Algebraic integer coverage result
C. BOND
>
> That proves there must exist d1 where it is an algebraic integer
> factor of 1 as it is a root of
>
> d^6 - 1488*d^4 + 4176*d^2 - 5.
>
The roots of this polynomial are:
d1,2 = +/-sqrt(496+244624/((1/2(241976581+10683sqrt(12415)i))^(1/3))+
(1/2(241976581+10683sqrt(12415)i))^(1/3))
d3,4 =
+/-sqrt(1/2(992-244624(1+sqrt(3)i)/((1/2(241976581+10683sqrt(12415)i))^(1/3))-
(1-sqrt(3)i)((1/2(241976581+10683sqrt(12415)i))^(1/3))
d5,6 =
+/-sqrt(1/2(992-244624(1-sqrt(3)i)/((1/2(241976581+10683sqrt(12415)i))^(1/3))-
(1+sqrt(3)i)((1/2(241976581+10683sqrt(12415)i))^(1/3))
Perhaps you would be good enough to match these roots with the
properties you claim they exhibit.
> I would hope that those making replies consider carefully before
> replying, but I do appreciate replies that address the mathematics.
We all hope that you do the same.
> I will continue to explain in detail as necessary.
>
> James Harris
--
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
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>The following shows that the irreducible polynomial
>
> d^6 - 1488*d^4 + 4176*d^2 - 5
>
>has roots that are algebraic integer factors of 1.
>
>I'll be using data from a post by a poster who called herself "Dot"
>
>messageid: <210420032353503610%d...@at.dot.com>
>
><Quote>
>We will work in the number field defined by the polynomial
>
>y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
> + 2460349402142038081821721600;
>
>
>We will let alpha be a root of this polynomial, and we will define
>algebraic integers d1, d2, d3, n1, n2, n3 in terms of alpha.
>(Taking alpha to be the real root that is approximately
> 154.451496990997802982236692688237765859998280922
>gives the numerical examples I gave earlier.
>
>
>
>MAGMA CODE:
>
>R<y>:=PolynomialRing(Rationals());
>
>f := y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
> + 2460349402142038081821721600;
>
>K<alpha>:=NumberField(f);
>
>
>// Define our coefficients:
>
>d1 := (1/1401502976449936390824092894020468134024616320)*
> (- 116858525730451586855252258697*alpha^11
> + 21546461434737374316017502217792*alpha^10
> + 22109717335034240282753967154849470*alpha^9
> - 4076616001448173117175959963381396768*alpha^8
> - 1647209069948549841509121657588818730783*alpha^7
> + 303715513906432298157776614831268988666880*alpha^6
> + 60404344550512455043182143206734511673314206*alpha^5
> - 11137524588944779726233177935687516957256627968*alpha^4
> - 1089475637849489503980551259022753852684824408489*alpha^3
> + 200881897616666532603687714008711443626111508690592*alpha^2
> + 7719396850213757345324975434862758616298359729245140*alpha
> - 1423343550253117792241514076181168733734615795436123200);
>
>d2 := (1/1401502976449936390824092894020468134024616320)*
> (- 116858525730451586855252258697*alpha^11
> - 21546461434737374316017502217792*alpha^10
> + 22109717335034240282753967154849470*alpha^9
> + 4076616001448173117175959963381396768*alpha^8
> - 1647209069948549841509121657588818730783*alpha^7
> - 303715513906432298157776614831268988666880*alpha^6
> + 60404344550512455043182143206734511673314206*alpha^5
> + 11137524588944779726233177935687516957256627968*alpha^4
> - 1089475637849489503980551259022753852684824408489*alpha^3
> - 200881897616666532603687714008711443626111508690592*alpha^2
> + 7719396850213757345324975434862758616298359729245140*alpha
> + 1423343550253117792241514076181168733734615795436123200);
>
>d3 := (1/65595009662544996294303701863730606291520)*
> (- 240064724443796467*alpha^11
> + 55326448296076857699530*alpha^9
> - 5309131403735081126863921733*alpha^7
> + 274489395789430590995126949847466*alpha^5
> - 8357164018995647094694127739446419219*alpha^3
> + 115535699440399080831946228351058377038940*alpha);
>
>
>n1 := (1/10683)*(-22*d1^4 + 33001*d1^2 - 394);
>n2 := (1/10683)*(-22*d2^4 + 33001*d2^2 - 394);
>n3 := (1/10683)*(-22*d3^4 + 33001*d3^2 - 394);
>
>// Verify that the coefficients are algebraic integers.
>// We verify that the following monic polynomials
>// evaluate to zero:
>
>n1^3 - 54*n1^2 + 390*n1 + 13;
>n2^3 - 54*n2^2 + 390*n2 + 13;
>n3^3 - 54*n3^2 + 390*n3 + 13;
>
>d1^6 - 1488*d1^4 + 4176*d1^2 - 5;
>d2^6 - 1488*d2^4 + 4176*d2^2 - 5;
>d3^6 - 1488*d3^4 + 4176*d3^2 - 5;
>
></Quote>
>*********************************************************************
>
>As it was kind of long I'll point out here that I'm finished with the
>main quote and will now quote from it for what follows.
>
>Ok, I'll use Dot's d1, which is
>
>d1 := (1/1401502976449936390824092894020468134024616320)*
> (- 116858525730451586855252258697*alpha^11
> + 21546461434737374316017502217792*alpha^10
> + 22109717335034240282753967154849470*alpha^9
> - 4076616001448173117175959963381396768*alpha^8
> - 1647209069948549841509121657588818730783*alpha^7
> + 303715513906432298157776614831268988666880*alpha^6
> + 60404344550512455043182143206734511673314206*alpha^5
> - 11137524588944779726233177935687516957256627968*alpha^4
> - 1089475637849489503980551259022753852684824408489*alpha^3
> + 200881897616666532603687714008711443626111508690592*alpha^2
> + 7719396850213757345324975434862758616298359729245140*alpha
> - 1423343550253117792241514076181168733734615795436123200);
>
>so
>
> 1401502976449936390824092894020468134024616320 d1 :=
>
> + 22109717335034240282753967154849470*alpha^9
> + 303715513906432298157776614831268988666880*alpha^6
> + 7719396850213757345324975434862758616298359729245140*alpha
> - 1423343550253117792241514076181168733734615795436123200
>
> - 116858525730451586855252258697*alpha^11
> + 21546461434737374316017502217792*alpha^10
>
> - 4076616001448173117175959963381396768*alpha^8
> - 1647209069948549841509121657588818730783*alpha^7
> + 60404344550512455043182143206734511673314206*alpha^5
> - 11137524588944779726233177935687516957256627968*alpha^4
> - 1089475637849489503980551259022753852684824408489*alpha^3
> + 200881897616666532603687714008711443626111508690592*alpha^2
>);
>
>Where I've grouped terms with a coefficient with a factor of 5.
>
>Note that because alpha is a root of
>
> y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
> + 2460349402142038081821721600
>
>there must exist an alpha which is not coprime to 5.
>
>Assuming that d1 is not coprime to 5, and alpha is not either and that
>they share non unit factors of 5 let's call any shared non unit
>factors of 5 between d1 and alpha, f.
So, the existence of f is an assumption this time?
Have you dropped the claim that if d1 is not coprime to 5, then there
must a nonunit which divides d1, 5, and alpha, all at the same time?
Because that was the claim that I was saying I was not clear about. If
rather than an assertion, it is merely an assumption, then there is no
problem. Of course, we must remember we are making this assumption.
So, it seems we are making the following assumption, presumably in
search of a contradiction:
ASSUMPTION: There exists an algebraic integer f, which is not a
unit, and which divides alpha, d1, and 5.
>From there I get
>
>
> 1401502976449936390824092894020468134024616320 d1 :=
>
> 22109717335034240282753967154849470*alpha^9
> + 303715513906432298157776614831268988666880*alpha^6
> + 7719396850213757345324975434862758616298359729245140*alpha
> - 1423343550253117792241514076181168733734615795436123200
>
> (alpha^2)(- 116858525730451586855252258697*alpha^9
> + 21546461434737374316017502217792*alpha^8
>
> - 4076616001448173117175959963381396768*alpha^6
> - 1647209069948549841509121657588818730783*alpha^5
> + 60404344550512455043182143206734511673314206*alpha^3
> - 11137524588944779726233177935687516957256627968*alpha^2
> - 1089475637849489503980551259022753852684824408489*alpha^1
> + 200881897616666532603687714008711443626111508690592)
>);
>
>and letting
>
> j = 22109717335034240282753967154849470*alpha^9
> + 303715513906432298157776614831268988666880*alpha^6
> + 7719396850213757345324975434862758616298359729245140*alpha
> - 1423343550253117792241514076181168733734615795436123200
>
>I have that j/5 would not be coprime to f because each of the
>coefficients next to an alpha has a factor of 5, so the alpha's would
>still be left, and the last coefficient
>
> - 1423343550253117792241514076181168733734615795436123200
>
>has a factor of 25, so it'd have a factor of f left.
