JSH Summery #4
- From: "Sulayman" <spamless@xxxxxxxxxxxx>
- Date: Sat, 11 Feb 2006 17:37:31 -0600
(These are orgional JSH postings to this group which I have compressed, so
we can get an overall view of what he is talking about)
............
Look at how close it is, with me figuring out a way to explain by using
standardized functions--in that they go to 0 when x=0--so that you can SEE
the 7 pop up, in only one factor, so that the mathematics clearly shows I'm
right. But if I don't argue these points, and keep going past the mistakes,
then people claiming that I am wrong, win, right? So then the future would
be changed because the flawed mathematical ideas would stay in place for
that much longer, and because they are flawed the science that depends on
the correct mathematics would not be developed, and civilization in the
future would be changed. Like, imagine if Newton or Gauss had faced
dedicated resistance that had witheld acceptance of their ideas. Our present
would not exist. To many of you I'm just some nut who won't shut-up, but to
future civilizations I am crucial to their existence. My success here is
necessary for their science and technology. Pay attention to the proof. All
the big picture issues aren't necessary to understand the mathematics, but
it's past time that some of you began to consider that there are very
important reasons why you are not to stop major discoveries. James Harris
...........
José Carlos Santos wrote: > jstevh@xxxxxxx wrote: > > > Like, imagine if
Newton or Gauss had faced dedicated resistance that > > had witheld
acceptance of their ideas. > > > > Our present would not exist. > > > > To
many of you I'm just some nut who won't shut-up, but to future > >
civilizations I am crucial to their existence. > > > > My success here is
necessary for their science and technology. > > Read these sentences again.
And now tell me: what would you call someone > else who would have written
them? > Doesn't matter. It's the truth. It's difficult for people right at
the point of a discovery to realize its impact over time. But here we have a
case where I've overturned over a hundred years worth of flawed mathematical
ideas. That sort of thing is not minor. > > Pay attention to the proof. > >
Which proof? > For those who don't know, the point of these discussions for
me is usually to work through drafts, where recently I've been looking for
improved exposition to try and break through the impasse with my research.
Thanks to the simplificity of the Decker example and the modified Decker
example I now use because they use quadratics versus cubics I feel like most
of you will get the proof now, and understand why the distributive property
is key. I have two ways for you to see the proof: 1. Read my post "JSH:
Simplifying exposition" 2. Go to my blog and read "A simpler example, some
quadratics" You can just go to http://mymath.blogspot.com/ or directly to
that post at
http://mymath.blogspot.com/2006/02/simpler-example-some-quadratics.html
which also is a fuller version anyway than my Usenet posting, as I use a
proper modified Decker equation to show the contradiction with standard
usage of Galois Theory and the theory of ideals. I really like using the
standardized functions which go to 0 at x=0, as many of you are probably
taught to do that sort of thing, so it's harder to dodge. Kind of wild--but
then again sort of obvious--when that 7 just pops up as a factor of the
constant term in a way that can only go one way by the distributive
property, even with the introduction of functions. Remember math is
important! Hiding the truth here just gets you a blackmark in history, and
shows you don't give a damn about those people in the future who need the
correct information found today. Past people accepted the truth, eventually,
or we wouldn't have our science and technology today. Now is your test.
James Harris
...........
I make mistakes. I come up with requirements or examples that I think are
great and then they don't work. But, if you believe in mathematics as an
area where the truth can be determined then the mistakes of one person
hardly matter as people WORKING TOGETHER can come to a conclusion which
follows from the axioms of mathematics. I now have a post where I hope I
covered all the bases as I'm not just here to disagree with people, or to
push some idea just because it's my ideas, but to figure out what is
mathematically correct. I kept pushing the distibutive property and in reply
posters kept talking about functions. And I pushed the distributive property
and they talked about functions and things degenerated badly when I asked a
poster if he was stupid when I was talking about constants and he kept
talking about functions! That bugged me. I shouldn't have asked if he was
stupid. And I decided I should take on the issue of functions head on.
Meanwhile as a research issue I've been puzzling over the Decker example and
my modifications to it, curious about the mathematical behavior, as some gut
feelings about it were wrong, and you can't make it behave. The issue for
those who are curious is that any time the last two coefficients have the
same prime factors and one of those prime factors is a prime factor of f,
it's trivial that ALL the roots must have factors in common with that
factor. With non-polynomial factorization that defaults you into a situation
where, oddly enough, you are forced to have a ring with fractions!!! The
mathematics doesn't allow anything else because of, well, because of the
distributive property. So, as usual, I work on more than one thing at one
time and I thank those posters who have gone to the effort of finding
counterexamples to my research conditions with the modified Decker examples.
Where do we go from here? Well, you may notice that besides losing it with
the poster who I asked if he was stupid, I'm rather calm, because the math
is easy, and explanations are easy, and why get excited? In contrast,
posters replying seem to keep getting excited. I say, get excited about
mathematics. It's all so exciting!!! And if you believe in mathematics at
all, then you should be able to explain things like my normalized equations
where the functions all go to 0 at x=0, which clearly show that the constant
terms don't allow anything but 7 to have multiplied through, in the ring of
algebraic integers, as then your dimensionless functions must lack
structure. If that doesn't make sense to you, read my previous thread. If it
still doesn't make sense to you, welcome to how easily simple ideas can get
complicated. My guess is that the mathematics in this area rapidly becomes
way to complicated for most people and far beyond the complexity levels of
today's mathematics. I call it level 2 or 2nd generation mathematics, where
I can hypothesize a 3rd generation mathematics, which has a complexity level
that for most that is, well, unimaginable. The 3rd generation mathematicians
haven't been born yet. Who knows what kind of minds they will have. James
Harris
...........
The goal of this thread is for me to cover all the points that have been
brought up by myself AND others who have argued with me over what I've
called the Decker example, which thankfully is a lot easier to play with as
it involves quadratics versus cubics. If I am wrong, a concise and very
objective review of all the mathematical arguments should reveal that I am
wrong, but of course, I don't think I am wrong, or I'd just say it! The
challenge for me now is to deliver on covering all the bases. Starting again
with the Decker equations, in the ring of algebraic integers, where the goal
is to remain in that ring, I have 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) +
7) = 7(25 x^2 + 30 x + 2) and 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) where the
a's are defined by a^2 - (x - 1)a + 7(x^2 + x) = 0 so you can determine that
at x=0, one of them goes to 0, while the other goes to -1, which shows that
at x=0, the factor of 7 is not split between the two factors (5a_1(x) + 7)
and (5a_2(x) + 7) but is a factor of just one, however, posters in arguing
with me, have claimed that case is a special case and that in general the 7
factors, where its factors can be considered functions of x, so let's use
w_1(x) and w_2(x) where w_1(x) w_2(x) = 7 and a_1(x) = w_1(x) b_1(x) and
a_2(x) = w_2(x) b_2(x) and divide 7 from both sides of the factorization
equation to get Q(x) = (5b_1(x) + w_2(x))(5b_2(x) + w_1(x)) but Q(x) is a
polynomial as Q(x) = 25 x^2 + 30 x + 2 and factorizations of polynomials
typically show factors of the constant term, while the factorization we now
have just shows functions, so I want to standardize it (I wanted to say
normalize it here but figure that's got other meanings that people prefer)
by considering values at x=0. Q(0) = ((5b_1(0) + w_2(0))(5b_2(0) + w_1(0)) =
2 where I need to choose one of the b's to go to 0, as one of the a's goes
to 0, and I'll arbitrarily pick b_1(0) = 0, which gives w_1(0) = 7, w_2(0) =
1 and b_2(0) = -1 so I can standardize with w_1(x) = w'_1(x) + 7 w_2(x) =
w'_2(x) + 1 b_2(x) = b'_2(x) - 1 and making those substitutions gives Q(x) =
(5b_1(x) + w'_2(x) + 1))(5b'_2(x) + w'_1(x) + 2) and in standardized form
the constant terms of the factorization are factors of the constant term of
the polynomial Q(x), which was accomplished by having all the functions go
to 0, when x=0. Now multiply back through by 7, using our standardized
values for w_1(x) and w_2(x), which gives 7 Q(x) = (5b_1(x)(w'_1(x) + 7) +
w'_2(x)(w'_1(x) + 7) + w'_1(x) + 7)(5b'_2(x)(w'_2(x) +1) + w'_1(x)(w'_2(x) +
1) + 2 w'_2(x) + 2) where I have the factorization in standardized form,
where at x=0, all the functions go to 0, and clearly you have that the first
factor (5b_1(x)(w'_1(x) + 7) + w'_2(x)(w'_1(x) + 7) + w'_1(x) + 7) has a
constant term that is 7, while clearly the second factor (5b'_2(x)(w'_2(x)
+1) + w'_1(x)(w'_2(x) + 1) + 2 w'_2(x) + 2) has a constant term that is 2.
(Remember though my choice of b_1(0) = 0 was arbitrary, as mathematically
it's not determined WHICH of the b's must go to 0, just that one does.) So
even with functions in, by the distributive proprty, it still remains that
case that 7 is a factor of just one. Now to the question of the form of the
w functions and their simple variant w' functions, as there is still a
problem to consider which is that ANY function of x will have some maximum
degree, like f(x) = x + 1 has a maximum degree of 1, while f(x) = x^2 + 2x +
1, has a maximum degree of 2, while f(x) = 2, has degree 0. However, it is
fairly easy to prove that the w functions and the simple variant w'
functions must have maximum degree 0, which just follows from solving for
the a's using a^2 - (x - 1)a + 7(x^2 + x) = 0 as you get a = ((x-1) +/-
sqrt((x-1)^2 - 28(x^2 + x)))/2 where you can see maximum degree 1, for BOTH
possible solutions, which means that the multiplication of the w's has not
raised or lowered the degree. That means that the w's are degree 0, or as I
like to say, they are dimensionless. In the ring of algebraic integers, they
are allowed to be dimensionless only if they are constants, which they are.
If that puzzles you consider the dimensionless function f(x) = (x+1)/(x+2)
which you'll note is in general outside the ring. However, as is well-known
at this point, at x=1, you DO have a special case where it can be shown that
for the w's to be constant, you cannot be in the ring of algebraic integers,
plus the other cases where with a^2 - (x - 1)a + 7(x^2 + x) = 0 irreducible
over Q, besides when x = 1 mod 7, you can show that neither of the a's has 7
as a factor IN THE RING OF ALGEBRAIC INTEGERS so you are forced out of the
ring. My thinking in this area had been that you could still work with the
Decker example and force it into a ring without fractions by simply avoiding
areas like x = 1 mod 7, but I tested that hypothesis with a generalization
of the Decker example, and then a modified generalization of that which some
posters nicely checked for me, and the equations never behaved. Simple
answer is that they can't be forced to remain in the ring of algebraic
integer nor in what I call the ring of objects, but that's secondary and for
me is a research issue. For the rest of you, if you wish to claim I'm wrong,
I'm just looking for a way that the constant terms of the standardized 7
Q(x) = (5b_1(x)(w'_1(x) + 7) + w'_2(x)(w'_1(x) + 7) + w'_1(x) +
7)(5b'_2(x)(w'_2(x) +1) + w'_1(x)(w'_2(x) + 1) + 2 w'_2(x) + 2) can be 7 and
2, by the distributive property, and 7 not have multiplied through ONLY ONE
of the factors. Notice here that the appeal to functions failed quickly when
I standardized them, as functions may be complicated in many ways,
potentially, but the issue here is not complexity, but mathematical
behavior. James Harris
..........
For YEARS now I've claimed I've found this massive error in the mathematical
field, which mathematicians are running from with huge negative consequences
for them when the world finds out, as they are fighting the progress of the
human race, and arguments have gone after arguments where it doesn't seem to
get resolved. Now there is a claim of counterexample to my claim of proof
using the distributive property with what I've called the generalized Decker
example. The claim looks rather solid. I say rather than debate that issue
at this point--and yes you can consider my not just giving up now to be
crank or crackpot behavior--I have put forth a simple challenge, which
should not be difficult, if I am actually wrong. I modified the generalized
Decker example in a special way: f Q(x) = f((x^2 + x)(5^2) + (-1 + 3x)(5) +
f) = f(25 x^2 + 40 x + 2) and f Q(x) = (5a_1(x) + f)(5a_2(x) + f) where the
a's are defined by a^2 - (3x - 1)a + f(x^2 + x) = 0 where I've removed the
ability to zero out the middle coefficient with an integer. That's it. I
know why that's significant. You may think I'm grasping. A counterexample
here proves me wrong with no more wiggle room. But, of course, since I know
I'm right, my real point is to emphasize to you how much evidence there is,
and later I can explain how the previous example allowed the appearance of a
counterexample, just from you being able to zero out the middle coefficient
with an integer. Mathematics is fascinating, and luckily for me, rigid in
its logic. Once you know the rules, you know what can or cannot happen.
Quite simply, no counterexample can be found with the modified equations.
James Harris
............
William Hughes wrote: > jstevh@xxxxxxx wrote: > > For YEARS now I've claimed
I've found this massive error in the > > mathematical field, which
mathematicians are running from with huge > > negative consequences for them
when the world finds out, as they are > > fighting the progress of the human
race, and arguments have gone after > > arguments where it doesn't seem to
get resolved. > > > > Now there is a claim of counterexample to my claim of
proof using the > > distributive property with what I've called the
generalized Decker > > example. > > > > The claim looks rather solid. I say
rather than debate that issue at > > this point--and yes you can consider my
not just giving up now to be > > crank or crackpot behavior--I have put
forth a simple challenge, which > > should not be difficult, if I am
actually wrong. > > > > I modified the generalized Decker example in a
special way: > > > > f Q(x) = f((x^2 + x)(5^2) + (-1 + 3x)(5) + f) > > =
f(25 x^2 + 40 x + 2) > > > > and > > > > f Q(x) = (5a_1(x) + f)(5a_2(x) + f)
0 > > > > where I've removed the ability to zero out the middle coefficientwhere the a's are defined by > > > > a^2 - (3x - 1)a + f(x^2 + x) =
with > > an integer. > > > > That's it. I know why that's significant. You
may think I'm grasping. > > > > A counterexample here proves me wrong with
no more wiggle room. > > Let x=2, f=-6 > > a^2 - 5a - 36 = (a-9)(a+4) >
-William Hughes Good example! Same problem as before in that 3x-1 has tobe coprime to f. But I know! I didn't SAY 3x-1 needs to be coprime to f. I
know, sounds like I'm just squirming, but hey, I make mistakes. Real
question is, what does the mathematics actually say, right? So I have a
modified proof exposition. For the moment though, to be fair, let's say it
looks like you're right and I'm wrong, but I have this puzzling LITTLE
argument that I'm going to put out there, where it would be nice to see what
wrong with it! I'm posting it in its own thread, of course. James Harris
..........
Rick Decker wrote: > jstevh@xxxxxxx wrote: > > > For YEARS now I've claimed
I've found this massive error in the > > mathematical field, which
mathematicians are running from with huge > > negative consequences for them
when the world finds out, as they are > > fighting the progress of the human
race, and arguments have gone after > > arguments where it doesn't seem to
get resolved. > > > > Now there is a claim of counterexample to my claim of
proof using the > > distributive property with what I've called the
generalized Decker > > example. > > > > The claim looks rather solid. I say
rather than debate that issue at > > this point--and yes you can consider my
not just giving up now to be > > crank or crackpot behavior--I have put
forth a simple challenge, which > > should not be difficult, if I am
actually wrong. > > > > I modified the generalized Decker example in a
special way: > > > > f Q(x) = f((x^2 + x)(5^2) + (-1 + 3x)(5) + f) > > =
f(25 x^2 + 40 x + 2) > > > That should be f(25 x^2 + 40 x + (f - 5)) > > >
and > > > > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > > > > where the a's are
defined by > > > > a^2 - (3x - 1)a + f(x^2 + x) = 0 > > > > where I've
removed the ability to zero out the middle coefficient with > > an integer.
grasping. > > > > A counterexample here proves me wrong with no more wiggleThat's it. I know why that's significant. You may think I'm
room. > > f = -77, x = 6, in case you missed the other thread. > > No, I
didn't miss it. I appreciate your energy though in trying to make sure!
Turns out I was wrong with my modification as I can't force 3x - 1 coprime
to f in general, and no modification ax + b with a and b integers can
prevent f from sharing factors at times which forces you out of rings that
don't have fractions. It turns out that is a characteristic of any quadratic
example, so my use of cubics ended up relying on the lowest degree where the
results will hold in a ring without any fractions, or, to be more precise,
my ring of objects. Good work keeping up on the counterexamples thing. I
appreciate it. James Harris
............
jstevh@xxxxxxx wrote: > Rick Decker wrote: > > jstevh@xxxxxxx wrote: > > > >
For YEARS now I've claimed I've found this massive error in the > > >mathematical field, which mathematicians are running from with huge > > >
negative consequences for them when the world finds out, as they are > > >
fighting the progress of the human race, and arguments have gone after > > >
arguments where it doesn't seem to get resolved. > > > > > > Now there is a
claim of counterexample to my claim of proof using the > > > distributive
property with what I've called the generalized Decker > > > example. > > > >
crank or crackpot behavior--I have put forth a simple challenge, which > > >The claim looks rather solid. I say rather than debate that issue at > >this point--and yes you can consider my not just giving up now to be > > >
should not be difficult, if I am actually wrong. > > > > > > I modified the
generalized Decker example in a special way: > > > > > > f Q(x) = f((x^2 +
x)(5^2) + (-1 + 3x)(5) + f) > > > = f(25 x^2 + 40 x + 2) > > > > > That
should be f(25 x^2 + 40 x + (f - 5)) > > > > > and > > > > > > f Q(x) =
(5a_1(x) + f)(5a_2(x) + f) > > > > > > where the a's are defined by > > > >
ability to zero out the middle coefficient with > > > an integer. > > > > >a^2 - (3x - 1)a + f(x^2 + x) = 0 > > > > > > where I've removed the
That's it. I know why that's significant. You may think I'm grasping. > >didn't miss it. I appreciate your energy though in trying to > make sure! >
f = -77, x = 6, in case you missed the other thread. > > > > > > No, IA counterexample here proves me wrong with no more wiggle room. > >
Turns out I was wrong with my modification as I can't force > > 3x - 1coprime to f > > in general, and no modification > > ax + b > > with a and b
integers can prevent f from sharing factors at times which > forces you out
of rings that don't have fractions. > Damn. That was dumb. I looked back at
my cubics. There is a way: Q(x) = f((x^2 + x)(5^2) + (-1 + fx)(5) + f) =
f(25 x^2 + (f+25)x + 2) and f Q(x) = (5a_1(x) + f)(5a_2(x) + f) where the
a's are defined by a^2 - (fx - 1)a + f(x^2 + x) = 0 and fx-1 is forced to be
coprime to f for all algebraic integer values. Kind of uglier though...hmmm.
