How I know, linchpin of my FLT proof
James Harris
claiming that I'm right with all these people posting disagreement,
and it seems to me that maybe explaining how might help.
You see, my short proof of Fermat's Last Theorem is flawless with all
the disputes from mathematicians attacking the following statement:
Given a factor g of a polynomial P(x), g=r+c, where c is a factor of
the constant term P(0) of the polynomial, given by the value of g at
x=0, and r=g-c.
That's it.
Now if people were arguing with you trying to attack that when you
know its truth proves your case that you have a short proof of
Fermat's Last Theorem, would you believe them?
With polynomial factors I can show the statement easily enough as, for
instance, g = x+2 is a factor of x^2 + 3x + 2, and here r=x, and c=2,
and of course, at x=0, g=2. With non-polynomial factors, it gets a
little harder, but not impossible, to show. Still, why should I have
to give examples, when it's so easy to prove?
In the face of my short proof of Fermat's Last Theorem, posters have
not disagreed with the truth of the statement with polynomial factors,
but have tried to cast doubt, disagree, or just ignore it for factors
in general.
Remember a polynomial factor is something like x+2, where you have x
with a positive *integer* degree. Another example of a polynomial
factor is sqrt(2)x^2 + 3x + 7^{1/3}. And again, it's a polynomial
because x has a positive *integer* degree. I generalized to a factor
of a polynomial.
All I did was make a generalized statement about *any* factor of a
polynomial, not just polynomial factors, and proved it, as it's easy
enough to do.
So yes, I have absolute certainty that my proof of FLT is correct
because the linchpin is that statement above.
When the mathematical proof of the statement is accepted, there is no
room for disagreement with my work.
Now for those of you who are logical and rational people, such a
statement should seem easily checkable.
Those of you who understand political realities might also understand
why mathematicians might try their best to fight a short proof of
Fermat's Last Theorem found by an outsider like myself, even when they
can be shown to be arguing against math and logic itself, with such a
simple statement.
It's about power. It's politics. Mathematicians have a society,
which they control. If they acknowledge the truth, it shakes up their
society, so mostly they ignore me, while a few make posts attacking.
For me it's just a weird experience to add to the rest, but it also
tells me a lot about how the real world works, and how easily people
can be convinced to disbelieve in things that should be obvious.
My guess is that many of you believe there's some "higher math" or
some special rule, or incredibly hard to understand math thing that
makes it not true that given a factor g of a polynomial P(x), g=r+c,
where c is a factor of the constant term, given by the value of g at
P(0), and r=g-c.
You may simply dismiss me without even paying attention to the
statement, or you may look at it and simply decide that you don't know
enough, that surely mathematicians couldn't fight such a thing, and if
they did, how could they get away with it?
So I can watch you. And I can see what you believe in, really. I can
see how you think, and how it's possible to control even large,
supposedly skeptical populations.
Yup, you can learn a lot about people with a little math.
James Harris
Sam Wormley
>
> You see, my short proof of Fermat's Last Theorem is flawless with all
> the disputes from mathematicians attacking the following statement:
>
> Given a factor g of a polynomial P(x), g=r+c, where c is a factor of
> the constant term P(0) of the polynomial, given by the value of g at
> x=0, and r=g-c.
>
> That's it.
Thanks for registering at
http://www.google.com/search?q=%22James+Harris%22+site%3Awww.crank.net
http://www.crank.net/harris.html
Uncle Al
>
> I remember a post where someone was wondering how I could keep
> claiming that I'm right with all these people posting disagreement,
> and it seems to me that maybe explaining how might help.
You are insane. You cannot counterargue fact, so you scream some more
as though reality never existed.
http://b5.sdvc.uwyo.edu/bab5/snds/reflecti.wav
You are loathsome and loud, http://w0rli.home.att.net/youare.swf
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>I remember a post where someone was wondering how I could keep
>claiming that I'm right with all these people posting disagreement,
>and it seems to me that maybe explaining how might help.
>
>You see, my short proof of Fermat's Last Theorem is flawless with all
>the disputes from mathematicians attacking the following statement:
>
>Given a factor g of a polynomial P(x), g=r+c, where c is a factor of
>the constant term P(0) of the polynomial, given by the value of g at
>x=0, and r=g-c.
>
>That's it.
Great, because that statement is vacuous nonsense as written.
What sort of object is g? If P(x) is a polynomial with coefficients in
R, i.e. an element of R[x], is g also an element of R[x]? Does
"factor" mean factor in R[x]? Does it mean "factor in R^R", the ring
of functions from R to R? Or in some subring thereof?
What sort of objects are r and c? In what ring is c meant to be a
factor of P(0)?
Is the decomposition given? The statement you wrote is grammatically
unclear, with no conclusion to the premises.
What sort of object is r?
Your statement can be fixed, of course, and it either becomes trivial
and inapplicable to the situation you have in hand, or too vacuous to
be of use.
For instance, one possibility would be:
THEOREM. Let R be a commutative ring with 1. Let P(x) in R[x], and let
g be a factor of P(x) in R[x]. Then there exist r in R[x] and c in R
such that g=r+c, and c divides the constant term of P(x) in R.
The proof of this follows trivially from the definition of the
operations in R[x]; however, this statement does not apply to the
situation you try to use your "statement" in, because in that case the
things you are playing with are not all polynomials in one variable.
So perhaps you mean:
"Let R be a commutative ring with 1. Let P(x) in R[x], and identify
P(x) with its natural image in R^R, the ring of all functions from R
to R, with pointwise operations. Let g in R^R be a factor of P(x) in
R[x]. Then there exists a constant c in R and a function r in R^R such
that c divides P(0) in R, and g=r+c."
In this case, again, there is little to do: since g is a factor of
P(x) in R^R, that means that there exists a function f:R->R such that
g*f = P(x). That is, for all a in R, g(a)*f(a)=P(a).
In particular, g(0)*f(0)=P(0); let c=g(0). Now define r:R->R as
r(a) = g(a)-c. Then r satisfies the conditions.
Unfortunatley, r is not necessarily a nice function, any more than g
is. In addition, this is also not applicable to the situation you have
in hand, since you do not deal with polynomials in a single variable.
So, you have a (possibly) trivial statement, which you have been
unable to state coherently and clearly in over 6 months, which is
nonetheless completely inapplicable to the situation you have in
hand. You never say "apply the lemma by setting P(x) equal to blah, g
equal to bleh, and so we conclude that r and c are blih and bloh."
[.rest deleted.]
======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A great
many people are staggered to this extend, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
The Last Danish Pastry
news:3c65f87.03080...@posting.google.com...
> I remember a post where someone was wondering how I could keep
your self-respect. Yes, it does seem to be an insoluble problem.
Maxim Stepin
> claiming that I'm right with all these people posting disagreement,
> and it seems to me that maybe explaining how might help.
>
> You see, my short proof of Fermat's Last Theorem is flawless with all
> the disputes from mathematicians attacking the following statement:
>
> Given a factor g of a polynomial P(x), g=r+c, where c is a factor of
> the constant term P(0) of the polynomial, given by the value of g at
> x=0, and r=g-c.
>
> That's it.
Not true.
Lots of people pointed that your definition of "objects" is curcular,
which obviously makes the whole proof invalid.
You never addressed those objections. Will you address them now?
I bet you won't.
Because when you see "objects" mentioned in a reply, you go suddenly blind.
William Hale
jst...@msn.com (James Harris) wrote:
> I remember a post where someone was wondering how I could keep
> claiming that I'm right with all these people posting disagreement,
> and it seems to me that maybe explaining how might help.
>
> You see, my short proof of Fermat's Last Theorem is flawless with all
> the disputes from mathematicians attacking the following statement:
>
> Given a factor g of a polynomial P(x), g=r+c, where c is a factor of
> the constant term P(0) of the polynomial, given by the value of g at
> x=0, and r=g-c.
>
> That's it.
Can't your statement be generalized as follows:
Given a factor or non-factor g of a polynomial P(x), g=r+c,
where c is a factor of the constant term P(0) of the polynomial,
given by the value of g at x=0, and r=g-c.
That is,
Given g and a polynomial P(x), then g=r+c,
where c is a factor of the constant term P(0) of the polynomial,
given by the value of g at x=0, and r=g-c.
-- Bill Hale
Ann
I like your style of self confidence and tolerance, it is the opposite
of dogmatic, it gives freedom to yourself (and the people). Honestly, I
see no reason why should you be disappointed. Imagine for a moment that
your proof of Fermat's theorem is "flawless" from every possible point
of view (really:), isn't that enough to make you happy? If I were you
I'd be the happiest person in the world (and on the Moon:). Or you
crave for some kind of social recognition, loads of money, or immortal
fame? Or maybe you desire the sweet intellectual property rights?
Luckily, your letters tell me otherwise.
Should it make you anxious if not everyone accepts your truth (words)?
Do we need to impose our dream on everyone? What for? How does it
matter whether the final proof will be five or thousand pages long?
(asuming you are not an environmentalist preserving the trees:) You
probably know that what kind of proofs are deemed acceptable
depends on the point in space-time, it is a convention which changes
and grows in complexity to keep pace with the expansion of the universe
(of words in our mind:). I think the point of your exercise was to gain
confidence in your abilities and to "know thyself". And you did quite
well here :)
One might say :) that I am speaking from a similar experience here. In
the course of the discussions in another newsgroup
(alt.dreams.castaneda, my home base:) I also had the chance to develop
a theory of my own. A strange one I admit but it made me happy. And I
did it spontaneously, in the last six months I took part in many
discussions (some of them were real battles; "Seraph: You do not truly
know someone until you fight them") and wrote thousand upon thousands
of pages until I found "the key, my key" to the way out of the matrix
(labyrinth of reason).
That is, I became an expert in word-crushing - that mysterious effect
of a word becoming less real in our mind and consciousness. Again the
universal laws of attraction and repulsion (from any Word) hold true:
adopting sequentially those two different points of view ('good' and
'bad') on the word make it invariably shake in our mind until we reach
a fancy state of sober detachment and an unexpected feeling of freedom.
Another finding was that spontaneous improvisation is one certain way
to beat reason, some of the 'treatises' (i.e. posts ranging from one to
five thousand lines:) I wrote in a day or two. Yet apart from some
annoying typos I can still read them with joy, if we are honest (with
oursouls) there is nothing to be ashamed of later.
But do you think I will seek recognition or claim property rights? Or
money, 'success' and personal profit? I couldn't care less about those
reasonable distractions (brakes), "I did what I did" as part of my
quest for the point of life (other abstract goal may also do the job)
and those writings I do not perceive as mine but as a product of
discussions, a joint work.
That general 'theory of words' (which I sub-consciously advertise
here:) helped me (the Sturdy Beggar) cleanse my sub-consciousness and
to free mysoul from the real chains of some 'unreal' polluters.
Words, basically. The most curious part is that after I found the point
of reason I followed he advice of Einstein and started asking
questions. And 'without noticing' I crushed all the major theories in
my mind, i.e. they became considerably less real. You are probably 100%
and more convinced that the 'genes', 'viruses', our gradual 'evolution'
from gases (to name just a few) are absolutely real and observable
facts, but I have found it was not so. The one certain thing I have
found them to be was words (or theories, i.e. galaxies of words).
I'll spare you the details of that disappointment, and only tell you
one of my greatest realizations: modern doctors are no healers but
long-Latin-words memorisers and masters of words. Dark words usually,
they first make us "hear the sound of inevitability", the rest is done
by our true belief. Perhaps, you know about the incredibly high
positive correlation between poor health (some call it chronic
disease:) and the visits to the doctor? Are you sure which way the
direction of causality goes?
More visits (doctors, medical books and TV 'information) lead to poorer
health or vice versa? Or is it an equivalence? Visits===>Diseases or
Illnesses===>Visits? Or is it more like:
Visits==>Disease==>Visit==>Chronic Disease==>Visit==>...===>Death
Visits <===> Diseases?
Medical words as splinters in our mind, can it be? Take it as a joke, a
serious one :) What matters to me is that the novel approach and other
'findings' of the theory of words helped me develop what I may call
ability for independent thinking. It seems a trivial quality (or an
inherited character trait:) but I have discovered a systematic way to
develop it. I found my key (which is by no means universal) and for
that I needed only choice, imagination, sobriety and laughter, loads of
it.
Basically, I believe that if Aristotle had known the word 'laughter' or
just finished his alleged "book on comedy" (the funny film "the name of
the rose") he would have found "the answer to the question" and
probably revoked some of his scholastic writings. Just as I did, I
found the splinter in my mind that was driving me mad. And I didn't
stop there :) On my way to 'Everest' I formulated (discovered or merely
reinvented the wheel:) some jolly rules of the Word and laws of
thermodynamic human nature. Like "Know thyself" first (and not start
with an escape in Nature outside:). The funniest of them was that: by
stalking others we eventually stalk ourselves. Another one was the
axiom of choice and its most general version: our personal preferences
for the words.
For example, 'nervous' doctors describe some 'psychic phenomena' as
brain states and neural networks, while we can, with the same success,
call them states of Buddha consciousness or Castanedian positions
(shifts) of the assemblage point. These are freely interchangeable
concepts aimed at explaining everything, even the magic perception of
reality (trance, dreams and many other real 'oddities'). If we distance
ourselves a bit we may realize instead that they are just words that
are not so easy to prove rigorously. Unless we cut a living brain and
stick the white disk electrodes and would it be enough? Even then we
may not see the "networks", though we may try to imagine them, that
usually works.
It is sometimes called a theoretical 'proof' of existence, like the
invisible rest of the super-stringed 10-dimensional universe in the
physical "theory of everything", that teeny-weeny point to which the
initially six-dimensional sub-universe has supposedly shrunk. So is
this infinitesimal dot real or not? Does it exist, is it there or is it
not?
'Both', say the masters of words, 'but it is a complex mystery, it is
there but you cannot find it, it's absolutely invisible, by assumption
and by definition' (in other words like a god). So catch it if you can.
The theory is truly untestable, you see, but nevertheless true :) Why?
Because 'we know it', it has to be so. Moreover :), the formulas fit
beautifully, therefore it can't be otherwise, the theoretical proof has
to be believed (when delivered, of course, for now they seem to be
waiting for something, like Newton, they are still in search of the
mathematics powerful enough to solve the equations).
Even more, that they cannot make a flea is immaterial to the beautiful
argument of the origin of the universe and species. The hard evidence
might be missing but we have to believe, because there isn't any better
alternative (explanation) around. I didn't find it convincing and found
an alternative explanation that works for me, probably you have found
another one and I am sure there exist many private answers (separate
Creation realities).
"It now seems plain to me that that theory ought to be vacated in
favour of a new and truer one...the Decent of Man from Higher Animals"
Mark Twain - "The Lowest Animal"
As you might have guessed even in this group the fight is one of words
and for words; which camp will impose its words on the rest of the
gullible hearts and minds. This is a strange consequence of the theory
of evolution: when we believe it we perceive (see) the world through
the eyes of the predator, everything is a threat: other "lower" and
less "favoured races" and species, invisible viruses, rival thought,
you name it. It is part of "the culture of fear". And quite
'naturally' scientists feel obliged to fight for the selection and
survival of their fittest dogma.
"Evolution is the law of policies: Darwin said it, Socrates endorsed
it, Cuvier proved it and established it for all time in his paper on
"The Survival of the Fittest." These are illustrious names, this is a
mighty doctrine: nothing can ever remove it from its firm base, nothing
dissolve it, but evolution." Mark Twain
"Evolution, Morpheus, evolution. Like the dinosaur." A sad "law of
policies" for I think everyone has the right to defend the matrix of
words in his or her mind. We protect the words we like, a basic human
right of freedom of speech, thought and religion (belief). When I
noticed that spiritual and religious WORDS were constantly under attack
I decided it was a good idea to give spirituality a hand.
In a way it is a choice for balance, I can hardly share the passionate
fanaticism with which some quasi-scientists seek to eradicate
spiritualism altogether: from schools, textbooks, press, news and
everyday life. I compare such a strategy to the Inquisition, it is like
a mirror image of it although the methods of the imposition of the
dogma have become much more refined and 'humane' (e.g. compulsory
secular education, college admission tests, worshipping the
technological 'progress' and machines in the media and ...). Whereas
the truth is that spiritualism and any other non-scientific thought has
the right to exist, it is as simple as that. It is called 'tolerance'.
I describe the tyrannical nature as anyone who tries to impose on
others his matrix of words (beliefs, thoughts, 'knowledge') by force.
I wouldn't like to be part of it, would you. On the other hand, I see
nothing wrong if you try to publish your proof somewhere, but do you
really need it? You've done it already in this group. That you are
faced with criticism is to be expected (we all have our flaws:),
especially in a "science.skeptics" group. Or is it
"skeptical.scientists", I am not sure :)
"I'd like to share a revelation that I've head during my time here":
nobody cares about my theory too, but how do I care. Just like you I
learned a lot about human nature (mine:), and most of all about 'human
ego'. Of course, we can choose to react differently (unhealthy) to the
criticism and become something like an angry scientist.
Do you know about the 'noble' character ("irrational temper") of
Newton? Indeed, at the end he managed to impose his truths on us for
some centuries ahead (in Leibniz and other cases - by pulling the
strings of the Royal Society of which he was a president), but how did
that make him 'happy' (a word that may very well be different from
'fame':)? Nervous breakdowns were following one after the other. In
the meanwhile, he learned the bible by heart and parrot it by rote,
alas, couldn't decipher the 'coded' Number. Something like that follows
inexorably when we lose touch with the virtuous mean (balance) between
No-ambition and Over-ambition. Even Aristotle had to learn it the hard
way, although it was him who formulated this rule of middle ground
happiness.
There are many truths ('even' in the 'certain' universe of mathematics)
and we are free to believe according to our choice, "the story ends, we
wake up in our bed and believe whatever we want to believe". We may
even choose to believe that the Word is no God, i.e. not 'Good'. And
not 'bad' either :), but something and somewhat in between. What and
where?
Best,
Ann
Mitchell
Usually discovered things are so simple that most don't want to
believe. I am not a mathematician but I see the same thing in physics.
It took a zero and a one to give birth to the infinite complexity
of mathematics. Only the great are looking for simplicity.
Everything thats true can be seen and therefore can be seen by everyone.
--------------- When the blind tell you you can't see!!!
Mitch
Maxim Stepin
Once again you choose not to address this issue.
Simple math question - does your "object ring" contain sqrt(2)?
I bet you don't know.
G. Frege
>
> You see, my short proof of Fermat's Last Theorem is flawless [...]
>
Great! If so, then why the hell don't you try to get it published in
some journal?
F.
Gib Bogle
Tut, tut. Don't you JSH does not answer questions?
Gib
Gib Bogle
<lots of good stuff>
This could be the beginning of a beautiful relationship.
Gib
|-|erc
> Ann wrote:
>
> <lots of good stuff>
I agree, and I also salute James even if I have read his posts
where's he's quite explicit the immortal fame, money and interview
with Opera are what he's after..
You can review my writings www.adamskingdom.com anytime Ann.
Sequel arriving soon,
Herc
The Last Danish Pastry
...
Simplicio: <disputation> 2+2=5
Salviati: <induction> 2+2=4
Simplicio: <conclusion> 2+2=5
Salviati: <theorem> 2+2=4
Simplicio: <reasoning> 2+2=5
Salviati: <discussion> 2+2=4
Simplicio: <ridicule> 2+2=5
Salviati: <rationality> 2+2=4
Simplicio: <inference> 2+2=5
Salviati: <logic> 2+2=4
Simplicio: <anger> 2+2=5
Salviati: <philosophy> 2+2=4
Simplicio: <debate> 2+2=5
Salviati: <syllogism> 2+2=4
Simplicio: <stupidity> 2+2=5
Salviati: <polemics> 2+2=4
...
...
...
Salviati: <disputation> 2+2=4
Simplicio: <induction> 2+2=5
Salviati: <conclusion> 2+2=4
Simplicio: <theorem> 2+2=5
Salviati: <reasoning> 2+2=4
Simplicio: <discussion> 2+2=5
Salviati: <ridicule> 2+2=4
Simplicio: <rationality> 2+2=5
Salviati: <inference> 2+2=4
Simplicio: <logic> 2+2=5
Salviati: <anger> 2+2=4
Simplicio: <philosophy> 2+2=5
Salviati: <debate> 2+2=4
Simplicio: <syllogism> 2+2=5
Salviati: <stupidity> 2+2=4
Simplicio: <polemics> 2+2=5
...
...
|-|erc
> Newton? Indeed, at the end he managed to impose his truths on us for
> some centuries ahead (in Leibniz and other cases - by pulling the
> strings of the Royal Society of which he was a president), but how did
> that make him 'happy' (a word that may very well be different from
> 'fame':)? Nervous breakdowns were following one after the other. In
> the meanwhile, he learned the bible by heart and parrot it by rote,
> alas, couldn't decipher the 'coded' Number. Something like that follows
> inexorably when we lose touch with the virtuous mean (balance) between
> No-ambition and Over-ambition. Even Aristotle had to learn it the hard
> way, although it was him who formulated this rule of middle ground
> happiness.
is one of my favorite things to write about how Newton spent most of his
time figuring out numerology of the bible, you have any sources for more
on this?
Herc
"Ann" <ao...@hotmail.com> wrote in
David C. Ullrich
>I remember a post where someone was wondering how I could keep
>claiming that I'm right with all these people posting disagreement,
>and it seems to me that maybe explaining how might help.
>
>You see, my short proof of Fermat's Last Theorem is flawless with all
>the disputes from mathematicians attacking the following statement:
>
>Given a factor g of a polynomial P(x), g=r+c, where c is a factor of
>the constant term P(0) of the polynomial, given by the value of g at
>x=0, and r=g-c.
>
>That's it.
Wrong on two counts. First, this is far from the only objection
people have. Second, nobody has objected to the statement
above - the objections have been to other versions of that
statement, which were either false or so vaguely worded as
to have an indeterminate truth value.
I kinda suspect you left something out, because the
statement
"Given a factor g of a polynomial P(x), g=r+c, where c is a factor of
the constant term P(0) of the polynomial, given by the value of g at
x=0, and r=g-c."
is utterly obvious (Pf: Let c = 1 and r = g - 1. QED.)
You meant to assert a little more about c and r, right?
************************
David C. Ullrich
Bhuvan
"James Harris" <jst...@msn.com> spewed forth in message
news:3c65f87.03080...@posting.google.com...
>
> So I can watch you. And I can see what you believe in, really. I can
> see how you think, and how it's possible to control even large,
> supposedly skeptical populations.
>
> Yup, you can learn a lot about people with a little math.
>
>
> James Harris
For a person who SO loathes the mathematical community, you sure spend a lot
of time and energy trying to get in!! You're such a loser!
Here's a hint of your pathetic nature: if you're listed in crank.com, you're
a loser!!!!
~Bhuvan
Will Twentyman
> I remember a post where someone was wondering how I could keep
> claiming that I'm right with all these people posting disagreement,
> and it seems to me that maybe explaining how might help.
>
> You see, my short proof of Fermat's Last Theorem is flawless with all
> the disputes from mathematicians attacking the following statement:
>
> Given a factor g of a polynomial P(x), g=r+c, where c is a factor of
> the constant term P(0) of the polynomial, given by the value of g at
> x=0, and r=g-c.
>
No one is attacking this statement. Just what comes *after* it.
> So yes, I have absolute certainty that my proof of FLT is correct
> because the linchpin is that statement above.
>
When no one argues with your "lynchpin", that's probably not where the
perceived problem is. Go back and look at the counter-arguments.
--
Will Twentyman
email: wtwentyman at copper dot net
Ann
it is like a riddle for me. First I will assume that your witty remark
was ironical, i.e. a genuine request for more 'information'. Hence,
"welcome to the real facts" (but in 'fact' again words:).
> On Tue, 5 Aug 2003, |-|erc wrote:
> > Ann wrote:
> > Do you know about the 'noble' character ("irrational temper") of
> > Newton? Indeed, at the end he managed to impose his truths on us for
> > some centuries ahead (in Leibniz and other cases - by pulling the
> > strings of the Royal Society of which he was a president), but how did
> > that make him 'happy' (a word that may very well be different from
> > 'fame':)? Nervous breakdowns were following one after the other. In
> > the meanwhile, he learned the bible by heart and parrot it by rote,
> > alas, couldn't decipher the 'coded' Number. Something like that follows
> > inexorably when we lose touch with the virtuous mean (balance) between
> > No-ambition and Over-ambition. Even Aristotle had to learn it the hard
> > way, although it was him who formulated this rule of middle ground
> > happiness.
>
> is one of my favorite things to write about how Newton spent most of his
> time figuring out numerology of the bible, you have any sources for more
> on this?
>
> Herc
Well, perhaps you would like to test me whether I could type 'Newton
biography' or 'Newton bible' in a search engine. There we are, I will
prove to you that I can, and shortly you will read the documented
evidence. Isn't it surprising that my 'intuition' (jolly smile here:) was
right; yet does it qualify for 'knowledge'?
First I'll spend 'a couple' of words on his 'happy' life and 'tolerant'
temper, for I believe that the many things we think we are in life and the
roles we play shouldn't be taken in isolation. Some positive psychologists
may try to convince us to divide oursouls into different lives
(professional, private, religious), supposedly that should make our life
easier. An illusion, these are intimately related and soon I'll deliver
the proof :)
Then I'll get to your point and address the (unpublished) project of his
lifetime - "the study of the ancient texts". Note that by some random
coincidence "he was fluent in Hebrew, Greek and Latin", a rare occurrence
in the universe of mathematics and physics :)
TRINITY: Dodge this!
Moreover, "Newton's writings on theological and biblical subjects alone
amount to about 1.3 million words, the equivalent of 20 of today's
standard length books." My comments will be either normal text
(unindented) or inside curly braces (or both).
Life & Character
Isaac Newton was born prematurely on Christmas
day 1642 (4 January 1643, New Style) in Woolsthorpe...
Much has been made of Newton's posthumous birth, his prolonged
:) separation from his mother, and his unrivaled hatred of his
stepfather. Until Hanna returned to Woolsthorpe in 1653 after the
death of her second husband, Newton was denied his mother's
attention, a possible clue to his complex character. Newton's
childhood was anything but happy, and throughout his life he
:) verged on emotional collapse, occasionally falling into violent
===> and vindictive attacks against friend and foe alike.
{What do we learn from here: "hatred", "complex character", "unhappy
childhood", "life verging on emotional collapse", "violent",
"vindictive".
Later we will encounter the words "pleasure; money; emotional breakdown;
furious; consumptive hatred; severe nervous disorder; deranged letters;
discomposure in head, or mind, or both; enjoyed power and worldly success;
he played it to his personal advantage; tyrannical and autocratic; his
control over the lives and careers of younger disciples was all but
absolute; marshaled all the forces at his command; secretly; dominate;
without rival; threatening to burn his mother and father; fame and
recognition; fear of criticism; his aim to humiliate Hooke in public was
abnormal; depression; mental illness; rage; irrational temper; the most
fearful, cautious and suspicious temper that his assistant has ever knew}
But the turning point in Newton's life came in June 1661 when he
left Woolsthorpe for Cambridge University. Here Newton entered a
new world, one he could eventually call his own. {the dream of
another "mad philosopher", an extrovert escape from himself in Nature
outside. Regretfully we cannot hide from oursoul anywhere in cosmos,
Aristotle tricked us here with the philosophy of science assumption of
independent and objective reality outside. While we don't know, the other
way round may also be worth exploring, i.e. starting instead from "know
thyself" and the dependent, subjective reality within. This was the noble
idea of Plato's "philosophic nature", first we cure ourselves and then the
'independent' universe.}
http://web.clas.ufl.edu/users/rhatch/pages/01-Courses/current-courses/08sr-newton.htm
How did Newton view his accomplishments in light of his belief in
God? Shortly before his death, he wrote, "I do not know what I may
appear to the world, but to myself I seem to have been only like a
boy, playing on the seashore and diverting myself in now and then
finding a smoother pebble or prettier seashell than ordinary,
while the great ocean of truth lay all undiscovered before me."
http://www.biblecodedigest.com/page.php/74
A joking remark:
{ "...one of the strongest motives that lead men to art and
science is escape from everyday life with its painful crudity and
hopeless dreariness, from the fetters of one's own ever-shifting
desires. A finely tempered nature longs to escape from the
personal life into the world of objective perception and thought."
Albert Einstein}
In 1678, Newton suffered a serious emotional breakdown, and in the
following year his mother died. Newton's response was to cut off
contact with others and engross himself in alchemical
research. These studies, once an embarrassment to Newton scholars,
were not misguided musings but rigorous investigations into the
hidden forces of nature.
Newton's later insights in celestial mechanics can be traced in
part to his alchemical interests {"nearly three decades of
alchemical research"}. By combining action-at-a-distance and
mathematics, Newton transformed the mechanical philosophy by
:) adding a mysterious but no less measurable quantity, gravitational
force.
{Surprise, surprise but later another master of words showed that in fact
there was no force, "it was not the spoon that bent but only ourself",
this "mysterious but no less measurable quantity" turned out an illusion
of our imperfect eyesight, "a byproduct of the bending space-time".}
...
Newton was so furious with Hooke that he threatened to suppress
Book III of the Principia altogether, finally denouncing science
as 'an impertinently litigious lady.' Newton calmed down and
finally consented to publication. But instead of acknowledging
:) Hooke's contribution Newton systematically deleted every possible
===> mention of Hooke's name. Newton's hatred for Hooke was
consumptive.
===> In 1693, however, Newton suffered a severe nervous disorder
{madness-fit}, not unlike his breakdown of 1677-1678. The cause is
open to interpretation... We only know Locke and Samuel Pepys
received strange and seemingly deranged letters {I'd love to read
them, wouldn't you} that prompted concern for Newton's
'discomposure in head, or mind, or both.'
Whatever the cause, shortly after his recovery Newton sought a new
position in London.
During his London years Newton enjoyed power and worldly
success...Newton was elected president of the Royal Society and
was annually reelected until his death... He was knighted in 1705.
{Dodge this:}
Although his creative years had passed, Newton continued to
exercise a profound influence on the development of science. In
:) ===> effect, the Royal Society was Newton's instrument, and he played
it to his personal advantage.
:) His tenure as president has been described as tyrannical and
autocratic, and his control over the lives and careers of younger
===> disciples was all but absolute.
Newton could not abide contradiction or controversy - his quarrels
with Hooke provide singular examples. But in later disputes, as
===> president of the Royal Society, Newton marshaled all the forces at
:) his command. For example, he published Flamsteed's astronomical
observations - the labor of a lifetime - without the author's
permission; and in his priority dispute with Leibniz concerning
!!! the calculus, Newton enlisted younger men to fight his war of
===> words,
{Let me repeat the last statement: "TO FIGHT HIS WAR OF WORDS". Is it so
impossible that all those scientific and religious battles between masters
of words (including recent wars) have this single purpose: control over
our credulous hearts and minds, fight for the gullible public opinion?
Media, TV and propaganda wars that pre-determine the final ('real':)
outcome. Notice, not necessarily a fight for the divine, independent,
objective 'truth' but for the private truth ('knowledge') of say, Newton -
"tyrant absolute".}
Newton enlisted younger men to fight his war of words, while
behind the lines he secretly directed charge and countercharge. In
the end, the actions of the Society were little more than
:) extensions of Newton's will, and until his death he dominated the
landscape of science without rival.
http://web.clas.ufl.edu/users/rhatch/pages/01-Courses/current-courses/08sr-newton.htm
When examining his sins at age nineteen, Isaac listed:-
===> "Threatening my father and mother Smith to burn them and the
house over them."
