In sci.physics, Virgil

wrote
on Fri, 17 Oct 2003 13:34:27 -0600
:
In article ,
(James Harris) wrote:



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).


The above cubic non-equation doesn't "give" anything.

Presumably, it gives three roots when set to 0, which
would have to be determined using standard methods.
If one assumes f and m to be integers, then the roots are
algebraic integers, by definition. I do not know offhand
whether the same could be said of the roots if f and m are
algebraic integers (although I for one don't see why not).

The product of these roots is divisible by f^2, although
one has to be careful: I don't think one can conclude that
two of the roots are divisible by f, as James appears to
claim elsewhere. Also, weird things happen in algebraic
integers regarding factorization, probably because of the
proliferation of units. Taking f=3, we can factorize f^2

9 = 3^(2/3) * 3^(2/3) * 3^(2/3)

as a perfectly valid product therein, and 3^(2/3) is
a root of x^3 - 9 = 0 and therefore an algebraic integer.

The factorization is not unique of course :-) ; one can
also write

9 = 3^(1/2) * 3^(3/4) * 3^(3/4)
9 = 3 * 3 * 1
9 = 3 * 3^(1/3) * 3^(2/3)

etc.

Mr. Harris does like to jump blindly over crevasses, it seems;
the territory should instead be carefully negotiated. :-)

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

In article ,
James Harris wrote:
(Arturo Magidin) wrote in message ...
In article ,
James Harris wrote:

[.snip.]

It's not possible. I'll put in values to help those readers confused
by symbols (though algebra is BASED on symbols) by letting u=2, f=13,
then you have

26/w_1(m)

and you don't have to be a rocket scientist to know that no function
w_1(m) that actually varies with m can exist such that 26/w_1(m) is an
algebraic integer for *all* integer m.


w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)}

I'm not saying that's the function in question (it is not), but there
is a function that "actually varies with m such that 26/w_1(m) is an
algebraic integer for *all* integer m." So the argument that no such
function can exist is simply bogus.

Well that example works, so now I'll say algebraic integer m, and it
blows up at m^2+1 = 0.

So let me see if I have this right:

(1) You claimed something was impossible under a given set of
conditions.

(2) When I proved that your claim was simply false, you changed the
set of conditions, and now argue that what I said was wrong.

Great.

So, first of all, you were wrong. You claimed no such function could
exist, but there's a function that did everything you required it to
do. And second of all, rather than admit you were wrong, you just
change the rules.

Here's an example that works for EVERY algebraic NUMBER m:

Step 1. Given an algebraic number m, let f(x) be the unique monic
polynomial with rational coefficients, irreducible over Q, which has m
as a root. Write it as:

f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0.

Note that a_0 must be different from 0.

Step 2. Write a_0, a rational number, as a_0 = r/s, with r and s
integers, r and s relatively prime.

Step 3. If r=1 or -1, let q = 1.

Step 4. If r is not equal to 1 or -1, then let q be the largest
rational prime that divides r.

Step 5. Let w(m) be a root of the polynomial

x^4 + 13qx^3 + qx + 13.


Then:

(a) w(m) is an algebraic integer.

(b) w(m) divides 13 in the ring of algebraic integers, since the
product of all the roots is 13, and every root is an algebraic
integer.

(c) w(m) is not constant: it takes different values at different
integers.

(d) w(m) takes each value a countably infinite number of times, so
w(m) cannot be given by a polynomial.


[.snip.]

What now, James? Are you going to change your claim again in order to
avoid the fact that you were just plain wrong?

For those confused by the discussion, I've pointed out that
mathematicians arguing with me have been arguing for a relation
relating the multiplicative inverse of one algebraic integer function
to another over all algebraic integers.

For those confused by the discussion, James claimed there could be NO
function (other than a constant function), from the integers to the
algebraic integers, with the property that at each integer we got an
algebraic integer which was a divisor of 13. James was wrong. He does
not like to admit he is wrong, so he's changing the subject.

For instance, with u=2, f=7

14/w_1(m) = r(m)

which would relate w_1(m) to r(m) for all algebraic integer m, which
is like claiming that xy=2 for all integer x, with an integer y.