This is fine.
[.snip.]
>Now letting
>
> k = (- 116858525730451586855252258697*alpha^9
> + 21546461434737374316017502217792*alpha^8
>
> - 4076616001448173117175959963381396768*alpha^6
> - 1647209069948549841509121657588818730783*alpha^5
> + 60404344550512455043182143206734511673314206*alpha^3
> - 11137524588944779726233177935687516957256627968*alpha^2
> - 1089475637849489503980551259022753852684824408489*alpha^1
> + 200881897616666532603687714008711443626111508690592)
>
>I note that k does not have factors of f (remember alpha does) because
>of the coefficient
>
> 200881897616666532603687714008711443626111508690592
This is also okay. If f divides both alpha and 5, then f cannot divide k.
>and now I can write
>
> 1401502976449936390824092894020468134024616320 d1 := j + alpha^2 k
>
>and consider that because
>
> 1401502976449936390824092894020468134024616320
>
>has a factor of 5, that is 5, of course, I can divide that off and
>consider
>
> j/5 + (alpha^2 k)/5
This is also fine, though it has not been explicitly justified that 5
divides alpha^2*k; however, since 5 divides both the left hand side and
j, it must divide alpha^2*k. So both summands are algebraic integers.
>and I've already noted above that j/5 still has factors of f, so now
>(alpha^2 k)/5 would be forced to have factors of f because of d1,
Yes, that's also correct.
>and
>remember from before that k does not have factors of f if alpha does
>(and it does).
Yes.
>However, consider what happens when we use a root alpha with
>
> y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
> + 2460349402142038081821721600;
>
>to get
>
> (alpha)^12 - 226446*(alpha)^10 + 21142596231*(alpha)^8 -
>1041903035971246*(alpha)^6 + 28575240325442631201*(alpha)^4 -
>413298187541059968292500*(alpha)^2 + 2460349402142038081821721600 = 0
>
>and notice that the last coefficient has a factor that is 25 and then
>notice the coefficient of (alpha)^2 has a factor of 25 as well.
>
>and if you divide through by alpha^2 you get
>
> (alpha)^10 - 226446*(alpha)^8 + 21142596231*(alpha)^6 -
>1041903035971246*(alpha)^4 + 28575240325442631201*(alpha)^2 -
>413298187541059968292500
> + 2460349402142038081821721600/(alpha^2) = 0
Since alpha^2 divides the last term, this is fine. All terms are
algebraic integers.
I think I glimpse what you are trying to say here, which is that if
gcd(alpha,5)=x, then x^2 must still divide
2460349402142038081821721600/(alpha^2),
since it divides all the other terms.
I'll pick this up below...
>so if there exists an algebraic integer h, such that
>
> alpha^2 h = 5^q
>
>where alpha and h are coprime to each other and q is a positive real
A ->real<- number? Surely you do not mean things like 5^{pi}. You must
mean a ->rational<- number.
But why should h exist? What makes you think that alpha^2 DIVIDES a
power of 5?
For example, you agree that some alphas must have factors of 2, right?
What if all of them do? Then h cannot exist at all. You need to write
something like
alpha^2*h = 5^q*m
for some integer m; you probably want m to be prime to 5. That's fine,
you can do that, by adjusting h as needed. And you probably also want,
not so much h prime to alpha, but h prime to the gcd of alpha and 5;
that is, that h does not contribute any x's to the procedure. (note
that f divides the gcd of alpha and 5, so this would still make h
coprime to f, in case that is what you wanted).
But note that we are again making an assumption:
ASSUMPTION 2: there exists an h, which is COPRIME to gcd(alpha,5), and
such that alpha^2*h = 5^q*m, with m coprime to 5 and q a positive
rational number.
>where q>1 then the last term would be coprime to 5.
I certainly did not see that; by doing some of the stuff that you
claimed was "unnecessary complication", I actually avoided your gappy
arguments about q and proved that no such h can exist, under the
assumption that f exists. Here is the argument:
I shall assume my correction, but note that the existence of h is
being assumed. It could be that no such h exists, even though alpha is
not coprime to 5...
So, basically, alpha^2 divides
2460349402142038081821721600 = 24603494021420380818217216*4*5^2.
So if we assume that alpha has a gcd of x with 5, say
alpha = x*y
5 = x*z
(and so f will be a divisor of x),
with y and z coprime, x not a unit, we also have that x^2*alpha^2
divides
24603494021420380818217216*4*5^2.
We have:
alpha^2*w = 5^2*4*24603494021420380818217216
for some algebraic integer w; therefore,
x^2*y^2*w = x^2*z^2*4*24603494021420380818217216
Cancelling the x's, we have
y^2*w = z^2*4*24603494021420380818217216
Since y and z are coprime, this proves that y divides, in the ring of
algebraic integers, an integer coprime to 5, and therefore, y is
coprime to 5. Therefore, y and x are coprime, in addition to y and z
being coprime.
We also know that x^2 still divides
2460349402142038081821721600/(alpha^2)
So, using the notation above, we have
w = 4*24603494021420380818217216*(x^2)*(z^2)/x^2*y^2
= [(4*24603494021420380818217216)/y^2]*(z^2/x^2)
with both fractions actually algebraic integers. Then w is a multiple
of x^2, x^2 divides z^2/x^2, and therefore, x^4 divides
z^2, so x^2 divides z.
That means that we can write 5 as 5 = xz = x*x^2*z', for some
algebraic integer z', and so that x^3 divides 5.
In fact, I think this settles the issue of f directly: you know that
alpha^2*k/5 is an algebraic integer, multiple of f (which divides
f. We know that k is coprime to 5 (and so to x), and that alpha=xy,
with x and y coprime. Then we have
x^2*y^2*k/5 = y^2*k/x*z' an algebraic integer,
but x is coprime to both y and k. This is impossible. No need to go on
with your q.
However, let us consider what it is we are contradicting: what we are
contradicting is the assumption 2 we made: that
ASSUMPTION 2: there exists an h, which is COPRIME to gcd(alpha,5), and
such that alpha^2*h = 5^q*m, with m coprime to 5 and q a positive
rational number.
Now, you say below:
> and
> alpha^2 h = 5
>
>must be true, as anything greater, like
>
> alpha^2 h = 5 sqrt(5)
>
>would create a contradiction,
This is not the only other alternative (assuming we fix the statement
by adding the m which is coprime to 5). In fact, this is the
alternative you just finished supposedly discarding. The alternative
to h existing would be that no such h exists: that is, that for any
algebraic integer h such that
alpha^2*h = 5*m with m an algebraic integer coprime to 5, it must be
that gcd(alpha,5) and h are not coprime.
This could happen, for example, if we had a situation like
alpha = (1-2i)*sqrt(1+2i); then no power of alpha is a rational power
of 5, and any h we find to make alpha^2*h = 5^q*m with m coprime to 5
would be non-coprime to alpha. (I am not saying alpha has this value,
I am saying that kind of situation could occur).
So I do not see how you have established the existence of h at all.
I have pointed out a couple of problems, it is probably best to stop
here, but let me add one more point. Assuming you have indeed
established that such an h would exist, coprime to gcd(alpha,5) such
that alpha^2*h = 5*m, with m coprime to 5, you continue:
> therefore
>
>
> j/5 + (alpha^2 k)/5
>
>cannot have factors of f, that is, it is coprime to f, which proves
>that d1 is coprime to f.
I think this is right: if you could find h as the above, then it would
follow, since alpha = x*y and y is coprime to x and 5, that 5 =
x^2*gcd(h,5), and gcd(h,5) is coprime to x. Therefore, even though j/5
is divisible by x, you would have that alpha^2*k/5 is not divisible by
x, which would give you indeed a contradiction.
What you would have contradicted (assuming we were agreed that the
existence of h was necessarily true as well), is the following
assumption:
ASSUMPTION: There exists an algebraic integer f, which is not a
unit, and which divides alpha, d1, and 5.
>Therefore, d1 is coprime to alpha with respect to factors of 5.
But you do realize that this is NOT the same as saying that d1 is
coprime to 5, right? Just that any common factors of d1 with 5 must be
coprime to alpha.
I will stop here, to see if we clear up the problem of the remaining
alternatives to the second assumption, namely to
ASSUMPTION 2: there exists an h, which is COPRIME to alpha, and
such that alpha^2*h = 5^q*m, with m coprime to 5 and q a positive
rational number.
======================================================================
"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
Virgil
the subject. it tends to pollute the NG.