{A 'noble' (Nobel) character and if he were to threaten me that way I will
also concede. To avoid his "sound of inevitability" I may accept to see
anything he tells me, the force of gravity, the fluxions, the
infinitesimal points, anything the dictator likes.
Another piece of evidence comes from Isaac's list of sins referred
to above. He lists one of his sins as:-
===> ... setting my heart on money, learning, and pleasure more than Thee ...
which tells us that Isaac must have had a passion for
learning. {not about himself, though, but reality outside, a
prudent and safe strategy that eradicates nasty risk. Why vivisect
ourselves when there are so many independent animals outside -
test objects for the 'progress of mankind' and divine research}
He headed the text with a Latin statement meaning "Plato is my
friend, Aristotle is my friend, but my best friend is truth"
showing himself a free thinker from an early stage.
{so far so good, the question is the attitude, what we do with this
independent and "free thinking". Do we use it for "money, pleasure",
success, fame and personal profit, do we try to impose it by force on
everybody else? Maybe, if we are very ambitious and nervous.}
He was always pulled in two directions, there was something in his
nature which wanted fame and recognition yet another side of him
:) feared criticism and the easiest way to avoid being criticised was
to publish nothing.
Certainly one could say that his reaction to criticism was
irrational, and certainly his aim to humiliate Hooke in public
because of his opinions was abnormal. However, perhaps because of
Newton's already high reputation, his corpuscular theory reigned
until the wave theory was revived in the 19th century.
Newton was at the height of his standing - seen as a leader of the
university and one of the most eminent mathematicians in the
world.
After suffering a second nervous breakdown in 1693, Newton retired
from research. The reasons for this breakdown have been discussed
by his biographers and many theories have been proposed: chemical
poisoning as a result of his alchemy experiments; frustration with
his researches; the ending of a personal friendship with Fatio de
Duillier, a Swiss-born mathematician resident in London; and
:)==> problems resulting from his religious beliefs.
Newton himself blamed lack of sleep but this was almost certainly
:) :) a symptom of the illness rather than the cause of it. There seems
little reason to suppose that the illness was anything other than
===> depression, a mental illness he must have suffered from throughout
most of his life, perhaps made worse by some of the events we have
just listed.
{A "depressed" but "very rich man", his youth goal of "setting his
heart on money" was achieved with great 'success'}
Newton decided to leave Cambridge to take up a government position
in London becoming Warden of the Royal Mint in 1696 and Master in
1699. However, he did not resign his positions at Cambridge until
1701. As Master of the Mint, adding the income from his estates,
===> we see that Newton became a very rich man.
Given the rage that Newton had shown throughout his life when
criticised, it is not surprising that he flew into an irrational
===> temper directed against Leibniz. ...
We have given details of this controversy in Leibniz's biography
and refer the reader to that article for details. Perhaps all that
is worth relating here is how Newton used his position as
President of the Royal Society.
:) In this capacity he appointed an "impartial" committee to decide
whether he or Leibniz was the inventor of the calculus. He wrote
the official report of the committee (although of course it did
:) :) not appear under his name) which was published by the Royal
:) :) Society, and he then wrote a review (again anonymously) which
appeared in the Philosophical Transactions of the Royal
Society. {Cunning, wasn't he.}
Newton's assistant Whiston had seen his rage at first hand. He
wrote:-
Newton was of the most fearful, cautious and suspicious temper
that I ever knew.
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Newton.html
Newton a theologist?
"Sir Isaac Newton and the Bible."
By Professor Arthur B. Anderson.
In 1705 he was knighted. From this time onwards he devoted much
of his leisure to theology, and wrote at great length on
prophecies and predictions, subjects which had always been of
interest to him.
In addition to his scientific work (Newton would have said as a
part of his scientific work), he devoted a substantial portion of
his enormous energy to the study of the Bible and Biblical texts
and history. He read the Bible daily throughout his life and
wrote over a million words of notes regarding his study of it.
Isaac Newton believed that the Bible is literally true in every
respect. Throughout his life, he continually tested Biblical
truth against the physical truths of experimental and theoretical
science. He never observed a contradiction. In fact, he viewed
his own scientific work as a method by which to reinforce belief
in Biblical truth.
He was a formidable Biblical scholar, was fluent in the ancient
languages, and had extensive knowledge of ancient history. He
believed that each person should read the Bible, and through that
reading, establish for himself an understanding of the universal
truth it contains.
Only one book of Newton's about the Bible was ever published. In
1733, six years after his death, J. Darby and T. Browne,
published "Observations Upon the Prophecies of Daniel and the
Apocalypse of St. John."
With his prodigious knowledge of ancient history and languages
and his unequaled mental powers, Isaac Newton is the best
qualified individual in this millennium to have written about the
===> prophecies. His study of the Book of Daniel began at the age of
:) 12 and continued to be a special interest throughout his
life. Moreover, he writes of the prophecies with a modesty that
indicates that he, himself, is in awe of the words he has been
given an opportunity to read.
Isaac Newton concluded that it is intended that Revelation will
be understood by very few until near the end of history, the time
of judgment, and the beginning of the everlasting kingdom of the
Saints of the Most High. Here is an excerpt from that book
{Note that I am merely quoting Newton's wisdom, it doesn't mean I like any
kind of 'dark' and apocalyptic prophesies, which by the way, scientists
make all the time: the inevitable death one day of the Earth, the Sun, the
stars and a bunch of other 'hypotheses'. This is part of "the culture of
fear", to threaten our soul with a ludicrous thought scenario and prevent
us from thinking, i.e. accept all the other words that follow. Back to
Newton:}
"This prophecy is called the Revelation, with respect to the
Scripture of Truth, which Daniel was commanded to shut up and
seal, till the time of the end. Daniel sealed it until the
time of the end, and until that time comes, the Lamb is
opening the seals: and afterward the two Witnesses prophesy
out of it a long time in sackcloth, before they ascend up to
heaven in a cloud. All of which is as much as to say, that
the prophecies of Daniel and John should not be understood
till the time of the end: but that some should prophesy out
of it in an afflicted and mournful state for a long time, and
that but darkly, so as to convert but few. But in the very
end, the Prophecy should be so far interpreted so as to
convince many. Then saith Daniel, many shall run to and fro,
and knowledge shall be increased.
For the Gospel must first be preached in all nations before
the great tribulation, and end of the world. The palm-
bearing multitude, which came out of this great tribulation,
cannot be innumerable out of all nations unless they be made
so by the preaching of the Gospel before it comes. There must
be a stone cut of the mountain without hands, before it can
fall on the toes of the Image, and become a great mountain
and fill the earth. An Angel must fly through the midst of
heaven with the everlasting Gospel to preach to all nations,
before Babylon falls, and the Son of man reaps his
harvest. The two prophets must ascend up to heaven in a
cloud, before the kingdoms of this world become the kingdoms
of Christ.
'Tis therefore a part of this Prophecy, that it should not be
understood before the last age of the world; and therefore it
makes for the credit of the Prophecy, that it is not yet
understood. But if the last age, the age of opening these
things, be now approaching, as by the great success of late
Interpreters it seems to be, we have more encouragement that
ever to look into these things.
If the general preaching of the Gospel be approaching, it is
for us and our posterity that these words mainly belong: In
the time of the end the wise shall understand, but none of
the wicked shall understand. Blessed is he that readeth, and
they that hear the words of this Prophecy, and keep those
things that are written therein (Daniel XII 4,10, Apoc. i
3)."
{And note the necessary pathos at the end, scientific books have
similar inspiring conclusions that have greatest effect on a
child's soul.}
In conclusion: Sir Isaac Newton was totally correct in his
Observations. If the greatest scientist who ever lived had no
===> problem believing the Bible, what excuse will evolutionists,
atheists, agnostics, or other so called men of science have on
Judgment Day!!
http://www.reformation.org/newton.html
Isaac Newton was a Christian who studied the Bible daily and
believed that God created everything, including the Bible. He
believed that the Bible was true in every respect. Throughout his
life he continually tested biblical truth against the physical
truths of experimental and theoretical science and never observed
a contradiction, according to his many biographers. Newton's
writings reflected his belief that his scientific work was a
method by which to reinforce belief in biblical truth. After he
completed his monumental Philosophiae Naturalis Principia
Mathematica, he began to devote more and more of his time to
researching the Bible, eventually writing a book he believed
unlocked the prophecies contained in Daniel and Revelation, two
Bible books he viewed as intertwined.
http://www.biblecodedigest.com/page.php/74
Yeah, he might have published just one book but wrote many others (pages)
like what we are doing now. Actually, I have yet to figure out what is
this urge to publish. Is it for the sweet intellectual property rights? Is
it real? Look at me for example, a letter consisting mostly of others'
quotes, what a shame :) Is there such a thing as "intellectual"
creativity?
"The secret to creativity is knowing how to hide your sources."
Albert Einstein
By the way, do you know his "sources"?
Note that the one religious book that was published appeared six years
after his death, I guess he didn't want to damage his scientific
reputation. The next Islamic quote shows that he was deep into the details
of the subject matter :)
SIR ISAAC NEWTON ON THE BIBLE
In 1690 Sir Isaac Newton (1642-1727) wrote a manuscript on the
corruption of the text of the New Testament concerning I John 5:7
and Timothy 3:16. It was entitled, "A Historical Account of Two
Notable Corruptions of Scripture." Due to the prevailing
environment against criticism, he felt it unwise to profess his
beliefs openly and felt that printing it in England would be too
dangerous. Newton sent a copy of this manuscript to John Locke
requesting him to have it translated into French for publication
in France. Two years later, Newton was informed of an attempt to
publish a Latin translation of it anonymously. However, Newton did
not approve of its availability in Latin and persuaded Locke to
take steps to prevent this publication.
http://cyberistan.org/islamic/newton1.html
"Newton's writings on theological and
biblical subjects alone amount to about 1.3 million words, the equivalent
of 20 of today's standard length books."
Other Researches.
Throughout his career Newton conducted research in theology and
history with the same passion that he pursued alchemy and
science. Although some historians have neglected Newton's
nonscientific writings, there is little doubt of his devotion to
these subjects, as his manuscripts amply attest. Newton's writings
===> on theological and biblical subjects alone amount to about 1.3
million words, the equivalent of 20 of today's standard length
books. Although these writings say little about Newtonian science,
:) :) they tell us a good deal about Isaac Newton.
{"His passion was to unite knowledge and belief", how does it sound. He
did it, actually, many of his 'beliefs' were our 'knowledge' for a couple
of gullible centuries after his death.}
Newton's research outside of science--in theology, prophecy, and
history--was a quest for coherence and unity. His passion was to
===> unite knowledge and belief, to reconcile the Book of Nature with
the Book of Scripture. But for all the elegance of his thought and
the boldness of his quest, the riddle of Isaac Newton remained. In
the end, Newton is as much an enigma to us as he was, no doubt, to
himself.
http://web.clas.ufl.edu/users/rhatch/pages/01-Courses/current-courses/08sr-newton.htm
And here is another 'great' man "economist John Maynard Keynes (pronounced
"Canes")" who seems to have read those million words of his genius
predecessor. According to him Newton "regarded the entire universe
(including the Bible) as a "cryptogram".
Two hundred years later, when Keynes became provost of Cambridge,
he discovered the box and spent years going over the papers, which
he estimated contained more than a million words about Newton's
Bible research...
Here is what Keynes had to say about Newton ...
... (Newton) looked on the whole universe and all that is in it
as a riddle, as a secret which could be read by applying pure
thought to certain evidence, certain mystic clues which God had
laid about the world to allow a sort of philosopher's treasure
hunt to the esoteric brotherhood. He believed that these clues
were to be found partly in the evidence of the heavens and in
the constitution of elements (and that is what gives the false
suggestion of his being an experimental natural philosopher),
but also partly in certain papers and traditions handed down by
the brethren in an unbroken chain back to the original cryptic
revelation in Babylonia. He regarded the universe as a
cryptogram set by the Almighty--just as he himself wrapt the
discovery of the calculus in a cryptogram when he communicated
with Leibnitz. By pure thought, by concentration of mind, the
riddle, he believed, would be revealed to the initiate.
He did read the riddle of the heavens. And he believed that by
the same powers of his introspective imagination he would read
the riddle of the Godhead, the riddle of past and future events
divinely fore-ordained, the riddle of the elements and their
constitution from an original undifferentiated first matter, the
riddle of health and of immortality. All would be revealed to
him if only he could persevere to the end, uninterrupted, by
himself, no one coming into the room, reading, copying,
testing - all by himself, no interruption for God's sake, no
disclosure, no discordant breakings in or criticism, with fear
and shrinking as he assailed these half-ordained, half-forbidden
things, creeping back into the bosom of the Godhead as into his
mother's womb. "Voyaging through strange seas of thought alone,"
:) not as Charles Lamb, "a fellow who believed nothing unless it
:) :) was as clear as the three sides of a triangle."
http://www.biblecodedigest.com/page.php/74
And what did Keynes himself believe in. Well, he was a practical man who
wanted us "all dead in the long run". It is the highest 'good' that all
our greedy economies grow indefinitely (and simultaneously) until the end
of infinity (or the burst of the bubble:). Another funny man:
http://www.u-turn.net/2-4/keynes.html
Anyway, from the above paragraphs it seems Newton had an interesting dream,
"he regarded the universe as a cryptogram" (a matrix, maybe:). Let us
compare it with Einstein's "universe", his introvert "task was to free
himself from this prison, this optical delusion of our consciousness":
"A human being is a part of a whole, called by us _universe_, a part
limited in time and space. He experiences himself, his thoughts and
feelings as something separated from the rest... a kind of optical
delusion of his consciousness. This delusion is a kind of prison for
us, restricting us to our personal desires and to affection for a few
persons nearest to us. Our task must be to free ourselves from this
prison by widening our circle of compassion to embrace all living
creatures and the whole of nature in its beauty."
Albert Einstein
Best,
Ann
Jim Ferry
> Hi James,
>
> I like your style of self confidence and tolerance, it is the opposite
> of dogmatic, it gives freedom to yourself (and the people). Honestly, I
> see no reason why should you be disappointed. Imagine for a moment that
> your proof of Fermat's theorem is "flawless" from every possible point
> of view (really:), isn't that enough to make you happy? If I were you
> I'd be the happiest person in the world (and on the Moon:). Or you
> crave for some kind of social recognition, loads of money, or immortal
> fame? Or maybe you desire the sweet intellectual property rights?
> Luckily, your letters tell me otherwise.
Imagine? His proof *is* flawless from every possible point of view.
No need to imagine it. It is we who are flawed.
You are right not to impugn his motives, however.
> Should it make you anxious if not everyone accepts your truth (words)?
> Do we need to impose our dream on everyone? What for? How does it
> matter whether the final proof will be five or thousand pages long?
> (asuming you are not an environmentalist preserving the trees:) You
> probably know that what kind of proofs are deemed acceptable
> depends on the point in space-time, it is a convention which changes
> and grows in complexity to keep pace with the expansion of the universe
> (of words in our mind:). I think the point of your exercise was to gain
> confidence in your abilities and to "know thyself". And you did quite
> well here :)
When you say "accepts your truth", you imply a subjective truth. Although
some truths may be subjective, this one is not: James's proof, indeed all
of his thoughts that I've ever encountered, are sublime, profound, objective
Truths.
And they are beautiful. This is more subjective, perhaps, but in James's
writings we see what are, according to St. Thomas Aquinas, the three
essential characteristics of beauty: integritas, consonantia, et claritas
(wholeness, harmony, and radiance). But, to allude to another saint, we
only see them as through a glass darkly.
> One might say :) that I am speaking from a similar experience here. In
> the course of the discussions in another newsgroup
> (alt.dreams.castaneda, my home base:) I also had the chance to develop
> a theory of my own. A strange one I admit but it made me happy. And I
> did it spontaneously, in the last six months I took part in many
> discussions (some of them were real battles; "Seraph: You do not truly
> know someone until you fight them") and wrote thousand upon thousands
> of pages until I found "the key, my key" to the way out of the matrix
> (labyrinth of reason).
James is misunderstood here because he is so out of place, so far above
this rabble of mathematicians -- like a Nagual in New Jersey, despised
and feared but in tune with powers beyond suburban comprehension.
> That is, I became an expert in word-crushing - that mysterious effect
> of a word becoming less real in our mind and consciousness. Again the
> universal laws of attraction and repulsion (from any Word) hold true:
> adopting sequentially those two different points of view ('good' and
> 'bad') on the word make it invariably shake in our mind until we reach
> a fancy state of sober detachment and an unexpected feeling of freedom.
Yes. The words are the problem. James's thoughts have the precision of
a laser, but to communicate with the masses he has to translate them into
our illogical, nebulous language. It must pain him like the sin of the
world pains God.
> Another finding was that spontaneous improvisation is one certain way
> to beat reason, some of the 'treatises' (i.e. posts ranging from one to
> five thousand lines:) I wrote in a day or two. Yet apart from some
> annoying typos I can still read them with joy, if we are honest (with
> oursouls) there is nothing to be ashamed of later.
James, too, is prolific in trying to pin down the ineffable with the blunt
beanbags of words.
I'm not trying to illuminate anything so complex, so I'll state it simply:
God is here on sci.math. And his name is James Harris.
--
| Jim Ferry | Center for Simulation |
+------------------------------------+ of Advanced Rockets |
| http://www.uiuc.edu/ph/www/jferry/ +------------------------+
| jferry@[delete_this]uiuc.edu | University of Illinois |
Nora Baron
> Last Theorem. There's no excuse that I'm ignored because I'm sure by
> now many of you have noticed that not only am I not ignored, but there
> are people who at times try to make objections that at least sound
> mathematical. Here you see mention of "objects" which are numbers in
> a higher ring than algebraic integers, where the object ring is also
> complete, whereas the ring of algebraic integers is flawed.
>
> I didn't realize early on that mathematicians had such a flaw as the
> problem I found with algebraic integers, but when it was clear that
> the ring was incomplete, I went ahead and found the complete ring,
> which I call the object ring, as is my right as discoverer.
>
> Some of you may not understand or still may find it hard to believe
> that the ring is incomplete, so I'm including the short argument,
> which proves that it is. I'll likely be including this argument in
> several posts, as there really is no more room for discussion, so I'm
> ready to put a sock in it.
>
> It is mathematics, after all.
>
> I'll use that result that I talked about before where here I'll have a
> factor g of the polynomial P(m) where
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f).
>
> It's in a special form so that I can factor it as
>
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> where the question is, how does f divide out?
>
> Clearly in the first expression P(m) is a multiple of f^2.
>
> Well, a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), so at m=0,
>
> a_1 a_2 a_3 = 0
>
> which means that at least one of the a's must equal 0.
>
Yes, this is OK when m = 0.
> So picking a_1 as one that goes to 0, I pick for my g,
>
> g = a_1 x + uf,
>
> so at m=0, g = uf, so c=uf, and r=g-c, so in general
>
> r = a_1 x.
>
Oddly enough I do not disagree with this either.
> You may wonder why mathematicians would argue against that g=r+c.
>
> Well, here's why.
>
> Going back to P(m), I notice that
>
> P(0)/f^2 = 3x u^2 + u^3 f = u^2(3x + uf)
>
> so if 3 and x are coprime to f, I have that P(0)/f^2 is coprime to f.
>
> More simply, you can look at it and see there are no more free f's to
> divide off, as the only remaining f you can see is blocked by 3x.
>
> So as g is a factor of P(m), dividing off f^2 divides off something
> from g, so let's call it w. Then I have
>
> g/w = r/w + c/w = a_1 x/w + uf/w
>
> and I know that g/w is a factor of P(m)/f^2, and I can just let m=0,
> again, and I get g/w = f/w, and as the constant term
>
> P(0)/f^2 = u^2(3x + uf)
>
> is coprime to f, I know that w = f.
>
> I know some want that w to be a factor of m, but consider
>
> g/w(m) = a_1 x/w(m) + uf/w(m) = a_1 x/w(m) + uf/w(0)
>
Oh jesus christ. I see what you're thinking. You
are thinking that uf/w(0) = c/w(0) is the constant term.
Oh, man!
You have gotten your variables confused. The constant
c is the constant term *with respect to the variable x*.
Thinking of g as a function of x, c = g(0).
But w is not a function of x. It is a function of m.
Thus w(0) means w(m = 0), not w(x = 0). There is no
reason whatsoever for that term
uf/w(0)
in your expression above, when you are referring to m <> 0.
In fact g is actually a function of two variables, m
and x. Or you could think of g as a family of first-degree
polynomials in the variable x, indexed by the variable m:
g(x, m) = a1(m)* + u*f.
The constant term c = u*f for all values of x and m.
g(x, m)/w(m) = (a1(m)/w(m)) * x + u*f/w(m),
and g(0, m) = u*f/w(m).
This last equation is the important one. For m <> 0,
you CANNOT say it is
g(0, m)/w(m) = u*f/w(0)
as you did above. The thing on the right is u*f/w(m).
Am I getting through to you on this?
I mean, a symptom here is that you wrote was
g/w(m) = a_1 x/w(m) + uf/w(m) = a_1 x/w(m) + uf/w(0)
You did *not* substitute 0 in for m everyplace, only in
that last term. Nor should you have done so. You were
right to leave it as w(m) in the term a_1 x/w(m). You
mistake was in putting in w(0) into that last term. Again:
w = w(m) is a *function of m*, NOT a *function of x*.
Bluntly put, you are making an understandable but pretty
dumb mistake.
> where from before I know that w(0) = f,
True, but when m <> 0, there is no reason
to write u*f/w(0). The constant term is g(x = 0),
not g(m = 0).
> so I have
>
> a_1 x/w(m) + uf/w(m) = a_1 x/w(m) + u
>
No - it just doesn't work that way. See above.
> and subtracting a_1 x/w(m) from both sides gives
>
> uf/w(m) = u, so f = w(m), proving that w(m) equals f for all m.
>
No - you have confused the two variables, x and m, and
befuddled yourself regarding the constant term - in general
you have no reason to replace uf/w(m) by uf/w(0).
> It's that simple. Now you can also see why mathematicians would argue
> against g=r+c if they're fighting an FLT proof, as if they admit it's
> true, it's all over for them, as there's no more room to argue.
>
> So what's the big deal? The big deal is that with g=r+c, c is
> *constant* and because it's a constant, you have that
>
The part about g = r + c, your 'lemma', is not the issue at
all. That is a triviality. You have misapplied it here by
confusing the variable m with the polynomial variable x, and
getting mixed up regarding the constant term.
> g/f = a_1 x/f + u
>
> is the new factor for all m, and not just m=0.
>
No - it doesn't follow, as noted above.
> So you see, if mathematicians admit that a factor g of a polynomial
> P(m) is g=r+c, splitting the factor up into a constant and potentially
> varying portion, then they are admitting that I am correct.
>
You are right about g = r + c, but that part is trivial
anyway. The rest of what you have is incorrect as explained
above.
> The battle is being fought out partly here on Usenet, on newsgroups
> where presumably people pride themselves on their rationality, their
> knowledge, and their ability to determine the truth.
>
Actually I am encouraged that you have at least finally
recognized that this part of your argument is a problem. Before
you were just waving your hands and saying that since the
factorization was of a certain form when m = 0, it must be
of that same form when m <> 0. Your present argument is an
attempt to bridge the gap. Obviously it fails.
It is academic anyway. There are proofs that your main
conclusion is wrong. These are clear, rigorous proofs and
you have not refuted any of them at all. What this means is that
not only does your central argument have an error, as
described above, but also that IT CANNOT BE FIXED. Here
as a reminder is one of several proofs:
===============================================================================
You claim that if you have a polynomial of the form
P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3,
where v = -1 + m*f^2, and m, u, and f are integers,
with f prime and m coprime to f, then P(x)/f^2
can be factored in the form
P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1]
where b1, b2, and b3 are algebraic integers.
I say not. Let m = 1, f = 5, and u = 1. Then
v = 24, and v^3 + 1 = 13825 = 25*553. It is
easily verified that
P(x)/f^2 = 553*x^3 - 72*x + 5.
If this is factored in the form [1] as claimed by
Harris, then -u/b1 = -1/b1 is a root of Q(x) = P(x)/f^2.
That is,
Q(-1/b1) = 553*(-1/b1)^3 - 72*(-1/b1) + 5 = 0.
Multiply through by b1^3:
5*b1^3 + 72*b1^2 - 553 = 0.
The expression on the left is a *non-monic*
polynomial in b1 with integer coefficients,
and it is *irreducible* over the rationals.
Therefore b1 cannot be an algebraic integer.
Therefore your claim is false.
===============================================================================
> How you all behave says a lot about what really goes on inside of
> peoples' heads, all over the world. You could consider it a huge
> reality show, and a test unlike any other in human history that was
> impossible to give before the arrival of the Internet and Usenet.
>
> Worldwide networks made this test possible.
>
> I bid you all what may be your first real welcome to the brave, new
> world.
>
I could do without the gratuitous pontificating, especially in
view of the rather dumb mistake you have made in your post.
Nora B.
>
> James Harris
Lorrill Buyens
rock and hollered:
>Hi James,
>
>I like your style of self confidence and tolerance, it is the opposite
>of dogmatic, it gives freedom to yourself (and the people). Honestly, I
>see no reason why should you be disappointed. Imagine for a moment that
>your proof of Fermat's theorem is "flawless" from every possible point
>of view (really:), isn't that enough to make you happy? If I were you
>I'd be the happiest person in the world (and on the Moon:). Or you
>crave for some kind of social recognition, loads of money, or immortal
>fame? Or maybe you desire the sweet intellectual property rights?
>Luckily, your letters tell me otherwise.
1. Messages on Usenet are called *posts*, not letters.
2. Three-hundred-line posts are seldom, if ever, read completely.
3. It's polite to indicate who said what in a much more complete fashion
than a breezy letterlike "Hi James."
4. James Harris is, was, and prob'ly always will be a fruitcake. HTH.
(followups set: sci.skeptic)
--
"No collection of individuals is less vindictive than
an audience at amateur theatricals."
- P. G. Wodehouse, _The Intrusion of Jimmy_
Gib Bogle
Much fruit of sense beneath is rarely found."
Alexander Pope
Gib Bogle
<reasoned refutations>
"And the wise men don't know how it fee-ee-ee-ee-ee-ee-eels
To be thick
As a brick"
Ian Anderson (Jethro Tull)
|-|erc
the numerology is evident under ordinary analysis.
This is done (one way) by analysing notable people in our society rather
than characters of the bible though.
Like what coincidence is it that Ronald Raegun introduced the
star wars program, a man called Ray Gun?
And why is Hawking, a king!! our smartest? What coincidence is it?
Why is Tiger Woods the best golfer?
What did the famous person named Di do?
The most famous billionaires name is Bill?
I have a statistical test that does a bottom figure for the accumulation
of coincidences at www.adamskingdom.com
I may redo the test so it shows more prominent examples over a wider
time frame..... anyway...
Thanks for your time, see its who you know not what you know,
and when my proof finally is verified, they won't just be words anymore.
Herc
"Ann" <ao...@hotmail.com> wrote in > Hi Herc,
|-|erc
************************************** INTRODUCTION
Jennifer's mating call was in her office, she blurts
out 'I'm 21', Japanese phoneme for me. In a teaching tute
Michelle was bragging about always finishing work early and
chatting and says 'I can do 2 things at once'. A week
later I figure out I can have any woman I choose and I
send a blank email, 'I can do more than 2 things at once'.
Next day at uni, she's with the in crowd, I'm sure everyone
knows, but she sits by herself behind me very still. She
knows I don't know if she knows.
I notice how much shorter she is, but she has a better
V taper than me. She's into phrenology and evolution,
chemistry major, noticed her iq when she describes her
black box hypothesis systems for year 12 pracs. Phone
her and get a date, at uni we notice each other in the
corner of our eyes, walk past and simultaneously say "Hi",
Japanese for yes. In a micro teaching session we
role play molecules and we end up holding a piece of
string bond between our fingers, she notices my index
finger has the only trimmed fingernail and smiles, my
first reassurance of destiny. Weeks of calls and no
date, she has a 6 year boyfriend (she's only 22), lucky guy
she's so sultry. Invites me to the final day drinks, chat
with a few under my belt but she leaves, never saw her
again, she wore a flower in her hair, thought it was for me.
Last month I moaned out aloud couple times at Jennifer just
as thunder roars, hardly think of Michelle, yesterday after
half hour silence I whisper 'Michelle' and a gust of wind shakes
the windows.
INTRODUCTION
Hello Michelle,
the news media doesn't determine if you are a prodogy, it is a
part of nature, you have to understand somewhat how fate works.
If it's my fate to marry someone, then you can deduce that no
matter what, in the mean time I cannot die. Does this make me
immortal, if I jump in front of a train will I survive? I won't but it
does mean that it is impossible for me to make the decision to
jump, for that is the event that would breach my fate. I can't tell
you for certain if you are a prodogy, but if it is your fate to see
me again then denying this will only further confound you, fate
cannot be escaped. Hating me for a time was just part of your
fate. The news media reflects whatever is in my
mind or my recent events, my life runs a parallel to the events
of the world, not just media, weather, people, machines, everything.
Can you interpret media and the environment? This may be only my
skill, but if I am right you share a focus of the world's events and there
will be some train of stimulus effecting you to come to me.
We are mortal like everyone we know, but now we cannot die, and
we the closest thing to being gods.
Below is my introduction to a book, I'm planning to compile my writings
together, and ask my father to do a biography, not for a few years, I
want my destiny to be realized first. We are similar, both broad -
great compliment, we are the two most intelligent of our species,
can you deny our possibility, the future of our race? I want to love you.
Most of all I want to hold you.
This is my true story, a story about the reality of prophecy, the truth
of fate. My fate or destiny, merely months away now, is a woman.
These writings are more about the underpinnings of the universe, to make
credible a conventionally unscientific belief. As an introduction to the
mysterious workings of purpose in a random universe, I will attempt
a story, a story of how this woman and I met.
First to introduce myself, a computer programmer and recreational
mathematician. This book is unlike any other, first I am not a writer
in the conventional sense, I fill notebook upon notebook with hazy
diagrams and brief descriptions I later cannot decipher myself, second,
as a mathematician all my writing serves a single purpose, to prove.
Take my words with an open mind and your understanding of the
world will change, for I will show you that numerology is a science.
Jennifer, my destiny, caught my attention with her name over two
years ago when I met her in a realty shop, and read her business
card, the name rhyming with a movie character I adored 'Jessica Six',
the movie - Logun's Run. What has a movie to do with prophecy,
indeed the prophecy of some couple? We are no ordinary couple but
I will leave this detail for the body of this book. Every couple
that comes together does so as each of their fate, but the driving
force bringing Jennifer and I together is of a new nature. You see
Jennifer is very beautiful, I can barely type to write - she has
bright green eyes, and I am very intelligent.
Only a few who have been close to me have insight to my intelligence,
I shouldn't boast, there are other people with much superior mental
skills to mine in areas, detail of the nature of my mind will again
be detailed in course. The movie Logun's Run, a fantastic movie,
refreshed my drive for monogamy, true love. There are many influences
acting upon us each day that can actually affect the type of decisions
we make, like me deciding to visit the realty store again. Movies,
songs, advertising, friends and strangers, everything in the world
can play influences, and anything in the world can play a part in
prophecy. The movie Logun's Run, the characters and actors names,
looks, birthdates, script, all run to a celestial clockwork, it
was my destiny to see the movie shortly before meeting Jennifer.
