Factorization lemma, example
James Harris
Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c
is a factor of the contant term P(0), while r=g-c, so it varies if g
varies.
That simple lemma not only anchors my paper, and my proof of Fermat's
Last Theorem, it can be used in lots of places. Now I'll use it with
the roots of
y^3 + 3y - 2
and the three roots are
y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3}
y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 -
(-1-sqrt(-3))/2(1+sqrt(2))^{1/3}
y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 -
(-1+sqrt(-3))/2(1+sqrt(2))^{1/3}
where the polynomial I'm interested in is
P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}.
Letting
g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1,
I have c=1, so r=g-1.
(Some may worry about my choice for c, I'll come back to that later.)
Now using the substitution g=r+1, with x=sqrt(2), with my roots, I
have
y_1 = r+1 - 1/(r+1)
y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1)
y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1)
so I have
y_1 = (r^2 + 2r)/(r+1)
y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1)
y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1)
which is
y_1 = (r^2 + 2r)/(r+1)
y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)]/(r+1)
y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)]/(r+1)
which proves that y_2 and y_3 are coprime to 2 because r^2/2 *should*
be an algebraic integer, though it provably is not, which is a
problem.
Now you may worry about picking c=1, when I had 1^{1/3}, but all that
happens with picking one of the others is that r becomes coprime to 2,
which I'll leave as an exercise for the reader.
So as I've said my work is irrefutable, and you may have notice people
avoiding that lemma which starts my paper, and has a key place on my
website.
Now you may know why and you may understand why I've been as upset as
I have.
It IS algebra, after all.
James Harris
David Bernier
James Harris wrote:
[...]
> So as I've said my work is irrefutable, and you may have notice people
> avoiding that lemma which starts my paper, and has a key place on my
> website.
>
> Now you may know why and you may understand why I've been as upset as
> I have.
>
> It IS algebra, after all.
It IS a garble, above all...
> James Harris
David Bernier
James Harris
The acts of desperation are fascinating. However you can't erase the
argument simply by deleting it out in your reply.
Those who are serious mathematicians can now see for themselves that
the ring of algebraic integers is flawed by considering the roots of
y^3 + 3y - 2, as it turns out that you only need to make the
substitution
r+1 = (sqrt(2)+1)^{1/3}
to *easily* and quickly lead to the result that only one of the roots
can have any non unit factors of 2, as I show in my original post, and
that root has a factor of 2 that is 2.
My warning to universities and mathematicians teaching at them
remains.
Some of you may face civil liabilities, maybe worse--angry
parents--when the full story comes out.
School is about to start, and if universities teach the flawed
mathematics they may also face federal authorities for deliberate
fraud.
Yup, some of you are looking at chaos as you scramble to figure out
how to prevent yourself from being legally liable, but I've warned for
some time now.
It's your fault if you ignored me.
It is mathematics after all.
James Harris
Guillaume
> parents--when the full story comes out.
Very witty, you're good. ;)
> School is about to start, and if universities teach the flawed
> mathematics they may also face federal authorities for deliberate
> fraud.
Ouch, school is just about to start, thanks to you. Hats off, buddy!
> Yup, some of you are looking at chaos as you scramble to figure out
> how to prevent yourself from being legally liable, but I've warned for
> some time now.
>
> It's your fault if you ignored me.
Guys: you've been warned. James told you how to get out of trouble,
but you won't listen.
> It is mathematics after all.
Oh, is it? As for me, I would call that "megalomania", but whatever. :)
James Harris
> My paper does rest fully on the following rather basic lemma:
Surprisingly to me I'm still seeing disputing posts.
I figured that pointing out some things I thought were obvious might
help.
> Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c
> is a factor of the contant term P(0), while r=g-c, so it varies if g
> varies.
>
> That simple lemma not only anchors my paper, and my proof of Fermat's
> Last Theorem, it can be used in lots of places. Now I'll use it with
> the roots of
>
> y^3 + 3y - 2
>
> and the three roots are
Notice that (1+sqrt(2))^{1/3} is a unit, as
(1+sqrt(2))^{1/3} (-1 + sqrt(2))^{1/3} = 1^{1/3}.
I was VERY surprised to see someone actually post questioning that
conclusion on the sci.math newsgroup. Remember r+1 is a unit, so you
have
y_1 = (r^2 + 2r)(-1 + sqrt(2))^{1/3}
y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1 + sqrt(2))^{1/3}
y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1 + sqrt(2))^{1/3}
and r^2/2 *should* be an algebraic integer, but provably is not.
However, notice that you run into a problem with r^2/2 not having a
factor of 2 that is 2.
Amazingly enough, you've been pushed out of the ring of algebraic
integers, as r^2/2 is provably NOT an algebraic integer.
However, notice that it is true that y_2 and y_3 must be coprime to 2.
The mathematics is actually fascinating, and luckily for me, I found
this simple demonstration.
> Now you may worry about picking c=1, when I had 1^{1/3}, but all that
> happens with picking one of the others is that r becomes coprime to 2,
> which I'll leave as an exercise for the reader.
>
> So as I've said my work is irrefutable, and you may have notice people
> avoiding that lemma which starts my paper, and has a key place on my
> website.
>
> Now you may know why and you may understand why I've been as upset as
> I have.
>
> It IS algebra, after all.
And as I pointed out before there have been people who have been
attacking algebra itself in posts. Now it still bothers me that I was
the one person standing up to defend algebra itself from what I
noticed.
But maybe the hard truth is that none of you actually believe in
mathematics itself, choosing instead to believe what other people
*tell* you is mathematical truth.