No, it's not like claiming that at all. Because the integers are a
UFD with a finite number of units, so there are only a finite number
of ways in which each integer can be expressed as a product of
integers.

The algebraic integer are NOT a UFD, have an infinite number of units,
and any algebraic integer can be expressed as a product of algebraic
integers in an infinite number of ways.

So it is not at all like claiming that.

It's that multiplicative inverse that blows away the objection as in
fact, w_1(m) is a constant with regard to m--independent of m--as I've
repeatedly shown.

As you've repeatedly -claimed-, but as has repeatedly been shown,
with EXPLICIT calculations, is not the case.

[.rest deleted.]

================================================== ====================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A great
many people are staggered to this extent, 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

(Nora Baron) wrote in message . com...
(James Harris) wrote in message om...
I've noted a problem in algebraic number theory with the inclusiveness
of the definition of algebraic integers as roots of monic polynomials
with integer coefficients.

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

First I have

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 the factorization

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).

My finding that only two of the a's have f as a factor without regard
to the value of m has been vigorously disputed.


It's incorrect. See below.

Mathematicians, you see, have other priorities than actual validity of
mathematics, as they have *social* issue. See below.

However, consider w_1(m), a factor of a_1 that is a factor of f, as
well as a function that varies with m, then it follows that

a_1 x + uf has w_1(m) as a factor,

so dividing through by w_1(m) gives

a_1 x/w_1(m) + uf/w_1(m)

but then uf/w_1(m) cannot in general be an algebraic integer as it's
not representable as a polynomial with a finite number of terms if
w_1(m) varies with m.


A totally off-the-wall, unjustified statement, and, as it
so happens, incorrect. But for now, if you want to claim
it is true, the shoe is on your foot: try to prove it.

I've introduced r(m), to handle the result of uf/w_1(m).

So the poster is requesting that I prove that

r(m) w_1(m) = uf, does not exist over the ring of algebraic integers.

I've concluded that using numbers for u and f, as they are
*independent* of m, helps, so let u=2, f=13.

Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m), so you have

xy=26

and now I'll chat further.

Then the question is, does their exist a multiplicative inverse in the
ring of algebraic integers for 26/w_1(m) for *all* algebraic integers
m?

The simple answer is that if w_1(m) varies with m, then it must vary
over an infinite number of algebraic integer values as m varies over
algebraic integers.

But w_1(m) must vary from 0 to infinity if it varies with m.

Replies I've seen have shied away from trying something like w_1(m) =
m+1 because most readers can immediately realize that 26/(m+1) or
anything like it, can't be an algebraic integer for all m.

Instead I've seen examples like 26^{1/m} or more trying, 26^{1/m^2+1},
but notice that because m^2+1 can equal 0, *in the ring of algebraic
integers*, you can get 26^{1/0} for the m's that are the roots of
m^2+1.

My guess is that some may be assuming that f is replaceable by some
function of m, but in fact, its independent of the value of m,

No, we are not assuming f is "replaceable by some function of m".
We assume f itself is *constant* with respect to m. However, the factorization
of a polynomial P(m) whose coefficients are functions of m, hence
dependent on m, is also in general dependent on m. We do *not* assume
that the corresponding factorization of f is constant with respect
to m.

So if you had any doubts about how low mathematicians could go,
consider that after all, they're trying to convince that you can have
one algebraic integer function defined by the multiplicative inverse
of another algebraic integer function over all algebraic integer m,
like with f=7, 14/w_1(m) = r(m).

It's like saying that you can have xy=2, where x and y are integers,
or algebraic integers, where x varies over the ring, and y remains in
it, though of course, I can just show that for x=5 that doesn't work.

Hmmm...that's why posters have tried to pick w_1(m) using exponential
functions.

Fascinating as they're showing how much they understand.

so it's
like 1/(x+1) which is also not representable by a polynomial if x is
an algebraic integer not equal to 0 or -2.

So the objection is refuted by the impossibility of uf/w_1(m) being an
algebraic integer,


Also incorrect and false. Remember what you said above:

"... consider w_1(m), a factor of a_1 that is a factor of f..."