It is difficult to explain several things together at once, I'm not
just intelligent in that I get everything right, and Jennifer is not just
beautiful in that she is flawless, like the shark that hasn't changed in
millions of years, we are the first perfect people.
It is difficult to understand fate, like the entire world is a directed
play, events come together with twists as if they were written before
hand. At certain times many things fit into place, and you must
wonder how the contributors made these things fit without some higher
purpose controlling their actions. I mentioned a movie character's
name that rhymed with Jennifer's, there is an actress I saw years
before meeting Jennifer who looks like her, the first time I saw the
actress on television I said 'I'm going to marry her'.
Many people can accept fate as part of our world, but don't grasp the
complexity involved. My life has had hundreds of occurrences which
all had to happen for my prophecy to be meaningful, seeing certain
movies, postcards, books, hearing certain songs, thinking certain
thoughts at the right time, doing particular things, all complex and all
laid out for me before I was born. The end result of one act of destiny
relies on the myriad of happenings throughout ones life, and all the
events that one interacts with. This imposes a rigid view of the world,
where every action is predetermined, perhaps there is some margin
of freedom in our actions that don't affect our destinies.
I have already lost most scientifically minded readers with my ideas
leading to divine purpose of furthering of our species, the accepted
view of random chemical, biological and social interaction responsible
for life is complete and sound. I have some thoughts on the overall
methodology of science to address the need for a more open view
to the nature of the universe. There are many unresolved issues in
current science, say you imagine a line from near your person and
extend it out towards the sky into space. Where does the line end
up? It goes past many stars and galaxies, then supposedly into
emptiness. Does the universe finish at a certain region, enclosed
within nothing, or extend infinitely? Most people imagine the big
bang as an event that manufactured matter into an empty universe.
Thirteen billion years ago there was nothing, twelve billion years ago
everything was created. This isn't true, there was no thirteen billion
years ago, time itself, together with what we know as the universe
started with the big bang. We don't readily comprehend that before
the big bang is meaningless because we live about a third of the way
into the universe's life, time may continue forever but supposedly
everything will just be a random soup, life won't exist anyway. The
way everything in the universe behaves is different both at different
times and at different scales, say for a star compared to an atom.
We don't know if the laws of physics will hold throughout time, they
may certainly be extended in future.
The progress of science is analogous to deciphering instructions, this
implies to me a period of construction, and even one day when we turn
it on. For instance, extra sensory perception has never been proven,
yet a moment may pass in the future that not only alters current scientific
views but alters our nature of existence, making esp conceivable. There
could be growth points as part of the universes life affecting the nature
of how things work.
Most people believe in some non scientific view, a standard
religion or their own beliefs, yet any examination or test of these views,
or indeed their audience by a skeptic refutes them. This makes me
think of a law for small particles that observation of exactly what the
particle is doing is impossible, called the uncertainty principle. The
act of observation interrupts the experiment. Could not this and
other laws of physics that work at micro scales manifest themselves into
our reality? The models of matter at small scales do not resemble
our tangible world, bizarre in that they accurately predict what
happens, but described as not what is really there. Could our visible
world also work in ways we can see but not comprehend?
In quantum physics many different actions of a particle
occur together at the same time, something we are unaccustomed
to in our observable world. Perhaps if these occurrences manifested
at observable scale, one of my parallel thoughts could be the location
of a person, if it were our destiny to meet then this thought would
become real. The bubble of non observation around an event could
explode to include human comprehension, it may be in our lifetimes
that testing of things like esp finds it valid. The universe may be
winding up to one set of laws, and after a pivotal moment, unwind
to another. I am not saying these things are true, merely that stringent
viewpoints of science are not always correct, and certainly not complete.
Scientists make principles from observations of their environment, it is
not mathematically sound for them to discount theories just because they
don't fit current theory. The line between fantasy and reality is blurred.
Reaching a complete understanding of the universe involves theories
of information, not matter. Our knowledge of events, plus facts that
transcend time, like a triangle always has three sides, the currently
unexplainable fact of our sentience and feeling are all part of the
universe. The fact that information now travels around our planet at
the speed of light is a part of the universe. Information takes many
forms, from a simple message in speech, to every activated neuron
on our retina, to a symbolic natural formation like a river running
into a sea, to the representation of a physical object.
The foundation of information is the mere number, it is
hard to define information but it works as a kind of parallel.
A complete theory of the universe would have to involve the parallels
or duality that occur in many ways: the non natural way small matter
behaves manifesting into reality, the environments solitary existence
interacting with its manifestation in our minds, the parallels of
broadcast fiction leading reality and describing history, the creation
of life from two lives coming together. The universe is inhabited with
social creatures, does not this make the universe a social entity?
Our journey into numerology focuses on sounds, is not a number
equally represented with a sound as with a scribble. Colours, substances,
music are all numerical in a sense, this sentence could replace each
letter with a number, and be ordered just like any counting number.
Prophecy and numerology go well together because they are about
order. First the classic numerology prose, allow me to introduce myself,
I am number seven. In fact, everything about me is a seven, birth day,
initial, adding up consonants in my name, adding up vowels. One day I
gave a pendant with 7 diamonds to my girlfriend who introduced me to
numerology, we sat at table 7 and later saw 7 birds fly by. Yes its
hardly an argument for numerology, but that's an example of how it can
work, I can only show you, not prove. Somehow my mind is the focus of
all events, my life has become an ongoing sequence of symbolic exchanges
of messages with our world. Another example is today I thought of something
to say to Jennifer about her eyes and a car drives by of remarkable metallic
green colour. There is a synchronized duality. If I ponder my gift and
roll two dice a seven appears, but I can't repeatedly roll sevens in front
of an audience, it's not my fate to prove my sevenness. The astonishing
flukes of my life only occur for purpose, I can flick a piece of paper
into place with precision, but this is a coordination outside of my body.
Ongoing examples don't prove anything, I will mention one last 'coincidence'
as part of this introduction, it occurred on the seventh of the seventh,
two thousand. I found an old CD that night, 'Reactivate 10' which made
me think of turning on my pursuit for Jennifer again, I should mention
Jennifer's number is 10. I repeatedly played a song, 'some people believe
in the 5th dimension, others think the sixth will show the way to the
seventh... ecstasy', then I stopped when I realized it finished on the
initial of my surname - X T C. I noticed the time, three thirty, being
half of seven, then I slept.
Sound plays an important role because prophecy is more powerful than
we can imagine, its not just some predicted story, it borders on the belief
that all of history is predetermined, and unraveling the past and future relies
on interpretation of our language, for in a sense language is not made by man, just
discovered. Having deciphered and written many computer programs, I am
accustomed to making up functional words, this has given me the ability
to decipher a fraction of spoken languages as being purposeful to the
nature of the universe. I will not take the storm out of the body of my book
now, it will better explain the keys and secrets that await us.
Phonetics is the primitive of language, and language is the tool
of meaning. In a purposeful universe words, names and symbols
are all linked in a complex network. Pythagarus didn't invent the
circle, most people can comprehend that it transcends time, but
the culmination of an entire language, the associated derivations
from people, places, animals and biology, chemistry, all jargon
and all symbolic objects, is difficult to believe that it is more than
a historically developed facet. Movie makers carefully make up
names to add a dimension to each character, the script is calculated,
very carefully at certain points to ensure flow in each scene and
everything ties together at the finish. Could not the universe itself
play out life, the story of the most advanced man, each scene
culminating into events to lead him to the most advanced woman?
The climax would be the progress of life, something we all know to
exist but not accommodated in any laws of physics.
The universe isn't a big space with moving particles, that
doesn't count for the existence of a concept, a desire.
it is intricate, a structure of meaning and purpose that
drives our physical world.
Such a pity, the wooden axle never able to bear its required load, the
tribe returned to the old technology, the proven rolling logs method,
the spark of ingenuity destined to remain one man's dream. We live
in a strange universe, we think of moving ourselves and other objects
and shortly after things happen, its a comfortable understanding, but
how does it work? Things didn't always happen, and things don't exist
when they are made. If the tribe failed in its invention, another tribe would
take its place, in a sense the wheel existed before it was made, and
will exist after they are all destroyed. We have one understanding of
ourselves with our subcomposition, but what we find aren't things that
move, physics describes matter like a magic show, vanishing, blurring,
jumping, doubling. Time is the mystery, on one hand a universe with
a beginning and ending of time, just a still object we flow through, and
another still world that we can't even touch, void of all events yet
visible, described as the platonic world, knowledge.
But things do have a temporal existence, our reality the physical world,
and there is only one universe, how it all blends together, this is my story.
Will Twentyman
> Will Twentyman <wtwen...@read.my.sig> wrote in message news:<3f2e749f_1@newsfeed>...
>
>>James Harris wrote:
>>
>>>I remember a post where someone was wondering how I could keep
>>>claiming that I'm right with all these people posting disagreement,
>>>and it seems to me that maybe explaining how might help.
>>>
>>>You see, my short proof of Fermat's Last Theorem is flawless with all
>>>the disputes from mathematicians attacking the following statement:
>>>
>>>Given a factor g of a polynomial P(x), g=r+c, where c is a factor of
>>>the constant term P(0) of the polynomial, given by the value of g at
>>>x=0, and r=g-c.
>>>
>>
>>No one is attacking this statement. Just what comes *after* it.
>
>
> of a polynomial P(x), g=r+c, where c is a factor of the constant term
And yet I keep failing to follow it, and when I explained why you keep
failing to explain it, even though I've repeatedly pointed out the post
where I stated where I don't follow the proof.
Randy Poe
> Hi, tis a genuine request, I actually have my own study that
> the numerology is evident under ordinary analysis.
>
> This is done (one way) by analysing notable people in our society rather
> than characters of the bible though.
>
> Like what coincidence is it that Ronald Raegun introduced the
> star wars program, a man called Ray Gun?
>
> And why is Hawking, a king!!
He's not a king.
> our smartest? What coincidence is it?
Zero. Why only apply the word "king" to the most well-known
person in one particular subfield of physics? What about
the rest of the sciences and other fields of human endeavor?
For instance...
>
> Why is Tiger Woods the best golfer?
>
You mean, why is Tiger Woods the "king" of golf? I don't
know, why? And how about before Tiger? And after? Was
Jack Nicklaus a failure because of the absence of "Wood"
in his name? Wouldn't "Wood" also suggest he should be
a carpenter or a woodcarver? What about other people
named "Woods"? Are they all good golfers?
> What did the famous person named Di do?
So being called "Di" implies that you'll die, and not
being called "Di" implies that you won't? Cool! I'm
golden!
> The most famous billionaires name is Bill?
Therefore you expect Bill Clinton and actor Bill
Pullman to become billionaires? What do you make of
Adnan Khoshogi?
Your numerology does serve a useful purpose in illustrating
in a very clear way the fundamental logical fallacy in
numerology (as well as most "psychic" phenomena, such as
John Edwards talking to the dead): In a large sample pool,
you can find a lot of random hits. If you ignore the misses
and only focus on the hits, your results show an "amazing"
number of successes.
Psychics such as Edwards succeed because there is a very
strong psychological effect of doing precisely that: remembering
the hits and forgetting the misses. I've read a number of
accounts from people actually at a John Edwards show as
opposed to watching the edited version on TV. If you
objectively study the unedited tapes or keep score
when you're sitting in the audience, you'll notice
things like "Is there significance with the letter
J? R? N? S? T? V? W? B? K?". In one amusing case he actually
went through 20 LETTERS before getting the hit, yet the
audience still was amazed that he got the letter EXACTLY
RIGHT! The subjective and objective results are completely
at odds.
I'm not going to tell you not to mess with numerology if it
provides you amusement. I will tell you that you really
should talk with a psychiatrist. They really can help
stop the torture you keep talking about. Wouldn't you like
to stop hearing those things or to turn the volume down?
Please consider getting help. None of us on the internet,
hundreds or thousands of miles away, can help you. But
there are people in your town who can.
- Randy
Maxim Stepin
news:<bgnltf$4mu$3...@lust.ihug.co.nz>...
Very good.
> What I've seen are repeated attempts at distractions.
That's what you think you see.
But in reality your definition of "object ring" is still circular, thus your whole FLT "proof" is
still invalid.
So you cannot say "nobody found an error in my FLT proof".
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>Gib Bogle <bo...@too.much.spam.ihug.co.nz> wrote in message news:<bgnltf$4mu$3...@lust.ihug.co.nz>...
Prove it. Use your definition and PROVE IT.
You have not provided a coherent definition of object. Many times I
have pointed out the problems with yoru definition, and you have NEVER
replied.
The definition that appears in your website is circular:
" Objects are members of commutative rings where any unit and its
multiplicative inverse are units in all possible commutative rings in
which either and all integers are members, where no member is a factor
of an object for which it is not a factor in all possible commutative
rings that include all integers in which it and that object are
members. "
If we drop the circularity by defining "object ring" as:
"An object ring is a commtuative ring R such that: (i) if u is a
unit in R, and v is its multiplicative inverse, then any
commutative ring containing the integers and either u or v must
satisfy that u (if u is there) is a unit; or v (if v is there) is
a unit. (ii) If a and b are elements of R such that a is a
factor of b in R, then in any commutative ring which contains the
integers, a, and b, also satisfies that a is a factor of b."
"An object is an element of an object ring."
This definition is not circular, but under this definition NOTHING is
an object and nothing is an object ring, because you have placed no
restrictions on the ring structures you are considering. A more
reasonable interpretation of what you mean, which I have offered many
times (and you have never replied) is:
"An object ring is a subring R of the complex numbers which
satisfies the following two properties:
(i) If u in R is a unit in R, then u is a unit in any
subring of C containing the integers and u; equivalently, u
is a unit in Z[u].
(ii) If a and b are in R, and a is a factor of b in R, then a is a
factor of b in any subring of C which contains the integers, a, and
b; equivalently, a is a factor of b in Z[a,b].
An object is a complex number which is an element of an object
ring."
Under this definition, the ONLY objects are the integers.
To prove that sqrt(2) is NOT an object under this proposed (and never
challenged) definition, simply note that in any ring containing
sqrt(2) and 2, 2 divides 2*sqrt(2); however, 2 does not divide
2*sqrt(2) in Z[2,2*sqrt(2)] = Z[2*sqrt(2)]. Therefore, sqrt(2) cannot
be an element of any object ring, and so sqrt(2) is NOT an object.
[.snip.]
======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A great
many people are staggered to this extend, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Bob Casanova
in sci.skeptic, posted by Gib Bogle
<bo...@too.much.spam.ihug.co.nz>:
>"Words are like leaves, and where they most abound
> Much fruit of sense beneath is rarely found."
>
>Alexander Pope
Well, that pretty much finishes the Britannica, not to
mention the OED. Marvel Comics, anyone?
--
Bob C.
Reply to Bob-Casanova @ worldnet.att.net
(without the spaces, of course)
"The most exciting phrase to hear in science,
the one that heralds new discoveries, is not
'Eureka!' but 'That's funny...'"
- Isaac Asimov
|-|erc
no silly, I'm not using a large sample at all, I'm getting hits with 50%
of the *small* sample of famous people.
I'm not getting 1000 people named bill and finding one that fits,
I'm going through all the most noted people, like we are a living
bible, and getting a very good correlation.
Of course the mapping doesn't work both ways, that's what makes
it stand out.
You were right with this comment :
What about
> the rest of the sciences and other fields of human endeavor?
because it works on the famous and notable, and I have gone through
lists of notable scientists and it works rather well for them, to the extent
you can give me a list of notable people, I can put the list of acheivements
on one side and list of their names on the other, give them to a primary school
class who don't know the people and they can match them up!
You see I'm not at fault here when I provide correctly sampled data and
the mechanism to analyse it. Here are 50 replies to me all verifiable in
google and if you score over 25 just guessing multiple choice if proves a
correlation greater than 9999/10000. Then its just a matter of using controls
against other people, get all the replies to you over the same period and
the statistic won't be 30/50 (99.999% confidence) it will be 12/50 (normal data).
Its really quite simple when you wake up from your dream that the government
doesn't want complete control of media and that 100,000 people knowing I'm
the truman in Townsville Queensland still not going public after 1 year is due to media
cover up as much as your apathy.
See as much as alt.atheism will argue like children these 4 out of 4 replies from
the people living in townsville say I am the truman are sarchastic, you should be a
bit smarter, several people have emailed me saying they checked out the links.
aus.tv know it
I'm from Townsville and YOU ARE the Truman!
http://tinyurl.com/iky5
I was in Townsville over the weekend, and I heard him.
Very spooky!
http://tinyurl.com/iky8
>phone someone in Townsville, half of you must know someone there,
>every day I go out people say "THERES THE TRUMAN"
I'm in Townsville. We're sick of you.
http://tinyurl.com/iky9
http://tinyurl.com/iky4
You rule Truman!
proof of numerology and the beginning of Adam seeing Eve after 5 years
due to your end of apathy is REALLY VERY SIMPLE.
--------------------------------------------------------------------------------
19 2003-02-02
Randi will test you when you properly apply to be tested. Sign up here:
http://www.randi.org/research/challenge.html
-----------------
Rich Shewmaker
CNote
Wanda
Rust
You see James Randi is rather famous because he gets people onto a stage
and sets them up to do magic shows, literally. He even offers to pay a
million dollars if they succeed in disproving classical science. He is
a [rich show-maker]. Now 2 people from sci.math after days of coersion
have already guessed
who the author of this post is. Is it really a one in 4 coincidence? Could
you guess it from 1000 names? And when I beat 1000 to one odds every
10 posts to me, then I'm not selecting them from a large sample am I. This list
is every second post to me, all contained within a 2 month period starting
1/1/2003.
--------------------------------------------------------------------------------
20 2003-02-02
It really all depends on the situation.
-----------------
Shanx
See You In Hell My Friend.
Someone
Greg Neill
--------------------------------------------------------------------------------
GO ON HAVE A GUESS YOURE ONLY SETTING GOD FREE
-----------------------------SAMPLE QUESTION-------------------------------------
2003-01-24
>crazy moon lovers!
>Herc
///////////
all you moon-huggers!
-----------------
Greg Neill
scribe2b
Roundtable
John L
--------------------------------------------------------------------------------
Examine what is written in each post, because these are all chronological
replies to me (covering the first 2 months of 2003), there is a paranormal
twist in each message. As I have attempted to explain, within my environment
reality is described for me.
The 4 names follow the post, 1 of them is the real author, in this example,
since the post contains a number of scribes ////// the answer is scribe2b.
To make the test much smaller, where possible only the authors comments
from the post are shown.
--------------------------------------------------------------------------------
1 2003-01-03
"|-|erc" wrote:
>"I CAN GUESS NAMES"
>
>this is another version :
>
>"I can guess peoples names by analysing what they reply to me
>in newsgroups providing their reply is not contrived"
>
>Doesn't matter what I say, 20 emails to ozskeptics saying
>half a dozen times "I can guess names", 5 emails to Randi
>saying "I can guess names", 10 emails to ukskeptics "I can
>guess names", 20 posts to sci.skeptics "I can guess names".
>
>"so can I", "not a testable claim", "doesn't say how", "doesn't
>say why", "doesn't say where".
>
>
>I give a possible test scenerio to all of you, you all say its
>up to us to make the test, look at the claim, according to
>that you can pass this, oh no, now you've changed your
>claim disqualified. Its a claim which I CAN prove, its
>not the proof mechanism, its a CLAIM. I can do it *somehow*
>not every which way but loose.
>
>I come here and say I can do paranormal, look a match question
>baker makes and you all refute me 50 different ways, I
>answer every one. after the fact you say, not 1 in 6. ok so
>I set up a test and guess what, get 1 in 6 again and this time
>its not an assertion after the fact, 1 in 36, oh now the judge
>is paranormal, what next?
>
>Your so entrenched with scientific method you can't say, what
>if we allow the skeptics clause even if we don't understand it.
>Science is made going against the grain.
>
>Herc
So if I get one of my kids to post under my name- because my name is
obvious (and only one kid, always the same one), you will be able to
guess *his* or *her* name, right? not the surname, but the first and
middle names?
-----------------
patty-anne-lea
Wally Anglesea
Matt Giwer
J.y.n.x
--------------------------------------------------------------------------------
2 2003-01-03
The bumblebee has, like a helicopter, insufficient wing area to fly.
My parrot can fly, but then he has wheels.
The wrens visiting my bird feedeing station don't have wheels
so I'm only fantasizing that they can fly, I suppose.
Just thinking out loud ...
Wood stumps warm me thrice,once while splitting them, once while they are on the
fire, and again when I read their posts to usenet. (Henry David Thoreau).
-----------------
NormDePloom
Edward Caruthers
malcolm burton
Lawrence & Bobbie
--------------------------------------------------------------------------------
3 2003-01-05
The problem is that everyone has the answers in advance.
That's not a test, that's a "demonstration." In a test,
the testee would be given names and articles and *not*
have access to the answers until *after* the test.
Nothing you described so far possesses this property.
-----------------
Chas
Xcott Craver
Roundtable
Odysseus
--------------------------------------------------------------------------------
4 2003-01-05
Gee Zedenk, that's real interesting. A couple of questions though. Where
did the "continuum" come from? Who or what lit the fuse that caused the
spark? What set in motion the totality that we call the vacuum of space and
the particles of matter that inhabit it?
LOL! At the end of it all, physics and science are no more than Bro'
Rabbit, and the mystery of God is Tar Baby :) No matter how much a man
rambles and babbles, he cannot escape the reality of a "First Cause".
Think on this Zedenk. Take the simplest of things, the Hydrogen atom. One
proton, and one electron orbiting it. Now apply the trinity theory to it.
The physical existence of the proton and the electron are the "Son". The
perpetual motion of the electron flying in it's orbit is the "Holy Spirit"
or will of God. The force that holds it forever in place and does not let
it collapse into the center or fly off into space is the "Father" or mind or
Law of God.
Now set this in your mind Zedenk. If even for a fraction of a second, any
of these 3 realities ceased to work, the universe would explode! Your
universe Zedenk, has nothing in it's existence to do with physical
properties or time or this dimension, it exists and continues to exist
because something unexplainable has not withdrawn the "Divine Power" that
glues it all together! Chaos is not a reasonable explanation for
never-ending "Law"!
C YA,
-----------------
Greg Neill
Wanda
sertec
Mitch Dickson
--------------------------------------------------------------------------------
5 2003-01-18
> I read that if you put something large and highly visible in your
> yard, say a bright tarp, you can release your birds and
> they can find their way home. Is this worth experimenting on?
I would have to say a big NO. A bird is gonna get confused and scared and
panic and just go! Not to mention preditors and such. Now, if you're
talking homeing pigeon, then maybe, but again, predators must be taken into
account.
-----------------
Someone
Mercury481
Lawrence & Bobbie
Tim Kozusko
--------------------------------------------------------------------------------
6 2003-01-20
Go for it.
-----------------
Lawrence & Bobbie
Ralph Hertle
J.y.n.x
The Pervert
--------------------------------------------------------------------------------
7 2003-01-23
:^) if any of you had a half open
Not me, mine is all the way open, and yours ?
You must not have seen my post advising that
typos should be ignored - not trolled.
Open your mind - if you can -
:^) Absolutely not. Intolerance is !
-----------------
Hold, I'll think of it in just a nanosecond or two
Ralph Hertle
John L
TheKid
--------------------------------------------------------------------------------
8 2003-01-24
|-|erc wrote:
>He said email him so maybe wont get your reply. The theory seemed
>alright, and even though SETI purposes you are right that doesn't
>discount the figure of only 100 planets in Milky Way that we could
>move to. How does a gas boil? boiler? Do gas particles really
>reach escape velocity? Anyway planets are becoming increasingly
>common even nearby so I doubt 100 billion stars would be near
>empty at all. Me, still working on my age halting serum, few centuries
>to enjoy the lower G on Mars then maybe retire for a few millenia
>enjoying the view from one of Saturns larger moons.
>
>Herc
>
The theory seemed alright but only assuming the assumptions he (we) makes. The
assumptions of the scientist at the SETI institute are most likely more
plausible that those by an unknown person in as obscure newsgroup who
frequently spams multiple groups with off the wall crackpot ideas. But hey,
what do I know.
-----------------
First Name
Hold, I'll think of it in just a nanosecond or two
Mercury481
Greg Evans
--------------------------------------------------------------------------------
9 2003-01-24
"|-|erc" wrote in message
>
> "tadchem" > wrote
> > Sound is a mechanical compression transmitted through a material.
> >
> > The speed with which a mechanical compression is transmitted through a
> > material is called the speed of sound.
> >
> > For copper this is about 3560 m/sec at 20° C.
> >
>
> but if you push a 3560 m rod of copper a foot, the other end
> will push out one foot in much less than a second. Sound is the
> transmission of vibration, not the compression of the structure.
>
> If you hit the rod with a sledge hammer, at the other end it would
> move and one second later you would hear the noise. Actually
> one second minus a very small amount for the structural compression.
>
> Herc
Thank you for better describing my thought
-----------------
Chris
raven1
sertec
Chas
--------------------------------------------------------------------------------
10 2003-01-24
///////////
all you moon-huggers!
-----------------
Greg Neill
scribe2b
Roundtable
John L
--------------------------------------------------------------------------------
11 2003-01-24
I've spent basically my whole life at 28 degrees lat. and have gotten quite
good at telling the time of day by the sun. In the past couple years I've
been to England three times and was baffled by the sun there - even knowing
to expect it. It was wild there in February actually being able to feel the
"day" getting longer. That was different.
Also different was the snow we got today. The Space Center had enough that
it piled up on cars.
-----------------
CNote
Mitch Dickson
Tim Kozusko
J.y.n.x
--------------------------------------------------------------------------------
12 2003-01-27
"The evidence based on metallurgical analysis of fractured surfaces
(produced by Geller) indicates that a paranormal
influence must have been operative in the
formation of the fractures."
Dr Wilbur Franklin (Physics Department,
Kent State University - U.S.A.)
"We have observed certain phenomena with
...
-----------------
G=EMC^2 Glazier
CNote
malcolm burton
Greg Neill
--------------------------------------------------------------------------------
13 2003-01-27
"|-|erc" wrote
> hey I'm the one solving dillemmas round here, contribute
> something useful yourself. its been well discussed b4 that
> odd binaries are allowed here.
No.
> formal rules such as attachments
> only in binaries named groups are easily programmed into
> news servers yet they dont actually exist do they?
>
> now which part of too bad didn't you get?
> Herc
Welcome to the brackish depths of my killfile, Sparky.
*plonk* ...problem solved.
-----------------
Rich Shewmaker
Ian
Greg Neill
scribe2b
--------------------------------------------------------------------------------
14 2003-01-31
thanks for the suggestions
the plastic was held by the magnet with a small metal superficial insert
the bottom of the tear drop was the same proportions as the sphere, and in
all was much heavier
I was considering the possibility of the teardrop wabbling and causing
turbulence on its first moments of fall, but wouldn't that have showed up as
a cubic term on the plot?
thanks
"|-|erc" wrote in message
> some ideas :
>
> how is a plastic sphere being held by a magnet?
>
> the magnetic field will remain after the voltage is cut
> and different shapes and materials will drop sooner
>
> the weight (also size) will influence the fall, a teardrop
> might be faster than a sphere in general but lighter
> material will fall slower in atmosphere in general aswell
>
> maybe your teardrop doesn't maintain vertical at low speeds
>
> Herc
>
>
>
> "" > wrote
> > Someone HAS to know what's going on.
> > If it helps, it's not anything wrong with our particular apparatus because
> > similar phenomenon happened with other groups on different workstations.
> > "" wrote in message
> > > Hi,
> > > I'm doing a first year university physics lab on free fall.. the apparatus
> > > is simple. An electromagnet positioned above two, moveable, photogates. When
> > > the voltage is cut, the object falls and the time interval between the two
> > > gates is measured on a timer accurate to 0.1 ms. We used two different
> > > objects to examine their "free fall", we used a plastic sphere and a
> > > streamlined (i.e. looks like a teardrop) steel object.
> > > We hypothesized that the plastic sphere would feel greater drag during its
> > > fall, and would thus, take longer to fall.
> > > However, when we performed the lab, we found that the plastic sphere was
> > > consistently falling approximately 5-15 ms faster than the steel (for
> > > heights of 50-200 cm).
> > > When we fitted the two sets of data our confusion grew as the plot for the
> > > plastic sphere had a cubic term, representing the 1/3 the drag coefficient,
> > > while the steel did not. (verified by chi-squared results).
> > >
> > > Can anyone explain this phenomenon?
> > > Thanks
-----------------
Shanx
Wanda
TheKid
raven1
--------------------------------------------------------------------------------
15 2003-02-01
|-|erc:
The picture looks great.
Interestingly, the ill-logic
of the basic design conguration
is more visible in the sketch
than in photos. The problems
would be that, numerous strong
points in the fuselage are needed
to support the heavy components
that are exterior to the main tank
and to resist and distribute high
local stresses that are due to the
high forces that are applied to the
structural components.
A vertical tower provides for the
inline stacking of components. The
symmetrical design of the USSR rocket
system would probably have far less
structural problems.
Neat sketch, however.
The unit spacing worked just fine.
Thanks,
_________________________________
|-|erc wrote:
> "" <> wrote
>
>>|-|erc:
>>
>>Please advise us regarding the font setting that
>>you used in making the picture. We can then select
>>the same font in order to see the picture as you
>>intended.
>>
>
>
> this is the original ascii art, I altered it for my default outlook font,
> but if you are adjusting font this one is better (any monospaced) :
>
>
>
> /\
> / \
> / \
> / \
> / \
> ^ /__________\ ^
> / \ | | / \
> /___\| /\ |/___\
> | || / \ || |
> | || /____\ || |
> | || // \\ || |
> | || //______\\ || |
> | || | || | || |
> |___|| | || | ||___|
> | || | || | || |
> | || | || | || |
> | || | || | || |
> | || | || | || |
> | ||/| || |\|| |
> | |/ | || | \| |
> | / | || | \ |
> | / | || | \ |
> | / |___||___| \ |
> |/ | | \|
> / |___||___| \
> (______|____||____|______)
> /___\ /_\/||\/_\ /___\
>
> // \\ // \\
> //// \\\\ //// \\\\
> //// //\\\\ \\\ //// //\\\\\\ \
> /// // // //\\\ \\//// //// / \\\ \\
> / / ///// / \ \\ /\\/// \// / // \\ \\ \
> / /// // // /// \\\ /\//\ /// / // \ \\\ \\
> / // / /// //// /\ //\\\\// // /\/ \\\\ \\ \\\
> /// /// //// // \\/\/ \/\//\\\/\/// \\ \\\ \\ \ \
> / // / ///// /////\ /\\\ /\/\\/\ // / /\\\ \ \\ \\\
> /// /// / /// / // /\\//\/ /\/\\ \// / /\ \\ \\\\ \ \\\
> // // ///// /// /\\/ /\/\//\\ \\\//\/// // /\\\\\\ \\ \\\
> ////// //// // // ///\\ /\/\\\ \//\/\///\ \//\\ \\\ \ \ \\\
> /// / ///// / /// /\\/\ //\\\\ \/\/\\//\/ /// \/\\ \\\\\ \\ \\\
> -Jas
-----------------
Rust
Ralph Hertle
Mercury481
PlanetaryMatrix
--------------------------------------------------------------------------------
16 2003-02-01
|-|erc wrote:
> thats really good, if you have a spare 30 secs download it , hollywood
> doesn't do much better.