If so, that's very sad.
James Harris
C. Bond
> I was VERY surprised to see someone actually post questioning that
> conclusion on the sci.math newsgroup. Remember r+1 is a unit, so you
> have
>
> y_1 = (r^2 + 2r)(-1 + sqrt(2))^{1/3}
>
> y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1 + sqrt(2))^{1/3}
>
> y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1 + sqrt(2))^{1/3}
>
> and r^2/2 *should* be an algebraic integer, but provably is not.
>
> However, notice that you run into a problem with r^2/2 not having a
> factor of 2 that is 2.
>
> Amazingly enough, you've been pushed out of the ring of algebraic
> integers, as r^2/2 is provably NOT an algebraic integer.
Forget about 'r'. It brings nothing to the table. The issue is whether y_1, y_2 and y_3 are coprime to 2.
> However, notice that it is true that y_2 and y_3 must be coprime to 2.
It is *not* true, as has been shown. y_1, y_2, and y_3 are all algebraic integers and NONE of them are
coprime to 2. At the risk of repeating a fact you will promptly ignore:
2 = y_1(y_1^2+3)
2 = y_2(y_2^2+3)
2 = y_3(y_3^2+3)
where, in the ring of algebraic integers, all sums and products of algebraic integers are algebraic
integers. Note that '2' and 'y_1^2+3' , etc. are all algebraic integers. Wake up man! What *is* your
problem?
Summary: your "proof" has led you to a false conclusion. Therefore your "proof" is not only refutable, it
is clearly flawed.
> But maybe the hard truth is that none of you actually believe in
> mathematics itself, choosing instead to believe what other people
> *tell* you is mathematical truth.
More likely they have identified the error in your lemma and *believe* the truth instead of believing your
error-ridden arguments.
--
A man of integrity identifies, acknowledges and corrects his errors. A man without it ignores, denies or
defends them.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
Will Twentyman
> My paper does rest fully on the following rather basic lemma:
>
> Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c
> is a factor of the contant term P(0), while r=g-c, so it varies if g
> varies.
This is not a lemma, it is a definition with a note.
Given g(x) a factor of the polynomial P(x), define c=g(0) and r(x)=g(x)-c.
Note: g(x)=r(x)+c.
No one is commenting on this because it is trivial.
>
> That simple lemma not only anchors my paper, and my proof of Fermat's
> Last Theorem, it can be used in lots of places. Now I'll use it with
> the roots of
>
> y^3 + 3y - 2
>
> and the three roots are
>
> y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3}
>
> y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 -
>
> (-1-sqrt(-3))/2(1+sqrt(2))^{1/3}
>
> y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 -
>
> (-1+sqrt(-3))/2(1+sqrt(2))^{1/3}
>
> where the polynomial I'm interested in is
>
> P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}.
>
> Letting
>
> g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1,
>
> I have c=1, so r=g-1.
>
> (Some may worry about my choice for c, I'll come back to that later.)
No, I worry that you aren't saying g(x)=(1+x)^(1/3) and r(x)=g(x)-1. I
know, you don't care about style, but it can make a difference.
>
> Now using the substitution g=r+1, with x=sqrt(2), with my roots, I
> have
I would say something like v=r(sqrt(2)) and use v instead of r in the
following.
>
> y_1 = r+1 - 1/(r+1)
>
> y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1)
>
> y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1)
>
> so I have
>
> y_1 = (r^2 + 2r)/(r+1)
>
> y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1)
>
> y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1)
>
> which is
>
> y_1 = (r^2 + 2r)/(r+1)
>
> y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)]/(r+1)
>
> y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)]/(r+1)
All of this is very nice until...
>
> which proves that y_2 and y_3 are coprime to 2 because r^2/2 *should*
> be an algebraic integer, though it provably is not, which is a
> problem.
What? How does this show that there are algebraic integers a and b such
that a*y_2+b*2=1? Similarly for y_3?
Also, you say they are coprime because r^2/2 "should" be an algebraic
integer, but isn't? What does this mean? Here's my interpretation:
"If r^2/2 is an algebraic integer, then y_2 and y_3 are coprime to 2.
However, r^2/2 is NOT an algebraic integer so I can't conclude anything
about whether y_2 or y_3 are coprime to 2 from the above argument."
The problem is that you cannot conclude anything from a false premise
and expect it to be accepted as true. It's not a problem in the math,
it's a problem in your logic.
>
> Now you may worry about picking c=1, when I had 1^{1/3}, but all that
> happens with picking one of the others is that r becomes coprime to 2,
> which I'll leave as an exercise for the reader.
>
> So as I've said my work is irrefutable, and you may have notice people
> avoiding that lemma which starts my paper, and has a key place on my
> website.
Your work says "If something was true, there would be a problem."
Fortunately, it's not true so there's no problem. Now if you can prove
"Something that *is* true causes a problem," we might pay more attention.
>
> Now you may know why and you may understand why I've been as upset as
> I have.
>
> It IS algebra, after all.
>
JSH: "If Godzilla is rampaging through New York, there is a huge problem."
Mathematicians: "Godzilla's not in New York, nor is he real."
JSH: "You don't understand, if GODZILLA IS RAMPAGING THROUGH NY, there
is a HUGE PROBLEM!"
Mathematicians: "Then it's a good thing he isn't."
JSH: "Call the national guard!"
Mathematicians: "No."
etc.