If w_1(m) is a "factor of f", that can only mean f/w_1(m) is
an algebraic integer, which of course implies that uf/w_1(m)
is an algebraic integer: this is your *assumption* here.

The assumption is itself a contradiction, in that it requires the
ability to define one algebraic integer function by the multiplicative
inverse of another algebraic integer function.

Basic algebra.

for all algebraic integers m, if w_1(m) varies with
m.

My hope is that posters who have been so successful in convincing
others that my argument is flawed will post concessions.



Absolutely not!

You are inching closer to the truth in this. You are seeing
how this can work. You are correct that w_1(m) is a factor of
f that depends on m. As noted above this implies that
uf/w_1(m) is an A.I., and a1(m)/w_1(m) is also. I think you
are beginning to see how this can happen.

Readers should note that attempt to push the impossible, one algebraic
integer function defined by the multiplicative inverse of another
algebraic integer function.

My guess is that seeing something like uf/w_1(m) may confuse some who
might believe that u and f are functions of m, so I like to show their
independence by tossing in values, like u=2, f=13, so you have
26/w_1(m).

I note a couple of other things. First, the polynomial in "a"
that you mention above:

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

Your claim is that two roots of this polynomial have a factor
that is f. This means that if r is one of those roots, then
r = f*c, where c is an algebraic integer. This implies that

f^3*c^3 + 3(-1 +m*f^2)*f^2*c^2 - f^2*(m^3*f^4 - 3*m*f^2 + 3*m) = 0,

or, factoring out f^2,

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


The polynomial in c on the left is non-monic and primitive
(if f is not a multiple of 3). If it is irreducible, then
c cannot be an algebraic integer. When f = 5 and m = 1, this
equation is

5*c^3 + 72*c^2 - 553 = 0,

and the polynomial in c is easily shown to be irreducible.
Therefore c cannot be an algebraic integer. This contradicts
your central claim: NONE of the roots of [#] can have a
factor that is 5 = f.


Second note: It may be worthwhile to see how irreducibility
is inextricably tied to factorization of roots. Assume that


Q(m) = x^2 + m*x + 30,

where m is an integer.

The constant term is divisible by 5 (also by 2 and 3, but I
will focus on 5 here).

For certain values of m, this polynomial is reducible, and the
corresponding roots are not both divisible by 5:

m = 11: r1 = 5, r2 = 6

m = 13: r1 = 10, r2 = 3

m = 17: r1 = 15, r2 = 2

m = 31: r1 = 30, r2 = 1.

In all these examples, obviously one root is *divisible* by 5
and the other is *coprime* to 5. And of course in all these
examples, the polynomial is reducible. This is parallel to
Harris's cubic when m = 0: two of the roots are divisible
by f and one is coprime to f. In this particular case his polynomial
[#] is reducible. In general it is not.

Now look at an other example. Say, m = 14. The
polynomial is irreducible, because the discriminant

D^2 = m^ - 4*30 = 76,

which is not a perfect square. The roots of the
polynomial a


r1 = (-14 + sqrt(76))/2 = -7 + sqrt(19) and

r2 = -7 - sqrt(19).

Clearly both r1 and r2 are algebraic integers.

Assume that r1 is a multiple of 5: r1 = 5*s1. Then

25*s1^2 + 70*s1 + 30 = 0.

Factor out 5:

[*] 5*s1^2 + 14*s1 + 6 = 0.

This happens to have discriminant D^2 = 76. [This is
not a coincidence!]. Thus the polynomial in[*] is
*also* irreducible. Therefore s1 cannot be an algebraic
integer.

Therefore r1 cannot be divisible by 5.

The same can similarly be shown for r2.

However, it is now easy to show that both r1 and r2 must both
be *non-coprime* to 5. For, suppose r1 is coprime to 5.
The fact that

r1*r2 = 5*6

would then imply that r2 is DIVISIBLE by 5, which
contradicts the result above that both r1 and r2 are
NOT divisible by 5.