I will humbly accept the compliment even though I am never humble.
http:www.sworld.org/artiv/fs.mpg is the next scene.
I was very surprised with how good it looked with almost no effort. Of course
comments on what is unrealistic are appreciated as I want to improve it.
-----------------
Tim Kozusko
Matt Giwer
Chris
Apostate
--------------------------------------------------------------------------------
17 2003-02-01
Yep, I took the data from both sources for 1 bar
-----------------
Someone
Lawrence & Bobbie
Odysseus
malcolm burton
--------------------------------------------------------------------------------
18 2003-02-01
"|-|erc" wrote in message
> > First of all I need a volunteer from the audience,
> >
> > give me the name of any rec newsgroup!
> >
>
>
> Dont be shy, say you there, any rec newsgroup, step
> right up and see real magic!
>
> Herc
You're doing it wrong.
Please immediately purchase a copy of Jim Cellini's DVD and view it five times
in a row. (Potty breaks are allowed; it's a long video.)
It'll be a good start for you. The rest of us are praying, burning white
candles, chanting, thinking positive thoughts, and/or invoking any number of
Chopra Quantum Whatevers to help bring about the release of volume two sometime
real soon now.
-----------------
Rust
John L
Greg Evans
Ben Sauvin
--------------------------------------------------------------------------------
19 2003-02-02
Randi will test you when you properly apply to be tested. Sign up here:
http://www.randi.org/research/challenge.html
-----------------
Rich Shewmaker
CNote
Wanda
Rust
--------------------------------------------------------------------------------
20 2003-02-02
It really all depends on the situation.
-----------------
Shanx
See You In Hell My Friend.
Someone
Greg Neill
--------------------------------------------------------------------------------
21 2003-02-02
If ever I actually found myself in that situation, I'd hold it upright,
with the intent of attacking my assailant's knife hand.
-----------------
cliff86
Rust
Shanx
NormDePloom
--------------------------------------------------------------------------------
22 2003-02-03
If you 'found yourself' in a knife fight, work with the grip that
comes fastest to you in an ambush situation. I prefer the reversed
grip for obvious reasons.
If you have more time, or are the aggressor, you probably want it
blade forward, edge out.
-----------------
Ian
Mercury481
Rich Shewmaker
Chas
--------------------------------------------------------------------------------
23 2003-02-04
This is fascinating stuff.. hopefully in a few decades I will be a Guru like
you guys :)
-----------------
Saad Malik
Chris
Matt Giwer
John L
--------------------------------------------------------------------------------
24 2003-02-05
|-|erc wrote:
> Yeah I know, I only posted here on a bet.
You lost, eh?
-----------------
cliff86
Greg Evans
beavith
scribe2b
--------------------------------------------------------------------------------
25 2003-02-05
"Should" and "do" are often very different things.
If you've been hard of hearing for a long time, you'll
probably understand when I claim that odd noises can be distracting.
I'm very comfortable driving around with my "natural" hearing, being
unable to hear all the various tickings and bangings and screechings
of my own car, or those around me.
I tried driving while wearing a hearing aid, once, and was
almost startled out of my skin when a truck right next to my good ear
let loose a blast of air as its brakes actuated.
In controverse argument, I offer the legions of people who
insist on driving while running their mouths on their stupid cell
phones.
I wear glasses. I do not know the numbers to describe my
visual deficit, but while I can operate comfortably without my glasses
when I'm not behind the wheel, there are very few circumstances under
which I would ever be caught behind a steering wheel without the
glasses.
-----------------
Ben Sauvin
First Name
Tim Kozusko
G=EMC^2 Glazier
--------------------------------------------------------------------------------
26 2003-02-06
"|-|erc" wrote
>
> "(Hildo) news multikabel> wrote > >
> > > Will a float in an aquarium go down a bit when the aquarium is put in
> > > a fast elevator that starts moving up? Why is this or isn't this?
> > > Thanks in anticipation.
> > > Mv
> >
> > As the elevator starts to move up the float will rise a little.
> > Then it will sink a little when the lift stops.
> > This is because the apparent gravity rises shortly as the elevator
> > accelerates upwards then returns to normal and it goes down a bit when the
> > elevator is decelerating.
> >
> > As the gravity rises the net upward force of the water on the float rises
>
> by the same amount as the increased force down from the weight of
> the float, cancelling out.
>
> Herc
>
Yes I agree with the theory. Still, I have actually seen what I described.
(At about 2g) but know I am doubting if I remember correctly. Maybe what I
saw whas the other way round.
-----------------
Terry Wilder
Edward Caruthers
(Hildo) news multikabel
Odysseus
--------------------------------------------------------------------------------
27 2003-02-07
its those added SRB's. ugh! i don't think its man-rated like the
saturn 5 was.
-----------------
See You In Hell My Friend.
sertec
beavith
Edward Caruthers
--------------------------------------------------------------------------------
28 2003-02-08
[expletives deleted] I did at first ... :(
Thanks for catching that ... I guess. ;)
OK, that should be between 81.6 and 93.3 million straight-line trips,
and OTOH the original statement implies a round trip to be about 1.03
million km. So while there's not as much of a discrepancy as my
regrettable error made me think, it's still large enough not to
require high-precision comparisons -- just a little care with
exponents. ;)
-----------------
Saad Malik
PlanetaryMatrix
CNote
Odysseus
--------------------------------------------------------------------------------
29 2003-02-10
At least we know why you are using a computer to communicate...you're not
allowed any sharp pointy things, such as pens, right?
-----------------
Wally Anglesea
beavith
raven1
malcolm burton
--------------------------------------------------------------------------------
30 2003-02-14
Well it is possible that the shape of a rain drop is not the best way to
enter the atmosphere.The reason is the front of the rain drop would
always be getting the friction(heat) Nature creates that shape when
drops of water are moving through the air at a top speed 0f 200 mph
Lets think about this idea of mine. Back to a round ball,and a round
ball inside this ball. The outer ball made to rotate,by
electro-magnetisum,and the inner ball(cabin) not rotating. This is done
very easy. Outer ball changing its surface by rotating is much like
passing your finger through a flame. There is a difference of high
temperature and the build up of heat. Bert
-----------------
Kurst
G=EMC^2 Glazier
Odysseus
Chas
--------------------------------------------------------------------------------
31 2003-02-15
If you are not trying to argue anything, why do you keep posting to me? I
never said you were a liar, so stop trying to convince me that you're not. I
really don't care if The Truman Show is based on your life, if you predicted
the upcoming war, or if your dick falls off, so please leave me out of any
further discussion.
-----------------
Xcott Craver
J.y.n.x
Saad Malik
Wanda
--------------------------------------------------------------------------------
32 2003-02-15
<PLONK>
-----------------
scribe2b
Greg Evans
Mitch Dickson
Lee S. Billings
--------------------------------------------------------------------------------
33 2003-02-16
It's also much farher away from the Sun than the moon is, so Titan
doesn't have as much of solar blasting that stripped closer planets.
-----------------
Impmon
gecko
scribe2b
Someone
--------------------------------------------------------------------------------
34 2003-02-21
Couldn't time be defined as the passage of an instance, no matter how small
that passing was? Wouldn't the passing of, let's say, one billion billion
billion billion billion billionth of a second taking a trillion years to
complete (because time itself was slowed down to such an extreme), still be
called time since an instant did pass? Just thinking... but the observer
would never know time was slowed to this extreme, since the observer most
certainly would also be slowed down, and time would seem normal to him.
Then you could really never know the age of the universe to any accuracy,
since the fluctuation of times' length could not be calculated and added to
the perceived time passage the observer experiences.
-----------------
G=EMC^2 Glazier
PlanetaryMatrix
Odysseus
gecko
--------------------------------------------------------------------------------
35 2003-02-21
Hmm which mint made the coin ?
"|-|erc" wrote
>
> "Richard" wrote
> > If I use my hand to push a coin across a table, can you pinpoint the cause of the
> > coins movement?
> >
> > I think you would say the fundamental cause is the decision to move the coin, or
> > would it? Could you ever pinpoint the cause?
>
> the decision happens 1 to 1.5 seconds before the movement, and can
> be detected, i.e. predicted.
>
> There are two classes of theories : your person is an idependant being that
> made this decision, probably some meta scientific training that made you
> think of it and inspired by an unresolved thirst for mothers milk.
>
> Alternatively your interactions within the universe are entwined within it and
> the cause was distributed beyond a single comprehension, perhaps the future
> outcomes of the coins movement has a greater effect, such as getting me to
> type upon it.
>
> Herc
-----------------
David H. Lipman
Kurst
Lee S. Billings
Mitch Dickson
--------------------------------------------------------------------------------
36 2003-02-22
what?
-----------------
malcolm burton
gecko
Edward Caruthers
Ben Sauvin
--------------------------------------------------------------------------------
37 2003-02-23
"|-|erc" wrote in message
"" > wrote
--Agreed. But there are lots of strange things going on due to
--network congestion caused by the worm.
--- You must think this, look you, that the worm will do his kind.
:-)
-- I refuse to write "LOL" since that always makes me think of someone
-- lolling round on a sofa, one arm trailing on the ground, eyes turned
-- upwards in their sockets, his tongue lolling out of his mouth, making
-- "lollll...lolllll...lollll" noises like a nutcase.
--
-- But I laughed out loud.
--
-- Without lolling.
- for years I thought the internet was a friendly place, all these
people
- sending me lots of love!
-
- Herc
Ahhhhhh.... ha ha ha ha ha ha ha!
That really made me laugh out loud.
Big hug, big kiss - after all, you deserve it.
-----------------
Roundtable
sertec
(Hildo) news multikabel
gecko
--------------------------------------------------------------------------------
38 2003-02-24
OOC: Eat a fig, figgy.
-----------------
cliff86
Saad Malik
Terry Wilder
Rust
--------------------------------------------------------------------------------
39 2003-02-24
Oh, so you're qualified/annointed/authorized to speak "for people as a
whole" now? Are you further saying that the opinions you expressed are not
what you really believe, but are merely immature bullshit assumed (for what
reason I'm not entirely sure) to be the opinion of the "people as a whole?"
> Its easy to sit in the top economic section of society and say everything
is dandy.
I wouldn't know about the top economic section of society, but it's also
easy to look at reality and say "Life is good." If it isn't for you, well,
that's a shame. And if you have to blame women or feminists for your
inadequacies, that, too, is a shame. (I lied about that last part.)
-----------------
TheKid
Mercury481
The Pervert
(Hildo) news multikabel
--------------------------------------------------------------------------------
40 2003-02-24
<snip schizophrenic ramble>
Someone's off their meds, I see.
-----------------
The Pervert
raven1
Lee S. Billings
scribe2b
--------------------------------------------------------------------------------
41 2003-02-24
Could be too much medication. Like Chief Wiggam said to Ralphie, "if
your nose starts bleeding it means you're picking it too much... or
not enough".
-----------------
Gary Rockley
PlanetaryMatrix
gecko
NormDePloom
--------------------------------------------------------------------------------
42 2003-02-24
Only under the right circumsatnces.
-----------------
Lee S. Billings
Lawrence & Bobbie
NormDePloom
PlanetaryMatrix
--------------------------------------------------------------------------------
43 2003-02-24
I was just thinking that it would be cool to put a pin hole camera in the
bow of a fast boat and drive it using VR glasses.
-----------------
Edward Caruthers
Mercury481
Ralph Hertle
First Name
--------------------------------------------------------------------------------
44 2003-02-25
You may know a lot about the cosmos but you can't spell to save your soul.
The word is "prodigy" not "prodogy".
-----------------
NormDePloom
TheKid
John L
First Name
--------------------------------------------------------------------------------
45 2003-02-26
Basically...
Your Off Topic posts are; unwarranted, annoyining and idiotic.
This is a discussion group on the use of Mr. David Harris' Pegasus mail and
utilities.
If you can't focus on that topic...GET LOST ! Your posts are more akin to spam.
Please go find a Psychiatry news forum, they may be able to help you with your
problem.
Are you taking your Prozac ? You should stay on the regiment...It'll help you
with your problem.
-----------------
Hold, I'll think of it in just a nanosecond or two
Mercury481
David H. Lipman
Greg Evans
--------------------------------------------------------------------------------
46 2003-02-26
"|-|erc" wrote in message
>
> > >
> > Ah yes, the things we learn when we least expect it. Feels good to take
> > control, doesn't it? *back rub hugs*
>
> back rub hugs, my favourite
> do 10 years of weight training and having strong arms means
> YOU do the massage!!
>
> Herc
Haa! Have been in the Air Force for 12 years, Have Rank, Guess who gives the
orders.
-----------------
(Hildo) news multikabel
Apostate
patty-anne-lea
See You In Hell My Friend.
--------------------------------------------------------------------------------
47 2003-02-27
This is a good point. We may not be further evolving, but that doesn't
mean
selection isn't going on.
In the last forty years, the welfare state has had some very powerful
positive side effects. Women have on the whole been able to shag who
they like, without need for a man's money. (The latest no fault
divorce laws allow women men's money basically regardless.)
I believe the meteoric rise in height is partly due to women's freedom
to shag about. Also the majority of people were poor in the 19th
century, and sleeping with rich men failed to work, as they could just
deny parentage.
One other thing you never hear these days is "Older men are more
attractive"
Obviously they never were, but with money coming from the state, women
no longer have to butter up the wealthy.
-----------------
Chris
Lee S. Billings
Ian
sertec
--------------------------------------------------------------------------------
48 2003-02-28
That would be nice if I knew where the top and bottom were..or even where
the left to right borders were..then I could do that. Not sure if there's
an easy way to find that out though with a triangle.
-----------------
scribe2b
Matt Giwer
cliff86
Chris
--------------------------------------------------------------------------------
49 2003-02-28
How about if we just called it the empty set?
-----------------
Rich Shewmaker
The Pervert
Terry Wilder
gecko
--------------------------------------------------------------------------------
50 2003-03-01
"|-|erc" <he...@adamskingdom.com> wrote in soc.atheism:
>> I merely asked how, IYO, mathematics relates to belief
>> or nonbelief in supernatual beings.
>
>because in mathematics you only believe what you KNOW is true.
And so it goes, possibly as long as the cosmos itself.
OK, Mr. Only Agnostics Know Which Side Is Up, what "belief" is it
that you are trying to inject into every atheist's veins, so you can claim
that we/they all believe things we don't know?
Are you prepared to believe that I know that I don't believe in any gawds?
Are you able to get a faint glimpse of the possibility that I could hold no
belief in any gawds, while not making any rash claims I can't demonstrate the
truth of (much less ask anyone else to believe as much or as little as I do)? Is
my not being so smug as to troll atheist groups chiding the posters for their
naïve faith in a strawman religious stance I posit for them, enough grounds for
you to count me among the question-begging 'atheist fundamentalists'?
Or have you even heard of agnostic atheists?
Does it make you feel all warm and secure, just before you fall off to sleep, to
know for sure that you haven't angered any gawds by scoffing at them?
Just in case, you know, you die before you wake?
-----------------
David H. Lipman
Apostate
malcolm burton
Chris
--------------------------------------------------------------------------------
Answers Below.
1 Wally Anglesea see angel
2 Edward Caruthers car
3 Xcott Craver no cot
4 Mitch Dickson first cause
5 Lawrence & Bobbie fence bird
6 J.y.n.x tempt, act, silence!
7 Hold, I'll think of it in just a nanosecond or two open mind
8 Mercury481 boiling atmosphere
9 sertec sir technical
10 Scribe2b scribes
11 Tim Kozusko mount koziosko
12 CNote see note
13 Greg Neill nil
14 Shanx show then thanks
15 Ralph Hertle alpha display hurtle, hurtle into space
16 Matt Giwer give away
17 Someone one bar
18 John L loo
19 Rich Shewmaker rich showmaker
20 See You In Hell My Friend. depends
21 Rust attacks metal
22 Chas chase
23 Saad Malik well said
24 Greg Evans even
25 Ben Sauvin save
26 \(Hildo\) news multikabel hill gravity
27 beavith animal
28 Odysseus odysey
29 malcolm burton button
30 G=EMC^2 Glazier glacier
31 Wanda wand dissapears
32 Lee S. Billings leave
33 Impmon impossible for man
34 PlanetaryMatrix space dimension
35 Kurst cursed
36 gecko primitive
37 Roundtable round
38 cliff86 jump off a cliff
39 The Pervert inadequate
40 raven1 raving
41 Gary Rockley rock
42 NormDePloom normal circumstances
43 First Name first person perspective
44 TheKid child
45 David H. Lipman tell off
46 patty-anne-lea pat
47 Ian I Agree
48 Chris nice
49 Terry Wilder empty will do
50 Apostate agnostic atheist
Mike Keith
>So it's impossible to find an error in a proof.
Of the many things you repeat ad nauseum, this is one of the stupidest.
Richard Henry
> However, you still seem to not understand what a mathematical proof
> is.
>
> It is a perfect argument that begins with a truth and proceeds by
> logical steps to a conclusion which then must be true.
>
> So it's impossible to find an error in a proof.
>
> However, a would-be discoverer *can* make errors in describing a
> proof, or think they see a proof where none exists, and potentially
> that can be found out by starting at the beginning of the "proof", and
> proceeding through it checking each step to make certain that it is a
> logical one.
So when someone finds an error in one of your steps, and points it out to
you, are you grateful?
Dmoranokst
>you, are you grateful?
No, he just calls them a liar. James acts like nobody but him understands
mathematics. He should go talk to a mathematician in person........oh wait, all
mathematicians are liars so why bother?
David Moran
Will Twentyman
> Will Twentyman <wtwen...@read.my.sig> wrote in message news:<3f2fd67e_3@newsfeed>...
Yes, I am attacking the proof that comes after it, as I said above.
> Consider
>
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f.
>
>
> Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization
>
> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
> where w_1 w_2 w_3 = f, and
>
> b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
>
> and at m=0
>
> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
>
> so two of the b's must equal 0, which means
>
> P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
>
> which is
>
> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
>
> proving that w_1 w_2 must equal 1, as f is coprime to 3 from before,
> which leaves b_3 = 3.
Notice that this depends on f coprime to three.
>
> Essentially objections now come down to claiming that the w's are
> dependent on m, but consider that w_1 w_2 = 1, when m=0, here where f
> is coprime to 3.
>
> But that was an arbitrary choice *I* made, so let f=3.
And now you look at the behavior when it isn't.
>
> Now w_1 w_2 = 3^{2/3} as long as m is coprime to 3, WITHOUT REGARD TO
> m.
You say that w_1 w_2 has a certain value as long as m is coprime to 3.
That is stating a dependence on m.
>
> So those posters who try to convince you that the w's are actually
> dependent on m, like being functions of m, must now also convince you
> that the w's make a decision, first looking to see if f=3 or have some
> non-unit factor in common with 3, and THEN they decide if they're
> dependent on m.
The w's are mathematical objects. They don't make decisions, they have
properties.
Maxim Stepin
> > So when someone finds an error in one of your steps, and points it out to
> > you, are you grateful?
>
> Well if it's something big your stomach drops, and it's like the whole
> world darkens. I've had times when I was cheerful, confident and
> maybe rather arrogant, and then at the next moment I felt like an
> abysmal failure, wondered why I was ever born, and desperately wanted
> to be rescued from the planet by friendly aliens.
And the fear of such emotions could certainly impair you judgment.
> Now if it's a minor error, then yeah, I'm immediately grateful.
And if someone points out a big error, you're immediately angry.
You can't handle the truth, it hurts you too much.
Arturo Magidin
[.snip.]
>Well I did find a problem with the definition of the object ring that
>I'd given, and I've updated it.
There have been at least two changes in recent memory, one sort of
announced, one done in silence. And then there was another change in
the past 36 hours, presumably what you are refering to here.
It is interesting that you DO NOT SAY what the problem you supposedly
found was. What was it? Why was it a problem? Did you find it
yourself, or is it the problem I've been pointing out for many months
that you had so diligently ignored?
The first change occurred back in January:
http://groups.google.com/groups?selm=3c65f87.0301131440.291430c6%40posting.google.com
Your definition at the time was that units in object rings had to have
complex norm equal to 1. You then modified the definition to:
"Objects are members of commutative rings where any unit and its
multiplicative inverse are units in all possible commutative rings in
which either are members, where no member is a factor
of an object for which it is not a factor in all possible commutative
rings in which it and that object are members."
and then changed it (without acknowledging any problems or announcing
the change) to:
"Objects are members of commutative rings where any unit and its
multiplicative inverse are units in all possible commutative rings in
which either and all integers are members, where no member is a
factor of an object for which it is not a factor in all possible commutative
rings that include all integers in which it and that object are
members."
throwing in the integers. You can even see that this was hasty, since
you simply threw in the phrase "that include all integers" in the
middle of a sentence to end up with an ungrammatical one.
I see now that you have a ->further<- change, done in the past 36
hours; your webpage now reads:
"The Object Ring is the set of all numbers where any member that is a
unit, i.e. factor of 1, and its multiplicative inverse are units in
all possible commutative rings in which either and all integers are
members, and where no non-unit member a is a factor of any two
integers that are coprime. "
You are still imprecise (it should most certainly NOT be "all numbers"
and "all possible commutative rings"; under that definition, NOTHING
is an object). Presumably you mean "all COMPLEX numbers" and "all
subrings of the complex numbers."
I note, however, that you have not defined any operations on this set,
nor proven that it is a ring under those operations.
This definition is certainly more inclusive than your previous one
(under which the ONLY objects were the integers). But your phrasing is
still too convoluted and unnecessary. First, you must specify if the
integers are "coprime" in the ring of integers, or in your object
ring.
Assuming the latter, then all complex numbers are the Object Ring,
since in that ring every two nonzero elements are coprime, and any
element is a unit.
So I assume you mean "two integers that are coprime in the ring of
integers." But in that case, the second clause is vacuous:
Suppose R is ANY subring of the complex numbers (hence, must contain
the integers), and let r be an element of R. Assume further that a and
b are two integers that are coprime (in the ring of integers), and
that r divides both a and b. The claim is that r is necessarily a
unit. This follows because if r divides both a and b, then it must
divide their integer gcd; but their gcd is 1, hence r divides 1, hence
r is a unit.
So your definition is reduced to:
"The Object Ring is the [subring of the complex numbers] satisfying
the condition that if u in the Object Ring is a unit in the Object
Ring, then it is a unit in Z[u], the smallest subring containing all
the integers and u."
Unfortunately, you have not proved that there is only one object ring,
nor have you proven that it is well defined. Also, it represents a
return to your attempted definition from some time ago. At the time, I
posted several proofs showing that if such a ring contains all
algebraic integers, and all its elements are algebraic, then it cannot
be anything OTHER than the ring of all algebraic integers. Here are
the proofs again:
THEOREM. Let R be any subring of C, and assume that all elements of R
are algebraic integers. If u in R is a unit, then u is a unit in Z[u],
the smallest subring of C containing the integers and u.
Proof. Since u is a unit in R, and all elements of R are algebraic
integers, it follows that 1/u is an algebraic integer. Let f(x) be the
minimal polynomial of 1/u. Then
f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1*x + a_0. Evaluating at u^{-1},
we have:
0 = f(u^{-1})
= u^{-n} + a_{n-1}u^{1-n} + ... + a_1u^{-1} + a_0.
Multiplying through by u^{n-1} we have
0 = u^{-1} + a_{n-1} + a_{n-1}u + ... + a_1*u^{n-2} + a_0*u^{n-1}.
Therefore,
u^{-1} = -(a_{n-1} + a_{n-1}u + ... + a_1*u^{n-2} + a_0*u^{n-1}),
and so lies in Z[u], as claimed. QED
THEOREM. Let R be a subring of the algebraic numbers, and assume that
R contains all algebraic integers. Assume further that if u is a unit
in R, then u is a unit in Z[u]. Then R is exactly equal to the ring of
all algebraic integers.
Proof. Assume not. Then R contains an algebraic number which is not an
algebraic integers; call it r. We may write r as a quotient of coprime
algebraic integers, since the ring of all algebraic integers is a
Bezout domain: r = a/b. Moreover, there exist algebraic integers x and
y such that ax+by = 1, again because the ring of all algebraic
integers is a Bezout domain, and a and b are coprime. And since r is
NOT an algebraic integer, we know that b is not a unit in the ring of
all algebraic integers.
Since R contains all algebraic integers, it contains x*r. It also
contains y. Therefore, it contains x*r + y. But
x*r + y = (xa)/b + y = (xa+yb)/b = 1/b.
Therefore, R contains 1/b, so b is a unit in R.
However, b is not a unit in Z[b], since Z[b] is a subring of the ring
of all algebraic integers, and we know that b is not a unit in that
larger ring. This contradicts the property of R, so we may conclude
that R does NOT contain any algebraic number which is not an algebraic
integers.
QED
Now, what about transcendental complex numbers? Then you run into some
serious problems. For example, the smallest subring of C containing
all algebraic integers and pi satisfies your conditions, because it is
isomorphic to the ring of polynomials with coefficients in the
algebraic integers. It's only units are the algebraic integers which
are units. But the same is true of the smallest subring of C
containing all algebraic integers and 1/pi. That would make both pi
and 1/pi into objects, but no subring of C may contain BOTH pi and
1/pi, and be an "object ring"! So, presumably, your definition is
meant to be applicable only to algebraic numbers in any case.
So, your definition is still unclear as written, and if some sense is
attempted to give to it, then the only conclusion that can follow is
that you have not defined anything new, contrary to your claims. You
assert (without proof) that "object" is a generalization of the
algebraic integers, yet if the notion includes all algebraic integers,
it cannot include anything else. As such, it is a poor generalization.
Arturo Magidin
>world darkens. I've had times when I was cheerful, confident and
>maybe rather arrogant, and then at the next moment I felt like an
>abysmal failure, wondered why I was ever born, and desperately wanted
>to be rescued from the planet by friendly aliens.
>
>
>After all, it's math. If you're wrong, you're wrong. There's no
>point hanging on to wishes when it comes to mathematics.
>
>Now if it's a minor error, then yeah, I'm immediately grateful.
Interesting. You have never acknowledged the following error,
nor shown any gratitude:
http://groups.google.com/groups?selm=b8v5t0%24c3e%241%40agate.berkeley.edu
(Claim about cubics; posted on May 3 2003 )
From: mag...@math.berkeley.edu (Arturo Magidin)
Newsgroups: alt.math.undergrad,alt.math,sci.math,sci.math.symbolic
Subject: Re: Algebraic integer coverage result
Date: Sat, 3 May 2003 01:26:56 +0000 (UTC)
Message-ID: <b8v5t0$c3e$1...@agate.berkeley.edu>
X-Comments: Although I post from my UC Berkeley account, I am a
Visiting Assistant Professor at the University of Montana.
In article <3c65f87.03050...@posting.google.com>,
James Harris <jst...@msn.com> wrote:
>jst...@msn.com (James Harris) wrote in message news:<3c65f87.03043...@posting.google.com>...
>
>The more I think about it, the more I'm sure that Magidin made a false
>statement about expressions in this post.
[.snip.]
>To see that imagine in the xy-plane with just two cubics that have two
>values reverse depending on two values along the x-axis. That can't
>happen unless the two cubics are crossing each other going in opposite
>directions. That is, one with a positive leading coefficient and the
>other with a negative one.
f(x) = x^3 - 6x + 6
g(x) = x^3 - 8x + 9
f(1) = 1 g(1) = 2
f(2) = 2 g(2) = 1.
*****************************************************************************
Nor have you ever acknowledged that your previous definition of
objects was circular and at best encompassed only the integers, even
though it has been pointed out to you for many months now.
|-|erc
Short Proof of Fermat's Last Theorem, it becomes another
Long Proof of Fermat's Last Theorem.
Herc
Lorrill Buyens
a rock and hollered:
>"Maxim Stepin" <maxim...@worldnet.att.net> wrote in message news:<bgrh1s$rbae6$1...@ID-57378.news.uni-berlin.de>...
>> "James Harris" <jst...@msn.com> wrote in message
>> news:3c65f87.03080...@posting.google.com...
>> > What I've seen are repeated attempts at distractions.
>>
>> That's what you think you see.
>>
>> But in reality your definition of "object ring" is still circular, thus your whole FLT "proof" is
>> still invalid.
>> So you cannot say "nobody found an error in my FLT proof".
>
>Well I did find a problem with the definition of the object ring that
>I'd given, and I've updated it.
>
>However, you still seem to not understand what a mathematical proof is.
>
>It is a perfect argument that begins with a truth and proceeds by
>logical steps to a conclusion which then must be true.
>
>So it's impossible to find an error in a proof.
Jimmy boy, I might know jack about math, but even I can tell that you're
talking through your hat.
KRamsay
(James Harris) writes:
|Clearly, they are constant in both cases with
|respect to m, without regard to the value of f. Which makes sense as
|f^2 is not a function of m, and it is what is being divided off.
It seems to be a common tendancy for people writing proofs (and not
always just amateurs) to use phrases like "clearly", "necessarily", "must
be true" and so on at just the spot where they're failing quite to make
an argument connect together.
We can tell you're essentially just bluffing and posturing here. Logical
arguments don't rely upon the reader's sense of surprise to close the
case. Now you write as if a function like (f-3)m were something absurd.
After all, if I set f=3 this becomes a constant. But then, miracle of
miracles, for f<>3 it "suddenly" depends on m. The argument you're
trying to sell us doesn't make any more sense.
I always used to think (when very young) that one should be able to
reach a point in a mathematical debate where you either agree, or
the losing side is left just arbitrarily denying one of the winning side's
points. It used to seem remarkable how long we could go with you
moving very slowly toward the truth, until you finally realized the latest
"wonder proof" was mistaken. Now I'm afraid you're seeming much
more like the person who simply asserts that the function he wants
to be constant must be constant, with no need to examine
demonstrations to the contrary.
Keith Ramsay
The Ghost In The Machine
<jst...@msn.com>
wrote
on 6 Aug 2003 16:29:04 -0700
<3c65f87.03080...@posting.google.com>:
> "Maxim Stepin" <maxim...@worldnet.att.net> wrote in message
> news:<bgrh1s$rbae6$1...@ID-57378.news.uni-berlin.de>...
> I'd given, and I've updated it.
>
> However, you still seem to not understand what a mathematical proof
> is.
>
> It is a perfect argument that begins with a truth and proceeds by
> logical steps to a conclusion which then must be true.
>
> So it's impossible to find an error in a proof.
Attempt at counterexample: I claim to prove that 1 = 2.
Let a = b = 1.
Then a^2 = ab.
a^2 - b^2 = ab - b^2.
(a+b)(a-b) = b(a-b)
Dividing by a-b we get
a+b = b.
1 = 2.
QED.
This is of course a claim of a proof only, and the error is
(hopefully) easily spotted. Many other claims have far more
obscure errors.
>
> However, a would-be discoverer *can* make errors in describing a
> proof, or think they see a proof where none exists, and potentially
> that can be found out by starting at the beginning of the "proof", and
> proceeding through it checking each step to make certain that it is a
> logical one.
I submit you have a claim. Has it at least been peer-reviewed? :-)
>
> James Harris
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Wayne Brown
> Have any of the other journals I've sent papers to actually
> peer-reviewed?
> I don't know.
It's unlikely that they could find one of your peers to conduct the
review. Now, if you could find a journal that uses ignorant, paranoid,
narcissistic drunks as reviewers, you might get a peer review...