--
Will Twentyman
email: wtwentyman at copper dot net
Will Twentyman
You keep saying *should*. Deal with the fact that it's not and draw
your conclusions from that. Either that or define more clearly *why*
you think it should be something it isn't.
>
> However, notice that you run into a problem with r^2/2 not having a
> factor of 2 that is 2.
Only through an artificial construction.
>
> Amazingly enough, you've been pushed out of the ring of algebraic
> integers, as r^2/2 is provably NOT an algebraic integer.
How is this amazing? You ran out of it headlong.
>
> However, notice that it is true that y_2 and y_3 must be coprime to 2.
Or not, as others have shown.
>
> The mathematics is actually fascinating, and luckily for me, I found
> this simple demonstration.
Which, like others before, does NOT work the way you want it to.
>
>
>>Now you may worry about picking c=1, when I had 1^{1/3}, but all that
>>happens with picking one of the others is that r becomes coprime to 2,
>>which I'll leave as an exercise for the reader.
>>
>>So as I've said my work is irrefutable, and you may have notice people
>>avoiding that lemma which starts my paper, and has a key place on my
>>website.
>>
>>Now you may know why and you may understand why I've been as upset as
>>I have.
>>
>>It IS algebra, after all.
>
>
> And as I pointed out before there have been people who have been
> attacking algebra itself in posts. Now it still bothers me that I was
> the one person standing up to defend algebra itself from what I
> noticed.
>
> But maybe the hard truth is that none of you actually believe in
> mathematics itself, choosing instead to believe what other people
> *tell* you is mathematical truth.
>
> If so, that's very sad.
>
>
> James Harris
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>My paper does rest fully on the following rather basic lemma:
>
>Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c
>is a factor of the contant term P(0), while r=g-c, so it varies if g
>varies.
Your "lemma" is nonsense as written, as has been explained to you many
times. A non-extensive list of the ambiguities and problems in the
'statement' above is:
(1) What kind of polynomial is P(x)? That is, it is an element of
which polynomial ring?
(2) What sort of object is g?
(3) In what ring are we working when we say that "g is a factor of
P(x)"?
(4) What sort of objects are c and r?
(5) What does it mean "c=g at x=0"?
(6) In what ring is c "a factor of the constant term P(0)"?
(7) What does it mean for g to vary?
These are only some of the problems with your statement. Any
"application" of the lemma rests entirely on the ambiguity and gaps of
your presentation.
======================================================================
"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
W. Dale Hall
James Harris wrote:
... a buncha stuff, leading to this error:
>>
>>Now using the substitution g=r+1, with x=sqrt(2), with my roots, I
>>have
>>
>> y_1 = r+1 - 1/(r+1)
>>
>> y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1)
>>
>> y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1)
Note how we have the roots of unity
z = exp(2 pi i /3) = (-1 + i sqrt(3))/2,
and
zbar = z^2 = (-1 - i sqrt(3))/2.
They are units, of course.
>>
>>so I have
>>
>> y_1 = (r^2 + 2r)/(r+1)
>>
>> y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1)
>>
>> y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1)
>>
What makes you think that removing a factor from (r^2 + 2r)
and multiplying it into (-1 + i sqrt(3))/2 is a legitimate step?
>>which is
>>
>> y_1 = (r^2 + 2r)/(r+1)
>>
>> y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)]/(r+1)
>>
>> y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)]/(r+1)
>>
>>which proves that y_2 and y_3 are coprime to 2 because r^2/2 *should*
>>be an algebraic integer, though it provably is not, which is a
>>problem.
>
No, you made an illegitimate step when you took the factor of 2 out
of r^2 + 2r and multiplied it into the factor (-1 + i sqrt(3))/2,
although the latter was already an algebraic integer.
You want to claim (foolishly) that y_2 is not an algebraic integer,
*not* that y_2 is coprime to 2. Your expression above has no
bearing on whether y_2 and 2 are coprime.
If you wanted to show y_2 and 2 to be coprime, you would need to
show two algebraic integers p and q for which
p y_2 + q 2 = 1.
However, y_2 DIVIDES 2, that is, there is already an algebraic integer
r for which
2 = r y_2.
Thus, in your desperation, you would be led here:
p y_2 + q 2 = 1,
p y_2 + q r y_2 = 1,
or
(p + q r) y_2 = 1.
Thus, y_2 must be a unit. Problem is, it can't be a unit, since it
is a root of x^3 + 3*x - 2.
>
> I was VERY surprised to see someone actually post questioning that
> conclusion on the sci.math newsgroup. Remember r+1 is a unit, so you
> have
>
> y_1 = (r^2 + 2r)(-1 + sqrt(2))^{1/3}
>
> y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1 + sqrt(2))^{1/3}
You would be less confused if you would learn to recognize roots
of unity, such as the primitive cube roots of 1:
exp(2 pi i/3) = (-1 + i sqrt(3))/2
and
exp(-2 pi i/3)= (-1 - i sqrt(3))/2.
When you do that, you realize that you robbed a factor from r^2 + 2r,
which is not divisible by 2 in the algebraic integers, only to multiply
it by (-1 + i sqrt(3))/2. That's like taking the product
1 * 5
and rewriting it:
(1/2) * 10.
Algebraically it's OK, but if you then went on to complain that 1/2
isn't an integer, it would be foolish, since it was your step that
produced the 1/2, not any property of the integers.
>
> y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1 + sqrt(2))^{1/3}
>
> and r^2/2 *should* be an algebraic integer, but provably is not.
>
> However, notice that you run into a problem with r^2/2 not having a
> factor of 2 that is 2.