Conclusion: for m = 14, the polynomial is irreducible,
and both roots have a nonunit algebraic integer factor
in common with 5.

This is true more generally. There is nothing special
about 14. Try, for example, m = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12,
14, 15, 16, 18, ... MOST values of m yield an irreducible
polynomial, and both of the roots will both share algebraic
integer factors in common with 5.

In general it is hard to write down exactly what these
factors are, even in the quadratic case. The important
thing to know about them here is that, as suggested above
by the w_1(m) notation, *they will be dependent on m*.

What's fascinating here is how casually posters will just start
talking, which I think actually usually works with a lot of readers
who don't even bother to read carefully.

Just remember that the base position here is that in the ring of
algebraic integers you can have one function defined by the
multiplicative inverse of another function.

Remember xy=26.


James Harris

"James Harris" wrote in message
m...
(Nora Baron) wrote in message
. com...
(James Harris) wrote in message
om...
I've noted a problem in algebraic number theory with the inclusiveness
of the definition of algebraic integers as roots of monic polynomials
with integer coefficients.

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

First I have

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 the factorization

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).

My finding that only two of the a's have f as a factor without regard
to the value of m has been vigorously disputed.


It's incorrect. See below.

Mathematicians, you see, have other priorities than actual validity of
mathematics, as they have *social* issue. See below.

However, consider w_1(m), a factor of a_1 that is a factor of f, as
well as a function that varies with m, then it follows that

a_1 x + uf has w_1(m) as a factor,

so dividing through by w_1(m) gives

a_1 x/w_1(m) + uf/w_1(m)

but then uf/w_1(m) cannot in general be an algebraic integer as it's
not representable as a polynomial with a finite number of terms if
w_1(m) varies with m.


A totally off-the-wall, unjustified statement, and, as it
so happens, incorrect. But for now, if you want to claim
it is true, the shoe is on your foot: try to prove it.

I've introduced r(m), to handle the result of uf/w_1(m).

So the poster is requesting that I prove that

r(m) w_1(m) = uf, does not exist over the ring of algebraic integers.

I've concluded that using numbers for u and f, as they are
*independent* of m, helps, so let u=2, f=13.

Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m), so you have

xy=26

and now I'll chat further.

Then the question is, does their exist a multiplicative inverse in the
ring of algebraic integers for 26/w_1(m) for *all* algebraic integers
m?

The simple answer is that if w_1(m) varies with m, then it must vary
over an infinite number of algebraic integer values as m varies over
algebraic integers.

But w_1(m) must vary from 0 to infinity if it varies with m.

Replies I've seen have shied away from trying something like w_1(m) =
m+1 because most readers can immediately realize that 26/(m+1) or
anything like it, can't be an algebraic integer for all m.

Instead I've seen examples like 26^{1/m} or more trying, 26^{1/m^2+1},
but notice that because m^2+1 can equal 0, *in the ring of algebraic
integers*, you can get 26^{1/0} for the m's that are the roots of
m^2+1.

My guess is that some may be assuming that f is replaceable by some
function of m, but in fact, its independent of the value of m,

No, we are not assuming f is "replaceable by some function of m".
We assume f itself is *constant* with respect to m. However, the
factorization
of a polynomial P(m) whose coefficients are functions of m, hence
dependent on m, is also in general dependent on m. We do *not* assume
that the corresponding factorization of f is constant with respect
to m.

So if you had any doubts about how low mathematicians could go,
consider that after all, they're trying to convince that you can have
one algebraic integer function defined by the multiplicative inverse
of another algebraic integer function over all algebraic integer m,
like with f=7, 14/w_1(m) = r(m).

It's like saying that you can have xy=2, where x and y are integers,
or algebraic integers, where x varies over the ring, and y remains in
it, though of course, I can just show that for x=5 that doesn't work.

Hmmm...that's why posters have tried to pick w_1(m) using exponential
functions.

Fascinating as they're showing how much they understand.

so it's
like 1/(x+1) which is also not representable by a polynomial if x is
an algebraic integer not equal to 0 or -2.