Oh well. James Harris
You have heard of the power of the pen, and now I'm going to make it
explicit for you what I am doing. Each day, each post where I remind that
you are fighting history, fighting against progress, and showing you are
just another group like so many others in human history who put their needs
above the needs of everyone else, is a weight of words meant to make sure
that people understand the real issues here. Because when you lose you will
whine. You will get upset with others when it's your faults. And you will
beg other people to not allow you to suffer for your own mistakes. Each day
I make these posts to remind people that you had every opportunity to do the
right thing.Each day I make these posts so that they understand that you
willfully put your needs and that of your own above those of everyone
else--that you put yourself above the world. These posts are meant to make
sure that people know that you were informed. That you were reminded that
the issues here are bigger than you are. That you knew that our world
depends on the truth, accurate knowledge and people not succeeding when they
fight against the truth for their own selfish needs, and that later when
they lose, they suffer the consequences--no matter how much they cry, or
complain, or say it's just not fair. James Harris
.....................................................................................................
I have a peer reviewed and published result which has gone by top
mathematicians with names many of you should recognize. No mathematicians of
any note has claimed an error in that argument, and I've explained it to a
few. BUT there are posters on Usenet who continually claim error and I keep
simplifying and explaining, where they keep talking about variables when I
talk about constants and the distributive property. It's social crap and
it's not remotely hard to see that it is and then it's not hard to figure
out why it's going on. I found a mistake, an error made over a hundred years
ago, by some clever techniques where I factor polynomials into
non-polynomial factors, and rather than admit the error, top mathematicians
just get quiet, while newsgroup people throw bogus arguments at it. There is
only one way you people are sitting calm in this situation which is by
holding on to the belief that no matter how much I explain, and no matter
how carefully I go over this, no one will listen as long as you all go along
with the lie that Galois Theory works as advertised and the theory of ideals
works as advertised. Maybe some of you feel it's bringing you all closer
together. But what you have is a cynicism against the human race as these
things have happened before. There's some discoverer with a result that
some group doesn't like because it upsets the status quo, so they fight it,
but then history shows that result to be critical to the continued progress
of humanity. Right now you people don't give a damn about humanity. You
can't think that far. All you can feel is the hurt and fear that your
precious ideas have been proven wrong, and then you feel comfort and
solidarity when you think you can all stand together and block the truth
from the world.
..........................................................................................................................
Stuff happens. So Dedekind missed something and you can't just define out
convergent infinite series so there's no basis for ideal theory. So what?
It's big, yeah, but there are upheavals in major fields, and so now there's
one in mathematics. That's not sad. That's an opportunity. What's sad is the
betrayal of the discipline by so many of you in a crucial test with an
argument so simple that there is just no way to be someone who loves
mathematics and run away from it. So what do you love? I think some of you
love complexity. The math ideas you know are complicated and hard to
understand which is the basis of their value to you. I think some of you
just love the idea of being into mathematics or admiring mathematicians
without regard to the reality of the hardness of the discipline or its
unforgiving nature. There are few names for the great mathematicians in
history. Now you know why. Most people cannot handle an area where the truth
is independent of their needs, or their society's needs. For many of you,
your society needs this result to not hold and that's all that matters to
you, so I get to argue with people over settled results, settled mathematics
that has even been peer reviewed and published. And history shows you will
lose, and you have to lose, as people like you have stood against so many
ideas that today are the foundations of our technological and scientific
world. People like you have fought against the theories and ideas that have
given us so much, and as mathematics is an important field, it is likely
that the correct mathematics, will give so much more. So you stand against
history, against humanity, against the good of the many, and it's so sad.
Here you have an opportunity to see mathematics for what it truly is, and
fall in love with mathematics for the right reasons. Because mathematics
gives you the power when you are right that one person can stand against the
entire world when it disagrees against experts and groups, against insults
and namecalling, against social pressure meant to deny the truth, because
mathematical proof is independent of all of that nonsense. History shows
that you people if you keep fighting will not only lose, but will lose that
much more greatly the longer you fight. But why be motivated out of fear of
the punishment? Why tell the truth about the mathematics because of the
consequences when the world learns you are its enemies? Why not tell the
truth because it's the truth? Don't make a sad story sadder. I challenge
some of you to look forward to a brighter future, and a world where the
mathematical foundations built today, will be part of the science of
tomorrow. James Harris
......................................................................
I've gone on in replies about how the argument I have relies at a key step
on the distributive property. I think it worth reminding at this point that
my argument that posters keep trying to find fault with, while crucially
ignoring the key step, is formally peer reviewed and was published in a math
journal. I want to re-interprete the conclusion that can be drawn by the
folding of that math journal in just a bit, but as some of you know, the
path to publication of a paper can be convoluted. My paper that was
published started out as some posts when I noticed posters arguing with more
heat over a particular piece of a short proof that I had for Fermat's Last
Theorem than any other. So I pulled out that piece and began posting on it,
and decided that I could focus on it alone, get it published, and with it
settled, come back to the issue of the full FLT proof. That was over three
years ago. An early draft of the paper went to Andrew Granville for
publication in the New York Journal of Mathematics, an electronic journal.
He replied to me that it was outside of his area, and I should check with
the chief editor, so I did. The chief editor, whose name escapes me, replied
that it was too short for their journal. I moved on. I just sent another
early draft to Barry Mazur, out of the blue and not to him for publication
in any journal, and he replied encouragingly and asked me some vague
question that I now forget what it was, but I answered him, and never heard
from him again. I kept at it, and concentrated on one journal because an
editor was from my alma mater Vanderbilt University. He rejected my paper at
that time, saying he didn't understand it. I forwarded him Barry Mazur's
email and he said he still didn't understand the paper, but offered that if
I were ever in Nashville, we could work it out on the chalkboard. So I made
it my business to drive to Nashville, which was a bit over a four hour trip
as I was then in the Atlanta area. I explained it all to him using his
chalkboard in his office, and afterward, he went home. Thanked me for an
enjoyable conversation, and that was it. I was furious and replied with some
heat, which didn't go over well. >From there the paper went to the Southwest
Journal of Pure and Applied Mathematics. They verified receipt and informed
me the paper was going out for formal peer review. They used two reviewers.
That journal had been around for about nine years. Months went by. I went
back to posting out of boredom and even at one point mentioned that I'd sent
the paper to that journal. As time went by, I worried that maybe they'd just
bury it, so I contacted them, even at one point noting I was an amateur, but
no, I didn't just come out and say that Usenet people hated me and hated the
argument. A Huckabee was encouraging and said it was in the process. After
nine months I was informed that I had a "nice paper" and it was published in
the electronic journal. I remember Huckabee replying to an effusive email
where I thanked them for publication, and he said that it didn't matter
about the source for a math paper as long as it was correct, even if it was
from a janitor, which I took to be an offhand mention to "Good Will
Hunting". So the paper came out, I began celebrating, informing friends and
family, and old school teachers, and then someone posted about it on the
sci.math newsgroup. Posters began hate filled rants as the newsgroup filled
with posts. They attacked the journal. They attacked the journal process.
Claimed that math papers were often wrong, in small journals. Lambasted the
editors. And some of them decided to mount an email campaign against the
paper, which succeeded, and the chief editor yanked it. A few months later
the journal quietly died. I want to re-interpret that death of the journal.
Think about it. Let's say that the editors went back to look carefully over
my paper, realized the basic argument WAS correct and realized that I'd
overturned Galois Theory and the theory of ideals--why continue? James
Harris ..............................................................
Decker's example can be generalized to help in finding rational solutions,
where you will find that the mathematics follows the distributive property:
f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) = f(25 x^2 + 30 x + 2)
and f Q(x) = (5a_1(x) + f)(5a_2(x) + f)
where the a's are defined by
a^2 - (x - 1)a + f(x^2 + x) = 0
and you can let f be a rational number. Now there will be posters who will
loudly declare coprimeness means nothing with rationals, but consider f(x^2
+ 3x + 2) = (fx + f)(x + 2) and solutions with rationals, as guess what? You
can STILL see that one factor is multiplied by f, even with rationals. If
you don't think so, play with that example with some rational f's and
rational x and see if the factor multiplied by f doesn't betray that it was.
Now then, if I am wrong, some rational f can be found with a rational x that
shows it. Like let f=32 with the generalized Decker example, and find some
rational solutions and see if that f gets split up. Or let f=1024 or
anything you want!!! You see, no counterexample exists, as the distributive
property is right!!! So posters here at best can loudly proclaim that Galois
Theory doesn't work with rationals but only with non-rationals, which is the
dodge because it actually doesn't work, but you can't see that with
non-rationals. The proof I've given relies on the distributive property.
Even lower rung mathematicians cannot be incapable of quickly seeing it MUST
be correct, but clearly as this impasse continues they are running. And them
running means they are hoping that none of you who are not already
established in careers, who are just learning as you're still in school,
will stop protecting them by ignoring this result. They are in the weak
positon of needing your protection so that they can teach you wrong
mathematical ideas, as if some students start protesting, they will collapse
like the cowards they are, running the other way, selling each other out to
protect themselves. First mathematicians on the block will be the ones who
are posters on sci.math, and their own will destroy their careers. People
like Magidin and Ullrich will be out of their universities so fast your head
will spin, as they are tossed to the wolves. So they sit and wait, checking
each day to see if any of you are breaking out of the wall of silent
acceptance, or irrational denial in the face of a simple proof that relies
on the distributive property at the point of dispute. That check may be the
best evidence against those mathematicians as they leave cyber clues to
prove they knew, but were checking to see if they could keep getting away
with lying. James Harris
..................................................................
William Hughes wrote: > jstevh@xxxxxxx wrote: > > Decker's example can be
generalized to help in finding rational > > solutions, where you will find
that the mathematics follows the distributive property:
f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f)
= f(25 x^2 + 30 x + 2)
and > > > > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > > > > where the a's are
defined by > > > > a^2 - (x - 1)a + f(x^2 + x) = 0 > > And your claim is
that one of the roots must always be divisible > by f and one coprime to
f. > > So take x=-2, f=-9 we get > > a^2 + 3a - 18 = (a+6)(a-3) > > The
two roots are -6 and 3, neither is divisible by -9. > > -William Hughes
That's no different from the special cases with the original Decker
example when x = 1 mod f, as then you have that sqrt(f) is a factor of
both. Well, at least you're trying. You need x-1 coprime to f. Go look,
you should still be able to find some rational examples, as remember, you
can use rational f, where the requirement becomes with f = b/a b/a - 1 =
(b-a)/a that b-a be coprime to f. James Harris
....................................................................
The points that are being argued over, are trivial, so why does the arguing
continue? I'll give you one of my favorite examples explaining, yet again,
how the distributive property is relevant: 7(x^2 + 3x + 2) = (7x + 7)(x + 2)
and you know by inspection which factor was multiplied by 7. Now let me
obscure things a bit with 7(x^2 + 3x + 2) = (a_1(x) + 7)(a_2(x) + 7) where
a_1(0) = 0, and a_2(0) = -5, and you can figure it out. The functions do not
change the distributive property. Nor does the value of x change which
factor was multiplied through by 7, and which was not. Notice too that if
you pick x=5, you will find that you have factors in common with 7 in BOTH,
but that doesn't change the basics. Those are the concepts. So what's
different with something like 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) =
7(25 x^2 + 30 x + 2) and the factorization using function 7 Q(x) = (5a_1(x)
+ 7)(5a_2(x) + 7) where the a's are defined by a^2 - (x - 1)a + 7(x^2 + x) =
0? If you are honest and look at it mathematically the only real difference
between that and my previous example is that the solution to the functions
obscures, which one has 7 as the factor. Otherwise, there is no meaningful
difference mathematically. I say, the distributive property STILL holds, and
it's basic mathematics, very old and respected and it should be respected
and conclusions that result from it, are valid. In contrast people argue
with me that factors of 7 somehow move around depending on the value of x,
so that the constants get multiplied by different values of 7, depending on
the value of x, where usually they don't talk that point. Listen carefully,
and you find these people talk functions, and stay away from the constants.
But let's go back to the simple example: 7(x^2 + 3x + 2) = (a_1(x) +
7)(a_2(x) + 7) with a_1(x) = 7x and a_2(x) = x - 5 you STILL HAVE FUNCTIONS,
but just have easy linear functions. Being a function does not make
something magical and does not give it magical powers to control constants
that are independent from variables. The difference between the simple
example and the more complex example is that with the simple example you can
SEE the value of the functions for any x, and knowing the factors of x gives
you the factors of the functions, while with the more complex example, you
can't always see directly. And why can't you? Because if you have solutions
for the a's from a^2 - (x - 1)a + 7(x^2 + x) = 0 with integer x, where the
result is irreducible over Q, and not the trivial case where x = 1 mod 7,
then you are blocked from directly looking at the factors by the ambiguity
of the square root. It's like with 1 + sqrt(4), if you couldn't evaluate
sqrt(4), how would you know that only one solution has 3 as a factor? With
1+sqrt(2) how do you know whether either solution has 3 as a factor?
TELLINGLY my arguments work when you can SEE the solutions as in getting
natural number results, which is why posters avoid that area. Remember, I
came back on Usenet talking about cases where you could see the answer with
my work. There is no evidence to the contrary. There is no mathematical
proof against these ideas as there can't be a disproof of the distributive
property. Claims that I am wrong go back to claims made by Dedekind over a
hundred years ago of proof, or are circular arguments claiming that results
in the ring of algebraic integers disprove my case. BUT despite anything
they may say here in arguing with me, or in attacking the merits of the
journal that published my paper, a paper over these arguments WAS PEER
REVIEWED AND PUBLISHED and no one can refute me mathematically as I have
mathematical proof. So instead they use social crap, and why? So that older
men who have built their careers on flawed ideas can stay comfortable on
your backs, depending on you accepting false mathematical ideas so that they
can keep their "accomplishments", their positions, and social status, while
you are screwed. How long can this impasse last? Probably until the
mathematicians currently in power retire. Once the people with the most to
lose from the truth, who spent most of their careers with the false
mathematical ideas retire, then the rumblings and bubblings of revolutionary
ideas can push through, and that's it, you are screwed. MUCH of what you are
learning now in number theory will just be tossed and will people even
bother to apologize to you? Will anyone consider your position now? How hard
it is? Will they care about the hours that you worked and toiled to learn
those ideas, all the homework, and assignments, and effort to get it down,
and then one day, the announcement comes--it's false. The world moves on,
and where are you? What then? What in the hell do you do then? What if they
say, oh, your degree isn't valid as the bulk of your degree is over
mathematical ideas shown to be false? And if you say, but how could I have
known? The world reminds you that the man who discovered the truth talked
about it VERY LOUDLY and PUBLICLY for years so you COULD HAVE KNOWN, but
chose to ignore the information, or were screwed by your fellow
mathematicians who COULD HAVE KNOWN but didn't talk about it. So your group
goes down, and takes you with it. If civilization is to continue, the truth
has to win in these situations. History shows that it will win. If you are
hoping on the failure of humanity I am here to tell you your hope is futile,
and I'll make that point later as well, as over 10,000 years of human
civilization has depended on groups like yours LOSING the fight against the
truth. The position of your society is ultimately against the future of
everyone else. And history also shows that there are LOTS of people like you
who get steamrolled, who never thought it was possible, never really saw it
coming, and when it happens they look to all these people to feel sorry for
them, and save them, and the world just keeps on turning. James Harris
......................................................................................
Rupert wrote: > jstevh@xxxxxxx wrote: > > The points that are being argued
over, are trivial, so why does the > > arguing continue? > > > > I'll give
you one of my favorite examples explaining, yet again, how > > the
distributive property is relevant: > > > > 7(x^2 + 3x + 2) = (7x + 7)(x + 2)
let > > me obscure things a bit with > > > > 7(x^2 + 3x + 2) = (a_1(x) +and you know by inspection which factor was multiplied by 7. Now
7)(a_2(x) + 7) > > > > where a_1(0) = 0, and a_2(0) = -5, and you can figure
it out. > > > > The functions do not change the distributive property. Nor
does the > > value of x change which factor was multiplied through by 7, and
which > > was not. Notice too that if you pick x=5, you will find that you
have > > factors in common with 7 in BOTH, but that doesn't change the
basics. > > > > Those are the concepts. > > > > So what's different with
something like > > > > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > > =
7(25 x^2 + 30 x + 2) > > > > and the factorization using function > > > > 7
Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > > > > where the a's are defined by > > >
a^2 - (x - 1)a + 7(x^2 + x) = 0? > > > > If you are honest and look at itmathematically the only real > > difference between that and my previous
example is that the solution to > > the functions obscures, which one has 7
as the factor. > > > > Why should either one have 7 as a factor? Take x=1,
for example, then > a=+/- sqrt(-14). The distributive property. What's key
in this thread is that there is NO MEANINGFUL DIFFERENCE between 7(x^2 + 3x
+ 2) = (a_1(x) + 7)(a_2(x) + 7) where a_1(0) = 0 and a_2(0) = -5 where the
a's are really strange bizarre functions or simple linear functions like
a_1(x) = 7x and a_2(x) = x-5. That's it. That's the point of the thread. Now
some of you are saying that depending on what kind of functions you have,
the distributive property holds or it doesn't hold. I say the distributive
property holds. Given the simple example with linear functions, do you deny
that the distributive property holds? I really doubt you will, as, of
course, if you try, you can just see that a_1(x) = 7x, and you may naively
believe that seeing is the proof. But being able to see that a_1(x) = 7x is
NOT the proof that given a_1(0) = 0, and a_2(0) = -5, with 7(x^2 + 3x + 2) =
(a_1(x) + 7)(a_2(x) + 7) it must be the case that one of the a's has 7 as a
factor. The only meaningful difference with more complicated examples, like
the Decker equations is that the radical obscures the functions for certain
non-rationals. That's it. That's also why with rational solutions you see
that everything follows the distributive property. So simply tossing out the
word "function" as if being functions means the distributive property goes
away does nothing. If you ponder this post, and are serious about
mathematics, then there will be no room for doubt, if you do so objectively.
I think it a true test of those who argue with me, to either explain how one
set of simple functions can obey the distributive property, while other more
complex ones can simply thumb their noses at it, and mathematics be
consistent. James Harris P.S. And you know that one of the a's has 7 as a
factor because of the distributive property.
...............................................................................