--
Wayne Brown | "When your tail's in a crack, you improvise
fwb...@bellsouth.net | if you're good enough. Otherwise you give
| your pelt to the trapper."
"e^(i*pi) = -1" -- Euler | -- John Myers Myers, "Silverlock"
Bob Day
message news:<2rdc01-...@lexi2.athghost7038suus.net>...
> > In sci.physics, James Harris
> > <jst...@msn.com>
> > wrote
> > on 6 Aug 2003 16:29:04 -0700
> > <3c65f87.03080...@posting.google.com>:
>
>
> > > However, you still seem to not understand what a mathematical proof
> > > is.
> > >
> > > It is a perfect argument that begins with a truth and proceeds by
> > > logical steps to a conclusion which then must be true.
> > >
> > > So it's impossible to find an error in a proof.
> >
> > Attempt at counterexample: I claim to prove that 1 = 2.
>
> understand what a mathematical proof is. A mathematical proof is a
> *perfect* argument, so no counterexample exists. Possibly you've been
> programmed by social conventions where claims of proof are called
> proofs.
>
> But it's like if I say I have proof that you are a dog.
>
> My *saying* I have proof does not create a proof.
>
> So if a person says they have proof you're a dog, does that prove that
> a proof can be in error?
>
> No, it's just that they're in error, and do not have proof you're a
> dog.
>
> If they did have proof you're a dog, then you'd be a dog.
Go away James. You're splitting hairs that: 1) don't need
to be split, and 2) don't illuminate any argument. Belaboring
pedantics such as this just tries peoples' patience.
-- Bob Day
Randy Poe
> The Ghost In The Machine <ew...@sirius.athghost7038suus.net> wrote in message news:<2rdc01-...@lexi2.athghost7038suus.net>...
> > <jst...@msn.com>
> > wrote
> > on 6 Aug 2003 16:29:04 -0700
> > <3c65f87.03080...@posting.google.com>:
>
>
> > > However, you still seem to not understand what a mathematical proof
> > > is.
> > >
> > > It is a perfect argument that begins with a truth and proceeds by
> > > logical steps to a conclusion which then must be true.
> > >
> > > So it's impossible to find an error in a proof.
> >
> > Attempt at counterexample: I claim to prove that 1 = 2.
>
> understand what a mathematical proof is. A mathematical proof is a
> *perfect* argument, so no counterexample exists.
Way too wide an opening there James. The obvious question is
"why do counterexamples to your proofs abound?" Could it
mean (gasp) your proof is less than perfect?
- Randy
Arturo Magidin
Arturo Magidin <mag...@math.berkeley.edu> wrote:
>In article <3c65f87.03080...@posting.google.com>,
>James Harris <jst...@msn.com> wrote:
>
> [.snip.]
>
>
>>Well I did find a problem with the definition of the object ring that
>>I'd given, and I've updated it.
>
>There have been at least two changes in recent memory, one sort of
>announced, one done in silence. And then there was another change in
>the past 36 hours, presumably what you are refering to here.
And yet another unannounced change has now occured.
Yesterday, the definition at
http://www.msnusers.com/AmateurMath/objectmathematic.msnw
was:
>"The Object Ring is the set of all numbers where any member that is a
>unit, i.e. factor of 1, and its multiplicative inverse are units in
>all possible commutative rings in which either and all integers are
>members, and where no non-unit member a is a factor of any two
>integers that are coprime. "
The definition right now (4:12 MDT) is:
"The Object Ring is the set of all numbers where 1 is the only member
that is both a unit, i.e. factor of 1, and an integer, where no
non-unit member is a factor of any two integers that are coprime. "
You are still being sloppy in saying "set of all numbers". I suspect
that you mean to restrict yourself to complex numbers, if not
ALGEBRAIC numbers, and to give this set the inherited structure. If
this is the case, then since -1 is both a unit and an integer in any
subring of the complex numbers, it looks like you have nothing, yet
again.
Oh, the second clause is still imprecise or empty. "Coprimeness" is a
property that depends on the ring, although implications in one
direction may hold. You probably mean "two integers that are coprime
in the ring of integers", in which case the condition is vacuously
true in any subring of C, as I noted yesterday.
Now, assuming you meant to say "1 and -1 are the only elements which
are both units and integers", then you still must prove that "Object
ring" under this definition specifies a unique such object.
Presumably, you want to say "largest subring of the complex numbers
such that...", because otherwise, the integers are The Object Ring,
but so is any subring of the ring of all algebraic integers. It would
be of paramount importance to make sure that it defines a unique
thing, if you are going to call refer to it by using the singular
definite article "the."
I am also pretty certain that this definition includes way too many
things that you do not want. But it is obvious that once again all you
are doing is trying to fix, by fiat, the problems that plagued your
original proof of two years ago.
I must, however, confess that I am flabbergasted at your brilliance:
here we have what, by your own account, is the key, central, germain,
touchstone, concept of your approach. And even though you have been
able to change the definition in significant ways over the past 8
months, yet your proof is so solid that changing this key definition
does not require you to change even a single word of the rest of your
developement to take into account these changes. Truly, a work of
genius.
Brian Quincy Hutchings
I often pretend to be my own college career counselor --
just kidding.
on second thought, this interface is new to me, so that
I'll just jump to the end by deleting this passage
from the 10year mission (henceforth, TTYM, which is actually short,
for "the 10y. Mission to Pummel Fermat et Son Fil
into an elementary lesson on differential equations, Yeeha),
and putting m thoughts on the matter, herein.
oops.
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03080...@posting.google.com>...
> The *mathematics* is straightforward.
>
> Readers should see that it's also clear that posters for social
> reasons are fighting the mathematical logic, which is fascinating
> behavior.
--les ducs d'Enron!
http://members.tripod.com/~american_almanac
James Harris
Hey, you're right. Good catch. I'll update the page.
James Harris
David C. Ullrich
You seem to have missed the rest of what he wrote. Lemme
remind you:
>> Now, assuming you meant to say "1 and -1 are the only elements which
>> are both units and integers", then you still must prove that "Object
>> ring" under this definition specifies a unique such object.
>> Presumably, you want to say "largest subring of the complex numbers
>> such that...", because otherwise, the integers are The Object Ring,
>> but so is any subring of the ring of all algebraic integers. It would
>> be of paramount importance to make sure that it defines a unique
>> thing, if you are going to call refer to it by using the singular
>> definite article "the."
>>
>> I am also pretty certain that this definition includes way too many
>> things that you do not want. But it is obvious that once again all you
>> are doing is trying to fix, by fiat, the problems that plagued your
>> original proof of two years ago.
>>
>> I must, however, confess that I am flabbergasted at your brilliance:
>> here we have what, by your own account, is the key, central, germain,
>> touchstone, concept of your approach. And even though you have been
>> able to change the definition in significant ways over the past 8
>> months, yet your proof is so solid that changing this key definition
>> does not require you to change even a single word of the rest of your
>> developement to take into account these changes. Truly, a work of
>> genius.
>James Harris
************************
David C. Ullrich
Arturo Magidin
Yeah, I noticed you said "a poster" had pointed out the rather silly
error. I also notice that you continue to state "all numbers" without
specifying whether you are considering only algebraic numbers, or in
fact "all numbers". If you are considering "all numbers", then what
you have need not even have operations defined: it would include
actual polynomials in any number of incompatible variables, and some
elements of positive characteristic. If you are going to update the
page, why not bother to fix the definition COMPLETELY?
Current definition states:
"The Object Ring is the set of all numbers where -1 and 1 are the only
members that are both a unit and an integer, where no non-unit member
is a factor of any two integers that are coprime. "
"Coprime" where? In the final product, or in the ring of integers?
It's possible that two integers are coprime in a larger ring without
being coprime in the integers.
The final clause is empty if you mean "coprime in the integers": in
ANY subring of the complex numbers, an element which divides two
integers which are relatively prime in the ring of integers must be a
unit. Putting it in the definition only obscures the latter.
Is your "object ring" unique? Is it a ring? You are using the singular
definite article, but you never prove there is only one subring of the
complex numbers which satisfies this condition: in fact, at least
EVERY subring of the algebraic integers does, and more besides, like
Z[pi], for example. And you only define it as a ->set<-, but you also
claim it is a ring. Since you are not specifying that it is a subring
of the complex numbers, what are the ring operations?
Does your ring contain multiplicative inverses for algebraic integers
which are NOT integers? Let f(x) be any monic irreducible cubic
with integer coefficients, |f(0)|>1, and assume moreover that its
discriminant is not a square in Q (so that the extension given by a
single root is not normal). Does your ring contain 1/r? It need not
cause any integer other than 1 and -1 to become invertible, but it
would still possibly cause problems with your congruences.
And let me repeat what I said before, which you removed without
addressing or acknowledging:
The Ghost In The Machine
<jst...@msn.com>
wrote
on 8 Aug 2003 11:16:29 -0700
<3c65f87.03080...@posting.google.com>:
> The Ghost In The Machine <ew...@sirius.athghost7038suus.net> wrote in message news:<2rdc01-...@lexi2.athghost7038suus.net>...
>> <jst...@msn.com>
>> wrote
>> on 6 Aug 2003 16:29:04 -0700
>> <3c65f87.03080...@posting.google.com>:
>
>
>> > However, you still seem to not understand what a mathematical proof
>> > is.
>> >
>> > It is a perfect argument that begins with a truth and proceeds by
>> > logical steps to a conclusion which then must be true.
>> >
>> > So it's impossible to find an error in a proof.
>>
>> Attempt at counterexample: I claim to prove that 1 = 2.
>
> understand what a mathematical proof is. A mathematical proof is a
> *perfect* argument, so no counterexample exists.
A mathematical proof is a sequence of steps. If those steps are
performed correctly and all assumptions are accounted for,
then it's a reasonably good proof. Perfect? I don't know how
to measure perfect.
Some interesting things happen when one changes the assumptions
though; the classical one is arguably Lobachevksy attempting to
find an absurdity by replacing the parallel postulate (an axiom),
and instead developing an entirely new geometric form, hyperbolic
geometry.
> Possibly you've been
> programmed by social conventions where claims of proof are called
> proofs.
>
> But it's like if I say I have proof that you are a dog.
>
> My *saying* I have proof does not create a proof.
>
> So if a person says they have proof you're a dog, does that prove that
> a proof can be in error?
It proves people can be in error when claiming proofs.
>
> No, it's just that they're in error, and do not have proof you're a
> dog.
>
> If they did have proof you're a dog, then you'd be a dog.
>
>> Let a = b = 1.
>>
>> Then a^2 = ab.
>>
>> a^2 - b^2 = ab - b^2.
>>
>> (a+b)(a-b) = b(a-b)
>>
>> Dividing by a-b we get
>
> And given that a=b that's an attempt at dividing by zero in the
> classic example.
Exactly.
>
> This example only seems to work by *human* error as human beings see
> 'a' and they see 'b' and think, different things, despite them being
> defined to be the same at the beginning.
>
> Doing the substitution a=b, ignoring the 1 for the moment gives
>
> a^2 = a^2
>
> a^2 - a^2 = a^2 - a^2
>
> (a+a)(a-a) = a(a-a)
>
> Dividing by a-a would be an error, as a-a=0.
>
> Now using the full substitution of a=b=1, you have
>
> 1^2 = 1^2
>
> 1^2 - 1^2 = 1^2 - 1^2
>
> (1+1)(1-1) = 1(1-1)
>
> and dividing by 1-1 would be an error as it equals 0.
Exactly. It's not a proof, merely a claim at one.
I claim that you claim to have a proof of Fermat's Last Theorem.
This is one reason why peer review is so important; while it
doesn't totally eliminate error, it at least allows for more
eyeballs to check for errors in the proof. The author,
presumably, then corrects those errors and republishes, or
abandons the effort. (I would think abandonment would be extremely
rare, unless the author, say, dies or something. :-) )
>
>> a+b = b.
>>
>> 1 = 2.
>>
>> QED.
>>
>> This is of course a claim of a proof only, and the error is
>> (hopefully) easily spotted. Many other claims have far more
>> obscure errors.
>
> Given a claim of proof, you can test it by determining if the argument
> begins with a truth, and proceeds by logical steps to a conclusion
> which then must be true.
>
> Unfortunately many people say "proof" when they mean claim of proof,
> so a lot of people believe that a math proof can be wrong, but they
> wouldn't believe that proof in any other context can be wrong, as then
> they realize it simply wasn't proof.
>
> If you have proof that someone committed a crime, then you have proof.
Crime commissions are in the legal realm. Of course one can
set up interesting logical problems a la Sherlock Holmes,
if one wishes.
The legal realm merely requires proof beyond a reasonable doubt.
Reason is used in both proofs and criminology.
>
> If it's not proof, then it's not proof.
>
> That when "math" is stuck next to "proof" some people suddenly think
> something changes is problematic, and may be why some can accept the
> possibility of error in a math proof.
I've found at least two errors in your proof submission.
The latter one is fatal; I can't work around it. The former's
effect on your proof is unclear.
Please fix. :-)
>
>> >
>> > However, a would-be discoverer *can* make errors in describing a
>> > proof, or think they see a proof where none exists, and potentially
>> > that can be found out by starting at the beginning of the "proof", and
>> > proceeding through it checking each step to make certain that it is a
>> > logical one.
>>
>> I submit you have a claim. Has it at least been peer-reviewed? :-)
>
> That's an interesting question and the answer is, I don't know. I
> have sent my work to math journals, and a paper is currently at a math
> journal, and I'm waiting to hear from them.
Well, we'll see; presumably they are reviewing it. :-)
>
> Have any of the other journals I've sent papers to actually
> peer-reviewed?
>
> I don't know.
>
> What is important to remember though, is that math society is a
> society, and I've already outlined the weird notion that a proof can
> be in error, where people actually believe that a *math proof* can be
> in error, when what they should realize is that a claim of proof can
> be in error.
>
> Math proofs are perfect, just like any other proof that actually is a
> proof.
>
> People, on the other hand, can say "proof" when in fact they don't
> have a proof.
And how do we know a proof is perfect?
>
> Just like someone can say you are a dog, claim they have proof, but it
> be nonsense.
>
> Hopefully I've cleared that issue up, and I've gone on about it
> because it was being questioned!!!
Questioning is part of the peer review process, methinks.
Obviously in Usenet the questioning is highly informal (and
occasionally peppered with insults, especially if alt.syntax.tactical
gets involved :-), or irrelevancies). I don't have a clue as to
how the more formal mathematical or scientific peer review process
works. I don't even know how the PhD peer review process works,
although I suspect one has to stand and defend his papers against
the attacks from his peers, presumably in a forum of, maybe 6 to 12
peers.
>
> Now here's a math proof. Those who doubt that fact can believe it's a
> claim of proof, but it's verified to be a proof by tracing the
> argument out.
>
> In this case, I begin with an expression. The expression exists, so
> that is the truth from which you start.
>
> Consider, in the ring of algebraic integers,
I'm assuming here that you are using a definition similar to
http://mathworld.wolfram.com/AlgebraicInteger.html
which defines an algebraic integer r (of degree N, if r satisfies
no "lower equation") as a solution to a polynomial
x^N + a_{N-1}x^{N-1} + ... + a_1x + a_0 = 0.
This gets a little weird, as algebraic integers of degree > 2
may not always be factorable in an elegant fashion.
We shall proceed ... carefully. :-)
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f).
>
> That is, I have the identity which defines P(m) in terms of various
> symbols, and it's all in the ring of algebraic integers, which means
> that the symbols can only represent numbers that are algebraic
> integers.
>
> Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization
>
> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
> where w_1 w_2 w_3 = f, and
>
> b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
>
> and at m=0
>
> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
>
> so two of the b's must equal 0, which means
>
> P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
>
> which is
>
> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
>
> proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves
> b_3 = 3.
Be careful here. You have proven that
P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
but you have *not* constrained u. If u is 0 things get ridiculous
and relatively uninteresting. If u is not 0 one can compute
P(0)/(u^2f^2) = (b_3 w_1 w_2 x + uf) = 3x + uf
and you've actually proven that b_3 w_1 w_2 = 3, if x != 0 (which
is also relatively uninteresting).
Such potholes are easily avoided of course (usually), but
your proof's real problem here is you leapt to the wrong conclusion
as you left out b_3.
Since (3 - 2sqrt(2)) * (3 + 2sqrt(2)) = 1 over the algebraic
integers, one also has to be careful about other conclusions
regarding this product as well. I'm not sure there's a
smallest algebraic integer > 0. (It's easy to prove the
set of algebraic integers "clusters" towards 1+ by considering
the equation y^n + 2; as n -> oo the primary root y = (2^(1/n))
tends to 1, and one gets an infinite subset. By replacing
y = (x + w) where w is any integer and grinding out the
resultant equation, one can show that the set of algebraic
integers "clusters" around any integer, including 0+. Therefore
there isn't a smallest positive algebraic integer.)
I don't know if this is a fatal flaw, but it is a problem.
>
> Now that was a lot of steps, but each was a logical one.
>
> First I introduced b_1, b_2, b_3, w_1, w_2, and w_3, which are defined
> by the factorization
>
> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
> then I set m=0, and used the definition of P(m) to get P(0).
>
> That told me that at m=0 two of the b's are 0, because then
>
> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
>
> where the "u^2" couldn't get there unless two of the b's are 0.
>
> Then using that result I get from
>
> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
> that
>
> P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
>
> and multiplying through by w_1 w_2 I have
>
> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f)
>
> which with
>
> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
>
> tells me that w_1 w_2 = 1, when m=0.
Again, where did b_3 go?
>
> Essentially objections to how f^2 divides off now come down to
> claiming that the w's are functions of m, but consider that w_1 w_2 =
> 1, when m=0, if f is coprime to 3.
>
> Now I'm focusing on what has been revealed to be an area of confusion.
> Apparently some people believe that when I divide off f^2 that it can
> divide off as a *function* of m, so that m=0 might be a special case.
It is possible to define f(x) = K, where K is an arbitrary constant.
Usually such functions are relatively uninteresting. Therefore
I fail to see why this is even a problem, let alone why people
would object thereto.
Of course your definition P(m) would more properly be defined
P(m,f,u,x), in certain contexts. In the computer engineering
realm you've basically defined a function/algorithmic procedure
P(m) with one parameter and three globals, one of which you're
attempting apparently to solve for (x). This isn't a real
big problem in mathematical circles, though.
> I'm now starting the argument to address that belief by noting again
> that w_1 w_2 = 1, when m=0, if f is coprime to 3. That is, when f
> doesn't have 3 as a factor.
>
> But that was an arbitrary choice, so let f=3.
>
> That is, I *said* f is coprime to 3 but in considering this
> possibility it's worth it to relax that restriction and now consider
> what would happen if it equals 3.
>
> Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.
>
> Seeing that is as simple as looking at
>
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f
>
> with f=3 as then you have
>
> P(m)/3^2 = (m^3 3^4 - 3m^2 3^2 + 3m) x^3 -
>
> 3(-1+m3^2 )x u^2 + 3u^3
>
> so *every* coefficient has a factor that is 3, as you can tell by
> looking.
>
> So with
>
> P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
> each of the b's and each of the w's has a factor that is 3^{1/3},
If we grind out this mess with f = 3, we get
P(m)/9 = (81 m^3 - 27 m^2 + 3m)x^3 - 3(-1+9m)xu^2 + 3u^3
so we can conclude that b_1 b_2 b_3 = (81 m^3 - 27 m^2 + 3m).
and w_1 w_2 w_3 = 3u^3.
We *cannot* conclude that all b's have an algebraic integer factor
3^{1/3} without additional information; for all I know
b_1 = (81 m^3 - 27 m^2 + 3m) and b_2 and b_3 are 1 -- unlikely,
admittedly, but theoretically possible.
Ditto for the w's.
You might as well divide P(m)/27 and compute b'_1, etc. and
w'_1, etc., as well. Whether this is useful is not clear to me.
This *is* a fatal flaw.
[rest snipped, as it requires reanalysis]
The Ghost In The Machine
<rpo...@yahoo.com>
wrote
on 8 Aug 2003 14:12:42 -0700
<585ab5d8.03080...@posting.google.com>:
Please describe one of these counterexamples; I'm mildly curious.
Admittedly, my previous post details at least two flaws; these
presumably can lead to some interesting counterexamples. Or
one can postulate f = u = 0 and generate some uninteresting ones. :-)
I've also noted that (3 - 2*sqrt(2)) * (3 + 2*sqrt(2)) = 1
is an interesting product of algebraic integers as well.
Obviously this can lead to some weird problems, as one
can prove positive integers have unique factorizations,
but algebraic integers do not:
1 = 1 * 1 = (3 - 2*sqrt(2)) * (3 + 2*sqrt(2))
= (4 - sqrt(15)) * (4 + sqrt(15))
= ...
= (n - sqrt(n^2 - 1)) * (n + sqrt(n^2 - 1))
etc.
Mr. Harris does jump to some interesting conclusions, though;
how does a * b * c = 3 in the algebraic integers yield
the requirement that each of a, b, and c has a factor
of 3^(1/3) (an algebraic integer of degree 3)? This particular
one puzzles me.
>
> - Randy
Tralfaz
"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
news:bh17lu$51u$1...@agate.berkeley.edu...
> In article <bguete$1cfc$1...@agate.berkeley.edu>,
> Arturo Magidin <mag...@math.berkeley.edu> wrote:
> >In article <3c65f87.03080...@posting.google.com>,
> >James Harris <jst...@msn.com> wrote:
> >
> > [.snip.]
> >
> >
> that is both a unit, i.e. factor of 1, and an integer, where no
> non-unit member is a factor of any two integers that are coprime. "
>
"set of all numbers" rational? real? algebraic? algebraic integers?
complex? Are the integers included (implied elsewhere on his webpage but
never explicitly shown)?
This is the best I can translate his definition.
Let S be the object ring.
1 e S, -1 (not e) S
If n =/= 1 and n e Z then if n e S then 1/n (not e) S
If A,B e Z and (A,B) = 1 and there does not exist x e S | xn = 1, if there
exists a, b e S | an = A and bn = B then n (not e) S
So the question is: what sort of structure does this set have? As defined,
it is obviously NOT a ring (-1) is excluded, so is the set closed under
addition and multiplication? Is it commutative?
-Tralfaz
Bill Dubuque
>
> Does your ring contain multiplicative inverses for algebraic integers
> which are NOT integers? Let f(x) be any monic irreducible cubic
> with integer coefficients, |f(0)|>1, and assume moreover that its
> discriminant is not a square in Q (so that the extension given by a
> single root is not normal). Does your ring contain 1/r? It need not
> cause any integer other than 1 and -1 to become invertible, but it
> would still possibly cause problems with your congruences.
There are also examples in quadratic fields, e.g. see below
Timothy Chow <tc...@math.mit.edu> wrote to sci.math.research on 1998/12/08:
>michael reid <re...@math.brown.edu> wrote:
>>Jim Propp asks:
>>>
>>> Does there exist an algebraic number w that is NOT an algebraic integer,
>>> but that nevertheless has the property that the only rational numbers
>>> in Z[w] are the rational integers?
>>
>> w = 1/(3 + sqrt(2)).
>>
>> w is integral at P = (3 - sqrt(2)), but not at P' = (3 + sqrt(2)).
>> it follows that Z[w] is integral everywhere, excepting at P',
>> and hence Q /\ Z[w] is integral everywhere, including at 7.
>
> This is a nice argument that is worth explaining in more detail.
> Among other things, it illustrates the power of localization.
>
> The Key Point is v_p(x) < 0 for some prime p in Q
> => v_P(x) < 0 for *every* prime P lying over p.
>
> Let F = Q(w) and let J be the ring of integers in F. We want to
> have v_p(x) >= 0 for every x in Z[w] and every rational prime p.
> We claim to achieve this, it suffices to choose, for each rational
> prime p, one prime P in J lying over p, and ensure that v_P(w) >= 0
> for all these P. For then v_P(x) >= 0 for all x in Z[w], and the
> above "Key Point" will force v_p(x) >= 0 should x happen to be in Q.
>
> On the other hand, we don't want to make v_P(w) >= 0 for *all*
> primes P in J, for then w would be an algebraic integer. So now
> it's clear what we need: we need to choose a prime p that splits
> in J and choose w such that v_P(w) < 0 for some prime lying
> over p and v_P'(w) >= 0 for some other prime P' over the same p.
> To find the simplest example, take the UFD: F = Q(r), r^2 = 2,
> and note that 7 splits into (3+r)(3-r) in J. Then it is
> easy to check that w = 1/(3+r) satisfies v_P(w) >= 0 for
> all primes P in J except P = (3+r)J, so it does the trick.
See also other posts in the thread containing the above post
http://groups.google.com/groups?selm=74k4r1%24sa0%40schubert.mit.edu
-Bill Dubuque
Randy Poe
<ew...@sirius.athghost7038suus.net> wrote:
>In sci.math, Randy Poe
><rpo...@yahoo.com>
>> Way too wide an opening there James. The obvious question is
>> "why do counterexamples to your proofs abound?" Could it
>> mean (gasp) your proof is less than perfect?
>
>Please describe one of these counterexamples; I'm mildly curious.
What I had in mind specifically involve the claims on certain numbers
being coprime to m, or is it f? There are several people out there who
keep giving examples of factorizations meeting all of James' rules,
followed by proof that the numbers he wants to be coprime are not.
>Mr. Harris does jump to some interesting conclusions, though;
>how does a * b * c = 3 in the algebraic integers yield
>the requirement that each of a, b, and c has a factor
>of 3^(1/3) (an algebraic integer of degree 3)? This particular
>one puzzles me.
I believe he is completely unaware of what "factor" really means, and
continues to rely on some sort of factoring-by-inspection argument.
Thus, if you have a*b*c = f^2*(other stuff), it is clear that the only
way to factor this is into f, f, and (other stuff), right? One of the
factors on the left must be f*(something), another must be
f*(something else). THAT is actually what I think his "exactly two
roots must be..." argument is all about.
Furthermore, he thinks that it doesn't matter what kind of numbers
these are. He really doesn't understand why people keep insisting that
the term "factor" be accompanied by a description of which ring you
are restricting to, and his response is that "any time you ask me what
the ring is, I'll just say 'algebraic integers'". I don't think he has
any concept what he means when he says "the ring is algebraic
integers". What ring are the polynomial coefficients drawn from?
"algebraic integers". What ring are the polynomials themselves in?
"algebraic integers". When you factor those polynomials into terms
like sqrt(x^3 + 5) with x a free variable, what ring are you factoring
over? "Algebraic integers, algebraic integers! If you need a ring it's
algebraic integers."
- Randy
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>Bill Dubuque <w...@nestle.ai.mit.edu> wrote in message news:<y8zsmoa...@nestle.ai.mit.edu>...
>> Arturo Magidin <mag...@math.berkeley.edu> wrote:
>> >
>> > Does your ring contain multiplicative inverses for algebraic integers
>> > which are NOT integers? Let f(x) be any monic irreducible cubic
>> > with integer coefficients, |f(0)|>1, and assume moreover that its
>> > discriminant is not a square in Q (so that the extension given by a
>> > single root is not normal). Does your ring contain 1/r? It need not
>> > cause any integer other than 1 and -1 to become invertible, but it
>> > would still possibly cause problems with your congruences.
>
>that would make an integer other than -1 or 1 a unit.
That's what you want it to do. Unfortunately, your definition still
has some serious problems, some unclear clauses, and you have yet to
produce a proof that it actually defines a well-determined thing.
The current definition:
The Object Ring is the set of all numbers where -1 and 1 are the
only members that are both a unit and an integer, where no non-unit
member is a factor of any two integers that are coprime.
Under this definition, there is no "object ring" because you are
allowing incompatible ring structures into your set. I've asked many
times before whether you mean your "numbers" to be taken ONLY from the
complex numbers, or perhaps from the ALGEBRAIC numbers. You have
failed to reply.
Let's assume, for the sake of discussion, that this is indeed the
case, and you mean your "Object Ring" to be a subring of the algebraic
numbers.
The last clause is unclear. "Coprimeness" is an ambient property: it
depends on the ring. So when you say "any two integers that are
coprime", you need to specify IN WHAT RING. I assume, but you have
never confirmed, that you mean "two integers that are coprime in the
ring of all integers."
If so, then the last clause has no content. The condition you give is
ALWAYS satisfied, in ANY ring containing the integers. It makes no
sense to include it.
So, let's assume that you MEANT:
The Object Ring is the [ring] of all [algebraic] numbers, with the
property that 1 and -1 are the only members that are both a unit
and an integer[.]
This still does NOT define a unique object: Z satisfies the condition;
so does any subring of the algebraic integers. So do a number of
orders of number fields, all of which contain algebraic numbers which
are not algebraic integers.
So let's assume that you meant
The Object Ring is the [largest subring of the algebraic] numbers
which has the property that 1 and -1 are the only members that are
both a unit and an integer[.]
Then you are still in trouble. Because there is no such "largest
subring", as Bill Dubuque noted in the post you were replying to. I
know the following is going to be over your head, but it is not for
your benefit anyway.
Take your favorite quadratic extension of Q, say Q[sqrt(2)]. Let p be
a rational prime which splits in Q[sqrt(2)], (p)=PQ, with P<>Q. Then the
localization of Z[sqrt(2)] obtained by inverting all the elements in P
satisfies your condition. So does the localization obtained by
inverting all the elements of Q. If we lift each of these, we obtain two
ring larger than the ring of all algebraic integers, call them R_P and
R_Q, each of which has the property you want. But the real problem,
and here's the kicker, is that any subring which contains BOTH R_P and
R_Q will contain a multiplicative inverse of p. So there is no
largest such object.
So right now, you have a definition which lists the properties you
want, but which is incoherent as written as has no actual referent.
In short: you cannot define things into existence. Your attempt at a
definition does not define anything, even when an attempt is made to
fix it.
Arturo Magidin
In article <3c65f87.03081...@posting.google.com>,
James Harris <jst...@msn.com> wrote:
>Bill Dubuque <w...@nestle.ai.mit.edu> wrote in message news:<y8zsmoa...@nestle.ai.mit.edu>...
>> >
>> > Does your ring contain multiplicative inverses for algebraic integers
>> > which are NOT integers? Let f(x) be any monic irreducible cubic
>> > with integer coefficients, |f(0)|>1, and assume moreover that its
>> > discriminant is not a square in Q (so that the extension given by a
>> > single root is not normal). Does your ring contain 1/r? It need not
>> > cause any integer other than 1 and -1 to become invertible, but it
>> > would still possibly cause problems with your congruences.
>
>that would make an integer other than -1 or 1 a unit.
That's what you want it to do. Unfortunately, your definition still
has some serious problems, some unclear clauses, and you have yet to
produce a proof that it actually defines a well-determined thing.
The current definition:
The Object Ring is the set of all numbers where -1 and 1 are the
only members that are both a unit and an integer, where no non-unit
member is a factor of any two integers that are coprime.
Under this definition, there is no "object ring" because you are
subring", as was implied in the post you were replying to. I know the
following is going to be over your head, but it is not for your
benefit anyway.