>
> Amazingly enough, you've been pushed out of the ring of algebraic
> integers, as r^2/2 is provably NOT an algebraic integer.
>
Repeating the description of that step won't save you. If you knew
what you were doing, you would have wanted to use steps that preserved
the integrality of the relevant terms. This step failed to do that.
Just because you can divide by 2 formally, that doesn't mean that it's a
legitimate step in your argument. Here, you divided a factor that
wasn't already divisible by 2.
Further, this argument is irrelevant to the quesion about co-primality.
> However, notice that it is true that y_2 and y_3 must be coprime to 2.
>
> The mathematics is actually fascinating, and luckily for me, I found
> this simple demonstration.
>
Fascinating to see what foolishness you care to dish out, perhaps.
>
>>Now you may worry about picking c=1, when I had 1^{1/3}, but all that
>>happens with picking one of the others is that r becomes coprime to 2,
>>which I'll leave as an exercise for the reader.
>>
>>So as I've said my work is irrefutable, and you may have notice people
>>avoiding that lemma which starts my paper, and has a key place on my
>>website.
>>
>>Now you may know why and you may understand why I've been as upset as
>>I have.
>>
>>It IS algebra, after all.
>
If it's so simple, why don't you do the math correctly? You aren't
drawing correct conclusions, even on the rare occasion you make some
correct manipulations.
>
> And as I pointed out before there have been people who have been
> attacking algebra itself in posts. Now it still bothers me that I was
> the one person standing up to defend algebra itself from what I
> noticed.
>
Show me how I've "attacked algebra itself". Please.
You have only stood up to proclaim falsehoods as truth, and to
decry those who refuse to let you get away with it.
> But maybe the hard truth is that none of you actually believe in
> mathematics itself, choosing instead to believe what other people
> *tell* you is mathematical truth.
>
Or, maybe the hard truth is that you are unable to tell whether you've
made mistakes, and so assume that such things never happen.
> If so, that's very sad.
>
I would say pitiful.
>
> James Harris
Dale.
Brian Quincy Hutchings
I mean, if you are to fulfill your ten-year mission
to unveil the simple algebraical proof, possibly akin to Fermat's,
any substantial pause at the 8-year milestone will acrue benefits,
probably in direct proportion to the period of such a respite.
(I mean, I don't see why it shouldnt' apply to mental exercise,
as it does to physical .-)
as for me, I don't see why I'm wasting my time, but
I might eventually learn how to complete the tetragon -- and
then some!
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03073...@posting.google.com>...
> > It IS algebra, after all.
> If so, that's very sad.
--les ducs d'Enron!
http://members.tripod.com/~american_almanac
Will Twentyman
>
>
> James Harris wrote:
>
> ... a buncha stuff, leading to this error:
>
>>>
>>> Now using the substitution g=r+1, with x=sqrt(2), with my roots, I
>>> have
>>>
>>> y_1 = r+1 - 1/(r+1)
>>>
>>> y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1)
>>>
>>> y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1)
>
>
> Note how we have the roots of unity
>
> z = exp(2 pi i /3) = (-1 + i sqrt(3))/2,
> and
> zbar = z^2 = (-1 - i sqrt(3))/2.
His 2(r+1) was meant to be in brackets.
y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/[2(r+1)]
y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/[2(r+1)]
James Harris
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03073...@posting.google.com>...
> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03073...@posting.google.com>...
> > My paper does rest fully on the following rather basic lemma:
>
> Surprisingly to me I'm still seeing disputing posts.
>
> I figured that pointing out some things I thought were obvious might
> help.
<deleted>
>
> y_1 = (r^2 + 2r)(-1 + sqrt(2))^{1/3}
>
> y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1 + sqrt(2))^{1/3}
>
> y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1 + sqrt(2))^{1/3}
The 2 should go beneath (-1-sqrt(-3)) as that has a factor of 2.
So what I have is
y_1 = (r^2 + 2r)(-1 + sqrt(2))^{1/3}
y_2 = [(r^2 + r)(-1+sqrt(-3))/2 + sqrt(-3)](-1 + sqrt(2))^{1/3}
y_3 = [(r^2 + r)(-1-sqrt(-3))/2 - sqrt(-3)](-1 + sqrt(2))^{1/3}
and I'm very disappointed but I guess I keep wishing for a simple resolution.
That wasn't it.
James Harris
James Harris
>
> Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c
> is a factor of the contant term P(0), while r=g-c, so it varies if g
> varies.
Simple enough, but unfortunately I made a mistake in previous posts at
least partly because of where I put a 2.
> That simple lemma not only anchors my paper, and my proof of Fermat's
> Last Theorem, it can be used in lots of places. Now I'll use it with
> the roots of
>
> y^3 + 3y - 2
>
> and the three roots are
>
> y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3}
>
> y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 -
>
> (-1-sqrt(-3))/2(1+sqrt(2))^{1/3}
>
> y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 -
>
> (-1+sqrt(-3))/2(1+sqrt(2))^{1/3}
Now here's why I keep running into a block, and it has to do with the
cuberoot operator. Depending on which way you take the cuberoot, you
rotate the solutions.
So if I say 1^{1/3} is 1, and get some value for y_1, then moving to
1^{1/3} = (-1+sqrt(-3))/2
will shift that value to whatever I'd picked before as y_2 or maybe
y_3.
There's no way around that using basic algebraic methods.
So when I try to find something using such methods that says that y_1
is the only one not coprime to 2, that can't work because taking the
cuberoot differently would mean it'd say the same about another root.
