For me there have been two perspectives as I work to figure out how to
explain the definition problem in mathematics with LOTS of opposition,
and I wonder about mathematicians so dedicated to attacking an
argument that is clearly correct.

I remind of that as I present what should finish their ability to
distract, as I've seen a strange and dedicated effort to ignore the
actual math, and simply toss up just about anything rather than face
the truth.

All variables are in the ring of algebraic integers unless otherwise
stated.

Let

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and let

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

so P(m) = f^2 R(m).

Now consider

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

Then it must be true that the following factorization exists

R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf)

where the b's are given by the following cubic:

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),

where a_3 = b_3, and at m=0, b_3 = 3.

PROOF:

From

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

at m=0, R(0) = u^2(3x + uf), and the cubic for the b's

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),

gives b^3 - 3b^2 = 0, which gives that two of the b's are 0, while the
third is 3.

That gives

R(0) = (0 x + u)(0 x + u)(3 x + uf) = u^2( 3x + uf)

as required.

No other primitive (non-primitive is like 2x+2 which has a factor of
2) cubic has the required roots, so the cubic

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m)

is the correct one.

Then it follows that two of the a's have a factor that is f.

However, it is possible to show that for certain integer values of f,
and integer values of m, the a's do not *in the ring of algebraic
integers* have f as a factor, which is a contradiction.

Mathematicians can continue to run from the truth, but they must keep
attacking algebra itself to do so.


James Harris

"James Harris" wrote in message
m...
For me there have been two perspectives as I work to figure out how to
explain the definition problem in mathematics with LOTS of opposition,
and I wonder about mathematicians so dedicated to attacking an
argument that is clearly correct.

I remind of that as I present what should finish their ability to
distract, as I've seen a strange and dedicated effort to ignore the
actual math, and simply toss up just about anything rather than face
the truth.

All variables are in the ring of algebraic integers unless otherwise
stated.

Let

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and let

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

so P(m) = f^2 R(m).

Now consider

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

Then it must be true that the following factorization exists

R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf)

where the b's are given by the following cubic:

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),

where a_3 = b_3, and at m=0, b_3 = 3.

PROOF:

From

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

at m=0, R(0) = u^2(3x + uf), and the cubic for the b's

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),

gives b^3 - 3b^2 = 0, which gives that two of the b's are 0, while the
third is 3.

That gives

R(0) = (0 x + u)(0 x + u)(3 x + uf) = u^2( 3x + uf)

as required.

No other primitive (non-primitive is like 2x+2 which has a factor of
2) cubic has the required roots, so the cubic

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m)

is the correct one.

Then it follows that two of the a's have a factor that is f.

However, it is possible to show that for certain integer values of f,
and integer values of m, the a's do not *in the ring of algebraic
integers* have f as a factor, which is a contradiction.

Mathematicians can continue to run from the truth, but they must keep
attacking algebra itself to do so.


James Harris

James,

Give us one number that proves your point. You've been asked to do so
more times than I can count, but you haven't. Why? Simple, you can't produce
such a number. If you could, I would have thought that you would have given
such a number. You are so adamant about fame, that you aren't rational; I
have difficulty following your claims, but have no problem understanding
others' claims. This is because I am trained in mathematics; not to the
extent of having a degree, but I have to wonder how much math you've
actually taken. Your standard rebuttal to an objection is just to restate it
as that will make it true. WE NEED PROOF! I can make a statement about my
field, meteorology, and some who are outside of my field might take it at
face value, but others in meteorology may look at it and ask why. If I can't
back it up with SPECIFIC reasons, why should they believe me? It seems as if
eight years ago, you thought you could flaunt this proof without any
objections. Then when people do object, you call them liars or accuse them
of being evil. It's time to grow up, James.

David Moran

James Harris wrote:

For me there have been two perspectives as I work to figure out how to
explain the definition problem in mathematics with LOTS of opposition,
and I wonder about mathematicians so dedicated to attacking an
argument that is clearly correct.

I remind of that as I present what should finish their ability to
distract, as I've seen a strange and dedicated effort to ignore the
actual math, and simply toss up just about anything rather than face
the truth.