[There was a fairly obvious problem in what I wrote before, so I've
cancelled that message and am posting a correction]
Take your favorite quadratic extension of Q. Let p be a rational prime
which splits in the ring of integers of K, (p)=PQ, with P<>Q. One can
then invert elements which are in P but not in Q to obtain a ring
larger than the ring of integers of K, which still satisfies the
condition that the only rational integers which are units are 1 and
-1. One can do a symmetric argument and do the same with elements in Q
but not in P. Then one can consider the subrings of the ring of all
algebraic numbers which include the algebraic integers, and EITHER the
elements inverted from P-Q, OR the elements inverted from Q-P, but
noth both. These rings will each, separately satisfy your conditions,
and be incomparable (neither one is contained in the other).
But the problem is that by choosing the elements carefully, one
obtains elements with the property that any ring which contains BOTH
those inverses will necessarily contain 1/p. For example, choose a
quadratic extension which is a PID, and let (p)=(a)(b), with (a)<>(b)
primes. Then take R[1/a], and R[1/b], where R is the ring of all
algebraic integers. Each of them satisfies the property you want, but
any ring which contains BOTH 1/a and 1/b will necessarily contain 1/p,
which should not be allowed in "The Object Ring."
So there is no largest subring of the complex numbers satisfying your
definition. There is no well-defined "Ring of Objects".
So right now, you have a definition which lists the properties you
want, but which is incoherent as written as has no actual referent.
In short: you cannot define things into existence. Your attempt at a
definition does not define anything, even when an attempt is made to
fix it.
======================================================================
C. Bond
[snip]
> I use the newsgroups for drafts, but typically I *try* to contact top
> mathematicians when it's beyond the draft stage. Often they wander
> off when things start getting really interesting though.
>
> I think they just don't want to have to acknowledge important results
> coming from someone like me.
How would you know? So far you haven't produced any 'important results'. In fact, you haven't even
produced a single number which you claim is 'missing' or 'left out' of the ring of algebraic
integers. This claim is the basis for your notion of a ring of objects, whatever that is, yet you
have never posted a single number which supports it.
Still dodging specifics, James? Give us a number which *should be* in the ring of algebraic
integers, but which is not. If there are no such numbers -- if they are simply 'unicorns in the zoo'
-- then there is no need for a ring of objects. If there are such numbers, produce and post one. (Not
an ambiguous expression with ill-defined symbols, but a specific number.)
--
There are two things you must never attempt to prove: the unprovable -- and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
Bill Dubuque
>
> Take your favorite quadratic extension of Q. Let p be a rational prime
> which splits in the ring of integers of K, (p)=PQ, with P<>Q. One can
> then invert elements which are in P but not in Q to obtain a ring
> larger than the ring of integers of K, which still satisfies the
> condition that the only rational integers which are units are 1 and -1.
That's the point of the example I posted.
> One can do a symmetric argument and do the same with elements in Q
> but not in P. Then one can consider the subrings of the ring of all
> algebraic numbers which include the algebraic integers, and EITHER the
> elements inverted from P-Q, OR the elements inverted from Q-P, but
> noth both. These rings will each, separately satisfy your conditions,
> and be incomparable (neither one is contained in the other).
>
> But the problem is that by choosing the elements carefully, one
> obtains elements with the property that any ring which contains BOTH
> those inverses will necessarily contain 1/p. For example, choose a
> quadratic extension which is a PID, and let (p)=(a)(b), with (a)<>(b)
> primes. Then take R[1/a], and R[1/b], where R is the ring of all
> algebraic integers. Each of them satisfies the property you want, but
> any ring which contains BOTH 1/a and 1/b will necessarily contain 1/p,
> which should not be allowed in "The Object Ring."
That's rather obvious. To wit: any ring that contains a non-integral
algebraic number w along with all its conjugates necessarily contains
a non-integral rational number, since the ring then contains all the
coefficients of the minimal polynomial of w, at least one of which
is a non-integral rational number. Thus once one allows adjunctions
of non-integral algebraic numbers there are restrictions on what
later extensions are possible if such extensions are required to not
contain any non-integral rational number. Contrast this with integral
extensions, where there are no restrictions whatsoever on just which
algebraic integers may be adjoined -- integral extensions never change
a nonunit in the base ring into a unit in the extension. That is one
of the key properties of integral extensions . It is of course an
obvious consequence of "lying-over" for an integral extension R < T
since: c nonunit => cR < P for a max ideal P, which is lain over by
prime Q in T which necessarily contains the extension cT, so cT < 1,
so c stays nonunit in T (one can also easily prove this directly by
massaging the the minimal poly of c). So for algebraic integers the
property of being unit or nonunit is absolute -- independent of the
ambient field containing the algebraic integer. Any enlargement of
the algebraic integers will fail to satisfy these crucial properties
(among others). In fact one can easily show that integral extensions
may be characterized in terms of the lying-over (LO) property and
the incomparable (INC) property, e.g. consider the following
THEOREM For commutative rings R < T the following are equivalent
(1) R < T is an integral extension
(2) A < B has INC and LO for all A,B with R < A < B < T
(3) A < A[u] has INC and LO for all A,B with R < A < T, all u in T
-Bill Dubuque
KRamsay
|There have been NO demonstrations to the contrary. If you have one,
|give it.
I've lost count of how many times they've presented a non-unit common
divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3
as usual are algebraic integers satisfying
65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5,
and likewise for a_2 and a_3 by substituting them for a_1. We'd be happy
to post the _calculation_ again if we thought you'd read it.
Your argument, on the other hand, always relies at the crucial point
on the reader sharing your feeling that the kind of "weird" behavior
the common divisors have as f and m vary just couldn't possibly
be so.
Keith Ramsay
James Harris
> In article <3c65f87.03080...@posting.google.com>, jst...@msn.com
> (James Harris) writes:
>
> |There have been NO demonstrations to the contrary. If you have one,
> |give it.
>
> I've lost count of how many times they've presented a non-unit common
> divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3
> as usual are algebraic integers satisfying
>
> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
>
> Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5,
> and likewise for a_2 and a_3 by substituting them for a_1. We'd be happy
> to post the _calculation_ again if we thought you'd read it.
Huh? Are you sure you're claiming that the non-monic primitive
irreducible over Q supports your case?
Oh, I know. You're probably talking about that argument that Arturo
Magidin tossed at me in a post that I just refuted.
Basically any such claims depend on the assertion that an algebraic
integer 'a' that is not a unit, must have a non-unit algebraic integer
factor.
That's a sneaky little argument as it's getting the reader to assume
something not proven, which is that 'a' actually has a non-unit
algebraic integer factor because it's not a unit.
> Your argument, on the other hand, always relies at the crucial point
> on the reader sharing your feeling that the kind of "weird" behavior
> the common divisors have as f and m vary just couldn't possibly
> be so.
>
> Keith Ramsay
Well if you're telling the truth, fill in the gap I've pointed out in
YOUR argument.
The gap is the assertion that given an algebraic integer 'a' which is
not a unit, that 'a' must have some non-unit factor *in the ring of
algebraic integers*.
To protect readers from thinking it's simple, I remind them of the
ring of evens, that is the set of even numbers, where you have 2 not a
factor of 6, and not a unit. But 2 and 6 don't have factors in the
ring.
That's because the ring doesn't have 1.
You have to go to a higher ring, in this case the ring of integers, to
escape that interesting little situation.
Now the problem is different in the ring of algebraic integers, but
the ring is screwed up, and the fix is again a higher ring.
What I want you to consider is the *possibility* that weird things can
happen with rings, when you're not careful enough, and that
mathematicians weren't careful enough with the ring of algebraic
integers.
James Harris
Will Twentyman
> Will Twentyman <wtwen...@read.my.sig> wrote in message news:<3f312e37$1_2@newsfeed>...
>
>>James Harris wrote:
>
>
> <deleted>
>
>>>>>Given acceptance of the intriguing little result that given a factor g
>>>>>of a polynomial P(x), g=r+c, where c is a factor of the constant term
>>>>>P(0), given by c=g at x=0, and r=g-c, the proof follows easily enough.
>>>>
>>>>And yet I keep failing to follow it, and when I explained why you keep
>>>>failing to explain it, even though I've repeatedly pointed out the post
>>>>where I stated where I don't follow the proof.
>>>
>>>
>>>Yet, in your previous reply you said:
>>>
>>>"No one is attacking this statement. Just what comes *after* it."
>>
>>Yes, I am attacking the proof that comes after it, as I said above.
>
>
> Hmmm...reading back I can see where I read you wrong there. So then
> it's settled that you accept the nifty statement that given a factor g
> f a polynomial P(x), g=r+c, where c is a factor of the constant term
> P(0), given by c=g at x=0, and r=g-c.
>
>
>>>Consider
>>>
>>> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>>>
>>>
>>>
>>>Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization
>>>
>>> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>>>
>>>where w_1 w_2 w_3 = f, and
>>>
>>> b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
>>>
>>>and at m=0
>>>
>>> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
>>>
>>>so two of the b's must equal 0, which means
>>>
>>> P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
>>>
>>>which is
>>>
>>> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
>>>
>>>which leaves b_3 = 3.
>>
>
>
> Yup.
Which means that all that follows is dependent on that coprimeness.
>>>Essentially objections now come down to claiming that the w's are
>>>dependent on m, but consider that w_1 w_2 = 1, when m=0, here where f
>>>is coprime to 3.
>>>
>>>But that was an arbitrary choice *I* made, so let f=3.
>>
>>And now you look at the behavior when it isn't.
>
>
> Well, f is a symbol, which represents a number, and typically I say
> it's a prime integer that is coprime to 3.
Here's the problem: Your work shows a result for certain values of f,
those that are coprime to 3. Taking a value other than those and
showing that it doesn't behave like it "should" when f is coprime
accomplishes nothing. You have two seperate cases. You cannot take
conclusions from one case and apply them to the other case.
>
> However, I found that some people like yourself apparently oddly
> believed that a constant factor f^2 could divide off as a function, so
> I consider what would happen if that is the case, as then you'd get a
> contradiction at f=3.
No contradiction. You are switching between cases. It should be no
surprise that f=3 generates different results. Your entire proof before
was dependent on f being coprime to 3.
>
> Imagine that w_1 and w_2 are functions of m.
>
> Then from the above, you have
>
> w_1(0)w_2(0) = 1.
>
> So you might have something like w_1 = h(m)+1, where h(0) = 0. But,
> it turns out that if f=3, you have instead w_1(0) = 3^{1/3}.
>
> So then, if w_1 is a function of m, it only is one if f does not equal
> 3 or have any non unit factors in common with 3, which is nonsensical.
Why is this nonsensical? What this means is that w_1 is a function of m
and f, and the value of f can make it be dependent or dependent on the
value of m.
>
> And readers, I find it odd to need to remind the poster that an
> equation doesn't just disappear because a variable is no longer
> coprime to a prime integer. Debating that it does is fascinating
> behavior on the part of the poster.
How can a variable be coprime to a prime integer? In what ring?
>>>Now w_1 w_2 = 3^{2/3} as long as m is coprime to 3, WITHOUT REGARD TO
>>>m.
>>
>>You say that w_1 w_2 has a certain value as long as m is coprime to 3.
>>That is stating a dependence on m.
>
>
> Oh yeah, I made a mistake. If f=3, then 3^3 divides off, so in fact
> there is no variation as m varies, each of the w's just has a factor
> that is 3^{1/3}.
>
> It's possibly confusing talking about it, but amazingly easy to show,
> as you have
>
> P(m)/3^2 = (m^3 3^4 - 3m^2 3^2 + 3m) x^3 -
>
> 3(-1+m3^2 )x u^2 + 3u^3
>
> where you can see that all the coefficients have a factor that is 3.
>
> The neat thing which I use by setting f=3, is that then it's trivial
> to show that the w's are constant valued.
>
> So it's impossible for the w's to be functions of m, so that when f is
> coprime to 3, w_1 w_2 can vary from being coprime to f, to not being
> coprime, as you seem to *wish*, and it's trivially shown.
There is something called a constant function, such as g(x)=2. It's a
function of x, just not very exciting.
>>>So those posters who try to convince you that the w's are actually
>>>dependent on m, like being functions of m, must now also convince you
>>>that the w's make a decision, first looking to see if f=3 or have some
>>>non-unit factor in common with 3, and THEN they decide if they're
>>>dependent on m.
>>
>>The w's are mathematical objects. They don't make decisions, they have
>>properties.
>
>
> Yup. So if they were functions of m, then they'd maintain that
> property, even when f=3.
Wrong. Things frequently have different properties in different
situations. If you think of them as functions of m *and* f, then it
becomes clear that for certain values of f they are constant functions
of m, and for other values of f they are variable functions of m.
>
> The *mathematics* is straightforward.
>
> Readers should see that it's also clear that posters for social
> reasons are fighting the mathematical logic, which is fascinating
> behavior.
>
Also note that you consistently try to apply the results from one case
to another case, without realizing that each case requires its own proof.
Example: "in the integers, 3 is coprime to 4. Therefor 8 must be
coprime to 4 as well. If you disagree, that must mean there is
something very strange going on with the integers!"
Analysis: "the jump from 3 to 8 is not mathematically valid, there is no
problem."
--
Will Twentyman
email: wtwentyman at copper dot net
Wayne Brown
> I use the newsgroups for drafts, but typically I *try* to contact top
> mathematicians when it's beyond the draft stage. Often they wander
> off when things start getting really interesting though.
I once was approached on a street corner in Washington, D.C. by a
very disheveled and agitated man. He was anxious to tell me all about
the numerous conspiracies and cover-ups that he knew about. After I
extricated myself and moved away, I saw that he was doing the same to
everyone else who passed his corner.
No doubt he was disappointed that I "wandered off when things got really
interesting."
> I think they just don't want to have to acknowledge important results
> coming from someone like me.
Yeah, that's the reason.
W. Dale Hall
James Harris wrote:
> kra...@aol.com (KRamsay) wrote in message news:<20030811015606...@mb-m26.aol.com>...
>
>>In article <3c65f87.03080...@posting.google.com>, jst...@msn.com
>>(James Harris) writes:
>>
>>|There have been NO demonstrations to the contrary. If you have one,
>>|give it.
>>
>>I've lost count of how many times they've presented a non-unit common
>>divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3
>>as usual are algebraic integers satisfying
>>
>> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
>>
>>Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5,
>>and likewise for a_2 and a_3 by substituting them for a_1. We'd be happy
>>to post the _calculation_ again if we thought you'd read it.
>
>
> Huh? Are you sure you're claiming that the non-monic primitive
> irreducible over Q supports your case?
>
> Oh, I know. You're probably talking about that argument that Arturo
> Magidin tossed at me in a post that I just refuted.
>
> Basically any such claims depend on the assertion that an algebraic
> integer 'a' that is not a unit, must have a non-unit algebraic integer
> factor.
>
The factor that has been provided,
r(a) = 8 a^2 - 4 a - 45
is
(1) an algebraic integer for any algebraic integer "a",
(2) a divisor of both "a" and 5 for
a = -(any of the ai's in the above
factorization of 65x^3 - 12x + 1)
The factorizations have already been given, but since you're so adept
at ignoring the evidence, I will provide them once again:
q(a) = 8 a^2 - 76 a - 185
r(a) = 8 a^2 - 4 a - 45
s(a) = 4 a^2 - 37 a - 104
Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the *negative*
of any of the ai's of the above factorization of 65x^3 - 12x + 1), the
following factorizations hold:
q(a) r(a) = 5
r(a) s(a) = a.
In fact, the minimal polynomial of this number (r(a) for -a = any of
the above ai's) is given as:
MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
The above facts prove that this "a" has a non-unit algebraic integer
as a factor.
> That's a sneaky little argument as it's getting the reader to assume
> something not proven, which is that 'a' actually has a non-unit
> algebraic integer factor because it's not a unit.
>
This is actually *much* easier to prove: the ring of algebraic
integers is closed under the extraction of roots of monic
polynomials with algebraic integer coefficients. In particular,
every equation:
x^n - a = 0
has algebraic integer roots, whenever "a" is an algebraic integer,
and n is a natural number.
Thus, if "a" is an algebraic integer, and n is any natural number,
the number
a^(1/n)
is an algebraic integer, irrespective of which of the n roots you take.
>
>>Your argument, on the other hand, always relies at the crucial point
>>on the reader sharing your feeling that the kind of "weird" behavior
>>the common divisors have as f and m vary just couldn't possibly
>>be so.
>>
>>Keith Ramsay
>
>
> Well if you're telling the truth, fill in the gap I've pointed out in
> YOUR argument.
>
No gap.
> The gap is the assertion that given an algebraic integer 'a' which is
> not a unit, that 'a' must have some non-unit factor *in the ring of
> algebraic integers*.
>
As I mentioned above, the factor "r" that I provided *is* an algebraic
integer, and in the case that "a" is (-1) times any of the ai's in the
factorization cited, I have provided its minimal polynomial. It is
simple to verify from that minimal polynomial that it's an algebraic
integer and that it's not a unit in the ring of algebraic integers.
> To protect readers from thinking it's simple, I remind them of the
> ring of evens, that is the set of even numbers, where you have 2 not a
> factor of 6, and not a unit. But 2 and 6 don't have factors in the
> ring.
>
> That's because the ring doesn't have 1.
>
> You have to go to a higher ring, in this case the ring of integers, to
> escape that interesting little situation.
>
> Now the problem is different in the ring of algebraic integers, but
> the ring is screwed up, and the fix is again a higher ring.
>
> What I want you to consider is the *possibility* that weird things can
> happen with rings, when you're not careful enough, and that
> mathematicians weren't careful enough with the ring of algebraic
> integers.
>
The problem is with your ignorance. The ring of algebraic integers has
none of the shortcomings you seem to want it to have. It is your method
which is at fault, since it forces you to believe a falsehood, namely
that the ai's in that above factorization are coprime to 5.
>
> James Harris
Arturo Magidin
>> In article <3c65f87.03080...@posting.google.com>, jst...@msn.com
>> (James Harris) writes:
>>
>> |There have been NO demonstrations to the contrary. If you have one,
>> |give it.
>>
>> I've lost count of how many times they've presented a non-unit common
>> divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3
>> as usual are algebraic integers satisfying
>>
>> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
>>
>> Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5,
>> and likewise for a_2 and a_3 by substituting them for a_1. We'd be happy
>> to post the _calculation_ again if we thought you'd read it.
>
>Huh? Are you sure you're claiming that the non-monic primitive
>irreducible over Q supports your case?
You are just saying words. 8a_1^2 + 4a_1 - 45 is not a "non-monic
primitive irreducible over Q". It's not even a polynomial. It's an
algebraic integer, just like 8*(4)^2 + 4(4) - 45 is an integer.
>Oh, I know. You're probably talking about that argument that Arturo
>Magidin tossed at me in a post that I just refuted.
>
>Basically any such claims depend on the assertion that an algebraic
>integer 'a' that is not a unit, must have a non-unit algebraic integer
>factor.
If a is not a unit, then a itself is a non-unit algebraic integer
factor of itself. That's because this ring has a 1.
If that's hard for you, think about integers: any integer other than
1, -1, is a factor of itself.
>That's a sneaky little argument as it's getting the reader to assume
>something not proven, which is that 'a' actually has a non-unit
>algebraic integer factor because it's not a unit.
>
>> Your argument, on the other hand, always relies at the crucial point
>> on the reader sharing your feeling that the kind of "weird" behavior
>> the common divisors have as f and m vary just couldn't possibly
>> be so.
>>
>> Keith Ramsay
>
>Well if you're telling the truth, fill in the gap I've pointed out in
>YOUR argument.
>
>The gap is the assertion that given an algebraic integer 'a' which is
>not a unit, that 'a' must have some non-unit factor *in the ring of
>algebraic integers*.
This is nonsense.
>To protect readers from thinking it's simple, I remind them of the
>ring of evens, that is the set of even numbers, where you have 2 not a
>factor of 6, and not a unit. But 2 and 6 don't have factors in the
>ring.
>
>That's because the ring doesn't have 1.
Are you claiming that the ring of algebraic integers does not have a
1?
Arturo Magidin
Arturo Magidin <mag...@math.berkeley.edu> wrote:
[.snip.]
>If that's hard for you, think about integers: any integer other than
>1, -1, is a factor of itself.
That, of course, should be "non-unit factor" or "nontrivial factor".
Arturo Magidin
>> In article <3c65f87.03080...@posting.google.com>, jst...@msn.com
>> (James Harris) writes:
>>
>> |There have been NO demonstrations to the contrary. If you have one,
>> |give it.
>>
>> I've lost count of how many times they've presented a non-unit common
>> divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3
>> as usual are algebraic integers satisfying
>>
>> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
>>
>> Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5,
>> and likewise for a_2 and a_3 by substituting them for a_1. We'd be happy
>> to post the _calculation_ again if we thought you'd read it.
>
>Huh? Are you sure you're claiming that the non-monic primitive
>irreducible over Q supports your case?
>
>Oh, I know. You're probably talking about that argument that Arturo
>Magidin tossed at me in a post that I just refuted.
>
>Basically any such claims depend on the assertion that an algebraic
>integer 'a' that is not a unit, must have a non-unit algebraic integer
>factor.
>
>That's a sneaky little argument as it's getting the reader to assume
>something not proven, which is that 'a' actually has a non-unit
>algebraic integer factor because it's not a unit.
If that's really what is behind it, then it's easy to prove: if a is
not a unit in a ring with 1, then a is a nonunit factor of a.
Again, remember:
DEFINITION. Let R be a commutative ring, and let a and b be elements
of R. Then "a is a factor of b" (in R) if and only if there exists an
element c in R such that a*c = b.
Since a=1*a, then a is a factor of a. Since a is not a unit, a is a
non-unit factor of a.
Surely you agree with ->that<-.
As to the "refutation", you sure talked a lot, but you did not
"refute" anything. Here's why you did not refute it:
You are demanding a proof that if a1, a2, a3 are algebraic integers
chosen so that 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1), then
each of a1, a2, a3 have a non unit factor in common with 5.
That is, we need to find algebraic integers r1,s1,q1,r2,s2,q2,r3,s3,q3
such that
(1) r1*s1 = a1; r1*q1 = 5 (so r1 is a factor in common of a1 and 5);
(2) r2*s2 = a2; r2*q2 = 5 (so r2 is a factor in common of a2 and 5);
(3) r3*s3 = a3; r3*q3 = 5 (so r3 is a factor in common of a3 and 5);
(4) none of r1, r2, r3 are units in the ring of algebraic integers.
Surely, you agree, that would necessarily show that a1 has a non unit
factor in common with 5 (namely r1); that a2 has a nonunit factor in
common with 5 (namely r2); and that a3 has a nonunit factori n common
with 5 (namely r3).
The values were given explicitly: Note that
-a1^3 - 12*a1^2 + 65 = 0
-a2^3 - 12*a2^2 + 65 = 0
-a3^3 - 12*a3^2 + 65 = 0.
Let r1 = 8*a1^2 + 4*a1 - 54
q1 = 8*a1^2 + 76a1 - 185
s1 = -(4a1^2 + 37a1 - 104)
r2 = 8*a2^2 + 4*a2 - 54
q1 = 8*a2^2 + 76a2 - 185
s1 = -(4a2^2 + 37a2 - 104)
r3 = 8*a3^2 + 4*a2 - 54
q1 = 8*a3^2 + 76a3 - 185
s1 = -(4a3^2 + 37a3 - 104)
Then a simple calculation using the fact that
-a1^3 - 12*a1^2 + 65 = 0;
-a2^3 - 12*a2^2 + 65 = 0;
-a3^3 - 12*a3^2 + 65 = 0;
shows that r1*q1 = r2*q2 = r3*q3 = 5.
Another simple calculation shows that
r1*s1 = a1
r2*s2 = a2
r3*s3 = a3.
(The explicit calculations may be found either in Dale's original
post,
http://groups.google.com/groups?selm=3F148963.6000800%40farir.com
or my more recent almost-verbatim-quote in
http://groups.google.com/groups?selm=bh6666%24a7e%241%40agate.berkeley.edu
)
Note also that all of r1, r2, r3, s1, s2, s3, q1, q2, a3 are algebraic
integers. This simply because they are obtained from algebraic
integers by addition and multiplication by integers and/or algebraic
integers.
So it is CERTAINLY the case that r1 is an algebraic integer which is a
common factor, IN THE RING OF ALGEBRAIC INTEGERS, of a1 and 5; that r2
is an algebraic integer which is a common factor IN THE RING OF
ALGEBRAIC INTEGERS, of a2 and 5; and that r3 is an algebraic integer
which is a common factor IN THE RING OF ALGEBRAIC INTEGERS, of a3 and
5. We have not, in any way, "left" or been "pushed out" of the ring of
algebraic integers.
Finally, we just need to verify that none of r1, r2, or r3 are units
in the ring of algebraic integers. If that is indeed the case, WE ARE
DONE. And we are done, because we already know they are common
factors, we only need to know they are not units in this ring.
The word "unit" is shorthand. It means nothing more and nothing less
than:
DEF. Let R be a ring with 1. An element a in R is a unit (in R) if and
only if there exists an element b IN R such that a*b=b*a = 1.
In a commutative ring, of course, it is enough to check a*b=1.
Then we have the following easy theorem, consequence of the theorem
you accepted last December after 10 months of denying it:
THEOREM. Let a<>0 be an algebraic integer, and let f(x) be a monic
polynomial with integer coefficients, irreducible over Q, which has a
as a root. Then a is a unit in the ring of all algebraic integers if
and only if the constant term of f(x) is either 1 or -1.
Proof. Let b be the unique complex number such that a*b=1. That is,
b=1/a. By definition, a is a unit in the ring of all algebraic
integers if, and only if, b is an algebraic integer.
Let
f(x) = x^n + ... + a_1*x + a_0.
Then f(a) = 0. Multiplying through by b^n, we have
0 = b^nf(a)
= b^n*a^n + ... + a_1*a*b^{n} + a_0*b^n
= 1 + ... = a_1*b^{n-1} + a_0*b^n.
Let g(x) = a_0*x^n + ... + a_{n-1}*x + 1.
That means that g(b) = 0. Since any factorization of g(x) will produce
a factorization of f(x) by substituting x^{-1} for x and clearing
denominators, we see that g(x) is an irreducible polynomial with
integer coefficients. Since the last term equals 1, it is therefore
primitive.
Therefore, b is an algebraic integer if and only if the leading term
of g(x) is either 1 or -1. Since the leading term of g(x) is a_0, that
says that b is an algebraic integer if and only if the constant term
of f(x) is 1 or -1.
IN summary:
a is an algebraic integer unit if and only if b is an algebraic
integer; b is an algebraic integer if and only if the constant term of
f(x) is 1 or -1. Therefore, a is an algebraic integer unit if and only
if the constant term of f(x) is 1 or -1.
QED
So one way to prove that none of r1, r2, r3 are units in the ring of
all algebraic integers is to exhibit a monic polynomial with integer
coefficients, irreducible over Q, with constant term different from 1
and -1, which has r1, r2, r3 as roots. (Not the only way: we could try
finding a different one for each ri, but it so happens they are all
roots of the same polynomial).
And indeed, as Dale noted, if we let
f(x) = x^3 - 969x^2 + 315x + 5,
it so happens that f(r1)=f(r2)=f(r3)=0. The irreducibility over Q of
f(x) may be checked by the rational root theorem: the only possible
roots are 1, -1, 5, and -5, none of them are roots, and so we are
done.
Now, you can dance around the issue all you want, James. You can
produce paragraph upon paragraph, post upon post, saying all sorts of
things about problems and sneakiness and how it is impossible to show
something or other.
But you cannot get around one very simple issue: we have EXPLICITLY
shown the factors, we have EXPLICITLY shown they are factors, we have
EXPLICITLY shown they are not units, we have EXPLICITLY shown they are
common factors.
We have EXPLICITLY produced the factors you claim cannot exist.
Anything else is sophistry.
(Though I say "we", the credit goes of course to Dale who actually
took the time to make the calculations and initial verifications).
Arturo Magidin
>to do with the word "unit". See below...
>
>> In fact, the minimal polynomial of this number (r(a) for -a = any of
>> the above ai's) is given as:
>>
>> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
>>
>> The above facts prove that this "a" has a non-unit algebraic integer
>> as a factor.
>
Why not? There is an algebraic integer, called here r(a), with the
following three properties:
(1) It is not a unit in the ring of all algebraic integers; you
apparently agree with that conclusion, but then engage in
sophistry about a supposed "problem" with it.
(2) There is an algebraic integer, called s(a), such that
r(a)*s(a)=a. So r(a) is a factor of a in the ring of algebraic
integers.
(3) There is an algebraic integer, called q(a), such that
r(a)*q(a)=5. So r(a) is a factor of 5 in the ring of algebraic
integers.
So: r(a) is (1) not a unit; (2) a factor of a; and (3) a factor of 5.
Why is it not the case that r(a) is a non-unit algebraic integer which
is a common factor of a and 5?
[.snip.]
>What you have is
>
> x^3 - 969 x^2 + 315 x + 5
>
>and what you want to do is convince readers that none of its roots can
>be coprime to 5.
There is no "convincing" that needs to be done. NONE Of the roots are
units, and ALL the roots are factors of 5. You're done.
>But you cannot do that,
Correction: James Harris cannot do that because he does not know what
"coprime" means.
> and instead have maintained that proving that
>none of the roots can be a unit proves that one of them must have a
>non-unit factor in common with 5 in the ring of algebraic integers.
They are each FACTORS of 5, and EACH non-units.
Do you agree or disagree with (all happening in a commutative ring
WITH 1):
(a) If a is a factor of b, then a and b have a common factor
(namely, a).
(b) If a is a factor of b and is not a unit, then a is a non-unit
common factor of a and b.
(c) If is a factor of b and is not a unit, then a and b have a
non-unit common factor (namely, a).
>I've just shot down that little trick, but you're squirming.
I do see a lot of squirming, but it's not coming from Dale.
[.snip.]
>The truth is that you're relying on the negative, which is that NONE
>of the roots of that expression can be units, to try and prove the
>positive, which is that then they all have a non-unit factor in common
>with 5 in the ring of algebraic integers.
You are completely, totally, utterly lost and confused.
The roots themselves are factors of 5. Each root is a factor of
itself, and a factor of 5. Therefore, each root is a common factor of
itself and 5. And since each of them is not a unit, then we have
EXHIBITED a non-unit common factor of each with 5. The roots are not
the original numbers we were interested in (the a's), they are the
common factors that have been produced.
[.snip.]
>> As I mentioned above, the factor "r" that I provided *is* an algebraic
>> integer, and in the case that "a" is (-1) times any of the ai's in the
>> factorization cited, I have provided its minimal polynomial. It is
>> simple to verify from that minimal polynomial that it's an algebraic
>> integer and that it's not a unit in the ring of algebraic integers.
>
>See readers? You can catch the trick here as notice the poster said
>"not a unit", and what I'm telling you is that these posters are
>working hard to convince YOU the reader.
Are you claiming that "non-unit" and "not a unit" are not the same
thing?
[.snip.]
W. Dale Hall
James Harris wrote:
> "W. Dale Hall" <mailt...@farir.com> wrote in message news:<3F37D13...@farir.com>...
>
> Hmmm...the poster has tried to dupe *you* the reader, and it all has
> to do with the word "unit". See below...
>
>
>>In fact, the minimal polynomial of this number (r(a) for -a = any of
>>the above ai's) is given as:
>>
>> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
>>
>>The above facts prove that this "a" has a non-unit algebraic integer
>>as a factor.
>
>
> No it doesn't. What you can prove is that if 'a' is coprime to 5 it
> can't be a unit in the ring of algebraic integers, but as I've said
> the ring of algebraic integers is screwed up.
>
No, no one can prove that. After all, a unit *is* coprime to every
number!