So you can't use such methods to pick.
Others seem to believe that because certain methods fail here *all* of
the roots must have some non-unit factor in common with 2.
However, I have a simple argument using a more complicated expression
which includes families of polynomials. But several people keep
arguing about it.
I've written a paper and it is currently at a journal.
And I'm waiting to hear further from them.
James Harris
C. Bond
> Oh well, I did make a mistake.
[snip]
> and I'm very disappointed but I guess I keep wishing for a simple resolution.
>
> That wasn't it.
>
> James Harris
Your problem is not just that you make mistakes, but that you defend them to the point of attacking your
critics. Shame on you.
--
There are two things you must never attempt to prove: the unprovable -- and the obvious.
W. Dale Hall
Yes, wishing is a good thing. Insisting that every scribble of yours is
superior to the best efforts of mathematicians over the ages is another.
I've said it before, and no doubt I'll say it again, but you would do
better if you actually learned a little bit of algebra. Not what you're
calling algebra (which is little more than the formal manipulation of
polynomials), but something about rings and fields. The basic facts
of rings and fields are not all that hard to understand, and elementary
Galois theory would take you a bit of work, but your efforts might be
rewarded by allowing you to see what it is you're attempting. Further,
you might actually come to understand why your particular approach is
unlikely to bear any real fruit.
I know it's fun to feel as though you're covering ground that has
never been touched on before. Your problem is that you sense that
this is the case whenever you stumble across something *you've*
never seen before. There is a huge difference between new *to you*
and new *in the history of mathematics*. Not a single one of your
purported discoveries fits the latter description of "new", and
it is to your discredit that you fail to realize that.
A little bit of education would help you recognize that there *are*
some general principles at work, and as long as your goal violates
those principles, it will remain unattainable. Beyond that rather
basic realization, having at your command some recognizeable
technique in so-called "higher" algebra will allow you to write
with fewer colloquialisms, less vague definitions, and certainly
greater reliability.
Less of this:
>
> My warning to universities and mathematicians teaching at them
> remains.
>
> Some of you may face civil liabilities, maybe worse--angry
> parents--when the full story comes out.
>
> School is about to start, and if universities teach the flawed
> mathematics they may also face federal authorities for deliberate
> fraud.
>
> Yup, some of you are looking at chaos as you scramble to figure out
> how to prevent yourself from being legally liable, but I've warned for
> some time now.
>
> It's your fault if you ignored me.
>
> It is mathematics after all.
>
would be a step in the right direction. You claim to have a
proof of something, and when folks deny that claim, or worse
yet, produce facts that refute your claims, it comes down to
these idle threats. You might suppose that such threats put
some form of fear into peoples hearts, but that supposition is
laughable. A threat that does not progress into action is simply
evidence that the person making the threat doesn't have the
will, or courage, or even the conviction of his beliefs, sufficient
to the task of making good on them.
Then, when it inevitably happens, you find the error, or actually
understand the counter-arguments, you further cement your well-
earned reputation as a fool. Is that really what you want?
Putting your picture up seems to have been a plea for people
to remember that you're human. I'll try to go easy on this,
but ...um, isn't EVERYONE human? You're human, I'm human,
ALL GOD'S CHILLUNS IS HUMAN!
What's the solution?
Don't threaten. Understand.
If you can't do that, then go ahead and threaten, but make sure
to follow through on those threats:
1. Sue the bastards!
2. Get your congressman to sponsor the Math Trials.
3. Go on the TV: the network news shows, Oprah,
Ricki Lake, Jerry Springer, Montel, there must
be a hundred of those issue-oriented shows for
folks to spew what they think is wrong with the
world. I know 60 minutes is always on the lookout
for a good scandal. This one should be a beaut:
MATH FRAUD PERPETRATED FOR CENTURIES!!!
4. Go on the public speaking circuit.
5. Convince someone other than yourself. Then have
that person convince two people, those two then
convince two apiece, and so forth. Pretty soon,
you'll have an actual MOVEMENT! Riot in the
streets, engage in sit-ins, run candidates for
public office (if only elected positions for
the local school board), attend math lectures
and ask questions during the Q&A at the end.
6. Prepare and teach a course at your local
community college, or in the evening division
of a local university. Develop a following.
Then proceed as in Step 5 above.
7. Write a book. Go on the obligatory book-signing
junket. Make sure to have the required cliches
so that you'll have good answers, and make good TV.
Or, you could exercise the alternative path:
1a. Continue along the same path you've done for 7 years.
2a. Periodically write up a note with some critical flaw,
and submit it to sci.math. Embrace the failure.
3a. Rant and rave and piss and moan about how you get
no respect. Take a page from Rodney Dangerfield.
>
> James Harris
The choice is yours.
Dale.
C. Bond
They do. This has been established in previous posts.
> However, I have a simple argument using a more complicated expression
> which includes families of polynomials. But several people keep
> arguing about it.
Because it is wrong.
> I've written a paper and it is currently at a journal.
>
> And I'm waiting to hear further from them.
>
> James Harris
I suppose you plan to consider what response you get from them? You certainly don't seem to make a habit of
that.
Will Twentyman
> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03073...@posting.google.com>...
>
Why do you respond to yourself more than to us?
Brian Quincy Hutchings
he must try to fit into the charicature of "four-eyes."
find a doctor of philosophy, and
get a prescription for a nice Club Med, Mister Monster Math.
or, at the least, make a hypothesis and try to break it, or
prove it. I read a great definition in the paper,
the other day, but I don't recall the example, humorous,
just that it linked "was necessary, but not sufficient."
that's all that you have to do,
grapple with those two words o'Leibniz.
