all.
That is the effect I was _hoping_ to produce. /O me stupidum/ :-)
> Or something even simpler:
>
> 7 = \sqrt[7] * \sqrt[7]
That will do!
M
JSH
> On Dec 1, 3:34 am, Mark Murray <w.h.o...@example.com> wrote:
>
> > JSH wrote:
> > > Now I've pointed this out many times over the years as I've repeatedly
> > > explained this error!
>
> > What you haven't figured out is that there are an infinite number of ways
> > of "hiding" the 7.
>
> > Example:
> > 7 = (1 - 2i)(7/2 + 14/5i)
>
> > Now see if you can divide the 7 out of either of the factors of the right
> > hand side. I dare you.
>
> > M
>
> I think you make a good point, but I don't think this is particularly
> a good example, in the sense that you could write
>
> 7 = (1 - 2i)(7/2 + 14/5i)
>
> 7*1 = 7*(1-2i)(1/2 + 2/5i)
>
> So you could still "divide off the 7."
You can always "divide off the 7" in ok rings, as you're multiplying
times it, as that's basic algebra.
Multiply 7 times x+1, and you get 7x+7, and hey, guess what? You can
divide it back off, and get x+1. Duh.
But the ring of algebraic integers is VERY special though as with:
7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
you cannot in general divide off that 7, where it can be hard to
understand--as maybe it's a subtle point?--but consider an example in
the ring of integers where you CAN:
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
To some that can seem too trivial as how can there be an issue over
"dividing off 7" when it's a constant and it's trivially multiplied as
can be seen on the left hand side, but I give the trivial example to
fully highlight how bizarre what happens with the ring of algebraic
integer is!
> Perhaps, a better example would be something like:
>
> 7 = (( -1-3i\sqrt[3])/2)((-1+3i\sqrt[3])/2)
>
> Here, you don't see the 7 on the right hand side at all.
Which is not the point.
> Or something even simpler:
>
> 7 = \sqrt[7] * \sqrt[7]
>
> Of course, I'm just using roots of monic binomials with integer
> coefficients whose constant term is 7 to produce these factorizations.
The point is that if you multiply somethings times something else,
like 7, then in ANY sensible ring you can divide the thing back off!!!
IN all your examples where you don't give a ring, if you settle on a
ring which is valid, you will find you can divide the 7 off.
But with the special example I've given you cannot IN THE RING OF
ALGEBRAIC INTEGERS.
That ring is unique! It's bizarrely insane. It will let you multiply
by the 7, but refuses to allow you to *generally* divide it off, if
you disagree, then demonstrate.
Divide the 7 off!
Mathematics is supposedly about what you can prove, right? So if you
know math, why can't you divide the 7 off?
James Harris
JSH
I think it's important to point out to readers that while Usenet
posters can be morons, real number theorists at actual universities
can't trot out the b.s. that you will see in these threads.
I know because I went back to my alma mater Vanderbilt University and
gave an earlier argument with cubics to Professor Ralph McKenzie, who
is also on the staff of the University of California at Berkeley, as
he's one of the leading American mathematicians.
He agreed with me. Wackiest experience I guess in this saga as I
explained on his chalkboard and he not only agreed with me on the
argument but shot down objections I'd been hearing on Usenet.
Math professors can't just trot out crap, unlike on Usenet where
posters can, and routinely do.
So what happened then? He went home.
That was years ago.
I was dumbfounded. Just like later I was shocked when my paper got
published and more shocked when some Usenet asses got it yanked with a
couple of emails!
It's not necessarily that your professors don't know of this problem.
Kind of hard to believe that a math journal went belly-up without the
mathematical community noticing.
And I am kind of loud if you hadn't noticed! I've been talking about
this problem for years.
Math is the easy part: no other known ring can "grab" a number like 7,
and hold it, which I also call entanglement.
And to see more evidence your community is on some level aware, do
this Google search (has to be Google):
algebraic integer entanglement
The guy just went home. So what could I do? I went back to my car
for the over 4 hour drive back to the Atlanta metro area--as I lived
there then, but now am in San Francisco--and all the while I'm just
dumbfounded.
Over 8 hours out of my life just on the drive to go back to my school
to talk to a "leading mathematician" and no arguments, no attacks on
my reasoning. He just went home.
Not like the movies. No Hollywood script here.
I explained the equations to a leading mathematician, and he agreed
with me, but then just went home.
James Harris