Let u be a unit algebraic integer, and let p be any algebraic integer.
Then there is an algebraic integer v for which uv = 1, and we get:
u v + 0 p = 1,
where 0 is zero.
This establishes the fact that a UNIT algebraic integer is coprime to
every algebraic integer, in the ring of algebraic integers. A very
slight rewording proves the result for a unit of any commutative
ring.
No wonder you think the ring of algebraic integers is screwed up!
Let's get the language straight. I have been clear in this that 'a'
refers to (-1) times any one of the coefficients ai in the factorization
65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
I claim that *each* of those numbers a has a non-unit algebraic integer
factor in common with 5.
PROOF: The above expressions q(a),r(a),s(a) all yield algebraic integers
for any algebraic integer a. When, in addition, a is a root of the
polynomial
x^3 - 12 x^2 + 65
then the products are as I mentioned above:
q(a) r(a) = 5
r(a) s(a) = a.
Your claim:
>
> No it doesn't. What you can prove is that if 'a' is
> coprime to 5 it can't be a unit in the ring of algebraic
> integers, but as I've said the ring of algebraic integers
> is screwed up.
amounts to this:
PARAPHRASE:
It doesn't matter whether anyone provides algebraic integers
q,r, and s for which
5 = q*r
a = r*s
it doesn't prove that a and 5 are not coprime.
However, it *does* prove that. Assume otherwise, that a is coprime to 5.
Then there exist algebraic integers u and v for which
a u + 5 v = 1
and the above factorizations show this:
rs u + qr v = 1
so
r( su + qv ) = 1,
and we deduce that r is a unit. However, it's not one of your fanciful
"Unit with no inverse" hallucinations. In fact, the above equation
provides us with an inverse for r. Based on its construction, it
*must be* an algebraic integer: u and v are specified to be algebraic
integers, and q,r,s were already specified as algebraic integers.
Thus, r must be an algebraic integer unit, in the ordinary sense:
an algebraic integer whose inverse
is *also* an algebraic integer.
However, I show below that the minimal polynomial for r precludes it
being a unit.
> Your algebraic manipulations will just keep running into that over and
> over again.
>
I've given the proof before, and it's only cut 'n' paste, so here it is,
once again:
Given:
q(x) = 8 x^2 - 76 x - 185
r(x) = 8 x^2 - 4 x - 45
s(x) = 4 x^2 - 37 x - 104
A quick bit of arithmetic will show the following:
q(x)*r(x) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8325
r(x)*s(x) = 32 x^4 - 312 x^3 - 864 x^2 + 2081 x + 4680
and:
(64 x + 128)*(x^3 - 12 x^2 + 65)
= 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320
(32 x + 72)*(x^3 - 12 x^2 + 65)
= 32 x^4 - 312 x^3 - 864 x^2 + 2080 x + 4680
and so, if you care to compare:
q(x)*r(x) = (64 x + 128)*p(x) + 5
r(x)*s(x) = (32 x + 72)*p(x) + x,
where p(x) = x^3 - 12 x^2 + 65. Note that this immediately shows
that for z = any root of p(x),
q(z)*r(z) = 5,
r(z)*s(z) = z.
To derive the minimal polynomial:
>> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
let mpr(x) = x^3 - 969 x^2 + 315 x + 5.
First, expand the polynomial mpr(r(x)):
mpr(r(x)) = (r(x))^3 - 969 (r(x))^2 + 315 (r(x)) + 5
= (8 x^2 - 4 x - 45)^3 - 969 (8 x^2 - 4 x - 45)^2
+ 315 (8 x^2 - 4 x - 45) + 5
= 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3
+ 731136 x^2 - 374400 x - 2067520
Next, multiply these two polynomials:
p(x) = x^3 - 12 x^2 + 65,
and
w(x) = 512 x^3 + 5376 x^2 - 5760 x - 31808
to get this:
p(x)*w(x) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3
+ 731136 x^2 - 374400 x - 2067520
Notice the equality
mpr(r(x)) = p(x)*w(x).
That means for every value of x, the polynomial you get by computing
r(x), then evaluating mpr(x) at that value, is equal to the product of
p(x) and w(x).
In short, if x is a root of p(x) (in particular, this is true for
x = -ai for any of your ai's),
mpr(r(x)) = 0.
Thus, the r's are roots of the polynomial
mpr(x) = x^3 - 969 x^2 + 315 x + 5
This cubic polynomial has integer coefficients, is monic,
and has no integer roots, thus it's irreducible. Its roots
cannot be units in the ordinary sense (i.e., with inverses
within the ring of algebraic integers), because the
constant term is a non-unit.
So, what can I conclude at this point?
1. each of the a's has an algebraic integer factor
in common with 5:
5 = q(a)*r(a)
a = r(a)*s(a).
where
q(a) = 8 a^2 - 76 a - 185
r(a) = 8 a^2 - 4 a - 45
s(a) = 4 a^2 - 37 a - 104
These values q(a),r(a),s(a) are all algebraic integers.
2. If a and 5 were coprime, then there would be an
inverse *in the ring of algebraic integers* for r(a).
3. The minimal polynomial mpr(x) for r(a) is:
mpr(x) = x^3 - 969 x^2 + 315 x + 5
>
>>>That's a sneaky little argument as it's getting the reader to assume
>>>something not proven, which is that 'a' actually has a non-unit
>>>algebraic integer factor because it's not a unit.
>>>
>>
>>This is actually *much* easier to prove: the ring of algebraic
>>integers is closed under the extraction of roots of monic
>>polynomials with algebraic integer coefficients. In particular,
>>every equation:
>>
>> x^n - a = 0
>
>
> That's not the equation you have.
>
Idiot! I'm showing that your alleged "something not proven", as I
quote:
... it's getting the reader to assume
something not proven, which is that 'a' actually has
a non-unit algebraic integer factor because it's not a
unit.
is hogwash!
EVERY NON-UNIT ALGEBRAIC INTEGER HAS NON-UNIT FACTORS
AMONG THE ALGEBRAIC INTEGERS!
I'll repeat that:
Take any algebraic integer K that is NOT a unit.
Then take the equation
x^n - K = 0.
and solve it. The roots are algebraic integers,
and they are ALSO non-unit factors of K.
What was it you claimed was unproven:
that 'a' actually has a non-unit algebraic integer
factor because it's not a unit.
What did I just find:
a non-unit algebraic integer factor K^(1/n), for
an ARBITRARY non-unit algebraic integer K.
Does that address this tiny little point?
Can we get off that one then? You're wrong, I'm right.
> What you have is
>
> x^3 - 969 x^2 + 315 x + 5
>
> and what you want to do is convince readers that none of its roots can
> be coprime to 5.
>
This is ridiculous. What was to be shown was this:
NONE OF THE COEFFICIENTS ai IN THE FACTORIZATION
65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
IS COPRIME TO 5.
I showed, for each a, two factorizations:
5 = q(a) r(a)
a = r(a) s(a).
The common factor r has minimal polynomial
x^3 - 969 x^2 + 315 x + 5
I don't have to show *ANYTHING* about that polynomial: its roots
are by definition algebraic integers, and the inverses of those
roots have minimal polynomial:
5 x^3 + 315 x^2 - 969 x + 1
and so none of them is a unit.
> But you cannot do that, and instead have maintained that proving that
> none of the roots can be a unit proves that one of them must have a
> non-unit factor in common with 5 in the ring of algebraic integers.
>
Idiot.
I have displayed the factor r(a), that you have
claimed does not exist.
That later polynomial is the minimal polynomial for the
common factor r(a), which shows that r(a) is not a unit.
Geez, it's just high school algebra, what's the confusion?
I have just exhibited the common factor. I have shown such a factor
for *each* of those roots. I have also shown the minimal polynomial
for those common factors, and used it to show that the common factor
is not a unit algebraic integer.
> I've just shot down that little trick, but you're squirming.
>
Please show me an error. You've had probably three months of looking
at the derivation, in one form or another, and *THAT* is all you can
come up with? I see that you could be confused, since all this is
just simple numbers, and anyone could be confused with numbers.
>
>>has algebraic integer roots, whenever "a" is an algebraic integer,
>>and n is a natural number.
>>
>>Thus, if "a" is an algebraic integer, and n is any natural number,
>>the number
>>
>> a^(1/n)
>>
>>is an algebraic integer, irrespective of which of the n roots you take.
>
>
> Duh, but you're not talking about such a simple expression as
>
> x^n - a = 0.
>
No, I was showing you that ANY non-unit algebraic integer has
non-unit algebraic integer factors, in fact will have infinitely
many of them.
> You're talking about
>
> x^3 - 969 x^2 + 315 x + 5 = 0
>
> and *repeatedly* posters like you try to fool readers on the
> newsgroups as you work to *convince* rather than get to the
> mathematical truth.
>
I have the truth up there. You are now acting to deny it.
That's OK, since we have been through YEARS of seeing you behave like
a little boy who has pooped his britches. "No, I don't have poop in
my britches, um, I grew a tail!"
> The truth is that you're relying on the negative, which is that NONE
> of the roots of that expression can be units, to try and prove the
> positive, which is that then they all have a non-unit factor in common
> with 5 in the ring of algebraic integers.
>
Hey, WHAT negative:
5 = q(a)*r(a)
a = r(a)*s(a).
Where's the negative? This is what I've shown, and you still can't see
through your own hatred of mathematicians to recognize that it plainly
shows your error.
> But that's beyond bogus as my point is that the ring of algebraic
> integers is screwed up to the extent that a root of that expression
> can be coprime to 5, and yet not be a unit.
>
Where did I claim something's not a unit? Not in the above pair of
expressions, in which the root is shown NOT to be coprime to 5.
No, it was in ANOTHER expression altogether:
x^3 - 969 x^2 + 315 x + 5
where I argue that none of the roots of THIS polynomial can be a unit.
... stuff deleted ...
>>
>>No gap.
>
>
> That's what you want to convince readers.
>
> However, simply asserting something that is not true does not fill the
> gap.
>
I made the following assertions. The proofs, up to verification of
a few polynomial expansions, can be seen above, but the interested
reader can produce these products explicitly, without any form of
influence on my part.
ASSERTION 1:
Given:
q(x) = 8 x^2 - 76 x - 185
r(x) = 8 x^2 - 4 x - 45
s(x) = 4 x^2 - 37 x - 104
we have
q(x)*r(x) = (64 x + 128)*p(x) + 5
r(x)*s(x) = (32 x + 72)*p(x) + x,
where p(x) = x^3 - 12 x^2 + 65.
Corollary: q(a)r(a) = 5,
r(a)s(a) = a
Therefore a and 5 have the common factor r(a) in the algebraic integers.
ASSERTION 2:
Given the polynomials
mpr(x) = x^3 - 969 x^2 + 315 x + 5
r(x) = 8 x^2 - 4 x - 45
p(x) = x^3 - 12 x^2 + 65.
w(x) = 512 x^3 + 5376 x^2 - 5760 x - 31808
we have:
mpr(r(x)) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3
+ 731136 x^2 - 374400 x - 2067520
and
p(x)w(x) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3
+ 731136 x^2 - 374400 x - 2067520
Therefore, for a = any root of p(x),
mpr(r(a)) = 0,
so mpr(x) is a polynomial for which r(a) is a root. mpr(x) is easily
seen to be irreducible, and so is the minimal polynomial of r(a).
Corollary: NONE OF THE FACTORS r(a) is a unit in the ring of algebraic
integers.
> Again, what you have is the negative--certain numbers aren't
> units--and you want the positive--that they all have to have non-unit
> factors in common with 5.
>
I have shown as much, but you refuse to look at it:
a and 5 have common factors in the ring of algebraic integers
at least one of those common factors (r(a)) is not a unit
So, you're confusing what has to be shown non coprime to 5, with what
has to be shown not to be a unit:
a is not coprime to 5
r(a) is not a unit.
> But you haven't proven the positive.
>
Of course I have, but you got your ears all plugged up
and refuse to listen.
> It's a gap.
>
> Denying it won't fill it, so quit being lazy and try to fill it.
>
You should say that to yourself once in a while.
>
>>>The gap is the assertion that given an algebraic integer 'a' which is
>>>not a unit, that 'a' must have some non-unit factor *in the ring of
>>>algebraic integers*.
>>>
>>
>>As I mentioned above, the factor "r" that I provided *is* an algebraic
>>integer, and in the case that "a" is (-1) times any of the ai's in the
>>factorization cited, I have provided its minimal polynomial. It is
>>simple to verify from that minimal polynomial that it's an algebraic
>>integer and that it's not a unit in the ring of algebraic integers.
>
>
> See readers? You can catch the trick here as notice the poster said
> "not a unit", and what I'm telling you is that these posters are
> working hard to convince YOU the reader.
>
What? Now you're claiming that I have attempted to fool some unnamed
class of readers by asserting something's not a unit?
Anyone can scroll back to verify *exactly* what I am claiming. I've
labelled it clearly.
>
>>>To protect readers from thinking it's simple, I remind them of the
>>>ring of evens, that is the set of even numbers, where you have 2 not a
>>>factor of 6, and not a unit. But 2 and 6 don't have factors in the
>>>ring.
>>>
>>>That's because the ring doesn't have 1.
>>>
>>>You have to go to a higher ring, in this case the ring of integers, to
>>>escape that interesting little situation.
>>>
>>>Now the problem is different in the ring of algebraic integers, but
>>>the ring is screwed up, and the fix is again a higher ring.
>>>
>>>What I want you to consider is the *possibility* that weird things can
>>>happen with rings, when you're not careful enough, and that
>>>mathematicians weren't careful enough with the ring of algebraic
>>>integers.
>>>
>>
>>The problem is with your ignorance. The ring of algebraic integers has
>>none of the shortcomings you seem to want it to have. It is your method
>>which is at fault, since it forces you to believe a falsehood, namely
>>that the ai's in that above factorization are coprime to 5.
>
>
> Now this poster is the one who tried to fool all of you, and now that
> I've caught him, he decides to hurl insults.
>
Please stop mischaracterizing a proof as an attempt at fraud.
Just because YOU couldn't follow the fairly simple arithmetic I've
provided, and just because it showed that everything you've been
doing for, what is it, 8 years now(?), has been for no good reason
at all?
You have perpetually misunderstood the mathematics you're attempting
to work with, and have, to date, NEVER accepted even the smallest
bit of responsibility for knowing what it is you're talking about.
It is not an insult to say that the problem with all this is that you
are ignorant of the subject. That is the simple truth.
Have I insulted you?
How?
By saying "you're ignorant of algebra"?
Can you demonstrate that you know some nontrivial algebraic topic, say
Galois theory?
What about any of the [other] standard theorems about fields and their
extensions?
To claim that you have "caught" me, you should at least provide some
statement that you take issue with. All I saw above was your claim that
I was making the argument:
"If 'a' isn't a unit, then 'a' can't be coprime to 5.
where my argument was this:
A. Here are two products: a,q,r,s are all algebraic integers:
5 = q*r
a = r*s
Then, unless 5 is a unit, then a and 5 have a non-unit
common factor in the ring of algebraic integers
B. Here is the minimal polynomial for r(a).
mpr(x) = x^3 - 969 x^2 + 315 x + 5
Thus, r(a) is not a unit
C. Thus, a and 5 share a non-unit algebraic integer factor.
> These posters are intellectually lazy, and they're rude.
>
Um, intellectually lazy? I think you just have a little file of all the
epithets that have come your way, and choose one that seems to fit.
Unfortunately, I have no grounds for claiming not to have been
intellectually lazy. Oh, I did find those factors you said were
impossible, didn't I? And that other cubic a couple of weeks ago,
where you said that no one could provide the monic polynomial for
which something was a root. I gave you that polynomial.
So, I guess that I'm intellectually lazy, yet can do what mankind
cannot possibly do, by your own estimation!
I'm SUPERMAN!!!! WOO HOOOO!!!
Mere mortals, step aside! I can solve your problems, even those
that mankind cannot possibly do, and not break a mental sweat.
SOOOO PPERRRRRR...MAAAAAAANNNNNNN!!!!!!
Yesssssssssssssssssss!!!!!!!!!!!
I gotta get one of those cape thingies, some blue tights, with the
red codpiece, er, maybe I can be Clark Kent? I never did like
those superhero gymnasts' costumes, capes, tiaras, big S on the
shirt. It always seemed a bit, er, sissy for my tastes.
Well, I have time to reflect on that one for awhile.
I have crushed your argument. It has nowhere to go, now that we can
all see how it predicts stuff that is plain wrong. It's fine for
symbols, you can push little a's and b's and u's and v's around
on your paper all day long, and once you come to plugging real
values in for the variables it says that the a's are coprime to
5. That's got to be embarassing.
Yeah, there's no insult on your side. At least I've been factually
correct: you ARE ignorant, and parade that ignorance around day after
day. In addition, I haven't been running around proclaiming to have
shown something to be true, all the while ignoring a direct calculation
that shows it to be false.
Who has been doing that?
Now that you're pretending to address the issue (since it won't go
away), who is it who is continuing to mis-label every thing, out of
some vain belief that if it fools the old JSHter, it'll fool someone
else on earth.
I think for these purposes, you will just have to settle on being
THE MOST GULLIBLE MAN ON EARTH.
No one else will ever believe you, and you will end up at the age of
50 a broken little man who couldn't let it go when it was clear he
was done.
> But they're not lazy when it comes to posting as they *keep* posting
> now don't they?
>
Let's do a post count for me vs. you
(source: Google, 1/1/2000 - 8/11/2003, sci.math only)
WDH: 693
JSH: 4180
if I count JSH-related posts, I have 322. About half
of what I contribute to sci.math goes to JSH nonsense.
Already I'm ashamed of myself. Boy, I'm just a-burnin'
up that internet, don't you think?
Next baseless accusation?
> I'm curious about what you, the reader, thinks is the reason for their
> persistence.
>
Why do YOU think I keep posting?
I've already given my reasons:
you are a punk
you leech of the help of others and treat your
benefactors shamefully.
you have an inflated sense of your own worth,
originality, talent, intelligence
you behave as though yours is the only
thought that's worth having
you bully others
you clog the newsgroup with ill will,
and lead other discussions into the ground
you despise what it is that I have spent
the better part of my life doing.
you cast aspersions at honorable people,
and make idle threats
you are a cynical abuser of the very
concept of intellectual debate,
you are a crybaby,
your useage of the English language is
substandard, even for a southerner.
you pretend to lecture those who are
your betters (in terms of comprehension,
intuition, and experience).
> Why do YOU think they keep posting?
>
>
> James Harris
Dale
C. Bond
[snip hysterical rant]
Is your nose starting to grow? If you had any credibility left, which you don't, your own posts over the
last few days would have gone a long way toward demolishing it. You continue to ignore those posts from
others which disprove your claim that the ring of algebraic integers is flawed. The flaw is entirely
between your ears.
Just to recap (for the record): Every root of a monic polynomial with integer coefficients divides the
constant term within the ring of algebraic integers. That is, the quotient obtained by dividing the
constant term by any root is an algebraic integer.
The above is easy to prove, and has been shown repeatedly by more than one poster, but so far seems
beyond your comprehension. Furthermore, you have not shown a single example of a number which *should be*
in the ring of algebraic integers, but which is *left out*. Tsk, tsk. That doesn't look good for you,
James. You have claimed repeatedly that such numbers exist, but you have never produced a single one!
At this point, what you have claimed is a 'linchpin of my FLT proof' has become a 'nail in your FLT
proof's coffin'.
KRamsay
(James Harris) writes:
|Well if you're telling the truth, fill in the gap I've pointed out in
|YOUR argument.
|
|The gap is the assertion that given an algebraic integer 'a' which is
|not a unit, that 'a' must have some non-unit factor *in the ring of
|algebraic integers*.
One obvious answer is that 'a' is a factor of itself.
If you mean nontrivial factor, then the square roots of 'a' are
non-unit factors of 'a'. If 'a' is an algebraic integer, then it's
a root of a monic polynomial with integer coefficients
x^n + c_1*x^{n-1} + ... + c_n = 0.
If b^2=a, then b is a root of the monic polynomial with integer
coefficients
x^{2n} + c_1*x^{2n-2} + ... + c_n = 0.
We know b is not a unit in the algebraic integers, because if b
were a unit then by definition there would be some algebraic
integer c such that bc=1. But then we would have 1=(bc)^2
=b^2c^2 = ac^2. Because c is an algebraic integer, then so is
c^2. That would imply a is a unit, but we assumed to start with
that a was not a unit.
If you want an explicit proof that if c is an algebraic integer then
so is c^2: suppose c is the root of a polynomial
x^m + d_1*x^{m-1} + ... + d_m = 0
with integer coefficients. Then we can group the even terms on one
side and the odd terms on the other:
d_m + d_{m-2}*x^2 + ... = -(d_{m-1} + d_{m-3}*x^2 + ...)*x.
Square both sides:
(d_m+d_{m-2}*x^2+...)^2 = (d_{m-1}+d_{m-3}*x^2+...)*x^2.
This is now a polynomial in x^2. Substituting y for x^2 we get
a monic polynomial with integer with integer coefficients which
has c^2 as a root.
|To protect readers from thinking it's simple, I remind them of the
|ring of evens, that is the set of even numbers, where you have 2 not a
|factor of 6, and not a unit. But 2 and 6 don't have factors in the
|ring.
|
|That's because the ring doesn't have 1.
Before reading this past year's discussion, I hadn't realized what
the definition of "coprime" was in such cases. So you sometimes
learn something new. (Not that this definition was ever important
to anything I had to do.)
|You have to go to a higher ring, in this case the ring of integers, to
|escape that interesting little situation.
|
|Now the problem is different in the ring of algebraic integers, but
|the ring is screwed up, and the fix is again a higher ring.
|
|What I want you to consider is the *possibility* that weird things can
|happen with rings, when you're not careful enough, and that
|mathematicians weren't careful enough with the ring of algebraic
|integers.
I'll do that if you agree to do something much easier. Consider the
possibility that "weird things" are causing the common factors which
you believe to be independent of m to depend upon m, but that
you weren't careful enough.
Keith Ramsay
Jesse F. Hughes
> your useage of the English language is
> substandard, even for a southerner.
Don't be an ass.
At least, not in the same sentence in which you misspell usage and
fail to capitalize "Southerner".
(Now, where's my requisite typo in this spelling flame?)
--
Jesse Hughes
"[I]t's the damndest thing. There's something wrong with every last
one of you, and I *never* thought that was a possibility. But now I
feel it's the only reasonable conclusion." --JSH sees some sorta light
Jesse F. Hughes
> "W. Dale Hall" <mailt...@farir.com> writes:
>
>> your useage of the English language is
>> substandard, even for a southerner.
>
> Don't be an ass.
>
> At least, not in the same sentence in which you misspell usage and
> fail to capitalize "Southerner".
>
> (Now, where's my requisite typo in this spelling flame?)
Oh, I see it. No quotes around "usage".
How come I can't see it *before* the post?
--
"If you *still* believe that [my proof is wrong], then I have to think
that your mind is limited [...], and it may be the case that not
everyone *can* achieve that, as the mental wiring may not be there for
the task." -- James Harris, on faculties needed to accept his proof.
Arturo Magidin
>> In article <3c65f87.03081...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>> >"W. Dale Hall" <mailt...@farir.com> wrote in message news:<3F37D13...@farir.com>...
>
>
>Now what are the actual expressions?
>
>They are
>
> q(a) r(a) = 5
> r(a) s(a) = a.
Which means that r(a) is a nonunit common factor of 5 and a.
Which is exactly what you claimed was impossible to exhibit.
That's it.
>where I simply note that r(a) may not have non-unit factors in common
>with 5,
r(a) IS a non-unit factor OF 5.
>which *should* make q(a) have a factor of 5 ***in the ring of
>algebraic integers***,
q(a) IS a non-unit factor of 5.
> but the ring of algebraic integers is screwed
>up.
Nonsense.
>In trying to prove that r(a) must have a non-unit factor in common
>with 5, you rely on the fact that it can't be a unit ***in the ring of
>algebraic integers***!!!
r(a) IS a nonunit, and a factor of 5: there it is above. There is
something that multiplied by r(a) equals 5. That's the DEFINITION of
being a factor. To do that, I do not use that r(a) is not a unit in
the ring.
Only AFTER proving that r(a) is a common factor of 5 and a do we show
that it is not a unit.
>The trick is to act as if proving the negative, proves that it must
>have some non-unit factor in common with 5, but there's a fascinating
>error with the ring of algebraic integers.
Apparently, the trick is to confuse yourself to a point where you
don't know what you are saying... At least, that seems to be what you
are doing.
>>
>> (2) There is an algebraic integer, called s(a), such that
>> r(a)*s(a)=a. So r(a) is a factor of a in the ring of algebraic
>> integers.
>
>Well, what if 'a' doesn't share non-unit factors in common with 5?
Are you blind? r(a) is a factor of a, and it is a factor of 5, and it
is not a unit. There is no "if".
>Then both r(a) and s(a) don't either.
r(a) is a factor of a, it is a factor of 5, and it is not a unit.
>That *should* leave q(a) with a factor of 5, but the ring is screwed
>up.
r(a) is a factor of 5, it is a factor of 5, and it is not a unit.
>In order to try and prove that it does have a factor of 5, you have to
>try and claim that it must because otherwise r(a) would be a unit.
Nonsense. I have shown that r(a) is a factor of a; that it is a factor
of 5, and that it is not a unit. That is the supposed thing that
nobody could do: exhibit an algebraic integer which is a factor of 5,
a factor of a, and not a unit.
>But the ring of algebraic integers is screwed up, which is my point.
>
>> (3) There is an algebraic integer, called q(a), such that
>> r(a)*q(a)=5. So r(a) is a factor of 5 in the ring of algebraic
>> integers.
>
>However, how do you know that r(a) must share non-unit factors in
>common with 5?
r(a) ITSELF is a nonunit factor of 5.
>Ultimately your claim must be that it's because it's not a unit in the
>ring of algebraic integers!!!
r(a) is a factor of 5, and it is not a unit. You are delusional.
>> So: r(a) is (1) not a unit; (2) a factor of a; and (3) a factor of 5.
>>
>> Why is it not the case that r(a) is a non-unit algebraic integer which
>> is a common factor of a and 5?
>
>For readers, these posters are working to *convince* not get to the
>bottom of the problem.
Answer the question: r(a) is (1) not a unit; (2) a factor of a; and
(3) a factor of 5.
Why is it not the case that r(a) is a non-unit algebraic integer which
is a common factor of a and 5?
>If they cared about the truth, then they only need consider my proof
>of the problem with algebraic integers.
Answer the question: r(a) is (1) not a unit; (2) a factor of a; and
(3) a factor of 5.
Why is it not the case that r(a) is a non-unit algebraic integer which
is a common factor of a and 5?
[.snip.]
>> [.snip.]
>>
>>
>> >What you have is
>> >
>> > x^3 - 969 x^2 + 315 x + 5
>> >
>> >and what you want to do is convince readers that none of its roots can
>> >be coprime to 5.
>>
>> There is no "convincing" that needs to be done. NONE Of the roots are
>> units, and ALL the roots are factors of 5. You're done.
>
>No, you have a gap. The gap is that now you need to prove that the
>roots share non-unit factors in common with 5 in the ring of algebraic
>integers.
The roots ARE non-unit factors of 5 in the ring of algebraic
integers. There is no gap.
>> >But you cannot do that,
>>
>> Correction: James Harris cannot do that because he does not know what
>> "coprime" means.
>
>Which is the semantic argument.
Do you know what semantic means? I don't think so.
>> > and instead have maintained that proving that
>> >none of the roots can be a unit proves that one of them must have a
>> >non-unit factor in common with 5 in the ring of algebraic integers.
>>
>> They are each FACTORS of 5, and EACH non-units.
>
>Which does NOT prove that they each have a non-unit factor in common
>with 5, because the ring of algebraic integers is flawed.
(1) You are assuming what you are claiming.
(2) Yes, it DOES prove that they each have a non-unit factor in common
with 5: THEMSELVES.
>> Do you agree or disagree with (all happening in a commutative ring
>> WITH 1):
>>
>> (a) If a is a factor of b, then a and b have a common factor
>> (namely, a).
>>
>> (b) If a is a factor of b and is not a unit, then a is a non-unit
>> common factor of a and b.
>>
>> (c) If is a factor of b and is not a unit, then a and b have a
>> non-unit common factor (namely, a).
>
>Hmmm...I can see that you're dedicated at working to convince the
>audience, so I'll give a demonstration to show them *how* you're
>working.
Answer the question. Do you agree or disagree with (a)? With (b)? With
(c)?
You are trying really hard to avoid any direct question, with specific
numbers, and instead rely on confusion. Why?
>Consider abc = 5, where a = sqrt(5), b=sqrt(5)(-1+sqrt(-3))/2,
>c=(-1-sqrt(-3))/2, but imagine that you're talking to someone who
>doesn't know about radicals, as all they know about are integers.
>
>Now neither 'a' nor 'b' is an integer, but this person is using d=ab,
>and they now see
>
> cd = 5
>
>and tell you that 'c' and 'd' must be factors of 5 ***in the ring of
>integers***.
>
>You say nope.
That would be because neither of them are integers.
>Well they come back, and argue, and point out that neither is a unit
>***in the ring of integers***.
>
>You say, yup, that is correct.
>
>Then they come back and say, well, cd = 5, neither is a unit in the
>ring, so they must be factors of 5.
>
>And you say, nope, they are not.
This example is completely inapplicable. Because we agree that:
r(a), s(a), and q(a) are ALL algebraic integers (which was not the
case in your example);
r(a) is an algebraic integer which is not a unit in the ring of
algebraic integers (which was not the case in your example);
r(a)*q(a)=5 and r(a)*s(a)=a.
And so, you agree that r(a) is an algebraic integer factor of 5 in the
ring of algebraic integers; that it is an algebraic integer factor of
a in the ring of algebraic integers; and that it is an algebraic
integer which is not a unit in the ring of algebraic integers.
So, why is it not the case that r(a) is an algebraic integer common
factor of 5 and a which is not a unit?
No more nonsense about the ring being "screwed up." Why is it not a
nonunit common factor?
>But the ring of algebraic integers is VERY screwed up, so that only
>gives you an idea.
No, you are screwed up. Nothing you have said even points to any
problems in the ring of algebraic integers, only in your understanding
of what a ring is, what "factor" means, and what "unit" means.
[.snip.]
>> >The truth is that you're relying on the negative, which is that NONE
>> >of the roots of that expression can be units, to try and prove the
>> >positive, which is that then they all have a non-unit factor in common
>> >with 5 in the ring of algebraic integers.
>>
>> You are completely, totally, utterly lost and confused.
>>
>> The roots themselves are factors of 5. Each root is a factor of
>> itself, and a factor of 5. Therefore, each root is a common factor of
>> itself and 5. And since each of them is not a unit, then we have
>> EXHIBITED a non-unit common factor of each with 5. The roots are not
>> the original numbers we were interested in (the a's), they are the
>> common factors that have been produced.
>>
>> [.snip.]
>
>And again the trick is that use of the word "unit" as in fact the
>factors are NOT units ***in the ring of algebraic integers***.
This is again nonsense.
>In a higher, more complete ring, one of them IS a unit, while two of
>them have a factor that is sqrt(5).
You have never defined "complete", but never mind.
Yes, there are plenty of LARGER rings where one is a unit and two have
a factor of sqrt(5). One example of such is Z[1/a1, 1/a2, 1/a3].