Will Twentyman <wtwen...@read.my.sig> wrote in message news:<3f280a3f$1_2@newsfeed>...
> > y_1 = (r^2 + 2r)(-1 + sqrt(2))^{1/3}
> >
> > y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1 + sqrt(2))^{1/3}
> >
> > y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1 + sqrt(2))^{1/3}
> >
> > and r^2/2 *should* be an algebraic integer, but provably is not.
>
> You keep saying *should*. Deal with the fact that it's not and draw
> your conclusions from that. Either that or define more clearly *why*
> you think it should be something it isn't.
>
> >
> > However, notice that you run into a problem with r^2/2 not having a
> > factor of 2 that is 2.
>
> Only through an artificial construction.
> > However, notice that it is true that y_2 and y_3 must be coprime to 2.
>
> Or not, as others have shown.
--les ducs d'Enron!
http://members.tripod.com/~american_almanac
Virgil
jst...@msn.com (James Harris) wrote:
> And as I pointed out before there have been people who have been
> attacking algebra itself in posts. Now it still bothers me that I was
> the one person standing up to defend algebra itself from what I
> noticed.
The image of JSH as a stand up guy defending any part of mathemetics
from the attacks of others is hysterical.
James Harris
>
> Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c
> is a factor of the contant term P(0), while r=g-c, so it varies if g
> varies.
>
> That simple lemma not only anchors my paper, and my proof of Fermat's
> Last Theorem, it can be used in lots of places. Now I'll use it with
> the roots of
>
> y^3 + 3y - 2
I find its roots fascinating.
> and the three roots are
>
> y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3}
>
> y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 -
>
> (-1-sqrt(-3))/2(1+sqrt(2))^{1/3}
>
> y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 -
>
> (-1+sqrt(-3))/2(1+sqrt(2))^{1/3}
As simple as they are you can't just take the roots and *look* at them
to tell whether or not any one of them is coprime to 2. It is just
impossible.
Yet I work that out and still I keep coming back, as the simplicity
and elegance of those roots keeps calling me.
Sure I know that the cuberoot is ambiguous, so depending on how you
take the cuberoot, you shift between the solutions, which is why you
can't prove whether or not one of them is coprime to 2, but still it
seems like there should be some way to see by looking.
But there isn't.
Given the impossibility of proving that none of them are coprime to 2,
you'd think that people objecting would sit back, but nope, they claim
Galois Theory eliminates a possibility.
But that can't be. Galois Theory is based on that shifting.
Actually, they can neither prove nor disprove the assertion with the
tools they're using.
> where the polynomial I'm interested in is
>
> P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}.
>
> Letting
>
> g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1,
>
> I have c=1, so r=g-1.
>
> (Some may worry about my choice for c, I'll come back to that later.)
>
> Now using the substitution g=r+1, with x=sqrt(2), with my roots, I
> have
>
> y_1 = r+1 - 1/(r+1)
>
> y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1)
>
> y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1)
>
> so I have
>
> y_1 = (r^2 + 2r)/(r+1)
>
> y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1)
>
> y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1)
<deleted>
So once again I found myself drawn to that simplicity, looking for
some way, some how to make the math definite to the eye--to make it
eliminate a possibility on sight.
But it doesn't. You can't look at those roots and tell.
There's just no way to limit the math in that way, and intellectually
I know why:
The math has the flexibility to handle several possibilities. For
some polynomials, none of roots would be coprime to integer factors of
the constant term, while for others, like y^3 + 3y - 2, several are
coprime.
But you have to use advanced techniques to prove it.
Still something keeps drawing me back to stare at those roots, looking
for what cannot be found--a limitation that I can see.
It's like in physics where you have quarks that can't be pulled out
one by one. You can determine they're there experimentally in pairs
or threesomes. You can bounce things off them, but you can't *see*
them.
And there's such a need to see.
James Harris
C. Bond
> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03073...@posting.google.com>...
> > My paper does rest fully on the following rather basic lemma:
> >
> > Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c
> > is a factor of the contant term P(0), while r=g-c, so it varies if g
> > varies.
> >
> > That simple lemma not only anchors my paper, and my proof of Fermat's
> > Last Theorem, it can be used in lots of places. Now I'll use it with
> > the roots of
> >
> > y^3 + 3y - 2
>
> I find its roots fascinating.
You are again answering your own posts with narcissistic monologues instead of answering your critics. If
you had any integrity, you would deal with the challenges not carry on conversations with yourself. Talk to
a mirror if you want, but why use a newgroup to talk to yourself.
> > and the three roots are
> >
> > y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3}
> >
> > y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 -
> >
> > (-1-sqrt(-3))/2(1+sqrt(2))^{1/3}
> >
> > y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 -
> >
> > (-1+sqrt(-3))/2(1+sqrt(2))^{1/3}
>
> As simple as they are you can't just take the roots and *look* at them
> to tell whether or not any one of them is coprime to 2. It is just
> impossible.
You don't have to just look. The observation that y_1 y_2 y_3 = 2 is sufficient to prove that each root is
a divisor of 2 in the ring of algebraic integers and, since none of these is a unit, none are coprime to 2.
> Yet I work that out and still I keep coming back, as the simplicity
> and elegance of those roots keeps calling me.
Didn't someone 'call you home' once? If so, how come you came back?
> Sure I know that the cuberoot is ambiguous, so depending on how you
> take the cuberoot, you shift between the solutions,
What the hell does that mean?
> which is why you
> can't prove whether or not one of them is coprime to 2,
Oh yes, you can. The product of the roots, y_1 y_2 y_3 equals 2. Hence, each of the roots divides 2 in the
ring of algebraic integers. This is because the quotient equals the product of the other two roots and the
product of two algebraic integers is an algebraic integer. Hence none of the roots are coprime to 2.
> but still it
> seems like there should be some way to see by looking.
>
> But there isn't.
So what? If mathematician's could prove everything by 'looking' they wouldn't need theorems and lemmas.
> Given the impossibility of proving that none of them are coprime to 2,
Hold it. This is proven by noting that the product of the roots, y_1 y_2 y_3, equals 2 and that each of the
roots divides 2 in the ring of algebraic integers -- meaning the quotient is an algebraic integer. Hence
none of the roots are coprime to 2. QED
> you'd think that people objecting would sit back, but nope, they claim
> Galois Theory eliminates a possibility.
I don't.
> But that can't be. Galois Theory is based on that shifting.
> Actually, they can neither prove nor disprove the assertion with the
> tools they're using.
Galois Theory isn't necessary in this case. 'Basic algebra' does the trick. You wouldn't want to attack
'basic algebra', would you?
> <deleted>
>
> So once again I found myself drawn to that simplicity, looking for
> some way, some how to make the math definite to the eye--to make it
> eliminate a possibility on sight.
>
> But it doesn't. You can't look at those roots and tell.
As has been shown above, it isn't necessary to 'look at those roots and tell'. None of them is coprime to
2. You have been refuted, defeated, vanquished, humiliated, crushed, conquered, exposed and undone. Take
down your false "proof" and acknowledge your errors.
> There's just no way to limit the math in that way, and intellectually
> I know why:
>
> The math has the flexibility to handle several possibilities. For
> some polynomials, none of roots would be coprime to integer factors of
> the constant term, while for others, like y^3 + 3y - 2, several are
> coprime.
None of them are coprime. The product of all the roots, y_1 y_2 y_3, is equal to 2. Each root divides 2
producing an algebraic integer quotient. Hence, in the ring of algebraic integers, none of the roots, count
'em again!, none is coprime to 2. Your claim is false.
> But you have to use advanced techniques to prove it.
You can't prove your claim at all, because it is false.
> Still something keeps drawing me back to stare at those roots, looking
> for what cannot be found--a limitation that I can see.
The limitation is between your ears, not in your eyes.
> It's like in physics where you have quarks that can't be pulled out
> one by one. You can determine they're there experimentally in pairs
> or threesomes. You can bounce things off them, but you can't *see*
> them.
No, it isn't like quarks. It is like an LSD trip.
> And there's such a need to see.
>
> James Harris
Open your eyes. You have been shown the way. To paraphrase Samuel Johnson: I can give you a solution, but I
can't give you an understanding.
Thomas Wasell
<jst...@msn.com> wrote:
[snip]
>The math has the flexibility to handle several possibilities. For
>some polynomials, none of roots would be coprime to integer factors of
>the constant term, while for others, like y^3 + 3y - 2, several are
>coprime.
(I assume your talking about _algebraic integers_.) The constant term of
y^3 + 3y - 2 is -2. If Y is _any_ root of y^3 + 3y - 2, then (-2)/Y is root
of x^3 + 3x^2 + 4. That's a monic polynomial, so (-2)/Y is an algebraic
integer, and therefore Y divides -2 in the ring of algebraic integers. So Y
in _not_ coprime to -2.
>But you have to use advanced techniques to prove it.
No, you don't. Solving cubics is not an "advanced technique".
>Still something keeps drawing me back to stare at those roots, looking
>for what cannot be found--a limitation that I can see.
>
>It's like in physics where you have quarks that can't be pulled out
>one by one. You can determine they're there experimentally in pairs
>or threesomes. You can bounce things off them, but you can't *see*
>them.
If that's not seeing (in physics), then what is?
>And there's such a need to see.
Yes. And (almost) everybody _does_.
--
Thomas Wasell | "I'd love to go out with you, but I'm having all my
was...@bahnhof.se | plants neutered."
James Dolan
c. bond <cb...@ix.netcom.com> wrote:
|> Sure I know that the cuberoot is ambiguous, so depending on how you
|> take the cuberoot, you shift between the solutions,
|
|What the hell does that mean?
it means that he's much closer to actually understanding something
about mathematics in this post than he has been in just about any of
his other posts. he's still horribly mixed up, but there's an
important truth that he almost seems close to understanding here.
|> But that can't be. Galois Theory is based on that shifting.
|> Actually, they can neither prove nor disprove the assertion with the
|> tools they're using.
|
|Galois Theory isn't necessary in this case.
galois theory is highly relevant and perhaps crucial here.
|> It's like in physics where you have quarks that can't be pulled out
|> one by one. You can determine they're there experimentally in pairs
|> or threesomes. You can bounce things off them, but you can't *see*
|> them.
|
|No, it isn't like quarks. It is like an LSD trip.
no, it's in fact very much like quarks, and in a surprisingly deep
way. but yes, it's also like an lsd trip.
--
[e-mail address jdo...@math.ucr.edu]
Wayne Brown
> I've written a paper and it is currently at a journal.
> And I'm waiting to hear further from them.
Why wait? Everyone already knows what's going to happen. They're
going to reject it, and you're going to make excuses. Why not start
rationalizing your failure *now* and get a head start?