In article
<3c65f87.03043...@posting.google.com>,
jst...@msn.com (James Harris) wrote:
> And I would divide out but I'd have to write a quick program to do so
> with such big numbers and I would rather just let people look at the
> last two digits.
>
> Now letting
>
> k = (- 116858525730451586855252258697*alpha^9
> + 21546461434737374316017502217792*alpha^8
>
> - 4076616001448173117175959963381396768*alpha^6
> - 1647209069948549841509121657588818730783*alpha^5
> + 60404344550512455043182143206734511673314206*alpha^3
> - 11137524588944779726233177935687516957256627968*alpha^2
> - 1089475637849489503980551259022753852684824408489*alpha^1
> + 200881897616666532603687714008711443626111508690592)
>
> I note that k does not have factors of f (remember alpha does) because
> of the coefficient
>
> 200881897616666532603687714008711443626111508690592
>
> and now I can write
>
> 1401502976449936390824092894020468134024616320 d1 := j + alpha^2 k
>
> and consider that because
>
> 1401502976449936390824092894020468134024616320
>
> has a factor of 5, that is 5, of course, I can divide that off and
> consider
>
> j/5 + (alpha^2 k)/5
>
> and I've already noted above that j/5 still has factors of f, so now
> (alpha^2 k)/5 would be forced to have factors of f because of d1, and
> remember from before that k does not have factors of f if alpha does
> (and it does).
>
> However, consider what happens when we use a root alpha with
>
> y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
> + 2460349402142038081821721600;
>
> to get
>
> (alpha)^12 - 226446*(alpha)^10 + 21142596231*(alpha)^8 -
> 1041903035971246*(alpha)^6 + 28575240325442631201*(alpha)^4 -
> 413298187541059968292500*(alpha)^2 + 2460349402142038081821721600 = 0
>
> and notice that the last coefficient has a factor that is 25 and then
> notice the coefficient of (alpha)^2 has a factor of 25 as well.
>
> and if you divide through by alpha^2 you get
>
> (alpha)^10 - 226446*(alpha)^8 + 21142596231*(alpha)^6 -
> 1041903035971246*(alpha)^4 + 28575240325442631201*(alpha)^2 -
> 413298187541059968292500
> + 2460349402142038081821721600/(alpha^2) = 0
>
> so if there exists an algebraic integer h, such that
>
> alpha^2 h = 5^q
>
> where alpha and h are coprime to each other and q is a positive real
> where q>1 then the last term would be coprime to 5.
>
> Therefore, q<=1, and from j/5 + (alpha^2 k)/5, it can't be less than
> 1, so q=1, and
>
> alpha^2 h = 5
>
> must be true, as anything greater, like
>
> alpha^2 h = 5 sqrt(5)
>
> would create a contradiction, therefore
>
>
> j/5 + (alpha^2 k)/5
>
> cannot have factors of f, that is, it is coprime to f, which proves
> that d1 is coprime to f.
>
> Therefore, d1 is coprime to alpha with respect to factors of 5.
>
> Now moving on to Dot's d3, I have
>
> d3 := (1/65595009662544996294303701863730606291520)*
> (- 240064724443796467*alpha^11
> + 55326448296076857699530*alpha^9
> - 5309131403735081126863921733*alpha^7
> + 274489395789430590995126949847466*alpha^5
> - 8357164018995647094694127739446419219*alpha^3
> + 115535699440399080831946228351058377038940*alpha);
>
> so, grouping as before,
>
> 65595009662544996294303701863730606291520 d3 :=
>
> 55326448296076857699530*alpha^9
> + 115535699440399080831946228351058377038940*alpha
>
> - alpha^2(240064724443796467*alpha^9
> + 5309131403735081126863921733*alpha^5
> - 274489395789430590995126949847466*alpha^3
> + 8357164018995647094694127739446419219*alpha);
>
> and notice that separating off 5 from both sides, or dividing off if
> you prefer, would still leave the right side with factors of 5 if
> alpha is not coprime to 5.
>
> Those factors of 5, would then be forced to be factors of d3, so I
> immediately have that d1, d2 and d3 are pairwise coprime.
>
> Now looking at the polynomial that defines the d's which is
>
> d^6 - 1488*d^4 + 4176*d^2 - 5
>
> I notice that conclusion is confirmed by the coefficient 4176, as it
> is coprime to 5.
>
> However alpha is a root of
>
> y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
> + 2460349402142038081821721600;
>
> and but its coefficient
>
> 413298187541059968292500
>
> prevents its roots from being pairwise coprime with respect to
> algebraic integer factors of 5.
>
> Proving that d1 and d2 are coprime to 5 if d3 is not, because d3 must
> always have all the non unit algebraic integer factors of 5 of alpha,
> but if any alpha had factors of d1 or d2 then d3 would have to have
> those as well, as shown, contradicting with d1, d2 and d3 being
> pairwise coprime.
>
> That proves there must exist d1 where it is an algebraic integer
> factor of 1 as it is a root of
>
> d^6 - 1488*d^4 + 4176*d^2 - 5.
>
>
> I would hope that those making replies consider carefully before
> replying, but I do appreciate replies that address the mathematics.
>
David C. Ullrich
>The following shows that the irreducible polynomial
>
> d^6 - 1488*d^4 + 4176*d^2 - 5
>
>has roots that are algebraic integer factors of 1.
You've been fuzzing out on the word "factor" again lately.
If you mean that the equation has roots which are factors
of 1 _in_ the ring of algebraic integers, then no, that's
not so. (And when you make statements that someone
who knows as little about these things as _I_ do can
easily prove wrong you're sinking pretty low, btw.)
[...]
>
>That proves there must exist d1 where it is an algebraic integer
>factor of 1 as it is a root of
>
> d^6 - 1488*d^4 + 4176*d^2 - 5.
>
>
>I would hope that those making replies consider carefully before
>replying, but I do appreciate replies that address the mathematics.
That's what you _say_, but it's clearly not so. I posted a
proof that this is wrong a few days ago - you ignored it.
>I will continue to explain in detail as necessary.
>
>
>James Harris
******************
David C. Ullrich
James Harris
> The following shows that the irreducible polynomial
>
> d^6 - 1488*d^4 + 4176*d^2 - 5
>
> has roots that are algebraic integer factors of 1.
>
>
> messageid: <210420032353503610%d...@at.dot.com>
<deleted>
Unfortunately the argument given did not resolve the issue.
Fortunately it IS resovable by considering the polynomial
d^6 - 1488*d^4 + 4176*d^2 - 5
directly as amazingly, in a bit of luck I guess,
d(d^2 + 2d + 1 - 1490d + 4175) = d^6 - 1488*d^4 + 4176*d^2
so
d(d^2 + 2d + 1 - 1490d + 4175) = 5, which means
d[{d + 1)^2 - 1490d + 4175] = 5
which forces any root plus 1 to have the algebraic integer factors of
any other root that is not coprime to 5.
But looking at the solutions to
d^6 - 1488*d^4 + 4176*d^2 - 5 = 0
with that information proves that at least one root must be coprime to
5.
Those solutions are
{{x -> 496 + 244624/(1/2*(241976581 +
10683*I*Sqrt[12415]))^(1/3) +
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
{x -> 496 - (122312*(1 + I*Sqrt[3]))/
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3) -
1/2*(1 - I*Sqrt[3])*
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
{x -> 496 - (122312*(1 - I*Sqrt[3]))/
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3) -
1/2*(1 + I*Sqrt[3])*
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)}}
where the last two subtracted from each other give
244624/(1/2*(241976581 +
10683*I*Sqrt[12415]))^(1/3) +
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)}
times I*Sqrt[3] proving that
496 + 244624/(1/2*(241976581 +
10683*I*Sqrt[12415]))^(1/3) +
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
must be coprime to 5.
Note that I've been forced to give demonstrations such as this for a
PROOF that I've had for about a year now.
I think it unfair and unfortunate that mathematics itself wasn't
accepted, and I fear that even a secondary demonstration like this one
may not sway mathematicians.
I did my best. I found a short proof of Fermat's Last Theorem, and
then people wanted more.
Now I've given more, and I don't know if it will be enough.
James Harris
James Harris
> The following shows that the irreducible polynomial
>
> d^6 - 1488*d^4 + 4176*d^2 - 5
>
> has roots that are algebraic integer factors of 1.
<deleted>
I'll note again that what I had didn't resolve the issue.
I'm posting again though because of a small mistake in a lot of my
recent posts as I shifted from
d^6 - 1488*d^4 + 4176*d^2 - 5
to
d^3 - 1488*d^2 + 4176*d - 5
as I didn't want to worry about 6 roots when 3 just differ by sign.
But I copied and posted the old polynomial in repeatedly.
An embarrassing error I admit.
Here's the correct demonstration with that minor fix.
Fortunately it IS resovable by considering the polynomial
d^3 - 1488*d^2 + 4176*d - 5
directly as amazingly, in a bit of luck I guess,
d(d^2 + 2d + 1 - 1490d + 4175) = d^3 - 1488*d^2 + 4176*d
so
d(d^2 + 2d + 1 - 1490d + 4175) = 5, which means
d[{d + 1)^2 - 1490d + 4175] = 5
which forces any root plus 1 to have the algebraic integer factors of
any other root that is not coprime to 5.
But looking at the solutions to
d^3 - 1488*d^2 + 4176*d - 5 = 0
C. BOND
[snip]
> Now I've given more, and I don't know if it will be enough.
One correct one would have been enough.
--
He isn't always right, but he's always sure!
--
http://www.crbond.com
James Harris
> James Harris wrote:
> [snip]
> > Now I've given more, and I don't know if it will be enough.
>
> One correct one would have been enough.
>
> --
> He isn't always right, but he's always sure!
I gave the correct argument last year. But false claims from others
pushed me into trying to find an alternate demonstration to support
one proof with another.
Now I've done that in a very short argument with a bit of algebra.
Note that Magidin in the past claimed to have proven using Galois
Theory that the roots of a monic irreducible like
d^3 - 1488*d^2 + 4176*d - 5
would all have to have non unit algebraic integer factors of 5. That
is, that none of the roots could be coprime to 5.
I mentioned that for him to be right would mean that Galois Theory was
wrong, but that didn't seem to go over very well, and what else could
I do?
Magidin started posting advanced mathematics that I admitted I didn't
understand, and no other mathematician came forward, so he could win,
easily, with fake math.
All he needed was to post with confidence that I was wrong, and put in
a lot of technical stuff. And that was about a YEAR ago, and now I'd
probably still be in trouble if I hadn't found an incredibly simple
demonstration that he's full of bunk, unless you wish to believe
instead that Galois Theory is flawed.
James Harris
Arturo Magidin
[.snip.]
>Fortunately it IS resovable by considering the polynomial
>
> d^3 - 1488*d^2 + 4176*d - 5
>
>directly as amazingly, in a bit of luck I guess,
>
>
> d(d^2 + 2d + 1 - 1490d + 4175) = d^3 - 1488*d^2 + 4176*d
>
>so
>
> d(d^2 + 2d + 1 - 1490d + 4175) = 5, which means
>
> d[{d + 1)^2 - 1490d + 4175] = 5
>
>which forces any root plus 1 to have the algebraic integer factors of
>any other root that is not coprime to 5.
Why? From the above, we know that 5/d is the algebraic integer
(d+1)^2 - 1490d + 4175.
So (d+1)^2 = 1490d - 4175 + (5/d).
Let v be an algebraic integer divisor of d. It divides 1490d, and it
divides 4175 (since d divides 5); why does it have to divide (5/d)?
Why does it have to divide (d+1)?
>But looking at the solutions to
>
> d^3 - 1488*d^2 + 4176*d - 5 = 0
>
>with that information proves that at least one root must be coprime to
>5.
>
>Those solutions are
>
> {{x -> 496 + 244624/(1/2*(241976581 +
> 10683*I*Sqrt[12415]))^(1/3) +
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
>
> {x -> 496 - (122312*(1 + I*Sqrt[3]))/
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3) -
> 1/2*(1 - I*Sqrt[3])*
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
>
> {x -> 496 - (122312*(1 - I*Sqrt[3]))/
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3) -
> 1/2*(1 + I*Sqrt[3])*
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)}}
Which one is the real and which two are the complex solutions? They
are all currently involving cubic roots of complex numbers.
>where the last two subtracted from each other give
>
> 244624/(1/2*(241976581 +
> 10683*I*Sqrt[12415]))^(1/3) +
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)}
Let's write the following:
A = 241976581 + 10683*sqrt(-12415).
zeta = (-1+sqrt(-3))/2
zeta^2 = (-1-sqrt(-3))/2
Your numbers become
x1 = 496 + 244624/[(1/2)*A^{1/3}] + [(1/2)A]^{1/3}
x2 = 496 + zeta^2*122312/[(1/2)A]^{1/3} + zeta*[(1/2)A]^{1/3}
x3 = 496 + zeta*122312/[(1/2)A]^{1/3} + zeta^2*[(1/2)A]^{1/3}
If we take x2 - x3, noting that zeta-zeta^2 = sqrt(-3), we have:
-sqrt(-3)*122312/[(1/2)A]^{1/3} + sqrt(-3)[(1/2)A]^{1/3}
You claim you get
244624/[(1/2)A]^{1/3} + [(1/2)A]^{1/3}
which is wrong; you forgot the factors of zeta and zeta^2 that were
multiplying.
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>"C. BOND" <cb...@ix.netcom.com> wrote in message news:<3EB04835...@ix.netcom.com>...
>> James Harris wrote:
>> [snip]
>> > Now I've given more, and I don't know if it will be enough.
>>
>> One correct one would have been enough.
>>
>> --
>> He isn't always right, but he's always sure!
>
>I gave the correct argument last year. But false claims from others
>pushed me into trying to find an alternate demonstration to support
>one proof with another.
Your logic is accurate as all your arguments have been.
What you are doing right now is trying to disprove a basic theorem of
Galois Theory. Even if you could succeed (thus proving that over the
past 150 years, hundreds of thousands of mathematicians who have
reviewed the arguments very carefully are wrong), this would not
suffice to "support your proof." It would only disprove ->my<- general
result that said that in MOST cases your conclusion fails.
Your proof is already dead, disproven by the explicit counterexample
of Bengt and Rupert of over a year ago, and by your own explicit
calculations of last week.
[.rest deleted.]
Arturo Magidin
[.snip.]
>Note that Magidin in the past claimed to have proven using Galois
>Theory that the roots of a monic irreducible like
>
> d^3 - 1488*d^2 + 4176*d - 5
>
>would all have to have non unit algebraic integer factors of 5. That
>is, that none of the roots could be coprime to 5.
You don't need Galois Theory to prove that. You only need the same
basic knowledge of polynomials as you needed for the theorem about
what kind of irreducible polynomials can have algebraic integer roots.
You know, that result hat you spend 10 months saying I was liar for
saying it was true, and that I was invoking "fake" mathematics, using
complicated language to obscure the lie.
The result that, when you finally bothered to look at, you agreed was
true.
THEOREM. Let f(x) be a monic polynomial with integer coefficients,
which is irreducible over Q. Then a root of f(x) is an algebraic
integer unit if and only if the constant term of f(x) is 1 or -1.
Proof. The definition of "algebraic integer unit" is "an algebraic
integer u such that 1/u is also an algebraic integer."
Let r be a root. If the constant term is 1 or -1, then r divides 1 or
-1 (in the ring of algebraic integers), so it is a unit.
Conversely, suppose that r is an algebraic integer unit. Let
s=1/r. Then s is an algebraic integer. Write
f(x) = x^n + ... + a_0.
0 = s^nf(r) = s^n(r^n+...+a_0)
= 1 + a_{n-1}s + ... + a_0s^n.
So s is a root of g(y) = a_0x^n+...+1; since g(y) is the polynomial we
get by taking f(1/y) and multiplying through by y^n, it follows from
irreducibility of f(x) that g(x) is irreducible.
Since s is the root of the irreducible primitive polynomial g(y) with
integer coefficients, it follows that the leading coefficient of g(y)
is either 1 or -1. So a_0=1 or a_0=-1. Therefore, if r is an algebraic
integer unit, then the constant term of f(x) is either 1 or -1. QED
COR. None of the roots of f(x)=x^3-1488*x^2 + 4176*d-5 are algebraic
integer units.
Proof. It is enough to show f(x) is irreducible. Since it is degree 3,
it is enough to show it does not have any rational roots. The only possible
rational roots of 1, -1, 5, and -5. None of them are roots, so f(x) is
irreducible. QED
No Galois Theory at all.
>I mentioned that for him to be right would mean that Galois Theory was
>wrong, but that didn't seem to go over very well, and what else could
>I do?
Learn some Galois Theory?
I mean, here you are. You admit that you do NOT know or understand ANY
Galois Theory. But somehow, you know that for me to be right, Galois
Theory has to be wrong. You don't know what Galois Theory says, but it
->has<- to be wrong, if I am right.
>Magidin started posting advanced mathematics
Actually, what I posted was NOT "advanced mathematics." It was the
very basics of Galois Theory, that are often taught on an upper
division undergraduate course.
> that I admitted I didn't
>understand, and no other mathematician came forward,
No other mathematician came forward to do what? Explain it to you?
That would be false. Lots of people tried, myself included. When you
said you were reading a book on Galois Theory, I and others offered to
help you with anything you didn't understand. But instead you started
saying how we were all idiots, because there was no map that could
possibly send the complex roots of an irreducible cubic to the real
root.
Or do you mean, no other mathematician came forward to explain to you
why I was wrong?
Have you considered the possibility that this was because I ->was
not<- wrong? That I was in fact correct?
> so he could win, easily, with fake math.
Does "fake math" mean "math that James Harris does not understand"?
Because if it does not mean that, what you have written here is false.
>All he needed was to post with confidence that I was wrong, and put in
>a lot of technical stuff.
For "technical stuff", read "mathematics." The very thing that you
claim the answers should contain.
> And that was about a YEAR ago,
And now you contradict yourself. In December, you agreed that there
was a serious problem with your argument. You now claim that the
problem is that the factorization result for polynomials over the
algebraic integers does not hold (though you have not been able to
show any error in that proof), but you are incorrect about that. Your
argument was wrong simply because you incorrectly claim that two of
the ai's must be multiples of f^j and the third must be coprime to
f. You derive this conclusion by incorrectly thinking that the case
m=0 is typical, and by mumbling some undefined nonsense about
"constant terms", "variables, not numbers", and how "terms" can be
"broken up" in sundry ways that are never defined or explained. Your
error, simply put, is in thinking that any common factor of a_i and f
must divide (x/f^j)^3 f^jy^3.
> and now I'd
>probably still be in trouble if I hadn't found an incredibly simple
>demonstration that he's full of bunk,
Your "incredibly simple demonstration" is, as usual, incorrect.
And, once again: EVEN IF YOU WERE CORRECT, it would not suffice to
show your argument for FLT is correct. It would only suffice to show
that my claim that the divisibility result is false for almost all
values of m were wrong. You would still have to prove it holds, and
you have already agreed with all the steps which show that it is false
when v=4. This is more than enough to show your argument is wrong,
regardless of any Galois Theory I ever used.
Brian Quincy Hutchings
you're driving Arturo nuts -- unless it's good fun for him. so,
please, stay away from the God-am computer,
til you've proven le premiere (IMHO) theoreme de Fermat, or
provided a counterexample.
thank you!
mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b8rsue$29gh$1...@agate.berkeley.edu>...
> You know, that result hat you spend 10 months saying I was liar for
> saying it was true, and that I was invoking "fake" mathematics, using
> complicated language to obscure the lie.
>
> The result that, when you finally bothered to look at, you agreed was
> true.
--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
James Harris
> In article <3c65f87.03050...@posting.google.com>,
> James Harris <jst...@msn.com> wrote:
> >"C. BOND" <cb...@ix.netcom.com> wrote in message news:<3EB04835...@ix.netcom.com>...
> >> James Harris wrote:
> >> [snip]
> >> > Now I've given more, and I don't know if it will be enough.
> >>
> >> One correct one would have been enough.
> >>
> >> --
> >> He isn't always right, but he's always sure!
> >
> >I gave the correct argument last year. But false claims from others
> >pushed me into trying to find an alternate demonstration to support
> >one proof with another.
>
> Your logic is accurate as all your arguments have been.
>
> What you are doing right now is trying to disprove a basic theorem of
> Galois Theory. Even if you could succeed (thus proving that over the
> past 150 years, hundreds of thousands of mathematicians who have
> reviewed the arguments very carefully are wrong), this would not
> suffice to "support your proof." It would only disprove ->my<- general
> result that said that in MOST cases your conclusion fails.
And now readers you see Magidin agreeing that Galois Theory itself is
under the gun, but I suggest to readers that Magidin's claims are more
likely to be false than Galois Theory.
But notice what kind of mathematician Magidin is demonstrating himself
to be.
He'd try to toss out Galois Theory first, and now suddenly he has a
general result that only worked in most cases.
> Your proof is already dead, disproven by the explicit counterexample
> of Bengt and Rupert of over a year ago, and by your own explicit
> calculations of last week.
>
> [.rest deleted.]
That is false. Rupert's supposed counterexample is no different than
what has been worked out recently, where I've shown that it doesn't
disprove my work.
If you're a mathematician give math, don't make false accusations.
There is no counterexample to my argument as there can't be.
A proof is a proof.
Now I played by rules Magidin and when you were right I admitted it.
Can't you do the same?
It's *mathematics* after all. Making false statements is futile.
James Harris
James Harris
> In article <3c65f87.03050...@posting.google.com>,
> James Harris <jst...@msn.com> wrote:
>
> [.snip.]
>
> >Note that Magidin in the past claimed to have proven using Galois
> >Theory that the roots of a monic irreducible like
> >
> > d^3 - 1488*d^2 + 4176*d - 5
> >
> >would all have to have non unit algebraic integer factors of 5. That
> >is, that none of the roots could be coprime to 5.
>
> You don't need Galois Theory to prove that. You only need the same
> basic knowledge of polynomials as you needed for the theorem about
> what kind of irreducible polynomials can have algebraic integer roots.
>
> You know, that result hat you spend 10 months saying I was liar for
> saying it was true, and that I was invoking "fake" mathematics, using
> complicated language to obscure the lie.
>
> The result that, when you finally bothered to look at, you agreed was
> true.
If that's the case then I was wrong.
See Magidin? It's easy. Just admit you're wrong when it has been
demonstrated mathematically!!!
> THEOREM. Let f(x) be a monic polynomial with integer coefficients,
> which is irreducible over Q. Then a root of f(x) is an algebraic
> integer unit if and only if the constant term of f(x) is 1 or -1.
>
> Proof. The definition of "algebraic integer unit" is "an algebraic
> integer u such that 1/u is also an algebraic integer."
>
> Let r be a root. If the constant term is 1 or -1, then r divides 1 or
> -1 (in the ring of algebraic integers), so it is a unit.
>
> Conversely, suppose that r is an algebraic integer unit. Let
> s=1/r. Then s is an algebraic integer. Write
<deleted>
As I guessed you use the *definition* which means your argument IS
circular as it has to be because what you have can't be a theorem,
when I've proven it false.
Now the result for the polynomial I gave is direct, easy, and not
refutable Magidin, so why continue arguing?
Is it because you don't actually believe in mathematics itself?
Or do you wish to be taken apart repeatedly in posts where you display
a lack of respect for mathematical truth?
Don't you get it Magidin? The point of mathematics is that it is NOT
a personal confrontation!!! What's true is true.
If it hurts your feelings that a particular result is true, that
doesn't change the result. It never will change the result.
Mathematics does NOT change with your moods. Mathematical truth does
not change from day-to-day, and it is not about politics, or whether
or not if something is true, you'll feel like crying.
Mathematics is about truth.
Now why don't you start telling the truth?
James Harris
Arturo Magidin
Amazing. You miss the point. Again.
EVEN IF YOU COULD DISPROVE ME OR GALOIS THEORY, THAT DOES NOT SUFFICE
TO ESTABLISH THAT YOUR PROOF IS CORRECT.
Your proof is WRONG, independent of ANYTHING I've done, because there
are explicit counterexamples to your claims. One of them you
calculated yourself two weeks ago!
And I suggest that it is more likely that you, who by your own
admission do not know anything about Galois Theory, are wrong, rather
than either me or Galois Theory.
>But notice what kind of mathematician Magidin is demonstrating himself
>to be.
>
>He'd try to toss out Galois Theory first, and now suddenly he has a
>general result that only worked in most cases.
"Suddenly"? Why, you incompetent twit. I have ->always<- said that the
result only applies when a specific value of m yields an IRREDUCIBLE
polynomial. That's why your example with (v^3+1)W^3 - 3vW + 1 with m=2
did not disprove what I said: because 2W^3-3W+1 is reducible.
[.snip.]
>> Your proof is already dead, disproven by the explicit counterexample
>> of Bengt and Rupert of over a year ago, and by your own explicit
>> calculations of last week.
>>
>> [.rest deleted.]
>
>That is false. Rupert's supposed counterexample is no different than
>what has been worked out recently, where I've shown that it doesn't
>disprove my work.
[personal insult removed]
Nonsense. You took the polynomial (v^3+1)W^3 - 3vW + 1. Yes?
with v = -1 + mf^{2j}
You agree that if your argument is correct, that means that whenever
you factor, for a specific value of m, this polynomial as
(v^3+1)W^3 - 3vW + 1 = (a1W+1)(a2W+1)(a3W+1),
with a1, a2, a3 algebraic integers, it would follow that exactly two
of the ai are divisible, in the ring of algebraic integers, by f^j.
You agree then that if your argument were correct, that would mean
that exactly two of a1/f^j, a2/f^j, a3/f^j are algebraic integers. You
agree then that if your argument is correct, exactly two of -a1/f^j,
-a2/f^j, -a3/f^j are algebraic integers.
You agree that when we set f=sqrt(5), m=1, j=1, so v=4, we have that
You also agree that -a1/sqrt(5), -a2/sqrt(5) and -a3/sqrt(5) are all
roots of the irreducible polynomial
5g^6 - 144 g^4 + 312 g^2 - 169
which you also agree has no algebraic integer roots.