But you have still not answered the question:
Since r(a) is a factor of 5, a factor of a, and not a unit, why is it
not a nonunit common factor of 5 and a?
[.snip.]
>For readers who don't realize how this ring business can be confusing,
>consider my example with the ring of evens. That ring doesn't have 1
>in it, as 1 is odd, so 2 and 6 do not share factors with each other.
The example is inapplicable. The ring of algebraic integers DOES have
a 1, so every element is a factor of itself.
[.snip.]
>Rather than be fascinated by it, and accept the mathematics, you can
>see posters like Arturo Magidin instead working to convince YOU.
>
>Why does he have to keep replying to me in posts?
I realize that you have absolutely no inkling about what could drive a
man of principle, but that is exactly why I reply to you: because I
have principles. And one of them is not to let mathematical nonsense
stand unchallenged when it can be refuted.
>Because I'm right. If he goes away, and isn't around to confuse
>readers, he probably realizes that eventually I might convince some of
>you to follow the mathematical logic.
I realize that someone like you cannot think of any other reason,
because I am not gaining fame, money, or women, by my actions, and
that is the stick by which you measure everything. Nonetheless, that
is not the case.
>That explains why posters like Arturo Magidin are so dedicated.
In you fantasy world only.
C. Bond
[snip]
> > >What you have is
> > >
> > > x^3 - 969 x^2 + 315 x + 5
> > >
> > >and what you want to do is convince readers that none of its roots can
> > >be coprime to 5.
> >
> > There is no "convincing" that needs to be done. NONE Of the roots are
> > units, and ALL the roots are factors of 5. You're done.
>
> No, you have a gap. The gap is that now you need to prove that the
> roots share non-unit factors in common with 5 in the ring of algebraic
> integers.
James, it has been pointed out many times that if 'a' divides 5 in the ring of algebraic integers (as do
all roots of monic polynomials with integer coefficients and constant term +/- 5), then the square roots,
cube roots, etc. of 'a' are non-unit algebraic integers which also divide 5 and clearly also divide 'a'.
Hence, if 'a' divides 5 in the ring of algebraic integers, then there are an *infinite* number of
algebraic integers which are factors in common with 'a' and 5. This is elementary algebra. Is it beyond
you? Can't you see that your conclusion is false and that the error is in *your* argument?
W. Dale Hall
James Harris wrote:
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bh94nn$27q9$1...@agate.berkeley.edu>...
>
... stuff deleted ...
>>>>
>>>> q(a) r(a) = 5
>>>> r(a) s(a) = a.
>>>
... stuff deleted ...
>>>>The above facts prove that this "a" has a non-unit algebraic integer
>>>>as a factor.
>>>
>>>No it doesn't.
>>
>>Why not? There is an algebraic integer, called here r(a), with the
>>following three properties:
>>
>> (1) It is not a unit in the ring of all algebraic integers; you
>> apparently agree with that conclusion, but then engage in
>> sophistry about a supposed "problem" with it.
>
>
> It is true that r(a) is not a unit in the ring of algebraic integers.
>
> Now what are the actual expressions?
>
> They are
>
> q(a) r(a) = 5
> r(a) s(a) = a.
>
> where I simply note that r(a) may not have non-unit factors in common
> with 5, which *should* make q(a) have a factor of 5 ***in the ring of
> algebraic integers***, but the ring of algebraic integers is screwed
> up.
Apparently, you have some inkling of what it means for two numbers to
share a factor. Allow me to propose a definition:
I'll assume I'm dealing solely with integral domains for
the present discussion.
In the ring R, an element s is called a factor of the element r
if there is an element u in R, for which
r = su
The element s is a *common factor* of the elements r and r',
if there are elements u and u' of R, for which
r = s, and r' = su'
Would you agree with these definitions, at least to the point that you
understand them?
I'll suppose so, and continue.
I have shown, and you appear to accept, these several facts:
r(a) is a factor of 5.
r(a) is a factor of a.
r(a) is a non-unit algebraic integer.
Now, you are claiming that the equations:
q r = 5
r s = a
may occur with r having *no* non-unit common factors with 5
in the ring of algebraic integers.
I claim that the number r *itself* serves as a non-unit factor
that r has in common with 5. It also serves as a non-unit factor
that a has in common with 5. This latter was the point of the
exercise.
Proof of claim:
The (above) definition of common factor:
s is a common factor of r and r' if there are
elements u, u' in R for which the following
hold:
r = s u
r' = s u'
In our case, the claim is that r(a) is a
common factor of r(a) and 5. By definition,
we need to find the algebraic integers u
and u' so that
r(a) = r(a)u
5 = r(a)u'
However, if u = 1, and u' = s(a), we have this:
r(a) = r(a)*1
5 = r(a)*s(a).
we already know that 1 and s(a) are algebraic integers,
so the result follows: r(a) is a common factor of
r(a) and 5.
>
> In trying to prove that r(a) must have a non-unit factor in common
> with 5, you rely on the fact that it can't be a unit ***in the ring of
> algebraic integers***!!!
>
Well, I've shown that r(a) is not a unit, and you have already agreed
to that. What's the problem?
> The trick is to act as if proving the negative, proves that it must
> have some non-unit factor in common with 5, but there's a fascinating
> error with the ring of algebraic integers.
>
That is your claim. You have not provided any evidence, and so may
not use it as a *basis* for your argument.
>
>> (2) There is an algebraic integer, called s(a), such that
>> r(a)*s(a)=a. So r(a) is a factor of a in the ring of algebraic
>> integers.
>
>
> Well, what if 'a' doesn't share non-unit factors in common with 5?
> Then both r(a) and s(a) don't either.
>
What? What do you think r(a) is? It's an algebraic integer, but not
a unit! Well, another proof:
r(a) is a common factor of a and 5 in the ring of
algebraic integers.
The above definition
s is a factor of r and r' if there exist
elements u and u' for which the following
equations are satisfied:
r = s u
r'= s u'
In the present case, we need to find u and u'
for which
a = r(a) u
5 = r(a) u'
but we have the values q(a), s(a), and know
that
a = r(a) s(a)
5 = r(a) q(a).
Thus r(a) is a common factor of a and 5 in the
ring of algebraic integers. Since r(a) is
not a unit, r(a) is a *non-unit* factor that
a and 5 share in the ring of algebraic integers.
> That *should* leave q(a) with a factor of 5, but the ring is screwed
> up.
>
No, r(a) is the common factor. You're just throwing out meaningless
objections to cover the fact that your argument is vacuous.
> In order to try and prove that it does have a factor of 5, you have to
> try and claim that it must because otherwise r(a) would be a unit.
>
NO ONE says that it has a factor of 5. 5 does NOT divide that number,
rather r(a) divides 5. Get the division relation straight, or at least
the language of what's a factor of what, before you start in making
such irrelevant claims.
> But the ring of algebraic integers is screwed up, which is my point.
>
Again, your argument is that my above derivations are incorrect due to
this "screwed up" nature of the algebraic integers. However, that's what
you are attempting to prove! It is not legitimate to assume the truth of
one's conclusion in the course of proof.
That is, you may not refer to this "defect" of the ring of algebraic
integers as a justification, during any argument that is attempting
to establish the existence of such a defect.
Further, I have *shown* the common factor, you have *agreed* that it
is a common factor, and you have also *agreed* that the common factor
is (1) an algebraic integer, and (2) not a unit.
In short, you agree to the particulars of the argument, yet you deny
the conclusion of the argument. Please explain how that tactic is a
legitimate one.
>
>> (3) There is an algebraic integer, called q(a), such that
>> r(a)*q(a)=5. So r(a) is a factor of 5 in the ring of algebraic
>> integers.
>
>
> However, how do you know that r(a) must share non-unit factors in
> common with 5?
>
I showed you.
> Ultimately your claim must be that it's because it's not a unit in the
> ring of algebraic integers!!!
>
So what? I showed that also.
>
>>So: r(a) is (1) not a unit; (2) a factor of a; and (3) a factor of 5.
>>
>>Why is it not the case that r(a) is a non-unit algebraic integer which
>>is a common factor of a and 5?
>
>
> For readers, these posters are working to *convince* not get to the
> bottom of the problem.
>
No, you refuse to admit to the conclusion of the argument, while
agreeing to the hypotheses as well as the logic of the argument.
You are abusing the process of argument.
> If they cared about the truth, then they only need consider my proof
> of the problem with algebraic integers.
>
Plus, you cast aspersions as to the ethics of those who argue against
you. If you had a mathematical argument, you wouldn't do that.
> Proofs don't duel. If they're correct, then they'd be able to find an
> error with my argument, but it has no error as the ring of algebraic
> integers IS screwed up, so instead they work to convince YOU the
> reader.
>
I have found the arithmetic that you say does not exist. Perhaps you
aren't aware, but it is completely standard in rebuttal of arguments
to display that the argument produces erroneous results.
I have shown that your claim (that the a's and 5 are coprime) is
incorrect. You, on the other hand, need to deny the facts, by denying
the fact that a divisor that is not a unit must then be a non-unit
divisor. The very nature of the contortions you apparently need to
go through exposes the poverty of your case.
>
>> [.snip.]
>>
>>
>>
>>>What you have is
>>>
>>> x^3 - 969 x^2 + 315 x + 5
>>>
>>>and what you want to do is convince readers that none of its roots can
>>>be coprime to 5.
>>
>>There is no "convincing" that needs to be done. NONE Of the roots are
>>units, and ALL the roots are factors of 5. You're done.
>
>
> No, you have a gap. The gap is that now you need to prove that the
> roots share non-unit factors in common with 5 in the ring of algebraic
> integers.
>
Duh.
5 = q(a) r(a)
a = r(a) s(a).
It's the polynomial
x^3 - 12 x^2 + 65
for which roots we are to demonstrate commonality with 5, not the
cubic you're displaying here. That cubic is used to prove that r(a)
is not a unit. However, you've seized on the word "unit" and can't
quite grasp what relation that has with the original problem.
>
>>>But you cannot do that,
>>
>>Correction: James Harris cannot do that because he does not know what
>>"coprime" means.
>
>
> Which is the semantic argument.
>
Right. In mathematics, it is not incorrect to call into play the very
definitions of what you're talking about. That way, a person can stay
on track, rather than meandering all over the place.
Put another way, the definition (which you've always pooh-poohed as
being some form of chicanery or circular argument) is an essential
ingredient in any proof. The definition tells what it *is* that you're
talking about, and a person ignores definitions at his peril.
>
>>>and instead have maintained that proving that
>>>none of the roots can be a unit proves that one of them must have a
>>>non-unit factor in common with 5 in the ring of algebraic integers.
>>
>>They are each FACTORS of 5, and EACH non-units.
>
>
> Which does NOT prove that they each have a non-unit factor in common
> with 5, because the ring of algebraic integers is flawed.
>
No, that is just what it does prove, and your use of your intended goal
"the ring of algebraic integers is flawed"
as a justification (i.e., "... because the ring is flawed" ) is an
illegitimate tactic: you assume what has not been proven.
>
>>Do you agree or disagree with (all happening in a commutative ring
>>WITH 1):
>>
>> (a) If a is a factor of b, then a and b have a common factor
>> (namely, a).
>>
>> (b) If a is a factor of b and is not a unit, then a is a non-unit
>> common factor of a and b.
>>
>> (c) If is a factor of b and is not a unit, then a and b have a
>> non-unit common factor (namely, a).
>
>
> Hmmm...I can see that you're dedicated at working to convince the
> audience, so I'll give a demonstration to show them *how* you're
> working.
>
> Consider abc = 5, where a = sqrt(5), b=sqrt(5)(-1+sqrt(-3))/2,
> c=(-1-sqrt(-3))/2, but imagine that you're talking to someone who
> doesn't know about radicals, as all they know about are integers.
>
> Now neither 'a' nor 'b' is an integer, but this person is using d=ab,
> and they now see
>
> cd = 5
>
> and tell you that 'c' and 'd' must be factors of 5 ***in the ring of
> integers***.
>
> You say nope.
This is a false comparison. Note how none of the numbers in your example
are actually in the ring you're discussing.
Are you claiming that none of the numbers q(a), r(a), and s(a) is an
algebraic integer? I didn't notice you doing so before.
>
> Well they come back, and argue, and point out that neither is a unit
> ***in the ring of integers***.
>
> You say, yup, that is correct.
>
> Then they come back and say, well, cd = 5, neither is a unit in the
> ring, so they must be factors of 5.
>
> And you say, nope, they are not.
>
> But the ring of algebraic integers is VERY screwed up, so that only
> gives you an idea.
>
Again, it's your belief of this "screwed up" business that is at the
same time your intended goal AND your justification for denying these
straightforward arguments.
>
>>>I've just shot down that little trick, but you're squirming.
>>
>>I do see a lot of squirming, but it's not coming from Dale.
>
>
> What mathematicians can do, because it's such an odd error, is keep
> casting doubt, and running away from the proof of the error.
>
You haven't found an error. Rather, you are straining at the concept
that a number could be a factor of itself.
> Probably for most of you the idea that you could have abc=5 where
> neither 'a', 'b', nor 'c' is a factor of 5, in the ring of algebraic
> integers, seems nonsensical.
>
Um, I showed that q(a), r(a), and s(a) are algebraic integers, unlike
your strawman argument where NONE of the elements belonged to the ring
under discussion. Not a legitimate comparison, unless you're claiming
that q,r,s are NOT in the ring.
> However, it's an esoteric problem in an esoteric branch of
> mathematics, which has been there for over a *hundred* years.
>
Yeah, back to the *esoteric* nature of arithmetic. Multiply the
polynomials, compare coefficients. Anyone can see I told the truth.
You still want to deny it.
> If mathematicians want to confuse most of you about it, they can.
>
Right. Pointing out SPECIFIC polynomials, providing the manipulations
to do, THAT is confusing, but talking about UBER polynomials, that is
just regular stuff.
>
>> [.snip.]
>>
>>
>>>The truth is that you're relying on the negative, which is that NONE
>>>of the roots of that expression can be units, to try and prove the
>>>positive, which is that then they all have a non-unit factor in common
>>>with 5 in the ring of algebraic integers.
>>
>>You are completely, totally, utterly lost and confused.
>>
>>The roots themselves are factors of 5. Each root is a factor of
>>itself, and a factor of 5. Therefore, each root is a common factor of
>>itself and 5. And since each of them is not a unit, then we have
>>EXHIBITED a non-unit common factor of each with 5. The roots are not
>>the original numbers we were interested in (the a's), they are the
>>common factors that have been produced.
>>
>> [.snip.]
>
>
> And again the trick is that use of the word "unit" as in fact the
> factors are NOT units ***in the ring of algebraic integers***.
>
Yes, apparently it has you confused.
> In a higher, more complete ring, one of them IS a unit, while two of
> them have a factor that is sqrt(5).
>
Oh, now I need to go to a ring where one of the roots is invertible?
Why? I found common factors without inverting any of the roots.
You have not proven what you claim, and are struggling in vain to
give it some sense.
>
>>>>As I mentioned above, the factor "r" that I provided *is* an algebraic
>>>>integer, and in the case that "a" is (-1) times any of the ai's in the
>>>>factorization cited, I have provided its minimal polynomial. It is
>>>>simple to verify from that minimal polynomial that it's an algebraic
>>>>integer and that it's not a unit in the ring of algebraic integers.
>>>
>>>See readers? You can catch the trick here as notice the poster said
>>>"not a unit", and what I'm telling you is that these posters are
>>>working hard to convince YOU the reader.
>>
>>Are you claiming that "non-unit" and "not a unit" are not the same
>>thing?
>>
>> [.snip.]
>
>
> Nope.
>
> For readers who don't realize how this ring business can be confusing,
> consider my example with the ring of evens. That ring doesn't have 1
> in it, as 1 is odd, so 2 and 6 do not share factors with each other.
>
> However, you go to the higher ring--integers--and they do.
>
Oh, where some element of 2Z becomes invertible? I don't think so.
> Mathematicians have this problem where they didn't think there was
> one, until I pushed it into the open.
>
Baloney
> Rather than be fascinated by it, and accept the mathematics, you can
> see posters like Arturo Magidin instead working to convince YOU.
>
"Fascination" does not do mathematics. It's a child looking at a
bubble, thinking it's magic. Mathematics must be built, and you
are not doing any building.
> Why does he have to keep replying to me in posts?
>
Oh, it's back to the "heads I win, tails you lose" argument:
Dig this, folks:
1. Magidin responds to JSH, and that means:
> Because I'm right. If he goes away, and isn't around to confuse
> readers, he probably realizes that eventually I might convince some of
> you to follow the mathematical logic.
>
> That explains why posters like Arturo Magidin are so dedicated.
>
> They need to hang around to keep YOU confused.
>
2. Magidin doesn't respond, and that means:
"See, I won!"
>
> James Harris
JSH is back to his denial of direct arguments, accusing his critics of
dishonesty, fraud, and intent to confuse the reader. No one can utter
a contrary argument to JSH, unless he is evil.
Dale.
Arturo Magidin
W. Dale Hall <wd_...@pacbell.net> wrote:
[.snip.]
[.newsgroups trimmed.]
>>>Why not? There is an algebraic integer, called here r(a), with the
>>>following three properties:
>>>
>>> (1) It is not a unit in the ring of all algebraic integers; you
>>> apparently agree with that conclusion, but then engage in
>>> sophistry about a supposed "problem" with it.
>>
>>
>> It is true that r(a) is not a unit in the ring of algebraic integers.
>>
>> Now what are the actual expressions?
>>
>> They are
>>
>> q(a) r(a) = 5
>> r(a) s(a) = a.
>>
>> where I simply note that r(a) may not have non-unit factors in common
>> with 5, which *should* make q(a) have a factor of 5 ***in the ring of
>> algebraic integers***, but the ring of algebraic integers is screwed
>> up.
>
>Apparently, you have some inkling of what it means for two numbers to
>share a factor. Allow me to propose a definition:
>
>I'll assume I'm dealing solely with integral domains for
>the present discussion.
>
> In the ring R, an element s is called a factor of the element r
> if there is an element u in R, for which
>
> r = su
>
> The element s is a *common factor* of the elements r and r',
> if there are elements u and u' of R, for which
>
> r = s, and r' = su'
Presumably, you meant "r=su".
>Would you agree with these definitions, at least to the point that you
>understand them?
>
>I'll suppose so, and continue.
>
>I have shown, and you appear to accept, these several facts:
>
> r(a) is a factor of 5.
> r(a) is a factor of a.
> r(a) is a non-unit algebraic integer.
>
>Now, you are claiming that the equations:
>
> q r = 5
> r s = a
>
>may occur with r having *no* non-unit common factors with 5
>in the ring of algebraic integers.
Perhaps an integer example would clarify James's confusion?
Suppose you want to show that the number 15 and 65 have a nonunit
integer factor in common. There are many ways to do so, but surely
James would agree that the following process accomplishes the result:
(1) 3, 5, and 13 are all integers;
(2) 13*5 = 65, so 5 is a factor of 15;
(3) 3*5 = 15, so 5 is a factor of 65;
(4) 5 is not a unit in the integers.
Now, this would seem to me to accomplish the objective: we have
explicitly exhibited 5 as a factor of both 15 and 65 in the integers
(by exhibiting other integers that multiplied by 5 yield the desired
numbers), and so it is a common factor.
James now is saying that we need to show that 5 "shares a non-unit
integer factor with 15" (or with 65). But he is confused. 5 itself was
exhibited as the nonunit common factor of 15 and 65. That's it.
W. Dale Hall
Jesse F. Hughes wrote:
> "W. Dale Hall" <mailt...@farir.com> writes:
>
>
>> your useage of the English language is
>> substandard, even for a southerner.
>
>
> Don't be an ass.
>
> At least, not in the same sentence in which you misspell usage and
> fail to capitalize "Southerner".
>
> (Now, where's my requisite typo in this spelling flame?)
>
Well, you caught me there.
My bad.
Dale.
W. Dale Hall
James Harris wrote:
> "W. Dale Hall" <mailt...@farir.com> wrote in message news:<3F38305E...@farir.com>...
>
>>James Harris wrote:
>
>
> <deleted>
>
>>>No it doesn't. What you can prove is that if 'a' is coprime to 5 it
>>>can't be a unit in the ring of algebraic integers, but as I've said
>>>the ring of algebraic integers is screwed up.
>>>
>>
>>No, no one can prove that. After all, a unit *is* coprime to every
>>number!
>
>
> That's true, but it doesn't prove that in a given ring two numbers
> that don't share unit factors *in the ring* are coprime, given your
> definition of "coprime".
So, why did you state that one *could* prove this, if it is false?
Actually, every element of a ring shares ALL unit factors with EVERY
OTHER element of the ring. Why would a person even talk about numbers
that DON'T share unit factors? That is literally talking about nothing!
>
> It's not complicated.
>
> Maybe it would help if you consider the ring of evens, where 2 and 6
> don't share non-unit factors in the ring because 1 and 3 are not in
> the ring of evens, but by *your* definition they are NOT coprime.
>
> Think about it.
>
Yes, I see that the rings I've been working with have a multiplicative
identity. Your ring doesn't, does it?
>
>>Let u be a unit algebraic integer, and let p be any algebraic integer.
>>Then there is an algebraic integer v for which uv = 1, and we get:
>>
>> u v + 0 p = 1,
>>
>>where 0 is zero.
>
>
> Yeah, where you're saying that it IS a unit.
>
I'm showing that a unit is coprime to everything. Duh.
> Remember, you've tried to use examples where numbers have been
> determined to NOT be units.
>
I'm showing that if you insist that the following setup:
5 = q(a)*r(a)
a = r(a)*s(a)
with 'a' and 5 being coprime, then it is *YOU* who is insisting
that r(a) is a unit.
> Going from there, you've jumped to concluding that they then must
> share non-unit factors *in the ring of algebraic integers* which you
> have not proven.
>
I have shown the non-unit factor: r(a). Can you understand that?
> Rather than actually work to figure out the math, you've instead
> engaged in chasing your tail, publicly through posts on these forums.
>
What? Are you saying that the factorization I've posted and provided
all the necessary calculation to verify that, is somehow inadequate?
Please explain. Show me the error.
That is, show how one of these is not correct:
1. 5 = q(a)*r(a)
2. a = r(a)*s(a)
3. r(a) is an algebraic integer
4. r(a) is NOT a unit.
>
>>This establishes the fact that a UNIT algebraic integer is coprime to
>>every algebraic integer, in the ring of algebraic integers. A very
>>slight rewording proves the result for a unit of any commutative
>>ring.
>
>
> It is true that a unit algebraic integer is coprime to every algebraic
> integer.
>
>
>>No wonder you think the ring of algebraic integers is screwed up!
>
>
> You're chasing your tail.
>
Sez you. You haven't found an error in my posted calculations, and
all you have as an explanation of why I'm wrong is to claim some
"problem" with the ring of algebraic integers. I've displayed
verifiable algebraic integers that achieve a factorization that
you have claimed (and apparently are continuing to claim) is
impossible. For a reason, you appeal to pixies.
>
>>Let's get the language straight. I have been clear in this that 'a'
>>refers to (-1) times any one of the coefficients ai in the factorization
>>
>> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
>>
>>I claim that *each* of those numbers a has a non-unit algebraic integer
>>factor in common with 5.
>
>
> But you do not prove it. I'll point out the gap in what you have that
> follows.
>
>
>>PROOF: The above expressions q(a),r(a),s(a) all yield algebraic integers
>>for any algebraic integer a. When, in addition, a is a root of the
>>polynomial
>>
>> x^3 - 12 x^2 + 65
>>
>>then the products are as I mentioned above:
>>
>> q(a) r(a) = 5
>> r(a) s(a) = a.
>>
>>Your claim:
>>
>> >
>> > No it doesn't. What you can prove is that if 'a' is
>> > coprime to 5 it can't be a unit in the ring of algebraic
>> > integers, but as I've said the ring of algebraic integers
>> > is screwed up.
>
>
> Yup.
>
>
>>amounts to this:
>>
>>PARAPHRASE:
>>
>> It doesn't matter whether anyone provides algebraic integers
>> q,r, and s for which
>>
>> 5 = q*r
>> a = r*s
>>
>> it doesn't prove that a and 5 are not coprime.
>
>
> That is correct, and in fact, for ONE of the three values for 'a' they
> are in fact coprime.
>
No, each of the values of 'a' has a corresponding factor r(a). That
factor is (1) an algebraic integer, (2) not a unit, and (3) also a
factor of 5. This is what you're disputing, although you have also
agreed that (1 - 3) is in fact correct.
>
>>However, it *does* prove that. Assume otherwise, that a is coprime to 5.
>>Then there exist algebraic integers u and v for which
>>
>> a u + 5 v = 1
>>
>>and the above factorizations show this:
>>
>> rs u + qr v = 1
>>
>>so
>> r( su + qv ) = 1,
>>
>>and we deduce that r is a unit.
>
>
> You do not "deduce" you claim, but you cannot prove.
>
You're not following the argument, possibly because you've forgotten
what it is I'm proving in that statement.
I'll be more explicit in *what* the claim is:
CLAIM: if a is coprime to 5 in the ring of algebraic integers,
and if there is an algebraic integer r, for which the following
two equations hold:
5 = qr
a = rs
for some algebraic integers q and s, then r must be a unit.
PROOF:
By the definition of "coprime" in the ring of algebraic
integers, and the assumption that a and 5 are coprime,
there are algebraic integers u and v such that:
a u + 5 v = 1.
Since a = rs, and 5 = qr, this equation can be written:
(rs) u + (qr) v = 1
factoring, we have:
r (su + qv) = 1.
Thus if w = su + qv, we have an algebraic integer w,
for which
r w = 1.
Therefore, r must be a unit in the ring of algebraic
integers.
> In fact, you *can* prove that it is not a unit.
>
Yes, I did that.
> If you assume that the ring of algebraic integers is ok, then you
> think you're done.
>
At no place did I "assume that the ring of algebraic integers is OK".
I have no definition of "OK" for rings, and in fact only used the
following features of that ring:
1. Definition of coprime
2. Substitution of equals for equals (a = rs, 5 = qr)
3. Associativity of multiplication
4. Distributivity of multiplication over addition.
5. Definition of unit.
> But you're chasing your tail, as the ring of algebraic integers is NOT
> ok, and in fact, in that situation where r *should* be a unit, it's
> NOT in the ring of algebraic integers, as there is a fascinating
> problem with the ring.
>
No, it is YOU who claims that r should be a unit, for a and 5 to be
coprime. I did not claim that; in fact, the proof shows that the
implication is:
(a and 5 coprime) & (r divides both a and r) ==> r is a unit.
I have stated that the proper conclusion of (r not a unit) is this:
a and 5 are NOT coprime.
>
>>However, it's not one of your fanciful
>>"Unit with no inverse" hallucinations. In fact, the above equation
>>provides us with an inverse for r. Based on its construction, it
>>*must be* an algebraic integer: u and v are specified to be algebraic
>>integers, and q,r,s were already specified as algebraic integers.
>
>
> <deleted>
>
> Why don't you show that inverse.
>
I just did: if a and 5 are coprime, choose the u and v that force
a u + 5 v = 1.
Then the inverse of r will be w = su + qv.
> Now that's bizarre as the poster has just attacked his own argument.
>
> Now I don't claim that r is a unit, as it's not *in the ring of
> algebraic integers*.
>
Idiot. I have already shown that r(a) = 8 a^2 - 4 a - 45
Therefore, since the algebraic integers form a ring (you still agree
to that, right?), r(a) must be an algebraic integer.
> It amazes me how often posters just throw out false statements.
>
Like what?
> Here the poster's claim is that a non-monic primitive irreducible over
> Q, which is what gives you the inverse for r in context, is an
> algebraic integer.
>
You are purposely misreading what I have written. I'll break it down
again for you. Please try to pay attention:
5 = q(a) r(a)
a = r(a) s(a).
q,r, and s are given by these formulas:
q(a) = 8 a^2 - 76 a - 185
r(a) = 8 a^2 - 4 a - 45
s(a) = 4 a^2 - 37 a - 104
Whenever a is a root of the cubic x^3 - 12 x^2 + 65, the above
factorizations are correct, and consist of algebraic integers.
Further, I have shown that if, despite the above factorization,
a and 5 are coprime (that is: IF YOUR CLAIM IS CORRECT), it must
follow that r(a) is a unit algebraic integer.
I also show that the minimal polynomial for r(a) is this:
MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
That shows that r(a) CANNOT be a unit.
The contradiction, which you're trying very hard to twist to
suit your own purposes, is that your claim (a coprime to 5)
cannot be true.
> Don't see it?
>
> Well consider the statements:
>
>
>>Then there exist algebraic integers u and v for which
>>
>> a u + 5 v = 1
>>
>>and the above factorizations show this:
>>
>> rs u + qr v = 1
>>
>>so
>> r( su + qv ) = 1,
>>
>>and we deduce that r is a unit.
>
>
> The poster wants to claim that r must share a non-unit factor in
> common with 5 by showing a contradiction by assuming it's a unit,
> which actually only shows that it can't be a unit in the ring of
> algebraic integers.
>
> So he should have just ended up chasing his tale.
>
> But he screwed up and managed to bite himself!
>
>
Yeah, right. If you could read and understand mathematical proofs,
you would have just said, "Oh, well, I was wrong there." Instead,
you're fixated on the fantasy that you've gone where no mathematician
has trod before. Fine, have your fantasy.
It's just not mathematics, nor is it supported by anything other than
your massively-flawed messing around with "uber polynomials".
>
> James Harris
Dale
Arturo Magidin
W. Dale Hall <mailt...@farir.com> wrote:
Nitpicking:
>What? Are you saying that the factorization I've posted and provided
>all the necessary calculation to verify that, is somehow inadequate?
>
>Please explain. Show me the error.
>
>That is, show how one of these is not correct:
>
> 1. 5 = q(a)*r(a)
> 2. a = r(a)*s(a)
> 3. r(a) is an algebraic integer
as are q(a) and s(a). An important point.
> 4. r(a) is NOT a unit.
Brian Quincy Hutchings
in the natural numbers. thank you!
"W. Dale Hall" <mailt...@farir.com> wrote in message news:<3F39291E...@farir.com>...
> Yes, I see that the rings I've been working with have a multiplicative
> identity. Your ring doesn't, does it?
what about m=0 ??... I haven't followed any of this, because
no-one wants to put any sort of a hypothesis up front,
to justify it (I mean, I *might*
be able to follow it). I do have one hypothesis:
necessarily, monsieur Harris, iff he exists
in the real world (sik), will not reply to this proof
of his lack of a proof, but
taht is not sufficient to dysprove the results
of his 10-year Mission!
nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.0308...@posting.google.com>...
> Notes:
>
> * 1. The Galois theory argument has been presented before.
>
> That the polynomial in [3],
>
> -m*k*f^2 + 3*v*y^2 + y^3
>
> is irreducible in general is easily seen by letting m = 1
> and f = 5, where it becomes
>
> y^3 + 72*y^2 - 13825,
>
> which is easily shown to be irreducible.
>
> *2. This argument does not say anything about whether e1,
> e2, and e3 are coprime to f.
>
> *3. This is the key new step in the argument.
>
> *4. The same argument is implied in Harris's "proof" of
> FLT, except there he is using "objects" rather than
> algebraic integers. However it has not been established
> whether objects are different from algebraic integers,
> and a number of basic theorems for objects which are
> needed for his argument have not been proven.
--Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac