mm-1081
===
Subject: JSH: About time
Now that IÕve revealed the odd and you could say esoteric
error in
core mathematics with such a short, and rather simple
argument, the
issue now is how long until mathematicians decide that 
theyÕd
rather
have correct mathematics versus the *belief* that they had
been
perfect in keeping error out of the collected body of work
that is
called mathematics.
My work is out there and rather easy to go over as can be
seen at the
Hong Konk math site:
See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
And I send people there because their allowal of the use of
LaTeX
makes for a *much* better presentation, and given the
*social* issues
IÕm facing, I need all the help I can get.
Some of you are now facing the reality of the human brain
versus any
fantasy you might have had about being completely rational.
Human
beings are NOT rational creatures but necessarily rely on
social
forces to determine what they believe.
You are creatures of society.
You may have believed that your mathematical knowledge was
based
completely on logic and rationality, but human beings donÕt
work that
way; itÕs built-in to your wiring NOT to work that way.
Some of you must learn to be more than human.
You must learn to be truly rational, for the first times in
your
lives.
So itÕs about time, as I wait, and wonder, how many of you
can handle
the truth.
And how many of you prefer the fantasy which was the world you
believed in, which actually never existed, except in your
imaginations; your wishes for a nicer world, where your
wishes matter.
James Harris
===
Subject: Re: About time
> Now that IÕve revealed the odd and you could say esoteric
error in
> core mathematics with such a short, and rather simple
argument, the
> issue now is how long until mathematicians decide that
theyÕd rather
> have correct mathematics versus the *belief* that they had
been
> perfect in keeping error out of the collected body of work
that is
> called mathematics.
> My work is out there and rather easy to go over as can be
seen at the
> Hong Konk math site:
> See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
> And I send people there because their allowal of the use of
LaTeX
> makes for a *much* better presentation, and given the
*social* issues
> IÕm facing, I need all the help I can get.
> Some of you are now facing the reality of the human brain
versus any
> fantasy you might have had about being completely rational.
Human
> beings are NOT rational creatures but necessarily rely on
social
> forces to determine what they believe.
> You are creatures of society.
> You may have believed that your mathematical knowledge was
based
> completely on logic and rationality, but human beings 
donÕt
work that
> way; itÕs built-in to your wiring NOT to work that way.
> Some of you must learn to be more than human.
> You must learn to be truly rational, for the first times in
your
> lives.
> So itÕs about time, as I wait, and wonder, how many of you
can handle
> the truth.
> And how many of you prefer the fantasy which was the world
you
> believed in, which actually never existed, except in your
> imaginations; your wishes for a nicer world, where your
wishes matter.
> James Harris
I want to know why you keep pointing to your Army service. Is
it because
you
want respect? If so, why do you expect respect when you 
donÕt
respect
others
when they did not disrespect you (i.e. calling people Ōevil
bastardsÕ).
David Moran
===
Subject: Re: JSH: About time
> Now that IÕve revealed the odd and you could say esoteric
error in
> core mathematics with such a short, and rather simple
argument, the
> issue now is how long until mathematicians decide that
theyÕd rather
> have correct mathematics versus the *belief* that they had
been
> perfect in keeping error out of the collected body of work
that is
> called mathematics.
> My work is out there and rather easy to go over as can be
seen at the
> Hong Konk math site:
OOPS! That should be Hong Kong.
> See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
> And I send people there because their allowal of the use of
LaTeX
> makes for a *much* better presentation, and given the
*social* issues
> IÕm facing, I need all the help I can get.
James Harris
===
Subject: Re: JSH: About time
you never do any justice to your would-be helpmeets
-- the entirety of your oft-cited Usenet audience,
one of the smaller qyoobiqles of the googolplex --
so that you allow yourself to make the same mistakes,
over & again. i.e. I didnÕt see any dyscussion
of a problem with anyone comprehending your equations,
because of ASCIIized algebra, although
the social issues seem to be relavent,
not to mention the help you can get.
anyway, when I went there,
Hong Kong was scratching its collective head
on your mathematics, after all.
> Now that IÕve revealed the odd and you could say esoteric
error in
> called mathematics.
> See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
>
> And I send people there because their allowal of the use of
LaTeX
> makes for a *much* better presentation, and given the
*social* issues
> IÕm facing, I need all the help I can get.
--le ducs dÕEnron!
http://larouchepub.com
===
Subject: Re: About time
> Now that IÕve revealed the odd and you could say esoteric
error in
> core mathematics with such a short, and rather simple
argument
The only esoteric things in math are things you donÕt
understand yet.
How exactly your result is going to change mathematics?
Maybe some big theorems are wrong? Or maybe some new
fascinating
theorems will be formulated? Who is going to do all that
exciting job?
Are you hoping that mathematicians will pick up where you
stopped and
move on? But why wait for them?
Why you stopped? Why not go all the way?
===
Subject: Re: JSH: About time
> Now that IÕve revealed the odd and you could say esoteric
error in
> core mathematics with such a short, and rather simple
argument, the
> issue now is how long until mathematicians decide that
theyÕd rather
> have correct mathematics versus the *belief* that they had
been
> perfect in keeping error out of the collected body of work
that is
> called mathematics.
> My work is out there and rather easy to go over as can be
seen at the
> Hong Konk math site:
Hong Konk? YouÕre sure it isnÕt Honk Honk?
> See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
> And I send people there because their allowal of the use of
LaTeX
> makes for a *much* better presentation, and given the
*social* issues
> IÕm facing, I need all the help I can get.
This gives me the chance to re-post a discussion of the LaTeX
website you mention above - maybe you would like to respond -
James Harris claims to prove, in
http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
that certain polynomials factor in a form which
contradicts other mathematical proofs. Specifically:
Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
- 3*(-1 +m*f^2)*x*u^2 + u^3*f),
where f is a prime, u is an integer coprime to f,
and m is an integer.
Assume P(m) is factored in the form
P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
where a1, a2, and a3 are algebraic integers.
Let g1 = (a1*x + u*f), g2 = (a2*x + u*f), g3 = (a3*x + u*f).
Note that P(0) = f^2*(3*x*u^2 + u^3*f).
Harris says:
... two of the aÕs go to 0 when m = 0 which
is also seen from the cubic defining the aÕs.
Then arbitrarily picking a1 and a2 as the ones
that go to 0 at m = 0, you have
g1 = u*f, g2 = u*f, g3 = 3*x + u*f.
But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
only two of the gÕs while with the third it is
blocked, as long as it [f] is coprime to 3 and
x, so assume it is, and assume as well that f
is coprime to u.
Then it follows from the constant terms that g1
and g2 each have a factor that is f.
Remember, the constant terms with respect to m
cannot vary as m varies, or they wouldnÕt be
constant terms, right?
Sounds like it makes sense, including that last bit,
doesnÕt it? P(0) is the evaluation of the polynomial
when m = 0, so it must be the constant term.
Of course all of what I quoted above except that last
paragraph pertains to what happens when m = 0. For
instance, saying that a1 = a2 = 0, that is true only when
m = 0. When m is nonzero, we know that a1 and a2 are also
nonzero.
So itÕs clear that a1 and a2 depend on m.
I donÕt think even James Harris disagrees with that.
So when he says g1 = u*f and g2 = u*f, that is true only
when m = 0. When m is not zero, you get, for example,
g1 = a1*x + u*f.
When Harris says constant term he does not mean
constant term with respect to x. He is dealing with what
he calls a nonpolynomial factorization. The variable here
not as P(x).
As noted above, a1 and a2 are dependent on m. Both a1
and a2 should really be written as a1(m) and a2(m).
A key step in HarrisÕs argument is this :
But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
only two of the gÕs while with the third it is
blocked
When Harris says blocked he means that f is not
a divisor of the third g.
Of course here he is obviously talking about m = 0,
though he would like to conclude this for m <> 0.
He is saying that when m = 0, the only way you are going
to factor out f^2 from the expression
(a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f)
is:
f^2*([a1(0)/f]*x + u)*([a2(0)/f]*x + u)*(a3(0)*x + u*f),
because a1(0) = a2(0) = 0 and a3(0) = 3:
f^2 * (u) * (u) * (3*x + u*f).
So far, just fine. This is all assuming m = 0.
There are infinitely many ways that f^2 can be written
as a product of three algebraic integers. Say, for
example, one of them is f^2 = f1 * f2 * f3. Assume that
none of f1, f2 or f3 is a unit in the algebraic integers.
Now choose an integer m <> 0. Suppose a1(m) is
divisible by, say, f1. Suppose that f/f1 is also an
algebraic integer. Then
a1(m)*x + u*f = f1*([a1(m)/f1]*x + u*f/f1).
And assume similar things for g2 and g3:
a2(m)*x + u*f = f2*([a2(m)/f2]*x + u*f/f2),
a3(m)*x + u*f = f3*([a2(m)/f3]*x + u*f/f3),
and all of a1(m)/f1, a2(m)/f2, a3(m)/f3, f/f1, f/f2, and
f/f3 are algebraic integers.
Putting all this together, one has
P(m)/f^2 = P(m)/(f1*f2*f3)
= ([a1(m)/f1]*x + u*f/f1) *
([a2(m)/f2]*x + u*f/f2) *
([a3(m)/f3]*x + u*f/f3).
The key thing here is that these factors f1, f2, and f3
may ALSO be dependent on m. When Harris has shown is that
when m = 0, f1 = f2 = f, and f3 = 1. Or you could say,
f1(0) = f2(0) = f and f3(0) = 1.
I donÕt claim that I have shown here that f^2 can be
factored so that all of the above fits together. What I
claim here is that Harris has not shown that it CANÕT
happen. As long as that gap is left open, he does not
have a proof.
To put it another way: Harris has not shown that f1(m) =
f2(m) = f and f3(m) = 1 in general. He has shown it only
when m = 0. Even though f1*f2*f3 = f^2, a constant with
respect to m, it is entirely possible that f1, f2, and f3 are
*dependent* on m. Why is this so hard to understand?
There are examples which show that no such proof for
m <> 0 can be obtained. For example, let f = 5, m = 1,
and u = 1. Then
P(m) = 25*(553 * x^3 - 72 * x + 5),
or
P(m)/25 = 553*x^3 - 72*x + 5.
Now suppose Harris is right. Then the factorization of
P(m)/f^2 must take the form
P(m)/25 = ([a1/5]*x + 1)*([a2/5]*x + 1)*([a3/5]*x + 5),
where a1/5 and a2/5 are *algebraic integers*.
Let b1 = a1/5. Then -1/b1 is a root of
553*x^3 - 72*x + 5 = 0. That is,
-553/b1^3 + 72/b1 + 5 = 0.
Multiplying through by b1^3, one obtains:
5*b1^3 + 72*b1^2 - 553 = 0.
The expression on the left is a *non-monic* polynomial in
b1 with integer coefficients, and it is *irreducible* over
the rationals.
Therefore b1 cannot be an algebraic integer [*].
That is, a1/5 cannot be an algebraic integer.
Therefore when m = 1, a1 = a1(m) is not divisible by f = 5
in the algebraic integers.
Therefore HarrisÕs claim is false.
Thus what we have here are two things:
(1) An example which shows that HarrisÕs claim
*cannot* be true, and
(2) An examination of his argument which points
to exactly where he has a gap.
The example is complete and straightforward. There are
no hidden tricks or bogus math statements. There is no
Galois theory [although there is a separate argument using
Galois theory which shows the same thing]. There is one
theorem [*] from elementary algebraic number theory, but this
theorem has been accepted as valid by Harris.
James HarrisÕs argument, by contrast, has a gap, a 
mysterious
unjustified step, and in the version quoted above, he
doesnÕt even state what that step is. He is claiming that
because a1(0) = 0 is divisible by f, it follows that a1(m)
is divisible by f for m <> 0. He says:
Remember, the constant terms with respect to m
cannot vary as m varies, or they wouldnÕt be
constant terms, right?
Of course this is true. It is a tautology. Constant
terms donÕt vary.
It can be a good idea, when you are losing an argument, to
state something which is obviously true, even though it has
nothing to do with your desired conclusion. It confuses
your attackers - all they can say is, well, yeah, youÕre
right about that.
What they *should* say is, SO WHAT?
Here, since a1(m) is not a constant function, why does
HarrisÕs statement tell you anything about divisibility of
a1(m) when m <> 0 ? That step is simply not there in
HarrisÕs argument. It is a gap. The counterexample shows
that it is more than a gap. It is an error which cannot be
fixed. The quote above buys you nothing. ItÕs 
not a
question about the constant term a1(0). ItÕs a question
about a1(m).
And, I think, there is another deeper undercurrent here.
When you factor a polynomial Q(x) into subpolynomials,
Q(x) = q1(x) * q2(x) * ... * qk(x),
the coefficients of q1, q2, ..., qk are independent of the
polynomial variable x. They are constants. Harris is
factoring the polynomial P(m) into 3 factors, but they are
(as he has reminded us many times) nonpolynomial factors.
What we expect with polynomial factors is not necessarily
true in HarrisÕs factorization: the coefficients 
a1(m),
a2(m), and a3(m) *cannot* be constant with respect to m.
{If they were, then a1(m) would be a1(0) = 0 for all m, and
we know that is false.} HarrisÕs argument would be in better
shape if they were. But that is, I think, part of his
imprecise sub-rosa thinking, and it has led him into an
error which he clings to like a drowning man clings to a
straw.
> Some of you are now facing the reality of the human brain
versus any
> fantasy you might have had about being completely rational.
Human
> beings are NOT rational creatures but necessarily rely on
social
> forces to determine what they believe.
You wish.
> You are creatures of society.
> You may have believed that your mathematical knowledge was
based
> completely on logic and rationality, but human beings 
donÕt
work that
> way; itÕs built-in to your wiring NOT to work that way.
*Your* wiring, maybe.
> Some of you must learn to be more than human.
> You must learn to be truly rational, for the first times in
your
> lives.
> So itÕs about time, as I wait, and wonder, how many of you
can handle
> the truth.
This is such patronizing drivel. I would think you would get
sick of writing it.
> And how many of you prefer the fantasy which was the world
you
> believed in, which actually never existed, except in your
> imaginations; your wishes for a nicer world, where your
wishes matter.
Arturo also had some speculation about where your thinking
went wrong. You wonÕt agree with him either.
Nora B.
> James Harris
===
Subject: Re: JSH: About time
> Now that IÕve revealed the odd and you could say esoteric
error in
> core mathematics with such a short, and rather simple
argument, the
> issue now is how long until mathematicians decide that
theyÕd rather
> have correct mathematics versus the *belief* that they had
been
> perfect in keeping error out of the collected body of work
that is
> called mathematics.
>
> My work is out there and rather easy to go over as can be
seen at the
> Hong Konk math site:
>
> Hong Konk? YouÕre sure it isnÕt Honk Honk?
Typo. It should be Hong Kong.
> See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
>
> And I send people there because their allowal of the use of
LaTeX
> makes for a *much* better presentation, and given the
*social* issues
> IÕm facing, I need all the help I can get.
>
> This gives me the chance to re-post a discussion of the
LaTeX
> website you mention above - maybe you would like to respond
-
Sure.
> James Harris claims to prove, in
> http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
> that certain polynomials factor in a form which
> contradicts other mathematical proofs. Specifically:
> Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
> - 3*(-1 +m*f^2)*x*u^2 + u^3*f),
> where f is a prime, u is an integer coprime to f,
> and m is an integer.
> Assume P(m) is factored in the form
> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
> where a1, a2, and a3 are algebraic integers.
> Let g1 = (a1*x + u*f), g2 = (a2*x + u*f), g3 = (a3*x + u*f).
> Note that P(0) = f^2*(3*x*u^2 + u^3*f).
> Harris says:
> ... two of the aÕs go to 0 when m = 0 which
> is also seen from the cubic defining the aÕs.
> Then arbitrarily picking a1 and a2 as the ones
> that go to 0 at m = 0, you have
> g1 = u*f, g2 = u*f, g3 = 3*x + u*f.
> But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
> only two of the gÕs while with the third it is
> blocked, as long as it [f] is coprime to 3 and
> x, so assume it is, and assume as well that f
> is coprime to u.
> Then it follows from the constant terms that g1
> and g2 each have a factor that is f.
> Remember, the constant terms with respect to m
> cannot vary as m varies, or they wouldnÕt be
> constant terms, right?
> Sounds like it makes sense, including that last bit,
> doesnÕt it? P(0) is the evaluation of the polynomial
> when m = 0, so it must be the constant term.
It does make sense. Basically I isolate terms of P(m) that are
*independent* of m.
Then I look at at the same terms for P(m)/f^2 and find that a
factor
of f^2 has been removed.
Logic dictates that the removal is *independent* of m, but
that is a
conclusion several posters wish to convince others is false,
so here I
am, once again, arguing about the obvious.
> Of course all of what I quoted above except that last
> paragraph pertains to what happens when m = 0. For
> instance, saying that a1 = a2 = 0, that is true only when
> m = 0. When m is nonzero, we know that a1 and a2 are also
> nonzero.
Readers need focus on the simple fact that with P(m)
considering P(0),
that is setting m=0, gives you terms that donÕt have m, so
they are
independent of it.
Stay focused on that fact as Nora Baron tries to take you for
a
ride.
> So itÕs clear that a1 and a2 depend on m.
> I donÕt think even James Harris disagrees with that.
> So when he says g1 = u*f and g2 = u*f, that is true only
> when m = 0. When m is not zero, you get, for example,
> g1 = a1*x + u*f.
> When Harris says constant term he does not mean
> constant term with respect to x. He is dealing with what
> he calls a nonpolynomial factorization. The variable here
> not as P(x).
>
> As noted above, a1 and a2 are dependent on m. Both a1
> and a2 should really be written as a1(m) and a2(m).
> A key step in HarrisÕs argument is this :
> But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
> only two of the gÕs while with the third it is
> blocked
> When Harris says blocked he means that f is not
> a divisor of the third g.
Readers remember, with P(m), I set m=0, that is, look at
P(0), to get
terms *independent* of m, as m has been set to 0, so itÕs
gone.
Then I do the same for P(m)/f^2, and notice that for those
terms
*independent* of m, a factor of f^2 has disappeared.
Now you can see that for P(0) I have
P(0) = u^2 f^2( 3x + uf)
and for P(0)/f^2 I have
P(0)/f^2 = u^2 ( 3x + uf)
so everything is simple enough.
> Of course here he is obviously talking about m = 0,
> though he would like to conclude this for m <> 0.
HereÕs where you have Nora Baron sneaking in something 
rather
dumb
mathematically.
If my point is isolating off m, so that I have terms
*independent* of
m, why canÕt I conclude that they are indeed independent of 
m?
ItÕs a bizarre thing to question and maybe many of you 
simply
canÕt
believe that a poster like Nora Baron would keep posting and
making
a fuss based on such a position.
But remember, I *isolate* independent factors of P(m), by
setting m=0,
so I donÕt have to worry about what value m has.
> He is saying that when m = 0, the only way you are going
> to factor out f^2 from the expression
> (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f)
> is:
> f^2*([a1(0)/f]*x + u)*([a2(0)/f]*x + u)*(a3(0)*x + u*f),
> because a1(0) = a2(0) = 0 and a3(0) = 3:
> f^2 * (u) * (u) * (3*x + u*f).
> So far, just fine. This is all assuming m = 0.
And consider that the poster Nora Baron is considering terms
which
have isolated out as INDEPENDENT of m, which is why I set
m=0, in the
first place!
ItÕs like this poster is going on and on about a rather 
simple
technique for pulling out factors independent of a particular
variable, as if itÕs actually sinister, where *really* 
IÕm
just trying
to fool people with a special case, as if things change when
m does
not equal 0.
But if these terms change as m changes, then theyÕre not
independent
of m, now are they?
Nora Baron is trying to refute algebra!!! It should be a news
ßash
around the math world that setting m=0, gives terms that are
STILL
dependent on m, which I call the Shadow m. It has
supernatural powers
and refuses to go away, even when set to 0.
> There are infinitely many ways that f^2 can be written
> as a product of three algebraic integers. Say, for
> example, one of them is f^2 = f1 * f2 * f3. Assume that
> none of f1, f2 or f3 is a unit in the algebraic integers.
> Now choose an integer m <> 0. Suppose a1(m) is
> divisible by, say, f1. Suppose that f/f1 is also an
> algebraic integer. Then
> a1(m)*x + u*f = f1*([a1(m)/f1]*x + u*f/f1).
> And assume similar things for g2 and g3:
> a2(m)*x + u*f = f2*([a2(m)/f2]*x + u*f/f2),
> a3(m)*x + u*f = f3*([a2(m)/f3]*x + u*f/f3),
> and all of a1(m)/f1, a2(m)/f2, a3(m)/f3, f/f1, f/f2, and
> f/f3 are algebraic integers.
> Putting all this together, one has
> P(m)/f^2 = P(m)/(f1*f2*f3)
> = ([a1(m)/f1]*x + u*f/f1) *
> ([a2(m)/f2]*x + u*f/f2) *
> ([a3(m)/f3]*x + u*f/f3).
> The key thing here is that these factors f1, f2, and f3
> may ALSO be dependent on m. When Harris has shown is that
> when m = 0, f1 = f2 = f, and f3 = 1. Or you could say,
> f1(0) = f2(0) = f and f3(0) = 1.
Which actually blows you out of the ring of algebraic
integers, and
worse provably, if the fÕs are *functions* of m, then f1, f2
and f3
have zeroÕs, at which point the equation would blow up.
Remember
dividing by 0 is a no-no.
BUT, remember this poster above has
> Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
> - 3*(-1 +m*f^2)*x*u^2 + u^3*f),
Well then, how can P(m)/f^2 introduce such a problem as
blowing up for
some value that where f1, f2 or f3 is 0?
Besides, if I find terms that are *independent* of m, then
they are
independent of m, so the same terms can be isolated in each
factor of
> = ([a1(m)/f1]*x + u*f/f1) *
> ([a2(m)/f2]*x + u*f/f2) *
> ([a3(m)/f3]*x + u*f/f3).
If you do so, youÕll find that the terms 
independent of m, for
*two*
factors, must have a factor that is f, while one does not.
Here Nora Baron needs you to believe that setting m=0 
doesnÕt
in
fact give you terms *independent* of m, but thatÕs just
mathematically
wrong.
But Nora Baron is working to convince you.
WhatÕs key here is the claim of dependency on m, but
remember, I set
m=0 to *remove* dependency on m, but now the Shadow m has
apparently
returned according to Nora Baron!!!
> I donÕt claim that I have shown here that f^2 can be
> factored so that all of the above fits together. What I
> claim here is that Harris has not shown that it CANÕT
> happen. As long as that gap is left open, he does not
> have a proof.
ThereÕs no gap, as itÕs just Nora Baron now 
switching to
trying to
show reasonable doubt, I guess.
However, as I emphasized repeatedly through this post, the
technique
of setting m=0, does actually work, as it shows factors
independent of
m, and allowing a poster like Nora Baron to question such
basic
algebra, is proof that many of you care more about society
than math.
After all, you *want* to believe her so you can believe IÕm
wrong.
ThatÕs a loser proposition, as itÕs math. If 
any of you had
sense
youÕd realize that believing some person just because it
makes you
feel good is NOT math.
> To put it another way: Harris has not shown that f1(m) =
> f2(m) = f and f3(m) = 1 in general. He has shown it only
> when m = 0. Even though f1*f2*f3 = f^2, a constant with
> respect to m, it is entirely possible that f1, f2, and f3
are
> *dependent* on m. Why is this so hard to understand?
Now you see that word dependent again, when IÕve used m=0 to
find
terms that are *independent*, which shows you immediately
that Nora
Baron is full of it.
ItÕs simply not sane for Nora Baron to repeatedly argue that
terms
independent of m, are in fact dependent on m, as itÕs a
rather wacky
contradiction.
I keep thinking people will laugh at such an argument, but
you donÕt.
You keep sitting quietly, like here where this poster Nora
Baron
just argued that terms independent of m--found by setting
m=0--might
in fact be dependent on m, which isnÕt mathematics at all.
Fascinating.
> There are examples which show that no such proof for
> m <> 0 can be obtained. For example, let f = 5, m = 1,
> and u = 1. Then
> P(m) = 25*(553 * x^3 - 72 * x + 5),
> or
> P(m)/25 = 553*x^3 - 72*x + 5.
> Now suppose Harris is right. Then the factorization of
> P(m)/f^2 must take the form
> P(m)/25 = ([a1/5]*x + 1)*([a2/5]*x + 1)*([a3/5]*x + 5),
> where a1/5 and a2/5 are *algebraic integers*.
> Let b1 = a1/5. Then -1/b1 is a root of
> 553*x^3 - 72*x + 5 = 0. That is,
> -553/b1^3 + 72/b1 + 5 = 0.
> Multiplying through by b1^3, one obtains:
> 5*b1^3 + 72*b1^2 - 553 = 0.
> The expression on the left is a *non-monic* polynomial in
> b1 with integer coefficients, and it is *irreducible* over
> the rationals.
> Therefore b1 cannot be an algebraic integer [*].
> That is, a1/5 cannot be an algebraic integer.
> Therefore when m = 1, a1 = a1(m) is not divisible by f = 5
> in the algebraic integers.
> Therefore HarrisÕs claim is false.
A hallmark of an error in core is the ability for *two*
different
sides to claim proof. Here the problem is a definition error 
in
core, as the definition of algebraic integers as roots of
*monic*
polynomials excludes numbers that should be included, though
roots of
non-monics.
You saw the proof above, which involves basic algebra, and
you saw the
position of Nora Baron which succincly is that terms
independent of
m, might be dependent on m, which is a direct contradiction.
> Thus what we have here are two things:
> (1) An example which shows that HarrisÕs claim
> *cannot* be true, and
The example shows how bad an error in core can be as it
allows people
to supposedly prove things not true.
Here you can figure out which side is right because my algebra
is
simpler, so you look at what youÕd have to throw out to
believe Nora
Baron and itÕs basic things like that setting m=0 with P(m)
gives
terms independent of m, which throws out *axioms* while IÕm
showing a
problem with a DEFINITION.
So trust Nora Baron and challenge axioms, or trust algebra,
and
modify the definition.
> (2) An examination of his argument which points
> to exactly where he has a gap.
That claim depends on the assertion that terms independent of
m might
be dependent on m, which is a direct contradiction.
> The example is complete and straightforward. There are
> no hidden tricks or bogus math statements. There is no
> Galois theory [although there is a separate argument using
> Galois theory which shows the same thing]. There is one
> theorem [*] from elementary algebraic number theory, but
this
> theorem has been accepted as valid by Harris.
> James HarrisÕs argument, by contrast, has a gap, a
mysterious
> unjustified step, and in the version quoted above, he
> doesnÕt even state what that step is. He is claiming that
> because a1(0) = 0 is divisible by f, it follows that a1(m)
> is divisible by f for m <> 0. He says:
> Remember, the constant terms with respect to m
> cannot vary as m varies, or they wouldnÕt be
> constant terms, right?
> Of course this is true. It is a tautology. Constant
> terms donÕt vary.
So why is this poster trying to claim that *maybe* they do?
> It can be a good idea, when you are losing an argument, to
> state something which is obviously true, even though it has
> nothing to do with your desired conclusion. It confuses
> your attackers - all they can say is, well, yeah, youÕre
> right about that.
> What they *should* say is, SO WHAT?
> Here, since a1(m) is not a constant function, why does
> HarrisÕs statement tell you anything about divisibility of
> a1(m) when m <> 0 ? That step is simply not there in
> HarrisÕs argument. It is a gap. The counterexample shows
> that it is more than a gap. It is an error which cannot be
> fixed. The quote above buys you nothing. ItÕs 
not a
> question about the constant term a1(0). ItÕs a question
> about a1(m).
No matter how many distractions the poster throws out at you,
the
simple fact is that the constant term is independent of m, I
look at
changes in the constant term, as that factor of f^2 goes
away, when
you have P(m)/f^2, so thereÕs no *rational* way to argue
dependency on
m.
> And, I think, there is another deeper undercurrent here.
> When you factor a polynomial Q(x) into subpolynomials,
> Q(x) = q1(x) * q2(x) * ... * qk(x),
> the coefficients of q1, q2, ..., qk are independent of the
> polynomial variable x. They are constants. Harris is
> factoring the polynomial P(m) into 3 factors, but they are
> (as he has reminded us many times) nonpolynomial factors.
> What we expect with polynomial factors is not necessarily
> true in HarrisÕs factorization: the 
coefficients a1(m),
> a2(m), and a3(m) *cannot* be constant with respect to m.
> {If they were, then a1(m) would be a1(0) = 0 for all m, and
> we know that is false.} HarrisÕs argument would be in 
better
> shape if they were. But that is, I think, part of his
> imprecise sub-rosa thinking, and it has led him into an
> error which he clings to like a drowning man clings to a
straw.
Notice how much Nora Baron has posted here.
ItÕs annoying to deal with such a poster.
> Some of you are now facing the reality of the human brain
versus any
> fantasy you might have had about being completely rational.
Human
> beings are NOT rational creatures but necessarily rely on
social
> forces to determine what they believe.
>
> You wish.
Well look at you Nora Baron here you are arguing that
variables
independent of m, might be dependent on m, which is a direct
contradiction, so itÕs an irrational position.
> You are creatures of society.
Nora Baron clearly is a member of math society who finds
trouble
with the idea that thereÕs a hundred year old 
definition error
in core
mathematics, so whatÕs the posterÕs solution? 
Throw out basic
algebra.
Yup, This poster canÕt handle the truth.
> You may have believed that your mathematical knowledge was
based
> completely on logic and rationality, but human beings 
donÕt
work that
> way; itÕs built-in to your wiring NOT to work that way.
>
> *Your* wiring, maybe.
Your wiring Nora Baron allows you to believe something that
directly
contradicts itself, which is what the belief that terms
independent of
m might in fact be dependent on m, which I think indicates a
limitation of that wiring.
Human beings can rather easily accept contradictions on that
level for
social reasons.
After all, people feed you and clothe you. It makes sense to
care
more about what society thinks, or what you think society
thinks than
something abstract like mathematical truth.
You canÕt eat a math proof.
> Some of you must learn to be more than human.
>
> You must learn to be truly rational, for the first times in
your
> lives.
>
> So itÕs about time, as I wait, and wonder, how many of you
can handle
> the truth.
>
> This is such patronizing drivel. I would think you would get
> sick of writing it.
Notice the social forces at work readers.
Human beings *need* social approval of truth before accepting
it.
> And how many of you prefer the fantasy which was the world
you
> believed in, which actually never existed, except in your
> imaginations; your wishes for a nicer world, where your
wishes matter.
>
> Arturo also had some speculation about where your thinking
> went wrong. You wonÕt agree with him either.
The argument at
http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
is simple enough that thereÕs no room for error.
James Harris
===
Subject: Re: JSH: About time
>> Now that IÕve revealed the odd and you could say esoteric
error
in
>> core mathematics with such a short, and rather simple
argument,
the
>> issue now is how long until mathematicians decide that
theyÕd
rather
>> have correct mathematics versus the *belief* that they had
been
>> perfect in keeping error out of the collected body of work
that
is
>> called mathematics.
>>
>> My work is out there and rather easy to go over as can be
seen at
the
>> Hong Konk math site:
>>
>> Hong Konk? YouÕre sure it isnÕt Honk Honk?
>Typo. It should be Hong Kong.
>> See <a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://mathd
b.math.cuhk.edu.hk/forum/e_show.php?msg=782</a>>
>> And I send people there because their allowal of the use
of LaTeX
>> makes for a *much* better presentation, and given the
*social*
issues
>> IÕm facing, I need all the help I can get.
>>
>> This gives me the chance to re-post a discussion of the
LaTeX
>> website you mention above - maybe you would like to
respond -
>Sure.
>> James Harris claims to prove, in
>> <a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://mathd
b.math.cuhk.edu.hk/forum/e_show.php?msg=782</a>> that certain
polynomials factor in a form which
>> contradicts other mathematical proofs. Specifically:
>> Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
>> - 3*(-1 +m*f^2)*x*u^2 + u^3*f),
>> where f is a prime, u is an integer coprime to f,
>> and m is an integer.
>> Assume P(m) is factored in the form
>> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
>> where a1, a2, and a3 are algebraic integers.
>> Let g1 = (a1*x + u*f), g2 = (a2*x + u*f), g3 = (a3*x +
u*f).
>> Note that P(0) = f^2*(3*x*u^2 + u^3*f).
>> Harris says:
>> ... two of the aÕs go to 0 when m = 0 which
>> is also seen from the cubic defining the 
aÕs.
>> Then arbitrarily picking a1 and a2 as the ones
>> that go to 0 at m = 0, you have
>> g1 = u*f, g2 = u*f, g3 = 3*x + u*f.
>> But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
>> only two of the gÕs while with the third it is
>> blocked, as long as it [f] is coprime to 3 and
>> x, so assume it is, and assume as well that f
>> is coprime to u.
>> Then it follows from the constant terms that g1
>> and g2 each have a factor that is f.
>> Remember, the constant terms with respect to m
>> cannot vary as m varies, or they wouldnÕt be
>> constant terms, right?
>> Sounds like it makes sense, including that last bit,
>> doesnÕt it? P(0) is the evaluation of the polynomial
>> when m = 0, so it must be the constant term.
>It does make sense. Basically I isolate terms of P(m) that
are
>*independent* of m.
By which you mean P(0).
>Then I look at at the same terms for P(m)/f^2 and find that a
factor
>of f^2 has been removed.
>Logic dictates that the removal is *independent* of m,
Removal here means that you divide P(m) by f^2. That does
not cause a problem in itself. It is really how f^2 is
distributed among the factors
ai*x + u*f
that is the issue.
You do not disagree that ai is dependent
on m. LetÕs write it as ai(m).
Therefore the way in which factors of f divide
ai(m) in general can be expected to depend on m.
You prove that for m = 0, f divides ai(0).
You think somehow that that proves f divides ai(m)
for all m.
LetÕs say f = 5 and ai(m) = sqrt(5) * m + 15.
Certainly f = 5 divides ai(0) = 15.
But f = 5 does not divide ai(1) = sqrt(5) + 15.
Here is another example. Say ai(m) = 5 * m, and f = 3.
The ai(0) = 0, and, as in your case, f divides ai(0).
However, if m = 1, ai(m) = 5, and f = 3 does not divide
a1(m) = 5. In fact f = 3 is relatively prime to 5,
even in the algebraic integers.
I am not saying that in your application, ai(m)
actually equals 5 * m. I am simply saying that your logic
breaks down. You have not shown that ai(m), as a
function of m, does not behave something like
ai(m) = 5 * m. You have nothing explicit whatsoever
about ai(m), EXCEPT when m = 0. But that clearly does
not tell you anything about how ai(m) behaves with
respect to f when m = 0. ThatÕs the point.
Put it another way. You have shown that ai(0)
is divisible by f. Since ai(m) is a function of
m which you probably cannot even write down,
you do not know anything about the divisibility
of ai(m) by f when m <> 0. It has NOTHING to do
with whether ai(0) or P(0) or whatever is
independent of ai(m), or your claim that you
have somehow removed the constant term. You
are badly confused on that point.
>but that is a
>conclusion several posters wish to convince others is false,
so here
I
>am, once again, arguing about the obvious.
WhatÕs obvious here is that you have leaped from
m = 0 to all other values of m with no justification.
>> Of course all of what I quoted above except that last
>> paragraph pertains to what happens when m = 0. For
>> instance, saying that a1 = a2 = 0, that is true only when
>> m = 0. When m is nonzero, we know that a1 and a2 are also
>> nonzero.
>>Readers need focus on the simple fact that with P(m)
considering
P(0),
>that is setting m=0, gives you terms that donÕt have m, so
they are
>independent of it.
>Stay focused on that fact as Nora Baron tries to take you
for a
>ride.
>> So itÕs clear that a1 and a2 depend on m.
>> I donÕt think even James Harris disagrees with that.
>> So when he says g1 = u*f and g2 = u*f, that is true only
>> when m = 0. When m is not zero, you get, for example,
>> g1 = a1*x + u*f.
>> When Harris says constant term he does not mean
>> constant term with respect to x. He is dealing with what
>> he calls a nonpolynomial factorization. The variable here
>> not as P(x).
>>
>> As noted above, a1 and a2 are dependent on m. Both a1
>> and a2 should really be written as a1(m) and a2(m).
>> A key step in HarrisÕs argument is this :
>> But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
>> only two of the gÕs while with the third it is
>> blocked
>> When Harris says blocked he means that f is not
>> a divisor of the third g.
>Readers remember, with P(m), I set m=0, that is, look at
P(0), to get
>terms *independent* of m, as m has been set to 0, so itÕs
gone.
Looking at m = 0 does not mean, in any sense that
m is gone.
>Then I do the same for P(m)/f^2,
No, you absolutely donÕt. You never really deal
with P(m).
When m = 0, a1*x + u*f = u*f. This is divisible
by f because a1 = 0 also.
When m <> 0, a1 is not equal to 0. You donÕt know
its value. In particular, you donÕt know that it
is divisible by f. Just SAYING that it is is not
sufficient.
> and notice that for those terms
>*independent* of m, a factor of f^2 has disappeared.
>Now you can see that for P(0) I have
>P(0) = u^2 f^2( 3x + uf)
>and for P(0)/f^2 I have
>P(0)/f^2 = u^2 ( 3x + uf)
>so everything is simple enough.
Yes: For m = 0. We agree on this. LetÕs
try to get past it. Divisibility of a1 by f
when m equals 0 does not tell you anything about
divisibility of a1 when m <> 0. a1 is dependent
on m in a very complicated way. You have never
tried writing a1 out as a function of m. It is
however fairly easy to show that for m = 1 and
f = 5, a1 CANNOT be divisible by f.
>> Of course here he is obviously talking about m = 0,
>> though he would like to conclude this for m <> 0.
>HereÕs where you have Nora Baron sneaking in something
rather dumb
>mathematically.
I am not sneaking anything in anywhere. Everything
is above board. What you see is what you get.
>If my point is isolating off m, so that I have terms
*independent* of
>m, why canÕt I conclude that they are indeed independent of
m?
Look. NO ONE is arguing here about P(0). That is not
at issue. We are arguing about P(m) when m <> 0.
The fact that P(0) does not change when m changes
does not imply that the factors of the form
ai*x + u*f
also do not change when m changes. In fact we KNOW
that ai must be a factor of m ! Why? Because when
m = 0, ai = 0. But when m <> 0, we know *for sure*
that ai is not zero [otherwise the cubic would have
degree 1]. Thus ai DEPENDS ON m. Therefore you cannot
assume that something about it which is true when m
equals 0 is also true when m <> 0.
>ItÕs a bizarre thing to question and maybe many of you
simply canÕt
>believe that a poster like Nora Baron would keep posting and
making
>a fuss based on such a position.
>But remember, I *isolate* independent factors of P(m), by
setting
m=0,
>so I donÕt have to worry about what value m has.
What you say about P(0) applies only when m = 0.
You are just confusing yourself with verbiage here
like isolate and independent factors. As noted
above, in the factors (ai*x + u*f), the numbers
ai are NOT CONSTANTS. They are dependent on m.
>> He is saying that when m = 0, the only way you are going
>> to factor out f^2 from the expression
>> (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f)
>> is:
>> f^2*([a1(0)/f]*x + u)*([a2(0)/f]*x + u)*(a3(0)*x + u*f),
>> because a1(0) = a2(0) = 0 and a3(0) = 3:
>> f^2 * (u) * (u) * (3*x + u*f).
>> So far, just fine. This is all assuming m = 0.
>And consider that the poster Nora Baron is considering terms
which
>have isolated out as INDEPENDENT of m, which is why I set
m=0, in the
>first place!
What you are saying is that showing that a1 is
divisible by f when m = 0 is sufficient to show that
a1 is divisible by f when m <> 0. It is just plain
not true.
>ItÕs like this poster is going on and on about a rather
simple
>technique for pulling out factors independent of a particular
>variable, as if itÕs actually sinister, where *really* 
IÕm
just
trying
>to fool people with a special case, as if things change when
m does
>not equal 0.
Actually I donÕt think you are trying to fool people. You
canÕt help doing that because you have fooled yourself.
>But if these terms change as m changes, then theyÕre not
independent
>of m, now are they?
Precisely. a1, for example, is dependent on m. It might
be, for all I know, a1(m) = sqrt(m) + cuberoot(m). Note
that when m = 0, a1(m) = 0 + 0 = 0. Note that when m = 64,
a1(m) = 8 + 4 = 12. If f = 5, then f divides a1(0) = 0.
But that does NOT imply that f divides a1(64) = 12. In fact
5 and 12 are relatively prime.
>Nora Baron is trying to refute algebra!!! It should be a news
ßash
>around the math world that setting m=0, gives terms that are
STILL
>dependent on m,
No one claims that. What you are saying is just
irrelevant. DonÕt call CNN quite yet unless you want
to be embarrassed on an international scale.
What you claim is that if a function has a certain
property when the argument is 0, then it has that
same property for all other arguments. For example,
say f(m) = 3*m + 7. Note that f(0) is divisible by
7. By your logic, f(m) would be divisible by 7 for all
other values of m. But then you try m = 1, and
find that f(1) = 10 is NOT divisible by 7. Therefore
your logic was faulty. You were right about f(0),
but it doesnÕt generalize to other values of m.
> which I call the Shadow m. It has supernatural powers
> and refuses to go away, even when set to 0.
You are deeply confused. You are digging a
deeper hole for yourself.
>> There are infinitely many ways that f^2 can be written
>> as a product of three algebraic integers. Say, for
>> example, one of them is f^2 = f1 * f2 * f3. Assume that
>> none of f1, f2 or f3 is a unit in the algebraic integers.
>> Now choose an integer m <> 0. Suppose a1(m) is
>> divisible by, say, f1. Suppose that f/f1 is also an
>> algebraic integer. Then
>> a1(m)*x + u*f = f1*([a1(m)/f1]*x + u*f/f1).
>> And assume similar things for g2 and g3:
>> a2(m)*x + u*f = f2*([a2(m)/f2]*x + u*f/f2),
>> a3(m)*x + u*f = f3*([a2(m)/f3]*x + u*f/f3),
>> and all of a1(m)/f1, a2(m)/f2, a3(m)/f3, f/f1, f/f2, and
>> f/f3 are algebraic integers.
>> Putting all this together, one has
>> P(m)/f^2 = P(m)/(f1*f2*f3)
>> = ([a1(m)/f1]*x + u*f/f1) *
>> ([a2(m)/f2]*x + u*f/f2) *
>> ([a3(m)/f3]*x + u*f/f3).
>> The key thing here is that these factors f1, f2, and f3
>> may ALSO be dependent on m. When Harris has shown is that
>> when m = 0, f1 = f2 = f, and f3 = 1. Or you could say,
>> f1(0) = f2(0) = f and f3(0) = 1.
>Which actually blows you out of the ring of algebraic
integers, and
>worse provably, if the fÕs are *functions* of m, then f1, 
f2
and f3
>have zeroÕs,
Ridiculous. Are you now saying all functions have zeros???
> at which point the equation would blow up. Remember
> dividing by 0 is a no-no.
>BUT, remember this poster above has
>> Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
>> - 3*(-1 +m*f^2)*x*u^2 + u^3*f),
>Well then, how can P(m)/f^2 introduce such a problem as
blowing up
for
>some value that where f1, f2 or f3 is 0?
Again, are you claiming that all functions have zeros???
>Besides, if I find terms that are *independent* of m, then
they are
>independent of m, so the same terms can be isolated in each
factor of
>> = ([a1(m)/f1]*x + u*f/f1) *
>> ([a2(m)/f2]*x + u*f/f2) *
>> ([a3(m)/f3]*x + u*f/f3).
>If you do so, youÕll find that the terms 
independent of m,
for *two*
>factors, must have a factor that is f, while one does not.
>Here Nora Baron needs you to believe that setting m=0
doesnÕt in
>fact give you terms *independent* of m, but thatÕs just
mathematically
>wrong.
For the n-th time: I have NO ARGUMENT with your
statement that a1(m) = 0 when m = 0. That is JUST
FINE. The problem is that a1(m) for m <> 0 is
NOT *independent* of m, and you need it to be
divisible by f when m <> 0. And proving that fact
for m = 0 gets you nowhere whatsoever when what you
need is divisibility of a1(m) when m <> 0.
I am beginning to think you are just *hopelessly*
confused on this. It is complicated greatly
by the fact that you want oh-so-desperately not
to be wrong. You are blinding yourself to the
obvious as has happened many times before.
>But Nora Baron is working to convince you.
No, I am working to convince YOU. I donÕt think
there is a single other reader out there, other that
you, that is actually confused about this. I am
not playing to the peanut gallery. This is strictly
an attempt to get you to understand. So far it is
a ßop.
>WhatÕs key here is the claim of dependency on m, but
remember, I set
>m=0 to *remove* dependency on m,
Setting m = 0 in no way removes dependency on m.
All you get when you set m = 0 are facts about the
m = 0 case. But you donÕt even need facts about
the m = 0 case. You need facts about the m = 1
m = 2, ..., m = 211, etc. cases. The m = 0 case
tells you NOTHING about the cases you actually need
to consider.
> but now the Shadow m has apparently
> returned according to Nora Baron!!!
Right! m = 0 is all well and good, and
your statements about that case are correct. What you
need are facts about all the other possible values
of m. You have proved nothing on that.
>> I donÕt claim that I have shown here that f^2 can be
>> factored so that all of the above fits together. What I
>> claim here is that Harris has not shown that it CANÕT
>> happen. As long as that gap is left open, he does not
>> have a proof.
>ThereÕs no gap, as itÕs just Nora Baron now 
switching to
trying to
>show reasonable doubt, I guess.
>However, as I emphasized repeatedly through this post, the
technique
>of setting m=0, does actually work, as it shows factors
independent
of
>m, and allowing a poster like Nora Baron to question such
basic
>algebra, is proof that many of you care more about society
than math.
>After all, you *want* to believe her so you can believe IÕm
wrong.
ThereÕs no need to grandstand here. I am not trying to
convince anyone else; just you.
>ThatÕs a loser proposition, as itÕs math. If 
any of you had
sense
>youÕd realize that believing some person just because it
makes you
>feel good is NOT math.
>> To put it another way: Harris has not shown that f1(m) =
>> f2(m) = f and f3(m) = 1 in general. He has shown it only
>> when m = 0. Even though f1*f2*f3 = f^2, a constant with
>> respect to m, it is entirely possible that f1, f2, and f3
are
>> *dependent* on m. Why is this so hard to understand?
>Now you see that word dependent again, when IÕve used m=0 
to
find
>terms that are *independent*, which shows you immediately
that Nora
>Baron is full of it.
LetÕs try again. We agree that a1 is a function of m.
Therefore a1*x + u*f is also a function of m. Suppose
that a1(m) = arctan(m) * 37. Note that a1(0) = 0.
Therefore a1(0) is divisible by f, no matter what f is.
Now consider a1(1). How do you know it too is divisible
by f ? Does knowing that a1(0) = 0 tell you anything
about this ??? HOW ? Is arctan(1) * 37 divisible by
f ? How do you know?
>ItÕs simply not sane for Nora Baron to repeatedly argue 
that
terms
>independent of m, are in fact dependent on m, as itÕs a
rather wacky
>contradiction.
>I keep thinking people will laugh at such an argument, but
you donÕt.
>You keep sitting quietly, like here where this poster Nora
Baron
>just argued that terms independent of m--found by setting
m=0--might
>in fact be dependent on m, which isnÕt mathematics at all.
>Fascinating.
>> There are examples which show that no such proof for
>> m <> 0 can be obtained. For example, let f = 5, m = 1,
>> and u = 1. Then
>> P(m) = 25*(553 * x^3 - 72 * x + 5),
>> or
>> P(m)/25 = 553*x^3 - 72*x + 5.
>> Now suppose Harris is right. Then the factorization of
>> P(m)/f^2 must take the form
> P(m)/25 = ([a1/5]*x + 1)*([a2/5]*x + 1)*([a3/5]*x + 5),
>> where a1/5 and a2/5 are *algebraic integers*.
>> Let b1 = a1/5. Then -1/b1 is a root of
>> 553*x^3 - 72*x + 5 = 0. That is,
>> -553/b1^3 + 72/b1 + 5 = 0.
>> Multiplying through by b1^3, one obtains:
>> 5*b1^3 + 72*b1^2 - 553 = 0.
>> The expression on the left is a *non-monic* polynomial in
>> b1 with integer coefficients, and it is *irreducible* over
>> the rationals.
>> Therefore b1 cannot be an algebraic integer [*].
>> That is, a1/5 cannot be an algebraic integer.
>> Therefore when m = 1, a1 = a1(m) is not divisible by f = 5
>> in the algebraic integers.
>> Therefore HarrisÕs claim is false.
>A hallmark of an error in core is the ability for *two*
different
>sides to claim proof. Here the problem is a definition error
in
>core, as the definition of algebraic integers as roots of
*monic*
>polynomials excludes numbers that should be included, though
roots of
>non-monics.
The definition is not defective. It clearly 
defines
a set of numbers.
You say some other numbers should be included. I believe
what you are saying is that you think there are algebraic
integers A and B such that A * B is not an algebraic integer.
That would not contradict the DEFINITION of algebraic integer.
That would instead contradict a THEOREM:
THEOREM: The set of algebraic integers is closed under
multiplication.
You should be asking: is this really a theorem? Can you
prove it? Have you seen a proof of it? Has anyone given a
proof of it here? Does it look to you like it is obviously
true? Was it known to Gauss or Dedekind?
It is part of what you need to conclude that the SET of
algebraic integers actually forms a RING. When you say that
th ealgebraic integers are not complete, I think you must
really be saying that they do not form a ring: they are
not closed under multiplication. Right?
It is not the DEFINITION that is a problem. The set
of numbers which are roots of monic polynomials with
integer coeffcients is clearly a well-defined SET. The
question you raise appears to be: is that SET a RING ?
>You saw the proof above, which involves basic algebra, and
you saw
the
>position of Nora Baron which succincly is that terms
independent of
>m, might be dependent on m, which is a direct contradiction.
HarrisÕs argument boils down to this: if a fact
concerning a function h(m) is true when m = 0, it
must be true for all other m, because h(0) is
independent of m.
Example: Let h(m) = 13*m + 11.
Note that h(0) is divisible by 11.
Now, (saith Harris) since h(0) is independent of
m, it must be true that h(m) is divisible by 11 for
all other values of m.
Unfortunately, h(1) = 24, which is not divisible
by 11.
So there must be SOMETHING WRONG with the logic
here. What could it be ???
>> Thus what we have here are two things:
>> (1) An example which shows that HarrisÕs claim
>> *cannot* be true, and
>The example shows how bad an error in core can be as it
allows people
>to supposedly prove things not true.
>Here you can figure out which side is right because my
algebra is
>simpler, so you look at what youÕd have to throw out to
believe Nora
>Baron and itÕs basic thin gs like that setting m=0 with 
P(m)
gives
>terms independent of m, which throws out *axioms* while IÕm
showing a
>problem with a DEFINITION.
See above. At length.
>So trust Nora Baron and challenge axioms, or trust algebra,
and
>modify the definition.
Better yet: forget about trusting Nora Baron. Read
the algebra and logic for yourself. Come to your own
conclusions.
>> (2) An examination of his argument which points
>> to exactly where he has a gap.
>That claim depends on the assertion that terms independent
of m might
>be dependent on m, which is a direct contradiction.
For the (N + 1)st time: yes, P(0) is independent of m.
I grant that it does not change when m changes. The
right question to ask next is: SO WHAT ? What you
have to consider is the factorization of P(m) for
*other* values of m. A typical factor in the
factorization is a1*x + u*f, where a1 is an algebraic
integer. u and f are integers also. The value
of the polynomial P(m) changes when m changes.
Therefore the factorization is not constant either;
the factorization changes when m changes. Of course
u and f are constant with respect to m, and x is a
variable independent of m. Therefore the only thing
that can change when m changes is a1. That is,
a1 = a1(m) is dependent on m. Whatever value is
has when m = 0, it has a different value when m = 1
or m = 2, etc.. When m = 0, it is divisible by f.
We agree on that. When m = 1, who knows? Its value
when m = 0 does not determine its value or divisibility
when m <> 0.
You just keep doing the same thing, disguising it
with different verbiage. You think that since a1(0)
is divisible by f, the same must be true for a1(m)
in general. This stuff about a1(0) being independent
of m, or your having isolated the constant term -
itÕs meaningless and irrelevant. You have shown
N O C O N N E C T I O N between a1(0) and
a1(m) for m <> 0.
>> The example is complete and straightforward. There are
>> no hidden tricks or bogus math statements. There is no
>> Galois theory [although there is a separate argument using
>> Galois theory which shows the same thing]. There is one
>> theorem [*] from elementary algebraic number theory, but
this
>> theorem has been accepted as valid by Harris.
>> James HarrisÕs argument, by contrast, has a gap, a
mysterious
>> unjustified step, and in the version quoted above, he
>> doesnÕt even state what that step is. He is claiming that
>> because a1(0) = 0 is divisible by f, it follows that a1(m)
>> is divisible by f for m <> 0. He says:
>> Remember, the constant terms with respect to m
>> cannot vary as m varies, or they wouldnÕt be
>> constant terms, right?
>> Of course this is true. It is a tautology. Constant
>> terms donÕt vary.
>So why is this poster trying to claim that *maybe* they do?
See above. Are you REALLY this dense ???
>> It can be a good idea, when you are losing an argument, to
>> state something which is obviously true, even though it has
>> nothing to do with your desired conclusion. It confuses
>> your attackers - all they can say is, well, yeah, youÕre
>> right about that.
>> What they *should* say is, SO WHAT?
>> Here, since a1(m) is not a constant function, why does
>> HarrisÕs statement tell you anything about divisibility 
of
>> a1(m) when m <> 0 ? That step is simply not there in
>> HarrisÕs argument. It is a gap. The counterexample shows
>> that it is more than a gap. It is an error which cannot be
>> fixed. The quote above buys you nothing. ItÕs 
not a
>> question about the constant term a1(0). ItÕs a question
>> about a1(m).
>No matter how many distractions the poster throws out at
you, the
>simple fact is that the constant term is independent of m, I
look at
>changes in the constant term, as that factor of f^2 goes
away, when
>you have P(m)/f^2, so thereÕs no *rational* way to argue
dependency
on
ItÕs staring you right in the face! P(m) / f^2 is clearly,
obviously DEPENDENT ON m !!!
>> And, I think, there is another deeper undercurrent here.
>> When you factor a polynomial Q(x) into subpolynomials,
>> Q(x) = q1(x) * q2(x) * ... * qk(x),
>> the coefficients of q1, q2, ..., qk are independent of the
>> polynomial variable x. They are constants. Harris is
>> factoring the polynomial P(m) into 3 factors, but they are
>> (as he has reminded us many times) nonpolynomial factors.
>> What we expect with polynomial factors is not necessarily
>> true in HarrisÕs factorization: the 
coefficients a1(m),
>> a2(m), and a3(m) *cannot* be constant with respect to m.
>> {If they were, then a1(m) would be a1(0) = 0 for all m, and
>> we know that is false.} HarrisÕs argument would be in
better
>> shape if they were. But that is, I think, part of his
>> imprecise sub-rosa thinking, and it has led him into an
>> error which he clings to like a drowning man clings to a
straw.
>Notice how much Nora Baron has posted here.
Yes. I thought if I explained it s-l-o-w-l-y
and carefully, you might get it. I was wrong.
>ItÕs annoying to deal with such a poster.
>> Some of you are now facing the reality of the human brain
versus
any
>> fantasy you might have had about being completely rational.
Human
>> beings are NOT rational creatures but necessarily rely on
social
>> forces to determine what they believe.
>>
>> You wish.
>Well look at you Nora Baron here you are arguing that
variables
>independent of m, might be dependent on m, which is a direct
>contradiction, so itÕs an irrational position.
>> You are creatures of society.
>Nora Baron clearly is a member of math society who finds
trouble
>with the idea that thereÕs a hundred year old 
definition
error in
core
>mathematics, so whatÕs the posterÕs solution? 
Throw out basic
>algebra.
No. HereÕs what you need to do. Stop fooling yourself
that whatÕs true when m = 0 is necessarily also true
for all other m (with no real proof - just confusing
statements about independence of the constant term,
etc). USE basic algebra. Next, stop pretending that
there is a problem in core, whatever core is. The
problem is in your own logic. If you had actually
uncovered a basic problem here, it would not be a problem
with a definition. It would be a problem involving
closure of a set under multiplication. ThatÕs moot
anyway, since your logic to get to that point is wrong.
>Yup, This poster canÕt handle the truth.
>> You may have believed that your mathematical knowledge was
based
>> completely on logic and rationality, but human beings
donÕt work
that
>> way; itÕs built-in to your wiring NOT to work that way.
>>
>> *Your* wiring, maybe.
>Your wiring Nora Baron allows you to believe something that
directly
>contradicts itself, which is what the belief that terms
independent
of
>m might in fact be dependent on m, which I think indicates a
>limitation of that wiring.
Nonsense. See above.
>Human beings can rather easily accept contradictions on that
level
for
>social reasons.
Now let the pontificating begin.
>After all, people feed you and clothe you. It makes sense to
care
>more about what society thinks, or what you think society
thinks than
>something abstract like mathematical truth.
>You canÕt eat a math proof.
>> Some of you must learn to be more than human.
>>
>> You must learn to be truly rational, for the first times in
your
>> lives.
>>
>> So itÕs about time, as I wait, and wonder, how many of 
you
can
handle
>> the truth.
>>
>> This is such patronizing drivel. I would think you would
get
>> sick of writing it.
>Notice the social forces at work readers.
The remark was directed solely at YOU, not at some
imaginary group of readers.
>Human beings *need* social approval of truth before
accepting it.
>> And how many of you prefer the fantasy which was the world
you
>> believed in, which actually never existed, except in your
>> imaginations; your wishes for a nicer world, where your
wishes
matter.
>>
>> Arturo also had some speculation about where your thinking
>> went wrong. You wonÕt agree with him either.
>The argument at
><a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://mathd
b.math.cuhk.edu.hk/forum/e_show.php?msg=782</a>is simple
enough that thereÕs no room for error.
Nothing is so simple that it does not leave
room for error. There is always infinite
room, though possibly a new record is being set
in this case.
Nora B.
>James Harris
===
Subject: Re: JSH: About time
>> Now that IÕve revealed the odd and you could say esoteric
error
> in
>> core mathematics with such a short, and rather simple
argument,
> the
>> issue now is how long until mathematicians decide that
theyÕd
> rather
>> have correct mathematics versus the *belief* that they had
been
>> perfect in keeping error out of the collected body of work
that
> is
>> called mathematics.
>>
>> My work is out there and rather easy to go over as can be
seen at
> the
>> Hong Konk math site:
>>
>>
>> Hong Konk? YouÕre sure it isnÕt Honk Honk?
>Typo. It should be Hong Kong.
>
>> See <a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://mathd
b.math.cuhk.edu.hk/forum/e_show.php?msg=782</a>>
>> And I send people there because their allowal of the use
of LaTeX
>> makes for a *much* better presentation, and given the
*social*
> issues
>> IÕm facing, I need all the help I can get.
>>
>>
>> This gives me the chance to re-post a discussion of the
LaTeX
>> website you mention above - maybe you would like to
respond -
>Sure.
>>
>> James Harris claims to prove, in
>>
>> <a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://mathd
b.math.cuhk.edu.hk/forum/e_show.php?msg=782</a>>
>> that certain polynomials factor in a form which
>> contradicts other mathematical proofs. Specifically:
>>
>>
>> Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
>> - 3*(-1 +m*f^2)*x*u^2 + u^3*f),
>>
>> where f is a prime, u is an integer coprime to f,
>> and m is an integer.
>>
>> Assume P(m) is factored in the form
>>
>> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
>>
>> where a1, a2, and a3 are algebraic integers.
>>
>> Let g1 = (a1*x + u*f), g2 = (a2*x + u*f), g3 = (a3*x +
u*f).
>>
>> Note that P(0) = f^2*(3*x*u^2 + u^3*f).
>>
>>
>> Harris says:
>>
>> ... two of the aÕs go to 0 when m = 0 which
>> is also seen from the cubic defining the 
aÕs.
>>
>> Then arbitrarily picking a1 and a2 as the ones
>> that go to 0 at m = 0, you have
>>
>> g1 = u*f, g2 = u*f, g3 = 3*x + u*f.
>>
>> But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
>> only two of the gÕs while with the third it is
>> blocked, as long as it [f] is coprime to 3 and
>> x, so assume it is, and assume as well that f
>> is coprime to u.
>>
>> Then it follows from the constant terms that g1
>> and g2 each have a factor that is f.
>>
>> Remember, the constant terms with respect to m
>> cannot vary as m varies, or they wouldnÕt be
>> constant terms, right?
>>
>>
>> Sounds like it makes sense, including that last bit,
>> doesnÕt it? P(0) is the evaluation of the polynomial
>> when m = 0, so it must be the constant term.
>It does make sense. Basically I isolate terms of P(m) that
are
>*independent* of m.
> By which you mean P(0).
Yes, as setting m=0 isolates terms that are *independent* of
m.
HereÕs whatÕs key, all in a row:
P(m) = g_1 g_2 g_3
P(0) = u^2 f^2 (3x + uf)
at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
P(0)/f^2 = u^2 (3x + uf)
and f is coprime to 3, x and u.
I focus on whatÕs independent of m for a reason.
ItÕs not really a subtle technique.
>Then I look at at the same terms for P(m)/f^2 and find that a
factor
>of f^2 has been removed.
>Logic dictates that the removal is *independent* of m,
> Removal here means that you divide P(m) by f^2. That does
> not cause a problem in itself. It is really how f^2 is
> distributed among the factors
> ai*x + u*f
> that is the issue.
Actually, what I do is note factors *independent* of m, so
they canÕt
suddenly gain a dependency on m, while this poster seems
fascinated
with the aÕs being themselves functions of m, as if that
destroys the
general principle.
At best itÕd mean you can also separate them out as well,
using the
same trick, which is to note their values when m=0, and it
turns out
that for only *two* of the aÕs, when m=0, they are 0 as 
well,
which
tells you something there.
Now then, the uf simply cannot lose an f as a function of m or
dependent on m, as itÕs *independent* of m, so the logic is
simple.
So this poster keeps making posts trying to refute the fact
that uf is
independent of m, as if somehow thereÕs someway that math is
illogical, but itÕs not.
> You do not disagree that ai is dependent
> on m. LetÕs write it as ai(m).
Yup, itÕs dependent on m, but notice it *still* is the case
that you
can isolate whatÕs independent of m by setting m=0, though
with the
aÕs, two of them then become 0, while one is 3x + uf, which
youÕll
notice is *independent* of m.
The principle is simple. WhatÕs fascinating is any person
trying to
argue against it, as this Nora Baron is doing.
> Therefore the way in which factors of f divide
> ai(m) in general can be expected to depend on m.
Your expectation--your intuition--is meaningless in the face
of
mathematical logic.
Your gut feeling is trash if it goes against the math.
Here the key term is uf, which is *independent* of m, so it
canÕt vary
on m.
However, this poster *wants* it to vary on m, so the poster
refuses to
accept the math and notice, now talks about whatÕs expected.
ItÕs a rather pathetic display which highlights what 
IÕve
said--human
beings are NOT rational, as they rely strongly on social
forces, even
with mathematics.
> You prove that for m = 0, f divides ai(0).
I isolate the terms independent on m, deliberately, so that I
can see
how they vary, without worrying about mÕs value, and it just
so
happens that the way to find those terms independent of m, is
to set
m=0, which is not rocket science.
> You think somehow that that proves f divides ai(m)
> for all m.
YouÕre showing a gross lack of logic as in fact I simply 
rely
on the
terms being independent of m, and the fact that P(0)/f^2
equals u^2(3x
+ uf) so if f is coprime to 3, x, and u, it must be the case
that the
factors that were g_1, g_2, and g_3 now no longer have f as a
factor
either.
Now since at m=0, remember thatÕs to get terms independent 
of
m,
g_1=uf, g_2=uf, but g_3 = 3x + uf, it stands to reason that f
divides
off of g_1 and g_2, based on the terms *independent* of m, so
the
value of m is not of consequence.
ThatÕs the point of isolating terms independent of m, so 
that
I now
they are independent of its value, which means that you 
canÕt
claim
that things change just because of mÕs value.
Clearly, logically, two of the gÕs have f as a factor,
without regard
to mÕs value, when f is coprime to 3, x, and u.
> LetÕs say f = 5 and ai(m) = sqrt(5) * m + 15.
> Certainly f = 5 divides ai(0) = 15.
> But f = 5 does not divide ai(1) = sqrt(5) + 15.
Which is why I focus on the terms *independent* of m.
> Here is another example. Say ai(m) = 5 * m, and f = 3.
> The ai(0) = 0, and, as in your case, f divides ai(0).
> However, if m = 1, ai(m) = 5, and f = 3 does not divide
> a1(m) = 5. In fact f = 3 is relatively prime to 5,
> even in the algebraic integers.
However, what I *actually* do is focus on those terms
INDEPENDENT of
m, so that I donÕt care what mÕs value is, and 
in fact it is
irrelevant, which is the point.
The technique for isolating those terms is to set m=0.
HereÕs whatÕs key, all in a row:
P(m) = g_1 g_2 g_3
P(0) = u^2 f^2 (3x + uf)
at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
P(0)/f^2 = u^2 (3x + uf)
and f is coprime to 3, x and u.
I focus on whatÕs independent of m for a reason.
ItÕs not really a subtle technique.
> I am not saying that in your application, ai(m)
> actually equals 5 * m. I am simply saying that your logic
> breaks down. You have not shown that ai(m), as a
> function of m, does not behave something like
> ai(m) = 5 * m. You have nothing explicit whatsoever
> about ai(m), EXCEPT when m = 0. But that clearly does
> not tell you anything about how ai(m) behaves with
> respect to f when m = 0. ThatÕs the point.
> Put it another way. You have shown that ai(0)
> is divisible by f. Since ai(m) is a function of
> m which you probably cannot even write down,
> you do not know anything about the divisibility
> of ai(m) by f when m <> 0. It has NOTHING to do
> with whether ai(0) or P(0) or whatever is
> independent of ai(m), or your claim that you
> have somehow removed the constant term. You
> are badly confused on that point.
ThatÕs against mathematical logic as if the terms that are
independent
of m, are indeed independent of m, then how can their factors
vary
with m?
Here you have u^2 f^2 (3x + uf) in one case, and u^2 (3x +
uf) as the
other case.
The are *independent* of m, and that remains true no matter
how much
you argue Nora Baron.
Now then, given that at m=0, those terms isolated out as
independent
of m, are for g_1 and g_2, uf, and for g_3, 3x + uf, logic
dictates
how the factors MUST distribute out.
Here the factors independent of m force a constraint on those
that are
dependent on m, in a way thatÕs fascinating and
understandable using
basic algebra.
ItÕs a neat trick.
>but that is a
>conclusion several posters wish to convince others is false,
so here
> I
>am, once again, arguing about the obvious.
>
> WhatÕs obvious here is that you have leaped from
> m = 0 to all other values of m with no justification.
Can you not understand that if terms are *independent* of m
then I can
leap to any values for m that I choose?
I can have m=2345403840, and it does NOT matter as those
terms are
independent of m.
So the justification is the independence from m.
That independence follows from the simple act of setting m=0
to
isolate out terms NOT dependent on m.
>> Of course all of what I quoted above except that last
>> paragraph pertains to what happens when m = 0. For
>> instance, saying that a1 = a2 = 0, that is true only when
>> m = 0. When m is nonzero, we know that a1 and a2 are also
>> nonzero.
>>Readers need focus on the simple fact that with P(m)
considering
> P(0),
>that is setting m=0, gives you terms that donÕt have m, so
they are
>independent of it.
>Stay focused on that fact as Nora Baron tries to take you
for a
>ride.
>> So itÕs clear that a1 and a2 depend on m.
>>
>> I donÕt think even James Harris disagrees with that.
>>
>> So when he says g1 = u*f and g2 = u*f, that is true only
>> when m = 0. When m is not zero, you get, for example,
>>
>> g1 = a1*x + u*f.
>>
>> When Harris says constant term he does not mean
>> constant term with respect to x. He is dealing with what
>> he calls a nonpolynomial factorization. The variable here
>> not as P(x).
>>
>> As noted above, a1 and a2 are dependent on m. Both a1
>> and a2 should really be written as a1(m) and a2(m).
>>
>> A key step in HarrisÕs argument is this :
>>
>> But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
>> only two of the gÕs while with the third it is
>> blocked
>>
>> When Harris says blocked he means that f is not
>> a divisor of the third g.
>Readers remember, with P(m), I set m=0, that is, look at
P(0), to get
>terms *independent* of m, as m has been set to 0, so itÕs
gone.
> Looking at m = 0 does not mean, in any sense that
> m is gone.
I can *give* whatÕs left as it is
P(0) = u^2 f^2 (3x + uf)
and yes, whether you wish to accept it or not, m IS gone.
>Then I do the same for P(m)/f^2,
> No, you absolutely donÕt. You never really deal
> with P(m).
Now I can just give P(0)/f^2.
P(0)/f^2 = u^2 (3x + uf)
and again, thereÕs no m, and now a factor of f^2 has gone
away.
> When m = 0, a1*x + u*f = u*f. This is divisible
> by f because a1 = 0 also.
Yes. But because IÕve isolated out terms independent of m,
and follow
them, they force a constraint on those terms dependent on m
*because*
they are independent of m.
> When m <> 0, a1 is not equal to 0. You donÕt know
> its value. In particular, you donÕt know that it
> is divisible by f. Just SAYING that it is is not
> sufficient.
> and notice that for those terms
>*independent* of m, a factor of f^2 has disappeared.
>Now you can see that for P(0) I have
>P(0) = u^2 f^2( 3x + uf)
>and for P(0)/f^2 I have
>P(0)/f^2 = u^2 ( 3x + uf)
>so everything is simple enough.
> Yes: For m = 0. We agree on this. LetÕs
> try to get past it. Divisibility of a1 by f
> when m equals 0 does not tell you anything about
> divisibility of a1 when m <> 0. a1 is dependent
> on m in a very complicated way. You have never
> tried writing a1 out as a function of m. It is
> however fairly easy to show that for m = 1 and
> f = 5, a1 CANNOT be divisible by f.
ItÕs possible to show because thereÕs a 
definition error in
core
mathematics, which allows for the *appearance* of two dueling
proofs.
But proofs donÕt duel.
The fix is to ignore the ßawed definition and 
rely on basic
axioms,
which reveal the truth.
Here IÕve pointed out repeatedly that the proof depends on
focusing on
terms independent of m, and those terms are found easily
enough by
setting m=0, with P(m), P(m)/f^2 and the factors of P(m),
g_1, g_2 and
g_3.
>> Of course here he is obviously talking about m = 0,
>> though he would like to conclude this for m <> 0.
>HereÕs where you have Nora Baron sneaking in something
rather dumb
>mathematically.
> I am not sneaking anything in anywhere. Everything
> is above board. What you see is what you get.
ItÕs dumb mathematically to insinuate that something
different will
happen for terms *independent* of m based on the value of m,
as if it
matter whether or not m=0 or not.
>If my point is isolating off m, so that I have terms
*independent* of
>m, why canÕt I conclude that they are indeed independent of
m?
> Look. NO ONE is arguing here about P(0). That is not
> at issue. We are arguing about P(m) when m <> 0.
> The fact that P(0) does not change when m changes
> does not imply that the factors of the form
> ai*x + u*f
> also do not change when m changes. In fact we KNOW
> that ai must be a factor of m ! Why? Because when
> m = 0, ai = 0. But when m <> 0, we know *for sure*
> that ai is not zero [otherwise the cubic would have
> degree 1]. Thus ai DEPENDS ON m. Therefore you cannot
> assume that something about it which is true when m
> equals 0 is also true when m <> 0.
Yes, the aÕs *are* dependent on m, but the terms that are 
NOT
force a
constraint on those that are because they arenÕt.
You have to follow *logic* which dicates that terms
independent of m
are indeed independent of m.
>ItÕs a bizarre thing to question and maybe many of you
simply canÕt
>believe that a poster like Nora Baron would keep posting and
making
>a fuss based on such a position.
>But remember, I *isolate* independent factors of P(m), by
setting
> m=0,
>so I donÕt have to worry about what value m has.
> What you say about P(0) applies only when m = 0.
> You are just confusing yourself with verbiage here
> like isolate and independent factors. As noted
> above, in the factors (ai*x + u*f), the numbers
> ai are NOT CONSTANTS. They are dependent on m.
However, setting m=0 necessarily gives you whatÕs NOT
dependent on m.
When you focus on whatÕs not dependent on mÕs 
value then you
are
forced into a conclusion where two of the gÕs have f as a
factor, when
f is coprime to 3, x and u, while one does not.
The math is rather simple and basic, if you will accept that
whatÕs
NOT dependent on the value of m, is in fact, not dependent on
the
value of m.
>> He is saying that when m = 0, the only way you are going
>> to factor out f^2 from the expression
>>
>> (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f)
>>
>> is:
>>
>> f^2*([a1(0)/f]*x + u)*([a2(0)/f]*x + u)*(a3(0)*x + u*f),
>>
>> because a1(0) = a2(0) = 0 and a3(0) = 3:
>>
>> f^2 * (u) * (u) * (3*x + u*f).
>>
>> So far, just fine. This is all assuming m = 0.
>And consider that the poster Nora Baron is considering terms
which
>have isolated out as INDEPENDENT of m, which is why I set
m=0, in the
>first place!
> What you are saying is that showing that a1 is
> divisible by f when m = 0 is sufficient to show that
> a1 is divisible by f when m <> 0. It is just plain
> not true.
No IÕm not saying that as what IÕm saying is 
that focusing on
those
terms that are independent on m, gives a conclusion which
forces
itself back upon those that are.
It does so because the terms independent of m are independent
of m.
Is it really so subtle of a point?
>ItÕs like this poster is going on and on about a rather
simple
>technique for pulling out factors independent of a particular
>variable, as if itÕs actually sinister, where *really* 
IÕm
just
> trying
>to fool people with a special case, as if things change when
m does
>not equal 0.
> Actually I donÕt think you are trying to fool people. You
> canÕt help doing that because you have fooled yourself.
How can I have fooled myself when the basic principle is that
whatÕs
independent of m is independent of m?
You, however, refuse to settle down but instead make VERY
long posts
where you keep dancing around that simple truth.
>But if these terms change as m changes, then theyÕre not
independent
>of m, now are they?
> Precisely. a1, for example, is dependent on m. It might
> be, for all I know, a1(m) = sqrt(m) + cuberoot(m). Note
> that when m = 0, a1(m) = 0 + 0 = 0. Note that when m = 64,
> a1(m) = 8 + 4 = 12. If f = 5, then f divides a1(0) = 0.
> But that does NOT imply that f divides a1(64) = 12. In fact
> 5 and 12 are relatively prime.
Your reasoning is specious as itÕs nothing like what I
actually do,
which involves considering the constant terms P(0) and
P(0)/f^2.
I note for readers that this poster clearly has little
intention of
actually being rational, but may simply believe that
continuing to
disagree, even after soundly being refuted is all thatÕs
necessary to
keep many of you convinced that IÕm wrong.
>Nora Baron is trying to refute algebra!!! It should be a news
> ßash
>around the math world that setting m=0, gives terms that are
STILL
>dependent on m,
> No one claims that. What you are saying is just
> irrelevant. DonÕt call CNN quite yet unless you want
> to be embarrassed on an international scale.
> What you claim is that if a function has a certain
> property when the argument is 0, then it has that
> same property for all other arguments. For example,
> say f(m) = 3*m + 7. Note that f(0) is divisible by
> 7. By your logic, f(m) would be divisible by 7 for all
> other values of m. But then you try m = 1, and
> find that f(1) = 10 is NOT divisible by 7. Therefore
> your logic was faulty. You were right about f(0),
> but it doesnÕt generalize to other values of m.
Your example is specious as I focus on whatÕs independent of
m, using
P(0) AND P(0)/f^2 as well as the factors g_1, g_2 and g_3.
ALL of that information, along with the requirement that f is
coprime
to 3, x and u is necessary for the conclusion.
Here this poster keeps acting as if factors independent of m
are still
somehow dependent on m, which is just not true.
> which I call the Shadow m. It has supernatural powers
> and refuses to go away, even when set to 0.
> You are deeply confused. You are digging a
> deeper hole for yourself.
How?
>> There are infinitely many ways that f^2 can be written
>> as a product of three algebraic integers. Say, for
>> example, one of them is f^2 = f1 * f2 * f3. Assume that
>> none of f1, f2 or f3 is a unit in the algebraic integers.
>>
>> Now choose an integer m <> 0. Suppose a1(m) is
>> divisible by, say, f1. Suppose that f/f1 is also an
>> algebraic integer. Then
>>
>> a1(m)*x + u*f = f1*([a1(m)/f1]*x + u*f/f1).
>>
>> And assume similar things for g2 and g3:
>>
>> a2(m)*x + u*f = f2*([a2(m)/f2]*x + u*f/f2),
>>
>> a3(m)*x + u*f = f3*([a2(m)/f3]*x + u*f/f3),
>>
>> and all of a1(m)/f1, a2(m)/f2, a3(m)/f3, f/f1, f/f2, and
>> f/f3 are algebraic integers.
>>
>> Putting all this together, one has
>>
>> P(m)/f^2 = P(m)/(f1*f2*f3)
>>
>> = ([a1(m)/f1]*x + u*f/f1) *
>> ([a2(m)/f2]*x + u*f/f2) *
>> ([a3(m)/f3]*x + u*f/f3).
>>
>> The key thing here is that these factors f1, f2, and f3
>> may ALSO be dependent on m. When Harris has shown is that
>> when m = 0, f1 = f2 = f, and f3 = 1. Or you could say,
>> f1(0) = f2(0) = f and f3(0) = 1.
>Which actually blows you out of the ring of algebraic
integers, and
>worse provably, if the fÕs are *functions* of m, then f1, 
f2
and f3
>have zeroÕs,
> Ridiculous. Are you now saying all functions have zeros???
Hmmm...how much math do you actually know Nora Baron?
> at which point the equation would blow up. Remember
> dividing by 0 is a no-no.
>BUT, remember this poster above has
>> Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
>> - 3*(-1 +m*f^2)*x*u^2 + u^3*f),
>Well then, how can P(m)/f^2 introduce such a problem as
blowing up
> for
>some value that where f1, f2 or f3 is 0?
> Again, are you claiming that all functions have zeros???
Answer my questions first, like, whatÕs your 
area of expertise?
And where were you trained as a mathematician?
>Besides, if I find terms that are *independent* of m, then
they are
>independent of m, so the same terms can be isolated in each
factor of
>> = ([a1(m)/f1]*x + u*f/f1) *
>> ([a2(m)/f2]*x + u*f/f2) *
>> ([a3(m)/f3]*x + u*f/f3).
>>If you do so, youÕll find that the terms 
independent of m,
for *two*
>factors, must have a factor that is f, while one does not.
>Here Nora Baron needs you to believe that setting m=0
doesnÕt in
>fact give you terms *independent* of m, but thatÕs just
> mathematically
>wrong.
> For the n-th time: I have NO ARGUMENT with your
> statement that a1(m) = 0 when m = 0. That is JUST
> FINE. The problem is that a1(m) for m <> 0 is
> NOT *independent* of m, and you need it to be
> divisible by f when m <> 0. And proving that fact
> for m = 0 gets you nowhere whatsoever when what you
> need is divisibility of a1(m) when m <> 0.
But I focus on whatÕs independent of m to draw my 
conclusion,
while
you keep trying to claim a dependency on m, as if itÕs
somehow beyond
your capacity to understand that setting m=0 is a technique
for
pulling out whatÕs not dependent on m.
I find your continuing argument odd, and I think 
itÕs time you
revealed your mathematical training.
It occurs to me that you may have none.
> I am beginning to think you are just *hopelessly*
> confused on this. It is complicated greatly
> by the fact that you want oh-so-desperately not
> to be wrong. You are blinding yourself to the
> obvious as has happened many times before.
WhatÕs your training?
>But Nora Baron is working to convince you.
> No, I am working to convince YOU. I donÕt think
> there is a single other reader out there, other that
> you, that is actually confused about this. I am
> not playing to the peanut gallery. This is strictly
> an attempt to get you to understand. So far it is
> a ßop.
YouÕre posting irrationally, and IÕm not that 
curious about
why, but
then again, it might help if you give your training in math.
>WhatÕs key here is the claim of dependency on m, but
remember, I set
>m=0 to *remove* dependency on m,
> Setting m = 0 in no way removes dependency on m.
> All you get when you set m = 0 are facts about the
> m = 0 case. But you donÕt even need facts about
> the m = 0 case. You need facts about the m = 1
> m = 2, ..., m = 211, etc. cases. The m = 0 case
> tells you NOTHING about the cases you actually need
> to consider.
ThatÕs not rational as rationally setting m=0 with P(m) and
its
factors gives me whatÕs not dependent on m.
> but now the Shadow m has apparently
> returned according to Nora Baron!!!
> Right! m = 0 is all well and good, and
> your statements about that case are correct. What you
> need are facts about all the other possible values
> of m. You have proved nothing on that.
But, how can that be possible if I focus on whatÕs
independent of m?
If I focus on whatÕs independent of m, then 
isnÕt it
independent of m?
HereÕs whatÕs key, all in a row:
P(m) = g_1 g_2 g_3
P(0) = u^2 f^2 (3x + uf)
at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
P(0)/f^2 = u^2 (3x + uf)
and f is coprime to 3, x and u.
I focus on whatÕs independent of m for a reason.
ItÕs not really a subtle technique.
>> I donÕt claim that I have shown here that f^2 can be
>> factored so that all of the above fits together. What I
>> claim here is that Harris has not shown that it CANÕT
>> happen. As long as that gap is left open, he does not
>> have a proof.
>ThereÕs no gap, as itÕs just Nora Baron now 
switching to
trying to
>show reasonable doubt, I guess.
>However, as I emphasized repeatedly through this post, the
technique
>of setting m=0, does actually work, as it shows factors
independent
> of
>m, and allowing a poster like Nora Baron to question such
basic
>algebra, is proof that many of you care more about society
than math.
>After all, you *want* to believe her so you can believe IÕm
wrong.
> ThereÕs no need to grandstand here. I am not trying to
> convince anyone else; just you.
<remainder deleted because of intriguing assertion>
Then email me instead of posting.
I dare you.
James Harris
===
Subject: Re: JSH: About time
>> Now that IÕve revealed the odd and you could say esoteric
error
> in
>> core mathematics with such a short, and rather simple
argument,
> the
>> issue now is how long until mathematicians decide that
theyÕd
> rather
>> have correct mathematics versus the *belief* that they had
been
>> perfect in keeping error out of the collected body of work
that
> is
>> called mathematics.
>>
I posted a reply to this message earlier, but later I noticed
something in your post that shed some light on your thinking.
So I am replying to that explicitly.
>
> By which you mean P(0).
> Yes, as setting m=0 isolates terms that are *independent*
of m.
> HereÕs whatÕs key, all in a row:
> P(m) = g_1 g_2 g_3
> P(0) = u^2 f^2 (3x + uf)
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
> P(0)/f^2 = u^2 (3x + uf)
> and f is coprime to 3, x and u.
> I focus on whatÕs independent of m for a reason.
> ItÕs not really a subtle technique.
>Then I look at at the same terms for P(m)/f^2 and find that a
factor
>of f^2 has been removed.
>Logic dictates that the removal is *independent* of m,
>
> Removal here means that you divide P(m) by f^2. That does
> not cause a problem in itself. It is really how f^2 is
> distributed among the factors
>
> ai*x + u*f
>
> that is the issue.
> Actually, what I do is note factors *independent* of m, so
they canÕt
> suddenly gain a dependency on m, while this poster seems
fascinated
> with the aÕs being themselves functions of m, as if that
destroys the
> general principle.
> At best itÕd mean you can also separate them out as well,
using the
> same trick, which is to note their values when m=0, and it
turns out
> that for only *two* of the aÕs, when m=0, they are 0 as
well, which
> tells you something there.
> Now then, the uf simply cannot lose an f as a function of m
or
> dependent on m, as itÕs *independent* of m, so the logic 
is
simple.
> So this poster keeps making posts trying to refute the fact
that uf is
> independent of m, as if somehow thereÕs someway that math 
is
> illogical, but itÕs not.
> You do not disagree that ai is dependent
> on m. LetÕs write it as ai(m).
> Yup, itÕs dependent on m, but notice it *still* is the 
case
that you
> can isolate whatÕs independent of m by setting m=0, though
with the
> aÕs, two of them then become 0, while one is 3x + uf, 
which
youÕll
> notice is *independent* of m.
Of course it is independent of m. To derive this you set
m = 0.
I am sure you are NOT saying that, for m <> 0,
the third term is necessarily 3*x + u*f. You are not
claiming that a3 is always 3, are you ?
And you are NOT saying that for m <> 0, a1 = a2 = 0.
So, sure, for m = 0, you can say all these things.
But so what? You cannot avoid dealing with the cases
where m <> 0. Those in fact are the cases of greatest
concern to you.
> The principle is simple. WhatÕs fascinating is any person
trying to
> argue against it, as this Nora Baron is doing.
> Therefore the way in which factors of f divide
> ai(m) in general can be expected to depend on m.
> Your expectation--your intuition--is meaningless in the
face of
> mathematical logic.
> Your gut feeling is trash if it goes against the math.
> Here the key term is uf, which is *independent* of m, so it
canÕt vary
Correct. When m = 0, g1(x) = a1*x + u*f = u*f.
But when m <> 0, you still want to claim that f factors out
of a1*x + u*f. If you carry out such a factorization, you get
f*([a1/f]*x + u).
This means that you need a1/f to be an algebraic integer.
Of course a1 is dependent on m. So we should write this as
a1(m)/f.
Now, for m = 0, a1(m) = 0, so the result *is* an algebraic
integer.
But *what you need* is that a1(m)/f is an algebraic integer
for m <> 0.
It really does you *no good at all* to know what it is when m
= 0.
So far as you know right now, a1(m) could be *anything* when
m <> 0.
There is no reason to assume it is divisible by f. It could be
divisible by non-unit FACTORS of f. {This is in fact true for
most values of m and f.}
> However, this poster *wants* it to vary on m, so the poster
refuses to
> accept the math and notice, now talks about whatÕs 
expected.
I donÕt want it to vary with m. It simply DOES. There is no
sense pretending it doesnÕt. Considering its value only when
m = 0 is not enough, and I cannot see why you thihk it is. You
need to know its value when m takes on *other* values. You do
not escape having to do this simply because P(0) is
independent
of m.
> ItÕs a rather pathetic display which highlights what 
IÕve
said--human
> beings are NOT rational, as they rely strongly on social
forces, even
> with mathematics.
> You prove that for m = 0, f divides ai(0).
> I isolate the terms independent on m, deliberately, so that
I can see
> how they vary, without worrying about mÕs value, and it
just so
> happens that the way to find those terms independent of m,
is to set
> m=0, which is not rocket science.
Investigating what happens when m = 0 in no way solves the
problem of what happens when m <> 0.
Think of it like this. You have focussed on m = 0, and
proved certain facts, e.g., a1(0) is divisible by f.
I could equally well focus on ANOTHER value of m: say, m = 1.
I would then be considering P(1). This too is independent of
m.
LetÕs go a little farther. Focus on m = 1 and u = 1 and f = 
5.
Again, I am just studying a special case which is independent
of m. m = 1 is just a constant value.
In this special case, it so happens that
P(m)/f^2 = 553*x^3 - 72*x + 5.
You know what happens next. There is a proof that if this P(m)
is factored in the form
P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
where a1, a2 and a3 are algebraic integers, then
NONE of a1, a2 or a3 is divisible by f = 5, and NONE are
relatively prime to f = 5!
Do you want to see that proof again?
Thus when I focus on m = 1 [that is, a constant value
of m, which is INDEPENDENT of m as a variable], I find that
what you claim is completely wrong.
> You think somehow that that proves f divides ai(m)
> for all m.
> YouÕre showing a gross lack of logic as in fact I simply
rely on the
> terms being independent of m, and the fact that P(0)/f^2
equals u^2(3x
> + uf) so if f is coprime to 3, x, and u, it must be the
case that the
> factors that were g_1, g_2, and g_3 now no longer have f as
a factor
> either.
Yes. WHEN m = 0. I think we can agree that you are NOT saying:
g1(m) = u*f when m = 0. Therefore g1(m) = u*f for all other
values of m.
No, I donÕt think you are that dumb. What you ARE saying,
however
is this:
g1(m) = u*f when m = 0. Therefore g1(m) is divisible by f
when m = 0. Therefore g1(m) is divisible by f for all other
values of m.
Yes, I DO think you are THAT dumb. And NO, I donÕt think 
your
argument amounts to any more than what I just said.
> Now since at m=0, remember thatÕs to get terms independent
of m,
> g_1=uf, g_2=uf, but g_3 = 3x + uf, it stands to reason that
f divides
> off of g_1 and g_2, based on the terms *independent* of m,
so the
> value of m is not of consequence.
What rank and utter nonsense. So to assess divisibility of
a function h(m) by an integer, all I ever need to worry about
is h(0), because h(0) is *independent* of m ??? Is this some
new mathematical principle, the principle of independence
or some such ???
> ThatÕs the point of isolating terms independent of m, so
that I now
> they are independent of its value, which means that you
canÕt claim
> that things change just because of mÕs value.
OF COURSE they can change! P(m) changes when m <> 0.
a1(m), a2(m), and a3(m) change. g1 = a1*x + u*f changes.
of the expression a1*x + u*f by f does NOT depend strictly
upon g1(0) = u*f. It also depends on a1(m), about which in
your application you know essentially nothing when m <> 0.
That is why it is not sufficient to consider only m = 0.
Nora B.
PS: ArturoÕs explicit factorization of P(m) was useful in
this also. You simply blew it off without trying to under-
stand it. Your pattern here is: if you get the sense that the
math does not support what you believe, you try very hard to
just brush it off and ignore it. That is not how one gets at
the truth in math. You have to consider things even when it
they are painful. You have so much vested interest here that
no one,
not even you yourself, should trust your judgement.
===
Subject: Re: JSH: About time
>> Now that IÕve revealed the odd and you could say esoteric
error
> in
>> core mathematics with such a short, and rather simple
argument,
> the
>> issue now is how long until mathematicians decide that
theyÕd
> rather
>> have correct mathematics versus the *belief* that they had
been
>> perfect in keeping error out of the collected body of work
that
> is
>> called mathematics.
>> I posted a reply to this message earlier, but later I
noticed
> something in your post that shed some light on your
thinking.
> So I am replying to that explicitly.
Well you gave a useful example in that reply and I suspect
IÕll be
explaining my replies to that for a while.
In any event, you earned a careful read through, so IÕll go
through
your post here carefully, and give a full reply.
>
> By which you mean P(0).
>
> Yes, as setting m=0 isolates terms that are *independent*
of m.
>
> HereÕs whatÕs key, all in a row:
>
> P(m) = g_1 g_2 g_3
>
> P(0) = u^2 f^2 (3x + uf)
>
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
>
> P(0)/f^2 = u^2 (3x + uf)
>
> and f is coprime to 3, x and u.
>
> I focus on whatÕs independent of m for a reason.
>
> ItÕs not really a subtle technique.
>
>
>Then I look at at the same terms for P(m)/f^2 and find that a
factor
>of f^2 has been removed.
>Logic dictates that the removal is *independent* of m,
>
> Removal here means that you divide P(m) by f^2. That does
> not cause a problem in itself. It is really how f^2 is
> distributed among the factors
>
> ai*x + u*f
>
> that is the issue.
>
> Actually, what I do is note factors *independent* of m, so
they canÕt
> suddenly gain a dependency on m, while this poster seems
fascinated
> with the aÕs being themselves functions of m, as if that
destroys the
> general principle.
>
> At best itÕd mean you can also separate them out as well,
using the
> same trick, which is to note their values when m=0, and it
turns out
> that for only *two* of the aÕs, when m=0, they are 0 as
well, which
> tells you something there.
>
> Now then, the uf simply cannot lose an f as a function of m
or
> dependent on m, as itÕs *independent* of m, so the logic 
is
simple.
>
> So this poster keeps making posts trying to refute the fact
that uf is
> independent of m, as if somehow thereÕs someway that math 
is
> illogical, but itÕs not.
>
> You do not disagree that ai is dependent
> on m. LetÕs write it as ai(m).
>
> Yup, itÕs dependent on m, but notice it *still* is the 
case
that you
> can isolate whatÕs independent of m by setting m=0, though
with the
> aÕs, two of them then become 0, while one is 3x + uf, 
which
youÕll
> notice is *independent* of m.
>
> Of course it is independent of m. To derive this you set
> m = 0.
> I am sure you are NOT saying that, for m <> 0,
> the third term is necessarily 3*x + u*f. You are not
> claiming that a3 is always 3, are you ?
No, I notice how the terms independent of m vary as I divide
off f^2
from P(m).
P(m) itself clearly has a factor that is f^2, as does P(0) as
itÕs
P(0) = u^2 f^2 (3x + uf).
As I have P(m) split up using g_1, g_2 and g_3, I can use
them at m=0,
to find that g_1 = uf, g_2 = uf, g_3 = 3x + uf.
BUT, dividing f^2 off, that is, using P(m)/f^2 means that the
constant
terms no longer can have f as a factor, if f is coprime to 3,
while I
toss in coprimeness to x and u as well, just in case.
So to get the factor of f out of what is *constant* with
respect to m,
it means that f has to divide off of g_1 and g_2.
That gives the correct answer of P(0)/f^2, which is
P(0)/f^2 = u^2 (3x + uf).
Your little example Nora Baron highlighted a way that there
can be
interaction between constant terms and a specific value of m,
so it
was definitely something that made me think.
> And you are NOT saying that for m <> 0, a1 = a2 = 0.
> So, sure, for m = 0, you can say all these things.
> But so what? You cannot avoid dealing with the cases
> where m <> 0. Those in fact are the cases of greatest
> concern to you.
The not so subtle point of my technique is to find 
whatÕs
independent
of m, and see what happens when I divide P(m) by f^2.
That *necessarily* happens without regard to m, except for a
particular case which you highlighted in your example in
another post.
Unfortunately for me, that case just makes explaining that
much
harder, but IÕll try.
> The principle is simple. WhatÕs fascinating is any person
trying to
> argue against it, as this Nora Baron is doing.
>
> Therefore the way in which factors of f divide
> ai(m) in general can be expected to depend on m.
>
> Your expectation--your intuition--is meaningless in the
face of
> mathematical logic.
>
> Your gut feeling is trash if it goes against the math.
>
> Here the key term is uf, which is *independent* of m, so it
canÕt vary
> Correct. When m = 0, g1(x) = a1*x + u*f = u*f.
That is, you can easily determine that g_1 has an independent
term uf.
Necessarily, that term does not vary as m varies, but just
sits there,
uninterested, and unaffected by mÕs value.
> But when m <> 0, you still want to claim that f factors out
> of a1*x + u*f. If you carry out such a factorization, you
get
> f*([a1/f]*x + u).
Actually, checking P(m)/f^2 gives that whatÕs independent of
m then is
u^2(3x + uf), as then
P(0)/f^2 = u^2(3x + uf).
Given that the independent term has been affected it stands
to reason
that the independent term of g_1 must also change, and here
as it has
a factor of f, which the P(0)/f^2 does NOT have, it must be
the case
that a factor of f has to divide off when f^2 is divided from
P(m).
Given that the terms *independent* of m have been affected,
it must be
the case the they change independent of the value of m.
Now a special case occurs if mf^2 = 1. And explaining it
thoroughly
is going to be troublesome, but IÕm thinking about how best 
to
approach it.
> This means that you need a1/f to be an algebraic integer.
Actually, I donÕt.
The term algebraic integer comes from an arbitrary definition,
which
defines algebraic integers as roots of monic polynomials with
integer
coefficients.
IÕve managed to show a problem with that 
definition as itÕs not
inclusive enough.
> Of course a1 is dependent on m. So we should write this as
> a1(m)/f.
> Now, for m = 0, a1(m) = 0, so the result *is* an algebraic
> integer.
> But *what you need* is that a1(m)/f is an algebraic integer
> for m <> 0.
No, I donÕt.
The term algebraic integer comes from an arbitrary definition,
which
defines algebraic integers as roots of monic polynomials with
integer
coefficients.
IÕve managed to show a problem with that 
definition as itÕs not
inclusive enough.
> It really does you *no good at all* to know what it is when
m = 0.
> So far as you know right now, a1(m) could be *anything*
when m <> 0.
> There is no reason to assume it is divisible by f. It could
be
> divisible by non-unit FACTORS of f. {This is in fact true
for
> most values of m and f.}
And yet again I explain that what I do is focus on the terms
*independent* of m, just so that I donÕt have to worry about
what
value m has, which I thought covered everything until your
example in
another post.
Checking P(m)/f^2 gives whatÕs independent of m, which is
u^2(3x +
uf), as then
P(0)/f^2 = u^2(3x + uf).
Given that the independent term has been affected it stands
to reason
that the independent term of g_1 must also change, and here
as it has
a factor of f, which the P(0)/f^2 does NOT have, it must be
the case
that a factor of f has to divide off when f^2 is divided from
P(m).
Remember, though g_1 = a_1 x + uf, I have that uf is
independent by
the simple technique of setting m=0 to figure out the
independent
terms of g_1, g_2 and g_3.
However, now I also know that thereÕs a special case with my
own
argument when mf^2 = 1.
> However, this poster *wants* it to vary on m, so the poster
refuses to
> accept the math and notice, now talks about whatÕs 
expected.
>
> I donÕt want it to vary with m. It simply DOES. There is 
no
> sense pretending it doesnÕt. Considering its value only 
when
> m = 0 is not enough, and I cannot see why you thihk it is.
You
> need to know its value when m takes on *other* values. You
do
> not escape having to do this simply because P(0) is
independent
> of m.
And there Nora Baron you reveal that you are lost on my point
as in
fact the terms *independent* of m do NOT vary with m.
Here what IÕm talking about is uf, which is the independent
term of
g_1, as it is independent from the value of m.
> ItÕs a rather pathetic display which highlights what 
IÕve
said--human
> beings are NOT rational, as they rely strongly on social
forces, even
> with mathematics.
>
> You prove that for m = 0, f divides ai(0).
>
> I isolate the terms independent on m, deliberately, so that
I can see
> how they vary, without worrying about mÕs value, and it
just so
> happens that the way to find those terms independent of m,
is to set
> m=0, which is not rocket science.
>
> Investigating what happens when m = 0 in no way solves the
> problem of what happens when m <> 0.
ThatÕs illogical Nora Baron.
If I have the independent term of g_1, is it NOT independent?
You keep focusing on the aÕs, when I keep pointing to 
whatÕs
independent of m.
Determining terms independent of m is easy enough, just set m
equal to
0.
> Think of it like this. You have focussed on m = 0, and
> proved certain facts, e.g., a1(0) is divisible by f.
ThatÕs irrelevant.
YouÕre obsessed with the aÕs when 
IÕm repeatedly telling you
to focus
on terms *independent* of m.
> I could equally well focus on ANOTHER value of m: say, m =
1.
> I would then be considering P(1). This too is independent
of m.
Hardly as the technique is to set m=0 to pull out terms
independent of
m, as otherwise theyÕd go to 0 with it.
> LetÕs go a little farther. Focus on m = 1 and u = 1 and f 
=
5.
> Again, I am just studying a special case which is
independent
> of m. m = 1 is just a constant value.
> In this special case, it so happens that
> P(m)/f^2 = 553*x^3 - 72*x + 5.
> You know what happens next. There is a proof that if this
P(m)
> is factored in the form
> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
> where a1, a2 and a3 are algebraic integers, then
> NONE of a1, a2 or a3 is divisible by f = 5, and NONE are
> relatively prime to f = 5!
> Do you want to see that proof again?
> Thus when I focus on m = 1 [that is, a constant value
> of m, which is INDEPENDENT of m as a variable], I find that
> what you claim is completely wrong.
Actually, checking P(m)/f^2 gives that whatÕs independent of
m then is
u^2(3x + uf), as then
P(0)/f^2 = u^2(3x + uf).
Given that the independent term has been affected it stands
to reason
that the independent term of g_1 must also change, and here
as it has
a factor of f, which the P(0)/f^2 does NOT have, it must be
the case
that a factor of f has to divide off when f^2 is divided from
P(m).
> You think somehow that that proves f divides ai(m)
> for all m.
>
> YouÕre showing a gross lack of logic as in fact I simply
rely on the
> terms being independent of m, and the fact that P(0)/f^2
equals u^2(3x
> + uf) so if f is coprime to 3, x, and u, it must be the
case that the
> factors that were g_1, g_2, and g_3 now no longer have f as
a factor
> either.
>
> Yes. WHEN m = 0. I think we can agree that you are NOT
saying:
> g1(m) = u*f when m = 0. Therefore g1(m) = u*f for all other
> values of m.
> No, I donÕt think you are that dumb. What you ARE saying,
however
> is this:
> g1(m) = u*f when m = 0. Therefore g1(m) is divisible by f
> when m = 0. Therefore g1(m) is divisible by f for all other
> values of m.
> Yes, I DO think you are THAT dumb. And NO, I donÕt think
your
> argument amounts to any more than what I just said.
Actually, checking P(m)/f^2 gives that whatÕs independent of
m then is
u^2(3x + uf), as then
P(0)/f^2 = u^2(3x + uf).
Given that the independent term has been affected it stands
to reason
that the independent term of g_1 must also change, and here
as it has
a factor of f, which the P(0)/f^2 does NOT have, it must be
the case
that a factor of f has to divide off when f^2 is divided from
P(m).
> Now since at m=0, remember thatÕs to get terms independent
of m,
> g_1=uf, g_2=uf, but g_3 = 3x + uf, it stands to reason that
f divides
> off of g_1 and g_2, based on the terms *independent* of m,
so the
> value of m is not of consequence.
>
> What rank and utter nonsense. So to assess divisibility of
> a function h(m) by an integer, all I ever need to worry
about
> is h(0), because h(0) is *independent* of m ??? Is this some
> new mathematical principle, the principle of independence
> or some such ???
YouÕre running yourself off the track Nora Baron. 
IÕll repeat
again.
Actually, checking P(m)/f^2 gives that whatÕs independent of
m then is
u^2(3x + uf), as then
P(0)/f^2 = u^2(3x + uf).
Given that the independent term has been affected it stands
to reason
that the independent term of g_1 must also change, and here
as it has
a factor of f, which the P(0)/f^2 does NOT have, it must be
the case
that a factor of f has to divide off when f^2 is divided from
P(m).
> ThatÕs the point of isolating terms independent of m, so
that I now
> they are independent of its value, which means that you
canÕt claim
> that things change just because of mÕs value.
>
> OF COURSE they can change! P(m) changes when m <> 0.
> a1(m), a2(m), and a3(m) change. g1 = a1*x + u*f changes.
Progress? If you begin to accept that uf is an independent
term of
g_1, then eventually it should start making sense to you.
Given that g_1, g_2 and g_3 have terms *independent* of m,
are they
not independent of m? If you accept that they are, then how
can you
later claim they are dependent on m based on your need to
have the gÕs
have varying factors of f?
The independent terms rule the roost here.
You want a dependency on m, so you keep avoiding that fact.
> of the expression a1*x + u*f by f does NOT depend strictly
> upon g1(0) = u*f. It also depends on a1(m), about which in
> your application you know essentially nothing when m <> 0.
> That is why it is not sufficient to consider only m = 0.
But you see Nora Baron I *want* to know what happens
INDEPENDENT of
m, so I use the technique of setting m=0 to isolate what is
independent.
Then I notice how they must change as I divide f^2 off of
P(m).
> Nora B.
> PS: ArturoÕs explicit factorization of P(m) was useful in
> this also. You simply blew it off without trying to under-
> stand it. Your pattern here is: if you get the sense that
the
> math does not support what you believe, you try very hard to
> just brush it off and ignore it. That is not how one gets at
> the truth in math. You have to consider things even when it
> they are painful. You have so much vested interest here
that no one,
> not even you yourself, should trust your judgement.
IÕve looked at the factorization before Nora Baron.
You assume too much.
Besides, I can read both of your posts and see what youÕre
avoiding,
which is the simplicity that terms independent of m are in
fact
independent of m.
The special case, revealed by your own example in another
post, occurs
if m has a value which allows other terms to interact with the
independent terms.
In the case of your example, that meant that a 1 in one of
your
factors, clearly an independent term, gets subtracted off
with m=1.
James Harris
===
Subject: Re: JSH: About time
>> Now that IÕve revealed the odd and you could say esoteric
error
> in
>> core mathematics with such a short, and rather simple
argument,
> the
>> issue now is how long until mathematicians decide that
theyÕd
> rather
>> have correct mathematics versus the *belief* that they had
been
>> perfect in keeping error out of the collected body of work
that
> is
>> called mathematics.
>>
> I posted a reply to this message earlier, but later I
noticed
> something in your post that shed some light on your
thinking.
> So I am replying to that explicitly.
>
>
> Well you gave a useful example in that reply and I suspect
IÕll be
> explaining my replies to that for a while.
See my reply to the most recent of your replies to that.
> In any event, you earned a careful read through, so IÕll 
go
through
> your post here carefully, and give a full reply.
There is quite a bit of repetition in your reply here. I
will try to shorten things up.
>
>
> I am sure you are NOT saying that, for m <> 0,
> the third term is necessarily 3*x + u*f. You are not
> claiming that a3 is always 3, are you ?
> No, I notice how the terms independent of m vary as I
divide off f^2
> from P(m).
> P(m) itself clearly has a factor that is f^2, as does P(0)
as itÕs
> P(0) = u^2 f^2 (3x + uf).
> As I have P(m) split up using g_1, g_2 and g_3, I can use
them at m=0,
> to find that g_1 = uf, g_2 = uf, g_3 = 3x + uf.
> BUT, dividing f^2 off, that is, using P(m)/f^2 means that
the constant
> terms no longer can have f as a factor, if f is coprime to
3, while I
> toss in coprimeness to x and u as well, just in case.
> So to get the factor of f out of what is *constant* with
respect to m,
> it means that f has to divide off of g_1 and g_2.
Right here is one place where you go wrong.
P(m) = f^2 * R(m), where R(m) is another polynomial whose
coefficients are not all divisible by f.
You appear to think that f^2 can be factored out of P(m)
in only one way. That is correct in one sense, not
correct in another.
You note that if m = 0, g1 = u*f, g2 = u*f and g3 = 3*x + u*f.
Therefore f factors out of g1 and g2 (when m = 0) but not
out of g3. Also correct.
Now your logic seems to be: since f^2 factors out of P in
only one way for ANY value of m, and since when m = 0, f
factors out of each of g1 and g2 and not out of g3, the same
must be true for ANY m. Is that your thinking?
Perhaps you are thinking there is some kind of underlying
continuity here. If f divides g1 = a1*x + u*f when m = 0,
then f must divide g1 when m = other values. There are
several problems with that. m is a *discrete*
variable. In your application it takes on only integer
values. It is itself not continuous.
Assume f = f1 * f2 * f3, where f1 divides g1, f2 divides g2,
and f3 divides g3. When m = 0, you have that f1 = f, f2 = f,
and f3 = 1. I believe that you think f1, f2, and f3 must
be constant. Of course my polynomial example in the other
post shows that this cannot be true. When m = 1,
f1 = f2 = f3 = f^{2/3}. This shows that f1, f2, and f3
are themselves *dependent* on m. You think that m = 1 is
an unusual exception, and that for all other values of
m, the factorization behaves like it does for m = 0:
that is, f1 = f2 = f, f3 = 1.
This is nothing but wishful thinking on your part. As
I noted in another post in this thread, the EXCEPTIONAL
case here is m = 0. m = 1 is more like the general case.
Any time the polynomial is irreducible, all of the
coefficients a1, a2, and a3 will have factors in common
with f. The case m = 0 is one of the rare cases where
the polynomial is NOT irreducible.
LetÕs take a simpler example. Let
Q(m) = 5*m*x^2 + 5*(m - 2)*x - 25.
Assume this is factored in the form
Q(m) = g1*g2 = (a1*x + 5)*(a2*x - 5).
Note that
Q(0) = 5*(x - 5).
Therefore
g1(0) = 0*x + 5, and
g2(0) = 1*x - 5, and
a1(0) = 0, a2(0) = 1.
Note that a1(0) is divisible by 5
and a2(0) = 1 is coprime to 5.
Now let m be ANY INTEGER other than 0.
Q(m) = 5*m*x^2 + 5*(m - 2)*x - 25
= 5*(m*x^2 - (m - 2)*x - 5)
The roots of m*x^2 - (m - 2)*x - 5 are
r1 = ((m - 2) + sqrt((m - 2)^2 + 20))/ (2*m)
r2 = ((m - 2) - sqrt((m - 2)^2 + 20))/ (2*m)
We know that
a1 = -5 / r1 and a2 = 5 / r2.
It is tedious, but straightforward, to show that
a1 and a2 are both algebraic integers which have
algebraic integer factors of 5. Here is how it goes:
a2 = -5*2*m*((m - 2) + sqrt((m - 2)^2 + 20))/20
= -m*((m - 2) + sqrt((m - 2)^2 + 20))/2.
Let u = ((m - 2) + sqrt((m - 2)^2 + 20))/2.
You can easily show that u is a root of:
v^2 - v*(m - 2) - 5 = 0.
The latter polynomial is irreducible for most values
of m (e.g., m = 2, 3, 5, .... Therefore its roots are
algebraic integer divisors of 5. Therefore a2 in general
also has algebraic integer factors of 5.
An exception is when m = 0. This should not be
surprising: note that when m = 0, the expressions
above for r1 and r2 have 0 in the denominator. This
is a special case where the quadratic formula fails,
since the polynomial is of degree 1, not quadratic.
And note how irreducibility of the polynomial in v
plays a central role here.
> That gives the correct answer of P(0)/f^2, which is
> P(0)/f^2 = u^2 (3x + uf).
> Your little example Nora Baron highlighted a way that there
can be
> interaction between constant terms and a specific value of
m, so it
> was definitely something that made me think.
Not enough, apparently. In the end you have talked yourself
back into the conclusion you want so very badly. And
you are just as wrong as before.
> And you are NOT saying that for m <> 0, a1 = a2 = 0.
>
> So, sure, for m = 0, you can say all these things.
>
> But so what? You cannot avoid dealing with the cases
> where m <> 0. Those in fact are the cases of greatest
> concern to you.
> The not so subtle point of my technique is to find 
whatÕs
independent
> of m, and see what happens when I divide P(m) by f^2.
Agreed, itÕs not subtle. You have consistently under-
estimated our understanding of your argument. There is
nothing subtle about it. ŌSubtleis meaningless 
applied
to something which is blatantly wrong.
> That *necessarily* happens without regard to m, except for a
> particular case which you highlighted in your example in
another post.
> Unfortunately for me, that case just makes explaining that
much
> harder, but IÕll try.
That case shows unambigously that a1 and a2 are not
always divisible by f, and that a3 is not always coprime
to f: even if you donÕt understand the rest of it.
> The principle is simple. WhatÕs fascinating is any person
trying to
> argue against it, as this Nora Baron is doing.
>
> Therefore the way in which factors of f divide
> ai(m) in general can be expected to depend on m.
>
> Your expectation--your intuition--is meaningless in the
face of
> mathematical logic.
>
> Your gut feeling is trash if it goes against the math.
>
> Here the key term is uf, which is *independent* of m, so it
canÕt
vary
>
> Correct. When m = 0, g1(x) = a1*x + u*f = u*f.
> That is, you can easily determine that g_1 has an
independent term uf.
Right. But it also has a term which is DEPENDENT on m:
namely, a1.
> Necessarily, that term does not vary as m varies, but just
sits there,
> uninterested, and unaffected by mÕs value.
True enough. But the same cannot be said of a1. And
divisibility of g1 = a1*x + u*f by f is NOT determined
solely by u*f, is it? You MUST take a1 into account also,
when m <> 0.
> But when m <> 0, you still want to claim that f factors out
> of a1*x + u*f. If you carry out such a factorization, you
get
>
> f*([a1/f]*x + u).
> Actually, checking P(m)/f^2 gives that whatÕs independent
of m then is
> u^2(3x + uf), as then
> P(0)/f^2 = u^2(3x + uf).
> Given that the independent term has been affected it stands
to reason
> that the independent term of g_1 must also change,
No, it absolutely does NOT stand to reason. The
independent term, u*f is NOT affected. I have never
claimed that. It is the term which DEPENDS on m, namely
a1, which is affected and causes the divisibility of g1
by f to change.
> and here as it has
> a factor of f, which the P(0)/f^2 does NOT have, it must be
the case
> that a factor of f has to divide off when f^2 is divided
from P(m).
Such utter nonsense. Do you think it is possible to
have two functions, A(y) and B(y), both not constant,
such that A(y)*B(y) is constant? Yes, you probably agree
that that can happen.
Now: do you think it is possible to have two functions,
A(y) and B(y), both integer-valued, such that A(y)* B(y)
is ALWAYS a multiple of 25, but it is also true that for
varying values of y, A(y) is sometimes divisible by 5,
sometimes not, and B(y) is sometimes divisible by 5,
sometimes not. Can that happen? Can you think of
examples? Would you like for me to give an example> Given
that the terms *independent* of m have been affected, it must
be
> the case the they change independent of the value of m.
> Now a special case occurs if mf^2 = 1. And explaining it
thoroughly
> is going to be troublesome, but IÕm thinking about how 
best
to
> approach it.
Yes, do that.
> This means that you need a1/f to be an algebraic integer.
> Actually, I donÕt.
When you are working in the ring of algebraic integers,
and you say something like a1 is divisible by integer f,
what you MUST mean is that a1/f is an algebraic integer.
If you DONÕT mean that, if you mean that a1/f could be
just any old algebraic NUMBER, then your whole argument
collapses into triviality.
> The term algebraic integer comes from an arbitrary
definition, which
> defines algebraic integers as roots of monic polynomials
with integer
> coefficients.
> IÕve managed to show a problem with that 
definition as itÕs
not
> inclusive enough.
No. The argument you have used to show this problem
is essentially your *present* argument, or one of your
previous equally incorrect arguments. To use the
conclusion of that argument at this point is assuming
what you hope to prove. At this point it is off limits.
> Of course a1 is dependent on m. So we should write this as
>
> a1(m)/f.
>
> Now, for m = 0, a1(m) = 0, so the result *is* an algebraic
> integer.
>
> But *what you need* is that a1(m)/f is an algebraic integer
> for m <> 0.
> No, I donÕt.
[mindless repetition deleted]
> It really does you *no good at all* to know what it is when
m = 0.
> So far as you know right now, a1(m) could be *anything*
when m <> 0.
> There is no reason to assume it is divisible by f. It could
be
> divisible by non-unit FACTORS of f. {This is in fact true
for
> most values of m and f.}
[more repetition deleted]
> However, now I also know that thereÕs a special case with
my own
> argument when mf^2 = 1.
> However, this poster *wants* it to vary on m, so the poster
refuses
to
> accept the math and notice, now talks about whatÕs 
expected.
>
>
> I donÕt want it to vary with m. It simply DOES. There is 
no
> sense pretending it doesnÕt. Considering its value only 
when
> m = 0 is not enough, and I cannot see why you thihk it is.
You
> need to know its value when m takes on *other* values. You
do
> not escape having to do this simply because P(0) is
independent
> of m.
> And there Nora Baron you reveal that you are lost on my
point as in
> fact the terms *independent* of m do NOT vary with m.
> Here what IÕm talking about is uf, which is the 
independent
term of
> g_1, as it is independent from the value of m.
Indubitably. But a1 is not. Is a1 not part of g1 also?
It is a function of m. LetÕs say that for some value of m,
say, m0, a1 = a1(m0) = sqrt(f). Then
g1 = g1(m0) = sqrt(f)*x + u*f.
Now: in this case, is f a factor of g1 ?
Do you have a proof that a1 = a1(m) CANNOT equal sqrt(f),
no matter what m is ?
NO. All you know is that a1 = 0 when m = 0. And
little good it does you.
> Investigating what happens when m = 0 in no way solves the
> problem of what happens when m <> 0.
> ThatÕs illogical Nora Baron.
> If I have the independent term of g_1, is it NOT
independent?
Yes, u*f is independent of m. But a1 is not.
> You keep focusing on the aÕs, when I keep pointing to 
whatÕs
> independent of m.
Yes, I KNOW! The trouble is, when you want to talk
about the divisibility of g1 by f, you must consider not
only the CONSTANT term u*f; you must also consider the
other stuff:
g1 = a1*x + u*f.
We both agree that a1 is dependent on m.
What if for m = 137, a1(m) = a1(137) = f^{1/11}?
If that happens, is a1*x + u*f divisible by f ?
Or have you somehow ruled out the possibility that
a1 = f^{1/11} for ALL values of m ? Not just for m = 0,
but for all other values? If so, I have not seen it.
> Determining terms independent of m is easy enough, just set
m equal to
> 0.
> Think of it like this. You have focussed on m = 0, and
> proved certain facts, e.g., a1(0) is divisible by f.
> ThatÕs irrelevant.
If itÕs irrelevant, why did you bother to do it? I 
donÕt
have any problem with this. You are now so rabid to deny
everything I say that you are now OBJECTING TO YOUR OWN STUFF!
> YouÕre obsessed with the aÕs when 
IÕm repeatedly telling
you to focus
> on terms *independent* of m.
YOU CANNOT IGNORE THE aÕs !!! They are part of the
functions g1, g2, and g3 ||| The divisibility of g1, etc.,
when m <> 0 by f is NOT determined solely by g1(0) = u*f. What
on earth makes you think that is true ? Have you actually
lost your marbles?
> I could equally well focus on ANOTHER value of m: say, m =
1.
> I would then be considering P(1). This too is independent
of m.
> Hardly as the technique is to set m=0 to pull out terms
independent of
> m, as otherwise theyÕd go to 0 with it.
It is NOT ENOUGH to consider only m = 0. All it tells
you about is divisibility at ONE POINT.
> LetÕs go a little farther. Focus on m = 1 and u = 1 and f 
=
5.
>
> Again, I am just studying a special case which is
independent
> of m. m = 1 is just a constant value.
>
> In this special case, it so happens that
>
> P(m)/f^2 = 553*x^3 - 72*x + 5.
>
> You know what happens next. There is a proof that if this
P(m)
> is factored in the form
>
> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
>
> where a1, a2 and a3 are algebraic integers, then
> NONE of a1, a2 or a3 is divisible by f = 5, and NONE are
> relatively prime to f = 5!
>
> Do you want to see that proof again?
>
> Thus when I focus on m = 1 [that is, a constant value
> of m, which is INDEPENDENT of m as a variable], I find that
> what you claim is completely wrong.
> Actually, checking P(m)/f^2 gives that whatÕs independent
of m then is
> u^2(3x + uf), as then
> P(0)/f^2 = u^2(3x + uf).
> Given that the independent term has been affected it stands
to reason
> that the independent term of g_1 must also change, and here
as it has
> a factor of f, which the P(0)/f^2 does NOT have, it must be
the case
> that a factor of f has to divide off when f^2 is divided
from P(m).
The same old tired drivel again. You didnÕt make even a
feeble attempt to respond to the example and you didnÕt
answer the question. Do you want to see the proof again
or not?
> You think somehow that that proves f divides ai(m)
> for all m.
>
> YouÕre showing a gross lack of logic as in fact I simply
rely on the
> terms being independent of m, and the fact that P(0)/f^2
equals
u^2(3x
> + uf) so if f is coprime to 3, x, and u, it must be the
case that the
> factors that were g_1, g_2, and g_3 now no longer have f as
a factor
> either.
>
>
> Yes. WHEN m = 0. I think we can agree that you are NOT
saying:
>
> g1(m) = u*f when m = 0. Therefore g1(m) = u*f for all other
> values of m.
>
> No, I donÕt think you are that dumb. What you ARE saying,
however
> is this:
>
> g1(m) = u*f when m = 0. Therefore g1(m) is divisible by f
> when m = 0. Therefore g1(m) is divisible by f for all other
> values of m.
>
> Yes, I DO think you are THAT dumb. And NO, I donÕt think
your
> argument amounts to any more than what I just said.
> Actually,
[repetition deleted]
> Now since at m=0, remember thatÕs to get terms independent
of m,
> g_1=uf, g_2=uf, but g_3 = 3x + uf, it stands to reason that
f divides
> off of g_1 and g_2, based on the terms *independent* of m,
so the
> value of m is not of consequence.
>
>
> What rank and utter nonsense. So to assess divisibility of
> a function h(m) by an integer, all I ever need to worry
about
> is h(0), because h(0) is *independent* of m ??? Is this some
> new mathematical principle, the principle of independence
> or some such ???
> YouÕre running yourself off the track Nora Baron. 
IÕll
repeat
> again.
> Actually,
[repetition deleted]
> ThatÕs the point of isolating terms independent of m, so
that I now
> they are independent of its value, which means that you
canÕt claim
> that things change just because of mÕs value.
>
>
> OF COURSE they can change! P(m) changes when m <> 0.
> a1(m), a2(m), and a3(m) change. g1 = a1*x + u*f changes.
> Progress? If you begin to accept that uf is an independent
term of
> g_1, then eventually it should start making sense to you.
Pigs will sprout wings at that point. Cows will
jump over the moon. People will ice-skate in hell.
> Given that g_1, g_2 and g_3 have terms *independent* of m,
are they
> not independent of m?
Oh, this is a nice question. I like this. HereÕs a
rephrasing: g1(0) is independent of m. Therefore isnÕt
g1(m) also independent of m?
HereÕs another rephrasing. Given that a has a constant
term, does it not follow that the function is constant ?
IsnÕt that obvious?
> If you accept that they are, then how can you
> later claim they are dependent on m based on your need to
have the gÕs
> have varying factors of f?
I have no such need. Things are as they are.
Follow the math.
> The independent terms rule the roost here.
They assuredly do not.
Say f(m) = m^2 + 7*m + 3.
The term which is independent of m is f(0) = 3.
It is divisible by 3.
Does it rule the roost?
Is f(m) divisible by 3 for all other values of m ?
What do you think? Just how far gone are you?
> You want a dependency on m, so you keep avoiding that fact.
I donÕt want it. It is just there. It is a fact of
life. You want to deny it, in spite of OVERWHELMING
MOUNTAINS OF EVIDENCE!
> of the expression a1*x + u*f by f does NOT depend strictly
> upon g1(0) = u*f. It also depends on a1(m), about which in
> your application you know essentially nothing when m <> 0.
> That is why it is not sufficient to consider only m = 0.
> But you see Nora Baron I *want* to know what happens
INDEPENDENT of
> m, so I use the technique of setting m=0 to isolate what is
> independent.
I know what you want. You can isolate stuff till you
are blue in the face. There is still other stuff. The
other stuff is out of your (and my) control. It varies
as m varies, and it affects divisibility of factors.
I think you must conceive of factorization as following
some kind of *fixed formula*. That is either not true, or
the formula is so complicated that you cannot tell how
it behaves for different values of m.
> Then I notice how they must change as I divide f^2 off of
P(m).
How f^2 divides off of P(m) does not determine how it
divides off of a *factorization* of P(m), when that
factorization itself depends on m.
> Nora B.
>
>
> PS: ArturoÕs explicit factorization of P(m) was useful in
> this also. You simply blew it off without trying to under-
> stand it. Your pattern here is: if you get the sense that
the
> math does not support what you believe, you try very hard to
> just brush it off and ignore it. That is not how one gets at
> the truth in math. You have to consider things even when it
> they are painful. You have so much vested interest here
that no one,
> not even you yourself, should trust your judgement.
> IÕve looked at the factorization before Nora Baron.
Not enough.
> You assume too much.
I assume you didnÕt read it deeply enough to understand
it. Otherwise you would not still be arguing.
> Besides, I can read both of your posts and see what youÕre
avoiding,
> which is the simplicity that terms independent of m are in
fact
> independent of m.
BULL.
> The special case, revealed by your own example in another
post, occurs
> if m has a value which allows other terms to interact with
the
> independent terms.
ItÕs NOT a special case. The special case is m = 0.
For most values of m, the polynomial is irreducible, and
you are forced then to conclude, as with my example above
P = 25*(553*x^3 - 72*x + 5),
that a1, a2, and a3 all have factors of f.
> In the case of your example, that meant that a 1 in one of
your
> factors, clearly an independent term, gets subtracted off
with m=1.
You donÕt have a satisfactory explanation even for this
case, let alone for the more general case. You STILL do
know have a genuine understanding of what is going on. You
badly need to take an algebra course.
Nora B.
PS: ItÕs true what I said above, isnÕt it? You 
think that
there is some fixed formula that gives the factorization of
your polynomial. You (and I) donÕt know what that formula
is, but that is the idea you have in your head: that there
is some formula, and when you factor f^2 out of P(m), it
follows that formula; and since you know how it factors
out when m = 0, and since there is a fixed formula somehere,
the factorization is independent of m, and it must
therefore factor out the same way for all m. This is really
the idea you have in your head, isnÕt it?
The concept that there might not BE such a formula has never
occurred to you. But in lots of math, there are no formulas.
There are algorithms, but not necessarily formulas. You
know that intimately yourself. Your prime counting function
is an algorithm. You cannot write down a nice closed-form
expression for pi(n).
Even if there were a formula, it might not be simple. It
could give widely differing factorizations for different
values of m. Formulas in general are not constant. If
there is a formula here, there is no reason to expect it
to be constant. Yet that too seems be part of your thinking.
It is enough, you think, to see how things factor when m = 0.
Then since the FORM of the factorization must be constant
[you claim] the same factorizations must apply when m <> 0.
Wrong, wrong, wrong. There is an algoritm for computing a1,
a2, and a3 (at least for cubics). It is complicated. As
Arturo has shown, it yields different results for
factorization
by f for different values of m. My example shows the same.
> James Harris
===
Subject: Re: JSH: About time
>> Now that IÕve revealed the odd and you could say esoteric
error
> in
>> core mathematics with such a short, and rather simple
argument,
> the
>> issue now is how long until mathematicians decide that
theyÕd
> rather
>> have correct mathematics versus the *belief* that they had
been
>> perfect in keeping error out of the collected body of work
that
> is
>> called mathematics.
>>
>> My work is out there and rather easy to go over as can be
seen at
> the
>> Hong Konk math site:
>>
>>
>It does make sense. Basically I isolate terms of P(m) that
are
>*independent* of m.
>
> By which you mean P(0).
> Yes, as setting m=0 isolates terms that are *independent*
of m.
> HereÕs whatÕs key, all in a row:
> P(m) = g_1 g_2 g_3
> P(0) = u^2 f^2 (3x + uf)
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
> P(0)/f^2 = u^2 (3x + uf)
> and f is coprime to 3, x and u.
> I focus on whatÕs independent of m for a reason.
> ItÕs not really a subtle technique.
Since the rest of your response is just repetition of
this section, I will deal with that and ignore most
of the rest.
P(m) is a function of m. You compute P(0) and note
that it is not divisible by f. You note as above,
that if
P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
then when m = 0, a1 = a2 = 0 and a3 = 3. You note
that when m = 0, a1 and a2 are divisible by f.
You then say at great length that in considering
m = 0, you are focussing on the part of P(m) which
is independent of m. True, in the sense that P(0)
does not change when m changes.
But after that your logic takes off into never-
never land. You seem to somehow think that because
you have dealt with the factorization of the constant
term P(0), it tells you something about P(m) when
m is not 0. But you never really establish a connection.
You just repeat over and over the bit about
independence.
Then I say, and you agree, that a1 (and a2 and a3)
are all functions of m: a1(m), a2(m), and a3(m).
We agree that you can say that a1(0) = 0.
What that means is that when you consider
g1(0) = a1(0)*x + u*f,
you can factor out f. What you get when you
factor out f is
g1(0) = u.
Again, this happens because a1(0) = 0.
Now you somehow make a great leap. You agree
that a1(m) depends on m, and that in general
a1(m) is not zero. But you think that one
property of a1(0) carries over to a1(m) for
other values of m: that it is divisible by f.
But you simply do not say why.
If your logic were correct it would apply
to other functions.
Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3.
Say Q(m) is factored in the form
Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f).
Now note Q(0)/f^2 = x + f. That is,
Q(0) = f * f * (x + f).
Therefore when m = 0, we can say a1 = 0, a2 = 0,
and a3 = 1.
Note that when m = 0, a1 and a2 are divisible by f.
Now by your logic, for values of m other than 0,
a1 and a2 must be divisible by f and a3 is relatively
prime to f.
Do you agree with this? This is a test of your
method. We need to know before we go to the next step.
Nora B.
===
Subject: Re: JSH: About time
<deleted> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
OOPS!!! In my previous post it should have been
a3 = h_1(m) - mh_2(m) + 1
where again h_1(1) has a factor that is f^{2/3} and h_2(1) =
1.
James Harris
===
Subject: Re: JSH: About time
<deleted>
Ok, I figured it out.
> If your logic were correct it would apply
> to other functions.
Yup.
> Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3.
> Say Q(m) is factored in the form
> Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f).
> Now note Q(0)/f^2 = x + f. That is,
> Q(0) = f * f * (x + f).
> Therefore when m = 0, we can say a1 = 0, a2 = 0,
> and a3 = 1.
> Note that when m = 0, a1 and a2 are divisible by f.
Yup.
> Now by your logic, for values of m other than 0,
> a1 and a2 must be divisible by f and a3 is relatively
> prime to f.
> Do you agree with this? This is a test of your
> method. We need to know before we go to the next step.
> Nora B.
The answer is that yes, but with the qualification that m not
equal 1,
but I have a better explanation for why than the wacky reply
when I
initially freaked out.
Look again at your
> Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f).
Here the constant terms for the factors are revealed to be
f, f and x + f
where it just so happens that for your a3 you have something
like--yup, I know some will probably not like this, but itÕs
the
reality--introducing h for the functions h(m),
a3 = h_1(m) - mx h_2(m) + x + f
where h_(1) has f^{2/3} as a factor and h_2(1)=1.
And yes, the same thing can happen with my argument, but it
requires
mf^2 = 1
but both m and f are integers in that argument, so that
condition
doesnÕt occur.
(If I didnÕt say they were before, well they are now.)
For the more adventurous, check Q(m) with m NOT equal to 1,
and yes,
you will find that *two* of the aÕs have a 
factor that is f.
It might be a fun exercise, assuming I didnÕt miss 
something.
If any of you out there think you can find fault with my
conclusion
here, please try.
It is math after all. And itÕs also a lot of fun.
James Harris
===
Subject: Re: JSH: About time
> <deleted> Ok, I figured it out.
See below.
> If your logic were correct it would apply
> to other functions.
> Yup.
> Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3.
>
> Say Q(m) is factored in the form
>
> Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f).
>
> Now note Q(0)/f^2 = x + f. That is,
>
> Q(0) = f * f * (x + f).
>
> Therefore when m = 0, we can say a1 = 0, a2 = 0,
> and a3 = 1.
>
> Note that when m = 0, a1 and a2 are divisible by f.
> Yup.
> Now by your logic, for values of m other than 0,
> a1 and a2 must be divisible by f and a3 is relatively
> prime to f.
>
> Do you agree with this? This is a test of your
> method. We need to know before we go to the next step.
>
> Nora B.
> The answer is that yes, but with the qualification that m
not equal 1,
Why should we accept that qualification ?
Would you like to see similar examples where the problem
occurs when m is, e.g., 7 ?
> but I have a better explanation for why than the wacky
reply when I
> initially freaked out.
> Look again at your
> Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f).
> Here the constant terms for the factors are revealed to be
> f, f and x + f
> where it just so happens that for your a3 you have something
> like--yup, I know some will probably not like this, but
itÕs the
> reality--introducing h for the functions h(m),
> a3 = h_1(m) - mx h_2(m) + x + f
> where h_(1) has f^{2/3} as a factor and h_2(1)=1.
Where the heck is this coming from? Certainly
I have introduced no such thing. This provides
absolutely no justification for your implied claim
here that your original argument will work with
your original polynomial, but not with *this*
polynomial.
Yes, you are right - some will probably not like
this. It just looks like smoke to cover up a
huge problem with your non-proof, invented on the
spot - like a bad magician, reaching into the
hat and pulling out - a dead rabbit.
> And yes, the same thing can happen with my argument, but it
requires
> mf^2 = 1
What? How so?
> but both m and f are integers in that argument, so that
condition
> doesnÕt occur.
> (If I didnÕt say they were before, well they are now.)
> For the more adventurous, check Q(m) with m NOT equal to 1,
and yes,
> you will find that *two* of the aÕs have a 
factor that is f.
Not so. There are two important points here, both related
to posts that Arturo made yesterday: one is fairly
straightforward, and the other requires some depth:
(1) The computation that Arturo gave for your original
polynomial, using CardanoÕs expressions for the
roots of your polynomial, showed almost exactly
the same thing that happens with my polynomial.
You did not pay attention to the details. You
really should have. He proved that all three
of a1, a2, and a3 have nonunit factors in common
with f, provided your polynomial is irreducible.
(2) When m = 1, my polynomial is irreducible over
the rationals (see more below on this). In fact
this is true for MOST values of m. Even more
interesting, as Arturo pointed out, m = 0 happens
to be one of the few values of m for which it is
NOT irreducible. The same is true for your original
polynomial.
So what ? you may ask.
The key is *symmetry*. When m = 1, my polynomial
reduces essentially to
z^3 - b^3.
For integer b, this is irreducible over the rationals.
The roots are:
r1 = b,
r2 = b*(cos(2*pi/3) + i*sin(2*pi/3)), and
r3 = b*(cos(4*pi/3) + i*sin(4*pi/3)).
You can *see* the symmetry! The roots are in the complex
plane, and you can permute them, r1 --> r2 --> r3 --> r1,
by simply rotating the plane through an angle of
2*pi/3.
Symmetry like this ALSO occurs with the roots of
other irreducible polynomials, but it is not as
simple or obvious, and it is not expressible as
a simple rotation. A very basic theorem in Galois
theory is that for irreducible polynomials, there
are field automorphisms which permute the roots. For
any pair of roots, say, r1 and r2, there is an
automorphism of the splitting field of the polynomial
which takes root r1 to root r2.
So what again? you may ask.
Let G be a Galois automorphism. It can be shown
that if r1 is a root which shares an algebraic
integer factor with, say, f, then G(r1) = r2 also shares
an algebraic integer factor with f.
You can see where this is going. If one root is
not coprime to f, then none of the other roots is
either.
Galois automorphisms are the *symmetries* of splitting
fields and their extensions. That is one of the
great central beauties of Galois theory, and one
of the reasons we consider Galois a mathematical
genius of the highest order.
[Someone with a name like Nora Baron has a special
appreciation of symmetries!]
Arturo implied all this, perhaps too tersely and
cryptically, when he spoke about irreducibility
and symmetry. Of course it went right over your head.
Arturo has the huge advantage over you of having studied this
kind of thing in grad school. He sees things at a
depth that you cannot, not because he is smarter, but
because he learned some theory. You could learn it as
well. It would save you a lot of wasted effort.
In your original polynomial, ArturoÕs calculations
showed exactly and unambiguously how factors of f
occurred in each one of the aÕs. This is exactly
parallel to what happens with my polynomial when m = 1.
You think for some reason that m = 1 is a special
case, and that that gets you off the hook. You could
not be more wrong. For MOST values of m, the
polynomials in question are irreducible. m = 1 is
NOT the special case.
THE SPECIAL CASE IS: m = 0! In that case you donÕt
have irreducibility. The Galois theorem does not
apply. You have spent virtually all of your time
dealing with the one case which does NOT generalize.
Focussing on this special case has led you badly
astray. The reasons it does not generalize, as you
can see if you understood even some of what I have
said here, are relatively deep. You can follow
ArturoÕs calculations with some effort, but to
really understand and appreciate what is going on you
need to know a little theory.
Your argument about independence of P(0) is simply bogus,
meaningless wordplay. It has no connection with the under-
lying symmetries of the irreducible cases. You donÕt get
something for nothing.
It is possible to show that you are wrong without resorting
to Galois theory. I have several times presented the
polynomial
25 * (553*x^3 - 72*x + 5)
i.e., your polynomial with m = 1, f = 5, u = 1. The polynomial
that a non-monic irreducible polynomial cannot have algebraic
integer roots, you deduce that none of the aiÕs is divisible
by 5.
Your quest is at an end, or at least at a turning point.
There is no error in core. You have underestimated the
importance of irreducibility, and you have deceived yourself
into thinking that a special case is representative of the
general case.
Yes, you could carry on and continue to press your bogus
logic, saying that my polynomial does not prove that your
argument is wrong because it is slightly different in some
way from YOUR polynomial. The point of it was, it had
all the essential characteristics, in terms of its behavior
at m = 0, that YOUR polynomial has. When you point out
differences, you must be sure that they are differences
which you actually USE in your proof. The silliness you cited
above about h_1(m) and h_2(m) has no place in your original
argument. It was made up ad hoc. It is cheating, in a sense -
not cheating me or others here - cheating the math.
> It might be a fun exercise, assuming I didnÕt miss
something.
You did miss something. You missed the *essential core of
the problem*.
More than a scare. I proved your method is wrong. Above
I have explained why.
Nora B.
> If any of you out there think you can find fault with my
conclusion
> here, please try.
> It is math after all. And itÕs also a lot of fun.
> James Harris
===
Subject: Re: JSH: About time
> <deleted>
> Ok, I figured it out.
> See below.
> If your logic were correct it would apply
> to other functions.
>
> Yup.
>
> Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3.
>
> Say Q(m) is factored in the form
>
> Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f).
>
> Now note Q(0)/f^2 = x + f. That is,
>
> Q(0) = f * f * (x + f).
>
> Therefore when m = 0, we can say a1 = 0, a2 = 0,
> and a3 = 1.
>
> Note that when m = 0, a1 and a2 are divisible by f.
>
> Yup.
>
> Now by your logic, for values of m other than 0,
> a1 and a2 must be divisible by f and a3 is relatively
> prime to f.
>
> Do you agree with this? This is a test of your
> method. We need to know before we go to the next step.
>
> Nora B.
>
> The answer is that yes, but with the qualification that m
not equal 1,
>
> Why should we accept that qualification ?
That math requires it.
> Would you like to see similar examples where the problem
> occurs when m is, e.g., 7 ?
Then itÕd shift appropriately.
> but I have a better explanation for why than the wacky
reply when I
> initially freaked out.
>
> Look again at your
>
> Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f).
>
> Here the constant terms for the factors are revealed to be
>
> f, f and x + f
>
> where it just so happens that for your a3 you have something
> like--yup, I know some will probably not like this, but
itÕs the
> reality--introducing h for the functions h(m),
>
> a3 = h_1(m) - mx h_2(m) + x + f
>
> where h_(1) has f^{2/3} as a factor and h_2(1)=1.
ThatÕs off as I noted in another post.
It should be
a3 = h_1(m) - m h_2(m) + 1
where h_1(1) has a factor that is f^{2/3} and h_2(m) = 1.
> Where the heck is this coming from? Certainly
> I have introduced no such thing. This provides
> absolutely no justification for your implied claim
> here that your original argument will work with
> your original polynomial, but not with *this*
> polynomial.
You need to solve for the aÕs with your expression.
> Yes, you are right - some will probably not like
> this. It just looks like smoke to cover up a
> huge problem with your non-proof, invented on the
> spot - like a bad magician, reaching into the
> hat and pulling out - a dead rabbit.
LOL. I figured itÕd give me problems, and like I 
said, you
came up
with a good example.
What you need to do now is solve for the aÕs.
> And yes, the same thing can happen with my argument, but it
requires
>
> mf^2 = 1
>
> What? How so?
Think about it.
> but both m and f are integers in that argument, so that
condition
> doesnÕt occur.
>
> (If I didnÕt say they were before, well they are now.)
>
> For the more adventurous, check Q(m) with m NOT equal to 1,
and yes,
> you will find that *two* of the aÕs have a 
factor that is f.
>
> Not so. There are two important points here, both related
> to posts that Arturo made yesterday: one is fairly
> straightforward, and the other requires some depth:
> (1) The computation that Arturo gave for your original
> polynomial, using CardanoÕs expressions for the
> roots of your polynomial, showed almost exactly
> the same thing that happens with my polynomial.
> You did not pay attention to the details. You
> really should have. He proved that all three
> of a1, a2, and a3 have nonunit factors in common
> with f, provided your polynomial is irreducible.
You wish.
> (2) When m = 1, my polynomial is irreducible over
> the rationals (see more below on this). In fact
> this is true for MOST values of m. Even more
> interesting, as Arturo pointed out, m = 0 happens
> to be one of the few values of m for which it is
> NOT irreducible. The same is true for your original
> polynomial.
> So what ? you may ask.
> The key is *symmetry*. When m = 1, my polynomial
> reduces essentially to
> z^3 - b^3.
> For integer b, this is irreducible over the rationals.
> The roots are:
> r1 = b,
> r2 = b*(cos(2*pi/3) + i*sin(2*pi/3)), and
> r3 = b*(cos(4*pi/3) + i*sin(4*pi/3)).
> You can *see* the symmetry! The roots are in the complex
> plane, and you can permute them, r1 --> r2 --> r3 --> r1,
> by simply rotating the plane through an angle of
> 2*pi/3.
> Symmetry like this ALSO occurs with the roots of
> other irreducible polynomials, but it is not as
> simple or obvious, and it is not expressible as
> a simple rotation. A very basic theorem in Galois
> theory is that for irreducible polynomials, there
> are field automorphisms which permute the roots. For
> any pair of roots, say, r1 and r2, there is an
> automorphism of the splitting field of the polynomial
> which takes root r1 to root r2.
> So what again? you may ask.
> Let G be a Galois automorphism. It can be shown
> that if r1 is a root which shares an algebraic
> integer factor with, say, f, then G(r1) = r2 also shares
> an algebraic integer factor with f.
> You can see where this is going. If one root is
> not coprime to f, then none of the other roots is
> either.
> Galois automorphisms are the *symmetries* of splitting
> fields and their extensions. That is one of the
> great central beauties of Galois theory, and one
> of the reasons we consider Galois a mathematical
> genius of the highest order.
> [Someone with a name like Nora Baron has a special
> appreciation of symmetries!]
So thatÕs your real name? Not a pseudonym?
> Arturo implied all this, perhaps too tersely and
> cryptically, when he spoke about irreducibility
> and symmetry. Of course it went right over your head.
> Arturo has the huge advantage over you of having studied
this
> kind of thing in grad school. He sees things at a
> depth that you cannot, not because he is smarter, but
> because he learned some theory. You could learn it as
> well. It would save you a lot of wasted effort.
> In your original polynomial, ArturoÕs calculations
> showed exactly and unambiguously how factors of f
> occurred in each one of the aÕs. This is exactly
> parallel to what happens with my polynomial when m = 1.
> You think for some reason that m = 1 is a special
> case, and that that gets you off the hook. You could
> not be more wrong. For MOST values of m, the
> polynomials in question are irreducible. m = 1 is
> NOT the special case.
> THE SPECIAL CASE IS: m = 0! In that case you donÕt
> have irreducibility. The Galois theorem does not
> apply. You have spent virtually all of your time
> dealing with the one case which does NOT generalize.
Well, you sure sound certain.
> Focussing on this special case has led you badly
> astray. The reasons it does not generalize, as you
> can see if you understood even some of what I have
> said here, are relatively deep. You can follow
> ArturoÕs calculations with some effort, but to
> really understand and appreciate what is going on you
> need to know a little theory.
>
> Your argument about independence of P(0) is simply bogus,
> meaningless wordplay. It has no connection with the under-
> lying symmetries of the irreducible cases. You donÕt get
> something for nothing.
The argument is, luckily for me, short enough that it is
machine
checkable.
How about this Nora Baron?
If a computer verifies my proof, will you believe it then?
James Harris
===
Subject: Re: JSH: About time
<deleted>
OOPS!!! My previous reply was just totally wacky.
Um, IÕll have to think about this one.
> If your logic were correct it would apply
> to other functions.
> Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3.
> Say Q(m) is factored in the form
> Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f).
> Now note Q(0)/f^2 = x + f. That is,
> Q(0) = f * f * (x + f).
> Therefore when m = 0, we can say a1 = 0, a2 = 0,
> and a3 = 1.
> Note that when m = 0, a1 and a2 are divisible by f.
Yeah, but when m=1, you have that each of the aÕs has a
factor that is
f^{2/3}.
Good example.
I must admit that IÕm ßustered for the moment.
> Now by your logic, for values of m other than 0,
> a1 and a2 must be divisible by f and a3 is relatively
> prime to f.
> Do you agree with this? This is a test of your
> method. We need to know before we go to the next step.
> Nora B.
There might not be a next step. I need to figure out if 
youÕve
shown
that I *have* been wrong.
Good job Nora Baron.
Let me think on this for a bit.
James Harris
===
Subject: Re: JSH: About time
>> Now that IÕve revealed the odd and you could say esoteric
error
> in
>> core mathematics with such a short, and rather simple
argument,
> the
>> issue now is how long until mathematicians decide that
theyÕd
> rather
>> have correct mathematics versus the *belief* that they had
been
>> perfect in keeping error out of the collected body of work
that
> is
>> called mathematics.
>>
>> My work is out there and rather easy to go over as can be
seen
at
> the
>> Hong Konk math site:
>>
>>
>It does make sense. Basically I isolate terms of P(m) that
are
>*independent* of m.
>
> By which you mean P(0).
>
> Yes, as setting m=0 isolates terms that are *independent*
of m.
>
> HereÕs whatÕs key, all in a row:
>
> P(m) = g_1 g_2 g_3
>
> P(0) = u^2 f^2 (3x + uf)
>
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
>
> P(0)/f^2 = u^2 (3x + uf)
>
> and f is coprime to 3, x and u.
>
> I focus on whatÕs independent of m for a reason.
>
> ItÕs not really a subtle technique.
>
> Since the rest of your response is just repetition of
> this section, I will deal with that and ignore most
> of the rest.
Good. I hate those overlong posts.
> P(m) is a function of m. You compute P(0) and note
> that it is not divisible by f. You note as above,
> that if
> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
> then when m = 0, a1 = a2 = 0 and a3 = 3. You note
> that when m = 0, a1 and a2 are divisible by f.
I use setting m=0 to find what isnÕt dependent 
on m.
> You then say at great length that in considering
> m = 0, you are focussing on the part of P(m) which
> is independent of m. True, in the sense that P(0)
> does not change when m changes.
Yup.
> But after that your logic takes off into never-
> never land. You seem to somehow think that because
> you have dealt with the factorization of the constant
> term P(0), it tells you something about P(m) when
> m is not 0. But you never really establish a connection.
> You just repeat over and over the bit about
> independence.
Well, no, as I divide f^2 off so that I consider P(m)/f^2.
The point is that then you have
P(0)/f^2 = u^2 (3x + uf) as the constant term.
Seems too simple readers?
Well, if P(m) = g_1 g_2 g_3, and you have that
at m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf
then what happens with P(m)/f^2?
LetÕs say you have w_1 w_2 w_3 = f^2, so that you have
g_1/w_1 = uf/w_1, g_2/w_2 = uf/w_2, and g_3/w_3 = (3x+uf)/w_3.
Now then how many of you find yourselves incapable of 
figuring
out
what w_1, w_2, and w_3 are here?
The answer is that w_1 = w_2 = f, while w_3 = 1, if f is
coprime to 3.
And thatÕs given by, what?
By looking at P(0)/f^2, of course.
The math is basic.
> Then I say, and you agree, that a1 (and a2 and a3)
> are all functions of m: a1(m), a2(m), and a3(m).
> We agree that you can say that a1(0) = 0.
> What that means is that when you consider
> g1(0) = a1(0)*x + u*f,
> you can factor out f. What you get when you
> factor out f is
> g1(0) = u.
> Again, this happens because a1(0) = 0.
That doesnÕt make sense. You should have g_1/f, or in your
notation
g1(0)/f = u.
Hmmm...I noted that the poster Nora Baron also deleted out my
questions about math training as IÕve began wondering if 
this
poster
has any.
Well sci.math newsgroup, yet another person may have been
taking us
all for a ride. After all, how much math knowledge has this
poster
ever shown?
Well, posters do that on Usenet.
> Now you somehow make a great leap. You agree
> that a1(m) depends on m, and that in general
> a1(m) is not zero. But you think that one
> property of a1(0) carries over to a1(m) for
> other values of m: that it is divisible by f.
> But you simply do not say why.
The point is that then you have
P(0)/f^2 = u^2 (3x + uf) as the constant term.
Seems too simple readers?
Well, if P(m) = g_1 g_2 g_3, and you have that
at m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf
then what happens with P(m)/f^2?
LetÕs say you have w_1 w_2 w_3 = f^2, so that you have
g_1/w_1 = uf/w_1, g_2/w_2 = uf/w_2, and g_3/w_3 = (3x+uf)/w_3.
Now then how many of you find yourselves incapable of 
figuring
out
what w_1, w_2, and w_3 are here?
The answer is that w_1 = w_2 = f, while w_3 = 1, if 3 is
coprime to f.
And thatÕs given by, what?
By looking at P(0)/f^2, of course.
The math is basic.
> If your logic were correct it would apply
> to other functions.
> Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3.
> Say Q(m) is factored in the form
> Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f).
> Now note Q(0)/f^2 = x + f. That is,
> Q(0) = f * f * (x + f).
Hmmm...youÕve shown how to force an m dependency on the
constant term
as if m=1, then
Q(0) = f^3, and NOT f^2(x + f).
Good job.
For that to be a problem with my work, youÕd need
mf^2 = 1.
> Therefore when m = 0, we can say a1 = 0, a2 = 0,
> and a3 = 1.
> Note that when m = 0, a1 and a2 are divisible by f.
> Now by your logic, for values of m other than 0,
> a1 and a2 must be divisible by f and a3 is relatively
> prime to f.
Hmmm...the insinuation being that my logic is not logic?
> Do you agree with this? This is a test of your
> method. We need to know before we go to the next step.
> Nora B.
Actually your example, besides giving readers a way to make a
constant
term dependent on the main variable, shows that you
understand how the
argument works.
Now then, if m does not equal 1, then you have
Q(0) = f^2(x + f)
and, yup, two of the aÕs MUST have f as a factor.
Try it Nora Baron, and see how powerful mathematics truly is.
ItÕs not about your belief as youÕre 
irrelevant.
ItÕs mathematics.
(Um, I have NOT checked as I trust the math. LetÕs see if
this poster
or any of the rest of you can trip me up here.)
If m *does* equal 1, then all of the aÕs have f^{2/3} as a
factor as
then
Q(0) = f^3.
James Harris
===
Subject: Re: JSH: About time
>> Now that IÕve revealed the odd and you could say esoteric
error
> in
>> core mathematics with such a short, and rather simple
argument,
> the
>> issue now is how long until mathematicians decide that
theyÕd
> rather
>> have correct mathematics versus the *belief* that they had
been
>> perfect in keeping error out of the collected body of work
that
> is
>> called mathematics.
>>
>> My work is out there and rather easy to go over as can be
seen
at
> the
>> Hong Konk math site:
>>
>>
>
>
>It does make sense. Basically I isolate terms of P(m) that
are
>*independent* of m.
>
> By which you mean P(0).
>
> Yes, as setting m=0 isolates terms that are *independent*
of m.
>
> HereÕs whatÕs key, all in a row:
>
> P(m) = g_1 g_2 g_3
>
> P(0) = u^2 f^2 (3x + uf)
>
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
>
> P(0)/f^2 = u^2 (3x + uf)
>
> and f is coprime to 3, x and u.
>
> I focus on whatÕs independent of m for a reason.
>
> ItÕs not really a subtle technique.
>
>
>
> Since the rest of your response is just repetition of
> this section, I will deal with that and ignore most
> of the rest.
> Good. I hate those overlong posts.
> P(m) is a function of m. You compute P(0) and note
> that it is not divisible by f. You note as above,
> that if
>
> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
>
> then when m = 0, a1 = a2 = 0 and a3 = 3. You note
> that when m = 0, a1 and a2 are divisible by f.
> I use setting m=0 to find what isnÕt dependent 
on m.
> You then say at great length that in considering
> m = 0, you are focussing on the part of P(m) which
> is independent of m. True, in the sense that P(0)
> does not change when m changes.
> Yup.
> But after that your logic takes off into never-
> never land. You seem to somehow think that because
> you have dealt with the factorization of the constant
> term P(0), it tells you something about P(m) when
> m is not 0. But you never really establish a connection.
> You just repeat over and over the bit about
> independence.
HereÕs an important answer from me, and readers should read
the lead
off paragraph from Nora Baron and my reply carefully.
> Well, no, as I divide f^2 off so that I consider P(m)/f^2.
> The point is that then you have
> P(0)/f^2 = u^2 (3x + uf) as the constant term.
> Seems too simple readers?
> Well, if P(m) = g_1 g_2 g_3, and you have that
> at m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf
> then what happens with P(m)/f^2?
> LetÕs say you have w_1 w_2 w_3 = f^2, so that you have
> g_1/w_1 = uf/w_1, g_2/w_2 = uf/w_2, and g_3/w_3 =
(3x+uf)/w_3.
> Now then how many of you find yourselves incapable of
figuring out
> what w_1, w_2, and w_3 are here?
> The answer is that w_1 = w_2 = f, while w_3 = 1, if f is
coprime to 3.
> And thatÕs given by, what?
> By looking at P(0)/f^2, of course.
> The math is basic.
> Then I say, and you agree, that a1 (and a2 and a3)
> are all functions of m: a1(m), a2(m), and a3(m).
> We agree that you can say that a1(0) = 0.
> What that means is that when you consider
>
> g1(0) = a1(0)*x + u*f,
>
> you can factor out f. What you get when you
> factor out f is
>
> g1(0) = u.
>
> Again, this happens because a1(0) = 0.
> That doesnÕt make sense. You should have g_1/f, or in your
notation
> g1(0)/f = u.
> Hmmm...I noted that the poster Nora Baron also deleted out
my
> questions about math training as IÕve began wondering if
this poster
> has any.
> Well sci.math newsgroup, yet another person may have been
taking us
> all for a ride. After all, how much math knowledge has this
poster
> ever shown?
> Well, posters do that on Usenet.
> Now you somehow make a great leap. You agree
> that a1(m) depends on m, and that in general
> a1(m) is not zero. But you think that one
> property of a1(0) carries over to a1(m) for
> other values of m: that it is divisible by f.
>
> But you simply do not say why.
> The point is that then you have
> P(0)/f^2 = u^2 (3x + uf) as the constant term.
> Seems too simple readers?
> Well, if P(m) = g_1 g_2 g_3, and you have that
> at m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf
> then what happens with P(m)/f^2?
> LetÕs say you have w_1 w_2 w_3 = f^2, so that you have
> g_1/w_1 = uf/w_1, g_2/w_2 = uf/w_2, and g_3/w_3 =
(3x+uf)/w_3.
> Now then how many of you find yourselves incapable of
figuring out
> what w_1, w_2, and w_3 are here?
> The answer is that w_1 = w_2 = f, while w_3 = 1, if 3 is
coprime to f.
> And thatÕs given by, what?
> By looking at P(0)/f^2, of course.
> The math is basic.
> If your logic were correct it would apply
> to other functions.
>
> Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3.
>
> Say Q(m) is factored in the form
>
> Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f).
>
> Now note Q(0)/f^2 = x + f. That is,
>
> Q(0) = f * f * (x + f).
> Hmmm...youÕve shown how to force an m dependency on the
constant term
> as if m=1, then
That is incorrect. And worse, IÕve spent enough time talking
about
terms INDEPENDENT of m that I shouldnÕt have made the 
mistake.
It surprises me that I panicked.
> Q(0) = f^3, and NOT f^2(x + f).
> Good job.
> For that to be a problem with my work, youÕd need
> mf^2 = 1.
That is true, and it forces m and f to be units or forces you
out of
the ring of algebraic integers, which is why itÕs not an
issue.
> Therefore when m = 0, we can say a1 = 0, a2 = 0,
> and a3 = 1.
>
> Note that when m = 0, a1 and a2 are divisible by f.
>
> Now by your logic, for values of m other than 0,
> a1 and a2 must be divisible by f and a3 is relatively
> prime to f.
> Hmmm...the insinuation being that my logic is not logic?
> Do you agree with this? This is a test of your
> method. We need to know before we go to the next step.
>
> Nora B.
> Actually your example, besides giving readers a way to make
a constant
> term dependent on the main variable, shows that you
understand how the
> argument works.
> Now then, if m does not equal 1, then you have
> Q(0) = f^2(x + f)
> and, yup, two of the aÕs MUST have f as a factor.
> Try it Nora Baron, and see how powerful mathematics truly
is.
> ItÕs not about your belief as youÕre 
irrelevant.
> ItÕs mathematics.
> (Um, I have NOT checked as I trust the math. LetÕs see if
this poster
> or any of the rest of you can trip me up here.)
> If m *does* equal 1, then all of the aÕs have f^{2/3} as a
factor as
> then
> Q(0) = f^3.
Wrong.
ThatÕs totally incorrect as the point is that the constant
term does
NOT change as m changes, and it doesnÕt here either.
Figuring out the problem with the example from Nora Baron
requires
getting the equation defining the aÕs, and my 
guess is that
itÕs a
non-monic.
The point of my argument is that INDEPENDENT terms are in fact
independent.
And nothing can change that reality.
James Harris
===
Subject: Re: JSH: About time
> HereÕs an important answer from me, and readers should 
read
the lead
> off paragraph from Nora Baron and my reply carefully.
>
> Now note Q(0)/f^2 = x + f. That is,
>
> Q(0) = f * f * (x + f).