Therefore, you agree that none of -a1/sqrt(5), -a2/sqrt(50,
-a3/sqrt(5) are algebraic integers.
So you agree that:
(1) If your argument were correct, then exactly two of -a1/sqrt(5),
-a2/sqrt(5), -a3/sqrt(5) are algebraic integers; and
(2) NONE of -a1/sqrt(5), -a2/sqrt(5), -a3/sqrt(5) are algebraic
integers.
Therefore, you should agree that your argument is wrong.
But somehow, you think this proves nothing...
>There is no counterexample to my argument as there can't be.
Except for the one that I just pointed out above, which YOU calculated
yourself.
>A proof is a proof.
Back to that argument, huh?
Well, you don't have a proof. You have (flawed) argument.
>
>Now I played by rules
What are those rules? Whenever some contradicts you, accuse him of
lying and threaten them with physical violence, loss of jobs, and
financial ruin?
Never look at a proof that is offered to you?
Never answer questions about your own argument, if you can help it?
What rules?
>Magidin and when you were right I admitted it.
>
>Can't you do the same?
I have. You've pointed out two mistakes over the last week when I was
analysing your most recent claims.
But your claims about units and about algebraic integer and about
divisibility are simply false. Whether you admit ir not.
>It's *mathematics* after all. Making false statements is futile.
You don't seem to get it, though.
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b8rsue$29gh$1...@agate.berkeley.edu>...
>> In article <3c65f87.03050...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>>
>> [.snip.]
>>
>> >Note that Magidin in the past claimed to have proven using Galois
>> >Theory that the roots of a monic irreducible like
>> >
>> > d^3 - 1488*d^2 + 4176*d - 5
>> >
>> >would all have to have non unit algebraic integer factors of 5. That
>> >is, that none of the roots could be coprime to 5.
>>
>> You don't need Galois Theory to prove that. You only need the same
>> basic knowledge of polynomials as you needed for the theorem about
>> what kind of irreducible polynomials can have algebraic integer roots.
>>
>> You know, that result hat you spend 10 months saying I was liar for
>> saying it was true, and that I was invoking "fake" mathematics, using
>> complicated language to obscure the lie.
>>
>> The result that, when you finally bothered to look at, you agreed was
>> true.
>
>If that's the case then I was wrong.
>
>See Magidin? It's easy. Just admit you're wrong when it has been
>demonstrated mathematically!!!
I do. You haven't demonstrated anything, though.
>> THEOREM. Let f(x) be a monic polynomial with integer coefficients,
>> which is irreducible over Q. Then a root of f(x) is an algebraic
>> integer unit if and only if the constant term of f(x) is 1 or -1.
>>
>> Proof. The definition of "algebraic integer unit" is "an algebraic
>> integer u such that 1/u is also an algebraic integer."
>>
>> Let r be a root. If the constant term is 1 or -1, then r divides 1 or
>> -1 (in the ring of algebraic integers), so it is a unit.
>>
>> Conversely, suppose that r is an algebraic integer unit. Let
>> s=1/r. Then s is an algebraic integer. Write
>
><deleted>
>
>As I guessed you use the *definition*
Duh.
What do you think a definition is?
It is like a macro. It is an explanation of what the word means. When
you write
"DEF. An "algebraic integer unit" is an algebraic integer r whose
multiplicative inverse is also an algebraic integer"
It means that, whenever you see the words "algebraic integer unit",
you can take them out, and replace them with "an algebraic integer
whose multiplicative inverse is also an algebraic integer."
THAT'S WHAT A DEFINITION IS.
It's shorthand.
>which means your argument IS circular
No, it does not mean that. It is amazing the depths of your ignorance
sometime.
So, if we say that "even integer" is defined to be "an integer which
is a multiple of 2", then the following is a "circular" proof, huh?
THEOREM. An integer is an even integer if and only if, when divided by
2, we have a remainder of 0.
Proof. Suppose that r is such that r = 2q+0 (remainder 0). Then r=2q,
so r is a multiple of 2.
Conversely, suppose r is a multiple of 2. Then r=2q= 2q+0. So the
remainder is 0. QED
So, that's an invalid proof, because I am using the definition of
"even integer", huh?
> as it has to be because what you have can't be a theorem,
>when I've proven it false.
Do you even know what a definition is?
You haven't proven in false.
>Now the result for the polynomial I gave is direct, easy, and not
>refutable Magidin, so why continue arguing?
Because it is WRONG, that's why.
>Don't you get it Magidin? The point of mathematics is that it is NOT
>a personal confrontation!!! What's true is true.
Interesting, coming from someone with so much obvious hate for
specific persons that confront his nonsense with mathematics, which he
then deletes rather than answering.
>Now why don't you start telling the truth?
Why don't you stop beating your wife?
Virgil
<3c65f87.03050...@posting.google.com>,
jst...@msn.com (James Harris) wrote:
> > What you are doing right now is trying to disprove a basic theorem of
> > Galois Theory. Even if you could succeed (thus proving that over the
> > past 150 years, hundreds of thousands of mathematicians who have
> > reviewed the arguments very carefully are wrong), this would not
> > suffice to "support your proof." It would only disprove ->my<- general
> > result that said that in MOST cases your conclusion fails.
>
> And now readers you see Magidin agreeing that Galois Theory itself is
> under the gun, but I suggest to readers that Magidin's claims are more
> likely to be false than Galois Theory.
>
> But notice what kind of mathematician Magidin is demonstrating himself
> to be.
>
> He'd try to toss out Galois Theory first, and now suddenly he has a
> general result that only worked in most cases.
James S Haris has worked hard in sci.math and several other
NGs to prove himself innumerate.
He now seems to be working equally hard to prov3e himself
illiterate.
He reads things in postings which are not there and does not
read the things that are there and there is no (mental)
health in him.
James also pretends to assume the virtues of those whom he
regards as opponents and to attribute his own vices to them.
There is no moral health in him either.
Virgil
> It's *mathematics* after all. Making false statements is futile.
>
>
> James Harris
Now James, do you think you are Borg?
Virgil
> As I guessed you use the *definition* which means your argument IS
> circular as it has to be because what you have can't be a theorem,
> when I've proven it false.
On the contrary. Absent having a good honest proof of some
statement, the best assurance we could possibly want for its
theoremhood is one of your disproofs of it.
David C. Ullrich
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b8rdoa$243k$1...@agate.berkeley.edu>...
>> In article <3c65f87.03050...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>> >"C. BOND" <cb...@ix.netcom.com> wrote in message news:<3EB04835...@ix.netcom.com>...
>> >> James Harris wrote:
>> >> [snip]
>> >> > Now I've given more, and I don't know if it will be enough.
>> >>
>> >> One correct one would have been enough.
>> >>
>> >> --
>> >> He isn't always right, but he's always sure!
>> >
>> >I gave the correct argument last year. But false claims from others
>> >pushed me into trying to find an alternate demonstration to support
>> >one proof with another.
>>
>> Your logic is accurate as all your arguments have been.
>>
>> What you are doing right now is trying to disprove a basic theorem of
>> Galois Theory. Even if you could succeed (thus proving that over the
>> past 150 years, hundreds of thousands of mathematicians who have
>> reviewed the arguments very carefully are wrong), this would not
>> suffice to "support your proof." It would only disprove ->my<- general
>> result that said that in MOST cases your conclusion fails.
>
>And now readers you see Magidin agreeing that Galois Theory itself is
>under the gun,
What's the expression? Of yeah: Whooosh...
HINT: When someone says "If you're right that would disprove
a basic theorem of Galois theory" that doesn't mean that
Galois theory is "under the gun". It's a polite way of saying
"you can't possibly be right, because what you're saying
would contradict a basic theorem of Galois theory".
>but I suggest to readers that Magidin's claims are more
>likely to be false than Galois Theory.
>
>But notice what kind of mathematician Magidin is demonstrating himself
>to be.
>
>He'd try to toss out Galois Theory first, and now suddenly he has a
>general result that only worked in most cases.
>
>
>> Your proof is already dead, disproven by the explicit counterexample
>> of Bengt and Rupert of over a year ago, and by your own explicit
>> calculations of last week.
>>
>> [.rest deleted.]
>
>That is false. Rupert's supposed counterexample is no different than
>what has been worked out recently, where I've shown that it doesn't
>disprove my work.
>
>If you're a mathematician give math, don't make false accusations.
>
>There is no counterexample to my argument as there can't be.
And here's _my_ proof of FLT:
Theorem: If x, y, z are non-zero integers and n > 2 is an integer
then x^n + y^n <> z^n.