All variables are in the ring of algebraic integers unless otherwise
stated.

You didn't finish anything, or prove anything. You merely restated your
flawed argument. Nor have you explained why you think (from other posts)
that the rational number 1/2 *should be* in the ring of algebraic
integers. As 1/2 is not the root of a monic polynomial with integer
coefficients, it clearly should *not* be in the ring of algebraic
integers. Nor have you provided any number you can prove *should be* in
the ring of algebraic integers, but which is not.

Your "proof" remains as error-ridden now as it has been since its
inception. Get a life -- this ain't it!

--
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

In article ,
(James Harris) wrote:

For me there have been two perspectives as I work to figure out how to
explain the definition problem in mathematics with LOTS of opposition,
and I wonder about mathematicians so dedicated to attacking an
argument that is clearly correct.


What ever became of your complaints to the FBI?

In article ,
"C. Bond" wrote:

You didn't finish anything, or prove anything. You merely restated your
flawed argument. Nor have you explained why you think (from other posts)
that the rational number 1/2 *should be* in the ring of algebraic
integers. As 1/2 is not the root of a monic polynomial with integer
coefficients, it clearly should *not* be in the ring of algebraic
integers. Nor have you provided any number you can prove *should be* in
the ring of algebraic integers, but which is not.

Asking for 1/2 to be included in the algebraic integers is really utter
stupidity - not that I would have expected anything else from Harris.

The algebraic numbers are an extension of the rational numbers: They are
the roots of polynomials of arbitrary high degree with integer
coefficients, whereas the rational numbers are the roots of polynomials
of degree 1. Restricting the leading coefficient to +/- 1 restricts the
rational numbers to the integers. It obviously restricts the algebraic
numbers in some way, and by analogy the resulting numbers are algebraic
integers. The point being that non-integer rational numbers are excluded!

"C. Bond" wrote in message ...
James Harris wrote:

For me there have been two perspectives as I work to figure out how to
explain the definition problem in mathematics with LOTS of opposition,
and I wonder about mathematicians so dedicated to attacking an
argument that is clearly correct.

I remind of that as I present what should finish their ability to
distract, as I've seen a strange and dedicated effort to ignore the
actual math, and simply toss up just about anything rather than face
the truth.

All variables are in the ring of algebraic integers unless otherwise
stated.

You didn't finish anything, or prove anything. You merely restated your
flawed argument. Nor have you explained why you think (from other posts)
that the rational number 1/2 *should be* in the ring of algebraic
integers.

I can't imaging a math argument that produces the word "should".
How to prove the statement "A should be B"?

I know how I can prove "A is B" or "A is not B", but not that weird one.


As 1/2 is not the root of a monic polynomial with integer
coefficients, it clearly should *not* be in the ring of algebraic
integers. Nor have you provided any number you can prove *should be* in
the ring of algebraic integers, but which is not.

Your "proof" remains as error-ridden now as it has been since its
inception. Get a life -- this ain't it!

--
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

(James Harris) wrote in message om...
For me there have been two perspectives as I work to figure out how to
explain the definition problem in mathematics with LOTS of opposition,
and I wonder about mathematicians so dedicated to attacking an
argument that is clearly correct.

I remind of that as I present what should finish their ability to
distract, as I've seen a strange and dedicated effort to ignore the
actual math, and simply toss up just about anything rather than face
the truth.

All variables are in the ring of algebraic integers unless otherwise
stated.

Let

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and let

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

so P(m) = f^2 R(m).

Now consider

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

Then it must be true that the following factorization exists

R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf)

where the b's are given by the following cubic:

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),


This is not correct. Note that a1/f = b1, or a1 = f*b1.
Putting this into the equation above that the a's
satisfy and simplifying, one obtains

f*b1^3 + 3*(-1 + m*f^2)*b1^2 - (m^3*f^4 - 3*m^2*f^2 + 3*m).

Clearly b2 is a root of the same polynomial. Since
b3 = a3, b3 satisfies a different equation:

b3^3 + 3*(-1+mf^2)b3^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) = 0.


where a_3 = b_3, and at m=0, b_3 = 3.

PROOF:

From

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

at m=0, R(0) = u^2(3x + uf), and the cubic for the b's

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),


... Not correct, as noted above.


gives b^3 - 3b^2 = 0, which gives that two of the b's are 0, while the
third is 3.


No. The correct equation for b1 and b2, when m = 0, is

f*b^3 - 3*b^2 = b^2(f*b - 3) = 0,

from which one may obtain b1 = b2 = 0. The correct
equation for b3, when m = 0, is

b^3 - 3*b^2 = b^2*(b - 3) = 0,

from which one may obtain b3 = 3.


That gives

R(0) = (0 x + u)(0 x + u)(3 x + uf) = u^2( 3x + uf)

as required.

No other primitive (non-primitive is like 2x+2 which has a factor of
2) cubic has the required roots, so the cubic

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m)

is the correct one.


As noted above, this is not correct.


Then it follows that two of the a's have a factor that is f.


It doesn't, except when m = 0. For example, if m = 1
and f = 5 and u = 1, the CORRECT equation for b1 and b2
becomes

5*b^3 + 72*b^2 - 553 = 0.

a primitive irreducible *non-monic* polynomial with integer
coefficients. This implies that b1, for example,
cannot be an algebraic integer. Since b1 = a1/f = a1/5,
this implies that a1 does not have a factor of f = 5.

The real problem here however is not that the equation for
the b's is incorrect. That part can be fixed as I noted
above. It is the completely unjustified leap
in the statement:

"Then it follows that two of the a's have a factor that is f."

This part really adds nothing to what Harris has argued
previously. Here he has a polynomial in x which has
coefficients that are functions of m. When m = 0, that
polynomial has two roots that are zero, hence two roots
which "have a factor that is f." He concludes *with no
justification whatsoever* that for any m 0, two of the
roots must "have a factor that is f". The example above
shows that this is necessarily false.

The gap in this argument is even more starkly obvious
than in previous Harris arguments, and the distance from
claim to counterexample is shorter.


However, it is possible to show that for certain integer values of f,
and integer values of m, the a's do not *in the ring of algebraic
integers* have f as a factor, which is a contradiction.


The part about the a's not having a factor of f is
correct. [Oddly enough, to my knowledge, Harris himself has
never given a proof of it.] If the conclusions above in
Harris's post were true, there would indeed be a
contradiction. Fortunately for mathematics Harris's
conclusions are wrong.


Mathematicians can continue to run from the truth, but they must keep
attacking algebra itself to do so.


It is interesting to see this dramatic change from two
days ago in the argument to show that two of the a's are
divisible by f. The nonsense about factors *independent of
m* has been left in the dust, and questions in the threads
dealing with that remain unanswered. Instead a new hastily
written thread is started which includes careless errors
and, at its heart, a big gap in the reasoning.

So who is running from the truth?

Nora B.


James Harris

C. Bond wrote:

Asking for 1/2 to be included in the algebraic integers is really utter
stupidity - not that I would have expected anything else from Harris.

The algebraic numbers are an extension of the rational numbers: They are
the roots of polynomials of arbitrary high degree with integer
coefficients, whereas the rational numbers are the roots of polynomials
of degree 1. Restricting the leading coefficient to +/- 1 restricts the
rational numbers to the integers. It obviously restricts the algebraic
numbers in some way, and by analogy the resulting numbers are algebraic
integers. The point being that non-integer rational numbers are excluded!

The "correct" definition of an algebraic integer can be motivated
much more convincingly. See my prior posts for classical methods
http://google.com/groups?selm=y8zllt...tle.ai.mit.edu
http://google.com/groups?selm=y8zr83...tle.ai.mit.edu

-Bill Dubuque

Followups to sci.math.

In sci.physics, James Harris

wrote
on 12 Oct 2003 10:13:34 -0700
:
For me there have been two perspectives as I work to figure out how to
explain the definition problem in mathematics with LOTS of opposition,
and I wonder about mathematicians so dedicated to attacking an
argument that is clearly correct.

I remind of that as I present what should finish their ability to
distract, as I've seen a strange and dedicated effort to ignore the
actual math, and simply toss up just about anything rather than face
the truth.

All variables are in the ring of algebraic integers unless otherwise
stated.

Let

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and let

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

so P(m) = f^2 R(m).

Now consider

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

Then it must be true that the following factorization exists

R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf)

where the b's are given by the following cubic:

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),

where a_3 = b_3, and at m=0, b_3 = 3.

An interesting problem for you. Setting m = 1, u = 1, f = 2,
one gets

P(1) = 28*x^3 - 36*x + 8, or
R(1) = P(1)/f^2 = 7*x^3 - 9*x + 2.

x=1 is clearly a root here. Dividing, we get

R(m) = (x-1) * (7*x^2 + 7*x - 2)

so the other two roots are 1/2 +/- sqrt(105)/14.

Your 'a-cubic' is given by C_a = a^3 + 9*a^2 - 28

By a little empirical testing (translation: I got lucky :-) ),
a_1 = -2; this gives C_a = (a+2)(a^2 + 7*a - 14),
which factors further into (a+2)(a+7/2-sqrt(105)/2)(a+7/2+sqrt(105)/2).

So far, not too bad; a_1 is clearly divisible by 2 (-2/2 is an
algebraic integer). The other two roots, however, are not, as

(x+7/4-sqrt(105)/4)*(x+7/4+sqrt(105)/4) = x^2+7/2x-7/2

Clearly, 7/4±sqrt(105)/4 are not algebraic integers;
therefore 7/2±sqrt(105)/2 are not divisible by 2.

Your 'b-cubic' is given by C_b = b^3 + 9*b^2 - 7.

It turns out this cubic is not zeroed by any of the following:

x = -2
x = -1
x = -1/2
x = -7/2+sqrt(105)/2
x = -7/4+sqrt(105)/4
x = -7/8+sqrt(105)/8
x = -7/2-sqrt(105)/2
x = -7/4-sqrt(105)/4
x = -7/8-sqrt(105)/8

which therefore means I've found a counterexample to your
"a_3 = b_3" hypothesis in your above statement.

(I don't know the actual roots of C_b. I'm not sure I need to.
Please verify the above computations, then fix your proof, James.)

[rest snipped]

--
#191,
It's still legal to go .sigless.

(Nora Baron) wrote in message . com...
(James Harris) wrote in message om...
For me there have been two perspectives as I work to figure out how to
explain the definition problem in mathematics with LOTS of opposition,
and I wonder about mathematicians so dedicated to attacking an
argument that is clearly correct.

I remind of that as I present what should finish their ability to
distract, as I've seen a strange and dedicated effort to ignore the
actual math, and simply toss up just about anything rather than face
the truth.

All variables are in the ring of algebraic integers unless otherwise
stated.

Let

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and let

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

so P(m) = f^2 R(m).

Now consider

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

Then it must be true that the following factorization exists

R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf)

where the b's are given by the following cubic:

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),


This is not correct. Note that a1/f = b1, or a1 = f*b1.
Putting this into the equation above that the a's
satisfy and simplifying, one obtains

f*b1^3 + 3*(-1 + m*f^2)*b1^2 - (m^3*f^4 - 3*m^2*f^2 + 3*m).

Clearly b2 is a root of the same polynomial. Since
b3 = a3, b3 satisfies a different equation:

b3^3 + 3*(-1+mf^2)b3^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) = 0.

Um, "Nora Baron" do you accept that there exists a SINGLE cubic which
has the b's as it's roots?

That is, ONE cubic with the roots b_1, b_2, and b_3?

If so, then your objection fails as provably that cubic must be
non-monic.

Do you understand?

where a_3 = b_3, and at m=0, b_3 = 3.

PROOF:

From

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

at m=0, R(0) = u^2(3x + uf), and the cubic for the b's

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),


... Not correct, as noted above.

