Penultimate test of objectors
James Harris
Consider the polynomial y^3 + 3y - 2 yet again. Its three roots are
algebraic integers, and multiply to give 2. Now divide off 2 from the
roots, and you have what *should* be three algebraic integer units.
BUT, if they are algebraic integers, then they must be roots of a
monic polynomial with integer coefficients.
But they are not, so no such polynomial can be found for those
particular numbers. Remember these are the roots of y^3 + 3y - 2 with
their factors in common with 2 divided off to give what *should* be
three units.
It occurs to me that I should have presented that test before, as in
retrospect if I were wrong, then someone could have ended the debate a
long time ago, simply by producing such a polynomial.
What I know is that the ring of algebraic integers is not complete, so
no such polynomial can be found. If you believe I'm wrong, then find
it.
Remember the very definition of algebraic integers puts a very simple
constraint: a monic polynomial with integer coefficients.
At least you already know what the first and last coefficients must
be: 1 or -1.
Now then, if I'm wrong, surely mathematicians can at least provide the
*degree* of the polynomial, or better yet, give the entire thing!!!
Have fun. It can't be found because it doesn't exist, as I've proven.
Now you can spend idle hours trying to do something impossible, or be
sensible and give up.
James Harris
Alan Morgan
James Harris <jst...@msn.com> wrote:
>It's amazing how simple the test is, and how important it is.
>
>Consider the polynomial y^3 + 3y - 2 yet again. Its three roots are
>algebraic integers, and multiply to give 2. Now divide off 2 from the
>roots, and you have what *should* be three algebraic integer units.
Why? What do you mean by "divide off 2 from the roots"? This could mean
either (r1 r2 r3)/2 or r1/2, r2/2, r3/2 or perhaps r1/a r2/b r3/c where
abc = 2.
If the first, I don't see what the three algebraic integer units are. If
the second, it isn't clear to me that any of the numbers must be algebraic
integers, let alone units. If the third, I have an uneasy feeling that
a=r1, b=r2, and c=r3, which doesn't really get you anywhere.
What are the three algebraic integer units that you claim *should* exist?
Alan
--
Defendit numerus
C. Bond
> It's amazing how simple the test is, and how important it is.
>
> Consider the polynomial y^3 + 3y - 2 yet again. Its three roots are
> algebraic integers, and multiply to give 2. Now divide off 2 from the
> roots, and you have what *should* be three algebraic integer units.
Why?
> BUT, if they are algebraic integers, then they must be roots of a
> monic polynomial with integer coefficients.
>
> But they are not, so no such polynomial can be found for those
> particular numbers. Remember these are the roots of y^3 + 3y - 2 with
> their factors in common with 2 divided off to give what *should* be
> three units.
Why?
> It occurs to me that I should have presented that test before, as in
> retrospect if I were wrong, then someone could have ended the debate a
> long time ago, simply by producing such a polynomial.
>
> What I know is that the ring of algebraic integers is not complete, so
> no such polynomial can be found. If you believe I'm wrong, then find
> it.
Why? If the ring of algebraic integers is not complete, it is *NOT* a
ring. There is no such thing as an 'incomplete ring'. But the algebraic
integers *DO* form a ring, as has been proven. (Some proofs are available
online.)
> Remember the very definition of algebraic integers puts a very simple
> constraint: a monic polynomial with integer coefficients.
>
> At least you already know what the first and last coefficients must
> be: 1 or -1.
>
> Now then, if I'm wrong, surely mathematicians can at least provide the
> *degree* of the polynomial, or better yet, give the entire thing!!!
You haven't proven that 'dividing off 2' (whatever that means) necessarily
produces units.
> Have fun. It can't be found because it doesn't exist, as I've proven.
> Now you can spend idle hours trying to do something impossible, or be
> sensible and give up.
>
> James Harris
Also it needn't be looked for because it is completely irrelevant. Your
polynomial: y^3+3y-2 has three roots, y_1, y_2 and y_3, which are all
divisors of 2 in the ring of algebraic integers and none of them are
units. Hence, your stated claim is wrong.
--
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
Uncle Al
>
> It's amazing how simple the test is, and how important it is.
>
> Consider the polynomial y^3 + 3y - 2 yet again. Its three roots are
> algebraic integers, and multiply to give 2. Now divide off 2 from the
> roots, and you have what *should* be three algebraic integer units.