--
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"
C. Bond
> in article <3f2a7bde...@ix.netcom.com>,
> c. bond <cb...@ix.netcom.com> wrote:
>
> |> Sure I know that the cuberoot is ambiguous, so depending on how you
> |> take the cuberoot, you shift between the solutions,
> |
> |What the hell does that mean?
>
> it means that he's much closer to actually understanding something
> about mathematics in this post than he has been in just about any of
> his other posts. he's still horribly mixed up, but there's an
> important truth that he almost seems close to understanding here.
Why not enlighten the rest of us then? Evidently James isn't quite up to
it, and you could perform a service by identifying the 'important truth he
almost seems close to understanding'.
> |> But that can't be. Galois Theory is based on that shifting.
> |> Actually, they can neither prove nor disprove the assertion with the
> |> tools they're using.
> |
> |Galois Theory isn't necessary in this case.
>
> galois theory is highly relevant and perhaps crucial here.
It may be relevant, but it certainly isn't necessary or crucial in proving
that the roots of a monic polynomial with integer coefficients are divisors
of the constant term within the ring of algebraic integers.
> |> It's like in physics where you have quarks that can't be pulled out
> |> one by one. You can determine they're there experimentally in pairs
> |> or threesomes. You can bounce things off them, but you can't *see*
> |> them.
> |
> |No, it isn't like quarks. It is like an LSD trip.
>
> no, it's in fact very much like quarks, and in a surprisingly deep
> way. but yes, it's also like an lsd trip.
Please provide an in-depth explanation. Why leave the less informed hanging
when you are able to clarify why it is that proving/disproving coprimality
within the ring of algebraic integers is like quarks?
> --
>
> [e-mail address jdo...@math.ucr.edu]
Looking forward to your follow-up, as I assume your post was not a 'hit and
run' contribution.
Arturo Magidin
>> My paper does rest fully on the following rather basic lemma:
>>
>> Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c
>> is a factor of the contant term P(0), while r=g-c, so it varies if g
>> varies.
>>
>> That simple lemma not only anchors my paper, and my proof of Fermat's
>> Last Theorem, it can be used in lots of places. Now I'll use it with
>> the roots of
>>
>> y^3 + 3y - 2
>
>I find its roots fascinating.
>
>> and the three roots are
>>
>> y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3}
>>
>> y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 -
>>
>> (-1-sqrt(-3))/2(1+sqrt(2))^{1/3}
>>
>> y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 -
>>
>> (-1+sqrt(-3))/2(1+sqrt(2))^{1/3}
>
>As simple as they are you can't just take the roots and *look* at them
>to tell whether or not any one of them is coprime to 2. It is just
>impossible.
No, it is not impossible. It is amazingly trivial, as Keith Ramsay
pointed out right after you posted them:
NONE OF THEM ARE COPRIME TO 2. THEY ARE ALL DIVISORS OF 2:
y_1*(y_1^2+3) = 2
y_2*(y_2^2+3) = 2
y_3*(y_3^2+3) = 2.
Surely you agree with all three of those expressions, since they are
all roots of y^3 + 3y - 2.
So they are all divisors of 2, and since none of them are units, they
are none of them coprime to 2. Simple, nay, trivial.
[.snip.]
>Given the impossibility of proving that none of them are coprime to 2,
it follows that red is blue, 0 is 1, and I am the Pope (who,
therefore, is not catholic).
>you'd think that people objecting would sit back, but nope, they claim
>Galois Theory eliminates a possibility.
Not even Galois theory is needed here. ->THIS<- one is trivial.
[.snip.]
Arturo Magidin
C. Bond <cb...@ix.netcom.com> wrote:
[.snip.]
Nitpicking...
The first time you made the point, you correctly stated:
>You don't have to just look. The observation that y_1 y_2 y_3 = 2 is sufficient to prove that each root is
>a divisor of 2 in the ring of algebraic integers and, since none of these is a unit, none are coprime to 2.
[.snip.]
Later, you said, in several different versions, things like:
>Oh yes, you can. The product of the roots, y_1 y_2 y_3 equals 2. Hence, each of the roots divides 2 in the
>ring of algebraic integers. This is because the quotient equals the product of the other two roots and the
>product of two algebraic integers is an algebraic integer. Hence none of the roots are coprime to 2.
The last statement only follows once you add "since none of them is a
unit [in the ring of algebraic integers]."
C. Bond
> In article <3F2A7BDE...@ix.netcom.com>,
> C. Bond <cb...@ix.netcom.com> wrote:
>
> [.snip.]
>
> Nitpicking...
>
> The first time you made the point, you correctly stated:
>
> >You don't have to just look. The observation that y_1 y_2 y_3 = 2 is sufficient to prove that each root is
> >a divisor of 2 in the ring of algebraic integers and, since none of these is a unit, none are coprime to 2.
>
> [.snip.]
>
> Later, you said, in several different versions, things like:
>
> >Oh yes, you can. The product of the roots, y_1 y_2 y_3 equals 2. Hence, each of the roots divides 2 in the
> >ring of algebraic integers. This is because the quotient equals the product of the other two roots and the
> >product of two algebraic integers is an algebraic integer. Hence none of the roots are coprime to 2.
>
> The last statement only follows once you add "since none of them is a
> unit [in the ring of algebraic integers]."
Thanks for the 'heads up'. I tried to make certain this was explicitly included after I realized I had left it
off one or more repititions of the same, or similar, argument. I don't take it as nitpicking, by the way. The
argument is so short that there is no reason (or excuse) for it to be incomplete.