And as I noted above, this poster seems to be lost on the idea that
given b_1, b_2 and b_3 there must exist a *single* cubic for which
they are the roots, as this poster tried to give TWO, which is rather
interesting nonsense.


gives b^3 - 3b^2 = 0, which gives that two of the b's are 0, while the
third is 3.


No. The correct equation for b1 and b2, when m = 0, is

f*b^3 - 3*b^2 = b^2(f*b - 3) = 0,

from which one may obtain b1 = b2 = 0. The correct
equation for b3, when m = 0, is

b^3 - 3*b^2 = b^2*(b - 3) = 0,

from which one may obtain b3 = 3.

Well "Nora Baron" do you understand that there must be a *single*
cubic out there, which will have the b's as roots?

If so, it cannot be non-monic as then it wouldn't work at m=0 for b_3
= 3, since two of the b's equal 0 then.

Which explains your dodge of trying to claim two cubics, when it's
basic algebra that a single cubic must exist that will work.

That gives

R(0) = (0 x + u)(0 x + u)(3 x + uf) = u^2( 3x + uf)

as required.

No other primitive (non-primitive is like 2x+2 which has a factor of
2) cubic has the required roots, so the cubic

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m)

is the correct one.


As noted above, this is not correct.

Then give the *single* correct cubic that works.

Remember, there's one for the a's, as

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

and you didn't try to give two cubics there, now did you?

Then it follows that two of the a's have a factor that is f.


It doesn't, except when m = 0. For example, if m = 1
and f = 5 and u = 1, the CORRECT equation for b1 and b2
becomes

5*b^3 + 72*b^2 - 553 = 0.

a primitive irreducible *non-monic* polynomial with integer
coefficients. This implies that b1, for example,
cannot be an algebraic integer. Since b1 = a1/f = a1/5,
this implies that a1 does not have a factor of f = 5.

But must there not exist a SINGLE cubic for which b_1, b_2 and b_3 are
roots?

Or do you wish to deny that fact "Nora Baron"?

The real problem here however is not that the equation for
the b's is incorrect. That part can be fixed as I noted
above. It is the completely unjustified leap
in the statement:

"Then it follows that two of the a's have a factor that is f."

This part really adds nothing to what Harris has argued
previously. Here he has a polynomial in x which has
coefficients that are functions of m. When m = 0, that
polynomial has two roots that are zero, hence two roots
which "have a factor that is f." He concludes *with no
justification whatsoever* that for any m 0, two of the
roots must "have a factor that is f". The example above
shows that this is necessarily false.

The gap in this argument is even more starkly obvious
than in previous Harris arguments, and the distance from
claim to counterexample is shorter.

Then why are you the one giving two cubics in one case, when you
didn't try to give two cubics for the a's?

Don't you understand enough algebra to know that a *single* cubic
exists for which b_1, b_2 and b_3 are the roots?

Or do you believe it doesn't exist?

Now then, given that it exists, do you understand that it cannot be
non-monic?

However, it is possible to show that for certain integer values of f,
and integer values of m, the a's do not *in the ring of algebraic
integers* have f as a factor, which is a contradiction.


The part about the a's not having a factor of f is
correct. [Oddly enough, to my knowledge, Harris himself has
never given a proof of it.] If the conclusions above in
Harris's post were true, there would indeed be a
contradiction. Fortunately for mathematics Harris's
conclusions are wrong.

Actually with a *single* cubic, the algebra shows quite clearly that I
can't be wrong, which is probably why you tossed out two cubics, as if
no one would realize that a single cubic will suffice for three
values.

Mathematicians can continue to run from the truth, but they must keep
attacking algebra itself to do so.


It is interesting to see this dramatic change from two
days ago in the argument to show that two of the a's are
divisible by f. The nonsense about factors *independent of
m* has been left in the dust, and questions in the threads
dealing with that remain unanswered. Instead a new hastily
written thread is started which includes careless errors
and, at its heart, a big gap in the reasoning.

That's not true. I just figured I could catch someone like you here
with this thread, and indeed, I caught you trying to foist two cubics
on people, when one will do.

So who is running from the truth?

Nora B.

Yup, "Nora Baron" is still running.


James Harris