Proof: The hypotheses imply that either x^n + y^n > z^n or
x^n + y^n < z^n. QED
There _is_ no counterexample to any of the statements in
my proof! And the theorem _does_ follow from those
statements. That doesn't make it a correct proof.
>A proof is a proof.
That's true. Here's another hint: "Nobody's found a
counterexample yet" is not the definition of "correct proof".
>Now I played by rules Magidin and when you were right I admitted it.
>
>Can't you do the same?
>
>It's *mathematics* after all. Making false statements is futile.
>
>
>James Harris
******************
David C. Ullrich
James Harris
The more I think about it, the more I'm sure that Magidin made a false
statement about expressions in this post.
I'll explain further below.
> The following shows that the irreducible polynomial
>
> d^6 - 1488*d^4 + 4176*d^2 - 5
>
> has roots that are algebraic integer factors of 1.
>
> I'll be using data from a post by a poster who called herself "Dot"
>
> messageid: <210420032353503610%d...@at.dot.com>
>
> <Quote>
> We will work in the number field defined by the polynomial
>
> y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
> + 2460349402142038081821721600;
>
>
> We will let alpha be a root of this polynomial, and we will define
> algebraic integers d1, d2, d3, n1, n2, n3 in terms of alpha.
> (Taking alpha to be the real root that is approximately
> 154.451496990997802982236692688237765859998280922
> gives the numerical examples I gave earlier.
>
>
>
> MAGMA CODE:
>
> R<y>:=PolynomialRing(Rationals());
>
> f := y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
> + 2460349402142038081821721600;
>
> K<alpha>:=NumberField(f);
>
Notice all that follow have can be considered as polynomials with
respect to alpha with a degree of 11, and they all have a negative
leading coefficient.
Now Magidin claimed that as you go through the 12 values for alpha,
where you have six absolute values, that the values for d1, d2 and d3
above change, and since they are roots to a given polynomial, there
must be a cyclic change. That is, d1 shifts to be what was d2 or d3
in another expression, and so on.
But as all the polynomials slope down from left to right that is
impossible.
They so slope because the leading coefficient is a negative number.
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.
So my original statement that the expressions for the d's give the
same value for each alpha must be true.
The question remains as to whether or not Magidin knowingly told a
falsehood about their values.
Also note that I haven't seen replies from any other mathematicians
correcting Magidin.
I can only assume that as I've feared mathematicians value their
social structure even over a short proof of Fermat's Last Theorem.
They may believe that as long as they keep silent about the truth,
none of you will ever believe me, and thereby, they will be able to
maintain status as it stands.
It's definitely odd, but remember, human beings are social
creatures--even mathematicians.
> n1 := (1/10683)*(-22*d1^4 + 33001*d1^2 - 394);
> n2 := (1/10683)*(-22*d2^4 + 33001*d2^2 - 394);
> n3 := (1/10683)*(-22*d3^4 + 33001*d3^2 - 394);
>
> // Verify that the coefficients are algebraic integers.
> // We verify that the following monic polynomials
> // evaluate to zero:
>
> n1^3 - 54*n1^2 + 390*n1 + 13;
> n2^3 - 54*n2^2 + 390*n2 + 13;
> n3^3 - 54*n3^2 + 390*n3 + 13;
>
> d1^6 - 1488*d1^4 + 4176*d1^2 - 5;
> d2^6 - 1488*d2^4 + 4176*d2^2 - 5;
> d3^6 - 1488*d3^4 + 4176*d3^2 - 5;
>
> </Quote>
> *********************************************************************
And the remainder of my argument follows as given and notice that it
proves what I've said it proves.
> d^6 - 1488*d^4 + 4176*d^2 - 5
>
> I notice that conclusion is confirmed by the coefficient 4176, as it
> is coprime to 5.
>
> However alpha is a root of
>
> y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
> + 2460349402142038081821721600;
>
> and but its coefficient
>
> 413298187541059968292500
>
> prevents its roots from being pairwise coprime with respect to
> algebraic integer factors of 5.
>
> Proving that d1 and d2 are coprime to 5 if d3 is not, because d3 must
> always have all the non unit algebraic integer factors of 5 of alpha,
> but if any alpha had factors of d1 or d2 then d3 would have to have
> those as well, as shown, contradicting with d1, d2 and d3 being
> pairwise coprime.
>
> That proves there must exist d1 where it is an algebraic integer
> factor of 1 as it is a root of
>
> d^6 - 1488*d^4 + 4176*d^2 - 5.
>
>
> I would hope that those making replies consider carefully before
> replying, but I do appreciate replies that address the mathematics.
>
Virgil
> 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.
You are assuming, on insufficient evidence, that you can
think.
You are assuming, on insufficient evidence, that you can
tell truth from falsehood.
> I'll explain further below.
Not now.
>
> > The following shows that the irreducible polynomial
> >
> > d^6 - 1488*d^4 + 4176*d^2 - 5
> >
> > has roots that are algebraic integer factors of 1.
Harris here used one of his favorite fudge words "factor".
If, as it appears, he means that there is root , r, of the
above equation for which 1/r is an algebraic integer, he is,
by definition, wrong.
If he means anything else, it is his responsibility to state
what he means unambiguously before attempting to prove it.
[snipped a mess of stuff as premature]
> > I would hope that those making replies consider carefully before
> > replying, but I do appreciate replies that address the mathematics.
> >
Without a clear and unambiguous definition of "factor",
which Harris has been remarkably coy about providing, It is
not clear what, if anything, he is trying to prove.
Absent such a definition, one need not bother to look at his
arguments.
> > I will continue to explain in detail as necessary.
Then explain in detail what you mean by your perpetually
undefined term "factor".
W. Dale Hall
James Harris 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.
>
> I'll explain further below.
>
>
... stuff deleted ...
>><Quote>
>>We will work in the number field defined by the polynomial
>>
>>y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
>> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
>> + 2460349402142038081821721600;
>>
>>
>>We will let alpha be a root of this polynomial, and we will define
>>algebraic integers d1, d2, d3, n1, n2, n3 in terms of alpha.
>>(Taking alpha to be the real root that is approximately
>> 154.451496990997802982236692688237765859998280922
>>gives the numerical examples I gave earlier.
>>
... stuff deleted ...
As far as I can tell, Magidin did NOT say that they move in a cyclic
fashion. In fact, that is plainly impossible, since there is no standard
ordering among the twelve roots of the equation for alpha. It is clear,
since the d's are roots of some 6th degree polynomial, that changes in
the choice of alpha will still leave the d's to be roots of the same
6th degree polynomial.
> But as all the polynomials slope down from left to right that is
> impossible.
>
This is not an argument that holds any water whatsoever. First, what
does it mean for an polynomial to "slope down from left to right"?
Presumably, you're saying that if I were to graph these, the graph
would be high on the left of the x-axis and low on the right of
the x-axis.
> They so slope because the leading coefficient is a negative number.
>
That's a horribly misleading way to picture the graph of an 11th degree
polynomial, but OK, put it that way if that's all you understand.
> 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.
>
I'm trying to picture what you have in mind, so bear with me.
"Imagine in the xy-plane with just two cubics that
have two values reverse depending on two values
along the x-axis"
OK, I have two cubics, that is 3rd-order polynomials in x.
I'll call them C1 and C2 for Cubic #1 and Cubic #2.
Now, I have two values "along" the x-axis: two values of x, got it:
I'll call them x1 and x2 for X value #1 and X value #2.
Now, the two cubics have two values reverse? Here's what I think you
mean:
C1(x1) = y1, C2(x1) = y2,
but
C1(x2) = y2, C2(x2) = y1.
You're claiming that can't happen unless the two cubics have opposite
leading coefficients? Hell, I can give them the SAME arrangement of
plus and minus signs and make that happen:
C1(x) = x^3 + x^2 - x + 1, C2(x) = x^3 + x^2 - 3*x + 2
The x-values that can't exist are these: x1 = 0, x2 = 1.
The y-values that can't exist are these: y1 = 1, y2 = 2.
These yield what you said can't exist:
C1(0) = 1 C2(0) = 2
C1(1) = 2 C2(1) = 1.
> So my original statement that the expressions for the d's give the
> same value for each alpha must be true.
>
Yeah, you sure got that insight thing happenin', don'cha?
> The question remains as to whether or not Magidin knowingly told a
> falsehood about their values.
>
The question remains as to whether you actually have anything
intelligent to say in *any* context.
> Also note that I haven't seen replies from any other mathematicians
> correcting Magidin.
>
Maybe it's because no one has bothered to check whether an 11th degree
polynomial always takes the same values, given the huge size of the
coefficients and the rather annoying nature of the quizmaster here.
> I can only assume that as I've feared mathematicians value their
> social structure even over a short proof of Fermat's Last Theorem.
>
Oh, well I can only assume something about your preferences regarding
farm animals, but politeness forbids me from revealing my assumptions.