So the objection is refuted by the impossibility of uf/w_1(m) being an
algebraic integer,


Also incorrect and false. Remember what you said above:

"... consider w_1(m), a factor of a_1 that is a factor of f..."

If w_1(m) is a "factor of f", that can only mean f/w_1(m) is
an algebraic integer, which of course implies that uf/w_1(m)
is an algebraic integer: this is your *assumption* here.

The assumption is itself a contradiction, in that it requires the
ability to define one algebraic integer function by the multiplicative
inverse of another algebraic integer function.

Basic algebra.
IF THIS IS BASIC ALGEBRA, WHY CAN'T I UNDERSTAND YOUR ARGUMENT? I UNDERSTAND
ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.

for all algebraic integers m, if w_1(m) varies with
m.

My hope is that posters who have been so successful in convincing
others that my argument is flawed will post concessions.



Absolutely not!

You are inching closer to the truth in this. You are seeing
how this can work. You are correct that w_1(m) is a factor of
f that depends on m. As noted above this implies that
uf/w_1(m) is an A.I., and a1(m)/w_1(m) is also. I think you
are beginning to see how this can happen.

Readers should note that attempt to push the impossible, one algebraic
integer function defined by the multiplicative inverse of another
algebraic integer function.

My guess is that seeing something like uf/w_1(m) may confuse some who
might believe that u and f are functions of m, so I like to show their
independence by tossing in values, like u=2, f=13, so you have
26/w_1(m).

I note a couple of other things. First, the polynomial in "a"
that you mention above:

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

Your claim is that two roots of this polynomial have a factor
that is f. This means that if r is one of those roots, then
r = f*c, where c is an algebraic integer. This implies that

f^3*c^3 + 3(-1 +m*f^2)*f^2*c^2 - f^2*(m^3*f^4 - 3*m*f^2 + 3*m) = 0,

or, factoring out f^2,

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


The polynomial in c on the left is non-monic and primitive
(if f is not a multiple of 3). If it is irreducible, then
c cannot be an algebraic integer. When f = 5 and m = 1, this
equation is

5*c^3 + 72*c^2 - 553 = 0,

and the polynomial in c is easily shown to be irreducible.
Therefore c cannot be an algebraic integer. This contradicts
your central claim: NONE of the roots of [#] can have a
factor that is 5 = f.


Second note: It may be worthwhile to see how irreducibility
is inextricably tied to factorization of roots. Assume that


Q(m) = x^2 + m*x + 30,

where m is an integer.

The constant term is divisible by 5 (also by 2 and 3, but I
will focus on 5 here).

For certain values of m, this polynomial is reducible, and the
corresponding roots are not both divisible by 5:

m = 11: r1 = 5, r2 = 6

m = 13: r1 = 10, r2 = 3

m = 17: r1 = 15, r2 = 2

m = 31: r1 = 30, r2 = 1.

In all these examples, obviously one root is *divisible* by 5
and the other is *coprime* to 5. And of course in all these
examples, the polynomial is reducible. This is parallel to
Harris's cubic when m = 0: two of the roots are divisible
by f and one is coprime to f. In this particular case his polynomial
[#] is reducible. In general it is not.

Now look at an other example. Say, m = 14. The
polynomial is irreducible, because the discriminant

D^2 = m^ - 4*30 = 76,

which is not a perfect square. The roots of the
polynomial a


r1 = (-14 + sqrt(76))/2 = -7 + sqrt(19) and

r2 = -7 - sqrt(19).

Clearly both r1 and r2 are algebraic integers.

Assume that r1 is a multiple of 5: r1 = 5*s1. Then

25*s1^2 + 70*s1 + 30 = 0.

Factor out 5:

[*] 5*s1^2 + 14*s1 + 6 = 0.

This happens to have discriminant D^2 = 76. [This is
not a coincidence!]. Thus the polynomial in[*] is
*also* irreducible. Therefore s1 cannot be an algebraic
integer.

Therefore r1 cannot be divisible by 5.

The same can similarly be shown for r2.

However, it is now easy to show that both r1 and r2 must both
be *non-coprime* to 5. For, suppose r1 is coprime to 5.
The fact that

r1*r2 = 5*6

would then imply that r2 is DIVISIBLE by 5, which
contradicts the result above that both r1 and r2 are
NOT divisible by 5.

Conclusion: for m = 14, the polynomial is irreducible,
and both roots have a nonunit algebraic integer factor
in common with 5.

This is true more generally. There is nothing special
about 14. Try, for example, m = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12,
14, 15, 16, 18, ... MOST values of m yield an irreducible
polynomial, and both of the roots will both share algebraic
integer factors in common with 5.

In general it is hard to write down exactly what these
factors are, even in the quadratic case. The important
thing to know about them here is that, as suggested above
by the w_1(m) notation, *they will be dependent on m*.

What's fascinating here is how casually posters will just start
talking, which I think actually usually works with a lot of readers
who don't even bother to read carefully.

Just remember that the base position here is that in the ring of
algebraic integers you can have one function defined by the
multiplicative inverse of another function.

Remember xy=26.


James Harris

"David Moran" wrote in message ...
"James Harris" wrote in message
m...
(Nora Baron) wrote in message

deleted

Also incorrect and false. Remember what you said above:

"... consider w_1(m), a factor of a_1 that is a factor of f..."

If w_1(m) is a "factor of f", that can only mean f/w_1(m) is
an algebraic integer, which of course implies that uf/w_1(m)
is an algebraic integer: this is your *assumption* here.

The assumption is itself a contradiction, in that it requires the
ability to define one algebraic integer function by the multiplicative
inverse of another algebraic integer function.

Basic algebra.
IF THIS IS BASIC ALGEBRA, WHY CAN'T I UNDERSTAND YOUR ARGUMENT? I UNDERSTAND
ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.

Unfortunately the people who *do* understand it are doing their best
to confuse everyone else, and you've done your part by basically
heckling me with posts.

If I didn't have mathematicians fighting me, then I could carefully go
through each and every detail to work it out so that everyone who has
a basic grasp of algebra can understand.

But first I need you to help me, by NOT letting posters like "Nora
Baron" and Arturo Magidin get away with muddying the waters.

Other posters need to begin challenging these people who have betrayed
you, after all, they've had many of you nodding along with something
as dumb as xy=2 with *both* x and y algebraic integers where x varies
over the algebraic integers which you get using u=1, f=2, and the
assumption of a w_1(m) as a factor of f that's a varying factor of
a_1, with x=w_1(m), y=uf/w_1(m).

It's posters like Arturo Magidin and "Nora Baron" who've helped make
mathematicians look like fools and patsies, willing to lie and
challenge mathematics itself either willingly, or from incompetence.


James Harris

In article (James Harris) writes:
(Nora Baron) wrote in message . com...
....
However, consider w_1(m), a factor of a_1 that is a factor of f, as
well as a function that varies with m, then it follows that

a_1 x + uf has w_1(m) as a factor,

so dividing through by w_1(m) gives

a_1 x/w_1(m) + uf/w_1(m)

but then uf/w_1(m) cannot in general be an algebraic integer as it's
not representable as a polynomial with a finite number of terms if
w_1(m) varies with m.

A totally off-the-wall, unjustified statement, and, as it
so happens, incorrect. But for now, if you want to claim
it is true, the shoe is on your foot: try to prove it.

I've introduced r(m), to handle the result of uf/w_1(m).

So the poster is requesting that I prove that

r(m) w_1(m) = uf, does not exist over the ring of algebraic integers.

I've concluded that using numbers for u and f, as they are
*independent* of m, helps, so let u=2, f=13.

Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m), so you have

xy=26

and now I'll chat further.

Then the question is, does their exist a multiplicative inverse in the
ring of algebraic integers for 26/w_1(m) for *all* algebraic integers
m?

The simple answer is that if w_1(m) varies with m, then it must vary
over an infinite number of algebraic integer values as m varies over
algebraic integers.

Why? Please prove that.

But w_1(m) must vary from 0 to infinity if it varies with m.

Why? Please prove that.