>
> BUT, if they are algebraic integers, then they must be roots of a
> monic polynomial with integer coefficients.
>
> But they are not, so no such polynomial can be found for those
> particular numbers. Remember these are the roots of y^3 + 3y - 2 with
> their factors in common with 2 divided off to give what *should* be
> three units.
What kind of muddled meaningless crap is this? Harris, you would have
to study to be certified an idiot. Now you've done it, you've
intellectually offended Uncle Al...
I cannot believe how incredibly stupid James Harris is. I mean
rock-hard stupid. Blazing hot mid-day sun on Mercury stupid. Surface
of Venus under 80 atmospheres of red hot carbon dioxide and sulfuric
acid vapor dehydrated for 300 million years rock-hard stupid. Stupid
so stupid that it goes way beyond the stupid we know into a whole
different sensorium of stupid. James Harris is trans-stupid stupid.
Meta-stupid. Stupid so collapsed upon itself that it is within its
own Schwarzschild radius. Black hole stupid. Stupid gotten so dense
and massive that no intellect can escape. Singularity stupid. James
Harris emits more stupid/second than our entire galaxy otherwise emits
stupid/year. Quasar stupid. Nothing else in the universe can be this
stupid. James Harris is an oozingly putrescent primordial fragment
from the original Big Bang of Stupid, a pure essence of stupid so
uncontaminated by anything else as to be beyond the laws of physics
that define maximally extrapolated hypergeometric n-dimensional
backgroundless stupid as we can imagine it. James Harris is Planck
stupid, a quantum foam of stupid, a vacuum decay of stupid, a grand
unified theory of stupid.
James Harris is the epiphany of stupid. James Harris is stooopid.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Guillaume
It doesn't look like you do.
Chuck Simmons
>
> Do you at least know what "penultimate" means?
>
> It doesn't look like you do.
Be nice. Harris has just had a couple of dismal days in sci.math where
his proofs and repairs of proofs have been shredded with careful
reasoning and counter example. He seems to have come here (sci.physics)
to lick his wounds.
Chuck
--
... The times have been,
That, when the brains were out,
the man would die. ... Macbeth
Chuck Simmons chr...@earthlink.net
C. Bond
At this point, I think you are being too kind. No one seems to be able get
his attention with a 2 by 4 to the head. It may take an anvil. By the way,
in my dictionary under the entry "stupid" they have a picture of James
Harris.
--
It takes a village to raise an idiot.
Ian H
"James Harris" <jst...@msn.com>
: Consider the polynomial y^3 + 3y - 2 yet again. Its three roots are
: algebraic integers, and multiply to give 2. Now divide off 2 from the
: roots, and you have what *should* be three algebraic integer units.
Beats me why you would think these need to be algebraic
INTEGERS. Even ordinary integers sometimes stop being
integers when you divide by 2.
David C. Ullrich
>It's amazing how simple the test is, and how important it is.
>
>Consider the polynomial y^3 + 3y - 2 yet again. Its three roots are
>algebraic integers, and multiply to give 2. Now divide off 2 from the
>roots, and you have what *should* be three algebraic integer units.
Why in the world would that be? If you mean that r/2 should be
an algebraic integer if r is a root of this polynomial, that's clearly
false, since r/2 is a root of 4y^2 + 3y - 1.
>BUT, if they are algebraic integers, then they must be roots of a
>monic polynomial with integer coefficients.
They're not algebraic integers. This would be a problem if there
was some reason they "should" be, but there is no reason they
should be. (Except for the fact that you _say_ they should be,
of course... I doubt that anyone finds that very compelling.)
>But they are not, so no such polynomial can be found for those
>particular numbers. Remember these are the roots of y^3 + 3y - 2 with
>their factors in common with 2 divided off to give what *should* be
>three units.
>
>It occurs to me that I should have presented that test before, as in
>retrospect if I were wrong, then someone could have ended the debate a
>long time ago, simply by producing such a polynomial.
>
>What I know is that the ring of algebraic integers is not complete, so
>no such polynomial can be found. If you believe I'm wrong, then find
>it.
>
>Remember the very definition of algebraic integers puts a very simple
>constraint: a monic polynomial with integer coefficients.
>
>At least you already know what the first and last coefficients must
>be: 1 or -1.