Yes, I did it again, went off with claims of easy proofs which turned out to
be wrong, as I fought it out over an approach that just didn't work. Big
deal. I don't doubt that as they've done in the past there will be posters
who will go on and on about me screwing up and boldly claiming I had these
dramatic proofs when I was just wrong. And admitting I am wrong! Wow, you
should see the heat that flows over my just coming out and saying that I
screwed up. I get made fun of for that as well, as if you people are
perfect. Nope. You just don't try hard enough. Sure, if you want to never
have to back-pedal and admit mistakes, you can work very hard to never make
them, look very carefully over every single argument you come up with--and
then just sit on them--as try as hard as you can, you will fail, and
screw-up if you dare to put it out there. I do dare. I love it. And I make a
lot of mistakes along the way. So yeah, the other side is to say though that
I screw up a lot. And I do. Lots of claims of proofs where I've had to
back-track as I was wrong. But that's extreme mathematics. You push the
limit. Forget about social crap so you don't care so much about the little
people who will get all hot and bothered about your mistakes. Also, when
you're arguing with people, emotion can get the best of you and you stop
thinking straight as you get defensive and look for any kind of explanation
that you think defends your case. In any event, I had some claims of proof
over the weekend that were just wrong. Went really wild Sunday with quite a
lot of crap and a lot of crap arguments where I was arguing positions that
just didn't work like I thought as I was happy as hell, telling myself how
brilliant these little proofs were. And they weren't proofs. James Harris
................................................
Oh, over the weekend I had a lot of trouble handling an example from a Rick
Decker of Hamilton College that paralleled my own non-polynomial
factorization argument with simpler quadratics where one had a special
feature. I made a lot of posts claiming proof where the arguments did not
prove what I thought they did when I posted them, and I was wrong. So yes,
those previous claims of proof made a few days ago, were crap. Here is the
correct answer to the Decker example. As a synoposis up front as I've had at
least one poster ask for something beforehand that explains what's going to
happen, I am going to use the Decker equations with the same approach I've
used with my own more complex cubic generating equations to show a complete
proof, which just so happens to shoot down standard usage of Galois Theory,
and the theory of ideals. The approach is basic, relies on simple algebraic
concepts, where the key step depends on the use of the distributive
property. In the ring of algebraic integers, I have from Decker that 7 Q(x)
= 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) = 7(25 x^2 + 30 x + 2) and 7 Q(x) =
(5a_1(x) + 7)(5a_2(x) + 7) where the a's are defined by a^2 - (x - 1)a +
7(x^2 + x) = 0. Given 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) it cannot be the
case that the 7's visible in the factorization both result from the 7
multiplied on the left side, as the constant term of Q(x) is coprime to 7,
since it is 2. So both constant terms--those terms constant with respect to
x--cannot be visible in the factorization. However, the actual constant
terms can be determined by clearing x out, letting x=0, I have that one of
the a's goes to 0, while the remaining one is -1, from a^2 + a = 0. And
looking at 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) I have that the actual
constant terms are 7 and 5(-1) + 7 = 2, but it's not mathematically
determined which of the two factors was multiplied by 7, and which one
actually is coprime to 7, but you know ONE must have been multiplied by 7,
from the distributive property and the constant terms. Note that the result
that only one of the factors was multiplied by 7 follows from the
distributive property, and noting the values of the constant terms. That
completes the proof. It's rather short as you can see. It contradicts with
standard ideas from Galois Theory because from that theory, if the a's are
non-rational and irreducible over Q, except for the special case where both
have sqrt(7) which I'll go over next, then by those ideas NEITHER of the a's
can have 7 as a factor. And it can be shown that neither does have 7 as a
factor in the ring of algebraic integers. But that then proves--as the
simple proof cannot be wrong or it wouldn't be a proof--that the ring of
algebraic integers is incomplete, and for those who don't understand why it
shows that, consider the analogy where considering evens as a ring, 2 and 6
are coprime in that ring because 3 is not even. Make sense? Similarly, 7
being coprime in the ring of algebraic integers, when it has been proven to
be a factor by a simple algebraic proof, shows that it is excluded because
its co-factor is not an algebraic integer. I would be interested in at least
some posters actually trying to answer the proof itself, and its steps,
especially the key step where I note that the constant factor, which is 7,
proves that 7 was multiplied through that particular factor. Now to the
sidepoint about special cases. It's not hard to explain mathematically. It's
easy to determine that at x=1 mod 7, both of the a's have sqrt(7) as a
factor. The explanation is simple enough, but it might help in understanding
to divide 7 from the factorization and solve for the solution where 7 is a
factor of only one, so Q(x) = (x^2 + x)(5^2) + (-1 + x)(5) + 7 = 25 x^2 + 30
x + 2 and Q(x) = (5b_1(x) + 1)(5a_2(x) + 7) where a_1(x) = 7b_1(x), so I
also have b_1(x) a_2(x) = x^2 + x and 5a_2(x) + 7(5)b_1(x) = (-1 + x)(5) so
I can solve out a_2(x), to get a_2(x) = (x^2 + x)/b_1(x), and substitute in
the second to get (x^2 + x)/b_1(x) + 7b_1(x) = -1 + x which gives 7
b_1(x)^2 - (x-1)b_1(x) + x^2 + x = 0 and the solution that b_1(x) = ((x-1)
+/- sqrt((x-1)^2 - 28(x^2 + x)))/14 and at x=1, I have have b_1(1) = +/-
sqrt(-14(5))/14 so it's just that at that point and it turns out at all
points where x = 1 mod 7, that 7 b_1(x) has sqrt(7) as a factor. So that
case actually says nothing about the earlier argument, which depends on the
distributive property. If you wish to dispute with only one factor having 7
as its factor, then you might consider the more general case: Q(x) =
(5b_1(x) + f_1)(5b_2(x) + f_2) where f_1 f_2 = 7, and b_1(x) f_2 = a_1(x)
and b_2(x) f_2 = a_2(x) and answer the question of, what mathematical reason
might there be for any particular choice for the f's? I've given a reason
based on the distributive property and the constant terms. If you dispute
that reason, you have an INFINITE number of possible algebraic integer f's
that are available, so how do you choose one configuration? Note again the
rest here is a side discussion noting how the cases at x = 1 mod 7 do not
change the basic argument, which relies on the distributive property. That
argument has gone through formal peer review and been published with the
more complicated expressions I used in my paper on non-polynomial
factorization, but then some social pressure was brought to bear agains the
journal and the paper was pulled, and a bit later the journal keeled over
and died. James Harris
...............................................................
Gib Bogle wrote: > jstevh@xxxxxxx wrote: > > > Oh, over the weekend I had a
lot of trouble handling an example from a > > Rick Decker of Hamilton
College that paralleled my own non-polynomial > > factorization argument
with simpler quadratics where one had a special > > feature. > > > > I made
a lot of posts claiming proof where the arguments did not prove > > what I
thought they did when I posted them, and I was wrong. > > > > So yes, those
previous claims of proof made a few days ago, were crap. > > > > Here is the
correct answer to the Decker example. > > I love it. Upmteen wrong
statements, repeated ad nauseam, and > accompanied by intemperate insults of
all who pointed out the errors, > and all finally conceded to be wrong. Now,
on the other hand, a true > statement. Yeah, right. The basic argument is
the same one as has been used for years. It has been formally peer
reviewed--and published in a math journal--and later the journal editors
retracted and the journal died. What happened over the weekend is that I
looked for another argument to go along with the one that has been checked
so thoroughly, as you people keep fighting me over the distributive
property, and I saw mirages. Big deal. In this thread I went back to the
tried-and-true where I guess people will again argue with me over the
freaking distributive property. Oh well. You losers don't make a lot of
public mistakes as you don't try hard enough. It's like batting. So what if
your average is high if you almost never go to the plate? And have such a
high average as you have few at bats and never play in the big games? Never
try, few failures, except that you are the failure. James Harris
.......................................................
William Hughes wrote: > jstevh@xxxxxxx wrote: > > Oh, over the weekend I had
a lot of trouble handling an example from a > > Rick Decker of Hamilton
College that paralleled my own non-polynomial > > factorization argument
with simpler quadratics where one had a special > > feature. > > > > I made
a lot of posts claiming proof where the arguments did not prove > > what I
thought they did when I posted them, and I was wrong. > > > > So yes, those
previous claims of proof made a few days ago, were crap. > > > > Here is the
correct answer to the Decker example. > > > > As a synoposis up front as
I've had at least one poster ask for > > something beforehand that explains
what's going to happen, I am going > > to use the Decker equations with the
same approach I've used with my > > own more complex cubic generating
equations to show a complete proof, > > which just so happens to shoot down
standard usage of Galois Theory, > > and the theory of ideals. > > > > The
approach is basic, relies on simple algebraic concepts, where the > > key
step depends on the use of the distributive property. > > > > In the ring of
algebraic integers, I have from Decker that > > > > 7 Q(x) = 7((x^2 +
x)(5^2) + (-1 + x)(5) + 7) > > = 7(25 x^2 + 30 x + 2) > > > > and > > > > 7
Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > > > > where the a's are defined by > > >
a^2 - (x - 1)a + 7(x^2 + x) = 0. > > > > Given > > > > 7 Q(x) = (5a_1(x)+
7)(5a_2(x) + 7) > > > > it cannot be the case that the 7's visible in the
factorization both > > result from the 7 multiplied on the left side, as the
constant term of > > Q(x) is coprime to 7, since it is 2. > > > > So both
constant terms--those terms constant with respect to x--cannot > > be
visible in the factorization. > > > > However, the actual constant terms can
be determined by clearing x out, > > letting x=0, I have that one of the a's
goes to 0, while the remaining > > one is -1, from > > > > a^2 + a = 0. > >
have that the actual constant terms are 7 and 5(-1) + 7 = 2, but it's > >And looking at > > > > 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) > > > > I
not mathematically determined which of the two factors was multiplied > > by
7, and which one actually is coprime to 7, but you know ONE must > > have
been multiplied by 7, from the distributive property and the > > constant
terms. > > > > No, you don't. . > > Your have > > k (5a_1(x) +7) = (b(x) +
1) > > with b(0)=0 > and you conclude that k must be 1/7 and b(x) must be
5a_1(x)/7. That's not correct. Key here is that it is not determinable WHICH
of the b's has been multiplied by 7, and which one has not. So you have 7
Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) where you only know that one has 7 as a
factor, and one does not, but there's no mathematical determination which
one. But given that 7 is clearly a factor of only one, based on the constant
terms, which is 7, for one and 2 for the other, it's shown that ONE of the
a's in general has 7 as a factor, though, as has been pointed out
repeatedly, an infinity of cases can be found where that is not true in the
ring of algebraic integers. > However, this > is just one possibility. > >
Try k = 1/(7(1+x^2)) then > > k(a(x) +7) = (5a_1(x)/(7(1+x^2) + 1/(1+x^2)) >
=( [5a_1(x)/(7(1+x^2)) + 1/(1+x^2) -1] + 1) > > so k = 1/(7(1+x^2)) and b(x)
= [5a_1(x)/(7(1+x^2)) + 1/(1+x^2) -1] > is another possibility. (There are
in fact an infinite number > of possibilities.) The constant term in both
cases is 7 before That's actually a good thing to consider when thinking
about your objections--an infinite number of possibilities--so how does the
math choose? The answer is that it is forced to go one way by the
distributive property. The functions here are only different from the
functions in the example 7(x^2 + 3x + 2) = (a_1(x) + 7)(a_2(x) + 7) where
a_1(x) = 7x and a_2(x) = x - 5, in that it is obscured WHICH function has 7
as a factor, except for rational cases. Now then, apply your exact same
objection to the example I've shown, and you'll see that it still appears to
work!!! Or do you disagree? If so, why can't you find a k with my simple
example just like before? > multiplication > by k and 1 after multiplication
by k. So looking at the constant term > cannot tell you which k was used. >
-William Hughes But your argument if true would also prove the same thingabout 7(x^2 + 3x + 2) = (a_1(x) + 7)(a_2(x) + 7) where a_1(x) = 7x and
a_2(x) = x - 5. I suggest to you that your argument is flawed. James Harris
...........................................................
Rick Decker wrote: > jstevh@xxxxxxx wrote: > > Oh, over the weekend I had a
lot of trouble handling an example from a > > Rick Decker of Hamilton
College that paralleled my own non-polynomial > > factorization argument
with simpler quadratics where one had a special > > feature. > > > > I made
a lot of posts claiming proof where the arguments did not prove > > what I
thought they did when I posted them, and I was wrong. > > > > So yes, those
previous claims of proof made a few days ago, were crap. > > > > Here is the
correct answer to the Decker example. > > > > As a synoposis up front as
I've had at least one poster ask for > > something beforehand that explains
what's going to happen, I am going > > to use the Decker equations with the
same approach I've used with my > > own more complex cubic generating
equations to show a complete proof, > > which just so happens to shoot down
standard usage of Galois Theory, > > and the theory of ideals. > > > > The
approach is basic, relies on simple algebraic concepts, where the > > key
step depends on the use of the distributive property. > > > > In the ring of
algebraic integers, I have from Decker that > > > > 7 Q(x) = 7((x^2 +
x)(5^2) + (-1 + x)(5) + 7) > > = 7(25 x^2 + 30 x + 2) > > > > and > > > > 7
Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > > > > where the a's are defined by > > >
a^2 - (x - 1)a + 7(x^2 + x) = 0. > > > > Given > > > > 7 Q(x) = (5a_1(x)+
7)(5a_2(x) + 7) > > > > it cannot be the case that the 7's visible in the
factorization both > > result from the 7 multiplied on the left side, as the
constant term of > > Q(x) is coprime to 7, since it is 2. > > > > So both
constant terms--those terms constant with respect to x--cannot > > be
visible in the factorization. > > > > However, the actual constant terms can
be determined by clearing x out, > > letting x=0, I have that one of the a's
goes to 0, while the remaining > > one is -1, from > > > > a^2 + a = 0. > >
have that the actual constant terms are 7 and 5(-1) + 7 = 2, but it's > >And looking at > > > > 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) > > > > I
not mathematically determined which of the two factors was multiplied > > by
7, and which one actually is coprime to 7, but you know ONE must > > have
been multiplied by 7, from the distributive property and the > > constant
terms. > > No. You keep tripping up on this point. Just because the product
of > two numbers, pq, is divisible by 7 in the algebraic integers, you >
cannot conclude that either p or q must be divisible by 7. That's > true in
the integers, but is not true in the algebraic integers. I conclude nothing
from the ring of algebraic integers, as I simply use the distributive
property, and often do that without specifying a ring, which often evokes
howls and questions like, in what ring? a(b+c) = ab +ac so knowing that the
term constant with respect to x, for ONE of the factors is 7, while the
other is 2, when 7 was multiplied across, I know that ONE of the functions
has a factor of 7, from the distributive property, without need of mention
of a ring. An example to explain the concept is 7(x^2 + 3x + 2) = (7x + 7)(x
+ 2) where the constant terms are 7 and 2, and the same argument proves that
7 was multiplied through the first factor. The only difference with your
more complex equations is that since the a's are roots of a quadratic
generator, their value is obscured by radicals EXCEPT when you have rational
solutions. But my point is that the distributive property still holds
whether you can see the answer directly or not. > > > > Note that the result
that only one of the factors was multiplied by 7 > > follows from the
distributive property, and noting the values of the > > constant terms. > >
isn'tThat completes the proof. It's rather short as you can see. > > It
a proof. > Then point out an error. The steps are clearly outlined, and the
most important step relies on the distributive property. Refuting that
position requires refuting the distributive property. I think you are wrong
and the distributive property is right. > Unsurprisingly, you've been
seduced by the rational integers. > There are only two rational integers x
that give rational integer > roots for > > r(a, x) = a^2 + (1 - x) a + 7(x^2
+ x) > I have been seduced by algebra. And I LIKE the distributive property
and refuse to give it up just because a lot of people argue with me,
attacking it, because the key step in the proof relies on it. > namely, > >
x = 0 r(a, x) = a^2 + a > x = -1 r(a, -1) = a^2 + 2 a > > In these two
cases, your conjecture is true, namely that one of the > roots is dividible
by 7 and the other is coprime to 7. > That's easy enough with 7 as the
factor, but you can generalize to f, and use ANY rational f to be able to
see the same result. The DISTRIBUTIVE PROPERY holds in all cases. >
Unfortunately, these are the ONLY values of x for which the roots > behave
as you want them to. > That is irrelevant as your example can be
generalized. But it's also irrelevant because I have a short proof. > For
example, let x = 3. We see that 7 Q(3) = 7(25(9) + 30(3) + 2) = > 7(317) and
r(a, 3) = a^2 - 2 a + 7(12) so > > a_i = (2 +/- sqrt(4 - 4(84)) / 2 = 1 +/-
sqrt(-83) > > and > > [5(1 + sqrt(-83)) + 7][5(1 - sqrt(-83)) + 7] > = [12 +
5 sqrt(-83)][12 - 5 sqrt(-83)] > = 144 + 25(83) > = 2219 = 7(317), as
expected. > > However, neither of 1 +/- sqrt(-83) is divisible by 7. If one
were, then > a = 7b for some algebraic integer b, but then we'd have > >
(7b)^2 - 2(7b) + 7(12) = 0 > > from [1] with x = 3, so > > 49 b^2 - 2(7) b +
7(12) = 0 > > and hence b would satisfy the monic irreducible polynomial
equation > > 7 b^2 - 2 b + 12 = 0. > > In addition, we can find explicit
nonunit common factors of a and 7, > although it takes a bit of work. We can
show that > > (4627 + 423 a)(4627 - 423 a) = 7^9 > > and > > (4627 + 423
a)(72192 - 512 a) = a^9 > > And since t = (4627 + 423 a)^{1/9) satisfies > >
t^{18} - 10100 t^{9} + 40353607 = 0 > > we have that t is an algebraic
integer nonunit common factor of a and 7, > establishing the fact that a and
7 are not coprime in the ring of > algebraic integers. Irrelevant to the
question of proof. I focus on a short algebraic proof for a reason, which is
that proof is supposed to be paramount. And I counter that the ring of
algebraic integers is flawed in that it is not complete, so you can get some
weird results that CONTRADICT the algebra. The only way to answer that claim
is to go to the proof, and show that it is not a proof. > > > > It
contradicts with standard ideas from Galois Theory because from that > >
theory, if the a's are non-rational and irreducible over Q, > > [You mean
"if r(a, x), considered as a polynomial in a, is irreducible > over Q."] > >
except for > > the special case where both have sqrt(7) which I'll goover
next, then > > by those ideas NEITHER of the a's can have 7 as a factor. > >
That's right, but you don't need Galois Theory to establish that. In >this example, except for the two cases mentioned above (x = 0, -1), > either
both roots will have 7 as a factor, or neither will. > > BTW, your "special
case" isn't. > You may not need Galois Theory but the result sinks current
usage. It also brings into question the theory of ideals. > > And it can be
shown that neither does have 7 as a factor in the ring of > > algebraic
integers. > > Yes. > > > > But that then proves--as the simple proof cannot
be wrong or it > > wouldn't be a proof--that the ring of algebraic integers
is incomplete, > > and for those who don't understand why it shows that,
consider the > > analogy where considering evens as a ring, 2 and 6 are
coprime in that > > ring because 3 is not even. > > > > Make sense?
Similarly, 7 being coprime in the ring of algebraic > > integers, when it
has been proven to be a factor by a simple algebraic > > proof, shows that
it is excluded because its co-factor is not an > > algebraic integer. > > Go
back to what I said above. If p and q are algebraic integers with > the
product pq divisible by 7, you cannot conclude that either of > p or q must
be divisible by 7. That's true in the rational numbers > but not true in the
algebraic integers. > > > > I would be interested in at least some posters
actually trying to > > answer the proof itself, and its steps, especially
the key step where I > > note that the constant factor, which is 7, proves
that 7 was multiplied > > through that particular factor. > > 7 is not a
"prime" in the ring of algebraic integers. That's where your > explanation
fails. > > > <snip> > It's not an explanation, it's a proof. I wish to
remind people that there is a PROOF and that proof is usually mostly ignored
so that people can start going on about the ring of algebraic integers and
results in that ring. BUT my point is that the the proof shows a problem
with the ring of algebraic integers, so why just keep going back to where I
claim there's a problem? I say you do that because you can't answer the
proof, refuse to acknowledge the truth, and have no where else to go. James
Harris .......................................................
marcus_b wrote: > > Oh, over the weekend I had a lot of trouble handling an
example from a > > Rick Decker of Hamilton College that paralleled my own
non-polynomial > > factorization argument with simpler quadratics where one
had a special > > feature. > > > > I made a lot of posts claiming proof
where the arguments did not prove > > what I thought they did when I posted
them, and I was wrong. > > > > So yes, those previous claims of proof made a
few days ago, were crap. > > > > Here is the correct answer to the Decker
example. > > > > As a synoposis up front as I've had at least one poster ask
for > > something beforehand that explains what's going to happen, I am
going > > to use the Decker equations with the same approach I've used with
my > > own more complex cubic generating equations to show a complete proof,
and the theory of ideals. > > > > The approach is basic, relies on simplewhich just so happens to shoot down standard usage of Galois Theory, >
algebraic concepts, where the > > key step depends on the use of the
distributive property. > > > > In the ring of algebraic integers, I have
from Decker that > > > > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > > =
7(25 x^2 + 30 x + 2) > > > > and > > > > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7)
0. > > Given > > > > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > > > > it cannotwhere the a's are defined by > > > > a^2 - (x - 1)a + 7(x^2 + x) =
be the case that the 7's visible in the factorization both > > result from
the 7 multiplied on the left side, as the constant term of > > Q(x) is
coprime to 7, since it is 2. > > > > So both constant terms--those terms
constant with respect to x--cannot > > be visible in the factorization. > >
Q(0)However, the actual constant terms can be determined by clearing x out,
letting x=0, I have that one of the a's goes to 0, while the remaining
one is -1, from > > > > a^2 + a = 0. > > > > And looking at > > > > 7
= (5a_1(0) + 7)(5a_2(0) + 7) > > > > I have that the actual constant terms
are 7 and 5(-1) + 7 = 2, > > but it's > > not mathematically determined
which of the two factors was multiplied > > by 7, and which one actually is
coprime to 7, but you know ONE must > > have been multiplied by 7, from the
distributive property and the > > constant terms. > > > > I need to fill in
a few details to understand your argument. Please > bear with me. > > Let's
go back to basics. You will agree that a key to your argument > is the
constant terms. > Yes. And they are 7 and 2. > Your notation implies that
a_1(x) and a_2(x) are functions of x. Yes, they are functions of x. > You
know that a_1(0) = 0 and a_2(0) = -1. These are the constant > terms of
these two functions. > Those are their values at 0. The constant terms are
the factors of the constant term of the polynomial--and I remind that the
polynomial is 7(25 x^2 + 30 x + 2) where you can see that the constant term
is 7(2). > Let f(x) be the maximal divisor of 7 which also divides a_1(x), >
in the algebraic integers. That is, f(x) is an algebraic integer > and
7/f(x) is an algebraic integer and a_1(x)/f(x) is an algebraic > integer. I
am not saying anything here about what f(x) actually > is. Ok. > > Define
g(x) similarly for a_2(x): that is, g(x) is an algebraic > integer and
7/g(x) is an algebraic integer and a_2(x)/g(x) is > an algebraic integer.
Ok. > Further it must be the case that f(x) g(x) = 7, because 7 > factors
out of the whole expression only once. > Yup. > Now consider > > (5 a_1(x) +
7)/f(x) = 5 a_1(x)/f(x) + 7/f(x). > > This is an algebraic integer because
each part is separately. > Ok. > Similarly, > > (5 a_2(x) + 7)/g(x) = 5
a_2(x)/g(x) + 7/g(x). > > Now, what you claim is that f(x) must equal 7 for
all x and > g(x) must equal 1 for all x, at least up to units. Huh? I claim
that the distributive property proves that one factor is multiplied through
by 7, while the other is not. And here it IS important to be extremely
precise about what I claim, as what you just said I claim, is kind of all
over the map in comparison, and notice, you make NO MENTION OF THE
DISTRIBUTIVE PROPERTY. I wish to emphasize that while my actual claim relies
on the distributive property. YOU MAKE NO MENTION OF THE DISTRIBUTIVE
PROPERTY. Let's cut to the chase. You will note that you can find algebraic
integers for f(x) and g(x), and I will say, so? The distributive property
still works, so what is the other possible explanation? I am curious to know
if you are willing to even concede that there is a possible explanation for
how 7 can multiply through by the distributive property, and yet you can
find f(x) and g(x) in the ring of algebraic integers. Can you think of a
way? > Here is how I think your reasoning goes, though you don't spell it >
all out here. > > The constant term of 5 a_2(x) + 7 must be > > 5 a_2(0) + 7
= 5*(-1) + 7 = 2. > > Therefore the constant term of > > (5 a_2(x) + 7)/g(x)
terms that one of the a's has been multiplied through by 7, while one hasmust be 2/g(x). No. The mathematics only tells you from the constant
not, but not which one. By the distributive property, there is only one
conclusion. Now then, how do you explain that f(x) and g(x) can be found
where they are algebraic integers, when it can be shown that if f(x)=7,
there are non-rational cases where they are not? I can explain it. If you
wish, you can try to avoid my questions in this area, and then I can come
back and explain exactly how. > > Now, since g(x) is a divisor of 7, it is
coprime to 2. So > the only possibility for g(x) is that it is a unit, i.e.,
it > is essentially equal to 1. And then you can conclude f(x) must > be 7.
theIs this a reasonable summary of your logic? > > Marcus > You left in
rest of what I said, and I'll truncate it in a bit, but I want to leave in
the next two paragraphs. > > Note that the result that only one of the
factors was multiplied by 7 > > follows from the distributive property, and
noting the values of the > > constant terms. > > > > That completes the
proof. It's rather short as you can see. > > My position relies on a(b+c) =
ab + ac and talk about units here or units there and what's true in the ring
of algebraic integers do not change the reality of the key step in the
proof. The ring of algebraic integers is a complex entity made more complex
by special rules which some of you are acting like you don't understand, so
you get results that contradict with what can be shown by proof. So with the
ring of algebraic integers you can contradict the distributive property. Fun
stuff, if you're a mathematician, a real one. But ultimately you have to
resolve the contradiction to show that mathematics IS consistent...if you're
a real mathematician. James Harris
......................................................
I've been working a bit at getting a disproof showing the problem with
Galois Theory that removes all possibility for disagreement that isn't just
clearly bogus grasping at failed ideas. Here it is. Again starting with the
equations given by Rick Decker of Hamilton College, In the ring of algebraic
integers, consider 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) = 7(25 x^2 +
30 x + 2) and 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) where the a's are defined
by a^2 - (x - 1)a + 7(x^2 + x) = 0 where Decker chose this example because
when x=1 mod 7, the middle term has 7 as a factor which he I guess thought
was a counterexample to my ideas because at x=0, the a's are 0 and 1,
indicating that 7 divides through only one of the expressions, except for
the special cases. The first step in the disproof is to focus on that last
quadratic and write it with the focus on x, which gives a non-monic
polynomial that becomes monic if 'a' has 7 as a factor: 7x^2 + (7-a)x + a^2
+ a = 0 Now choose an integer value for a, like a=3, so that you have a
non-monic polynomial irreducible over Q, and therefore, you have that x
cannot be an algebraic integer. But because 3 is coprime to 7 I know that
the other solution for 'a' MUST have 7 as a factor, right? But now I just go
back to a^2 - (x - 1)a + 7(x^2 + x) = 0 and note that for any solution for x
that resulted from my previous equation, where again, I know that x cannot
be an algebraic integer there is now NO way that x can be a ratio of
algebraic integers with a denominator wiht only SOME factors in common with
7, and that quadratic have a root that has 7 as a factor. It's a
mathematical impossibility. So there is a direct contradiction with the
result from the ring of algebraic integers, as with one integer 'a' such
that a^2 + a is coprime to 7, it MUST be the case that the other 'a' has 7
as a factor, while that is impossible in the ring of algebraic integers,
while there is no meaningful way for that 'a' to be like some kind of
fraction i.e. a ratio of algebraic integers where the numerator and
denominator are actually coprime. However, it CAN be written as a ratio of
algebraic integers where the numerator and denominator are coprime in the
ring of algebraic integers, which directly shows the apparent contradiction,
which is resolved by recognizing the ring of algebraic integers is
incomplete. Denial of the mathematics does not change anything. Bogus math
ideas that DO NOT WORK are not worth fighting for. I hope that some of you
choose to be mathematicians versus people who just claim to be
mathematicians. Each day that you let the false mathematical ideas reign
supreme is another day you take away time with the correct and more powerful
ideas from humanity. You are attacking the future of humanity, and fighting
against correct and powerful mathematical ideas to hold on to bogus ones
that DO NOT ACTUALLY WORK, which can be proven wrong with trivial algebra!!!
Why do it? James Harris.
.........................................................
William Hughes wrote: > jstevh@xxxxxxx wrote: > > I've been working a bit at
getting a disproof showing the problem with > > Galois Theory that removes
all possibility for disagreement that isn't > > just clearly bogus grasping
at failed ideas. > > > > Here it is. > > What is this, the fifth in 24
hours? > So? This exposition leaves no room for arguing, which is why I
guess you started with social crap. Who cares how long it takes me to
explain in a way that no one can refute, once I achieve the goal? See below.
The goal is achieved. > > > > > Again starting with the equations given by
Rick Decker of Hamilton > > College, In the ring of algebraic integers,
consider > > > > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > > = 7(25 x^2
+ 30 x + 2) > > > > and > > > > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > > > >
where the a's are defined by > > > > a^2 - (x - 1)a + 7(x^2 + x) = 0 > > > >
where Decker chose this example because when x=1 mod 7, the middle term > >
has 7 as a factor which he I guess thought was a counterexample to my > >
ideas because at x=0, the a's are 0 and 1, indicating that 7 divides > >
through only one of the expressions, except for the special cases. > > > >
The first step in the disproof is to focus on that last quadratic and > >
write it with the focus on x, which gives a non-monic polynomial that > >
becomes monic if 'a' has 7 as a factor: > > > > 7x^2 + (7-a)x + a^2 + a = 0
non-monic polynomial irreducible over Q, and therefore, you have that x > >Now choose an integer value for a, like a=3, so that you have a > >
cannot be an algebraic integer. > > > > But because 3 is coprime to 7 I know
that the other solution for 'a' > > MUST have 7 as a factor, right? > > > >
But now I just go back to > > > > a^2 - (x - 1)a + 7(x^2 + x) = 0 > > > >
and note that for any solution for x that resulted from my previous > >
equation, where again, I know that x cannot be an algebraic integer > >
there is now NO way that x can be a ratio of algebraic integers with a > >
denominator wiht only SOME factors in common with 7, and that quadratic > >
have a root that has 7 as a factor. > > > > Why? This is not clear at all?
Are you still using, "if a doesn't > divide b and a doesn't divide c, then a
doesn't divide (b+c)"? > > > -William Hughes Choose an 'a' that is an
integer, such that a^2 + a is coprime to 7, for instance, let one of the a's
equal 3. Then 7x^2 + (7-a)x + a^2 + a = 0 will be irreducible over Q,
proving that x cannot be an algebraic integer. Now let x = d/c, where d and
c are pairwise coprime and algebraic integers, and make the substitution
into a^2 - (x - 1)a + 7(x^2 + x) = 0 which gives a^2 - (d/c - 1)a +
7(d^2/c^2 + d/c) = 0 which is a^2 - (d/c - 1)a + (7(d^2 + cd))/c^2 = 0 and
if c is not coprime to 7, then neither solution for 'a' can have 7 as a
factor, but given the integer 'a' previously chosen, it MUST be the case
that 7 is a factor of the other one, and the contradiction is shown. You
see, if c has just some factors in common with 7 while being coprime to
others then those factors will divide off from 7, leaving them unavailable
for the a's, but with one of the a's coprime to 7, like if it's 3, the other
of the a's MUST have 7 itself as a factor. Trivial it's so easy. You see, c
must be a unit, but provably cannot be a unit in the ring of algebraic
integers, proving quite simply that that ring is incomplete. There is no
room left for argument. The ring of algebraic integers has been shown to be
incomplete, disproving the standard usage of Galois Theory, and showing the
theory of ideals is flawed. It's so simple that it's hard to imagine what
would convince you if that doesn't. James Harris
.................................................................
Let's just say for the sake of argument that mathematicians just came out
and acknowledged that through accidents of history and subtlety of some
difficult concepts an erroneous group of techniques became dominant in
number theory. That news would make headlines around the world. Now though,
there is quiet, so let's look at the other way this can work out, as
mathematicians can instead try to ignore the result. Then, as history shows,
the result will emerge eventually as there is over 10,000 years of human
civilization where these kinds of battles have played out, and the side
opposing the truth, has always lost. If the story emerges within a couple of
years then clearly mathematicians will have been gambling on it not coming
out within their careers, and will have lost that gamble. Why is it a
gamble? Well these things have happened before. In the physics field
recently there was the emergence of quantum mechanic and relativity, where
adjustments had to be made with dramatic consequences for the entire world.
The physicists absorbed the impact, fought a bit, yes, but quickly came on
the side of truth. If mathematicians do not, then they will still lose, but
they also get a much darker story, and those mathematicians who are in
powerful positions when the story comes out, will probably take the worst of
it. Social castigation. Their pictures in the papers. Reporters hounding
them with hard questions. And it won't end during their lifetimes. These
stories are huge on a scale that's hard to comprehend. They keep coming but
people never see them coming. The stories get bigger with time, but while
you're living in it, it can seem unimaginable. Simple self-preservation
would make some of you sing like canaries at this point if you had any sense
of what is going to come, while otherwise you have to hold on to the belief
that you can play the odds, play for time, and hope that the story stays
buried long enough for you to have a long career in mathematics, retire, and
die with no one ever knowing the truth. Let the future handle it, you may
think. But you will not get that time. I'll make sure you don't get that
time. You will not get to grow old and die with the world thinking you're
something you're not. The story will come out before then and instead you'll
be castigated by world society. I'll paint you for what you are--a dangerous
element in society fighting against the foundations of society and
technological progress--and remind that if people like you ever succeed then
our Progress as a species, comes to a screeching halt. What if physicists
had tried and succeeded at what some of you may now think you can succeed
at? What if they'd shut down Einstein's work, ignored his papers? What if
they'd fought quantum mechanics tooth-and-nail? What if they'd blocked the
knowledge? Well, I wouldn't be typing this up on this computer as computers
wouldn't be here. Or maybe we'd have some kind of clunky mechanical
computer, but would we have had the transistor, and the technological
revolution? Or might we have physicists fighting to explain odd behavior
within the framework of the old knowledge, vigorously attacking "cranks" and
"crackpots" who attempted to push through the ideas of quantum mechanics?
You will not succeed in blocking the knowledge. But if you attempt to do so,
when you are broken, you lose so much. So why bother? Why not just tell the
truth now? Why fight the future? James Harris
.......................................................
Jesse F. Hughes wrote: > jstevh@xxxxxxx writes: > > > Then, as history
shows, the result will emerge eventually as there > > is over 10,000 years
of human civilization where these kinds of > > battles have played out, and
the side opposing the truth, has always > > lost. > > How do you *know* that
the side oppressing the truth has always lost? > You'd never hear about the
battle if they'd won. > The mathematical ideas I have are likely to be
important in the sciences. We know the wrong side hasn't won because we have
the scientific progress to show it. Mathematics has been the backbone of the
sciences. There are rumblings of problems with the current mathematical
ideas from the world of physics already. My ideas in areas like
non-polynomial factorization may have practical importance in quantum
mechanics, for instance, and could lead to major advances in our
understanding of our physical world. So, yeah, I thought for a while that
maybe the wrong side could win and you'd have some poor slob who had these
important ideas that were lost to history, but somehow there seems to be an
arrow of Progress that doesn't allow that to actually happen, though I guess
you could claim that our technology could be light years ahead of where it
is now, if some person way back hadn't had their ideas quashed. I say, the
world seems to barely be able to handle where we are now, let alone us being
even further ahead, so everything seems to happen in its own time according
to some super plan that cannot be beaten. James Harris
............................................
In the ring of algebraic integers, consider 7 Q(x) = 7((x^2 + x)(5^2) + (-1
+ x)(5) + 7) = 7(25 x^2 + 30 x + 2) and 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7)
where the a's are defined by a^2 - (x - 1)a + 7(x^2 + x) = 0 where Decker
chose this example because when x=1 mod 7, the middle term has 7 as a factor
which he I guess thought was a counterexample to my ideas because at x=0,
the a's are 0 and 1, indicating that 7 divides through only one of the
expressions, except for the special cases. The first step in the disproof is
to focus on that last quadratic and write it with the focus on x, which
gives a non-monic polynomial that becomes monic if 'a' has 7 as a factor:
7x^2 + (7-a)x + a^2 + a = 0 as solving for x gives x = ((a-7) +/-
sqrt(-27a^2 - 42a + 49))/14 where you can see by inspection that if 'a' has
7 as a factor, then x is an algebraic integer. Notice the 14 in the
denominator gives you 2 and 7, where both have to divide off for x to be in
the ring of algebraic integers. But let w_1 w_2 = 7, where w_1 and w_2 are
those factors where each is non-unit, and a = w_1 b as then you have x =
((b - w_2) +/- sqrt(-27b^2 - 6w_2b + w_2^2))/2w_2 showing that w_2 is left
in the denominator, but for x to be an algebraic integer with w_2 a
non-unit, it MUST divide back through the numerator, where it would need do
so by sign. For instance, then x = ((b - w_2) + sqrt(-27b^2 - 6w_2b +
w_2^2))/2w_2 might be an algebraic integer, while x = ((b - w_2) -
sqrt(-27b^2 - 6w_2b + w_2^2))/2w_2 would not, but that's impossible by
symmetry rules, proving that x cannot be an algebraic integer for that case.
What happens then whenboth of the a's appear to have non-unit factors in the
ring of algebraic integers in common with 7? Well, for one of them the
factor is algebraically a unit, so it simply divides through BOTH cases,
without regard to sign, leaving an algebraic integer x. Of course, if some
of you wish to disagree with me on this point, then feel free, but notice
we're at the end of the proof. I am curious to see if any will attack the
idea that symmetry rules prevent x from being an algebraic integer. That's
because those ideas should be familiar. James Harris
.........................................................