>
> Hmmm...youÕve shown how to force an m dependency on the
constant term
> as if m=1, then
> That is incorrect. And worse, IÕve spent enough time
talking about
> terms INDEPENDENT of m that I shouldnÕt have made the
mistake.
> It surprises me that I panicked.
> Q(0) = f^3, and NOT f^2(x + f).
Yes, this last line of yours is incorrect.
>
> Good job.
>
> For that to be a problem with my work, youÕd need
>
> mf^2 = 1.
> That is true,
Why is it true? In your own polynomial, m*f^2 is part
of the coefficient of the x^3 term. No different here
with my polynomial.
> and it forces m and f to be units or forces you out of
> the ring of algebraic integers, which is why itÕs not an
issue.
>
> Therefore when m = 0, we can say a1 = 0, a2 = 0,
> and a3 = 1.
>
> Note that when m = 0, a1 and a2 are divisible by f.
>
> Now by your logic, for values of m other than 0,
> a1 and a2 must be divisible by f and a3 is relatively
> prime to f.
>
> Hmmm...the insinuation being that my logic is not logic?
>
> Do you agree with this? This is a test of your
> method. We need to know before we go to the next step.
>
> Nora B.
>
> Actually your example, besides giving readers a way to make
a constant
> term dependent on the main variable, shows that you
understand how the
> argument works.
>
> Now then, if m does not equal 1, then you have
>
> Q(0) = f^2(x + f)
This is certainly true as it stands. But did you mean to say
Q(m) here ?
>
> and, yup, two of the aÕs MUST have f as a factor.
>
> Try it Nora Baron, and see how powerful mathematics truly
is.
>
I think you mean, try Q(m) for m <> 1. You think
I will find that two of the aÕs have f as a 
factor.
OK, IÕll try it. The polynomial is
Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3.
LetÕs try f = 7 and m = 2, for example:
Q(m) = 98*x^3 - 49*x + 7^3
= 49*(2*x^3 - x + 7).
Now suppose this is factored in the form
Q(m)/49 = ((a1/7)*x + 1)*((a2/7)*x + 1)*(a3*x + 7),
where a1/7 and a2/7 are algebraic integers.
Let b1 = a1/7. Then -1/b1 is a root of
2*x^3 - x + 7 = 0.
This implies
7*b1^3 + b1^2 - 2 = 0.
This is a *non-monic* polynomial with *integer
coefficients*, and it is *irreducible*. Therefore
b1 cannot be an algebraic integer. Therefore a1/7
cannot be an algebraic integer. The same argument
applies to a2/7.
Therefore it is not true that, in the
factorization of Q(m), a1 or a2 can have
f = 7 as a factor.
Recall that this was with m = 2.
So the behavior of Q(m) when m = 2
is quite similar to its behavior when m = 1.
You can show with a little more work that
EACH of a1, a2, and a3 must share nonunit factors
with f = 7, just as in the case m = 1, except
when m = 2 those factors are not all the same.
There is really nothing special about the
behavior of the factorizations for m = 1 or
m = 2. You will find the same thing for MOST
integers m. The reason: for most m, the polynomial
is irreducible. You see what a key role irreducibility
plays here. Of course, when m = 0, the polynomial
is of degree 1 and IS reducible. Thus it is m = 0
which is the special case, not m = 1.
> ItÕs not about your belief as youÕre 
irrelevant.
>
> ItÕs mathematics.
>
> (Um, I have NOT checked as I trust the math. LetÕs see if
this poster
> or any of the rest of you can trip me up here.)
>
Done. See above.
> If m *does* equal 1, then all of the aÕs have f^{2/3} as a
factor as
> then
>
> Q(0) = f^3.
> Wrong.
Of course itÕs wrong. Q(0) = f^2*(x + f), as I noted above.
> ThatÕs totally incorrect as the point is that the constant
term does
> NOT change as m changes, and it doesnÕt here either.
Also true.
> Figuring out the problem with the example from Nora Baron
requires
> getting the equation defining the aÕs, and my 
guess is that
itÕs a
> non-monic.
For m <> 1, good guess. See above.
> The point of my argument is that INDEPENDENT terms are in
fact
> independent.
Hard to argue with that.
> And nothing can change that reality.
ItÕs a tautology. But here it doesnÕt buy you 
squat,
as often happens with tautologies.
Nora B.
> James Harris
===
Subject: Re: JSH: About time
> P(m) = g_1 g_2 g_3
So at least one of g_1, g_2, g_3 is dependent on m? If so,
why not
indicate it.
YouÕre one hell of a sloppy mathematician. IÕm 
firmly
convinced that if
you stuck all dependencies and quantors in your proof that
youÕre
leaving out, you yourself would find the error.
V.
V.
===
Subject: Re: JSH: About time
> P(m) = g_1 g_2 g_3
> So at least one of g_1, g_2, g_3 is dependent on m? If so,
why not
> indicate it.
How does that change focusing on whatÕs constant by setting
m=0?
> YouÕre one hell of a sloppy mathematician. 
IÕm firmly
convinced that if
> you stuck all dependencies and quantors in your proof that
youÕre
> leaving out, you yourself would find the error.
> V.
> V.
IÕm NOT a mathematician. IÕm a discoverer 
facing
irrationality from
mathematicians refusing to accept a very basic and simple
argument,
apparently because they care less about math than social
realities.
Here the fear appears to be on the consequences of the truth
getting
out about an over hundred year old definition error at the
heart of
the body of work, which represents the efforts of many past
discoverers, routinely called mathematics.
TodayÕs mathematicians appear to be merely scholars now
determined
to hold on to their fantasy of perfect work by those
discoverers from
the past. Which isnÕt even fun.
James Harris
===
Subject: Re: JSH: About time
> P(m) = g_1 g_2 g_3
> So at least one of g_1, g_2, g_3 is dependent on m? If so,
why not
> indicate it.
> How does that change focusing on whatÕs constant by 
setting
m=0?
Focusing on m=0 is good. But what exactly happens when m is
not 0?
===
Subject: Re: JSH: About time
>> Now that IÕve revealed the odd and you could say esoteric
error
> in
>> core mathematics with such a short, and rather simple
argument,
> the
>> issue now is how long until mathematicians decide that
theyÕd
> rather
>> have correct mathematics versus the *belief* that they had
been
>> perfect in keeping error out of the collected body of work
that
> is
>> called mathematics.
> My work is out there and rather easy to go over as can be
seen at
> the
>> Hong Konk math site:
>> Hong Konk? YouÕre sure it isnÕt Honk Honk?
>Typo. It should be Hong Kong.
>> See <a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://math
db
.math.cuhk.edu.hk/forum/e_show.php?msg=782</a> And I send
people there because their allowal of the use of LaTeX
>> makes for a *much* better presentation, and given the
*social*
> issues
>> IÕm facing, I need all the help I can get.
>> This gives me the chance to re-post a discussion of the
LaTeX
>> website you mention above - maybe you would like to
respond -
>Sure.
> James Harris claims to prove, in
> <a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://math
db
.math.cuhk.edu.hk/forum/e_show.php?msg=782</a> that certain
polynomials factor in a form which
>> contradicts other mathematical proofs. Specifically:
>> Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
>> - 3*(-1 +m*f^2)*x*u^2 + u^3*f),
> where f is a prime, u is an integer coprime to f,
>> and m is an integer.
> Assume P(m) is factored in the form
> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
> where a1, a2, and a3 are algebraic integers.
> Let g1 = (a1*x + u*f), g2 = (a2*x + u*f), g3 = (a3*x + u*f).
> Note that P(0) = f^2*(3*x*u^2 + u^3*f).
>> Harris says:
> ... two of the aÕs go to 0 when m = 0 which
>> is also seen from the cubic defining the 
aÕs.
> Then arbitrarily picking a1 and a2 as the ones
>> that go to 0 at m = 0, you have
> g1 = u*f, g2 = u*f, g3 = 3*x + u*f.
> But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
>> only two of the gÕs while with the third it is
>> blocked, as long as it [f] is coprime to 3 and
>> x, so assume it is, and assume as well that f
>> is coprime to u.
> Then it follows from the constant terms that g1
>> and g2 each have a factor that is f.
> Remember, the constant terms with respect to m
>> cannot vary as m varies, or they wouldnÕt be
>> constant terms, right?
>> Sounds like it makes sense, including that last bit,
>> doesnÕt it? P(0) is the evaluation of the polynomial
>> when m = 0, so it must be the constant term.
>It does make sense. Basically I isolate terms of P(m) that
are
>*independent* of m.
> By which you mean P(0).
> Yes, as setting m=0 isolates terms that are *independent*
of m.
> HereÕs whatÕs key, all in a row:
> P(m) = g_1 g_2 g_3
> P(0) = u^2 f^2 (3x + uf)
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
> P(0)/f^2 = u^2 (3x + uf)
> and f is coprime to 3, x and u.
> I focus on whatÕs independent of m for a reason.
> ItÕs not really a subtle technique.
>Then I look at at the same terms for P(m)/f^2 and find that a
factor
>of f^2 has been removed.
>Logic dictates that the removal is *independent* of m,
> Removal here means that you divide P(m) by f^2. That does
> not cause a problem in itself. It is really how f^2 is
> distributed among the factors
> ai*x + u*f
> that is the issue.
> Actually, what I do is note factors *independent* of m, so
they canÕt
> suddenly gain a dependency on m, while this poster seems
fascinated
> with the aÕs being themselves functions of m, as if that
destroys the
> general principle.
> At best itÕd mean you can also separate them out as well,
using the
> same trick, which is to note their values when m=0, and it
turns out
> that for only *two* of the aÕs, when m=0, they are 0 as
well, which
> tells you something there.
> Now then, the uf simply cannot lose an f as a function of m
or
> dependent on m, as itÕs *independent* of m, so the logic 
is
simple.
> So this poster keeps making posts trying to refute the fact
that uf is
> independent of m, as if somehow thereÕs someway that math 
is
> illogical, but itÕs not.
> You do not disagree that ai is dependent
> on m. LetÕs write it as ai(m).
> Yup, itÕs dependent on m, but notice it *still* is the 
case
that you
> can isolate whatÕs independent of m by setting m=0, though
with the
> aÕs, two of them then become 0, while one is 3x + uf, 
which
youÕll
> notice is *independent* of m.
> The principle is simple. WhatÕs fascinating is any person
trying to
> argue against it, as this Nora Baron is doing.
> Therefore the way in which factors of f divide
> ai(m) in general can be expected to depend on m.
> Your expectation--your intuition--is meaningless in the
face of
> mathematical logic.
> Your gut feeling is trash if it goes against the math.
> Here the key term is uf, which is *independent* of m, so it
canÕt vary
> on m.
> However, this poster *wants* it to vary on m, so the poster
refuses to
> accept the math and notice, now talks about whatÕs 
expected.
> ItÕs a rather pathetic display which highlights what 
IÕve
said--human
> beings are NOT rational, as they rely strongly on social
forces, even
> with mathematics.
> You prove that for m = 0, f divides ai(0).
> I isolate the terms independent on m, deliberately, so that
I can see
> how they vary, without worrying about mÕs value, and it
just so
> happens that the way to find those terms independent of m,
is to set
> m=0, which is not rocket science.
> You think somehow that that proves f divides ai(m)
> for all m.
> YouÕre showing a gross lack of logic as in fact I simply
rely on the
> terms being independent of m, and the fact that P(0)/f^2
equals u^2(3x
> + uf) so if f is coprime to 3, x, and u, it must be the
case that the
> factors that were g_1, g_2, and g_3 now no longer have f as
a factor
> either.
> Now since at m=0, remember thatÕs to get terms independent
of m,
> g_1=uf, g_2=uf, but g_3 = 3x + uf, it stands to reason that
f divides
> off of g_1 and g_2, based on the terms *independent* of m,
so the
> value of m is not of consequence.
> ThatÕs the point of isolating terms independent of m, so
that I now
> they are independent of its value, which means that you
canÕt claim
> that things change just because of mÕs value.
> Clearly, logically, two of the gÕs have f as a factor,
without regard
> to mÕs value, when f is coprime to 3, x, and u.
> LetÕs say f = 5 and ai(m) = sqrt(5) * m + 15.
> Certainly f = 5 divides ai(0) = 15.
> But f = 5 does not divide ai(1) = sqrt(5) + 15.
> Which is why I focus on the terms *independent* of m.
> Here is another example. Say ai(m) = 5 * m, and f = 3.
> The ai(0) = 0, and, as in your case, f divides ai(0).
> However, if m = 1, ai(m) = 5, and f = 3 does not divide
> a1(m) = 5. In fact f = 3 is relatively prime to 5,
> even in the algebraic integers.
> However, what I *actually* do is focus on those terms
INDEPENDENT of
> m, so that I donÕt care what mÕs value is, 
and in fact it is
> irrelevant, which is the point.
> The technique for isolating those terms is to set m=0.
> HereÕs whatÕs key, all in a row:
> P(m) = g_1 g_2 g_3
> P(0) = u^2 f^2 (3x + uf)
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
> P(0)/f^2 = u^2 (3x + uf)
> and f is coprime to 3, x and u.
> I focus on whatÕs independent of m for a reason.
> ItÕs not really a subtle technique.
> I am not saying that in your application, ai(m)
> actually equals 5 * m. I am simply saying that your logic
> breaks down. You have not shown that ai(m), as a
> function of m, does not behave something like
> ai(m) = 5 * m. You have nothing explicit whatsoever
> about ai(m), EXCEPT when m = 0. But that clearly does
> not tell you anything about how ai(m) behaves with
> respect to f when m = 0. ThatÕs the point.
> Put it another way. You have shown that ai(0)
> is divisible by f. Since ai(m) is a function of
> m which you probably cannot even write down,
> you do not know anything about the divisibility
> of ai(m) by f when m <> 0. It has NOTHING to do
> with whether ai(0) or P(0) or whatever is
> independent of ai(m), or your claim that you
> have somehow removed the constant term. You
> are badly confused on that point.
> ThatÕs against mathematical logic as if the terms that are
independent
> of m, are indeed independent of m, then how can their
factors vary
> with m?
> Here you have u^2 f^2 (3x + uf) in one case, and u^2 (3x +
uf) as the
> other case.
> The are *independent* of m, and that remains true no matter
how much
> you argue Nora Baron.
> Now then, given that at m=0, those terms isolated out as
independent
> of m, are for g_1 and g_2, uf, and for g_3, 3x + uf, logic
dictates
> how the factors MUST distribute out.
> Here the factors independent of m force a constraint on
those that are
> dependent on m, in a way thatÕs fascinating and
understandable using
> basic algebra.
> ItÕs a neat trick.
>but that is a
>conclusion several posters wish to convince others is false,
so here
> I
>am, once again, arguing about the obvious.
> WhatÕs obvious here is that you have leaped from
> m = 0 to all other values of m with no justification.
> Can you not understand that if terms are *independent* of m
then I can
> leap to any values for m that I choose?
> I can have m=2345403840, and it does NOT matter as those
terms are
> independent of m.
> So the justification is the independence from m.
> That independence follows from the simple act of setting
m=0 to
> isolate out terms NOT dependent on m.
>> Of course all of what I quoted above except that last
>> paragraph pertains to what happens when m = 0. For
>> instance, saying that a1 = a2 = 0, that is true only when
>> m = 0. When m is nonzero, we know that a1 and a2 are also
>> nonzero.
>>Readers need focus on the simple fact that with P(m)
considering
> P(0),
>that is setting m=0, gives you terms that donÕt have m, so
they are
>independent of it.
>Stay focused on that fact as Nora Baron tries to take you
for a
>ride.
>> So itÕs clear that a1 and a2 depend on m.
> I donÕt think even James Harris disagrees with that.
> So when he says g1 = u*f and g2 = u*f, that is true only
>> when m = 0. When m is not zero, you get, for example,
> g1 = a1*x + u*f.
> When Harris says constant term he does not mean
>> constant term with respect to x. He is dealing with what
>> he calls a nonpolynomial factorization. The variable here
>> not as P(x).
> As noted above, a1 and a2 are dependent on m. Both a1
>> and a2 should really be written as a1(m) and a2(m).
> A key step in HarrisÕs argument is this :
> But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
>> only two of the gÕs while with the third it is
>> blocked
> When Harris says blocked he means that f is not
>> a divisor of the third g.
>Readers remember, with P(m), I set m=0, that is, look at
P(0), to get
>terms *independent* of m, as m has been set to 0, so itÕs
gone.
> Looking at m = 0 does not mean, in any sense that
> m is gone.
> I can *give* whatÕs left as it is
> P(0) = u^2 f^2 (3x + uf)
> and yes, whether you wish to accept it or not, m IS gone.
>Then I do the same for P(m)/f^2,
> No, you absolutely donÕt. You never really deal
> with P(m).
> Now I can just give P(0)/f^2.
> P(0)/f^2 = u^2 (3x + uf)
> and again, thereÕs no m, and now a factor of f^2 has gone
away.
> When m = 0, a1*x + u*f = u*f. This is divisible
> by f because a1 = 0 also.
> Yes. But because IÕve isolated out terms independent of m,
and follow
> them, they force a constraint on those terms dependent on m
*because*
> they are independent of m.
> When m <> 0, a1 is not equal to 0. You donÕt know
> its value. In particular, you donÕt know that it
> is divisible by f. Just SAYING that it is is not
> sufficient.
> and notice that for those terms
>*independent* of m, a factor of f^2 has disappeared.
>Now you can see that for P(0) I have
>P(0) = u^2 f^2( 3x + uf)
>and for P(0)/f^2 I have
>P(0)/f^2 = u^2 ( 3x + uf)
>so everything is simple enough.
> Yes: For m = 0. We agree on this. LetÕs
> try to get past it. Divisibility of a1 by f
> when m equals 0 does not tell you anything about
> divisibility of a1 when m <> 0. a1 is dependent
> on m in a very complicated way. You have never
> tried writing a1 out as a function of m. It is
> however fairly easy to show that for m = 1 and
> f = 5, a1 CANNOT be divisible by f.
> ItÕs possible to show because thereÕs a 
definition error in
core
> mathematics, which allows for the *appearance* of two
dueling proofs.
> But proofs donÕt duel.
> The fix is to ignore the ßawed definition 
and rely on basic
axioms,
> which reveal the truth.
> Here IÕve pointed out repeatedly that the proof depends on
focusing on
> terms independent of m, and those terms are found easily
enough by
> setting m=0, with P(m), P(m)/f^2 and the factors of P(m),
g_1, g_2 and
> g_3.
>> Of course here he is obviously talking about m = 0,
>> though he would like to conclude this for m <> 0.
>HereÕs where you have Nora Baron sneaking in something
rather dumb
>mathematically.
> I am not sneaking anything in anywhere. Everything
> is above board. What you see is what you get.
> ItÕs dumb mathematically to insinuate that something
different will
> happen for terms *independent* of m based on the value of
m, as if it
> matter whether or not m=0 or not.
>If my point is isolating off m, so that I have terms
*independent* of
>m, why canÕt I conclude that they are indeed independent of
m?
> Look. NO ONE is arguing here about P(0). That is not
> at issue. We are arguing about P(m) when m <> 0.
> The fact that P(0) does not change when m changes
> does not imply that the factors of the form
> ai*x + u*f
> also do not change when m changes. In fact we KNOW
> that ai must be a factor of m ! Why? Because when
> m = 0, ai = 0. But when m <> 0, we know *for sure*
> that ai is not zero [otherwise the cubic would have
> degree 1]. Thus ai DEPENDS ON m. Therefore you cannot
> assume that something about it which is true when m
> equals 0 is also true when m <> 0.
> Yes, the aÕs *are* dependent on m, but the terms that are
NOT force a
> constraint on those that are because they arenÕt.
> You have to follow *logic* which dicates that terms
independent of m
> are indeed independent of m.
>ItÕs a bizarre thing to question and maybe many of you
simply canÕt
>believe that a poster like Nora Baron would keep posting and
making
>a fuss based on such a position.
>But remember, I *isolate* independent factors of P(m), by
setting
> m=0,
>so I donÕt have to worry about what value m has.
> What you say about P(0) applies only when m = 0.
> You are just confusing yourself with verbiage here
> like isolate and independent factors. As noted
> above, in the factors (ai*x + u*f), the numbers
> ai are NOT CONSTANTS. They are dependent on m.
> However, setting m=0 necessarily gives you whatÕs NOT
dependent on m.
> When you focus on whatÕs not dependent on 
mÕs value then
you are
> forced into a conclusion where two of the gÕs have f as a
factor, when
> f is coprime to 3, x and u, while one does not.
> The math is rather simple and basic, if you will accept
that whatÕs
> NOT dependent on the value of m, is in fact, not dependent
on the
> value of m.
>> He is saying that when m = 0, the only way you are going
>> to factor out f^2 from the expression
> (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f)
> is:
> f^2*([a1(0)/f]*x + u)*([a2(0)/f]*x + u)*(a3(0)*x + u*f),
> because a1(0) = a2(0) = 0 and a3(0) = 3:
> f^2 * (u) * (u) * (3*x + u*f).
> So far, just fine. This is all assuming m = 0.
>And consider that the poster Nora Baron is considering terms
which
>have isolated out as INDEPENDENT of m, which is why I set
m=0, in the
>first place!
> What you are saying is that showing that a1 is
> divisible by f when m = 0 is sufficient to show that
> a1 is divisible by f when m <> 0. It is just plain
> not true.
> No IÕm not saying that as what IÕm saying is 
that focusing
on those
> terms that are independent on m, gives a conclusion which
forces
> itself back upon those that are.
> It does so because the terms independent of m are
independent of m.
> Is it really so subtle of a point?
>ItÕs like this poster is going on and on about a rather
simple
>technique for pulling out factors independent of a particular
>variable, as if itÕs actually sinister, where *really* 
IÕm
just
> trying
>to fool people with a special case, as if things change when
m does
>not equal 0.
> Actually I donÕt think you are trying to fool people. You
> canÕt help doing that because you have fooled yourself.
> How can I have fooled myself when the basic principle is
that whatÕs
> independent of m is independent of m?
> You, however, refuse to settle down but instead make VERY
long posts
> where you keep dancing around that simple truth.
>But if these terms change as m changes, then theyÕre not
independent
>of m, now are they?
> Precisely. a1, for example, is dependent on m. It might
> be, for all I know, a1(m) = sqrt(m) + cuberoot(m). Note
> that when m = 0, a1(m) = 0 + 0 = 0. Note that when m = 64,
> a1(m) = 8 + 4 = 12. If f = 5, then f divides a1(0) = 0.
> But that does NOT imply that f divides a1(64) = 12. In fact
> 5 and 12 are relatively prime.
> Your reasoning is specious as itÕs nothing like what I
actually do,
> which involves considering the constant terms P(0) and
P(0)/f^2.
> I note for readers that this poster clearly has little
intention of
> actually being rational, but may simply believe that
continuing to
> disagree, even after soundly being refuted is all thatÕs
necessary to
> keep many of you convinced that IÕm wrong.
>Nora Baron is trying to refute algebra!!! It should be a news
> ßash
>around the math world that setting m=0, gives terms that are
STILL
>dependent on m,
> No one claims that. What you are saying is just
> irrelevant. DonÕt call CNN quite yet unless you want
> to be embarrassed on an international scale.
> What you claim is that if a function has a certain
> property when the argument is 0, then it has that
> same property for all other arguments. For example,
> say f(m) = 3*m + 7. Note that f(0) is divisible by
> 7. By your logic, f(m) would be divisible by 7 for all
> other values of m. But then you try m = 1, and
> find that f(1) = 10 is NOT divisible by 7. Therefore
> your logic was faulty. You were right about f(0),
> but it doesnÕt generalize to other values of m.
> Your example is specious as I focus on whatÕs independent
of m, using
> P(0) AND P(0)/f^2 as well as the factors g_1, g_2 and g_3.
> ALL of that information, along with the requirement that f
is coprime
> to 3, x and u is necessary for the conclusion.
> Here this poster keeps acting as if factors independent of
m are still
> somehow dependent on m, which is just not true.
> which I call the Shadow m. It has supernatural powers
> and refuses to go away, even when set to 0.
> You are deeply confused. You are digging a
> deeper hole for yourself.
> How?
>> There are infinitely many ways that f^2 can be written
>> as a product of three algebraic integers. Say, for
>> example, one of them is f^2 = f1 * f2 * f3. Assume that
>> none of f1, f2 or f3 is a unit in the algebraic integers.
> Now choose an integer m <> 0. Suppose a1(m) is
>> divisible by, say, f1. Suppose that f/f1 is also an
>> algebraic integer. Then
> a1(m)*x + u*f = f1*([a1(m)/f1]*x + u*f/f1).
> And assume similar things for g2 and g3:
> a2(m)*x + u*f = f2*([a2(m)/f2]*x + u*f/f2),
> a3(m)*x + u*f = f3*([a2(m)/f3]*x + u*f/f3),
> and all of a1(m)/f1, a2(m)/f2, a3(m)/f3, f/f1, f/f2, and
>> f/f3 are algebraic integers.
> Putting all this together, one has
> P(m)/f^2 = P(m)/(f1*f2*f3)
> = ([a1(m)/f1]*x + u*f/f1) *
>> ([a2(m)/f2]*x + u*f/f2) *
>> ([a3(m)/f3]*x + u*f/f3).
> The key thing here is that these factors f1, f2, and f3
>> may ALSO be dependent on m. When Harris has shown is that
>> when m = 0, f1 = f2 = f, and f3 = 1. Or you could say,
>> f1(0) = f2(0) = f and f3(0) = 1.
>Which actually blows you out of the ring of algebraic
integers, and
>worse provably, if the fÕs are *functions* of m, then f1, 
f2
and f3
>have zeroÕs,
> Ridiculous. Are you now saying all functions have zeros???
> Hmmm...how much math do you actually know Nora Baron?
> at which point the equation would blow up. Remember
> dividing by 0 is a no-no.
>BUT, remember this poster above has
>> Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
>> - 3*(-1 +m*f^2)*x*u^2 + u^3*f),
>Well then, how can P(m)/f^2 introduce such a problem as
blowing up
> for
>some value that where f1, f2 or f3 is 0?
> Again, are you claiming that all functions have zeros???
> Answer my questions first, like, whatÕs your 
area of
expertise?
> And where were you trained as a mathematician?
That is irrelevant as you are not a trained mathematician
either.
>Besides, if I find terms that are *independent* of m, then
they are
>independent of m, so the same terms can be isolated in each
factor of
>> = ([a1(m)/f1]*x + u*f/f1) *
>> ([a2(m)/f2]*x + u*f/f2) *
>> ([a3(m)/f3]*x + u*f/f3).