So if you had any doubts about how low mathematicians could go,
consider that after all, they're trying to convince that you can have
one algebraic integer function defined by the multiplicative inverse
of another algebraic integer function over all algebraic integer m,
like with f=7, 14/w_1(m) = r(m).

It's like saying that you can have xy=2, where x and y are integers,
or algebraic integers, where x varies over the ring, and y remains in
it, though of course, I can just show that for x=5 that doesn't work.

Not at all. The range of w_1(m) may be severely limited, while the
range of x is not limited. It is similar to saying that you can have
xy = 2, where x ranges over the algebraic integer factors of 2, and so y
also ranges over the algebraic integer factors of 2. 5 is now not
a counterexample because it is not an algebraic integer factor of 2.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

"James Harris" wrote in message
m...
"David Moran" wrote in message
...
"James Harris" wrote in message
m...
(Nora Baron) wrote in message

deleted

Also incorrect and false. Remember what you said above:

"... consider w_1(m), a factor of a_1 that is a factor of f..."

If w_1(m) is a "factor of f", that can only mean f/w_1(m) is
an algebraic integer, which of course implies that uf/w_1(m)
is an algebraic integer: this is your *assumption* here.

The assumption is itself a contradiction, in that it requires the
ability to define one algebraic integer function by the multiplicative
inverse of another algebraic integer function.

Basic algebra.
IF THIS IS BASIC ALGEBRA, WHY CAN'T I UNDERSTAND YOUR ARGUMENT? I
UNDERSTAND
ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.

Unfortunately the people who *do* understand it are doing their best
to confuse everyone else, and you've done your part by basically
heckling me with posts.

If I didn't have mathematicians fighting me, then I could carefully go
through each and every detail to work it out so that everyone who has
a basic grasp of algebra can understand.

But first I need you to help me, by NOT letting posters like "Nora
Baron" and Arturo Magidin get away with muddying the waters.

Other posters need to begin challenging these people who have betrayed
you, after all, they've had many of you nodding along with something
as dumb as xy=2 with *both* x and y algebraic integers where x varies
over the algebraic integers which you get using u=1, f=2, and the
assumption of a w_1(m) as a factor of f that's a varying factor of
a_1, with x=w_1(m), y=uf/w_1(m).

It's posters like Arturo Magidin and "Nora Baron" who've helped make
mathematicians look like fools and patsies, willing to lie and
challenge mathematics itself either willingly, or from incompetence.


James Harris

It's you who are subjecting yourself to the heckling. If you want the
heckling to stop, either stop posting, or stop the personal attacks. I can
think of a much better way to do what you're doing, but will you listen to
me? Probably not.

David Moran

"Dik T. Winter" wrote in message ...
In article (James Harris) writes:
(Nora Baron) wrote in message . com...
...
However, consider w_1(m), a factor of a_1 that is a factor of f, as
well as a function that varies with m, then it follows that

a_1 x + uf has w_1(m) as a factor,

so dividing through by w_1(m) gives

a_1 x/w_1(m) + uf/w_1(m)

but then uf/w_1(m) cannot in general be an algebraic integer as it's
not representable as a polynomial with a finite number of terms if
w_1(m) varies with m.

A totally off-the-wall, unjustified statement, and, as it
so happens, incorrect. But for now, if you want to claim
it is true, the shoe is on your foot: try to prove it.

I've introduced r(m), to handle the result of uf/w_1(m).

So the poster is requesting that I prove that

r(m) w_1(m) = uf, does not exist over the ring of algebraic integers.

I've concluded that using numbers for u and f, as they are
*independent* of m, helps, so let u=2, f=13.

Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m), so you have

xy=26

and now I'll chat further.

Then the question is, does their exist a multiplicative inverse in the
ring of algebraic integers for 26/w_1(m) for *all* algebraic integers
m?

The simple answer is that if w_1(m) varies with m, then it must vary
over an infinite number of algebraic integer values as m varies over
algebraic integers.

Why? Please prove that.

It has to do with continuity and slope. If you could have w_1(m) and
w_1(m') equal when m does not equal m' then at that point you'd have
infinite slope or a discontinuity.