William Hughes wrote: > jstevh@xxxxxxx wrote: > > In the ring of algebraic
integers, consider > > > > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > >
= 7(25 x^2 + 30 x + 2) > > > > and > > > > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) +
7) > > > > where the a's are defined by > > > > a^2 - (x - 1)a + 7(x^2 + x)
= 0 > > > > > And we have many many cases where neither root is divisible >
by 7 nor coprime to 7. In what ring? > > E.g. Take x=2 > > We get > > a^2 -
a + 42 = 0 > > (note the middle term does not have 7 as a factor) > > The
roots are > > a_1 = (1+sqrt(-167))/2 and a_2=(1+sqrt(-167))/2 > > Both a_1
and a_2 are roots of > > 49a^2 -7a + 42 =0 > > and hence of > > 7a^2 -a + 6
=0 > > But the last is primitive, non-monic, irreducible over Q and > has
integer coefficients. Thus a_1/7 and a_2/7 are not > algebraic integers, so
neither a_1 nor a_2 is divisible > by 7. But a1*a2=42 so neither a_1 nor a_2
is coprime > to 7. > > -William Hughes In what ring? James Harris
.............................................
William Hughes wrote: > jstevh@xxxxxxx wrote: > > William Hughes wrote: > >
jstevh@xxxxxxx wrote: > > > > In the ring of algebraic integers, considerx^2 + 30 x + 2) > > > > > > > > and > > > > > > > > 7 Q(x) = (5a_1(x) +
7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > > > > = 7(25
7)(5a_2(x) + 7) > > > > > > > > where the a's are defined by > > > > > > > >
a^2 - (x - 1)a + 7(x^2 + x) = 0 > > > > > > > > > > > > > And we have many
many cases where neither root is divisible > > > by 7 nor coprime to 7. > >
factor) > > > > > > The roots are > > > > > > a_1 = (1+sqrt(-167))/2 andIn what ring? > > > > > > > > E.g. Take x=2 > > > > > > We get > > > >
a^2 - a + 42 = 0 > > > > > > (note the middle term does not have 7 as a
a_2=(1+sqrt(-167))/2 > > > > > > Both a_1 and a_2 are roots of > > > > > >
49a^2 -7a + 42 =0 > > > > > > and hence of > > > > > > 7a^2 -a + 6 =0 > > >
hasBut the last is primitive, non-monic, irreducible over Q and > > >
integer coefficients. Thus a_1/7 and a_2/7 are not > > > algebraic integers,
so neither a_1 nor a_2 is divisible > > > by 7. But a1*a2=42 so neither a_1
nor a_2 is coprime > > > to 7. > > > > > > -William Hughes > > > > In what
ring? > > > jstevh@xxxxxxx wrote: > In the ring of algebraic integers,
consider > > -William Hughes Yup. Now let's focus on the solution for x.
7x^2 + (7-a)x + a^2 + a = 0 so solving for x gives x = ((a-7) +/-
sqrt(-27a^2 - 42a + 49))/14 where the original in the Decker example has
a^2 - (x - 1)a + 7(x^2 + x) = 0 so you can just pick an x, solve for the a's
using the quadratic formula and plug that back into the solution for x. I
pick x=3. That gives a = 1 + sqrt(-83). Working it out, hopefully I got the
algebra right, I get x = (-6 + sqrt(-83) +/- sqrt(2221 - 96sqrt(83))/14 and
I can multiply top and bottom by (-6 + sqrt(-83) -/+ sqrt(2221 - 96sqrt(83))
use x=3, and simplify to get 1 = (-54 + 2sqrt(-83)/((-6 + sqrt(-83) -/+
sqrt(2221 - 96sqrt(83))) so (-6 + sqrt(-83) -/+ sqrt(2221 - 96sqrt(83)))
= -54 + 2sqrt(-83) and the result by symmetry that -54 + 2sqrt(-83) must
have 7 as a factor, which it provably does not in the ring of algebraic
integers, so there is the appearance of a contradiction. Hopefully I got all
the algebra right. Comments? Note that for x to be an algebraic integer it
must be the case that 14 divides through x = ((a-7) +/- sqrt(-27a^2 - 42a +
49))/14 or you have what is the equivalent of a fraction, but with
non-rationals. Oddly enough, you can just pick some x, like I did and work
through to find as I have that you get a contradiction with the ring of
algebraic integers, when, of course x is an algebraic integer. Fun, eh? If
you like contradictory mathematical ideas, run yourself in circles. I like
mathematical ideas that work. James Harris
..............................................
William Hughes wrote: > jstevh@xxxxxxx wrote: > > William Hughes wrote: > >
jstevh@xxxxxxx wrote: > > > > William Hughes wrote: > > > > >jstevh@xxxxxxx wrote: > > > > > > In the ring of algebraic integers,
consider > > > > > > > > > > > > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) +
7) > > > > > > = 7(25 x^2 + 30 x + 2) > > > > > > > > > > > > and > > > > >
where the a's are defined by > > > > > > > > > > > > a^2 - (x - 1)a + 7(x^27 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > > > > > > > > > > > >
+ x) = 0 > > > > > > > > > > > > > > > > > > > > > And we have many many
cases where neither root is divisible > > > > > by 7 nor coprime to 7. > > >
(note the middle term does not have 7 as a factor) > > > > > > > > > > TheIn what ring? > > > > > > > > > > > > > > E.g. Take x=2 > > > > >
We get > > > > > > > > > > a^2 - a + 42 = 0 > > > > > > > > > >
roots are > > > > > > > > > > a_1 = (1+sqrt(-167))/2 and
a_2=(1+sqrt(-167))/2 > > > > > > > > > > Both a_1 and a_2 are roots of > > >
non-monic, irreducible over Q and > > > > > has integer coefficients. Thus49a^2 -7a + 42 =0 > > > > > > > > > > and hence of > > > > >
7a^2 -a + 6 =0 > > > > > > > > > > But the last is primitive,
a_1/7 and a_2/7 are not > > > > > algebraic integers, so neither a_1 nor a_2
is divisible > > > > > by 7. But a1*a2=42 so neither a_1 nor a_2 is coprime
ring? > > > > > > > jstevh@xxxxxxx wrote: > > > In the ring of algebraicto 7. > > > > > > > > > > -William Hughes > > > > > > > > In what
integers, consider > > > > > > -William Hughes > > > > Yup. Now let's focus
on the solution for x. > > > > 7x^2 + (7-a)x + a^2 + a = 0 > > > > so
solving for x gives > > > > x = ((a-7) +/- sqrt(-27a^2 - 42a + 49))/14 > > >
where the original in the Decker example has > > > > a^2 - (x - 1)a +7(x^2 + x) = 0 > > > > so you can just pick an x, solve for the a's using
the quadratic > > formula and plug that back into the solution for x. > > >
I pick x=3. That gives a = 1 + sqrt(-83). > > > > Working it out,hopefully I got the algebra right, I get > > > > x = (-6 + sqrt(-83) +/-
sqrt(2221 - 96sqrt(83))/14 > > > That should be > > x = (-6 + sqrt(-83) +/-
sqrt(2221 - 96sqrt(-83)))/14 > > (increment the Oops counter). No need, as I
just left off a parenthesis on the end which doesn't change anything. > >
Note that > > 3 = (-6 + sqrt(-83) + sqrt(2221 - 96sqrt(-83)))/14 > > So one
of the solutions is x=3 to nobody's surprise. > Well, yeah, that's ONE
solution, but you cheat! You deleted out the entire argument which shows
another solution, and then just state what you want, after trying to make
the trivial leaving off of a parentheses into something it's not. Here's the
argument again, which you deleted out, with the fix: I pick x=3. That gives
a = 1 + sqrt(-83). Working it out, hopefully I got the algebra right, I get
x = (-6 + sqrt(-83) +/- sqrt(2221 - 96sqrt(83)))/14 and I can multiply top
and bottom by (-6 + sqrt(-83) -/+ sqrt(2221 - 96sqrt(83)) use x=3, and
simplify to get 1 = (-54 + 2sqrt(-83)/((-6 + sqrt(-83) -/+ sqrt(2221 -
96sqrt(83))) so (-6 + sqrt(-83) -/+ sqrt(2221 - 96sqrt(83))) = -54 +
2sqrt(-83) and the result by symmetry that -54 + 2sqrt(-83) must have 7 as a
factor, which it provably does not in the ring of algebraic integers, so
there is the appearance of a contradiction. Hopefully I got all the algebra
right. Comments? And no just deleting everything out to proclaim your own
solution! You people need to start following some basic rules. James Harris
.................................................
marcus_b wrote: > William Hughes wrote ... > > [snip] > > a^2 - (x - 1)a +
7(x^2 + x) = 0 > > And we have many many cases where neither root is
divisible > by 7 nor coprime to 7. > > E.g. Take x=2 > > We get > > a^2 - a
+ 42 = 0 > > (note the middle term does not have 7 as a factor) > > The
roots are > > a_1 = (1+sqrt(-167))/2 and a_2=(1+sqrt(-167))/2 >
----------------------------------------------------------------- > >[snip] > > It was noted in a long-distant post by Keith Ramsay that > > a_1
= (1 + sqrt(-167))/2 is such that > > a_1^11 = (44555 - 222*a_1)*(-12882 -
2017*a_1) and > > 7^11 = (44555 - 222*a_1)*(44555 + 222*a_1). > > Therefore
a_1^11 and 7^11 share an algebraic integer factor, > namely > > 44555 -
222*a_1. > > Therefore a_1 and 7 share the algebraic integer factor > >
(44555 - 222*a_1)^(1/11). > > Note that 44555 is divisible by 7, but a_1 is
not (in the > algebraic integers). Therefore this common algebraic integer >
factor is not divisible by 7. > It's not divisible by 7 in the ring of
algebraic integers. You keep going in circles around that central point. My
counter would be that one solution of the 11 generated by that radical
expression as it is taking the 11th root, does have 7 as a factor. But
rather than argue that point let me show you the simple way to see a
contradiction if I'm wrong, as I just figured it out. > Which proves very
explicitly that Harris is wrong. The algebraic > integers are NOT
"incomplete". > > Marcus They are, as this simple algebra shows. I have from
before a^2 - (x - 1)a + 7(x^2 + x) = 0 from which I can easily get 7x^2 +
(7-a)x + a^2 + a = 0 and this time I'll choose an integer value for a, like
a=3, so that I have a non-monic polynomial irreducible over Q for x. But
because 3 is coprime to 7 I know that the other solution for 'a' MUST have 7
as a factor, right? But now I just go back to a^2 - (x - 1)a + 7(x^2 + x) =
0 and note that for any solution for x, remember x is NOT an algebraic
integer, where x as a ratio of algebraic integers has factors in common with
7, while being coprime to some factors of 7, it is NOT possible that the a's
have 7 as a factor, which is a direct contradiction with one of the a's
being 3, and thus forcing the other of the a's to have 7 as a factor.
Interestingly enough, there is no meaningful way to consider the second a
which is NOT an algebraic integer to be a fraction, so the ring of algebraic
integers is easily shown to be incomplete as I've said. Done arguing yet?
Why fight for ideas that can so easily be shown to be mathematically
incorrect? What do you think, you get some prize for using bogus math? James
Harris ..........................................
marcus_b wrote: > >marcus_b wrote: > >> William Hughes wrote ... > > >
[snip] > > >> a^2 - (x - 1)a + 7(x^2 + x) = 0 > > >> And we have many many
cases where neither root is divisible > >> by 7 nor coprime to 7. > > >>
E.g. Take x=2 > > >> We get > > >> a^2 - a + 42 = 0 > > >> (note the middle
term does not have 7 as a factor) > > >> The roots are > > >> a_1 =
(1+sqrt(-167))/2 and a_2=(1+sqrt(-167))/2 > >
----------------------------------------------------------------- > > >[snip] > > >> It was noted in a long-distant post by Keith Ramsay that > >
222*a_1)*(-12882 - 2017*a_1) and > > >> 7^11 = (44555 - 222*a_1)*(44555 +a_1 = (1 + sqrt(-167))/2 is such that > > >> a_1^11 = (44555 -
222*a_1). > > >> Therefore a_1^11 and 7^11 share an algebraic integer
factor, > >> namely > > >> 44555 - 222*a_1. > > >> Therefore a_1 and 7 share
the algebraic integer factor > > >> (44555 - 222*a_1)^(1/11). > > >> Note
that 44555 is divisible by 7, but a_1 is not (in the > >> algebraic
integers). Therefore this common algebraic integer > >> factor is not
divisible by 7. > > >It's not divisible by 7 in the ring of algebraic
integers. > > Yes - that's what I said. Neither a_1 nor a_2 is divisible >
by 7, and neither is coprime to 7. Since I gave an explicit > common factor,
this is not a case where you depend on > Dedekind's theorem (on ideals) to
show that a common factor > exists. Therefore your claim that you have found
a problem > with the theory of ideals does not apply. > But I've proven that
the ring of algebraic integers is incomplete. The belief that it is complete
relies on a false claim of proof using the theory of ideals made by
Dedekind, brining the theory of ideals into question. > >You keep going in
circles around that central point. > > >My counter would be that one
solution of the 11 generated by that > >radical expression as it is taking
the 11th root, does have 7 as a > >factor. > > All you need to consider here
is a_1. Since it does not have 7 > as a factor and it is not coprime to 7,
and since a_1 * a_2 is > divisible by 7, both a_1 and a_2 are not coprime to
7 and not > divisible by 7. > That's the case IN THE RING OF ALGEBRAIC
INTEGERS where I can show that you get a contradiction with that ring, as I
did in reply to you, quite simply. You can do the same thing with 2 and 6 in
the ring of evens, where one person can claim that 2 is not coprime to 6,
while the other keeps claiming it is, while they remain in the ring of
evens. You cannot disprove my claims by relying on results in the ring of
algebraic integers, any more than someone could prove that 2 is coprime to 6
by remaining in the ring of evens. It just doesn't fly. > >But rather than
argue that point let me show you the simple way to see > >a contradiction if
I'm wrong, as I just figured it out. > > >> Which proves very explicitly
that Harris is wrong. The algebraic > >> integers are NOT "incomplete". > >
before > > >a^2 - (x - 1)a + 7(x^2 + x) = 0 > > >from which I can easily getMarcus > > >They are, as this simple algebra shows. > > >I have from
value for a, like a=3, so that I > >have a non-monic polynomial irreducible7x^2 + (7-a)x + a^2 + a = 0 > > >and this time I'll choose an integer
over Q for x. > > >But because 3 is coprime to 7 I know that the other
solution for 'a' > >MUST have 7 as a factor, right? > > No - you are now
considering this as a polynomial in "x". You > chose a value for "a", and
then you solve for "x". You have to > decide one at a time which variables
you are treating as constants > and which as 'unknowns'. Here you are
treating "x" as the unknown, > and "a" as a constant. > That's not
mathematics! Obviously if one of the choices for 'a' is 3, then there is a
resultant x, which must fulfil the conditions of all the equations. > > >But
now I just go back to > > >a^2 - (x - 1)a + 7(x^2 + x) = 0 > > >and note
that for any solution for x, remember x is NOT an algebraic > >integer,
where x as a ratio of algebraic integers has factors in common > >with 7,
while being coprime to some factors of 7, it is NOT possible > >that the a's
have 7 as a factor, which is a direct contradiction with > >one of the a's
being 3, and thus forcing the other of the a's to have 7 > >as a factor. > >
No - you're just confusing yourself on this. You either choose > x and solve
for a, or you choose a and solve for x. You don't get > to do both at the
same time. > Um, do you think that if you do one you don't do the other?
That is, given a^2 - (x - 1)a + 7(x^2 + x) = 0 and a choice for 'a', you are
claiming that you can't use that solution with 7x^2 + (7-a)x + a^2 + a = 0
the same exact equation, just written in a slightly different way? They are
the SAME equation, just written two different ways!!! With the second and a
choice for one of the a's that is an integer and coprime to 7, where a^2 + a
is coprime to 7, solutions for x must fit the same equation. I just wrote
the same equation two different ways, but it's the same damn equation! If
math people are going to go this far then I wonder how there can be any
hope. You throw algebra away as if it were trash. James Harris
.....................................................
marcus_b wrote: > jstevh@xxxxxxx wrote: > > marcus_b wrote: > > > >marcus_b
wrote: > > > >> William Hughes wrote ... > > > > > > > [snip] > > > > > > >>
a^2 - (x - 1)a + 7(x^2 + x) = 0 > > > > > > >> And we have many many cases
where neither root is divisible > > > >> by 7 nor coprime to 7. > > > > > >
The roots are > > > > > > >> a_1 = (1+sqrt(-167))/2 and a_2=(1+sqrt(-167))/2E.g. Take x=2 > > > > > > >> We get > > > > > > >> a^2 - a + 42 = 0 > >
(note the middle term does not have 7 as a factor) > > > > > > >>
Ramsay that > > > > > > >> a_1 = (1 + sqrt(-167))/2 is such that > > > > > >----------------------------------------------------------------- > > > >
[snip] > > > > > > >> It was noted in a long-distant post by Keith
(44555 - 222*a_1)*(44555 + 222*a_1). > > > > > > >> Therefore a_1^11 anda_1^11 = (44555 - 222*a_1)*(-12882 - 2017*a_1) and > > > > > > >> 7^11 =
7^11 share an algebraic integer factor, > > > >> namely > > > > > > >>
44555 - 222*a_1. > > > > > > >> Therefore a_1 and 7 share the algebraic
integer factor > > > > > > >> (44555 - 222*a_1)^(1/11). > > > > > > >> Note
that 44555 is divisible by 7, but a_1 is not (in the > > > >> algebraic
integers). Therefore this common algebraic integer > > > >> factor is not
divisible by 7. > > > > > > >It's not divisible by 7 in the ring of
algebraic integers. > > > > > > Yes - that's what I said. Neither a_1 nor
a_2 is divisible > > > by 7, and neither is coprime to 7. Since I gave an
explicit > > > common factor, this is not a case where you depend on > > >
Dedekind's theorem (on ideals) to show that a common factor > > > exists.