> They may believe that as long as they keep silent about the truth,
> none of you will ever believe me, and thereby, they will be able to
> maintain status as it stands.
>
Take your conspiracy theories and deal with them in the privacy of
your own room, please.
> It's definitely odd, but remember, human beings are social
> creatures--even mathematicians.
>
But not JSH. Not a bone of jealousy, nor a drop of bitterness
in that sweetie. Nosirree. He's just as kind as the morning sun,
as gentle as the dew on the grass.
>
>>n1 := (1/10683)*(-22*d1^4 + 33001*d1^2 - 394);
>>n2 := (1/10683)*(-22*d2^4 + 33001*d2^2 - 394);
>>n3 := (1/10683)*(-22*d3^4 + 33001*d3^2 - 394);
>>
>>// Verify that the coefficients are algebraic integers.
>>// We verify that the following monic polynomials
>>// evaluate to zero:
>>
>>n1^3 - 54*n1^2 + 390*n1 + 13;
>>n2^3 - 54*n2^2 + 390*n2 + 13;
>>n3^3 - 54*n3^2 + 390*n3 + 13;
>>
>>d1^6 - 1488*d1^4 + 4176*d1^2 - 5;
>>d2^6 - 1488*d2^4 + 4176*d2^2 - 5;
>>d3^6 - 1488*d3^4 + 4176*d3^2 - 5;
>>
>></Quote>
>>*********************************************************************
>
>
> And the remainder of my argument follows as given and notice that it
> proves what I've said it proves.
>
... no, I think I'll pass. Your argument that
the d's must all be independent of which alpha
is selected was a stinking heap of poo, and I
thought a stinking heap of poo with all those
peanuts and bits of corn would be even less nice
to deal with.
... stuff deleted ...
>>Now looking at the polynomial that defines the d's which is
>>
>> d^6 - 1488*d^4 + 4176*d^2 - 5
>>
>>I notice that conclusion is confirmed by the coefficient 4176, as it
>>is coprime to 5.
>>
>>However alpha is a root of
>>
>> y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
>> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
>> + 2460349402142038081821721600;
>>
>>and but its coefficient
>>
>> 413298187541059968292500
>>
>>prevents its roots from being pairwise coprime with respect to
>>algebraic integer factors of 5.
>>
>>Proving that d1 and d2 are coprime to 5 if d3 is not, because d3 must
>>always have all the non unit algebraic integer factors of 5 of alpha,
>>but if any alpha had factors of d1 or d2 then d3 would have to have
>>those as well, as shown, contradicting with d1, d2 and d3 being
>>pairwise coprime.
>>
You haven't proven anything of the sort.
>>That proves there must exist d1 where it is an algebraic integer
>>factor of 1 as it is a root of
>>
>> d^6 - 1488*d^4 + 4176*d^2 - 5.
>>
>>
>>I would hope that those making replies consider carefully before
>>replying, but I do appreciate replies that address the mathematics.
>>
>>I will continue to explain in detail as necessary.
>>
How about those factors of 1, for which the remaining factor is not
an algebraic integer? How is it that you define factor, anyhow?
What *is* the constraint on the remaining factor, if A is a
factor of B?
If there isn't any, doesn't that make every nonzero number a
factor of every other nonzero number? Doesn't that invalidate
your whole argument anyhow?
If there *is* some constraint on what these non-integer
factors can be, then why don't you spell it out?
Explain or quit.
I doubt you have the courage to do either, and doubt that you
have the intelligence to even consider learning Galois theory,
which (if you actually managed to learn anything) would set you
straight, in a hurry.
>>
>>James Harris
>
Dale
Arturo Magidin
>
>The more I think about it, the more I'm sure that Magidin made a false
>statement about expressions in this post.
Perhaps if rather than just thinking you tried to learn some
mathematics, things might be different.
James, for the n-th time, DO NOT ATTRIBUTE TO ME THINGS UNLESS YOU CAN
QUOTE ME SAYING THEM.
You are giving your ->interpretation<- of what I said. In particular,
I NEVER SAID "there must be a cyclic change."
What I said is that if you used the fixed expressions above, then,
choosing different values of alpha (that still are roots of the
original polynomial) will yield different different values of d1, d2,
and d3. I also noted that since there are six possible values for d1,
and twelve possible values for alpha, each value of d1 will be
obtained from two values of alpha.
> That is, d1 shifts to be what was d2 or d3
>in another expression, and so on.
>
>But as all the polynomials slope down from left to right that is
>impossible.
Nonsense. Just because the leading coefficient is negative does not
mean the polynomials are decreasing functions.
>They so slope because the leading coefficient is a negative number.
Great. So, for example, the polynomial f(x)= -x^3 + 5x + 10 must slope down
from left to right. The leading coefficient is a negative number, and
the leading term is of odd degree.
Excellent. Too bad that f(-1) = 1 - 5 + 10 = 6, and f(1) =
-1+5+10=14. I guess 6 must be larger than 14, after all.
Don't you even know any basic calculus? YOu have a polynomial of
degree 11. The derivative is a polynomial of degree 10. If it has no
repeated roots, then the derivative will change signs ten times, so
the original function will change from increasing to decreasing or
vice versa ten times.
If it has repeated roots, then the number of sign changes will be
smaller, but unless it has only roots which are of even multiplicity,
the polynomial will change signs at least once.
>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.
>
>So my original statement that the expressions for the d's give the
>same value for each alpha must be true.
It's false. Your "argument" here is nonsense. As usual. IN addition,
there were plenty of other problems with your expressions, problems
that I raised and you never addressed.
Arturo Magidin
>
>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.
Amazing stuff. It's not like this is calculus. It's not like we know
that the derivative of a cubic is a quadratic; and that the cubic is
increasing where the derivative is positive, decreasing where it is
negative; so that if the derivative does not have a repeated root,
then the derivative will be sometimes positive and sometimes negative,
so the cubic will change from increasing to decreasing to increasing
(if it has positive leading coefficient), or from decreasing to
increasing to decreasing (if it has negative leading coefficient);
unless the derivative has a repeated root.
For the derivative f'(x) = a(x-r)^2 to have a repeated root, we must
have
f'(x) = ax^2 - 2rax + r^2a
f(x) = (a/3)x^3 - rax^2 + r^2ax + C
giving a very special form for the function.
With a polynomial of degree 11, the derivative is degree 10. Unless
every root of that derivative is of even degree, we expect the
derivative to change signs at least twice, so the function will, in
general, change at least once from decreasing to increasing, and at
least once from increasing to decreasing.
It's not like this is first semester calculus, so I can understand why
James would be so sure he was right. This is complicated mathematics,
which only an expert can be expected to know or understand.
David Moran
"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
news:b8v5t0$c3e$1...@agate.berkeley.edu...
calculus. Anyone who has had differential calculus would understand this.
David Moran
Dave Rusin
David C. Ullrich <ull...@math.okstate.edu> wrote:
>And here's _my_ proof of FLT:
>
>Theorem: If x, y, z are non-zero integers and n > 2 is an integer
>then x^n + y^n <> z^n.
>
>Proof: The hypotheses imply that either x^n + y^n > z^n or
>x^n + y^n < z^n. QED
Say, does this proof use your axiom,
(p <-> q) -> r <-> (p & q -> r) & ( ~p & ~q -> r) (Axiom)
I just want to get this FLT thing down to an atomic level.
David C. Ullrich
wrote:
I think the appropriate response to this is to post the
headers from your post:
Path:
mindspring!stamper.news.atl.earthlink.net!stamper.news.pas.earthlink.net!newsfeed2.earthlink.net!newsfeed.earthlink.net!newsfeed.news2me.com!canoe.uoregon.edu!logbridge.uoregon.edu!vixen.cso.uiuc.edu!husk.cso.niu.edu!news!rusin
From: ru...@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: alt.math.undergrad,alt.math,sci.math,sci.math.symbolic
Subject: Re: Algebraic integer coverage result
Date: 4 May 2003 05:08:39 GMT
Organization: Northern Illinois Univ., Dept. of Mathematical Sciences
Lines: 13
Message-ID: <b9278n$oqh$1...@news.math.niu.edu>
References: <3c65f87.03043...@posting.google.com>
<b8rdoa$243k$1...@agate.berkeley.edu>
<3c65f87.03050...@posting.google.com>
<jql5bv4s1o1eio48h...@4ax.com>
NNTP-Posting-Host: vesuvius.math.niu.edu
X-Trace: news.math.niu.edu 1052024919 25425 131.156.3.93 (4 May 2003
05:08:39 GMT)
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Xref: mindspring alt.math.undergrad:19455 alt.math:1219
sci.math:647686 sci.math.symbolic:44668
Note the ".edu" - I think that's very telling. Also the
Organization: header.
******************
David C. Ullrich