I'm not interested in explaining basic mathematics but at least
hopefully I can give other readers some sense of the frustration I've
had to handle dealing with posters who are so mathematically ignorant,
while fanatically replying to my posts negatively.

But w_1(m) must vary from 0 to infinity if it varies with m.

Why? Please prove that.

It turns out that you need the the absolute value, like r(m)r*(m), and
with it it's possible to show that for an algebraic integer function
that varies as m varies--a continuous function--as m varies over all
of algebraic integer r(m)r*(m) must vary from 0 to positive or
negative infinity.

So if you had any doubts about how low mathematicians could go,
consider that after all, they're trying to convince that you can have
one algebraic integer function defined by the multiplicative inverse
of another algebraic integer function over all algebraic integer m,
like with f=7, 14/w_1(m) = r(m).

It's like saying that you can have xy=2, where x and y are integers,
or algebraic integers, where x varies over the ring, and y remains in
it, though of course, I can just show that for x=5 that doesn't work.

Not at all. The range of w_1(m) may be severely limited, while the
range of x is not limited. It is similar to saying that you can have
xy = 2, where x ranges over the algebraic integer factors of 2, and so y
also ranges over the algebraic integer factors of 2. 5 is now not
a counterexample because it is not an algebraic integer factor of 2.

Yeah you can *say* just about anything, but mathematically that
statement is bull****, and for me to have made the discoveries I've
made and be stuck because a lot of posters can get away with bull****
is just ****ing me off.


James Harris

In article ,
(James Harris) wrote:

I'm not interested in explaining basic mathematics but at least
hopefully I can give other readers some sense of the frustration I've
had to handle dealing with posters who are so mathematically ignorant,
while fanatically replying to my posts negatively.

In other words, you have not the slightest idea what you are talking
about.

"David Moran" wrote in message ...
"James Harris" wrote in message
m...
"David Moran" wrote in message
...
"James Harris" wrote in message
m...
(Nora Baron) wrote in message

deleted

Also incorrect and false. Remember what you said above:

"... consider w_1(m), a factor of a_1 that is a factor of f..."

If w_1(m) is a "factor of f", that can only mean f/w_1(m) is
an algebraic integer, which of course implies that uf/w_1(m)
is an algebraic integer: this is your *assumption* here.

The assumption is itself a contradiction, in that it requires the
ability to define one algebraic integer function by the multiplicative
inverse of another algebraic integer function.

Basic algebra.
IF THIS IS BASIC ALGEBRA, WHY CAN'T I UNDERSTAND YOUR ARGUMENT? I
UNDERSTAND
ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.

Unfortunately the people who *do* understand it are doing their best
to confuse everyone else, and you've done your part by basically
heckling me with posts.

If I didn't have mathematicians fighting me, then I could carefully go
through each and every detail to work it out so that everyone who has
a basic grasp of algebra can understand.

But first I need you to help me, by NOT letting posters like "Nora
Baron" and Arturo Magidin get away with muddying the waters.

Other posters need to begin challenging these people who have betrayed
you, after all, they've had many of you nodding along with something
as dumb as xy=2 with *both* x and y algebraic integers where x varies
over the algebraic integers which you get using u=1, f=2, and the
assumption of a w_1(m) as a factor of f that's a varying factor of
a_1, with x=w_1(m), y=uf/w_1(m).

It's posters like Arturo Magidin and "Nora Baron" who've helped make
mathematicians look like fools and patsies, willing to lie and
challenge mathematics itself either willingly, or from incompetence.


James Harris

It's you who are subjecting yourself to the heckling. If you want the
heckling to stop, either stop posting, or stop the personal attacks. I can
think of a much better way to do what you're doing, but will you listen to
me? Probably not.

David Moran

Readers here's what I was replying to from this *same* poster, which
I'm copying down from this post.

IF THIS IS BASIC ALGEBRA, WHY CAN'T I UNDERSTAND YOUR ARGUMENT? I
UNDERSTAND
ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.

So I reply thinking that maybe the guy is serious, and you can see
what he said in return, and notice his arrogance and lack of social
graces.

It's like I'm dealing with children with VERY bad manners.


James Harris