Therefore your claim that you have found a problem > > > with the theory of
ideals does not apply. > > > > > > > But I've proven that the ring of
algebraic integers is incomplete. > > > > The belief that it is complete
relies on a false claim of proof using > > the theory of ideals made by
Dedekind, brining the theory of ideals > > into question. > > > > > >You
keep going in circles around that central point. > > > > > > >My counter
would be that one solution of the 11 generated by that > > > >radical
expression as it is taking the 11th root, does have 7 as a > > > >factor. >
divisible by 7, both a_1 and a_2 are not coprime to 7 and not > > >All you need to consider here is a_1. Since it does not have 7 >
as a factor and it is not coprime to 7, and since a_1 * a_2 is > > >
divisible by 7. > > > > > > > That's the case IN THE RING OF ALGEBRAIC
INTEGERS where I can show that > > you get a contradiction with that ring,
as I did in reply to you, quite > > simply. > > > > You can do the same
thing with 2 and 6 in the ring of evens, where one > > person can claim that
2 is not coprime to 6, while the other keeps > > claiming it is, while they
remain in the ring of evens. > > > > You cannot disprove my claims by
relying on results in the ring of > > algebraic integers, any more than
someone could prove that 2 is coprime > > to 6 by remaining in the ring of
evens. > > > > It just doesn't fly. > > > > > >But rather than argue that
point let me show you the simple way to see > > > >a contradiction if I'm
wrong, as I just figured it out. > > > > > > >> Which proves very explicitly
that Harris is wrong. The algebraic > > > >> integers are NOT "incomplete".
= 0 > > > > > > >and this time I'll choose an integer value for a, like a=3,Marcus > > > > > > >They are, as this simple algebra shows.
I have from before > > > > > > >a^2 - (x - 1)a + 7(x^2 + x) = 0
from which I can easily get > > > > > > >7x^2 + (7-a)x + a^2 + a
so that I > > > >have a non-monic polynomial irreducible over Q for x. > > >
considering this as a polynomial in "x". You > > > chose a value for "a",But because 3 is coprime to 7 I know that the other solution for 'a'
MUST have 7 as a factor, right? > > > > > > No - you are now
and then you solve for "x". You have to > > > decide one at a time which
variables you are treating as constants > > > and which as 'unknowns'. Here
you are treating "x" as the unknown, > > > and "a" as a constant. > > > > >
the equations. > > > > > > > > >But now I just go back to > > > > > > >a^2 -That's not mathematics! Obviously if one of the choices for 'a' is 3, >then there is a resultant x, which must fulfil the conditions of all > >
(x - 1)a + 7(x^2 + x) = 0 > > > > > > >and note that for any solution for x,
remember x is NOT an algebraic > > > >integer, where x as a ratio of
algebraic integers has factors in common > > > >with 7, while being coprime
to some factors of 7, it is NOT possible > > > >that the a's have 7 as a
factor, which is a direct contradiction with > > > >one of the a's being 3,
and thus forcing the other of the a's to have 7 > > > >as a factor. > > > >
and solve for a, or you choose a and solve for x. You don't get > > > to doNo - you're just confusing yourself on this. You either choose > > > x
both at the same time. > > > > > > > Um, do you think that if you do one you
don't do the other? > > > > That is, given > > > > a^2 - (x - 1)a + 7(x^2 +
x) = 0 > > > > and a choice for 'a', you are claiming that you can't use
that solution > > with > > > > 7x^2 + (7-a)x + a^2 + a = 0 > > > > the same
exact equation, just written in a slightly different way? > > > > They are
the SAME equation, just written two different ways!!! > > > > With the
second and a choice for one of the a's that is an integer and > > coprime to
7, where a^2 + a is coprime to 7, solutions for x must fit > > the same
equation. > > > > I just wrote the same equation two different ways, but
it's the same > > damn equation! > > > > If math people are going to go this
far then I wonder how there can be > > any hope. > > > > You throw algebra
away as if it were trash. > > > > You know, the bottom line on this is the
following. You say > that in the algebraic integers, one root of > > a^2 - a
+ 42 = 0 > > *should* have 7 as a factor, while the other is coprime to 7.
And that's just NOT true, as it's possible to prove that they do not. I can
then show a contradiction with the ring of algebraic integers, which is
equivalent to showing a contradiction with someone in the ring of evens
claiming that 2 is coprime to 6. The analogy should not escape you. > Yet
you know, and we all know, that a theorem of Dedekind > and a theorem in
Galois theory both say that that is not true. So > you concluded that
Dedekind's theorem is false and Galois theory > is wrong. You also concluded
that the algebraic integers are > incomplete in the sense that they do not
permit a factorization > of the kind that you want. > That's not true. I
prove that you get the appearance of a contradiction if you assume that the
ring of algebraic integers is complete. That proof is so trivial, you just
ignored it in your reply, though at least you didn't delete it out! By
picking an 'a' that is an *integer* and coprime to 7, like 3, I can easily
show a contradiction with standard teaching, and I see you are just
blissfully trying to ignore the proof. I have a question for you if you wish
to maintain that you can only use those equations once, how do you suppose
that knowing 3 is one of the a's, you would get the other one? > But I
showed explicitly, without any appeal to either Dedekind's > theorem or
Galois theory, that neither root is coprime to 7 and > neither root is
divisible by 7. Both roots have algebraic integer > factors in common with
7. Factors of 7 are distributed across > both roots. > That is true IN THE
RING OF ALGEBRAIC INTEGERS so around and around we go. Just like if I were
arguing with someone claiming that 2 is coprime to 6 who stayed in the ring
of evens, repeating over and over again that 2 is coprime to 6 as they can
prove--but needing me to keep reminding that the ring is evens!!! It's
exactly the same thing. > You say this cannot happen in the algebraic
integers. But I No I DO NOT!! Why make a false claim? The very point I make
is that it's true in the ring of algebraic integers proving that ring is
incomplete!!! Can you hear yourself? Take away your personal feelings.
Forget about me. This is not personal. What's correct is what's important.
Focus on the math, and not the social stuff. > showed it can. You can verify
it by just carrying out the arithmetic. > It is just a matter of brute
calculation, just as true as 2 + 2 = 4. > > Therefore what you claimed is
wrong. Therefore you have an > error which you have not yet recognized.
Getting mad at me > for pointing this out is rather unproductive, don't you
think? > Shouldn't you be trying to find the mistake in your reasoning? > >
Marcus > I'm not mad at you. I am a bit exasperated yes, but look at what
you're doing! In one reply you claim I can't use the same equation twice,
making up some bizarre non-math rule. Here you claim that I make a claim
that I do not, relying on proof of something being true in the ring of
algebraic integers when I repeatedly explain why that's specious, and you
end by just declaring that I am wrong. Where in your training as a math
student did they teach you to make up things that are not true, deny
mathematical proof, and ultimately just state something that you should
realize is mathematically false--because the proof is in front of you and
EASY--but instead you just go with your own opinion as if mathematics were
trash? Is mathematics of no real importance to you? Is it just junk with no
meaning or great purpose? Do you see math as garbage? If not, then why lie?
I am serious. If you actually value mathematics, why lie about it? James
Harris ..............................................
Now I'm going to go over the Decker example and how it refutes the standard
teaching of Galois Theory, and it turns out that it shows the theory of
ideals to be flawed as well, but it's more convoluted to explain why that's
true and I'll leave that explanation as it is somewhat technical to others.
This example because of the use of simple quadratics takes away the
objections of posters who have inexplicably refused to accept the
distributive property in proofs I've shown with my own more complicated
expressions. And also I'm starting with equations put up by a Rick Decker,
so it's not like I picked them personally. The attempt will be to have
everthing below in the ring of algebraic integers, so consider that the
ring, but notice at a key point, we will be forced out of that ring.
Consider 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) = 7(25 x^2 + 30 x + 2)
and 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) where the a's are defined by a^2 -
(x - 1)a + 7(x^2 + x) = 0 where Decker chose this example because at x=1,
the middle term goes to 0 which he I guess thought was a counterexample to
my ideas because at x=0, the a's are 0 and 1, indicating that 7 divides
through only one of the expressions, but clearly it does not at x=1, but
that special case doesn't change the refutation so I'll leave out the
explanation of it The following disproof focuses directly on Galois Theory
while the relation to the theory of ideals is more complex. The first step
in the disproof is to focus on that last quadratic a^2 - (x - 1)a + 7(x^2 +
x) = 0 and write it with the focus on x, which gives a non-monic polynomial
that importantly becomes monic if 'a' has 7 as a factor: 7x^2 + (7-a)x + a^2
+ a = 0 and using a = 7b, you get 7x^2 + (7 - 7b)x + 49b^2 + 7b = 0 and
dividing off 7, gives you x^2 + (1 - b)x + 7b^2 + b = 0 so for ANY algebraic
integer 'a' that has 7 as a factor, you get an algebraic integer x. Also,
notice that 7b^2 in the expression so that if you focus on 'b' you have a
non-monic polynomial and as before, 'b' cannot be an algebraic integer if
that polynomial has integer coefficients and is irreducible over Q. BUT
a^2 - (x - 1)a + 7(x^2 + x) = 0 means that for ANY algebraic integer x, 'a'
is an algebraic integer, but if 'a' does not have 7 as a factor 7x^2 +
(7-a)x + a^2 + a = 0 will not allow an algebraic integer x, if 'a' is an
integer and the quadratic is irreducible over Q. So you are forced out of
the ring of algebraic integers. Algebraically you still have x^2 + (1 - b)x
+ 7b^2 + b = 0 but the 'b' is outside of the ring of algebraic integers for
certain algebraic integer values of x. To save Galois Theory and the theory
of ideals, the attempted defense of those ideas at this point would require
that for some non-rational 'a' with partial factors of 7, as it is a root of
a^2 - (x - 1)a + 7(x^2 + x) = 0 with an integer x other than x=1, you would
get a polynomial with an integer solution for x from 7x^2 + (7-a)x + a^2 + a
= 0 which would remain non-monic because the factors of 7 not in common with
'a' would remain on the leading coefficient. However, you can simply pick
some x, like x=3, and solve for the a's, and then make the substitution in
which will give a seemingly non-monic expression of some complexity as it
will have square root radicals, but you can use basic algebraic
manipulations to remove the square roots, giving you a quartic, where you
will find that the leading coefficient just divides off, as one of the roots
will be 3, so it will also be reducible over Q. So Galois Theory as a way to
determine where factors go is refuted, and it is proven that the ring of
algebraic integers is incomplete, so that some numbers are excluded simply
because they are not roots of monic polynomials with integer coefficients,
which leads to the conclusion that the theory of ideals is flawed. This post
marks the end of an era in the world of mathematics. James Harris
........................................
...
José Carlos Santos wrote: > jstevh@xxxxxxx wrote: > > > This post marks the
end of an era in the world of mathematics. > > When will the rest of the
world notice that? > > Best regards, > > Jose Carlos Santos A couple of
days? I'm not sure, but it's likely to be quick. The problem with my
previous examples were things like very complicated expressions and the
ability of people to find areas where they could I guess think they saw
reason for doubt. Decker's example removes a lot of those areas and relies
on quadratics which are easier for people. With that said, the problem for
professors who may think that they can go on with the other teaching is that
the story can travel to their students who will be confused and hurt if men
they admire and think a lot of, try to ignore one of the biggest stories in
math history, and teach them flawed information. So it's likely that the
argument will fly around the world. People may sit for a while, but
mathematics is a difficult disciplne. Even if they try, math students will
find it hard to work at learning ideas that more and more of them know do
not work. The odd thing is that some probably will try. Here's a better
ending than in my original post: Taking 7x^2 + (7-a)x + a^2 + a = 0 I need
to actually solve for x using the quadratic formula: x = ((a-7) +/-
sqrt(a^2 - 14a + 49 -28a^2 - 28a))/14 so x = ((a-7) +/- sqrt(-27a^2 - 42a +
49))/14 where now it's still clear that with a = 7b, you end up with an
algebraic integer x, while if 'a' is coprime to 7 or, following standard
teaching of Galois Theory, just has some factors in common with 7, then x is
not an algebraic integer. That's neat, ties it all up and leave no room for
mathematical doubt or arguments about functions and constants. Telling tests
of character are coming for many of you in the next few hours and days.
James Harris ..................................................
David Moran wrote: > <jstevh@xxxxxxx> wrote in message >
news:1139086835.818988.158310@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > > Now I'm
going to go over the Decker example and how it refutes the > > standard
teaching of Galois Theory, and it turns out that it shows the > > theory of
ideals to be flawed as well, but it's more convoluted to > > explain why
that's true and I'll leave that explanation as it is > > somewhat technical
to others. > > <snip crap> > > Once again, how do you know how Galois Theory
is taught if you've not had a > formal course in it? I can't argue on how
it's taught because I know very > little about the subject. However, if you
were to strike up an argument > about differential equations, for example, I
feel I could hold my own as > that's my main area of interest. > > Dave I've
been arguing over these areas with sci.math posters who have gone into
detail about how my ideas contradict with how Galois Theory is used. Some of
them are supposedly experts in the area, for instance, I think that it's
actually Arturo Magidin's area of expertise, and that he has a Ph.D in the
area from the University of California at Berkeley. James Harris
..............................................
David Moran wrote: > <jstevh@xxxxxxx> wrote in message >
news:1139112273.914242.164390@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > > David
Moran wrote: > >> <jstevh@xxxxxxx> wrote in message > >>
news:1139086835.818988.158310@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > >> > Now I'm
going to go over the Decker example and how it refutes the > >> > standard
teaching of Galois Theory, and it turns out that it shows the > >> > theory
of ideals to be flawed as well, but it's more convoluted to > >> > explain
why that's true and I'll leave that explanation as it is > >> > somewhat
technical to others. > >> > >> <snip crap> > >> > >> Once again, how do you
know how Galois Theory is taught if you've not had > >> a > >> formal course
in it? I can't argue on how it's taught because I know very > >> little
about the subject. However, if you were to strike up an argument > >> about
differential equations, for example, I feel I could hold my own as > >>
that's my main area of interest. > >> > >> Dave > > > > I've been arguing
over these areas with sci.math posters who have gone > > into detail about
how my ideas contradict with how Galois Theory is > > used. > > > > Some of
them are supposedly experts in the area, for instance, I think > > that it's
actually Arturo Magidin's area of expertise, and that he has > > a Ph.D in
the area from the University of California at Berkeley. > > > > > > James
Harris > > > > But, do you know anything about it yourself? I don't think
you can argue > about something you don't know about. > > Dave I don't claim
expertise in Galois Theory, but I have checked to verify the basic claims of
posters who have cited Galois Theory in disagreeing with me. The results I
have give no room for doubt about the applicability of Galois Theory, which
is not totally invalidated--just the conclusions about where factors go. The
more damaging impact is on the theory of ideals, where I have even less
expertise, but I think my central point invalidates the concept of a prime
ideal, showing it has no mathematical basis. James Harris
..................................................
William Hughes wrote: > jstevh@xxxxxxx wrote: > > Now I'm going to go over
the Decker example and how it refutes the > > standard teaching of Galois
Theory, and it turns out that it shows the > > theory of ideals to be flawed
as well, but it's more convoluted to > > explain why that's true and I'll
leave that explanation as it is > > somewhat technical to others. > > > >
This example because of the use of simple quadratics takes away the > >
objections of posters who have inexplicably refused to accept the > >
distributive property in proofs I've shown with my own more complicated > >
expressions. > > > > And also I'm starting with equations put up by a Rick
Decker, so it's > > not like I picked them personally. > > > > The attempt
will be to have everthing below in the ring of algebraic > > integers, so
consider that the ring, but notice at a key point, we will > > be forced out
of that ring. > > > > Consider > > > > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 +
x)(5) + 7) > > = 7(25 x^2 + 30 x + 2) > > > > and > > > > 7 Q(x) = (5a_1(x)
+ 7)(5a_2(x) + 7) > > > > where the a's are defined by > > > > a^2 - (x -
1)a + 7(x^2 + x) = 0 > > > > where Decker chose this example because at x=1,
the middle term goes to > > 0 which he I guess thought was a counterexample
to my ideas because at > > x=0, the a's are 0 and 1, indicating that 7
divides through only one of > > the expressions, but clearly it does not at
x=1, but that special case > > doesn't change the refutation so I'll leave
out the explanation of it > > > > The following disproof focuses directly on
Galois Theory while the > > relation to the theory of ideals is more
complex. > > > > The first step in the disproof is to focus on that last
quadratic > > > > a^2 - (x - 1)a + 7(x^2 + x) = 0 > > > > and write it with
the focus on x, which gives a non-monic polynomial > > that importantly
becomes monic if 'a' has 7 as a factor: > > > > 7x^2 + (7-a)x + a^2 + a = 0
a= -2. Note further that this has non-integer > coefficients if a is not aNote that this has integer coefficients and is reducible > over Q if
rational integer and in this case > the question of irreducibility does not
arise. (Though if a is a > solution > to a^2 - (x - 1)a + 7(x^2 + x) = 0 for
x an algebraic integer, > 7x^2 + (7-a)x + a^2 + a = 0 has a algebraic
integer root). > > > > > and using a = 7b, you get > > And here b is in
general not an algebraic integer. > > > > > 7x^2 + (7 - 7b)x + 49b^2 + 7b =
0 > > > > and dividing off 7, gives you > > > > x^2 + (1 - b)x + 7b^2 + b =
0 > > > > so for ANY algebraic integer 'a' that has 7 as a factor, you get
an > > algebraic integer x. > > True > . > > > > Also, notice that 7b^2 in
the expression so that if you focus on 'b' > > you have a non-monic
polynomial and as before, 'b' cannot be an > > algebraic integer if that
polynomial has integer coefficients and is > > irreducible over Q. > > > > >
True, however, as noted above, if a is a solution to > a^2 - (x - 1)a +
7(x^2 + x) = 0, and is not a rational integer, > the polynomial x^2 + (1 -
b)x + 7b^2 + b = 0 does not have > integer coefficients, but does have at
least one algebraic integer > root. (Remember, algebraic integers can be
roots of other > polynomials than monic polynomials with integer
coefficients). > > > BUT > > > > a^2 - (x - 1)a + 7(x^2 + x) = 0 > > > >
means that for ANY algebraic integer x, 'a' is an algebraic integer, > > but
if 'a' does not have 7 as a factor > > > > 7x^2 + (7-a)x + a^2 + a = 0 > > >
will not allow an algebraic integer x, if 'a' is an integer and the > >quadratic is irreducible over Q. > > > > True, but no-one has claimed the
opposite. And clearly you > cannot get such an 'a' by solving a^2 - (x - 1)a
+ 7(x^2 + x) = 0 > where x is an algebraic integer. > > > So you are forced
out of the ring of algebraic integers. > > > > Algebraically you still have
ring of algebraic integers for certain > > algebraic integer values of x. >x^2 + (1 - b)x + 7b^2 + b = 0 > > > > but the 'b' is outside of the
7(x_1^2Let x_1 be an algebraic integer > > Let > > a^2 - (x_1 - 1)a +
+ x_1) = 0 > > Then > > 7x_1^2 + (7-a)x_1 + a^2 + a = 0 > > so x_1 is a root
of > > 7x^2 + (7-a)x + a^2 + a = 0 > > Now let gcd(7,a) = r, 7=rs, a=rt,
gcd(s,t) =1, s non unit > > We get x_1 is a root of > > sx^2 + (s-t)x + rt^2
+ t = 0 > > So sx^2 + (s-t)x + rt^2 + t = 0 has an algebraic integer root. >
The fact that this polynomial is not monic is of no importance. > It doesn't
have integer coefficients. > It's easier and more direct to simply solve for
x in terms of the a's: x = ((a-7) +/- sqrt(-27a^2 - 42a + 49))/14 Here you
can just see by inspection that if 'a' has 7 as a factor, then 7 divides
through the numerator and the denominator. BUT, if 'a' only shares some
non-unit factors with 7 while being coprime to others, those factors will
divide off, leaving the others in the denominator. That 14 in the
denominator is key here. So x cannot in that case be an algebraic integer,
as you need a difference in divisibility by sign, which proves that it's
impossible for x to be an algebraic integer, if there are shared factors.