>
>Now then, if I'm wrong, surely mathematicians can at least provide the
>*degree* of the polynomial, or better yet, give the entire thing!!!
>
>Have fun. It can't be found because it doesn't exist, as I've proven.
> Now you can spend idle hours trying to do something impossible, or be
>sensible and give up.
>
>
>James Harris
************************
David C. Ullrich
Wayne Brown
> Guillaume wrote:
>>
>> Do you at least know what "penultimate" means?
>>
>> It doesn't look like you do.
> Be nice. Harris has just had a couple of dismal days in sci.math where
> his proofs and repairs of proofs have been shredded with careful
> reasoning and counter example. He seems to have come here (sci.physics)
> to lick his wounds.
Those "couple of dismal days in sci.math" were proceeded by a few dismal
days in alt.fiction.original. During that period he convinced many in
that newsgroup (who had no prior knowledge of him) that he's as big a
jerk as the residents of sci.math think he is.
--
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"
Arturo Magidin
>
>Consider the polynomial y^3 + 3y - 2 yet again. Its three roots are
>algebraic integers, and multiply to give 2. Now divide off 2 from the
>roots, and you have what *should* be three algebraic integer units.
This is more confused than usual...
The three roots of y^3+3y-2 are all divisors of 2. Presumably, you
meant "divide off their common factor with 2", not "divide 2". that
would be like saying that since 2 divides 4, dividing 4 off 2 (i.e.,
taking 2/4=1/2) should give a unit.
So, let y_1, y_2, y_3 be the three roots. Since we all agree,
presumably, that y_1*y_2*y_3 = 2, one may choose the common factors of
y_i with 2 to be y_i itself. Dividing it off, we get y_i/y_i = 1,
which is, of course, a unit.
>BUT, if they are algebraic integers, then they must be roots of a
>monic polynomial with integer coefficients.
Indeed: x-1 works.
>But they are not,
But they are: they can all be chosen to be equal to 1.
> so no such polynomial can be found for those
>particular numbers. Remember these are the roots of y^3 + 3y - 2 with
>their factors in common with 2 divided off to give what *should* be
>three units.
Again, a confusion. There is no well defined "factor in common with
2"; it is only defined up to associates; meaning that if r is a common
factor of y_i with 2, then so is r*u for any unit u. So you cannot
speak aboout the common factors as if they were somehow unique. In the
integers we can do that, because there are only two greatest common
divisors for any pair of nonzero elements, and one is the negative of
the other; so we may, by fiat, choose the positive one.
But in the algebraic integers there is no such choice.
Again, since each y_i divides 2, the gcd may be chosen to be y_i
itself. Dividing it off we get y_i/y_i=1, and 1 is a root of x-1, a
monic polynomial with integer coefficients.
Again, it seems that what you are trying to claim is something you
attempted many times and abandoned about 6 months ago: you think there
is an irreducible monic polynomial with integer coefficients, which
has some roots which are units in the ring of algebraic integers, and
some which are not.
But this is false. If you have an irreducible monic polynomial with
integer coefficients, then either all the roots are units in the ring
of algebraic integers, or none of them are. And it is very easy to see
which case you are in: they are all units if and only if the constant
term of the monic irreducible is either 1 or -1.
>Now then, if I'm wrong, surely mathematicians can at least provide the
>*degree* of the polynomial, or better yet, give the entire thing!!!
Degree 1. x-1. None of y_1, y_2, y_3 are units, they are all divisors
of 2, so none of them are coprime to 2.
>Have fun. It can't be found because it doesn't exist, as I've proven.
This must be a meaning of the word "proven" that is in direct
opposition to the commonly accepted meaning. I cannot say I have never
encountered it before, having read many of yours posts.
======================================================================
"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 Ghost In The Machine
<n...@my.real.address>
wrote
on Fri, 1 Aug 2003 16:57:30 +1200
<10597139...@clint.its.waikato.ac.nz>:
AIUI an algebraic integer is the root of an equation such as the above,
with a leading coefficient of 1 and integer coefficients. By
necessity the roots must multiply to 2, as the equation is
equivalent to
(x - r1)(x - r2) ... (x - rn)
which multiplies out to
x^n + ... + r1r2...rn
Where JSH makes his error is that he assumes all roots have a
factor of 2, which is clearly incorrect; the equation
x^2 - 3x + 2
has the roots 1 and 2, for example.
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.