Note that the only exception is when sqrt(7) is the factor of both the a's
which occurs when x= 1 mod 7. James Harris
.....................................................
William Hughes wrote: > jstevh@xxxxxxx wrote: > > William Hughes wrote: > >
jstevh@msn.com wrote: > > > > Now I'm going to go over the Decker example and how it
refutes the > > > > standard teaching of Galois Theory, and it turns out
that it shows the > > > > theory of ideals to be flawed as well, but it's
more convoluted to > > > > explain why that's true and I'll leave that
explanation as it is > > > > somewhat technical to others. > > > > > > > >
This example because of the use of simple quadratics takes away the > > > >
objections of posters who have inexplicably refused to accept the > > > >
distributive property in proofs I've shown with my own more complicated > >
upexpressions. > > > > > > > > And also I'm starting with equations put
by a Rick Decker, so it's > > > > not like I picked them personally. > > > >
algebraicThe attempt will be to have everthing below in the ring of
will > > > > be forced out of that ring. > > > > > > > > Consider > > > > >integers, so consider that the ring, but notice at a key point, we
x7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > > > > = 7(25 x^2 + 30
+ 2) > > > > > > > > and > > > > > > > > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7)
+ 7(x^2 + x) = 0 > > > > > > > > where Decker chose this example because atwhere the a's are defined by > > > > > > > > a^2 - (x - 1)a
x=1, the middle term goes to > > > > 0 which he I guess thought was a
counterexample to my ideas because at > > > > x=0, the a's are 0 and 1,
indicating that 7 divides through only one of > > > > the expressions, but
clearly it does not at x=1, but that special case > > > > doesn't change the
refutation so I'll leave out the explanation of it > > > > > > > > The
following disproof focuses directly on Galois Theory while the > > > >
relation to the theory of ideals is more complex. > > > > > > > > The first
step in the disproof is to focus on that last quadratic > > > > > > > >
a^2 - (x - 1)a + 7(x^2 + x) = 0 > > > > > > > > and write it with the focus
on x, which gives a non-monic polynomial > > > > that importantly becomes
monic if 'a' has 7 as a factor: > > > > > > > > 7x^2 + (7-a)x + a^2 + a = 0
coefficients if a is not a rational integer and in this case > > > theNote that this has integer coefficients and is reducible
over Q if a= -2. Note further that this has non-integer > > >
question of irreducibility does not arise. (Though if a is a > > > solution
7x^2 + (7-a)x + a^2 + a = 0 has a algebraic integer root). > > > > > > > > >to a^2 - (x - 1)a + 7(x^2 + x) = 0 for x an algebraic integer, > > >
algebraic integer. > > > > > > > > > > > 7x^2 + (7 - 7b)x + 49b^2 + 7b = 0 >and using a = 7b, you get > > > > > > And here b is in general not an
+and dividing off 7, gives you > > > > > > > > x^2 + (1 - b)x
7b^2 + b = 0 > > > > > > > > so for ANY algebraic integer 'a' that has 7 as
a factor, you get an > > > > algebraic integer x. > > > > > > True > > > . >
on 'b' > > > > you have a non-monic polynomial and as before, 'b' cannot beAlso, notice that 7b^2 in the expression so that if you focus
an > > > > algebraic integer if that polynomial has integer coefficients and
is > > > > irreducible over Q. > > > > > > > > > > > > > True, however, as
noted above, if a is a solution to > > > a^2 - (x - 1)a + 7(x^2 + x) = 0,
and is not a rational integer, > > > the polynomial x^2 + (1 - b)x + 7b^2 +
b = 0 does not have > > > integer coefficients, but does have at least one
algebraic integer > > > root. (Remember, algebraic integers can be roots of
other > > > polynomials than monic polynomials with integer coefficients). >
a^2 + a = 0 > > > > > > > > will not allow an algebraic integer x, if 'a' ismeans that for ANY algebraic integer x, 'a' is an algebraic integer, >BUT > > > > > > > > a^2 - (x - 1)a + 7(x^2 + x) = 0 > > > > > >
but if 'a' does not have 7 as a factor > > > > > > > > 7x^2 + (7-a)x +
an integer and the > > > > quadratic is irreducible over Q. > > > > > > > >
get such an 'a' by solving a^2 - (x - 1)a + 7(x^2 + x) = 0 > > > where x isTrue, but no-one has claimed the opposite. And clearly you > > > cannot
an algebraic integer. > > > > > > > So you are forced out of the ring of
algebraic integers. > > > > > > > > Algebraically you still have > > > > > >
the ring of algebraic integers for certain > > > > algebraic integer valuesx^2 + (1 - b)x + 7b^2 + b = 0 > > > > > > > > but the 'b' is outside of
of x. > > > > > > > > > > Let x_1 be an algebraic integer > > > > > > Let >
7x_1^2 + (7-a)x_1 + a^2 + a = 0 > > > > > > so x_1 is a root of > > > > > >a^2 - (x_1 - 1)a + 7(x_1^2 + x_1) = 0 > > > > > > Then > > > > >
7x^2 + (7-a)x + a^2 + a = 0 > > > > > > Now let gcd(7,a) = r, 7=rs, a=rt,
gcd(s,t) =1, s non unit > > > > > > We get x_1 is a root of > > > > > > sx^2
+ (s-t)x + rt^2 + t = 0 > > > > > > So sx^2 + (s-t)x + rt^2 + t = 0 has an
algebraic integer root. > > > The fact that this polynomial is not monic is
of no importance. > > > It doesn't have integer coefficients. > > > > > > >
It's easier and more direct to simply solve for x in terms of the a's: > > >
x = ((a-7) +/- sqrt(-27a^2 - 42a + 49))/14 > > > > Here you can just seeby inspection that if 'a' has 7 as a factor, then > > 7 divides through the
numerator and the denominator. > > > > BUT, if 'a' only shares some non-unit
factors with 7 while being > > coprime to others, those factors will divide
off, leaving the others in > > the denominator. > > > > So what. If t
divides u and t divides v then t divides (u+v). > However it is possible to
have t does not divide u and t does > not divide v but t does divide (u+v).
Yup. > > E.g 2 divides (1+ sqrt(5)) in the ring of algebraic integers, even
though 2 does not divide 1 and 2 does not divide sqrt(5). > So((1+sqrt(5))/2) and ((1-sqrt(5))/2) are both algebraic integers. >
-William Hughes Right. They BOTH have to have it as a factor. What x =((a-7) +/- sqrt(-27a^2 - 42a + 49))/14 shows is that if 'a' has some factors
in common with 7, while not having 7 as a factor, then x cannot be an
algebraic integer as those factors cannot divide out for both solutions.
What you need here is, say, for x = ((a-7) + sqrt(-27a^2 - 42a + 49))/14 to
have those factors divide out, while x = ((a-7) - sqrt(-27a^2 - 42a +
49))/14 would not be an algebraic integer because they don't, but how would
you pick? That's why it's important that 7x^2 + (7-a)x + a^2 + a = 0 remains
non-monic if 'a' doesn't have 7 as a factor. James Harris
......................................................
Jesse F. Hughes wrote: > jstevh@xxxxxxx writes: > > > So Galois Theory as a
way to determine where factors go is refuted, and > > it is proven that the
ring of algebraic integers is incomplete[...] > > James, I know you're a
busy guy and all, but I am still waiting for a > definition of "complete"
here. What does it mean when you say that > the ring of algebraic integers
is not complete? Which of Dedekind's > theorems is thereby refuted? And how
does this contradict Galois > theory? > > Some of us are a bit slow here and
could use even simpler > explanations. > Galois Theory has been used to make
claims about factors and how they distribute. When it comes down to
particulars though claims relevant to this discussion generally are based on
whether or not a factor is a factor in the ring of algebraic integers. With
the Decker example you can see how 7 as a factor of one of the a's works for
quite a few cases, except when x = 1 mod 7 (I've been saying when x=1 but
that's the correct requirement) and when the equation defining the a's is
irreducible over Q, as then neither of the a's can have 7 as a factor in the
ring of algebraic integers. Completeness or lack of it is best shown with an
example, where my favorite is to give 2 and 6 considering evens a ring, and
note that 2 is coprime to 6 IN THAT RING because 3 is not even. Similarly,
numbers are excluded from the ring of algebraic integers because they are
not roots of monic polynomials with integer coefficients, which can lead to
apparent contradictions, where you can appear to prove two different and
contradictory things. Back to my 2 and 6 example, it's like being able to
prove that 2 is not coprime to 6, but then someone coming back at you that
it IS coprime to 6, in the ring of evens. You have a contradiction unless
you understand that 3 is excluded so you are both right. I can prove 7 as a
factor in a way I call algebraically which is equivalent to being in the
ring of objects, which I defined. Dedekind claimed to have a proof depending
on the theory of ideals which would it turns out prove that the ring of
algebraic integers is complete, and cannot have the problem I outline, like
with evens where 2 and 6 are coprime. Here completenes means that you
couldn't algebraically contradict with a coprimeness result in the ring of
algebraic integers if it were complete. So that coprimeness in that ring
means coprimeness in general, versus like with evens, coprimeness in the
ring, merely meaning you may have an element that is not in the ring, like 3
is not even so 2 is coprime to 6. The Dedekind proof claim that I've seen
cited on this newsgroup by William Hughes, says that he relies on the theory
of ideals. So, if that is true, the theory of ideals has been shown to be
invalid. All of that from a simple quadratic example. James Harris
...............................................
Few of you may have noticed a posting by a Rick Decker--a professor last
time I checked, at Hamilton College--meant to refute my ideas where he put
up a simple quadratic example. But instead his example allows me to explain
with a simple quadratic versus those complicated expressions I've been
using, and directly show Galois Theory failing. I want you all to consider
that no matter how much you might believe in a particular mathematical idea
it has no real value if it does not work. It might seem strange to face a
situation where a lot of people believed in some idea that turned out not to
work, but history shows it has happened many times before. I just posted a
reply in another thread where I gave the simple refutation of Galois Theory
more directly. But I first talked through a key part of that refutation,
yesterday. So then, why do I have to personally push the information with
something so huge? Maybe no one noticed? But what if they did? What if they
know it's right, and have decided to just sit quietly on the information?
That's an important point, as my fear is that some of you may not realize
that there are actually people who finding out I am right, would work to
keep the false ideas popular, and in use, and to do so they have to ignore
what can be easily shown. Decker himself is duty bound to not only
acknowledge the clear information given by his own example, but as he said
he would, work now to make it public knowledge that Galois Theory has been
shown to not work and be a vocal supporter of my research versus a critic.
The refutation is now simpler, leaving fewer areas for denial for posters
trying to argue with me, but history shows that people can fight quite
vigorously to protect their self image when they have been shown to be
wrong. But I want you to consider carefully that you are better off knowing
that truth, and that mathematics is a lot more useful, fun and important,
when it actually works versus just being something that a lot of people
think works. James Harris ..............................................
Marty wrote: > <jstevh@xxxxxxx> wrote in message >
news:1139077253.429594.287810@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > > Few of you
may have noticed a posting by a Rick Decker--a professor > > last time I
checked, at Hamilton College--meant to refute my ideas > > where he put up a
simple quadratic example. > > > > But instead his example allows me to
explain with a simple quadratic > > versus those complicated expressions
I've been using, and directly show > > Galois Theory failing. > > You are
quite WRONG, but you already know that. > >
http://mathworld.wolfram.com/FundamentalTheoremofGaloisTheory.html >
http://www.cs.amherst.edu/~dac/galois.html >
http://www.math.uni-bonn.de/people/gata/ >
http://www.davidson.edu/academic/math/swallow/ExpGalThWeb/ Citations don't
change algebra. The key point about Decker's example is that it relies on
very basic quadratic expressions, and I think it telling that you make an
appeal to authority to try and counter algebra. People make mistakes.
Mathematicians will not be the first group of people to hold on to wrong
ideas for over a hundred years. But a mathematical proof is perfect. Forget
the people, focus on the math. James Harris
................................................
David Moran wrote: > > But a mathematical proof is perfect. > > Not always.
The logic could be flawed, leading to an incorrect conclusion. > > Dave Then
it's not a proof. For some reason the word "proof" has been abused in math
society to also mean claim of proof, when that's just not useful. A
mathematical proof is a perfect argument. Claims of proof can be called
that, just like in other areas, for instance, if a prosecutor claims proof
that you committed a crime, but it turns out that prosecutor does not have
proof, then it's not failed proof! It's lack of proof. That is, there never
was actually proof. It's not like there was proof that turned out to not be
proof. See what I mean? An actual mathematical proof is a perfect argument,
without any error. Someone can claim proof, but if their argument is shown
to be flawed, then it's not proof. James Harris
..............................................
William Hughes wrote: > jstevh@xxxxxxx wrote: > > So yeah, you can take the
expression > > > > a^3 + 3(-1+xf^2)a^2 - f^2(x^3 f^4 - 3x^2 f^2 + 3x) = 0 >
likeand instead of focusing on the a's multiply it out and focus on x
non-monic where the leading coefficient is f^2. > > > > Actually -f^6 (oryou normally do anyway to get a polynomial P(x). > > > > It will be
f^6 if you multiply through by -1) > Right. I was miistaken. > > But, it
will also be clear that if 'a' has f as a factor, the f^2 will > > divide
off, > > The leading coefficient of P(x) is f^6 > The constant term of P(x)
is 3a^2-a^3 > > Even if 'a' has f as a factor the leading coefficient will
not divide > off and we will not be left with a monic polynomial. > > >
giving you a monic polynomial, and thus, an algebraic > > integer value for
x. > > > > No. But in any case while a monic polynomial would insure an >
algebraic integer x, it is not true that an algebraic integer x cannot >
satisfy a non-monic polynomial. You have made this error > before. >
-William Hughes Here's a better example where the initial expressionswere
given by a Rick Decker, interestingly enough, as part of an attempt to
refute my mathematical ideas. Instead he's given me to tools to explain them
simply. The following works by focusing on simple quadratics given by that
poster where I can directly consider what happens with one of them having 7
as a factor, and show that except for a special case that possibility is
required while the predictions of Galois Theory are impossible. With 7 Q(x)
= 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) = 7(25 x^2 + 30 x + 2) and 7 Q(x) =
(5a_1(x) + 7)(5a_2(x) + 7) the a's are defined by a^2 - (x - 1)a + 7(x^2 +
x) = 0 where Decker chose this example because at x=1, the middle term goes
to 0 which he I guess thought was a counterexample to my ideas because at
x=0, the a's are 0 and 1, indicating that 7 divides through only one of the
expressions. That can be explained but more importantly, so I'll leave it
for now, the Decker example can be used to simply refute Galois Theory, and
the theory of ideals. The following disproof focuses directly on Galois
Theory while the relation to the theory of ideals is more complex. The first
step in the disproof is to focus on that last quadratic and write it with
the focus on x, which gives a non-monic polynomial that becomes monic if 'a'
has 7 as a factor: 7x^2 + (7-a)x + a^2 + a = 0 and using a = 7b, you get
7x^2 + (7 - 7b)x + 49b^2 + 7b = 0 and dividing off 7, gives you x^2 + (1 -
b)x + 7b^2 + b = 0 so for ANY algebraic integer 'a' that has 7 as a factor,
you get an algebraic integer x. Also, notice that 7b^2 in the expression so
that if you focus on 'b' you have a non-monic polynomial and as before, 'b'
cannot be an algebraic integer if that polynomial has integer coefficients
and is irreducible over Q. BUT a^2 - (x - 1)a + 7(x^2 + x) = 0 means that
for ANY algebraic integer x, 'a' is an algebraic integer, but if 'a' does
not have 7 as a factor 7x^2 + (7-a)x + a^2 + a = 0 will not allow an
algebraic integer x, if 'a' is an integer and the quadratic is irreducible
over Q. To save Galois Theory and the theory of ideals, the attempted
defense of those ideas at this point would require that for some
non-rational 'a' with partial factors of 7, as it is a root of a^2 - (x -
1)a + 7(x^2 + x) = 0 with an integer x other than x=1, you would get a
polynomial with an integer solution for x from 7x^2 + (7-a)x + a^2 + a = 0
which would remain non-monic because the factors of 7 not in common with 'a'
would remain on the leading coefficient. But that also means that you would
have to have solutions for x that would work for that 'a' which are
themselves not algebraic integers because the polynomial is non-monic.
However, assume that x is an integer, so with a^2 - (x - 1)a + 7(x^2 + x) =
0 7(x^2 + x) = 7M where M is some integer, and you get two solutions for x,
consistent with there being two values of x for any given 'a' that will
work, but that quadratic is monic! And it cannot be non-monic for any
solutions where x is an integer, removing the possibility needed for Galois
Theory and the theory of ideals as that means that 7x^2 + (7-a)x + a^2 + a =
0 cannot map an algebraic integer solution for the a's to a non-algebraic
integer solution for x, which takes away the possibility of 'a' having a
partial factor of 7. I think what I have is wordy, but for the
mathematically sophisticated the short answer is that you can take 7 Q(x) =
7((x^2 + x)(5^2) + (-1 + x)(5) + 7) = 7(25 x^2 + 30 x + 2) and 7 Q(x) =
(5a_1(x) + 7)(5a_2(x) + 7) where the a's are defined by a^2 - (x - 1)a +
7(x^2 + x) = 0 and focus on that last to show that 'a' must have 7 as a
factor or be coprime to 7 except for the special case at x=1. James Harris
..................................................
José Carlos Santos wrote: > jstevh@xxxxxxx wrote: > > > The following
disproof focuses directly on Galois Theory while the > > relation to the
theory of ideals is more complex. > > > > The first step in the disproof is
to focus on that last quadratic and > > write it with the focus on x, which
gives a non-monic polynomial that > > becomes monic if 'a' has 7 as a
factor: > > It has 7 as a factor *in which ring*? I suppose that you're
working with > the ring of algebraic integers. Am I right? > > > 7x^2 +
(7-a)x + a^2 + a = 0 > > > > and using a = 7b, you get > > > > 7x^2 + (7 -
7b)x + 49b^2 + 7b = 0 > > > > and dividing off 7, gives you > > > > x^2 +
(1 - b)x + 7b^2 + b = 0 > > > > so for ANY algebraic integer 'a' that has 7
as a factor, you get an > > algebraic integer x. > > True, _x_ must be an
algebraic integer. Not obvious, but true. > Yes. So that is the one route to
always getting algebraic integers, which is to have algebraic integer b,
with a=7b, as that will also give you an algebraic integer x, no matter
what. > > Also, notice that 7b^2 in the expression so that if you focus on
'b' > > you have a non-monic polynomial and as before, 'b' cannot be an > >
algebraic integer if that polynomial has integer coefficients and is > >
irreducible over Q. > > Indeed. But then you're saying that _b_ is a factor
of _a_ in the ring > of algebraic integers *and* that _b_ does not belong to
that ring. Don't > you see a contradiction here? > Ah, but that's not the
conclusion that follows mathematically! Mathematically the proper conclusion
is that 7 is in general a factor of one of the a's, except at the special
case at x=1, but b is not always an algebraic integer given algebraic
integer x. So with x^2 + (1 - b)x + 7b^2 + b = 0 you still get the monic
polynomial, but--and this is the crucial point--b cannot be an algebraic
integer for those cases where neither of the a's can have 7 as a factor in
the ring of algebraic integers. So you can see the problem with
incompleteness of the ring of algebraic integers. The 7 factor is
algebraically there, but because of a technicality, it is not a factor in
the ring of algebraic integers, which is something best explained I think by
considering 2 and 6 in the ring of evens, as because 3 is not even, 2 is not
a factor of 6, in that ring. And I want to emphasize again that there is no
reason to fight for mathematical ideas that DO NOT WORK, so the best thing
here is to work through the explanations so that everyone can understand
them readily, and get to the hard process of ending the current teaching of
Galois Theory and the theory of ideals. That teaching will be replaced with
correct and extremely powerful mathematical ideas. James Harris
...............................................
José Carlos Santos wrote: > jstevh@xxxxxxx wrote: > > >>> The following
disproof focuses directly on Galois Theory while the > >>> relation to the
theory of ideals is more complex. > >>> > >>> The first step in the disproof
is to focus on that last quadratic and > >>> write it with the focus on x,
which gives a non-monic polynomial that > >>> becomes monic if 'a' has 7 as
a factor: > >> It has 7 as a factor *in which ring*? I suppose that you're
working with > >> the ring of algebraic integers. Am I right? > > You did
not answer this question! > I didn't notice it before. The answer is, the
ring of algebraic integers. > >>> 7x^2 + (7-a)x + a^2 + a = 0 > >>> > >>>
and using a = 7b, you get > >>> > >>> 7x^2 + (7 - 7b)x + 49b^2 + 7b = 0 >
b = 0 > >>> > >>> so for ANY algebraic integer 'a' that has 7 as a factor,and dividing off 7, gives you > >>> > >>> x^2 + (1 - b)x + 7b^2 +
you get an > >>> algebraic integer x. > >> True, _x_ must be an algebraic
integer. Not obvious, but true. > >> > > > > Yes. So that is the one route
to always getting algebraic integers, > > which is to have algebraic integer
b, with a=7b, as that will also give > > you an algebraic integer x, no
matter what. > > That's an important point I made showing an infinite number
of solutions following the theory I have, which is that 7 is a factor of
only one of the a's in general, except for the special case at x=1. The
reason you're pushed out of the ring of algebraic integers when x is an
integer is because of the rule that algebraic integers are roots of monic
polynomials with integer coefficients. That arbitrary rule is like the rule
that evens have 2 as a factor, in the ring of integers, so that if you go to
the ring of evens, the rule means that 2 is NOT a factor of 6 because 3 is
not in the ring. Rules can create special situations, which the algebra just
ignores, so, for instance, algebraically 2 is a factor of 6. Some of you may
have noticed I talk about "algebraically" having a factor a lot, and that
may not seem like a rigorous idea, but it can be shown to being equivalent
to being in a ring that I call the ring of objects. > >>> Also, notice that
7b^2 in the expression so that if you focus on 'b' > >>> you have a
non-monic polynomial and as before, 'b' cannot be an > >>> algebraic integer
if that polynomial has integer coefficients and is > >>> irreducible over Q.
Don't > >> you see a contradiction here? > >> > > > > Ah, but that's not theIndeed. But then you're saying that _b_ is a factor of _a_ in the ring
of algebraic integers *and* that _b_ does not belong to that ring.
conclusion that follows mathematically! > > No? Exactly where is my mistake?
I show, and it's trivial to see that it's true, that given an algebraicinteger b, with a = 7b you are guaranteed to have an algebraic integer x.
BUT with 7x^2 + (7-a)x + a^2 + a = 0 if 'a' does not have 7 as a factor,
then you are forced into having a non-monic polynomial. At least one
solution of that polynomial MUST be out of the ring of algebraic integers.
Therefore, it follows logically that if you use a = 7b, and have x^2 + (1 -
b)x + 7b^2 + b = 0 that b cannot be in the ring of algebraic integers for
those cases where 'a' does not have 7 as a factor in the ring of algebraic
integers. So you're pushed out of that ring. > > Mathematically the proper
conclusion is that 7 is in general a factor > > of one of the a's, except at
the special case at x=1, but b is not > > always an algebraic integer given
algebraic integer x. > > Where did you get the idea that "7 is in general a
factor of one of the > a's, except at the special case at x=1"? Take x = 2,
for instance. Then > a = 1/2 +/- i*sqrt(167)/2. The minimal polynomial of
a/7 is then > 7z^2 - z + 6. Therefore, a/7 is *not* an algebraic integer and
so 7 is > not a factor of any of the two a's in the ring of algebraic
integers. > > Best regards, > > Jose Carlos Santos It is true that for those
values that 7 is NOT a factor of either of the a's in the ring of algebraic
integers, but algebraically it can be shown that 7 is in general a factor of
one of the a's except for the special case at x=1. That may sound
contradictory, but again the example with 2 and 6 in the ring of evens shows
exactly how it can happen if some numbers are excluded by a special rule.
With the evens 2 and 6 are coprime in the ring of evens because 3 is not
even. With algebraic integers numbers are excluded because they are not
roots of monic polynomials with integer coefficients. James Harris
.....................................
José Carlos Santos wrote: > jstevh@xxxxxxx wrote: > > >>>>> The first step
in the disproof is to focus on that last quadratic and > >>>>> write it with
the focus on x, which gives a non-monic polynomial that > >>>>> becomes
monic if 'a' has 7 as a factor: > >>>> It has 7 as a factor *in which ring*?
I suppose that you're working with > >>>> the ring of algebraic integers. Am
I right? > >> You did not answer this question! > >> > > > > I didn't notice
it before. The answer is, the ring of algebraic > > integers. > > OK. Thanks
for making that clear. > > >>>>> 7x^2 + (7-a)x + a^2 + a = 0 > >>>>> > >>>>>
and using a = 7b, you get > >>>>> > >>>>> 7x^2 + (7 - 7b)x + 49b^2 + 7b = 0
b)xand dividing off 7, gives you > >>>>> > >>>>> x^2 + (1 -
+ 7b^2 + b = 0 > >>>>> > >>>>> so for ANY algebraic integer 'a' that has 7
as a factor, you get an > >>>>> algebraic integer x. > >>>> True, _x_ must
be an algebraic integer. Not obvious, but true. > >>>> > >>> Yes. So that is
the one route to always getting algebraic integers, > >>> which is to have
algebraic integer b, with a=7b, as that will also give > >>> you an
algebraic integer x, no matter what. > >>> > > > > That's an important point
I made showing an infinite number of > > solutions following the theory I
have, which is that 7 is a factor of > > only one of the a's in general,
except for the special case at x=1. > > > > The reason you're pushed out of
the ring of algebraic integers when x > > is an integer is because of the
rule that algebraic integers are roots > > of monic polynomials with integer
coefficients. > > That's not a rule. That's what the expression "algebraic
integer" means. > Calling it a "rule" is as silly as saying that there's a
rule that says > that any rational number must be the quotient of two
integers. > Not exactly. The rule for evens is that 2 is a factor of any
number that is even. Now that is also the definition of evens. But it
doesn't change it from being a rule that for evens 2 is a factor. Your
example that any rational number must be the quotient of two integers is
more like just a definition in that it doesn't quite make sense as a rule.
In contrast, the rule is that to be an algebraic integer a number must be a
root of a monic polynomial with integer coefficients, which importantly DOES
NOT FOLLOW from simply defining algebraic integers to be roots of monic
polynomials with integer coefficients. It has to be proven, and the proof is
easy enough. But the definition of algebraic integers does not directly give
the rule, as the rule has to be proven as resulting from that definition. >
That arbitrary rule is like the rule that evens have 2 as a factor, in >the ring of integers, so that if you go to the ring of evens, the rule > >
means that 2 is NOT a factor of 6 because 3 is not in the ring. > > The term
"even" is defined in the ring of integers. Have you ever seen > it defined
for the ring of even numbers? Where? > The example is to show how an
exclusionary rule can cause apparent contradictions. So 2 is coprime to 6 in
a ring made up only of evens because 3 is excluded from that ring because
it's not even. Similarly the exclusionary rule that numbers must be roots of
monic polynomials with integer coefficients excludes numbers from the ring
of algebraic integers giving the possibility of apparent contradictions, as
I've repeatedly shown. So I prove that 7 is a factor and then one of you
comes back and claims that it's not a factor, but your claim is in the ring
of algebraic integers. Similarly I could prove that 2 is a factor of 6, and
someone in the ring of evens could truthfully proclaim that false, as 2 is
coprime to 6 in THAT ring. Round and round you can go with such discussions
where one person can prove a factor, while others in a ring with a special
rule can proclaim that is not a factor. > > Rules can create special
situations, which the algebra just ignores, > > so, for instance,
algebraically 2 is a factor of 6. > > > > Some of you may have noticed I
talk about "algebraically" having a > > factor a lot, and that may not seem
like a rigorous idea, but it can be > > shown to being equivalent to being
in a ring that I call the ring of > > objects. > > Which is...? > I'm not
going into the ring of objects now as that is a side discussion. For the
purposes of this discussion consider the possibility that there is a ring
that does not have the problem of the ring of algebraic integers in that you
cannot appear to prove a contradiction i.e. that a number does and does not
have a given factor. The key here is that I can algebraically prove a
factor, and find that in the ring of algebraic integers that number is not a
factor, just like you can prove 2 is a factor of 6, while someone else can
claim you are wrong by relying on the ring of evens. > >>>>> Also, notice
that 7b^2 in the expression so that if you focus on 'b' > >>>>> you have a
non-monic polynomial and as before, 'b' cannot be an > >>>>> algebraic
integer if that polynomial has integer coefficients and is > >>>>>
irreducible over Q. > >>>> Indeed. But then you're saying that _b_ is a
factor of _a_ in the ring > >>>> of algebraic integers *and* that _b_ does
not belong to that ring. Don't > >>>> you see a contradiction here? > >>>> >
Exactly where is my mistake? > >> > > > > I show, and it's trivial to seeAh, but that's not the conclusion that follows mathematically! > >> No?
that it's true, that given an algebraic > > integer b, with > > > > a = 7b >
factor, then you are forced into having a > > non-monic polynomial. > > > >you are guaranteed to have an algebraic integer x. > > > > BUT with >
7x^2 + (7-a)x + a^2 + a = 0 > > > > if 'a' does not have 7 as a
At least one solution of that polynomial MUST be out of the ring of > >
algebraic integers. > > Really? Which root of the polynomial 2x - 2 does not
belong to the ring > of algebraic integers? > The polynomial is non-monic,
so you know that at least one of its solutions is outside of the ring of
algebraic integers, if 'a' does not have 7 as a factor. That's basic. For
example, with 2x^2 + 3x + 1 = (2x + 1)(x + 1) notice that one root of x
is -1/2 which is not an algebraic integer. > > Therefore, it follows
logically that if you use a = 7b, and have > > > > x^2 + (1 - b)x + 7b^2 + b
= 0 > > > > that b cannot be in the ring of algebraic integers for those
cases > > where 'a' does not have 7 as a factor in the ring of algebraic > >
integers. > > > > So you're pushed out of that ring. > > What makes you
believe that a root of a non-monic polynomial whose > coefficients are
algebraic integers cannot be an algebraic integer? > See my counter-example
above. > This is getting tedious. I didn't say that it couldn't be an
algebraic integer. I noted that ONE of them couldn't be, and you can see my
example. James Harris
............................................................
José Carlos Santos wrote: > jstevh@xxxxxxx wrote: > > >> That's not a rule.
That's what the expression "algebraic integer" means. > >> Calling it a
"rule" is as silly as saying that there's a rule that says > >> that any
rational number must be the quotient of two integers. > > > > Not exactly.
The rule for evens is that 2 is a factor of any number > > that is even. > >
Actually, that's the definition, not a rule. > It's a definition and a rule.
from being a rule that for evens 2 is a factor. > > > > Your example thatNow that is also the definition of evens. But it doesn't change it > >
any rational number must be the quotient of two > > integers is more like
just a definition in that it doesn't quite make > > sense as a rule. > >
Exactly what happens with the concept of "algebraic integer"! > > > In
contrast, the rule is that to be an algebraic integer a number must > > be a
root of a monic polynomial with integer coefficients, which > > importantly
DOES NOT FOLLOW from simply defining algebraic integers to > > be roots of
monic polynomials with integer coefficients. > > Fantastic! You wrote that
if one defines "algebraic integers" to be > roots of monic polynomials with
integer coefficients, it DOES NOT FOLLOW > that an algebraic integer must be
a root of a monic polynomial with > integer coefficients. I would not have
believed it if I hadn't seen it > myself. :-) > It takes a proof that any
algebraic integer must be the root of some monic polynomial with integer
coefficients. The rule then is at that point set, by the proof, but not by
definition. That is, the definition of algebraic integers can be shown to
lead to the rule. > > It has to be proven, and the proof is easy enough. > >
Are you familiar with the concept of "tautology"? > Yup. I can see that
you're kind of clueless here, which will be revealed by your next comments.
But how could you get lost on such simple points? > > The example is to show
how an exclusionary rule can cause apparent > > contradictions. > > They are
"apparent" for you alone. > I doubt it. > > So 2 is coprime to 6 in a ring
made up only of evens because 3 is > > excluded from that ring because it's
not even. > > Define "coprime" in the ring of even numbers. > Not sharing
factors within the ring. That is, 2 and 6 do not share a factor in the ring
of evens. Understand? > >>>>>>> Also, notice that 7b^2 in the expression so
that if you focus on 'b' > >>>>>>> you have a non-monic polynomial and as
before, 'b' cannot be an > >>>>>>> algebraic integer if that polynomial has
integer coefficients and is > >>>>>>> irreducible over Q. > >>>>>> Indeed.
But then you're saying that _b_ is a factor of _a_ in the ring > >>>>>> of
algebraic integers *and* that _b_ does not belong to that ring. Don't >
theyou see a contradiction here? > >>>>>> > >>>>> Ah, but that's not
conclusion that follows mathematically! > >>>> No? Exactly where is my
mistake? > >>>> > >>> I show, and it's trivial to see that it's true, that
given an algebraic > >>> integer b, with > >>> > >>> a = 7b > >>> > >>> you
are guaranteed to have an algebraic integer x. > >>> > >>> BUT with > >>> >
factor, then you are forced into having a > >>> non-monic polynomial. > >>>7x^2 + (7-a)x + a^2 + a = 0 > >>> > >>> if 'a' does not have 7 as a
does not belong to the ring > >> of algebraic integers? > >> > > > > TheAt least one solution of that polynomial MUST be out of the ring of >algebraic integers. > >> Really? Which root of the polynomial 2x - 2
polynomial is non-monic, so you know that at least one of its > > solutions
is outside of the ring of algebraic integers, if 'a' does not > > have 7 as
a factor. > > The polynomial 2x - 2 is non-monic. Can you please tell me
which of its > roots is not an algebraic integer? You have a factor of 2,
which isn't applicable to the polynomial I was talking about. Your comment
is like I should explain every tiny little detail as if you can't look at
7x^2 + (7-a)x + a^2 + a = 0 and see that it does NOT have 2 as a factor. > >
That's basic. > > > > For example, with > > > > 2x^2 + 3x + 1 = (2x + 1)(x +
1) > > > > notice that one root of x is -1/2 which is not an algebraic
integer. > > Again, can you please tell me which of the roots of the
polynomial > 2x - 2 is not an algebraic integer? > > > This is getting
tedious. > > I agree! > You're being deliberately obtuse. > > I didn't say
that it couldn't be an algebraic integer. I noted that > > ONE of them
couldn't be, and you can see my example. > > Which of the roots of the
polynomial 2x - 2 cannot be algebraic integer? > > Best regards, > > Jose
Carlos Santos Your behavior is clearly meant to be simply aggravating as
there is no way you can sensibly think that 2x - 2 is applicable when the
expression in question is 7x^2 + (7-a)x + a^2 + a = 0 where there is no
constant factor visible. That kind of behavior isn't clever, so it's hard to
imagine what you think you gain. I guess you just want to be aggravating. So
childish. James Harris
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José Carlos Santos wrote: > jstevh@xxxxxxx wrote: > > >>> Now that is also
the definition of evens. But it doesn't change it > >>> from being a rule
that for evens 2 is a factor. > >>> > >>> Your example that any rational
number must be the quotient of two > >>> integers is more like just a
definition in that it doesn't quite make > >>> sense as a rule. > >> Exactly
what happens with the concept of "algebraic integer"! > >> > >>> In
contrast, the rule is that to be an algebraic integer a number must > >>> be
a root of a monic polynomial with integer coefficients, which > >>>
importantly DOES NOT FOLLOW from simply defining algebraic integers to > >>>
be roots of monic polynomials with integer coefficients. > >> Fantastic! You
wrote that if one defines "algebraic integers" to be > >> roots of monic
polynomials with integer coefficients, it DOES NOT FOLLOW > >> that an
algebraic integer must be a root of a monic polynomial with > >> integer
coefficients. I would not have believed it if I hadn't seen it > >> myself.
:-) > >> > > > > It takes a proof that any algebraic integer must be the
root of some > > monic polynomial with integer coefficients. > > Even if
"algebraic integer" means "root of some monic polynomial with > integer
coefficients? > Yup. I'll take an example from your playbook, and define
Santos numbers as roots of non monic polynomials, so 2x + 2 = 0 is, by that
definition, a Santos number. But x=1 is also the root of polynomials that
are NOT monic, so the definition doesn't create the rule that a Santos
number must be the root of only non monic polynomials. You're way past
tiresome, as you just keep arguing about dumb stuff. The "pure math" that
comes from the ideas I've shot down has been impractical, with
mathematicians saying maybe, someday it might be useful, but I say, it has
never been practical, because it's wrong (with some questions now in my mind
about the use of group theory in physics). Yet, I give proof, and you people
fight the proof. I get proof published and you fight publication. I come
back with examples showing integers behaving as predicted by my research,
and you try to ignore that, so I use an example one of you came up with, to
finally crush any semblance of a mathematical objection, and you still fight
the mathematical truth. What DO you believe in? James Harris
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