>>If you do so, youÕll find that the terms 
independent of m,
for *two*
>factors, must have a factor that is f, while one does not.
>Here Nora Baron needs you to believe that setting m=0
doesnÕt in
>fact give you terms *independent* of m, but thatÕs just
> mathematically
>wrong.
> For the n-th time: I have NO ARGUMENT with your
> statement that a1(m) = 0 when m = 0. That is JUST
> FINE. The problem is that a1(m) for m <> 0 is
> NOT *independent* of m, and you need it to be
> divisible by f when m <> 0. And proving that fact
> for m = 0 gets you nowhere whatsoever when what you
> need is divisibility of a1(m) when m <> 0.
> But I focus on whatÕs independent of m to draw my
conclusion, while
> you keep trying to claim a dependency on m, as if itÕs
somehow beyond
> your capacity to understand that setting m=0 is a technique
for
> pulling out whatÕs not dependent on m.
> I find your continuing argument odd, and I think 
itÕs time
you
> revealed your mathematical training.
> It occurs to me that you may have none.
> I am beginning to think you are just *hopelessly*
> confused on this. It is complicated greatly
> by the fact that you want oh-so-desperately not
> to be wrong. You are blinding yourself to the
> obvious as has happened many times before.
> WhatÕs your training?
>But Nora Baron is working to convince you.
> No, I am working to convince YOU. I donÕt think
> there is a single other reader out there, other that
> you, that is actually confused about this. I am
> not playing to the peanut gallery. This is strictly
> an attempt to get you to understand. So far it is
> a ßop.
> YouÕre posting irrationally, and IÕm not 
that curious about
why, but
> then again, it might help if you give your training in math.
>WhatÕs key here is the claim of dependency on m, but
remember, I set
>m=0 to *remove* dependency on m,
> Setting m = 0 in no way removes dependency on m.
> All you get when you set m = 0 are facts about the
> m = 0 case. But you donÕt even need facts about
> the m = 0 case. You need facts about the m = 1
> m = 2, ..., m = 211, etc. cases. The m = 0 case
> tells you NOTHING about the cases you actually need
> to consider.
> ThatÕs not rational as rationally setting m=0 with P(m) 
and
its
> factors gives me whatÕs not dependent on m.
> but now the Shadow m has apparently
> returned according to Nora Baron!!!
> Right! m = 0 is all well and good, and
> your statements about that case are correct. What you
> need are facts about all the other possible values
> of m. You have proved nothing on that.
> But, how can that be possible if I focus on whatÕs
independent of m?
> If I focus on whatÕs independent of m, then 
isnÕt it
independent of m?
> HereÕs whatÕs key, all in a row:
> P(m) = g_1 g_2 g_3
> P(0) = u^2 f^2 (3x + uf)
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
> P(0)/f^2 = u^2 (3x + uf)
> and f is coprime to 3, x and u.
> I focus on whatÕs independent of m for a reason.
> ItÕs not really a subtle technique.
>> I donÕt claim that I have shown here that f^2 can be
>> factored so that all of the above fits together. What I
>> claim here is that Harris has not shown that it CANÕT
>> happen. As long as that gap is left open, he does not
>> have a proof.
>ThereÕs no gap, as itÕs just Nora Baron now 
switching to
trying to
>show reasonable doubt, I guess.
>However, as I emphasized repeatedly through this post, the
technique
>of setting m=0, does actually work, as it shows factors
independent
> of
>m, and allowing a poster like Nora Baron to question such
basic
>algebra, is proof that many of you care more about society
than math.
>After all, you *want* to believe her so you can believe IÕm
wrong.
> ThereÕs no need to grandstand here. I am not trying to
> convince anyone else; just you.
> <remainder deleted because of intriguing assertion> Then
email me instead of posting.
> I dare you.
> James Harris
===
Subject: Re: JSH: About time
Visiting Assistant Professor at the University of Montana.
> Now that IÕve revealed the odd and you could say esoteric
error
> core mathematics with such a short, and rather simple
argument,
>the
> issue now is how long until mathematicians decide that
theyÕd
>rather
> have correct mathematics versus the *belief* that they had
been
> perfect in keeping error out of the collected body of work
that
> called mathematics.
>
> My work is out there and rather easy to go over as can be
seen at
>the
> Hong Konk math site:
>
>
> Hong Konk? YouÕre sure it isnÕt Honk Honk?
>>Typo. It should be Hong Kong.
> See <a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://mathd
b.math.cuhk.edu.hk/forum/e_show.php?msg=782</a>
> And I send people there because their allowal of the use of
LaTeX
> makes for a *much* better presentation, and given the
*social*
>issues
> IÕm facing, I need all the help I can get.
>
>
> This gives me the chance to re-post a discussion of the
LaTeX
> website you mention above - maybe you would like to respond
-
>>Sure.
>
> James Harris claims to prove, in
>
> <a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://mathd
b.math.cuhk.edu.hk/forum/e_show.php?msg=782</a>
> that certain polynomials factor in a form which
> contradicts other mathematical proofs. Specifically:
>
>
> Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
> - 3*(-1 +m*f^2)*x*u^2 + u^3*f),
>
> where f is a prime, u is an integer coprime to f,
> and m is an integer.
>
> Assume P(m) is factored in the form
>
> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
>
> where a1, a2, and a3 are algebraic integers.
>
> Let g1 = (a1*x + u*f), g2 = (a2*x + u*f), g3 = (a3*x + u*f).
>
> Note that P(0) = f^2*(3*x*u^2 + u^3*f).
>
>
> Harris says:
>
> ... two of the aÕs go to 0 when m = 0 which
> is also seen from the cubic defining the aÕs.
>
> Then arbitrarily picking a1 and a2 as the ones
> that go to 0 at m = 0, you have
>
> g1 = u*f, g2 = u*f, g3 = 3*x + u*f.
>
> But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
> only two of the gÕs while with the third it is
> blocked, as long as it [f] is coprime to 3 and
> x, so assume it is, and assume as well that f
> is coprime to u.
>
> Then it follows from the constant terms that g1
> and g2 each have a factor that is f.
>
> Remember, the constant terms with respect to m
> cannot vary as m varies, or they wouldnÕt be
> constant terms, right?
>
>
> Sounds like it makes sense, including that last bit,
> doesnÕt it? P(0) is the evaluation of the polynomial
> when m = 0, so it must be the constant term.
>>It does make sense. Basically I isolate terms of P(m) that
are
>>*independent* of m.
> By which you mean P(0).
>>Then I look at at the same terms for P(m)/f^2 and find that
a factor
>>of f^2 has been removed.
>>Logic dictates that the removal is *independent* of m,
> Removal here means that you divide P(m) by f^2. That does
>not cause a problem in itself. It is really how f^2 is
>distributed among the factors
> ai*x + u*f
>that is the issue.
> You do not disagree that ai is dependent
>on m. LetÕs write it as ai(m).
> Therefore the way in which factors of f divide
>ai(m) in general can be expected to depend on m.
> You prove that for m = 0, f divides ai(0).
> You think somehow that that proves f divides ai(m)
>for all m.
> LetÕs say f = 5 and ai(m) = sqrt(5) * m + 15.
> Certainly f = 5 divides ai(0) = 15.
> But f = 5 does not divide ai(1) = sqrt(5) + 15.
> Here is another example. Say ai(m) = 5 * m, and f = 3.
>The ai(0) = 0, and, as in your case, f divides ai(0).
>However, if m = 1, ai(m) = 5, and f = 3 does not divide
>a1(m) = 5. In fact f = 3 is relatively prime to 5,
>even in the algebraic integers.
> I am not saying that in your application, ai(m)
>actually equals 5 * m. I am simply saying that your logic
>breaks down. You have not shown that ai(m), as a
>function of m, does not behave something like
>ai(m) = 5 * m. You have nothing explicit whatsoever
>about ai(m), EXCEPT when m = 0. But that clearly does
>not tell you anything about how ai(m) behaves with
>respect to f when m = 0. ThatÕs the point.
> Put it another way. You have shown that ai(0)
>is divisible by f. Since ai(m) is a function of
>m which you probably cannot even write down,
>you do not know anything about the divisibility
>of ai(m) by f when m <> 0. It has NOTHING to do
>with whether ai(0) or P(0) or whatever is
>independent of ai(m), or your claim that you
>have somehow removed the constant term. You
>are badly confused on that point.
In the specific case at issue, where the values of a1, a2, a3
are just
the roots of a cubic, I once worked out the explicit formulas
of a1,
a2, and a3 in terms of m and f. Maybe that will help to fix 
the
discussion?
I got the formulas from the Cardano formulas, applied to
x^3 -3vx^2 + (v^3+1)
where v = -1+mf^2, which I believe is the case James is
dealing
with. If it is not, then of course my calculations are
useless for the
purposes of giving something to fix on.
a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
a2 = m*f^2-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
+ [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
a3 = m*f^2-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
+ [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
where
C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
+ 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2.
D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2
(IÕm pretty sure I did not make any mistakes, but if anyone
spots
any, let me know).
So, what happens when m=0?
When m=0, you get C=D=0, so
a1 = m*f^{2}-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
= -1 + (1/2^{1/3})*[cuberoot(-2) + cuberoot(-2)].
= -1 + 2^{-1/3}*(-2^{1/3}-2^{1/3})
= -1 + (-1) + (-1) = 3.
a2 = m*f^{2}-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
+ [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
= -1 + [(-1+sqrt(-3))/2]*2^{-1/3}(-2^{1/3}) +
[(-1-sqrt(-3))/2]*2^{-1/3}(-2^{1/3})
= -1 - (-1+sqrt(-3))/2 - (-1-sqrt(-3))/2
= -1 +(1/2) - sqrt(-3)/2 + (1/2) + sqrt(-3)/2
= 0.
a3 = m*f^{2}-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
+ [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
= -1 + [(-1+sqrt(-3))/2]*2^{-1/3}(-2^{1/3}) +
[(-1-sqrt(-3))/2]*2^{-1/3}(-2^{1/3})
= -1 - (-1+sqrt(-3))/2 - (-1-sqrt(-3))/2
= -1 +(1/2) - sqrt(-3)/2 + (1/2) + sqrt(-3)/2
= 0.
So, if we write
g_1(m) = [g_1(m) - g_1(0)] + g_1(0)
= [g_1(m) - (3x+1)] + (3x+1)
this is supposed to somehow tells us something about g_1(m) in
general. Specifically, about how it is divisible by f.
Can we get that
a1 = m*f^{2}-1 +
(1/2^{1/3})*(cuberoot(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
+ 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^{2}.
D = m^3*f^{6} - 3*m^2*f^{4} + 3*m*f^{2}
is necessarily a multiple of f whenever m is not equal to
zero? C and
D are multiples of f, but look at the expression.
C is a multiple of f^2 but, if f is coprime to 12m (which
cannot occur
when m is zero, but can happen for other values), then C/f^2
is
coprime to f. D is a multiple of f^2, and if it is coprime to
3m then
D/f^2 is coprime to f. So sqrt(C) is a multiple of f, but
sqrt(C)/f is
coprime to f. So D+sqrt(C) would be divisible by f, and we
would
have (D+sqrt(C))/f = (D/f) + (sqrt(C)/f), which would be
coprime to
f. If 2 is coprime to f, then D + sqrt(C)-2 is coprime to f,
in which
case cuberoot(D-2+sqrt(C)) is coprime to f, as is
cuberoot(D-2-sqrt(C)).
So then a1 is: a multiple of f, minus 1, plus two things
which are
coprime to f added together. So you have a multiple of f plus
three
things which are coprime to f. That could be coprime to f or
not
coprime to f, and there is no reason to believe that it would
necessarily be divisible to f in the latter case.
Why do you conclude that this is ALWAYS coprime to f? The sum
of two
things coprime to f could be non-coprime to f, as the trivial
example
of 3 and 5 show with respect to 2.
And say instead that f is equal to 2, and m is odd. Then
D-2+sqrt(C)
is a multiple of 2; D is a multiple of 4, and D/4 is coprime
to 2. C
is a multiple of 16, and C/16 is coprime to 2. So sqrt(C) is a
multiple of 4, and sqrt(C)/4 is coprime to 2. So D-2+sqrt(C)
is a
multiple of 2, and (D-2+sqrt(C))/2 = (D/2) + (sqrt(C)/2) - 1,
which is
coprime to 2. So cuberoot(D-2+sqrt(C)) is a multiple of
2^{1/3}, and
cuberoot(D-2+sqrt(C))/2^{1/3} is coprime to 2. So after
dividing by
2^{1/3} you get something which is coprime to 2, so you end
up with a
multple of f minus 1, plus something which is coprime to 2,
plus
something which is coprime to 2 (same argument holds for
D-2-sqrt(C)).
If they were all integers, you would have something even,
minus 1,
plus something odd plus something odd, so you would indeed
conclude
that it is odd (coprime to 2). But here you could have the
sum of two
things which are coprime to 2 once again being coprime to 2,
or the
sum of two things coprime to 2 being not coprime to 2 but not
a
multiple of 2. Yet you think we can always conclude that it
is coprime
to 2. Why?
And why can you conclude that
a2 = m*f^{2}-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
+ [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
will be divisible by f, just because when m=0 you got 0? Just
because,
the -1?
They cannot be expressed as (something)*f + a2(0), as you
seem to
think they can. If f=2, again we have that
cuberoot(D-2+sqrt(C))/2^{1/3} should be coprime to 2, so now
you have
the sum of m*f^2 and THREE things coprime to 2. If they were
integers
you would conclude that the resulting number is an even minus
1 plus
two odds, which would make it odd again, yet here you are
concluding
that they will be even. Why?
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 manÕs 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
magidin@math.berkeley.edu
===
Subject: Re: JSH: About time
> Now that IÕve revealed the odd and you could say esoteric
error
> in
> core mathematics with such a short, and rather simple
argument,
> the
> issue now is how long until mathematicians decide that
theyÕd
> rather
> have correct mathematics versus the *belief* that they had
been
> perfect in keeping error out of the collected body of work
that
> is
> called mathematics.
>
> My work is out there and rather easy to go over as can be
seen at
> the
> Hong Konk math site:
>
>
> Hong Konk? YouÕre sure it isnÕt Honk Honk?
>Typo. It should be Hong Kong.
>>
> See <a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://mathd
b.math.cuhk.edu.hk/forum/e_show.php?msg=782</a>
> And I send people there because their allowal of the use of
LaTeX
> makes for a *much* better presentation, and given the
*social*
> issues
> IÕm facing, I need all the help I can get.
>
>
> This gives me the chance to re-post a discussion of the
LaTeX
> website you mention above - maybe you would like to respond
-
>Sure.
>>
> James Harris claims to prove, in
>
> <a
href=http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>
http://mathd
b.math.cuhk.edu.hk/forum/e_show.php?msg=782</a>
> that certain polynomials factor in a form which
> contradicts other mathematical proofs. Specifically:
>
>
> Let P(m) = f^2*(m^3*f^4 -3*m^2*f^2 + 3*m)*x^3
> - 3*(-1 +m*f^2)*x*u^2 + u^3*f),
>
> where f is a prime, u is an integer coprime to f,
> and m is an integer.
>
> Assume P(m) is factored in the form
>
> P(m) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
>
> where a1, a2, and a3 are algebraic integers.
>
> Let g1 = (a1*x + u*f), g2 = (a2*x + u*f), g3 = (a3*x + u*f).
>
> Note that P(0) = f^2*(3*x*u^2 + u^3*f).
>
>
> Harris says:
>
> ... two of the aÕs go to 0 when m = 0 which
> is also seen from the cubic defining the aÕs.
>
> Then arbitrarily picking a1 and a2 as the ones
> that go to 0 at m = 0, you have
>
> g1 = u*f, g2 = u*f, g3 = 3*x + u*f.
>
> But P(0)/f^2 = 3*x*u^2 + u^3*f as f divides off
> only two of the gÕs while with the third it is
> blocked, as long as it [f] is coprime to 3 and
> x, so assume it is, and assume as well that f
> is coprime to u.
>
> Then it follows from the constant terms that g1
> and g2 each have a factor that is f.
>
> Remember, the constant terms with respect to m
> cannot vary as m varies, or they wouldnÕt be
> constant terms, right?
>
>
> Sounds like it makes sense, including that last bit,
> doesnÕt it? P(0) is the evaluation of the polynomial
> when m = 0, so it must be the constant term.
>It does make sense. Basically I isolate terms of P(m) that
are
>>*independent* of m.
>> By which you mean P(0).
>>Then I look at at the same terms for P(m)/f^2 and find that
a factor
>>of f^2 has been removed.
>Logic dictates that the removal is *independent* of m,
> Removal here means that you divide P(m) by f^2. That does
>not cause a problem in itself. It is really how f^2 is
>distributed among the factors
> ai*x + u*f
>that is the issue.
> You do not disagree that ai is dependent
>on m. LetÕs write it as ai(m).
> Therefore the way in which factors of f divide
>ai(m) in general can be expected to depend on m.
> You prove that for m = 0, f divides ai(0).
> You think somehow that that proves f divides ai(m)
>for all m.
> LetÕs say f = 5 and ai(m) = sqrt(5) * m + 15.
> Certainly f = 5 divides ai(0) = 15.
> But f = 5 does not divide ai(1) = sqrt(5) + 15.
> Here is another example. Say ai(m) = 5 * m, and f = 3.
>The ai(0) = 0, and, as in your case, f divides ai(0).
>However, if m = 1, ai(m) = 5, and f = 3 does not divide
>a1(m) = 5. In fact f = 3 is relatively prime to 5,
>even in the algebraic integers.
> I am not saying that in your application, ai(m)
>actually equals 5 * m. I am simply saying that your logic
>breaks down. You have not shown that ai(m), as a
>function of m, does not behave something like
>ai(m) = 5 * m. You have nothing explicit whatsoever
>about ai(m), EXCEPT when m = 0. But that clearly does
>not tell you anything about how ai(m) behaves with
>respect to f when m = 0. ThatÕs the point.
> Put it another way. You have shown that ai(0)
>is divisible by f. Since ai(m) is a function of
>m which you probably cannot even write down,
>you do not know anything about the divisibility
>of ai(m) by f when m <> 0. It has NOTHING to do
>with whether ai(0) or P(0) or whatever is
>independent of ai(m), or your claim that you
>have somehow removed the constant term. You
>are badly confused on that point.
I point out to reader that the simple principle IÕm using is
that
setting m=0 shows terms *without* a dependency on m, which is
not
complicated.
Notice how much effort this poster puts in trying to refute
that
simple principle, which follows from rather basic algebra.
> In the specific case at issue, where the values of a1, a2,
a3 are just
> the roots of a cubic, I once worked out the explicit
formulas of a1,
> a2, and a3 in terms of m and f. Maybe that will help to fix
the
> discussion?
> I got the formulas from the Cardano formulas, applied to
> x^3 -3vx^2 + (v^3+1)
> where v = -1+mf^2, which I believe is the case James is
dealing
> with. If it is not, then of course my calculations are
useless for the
> purposes of giving something to fix on.
> a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
> a2 = m*f^2-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
> + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
> a3 = m*f^2-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
> + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
> where
> C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
> + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2.
> D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2
> (IÕm pretty sure I did not make any mistakes, but if 
anyone
spots
> any, let me know).
> So, what happens when m=0?
Well, for one thing, you show whatÕs *independent* of m,
which is the
entire point.
For those who wonder, what I do is find whatÕs 
independent of
m for
P(m), and then later look at whatÕs independent of m for
P(m)/f^2.
It just so happens that to isolate whatÕs independent of m, 
I
set m=0,
which isnÕt such an impossible thing to figure 
out, now is it?
> When m=0, you get C=D=0, so
> a1 = m*f^{2}-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
> = -1 + (1/2^{1/3})*[cuberoot(-2) + cuberoot(-2)].
> = -1 + 2^{-1/3}*(-2^{1/3}-2^{1/3})
> = -1 + (-1) + (-1) = 3.
>
> a2 = m*f^{2}-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
> + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
> = -1 + [(-1+sqrt(-3))/2]*2^{-1/3}(-2^{1/3}) +
[(-1-sqrt(-3))/2]*2^{-1/3}(-2^{1/3})
> = -1 - (-1+sqrt(-3))/2 - (-1-sqrt(-3))/2
> = -1 +(1/2) - sqrt(-3)/2 + (1/2) + sqrt(-3)/2
> = 0.
> a3 = m*f^{2}-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
> + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
> = -1 + [(-1+sqrt(-3))/2]*2^{-1/3}(-2^{1/3}) +
[(-1-sqrt(-3))/2]*2^{-1/3}(-2^{1/3})
> = -1 - (-1+sqrt(-3))/2 - (-1-sqrt(-3))/2
> = -1 +(1/2) - sqrt(-3)/2 + (1/2) + sqrt(-3)/2
> = 0.
> So, if we write
> g_1(m) = [g_1(m) - g_1(0)] + g_1(0)
> = [g_1(m) - (3x+1)] + (3x+1)
> this is supposed to somehow tells us something about g_1(m)
in
> general. Specifically, about how it is divisible by f.
I check for P(m) and P(m)/f^2 for their factors g_1, g_2 and
g_3, and
note that in one case you have u^2 f^2 (3x + uf) for P(0),
and in the
other you have u^2 (3x + uf) for P(0)/f^2, which necessarily
means
that the factors that were g_1, g_2, and g_3 now donÕt have 
f
as a
factor.
Given that at m=0, g_1 = uf, g_2 = uf, but g_3 = 3x + uf, it
stands to
reason that the terms independent of m, in g_1 and g_2 are
the ones
that have to lose the f, which requires that g_1 and g_2 have
f as a
factor, as otherwise, if f divides off dependent on m, then
it canÕt
be true that uf is independent of m, when it must be as I get
it by
setting m=0.
ThatÕs it.
That then tells you something about the aÕs, based on the
distributive
property.
> Can we get that
> a1 = m*f^{2}-1 +
(1/2^{1/3})*(cuberoot(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
> C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
> + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^{2}.
> D = m^3*f^{6} - 3*m^2*f^{4} + 3*m*f^{2}
> is necessarily a multiple of f whenever m is not equal to
zero? C and
> D are multiples of f, but look at the expression.
> C is a multiple of f^2 but, if f is coprime to 12m (which
cannot occur
> when m is zero, but can happen for other values), then
C/f^2 is
> coprime to f. D is a multiple of f^2, and if it is coprime
to 3m then
> D/f^2 is coprime to f. So sqrt(C) is a multiple of f, but
sqrt(C)/f is
> coprime to f. So D+sqrt(C) would be divisible by f, and we
would
> have (D+sqrt(C))/f = (D/f) + (sqrt(C)/f), which would be
coprime to
> f. If 2 is coprime to f, then D + sqrt(C)-2 is coprime to
f, in which
> case cuberoot(D-2+sqrt(C)) is coprime to f, as is
> cuberoot(D-2-sqrt(C)).
> So then a1 is: a multiple of f, minus 1, plus two things
which are
> coprime to f added together. So you have a multiple of f
plus three
> things which are coprime to f. That could be coprime to f
or not
> coprime to f, and there is no reason to believe that it
would
> necessarily be divisible to f in the latter case.
> Why do you conclude that this is ALWAYS coprime to f? The
sum of two
> things coprime to f could be non-coprime to f, as the
trivial example
> of 3 and 5 show with respect to 2.
I look at terms that are *independent* of m, and it makes
things easy.
> And say instead that f is equal to 2, and m is odd. Then
D-2+sqrt(C)
> is a multiple of 2; D is a multiple of 4, and D/4 is
coprime to 2. C
> is a multiple of 16, and C/16 is coprime to 2. So sqrt(C)
is a
> multiple of 4, and sqrt(C)/4 is coprime to 2. So
D-2+sqrt(C) is a
> multiple of 2, and (D-2+sqrt(C))/2 = (D/2) + (sqrt(C)/2) -
1, which is
> coprime to 2. So cuberoot(D-2+sqrt(C)) is a multiple of
2^{1/3}, and
> cuberoot(D-2+sqrt(C))/2^{1/3} is coprime to 2. So after
dividing by
> 2^{1/3} you get something which is coprime to 2, so you end
up with a
> multple of f minus 1, plus something which is coprime to 2,
plus
> something which is coprime to 2 (same argument holds for
D-2-sqrt(C)).
> If they were all integers, you would have something even,
minus 1,
> plus something odd plus something odd, so you would indeed
conclude
> that it is odd (coprime to 2). But here you could have the
sum of two
> things which are coprime to 2 once again being coprime to
2, or the
> sum of two things coprime to 2 being not coprime to 2 but
not a
> multiple of 2. Yet you think we can always conclude that it
is coprime
> to 2. Why?
> And why can you conclude that
> a2 = m*f^{2}-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
> + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
> will be divisible by f, just because when m=0 you got 0?
Just because,
out the -1?
> They cannot be expressed as (something)*f + a2(0), as you
seem to
> think they can. If f=2, again we have that
> cuberoot(D-2+sqrt(C))/2^{1/3} should be coprime to 2, so
now you have
> the sum of m*f^2 and THREE things coprime to 2. If they
were integers
> you would conclude that the resulting number is an even
minus 1 plus
> two odds, which would make it odd again, yet here you are
concluding
> that they will be even. Why?
You can conclude that those terms which are independent of m,
are
indeed independent of m, and with that conclusion there is no
mathematical objection.
James Harris
===
Subject: Re: JSH: About time
Visiting Assistant Professor at the University of Montana.
[.snip.]
>I point out to reader that the simple principle IÕm using 
is
that
>setting m=0 shows terms *without* a dependency on m, which
is not
>complicated.
And people point out that the problem is what you attempt to
CONCLUDE
from this. You seem to be claiming that g(m)-g(0) is always a
multiple
of f, but you have never proven this.
[.snip.]
>> In the specific case at issue, where the values of a1, a2,
a3 are just
>> the roots of a cubic, I once worked out the explicit
formulas of a1,
>> a2, and a3 in terms of m and f. Maybe that will help to fix
the
>> discussion?
>> I got the formulas from the Cardano formulas, applied to
>> x^3 -3vx^2 + (v^3+1)
>> where v = -1+mf^2, which I believe is the case James is
dealing
>> with. If it is not, then of course my calculations are
useless for the
>> purposes of giving something to fix on.
>> a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
>> a2 = m*f^2-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
>> a3 = m*f^2-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
>> where
>> C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
>> + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2.
>> D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2
>> (IÕm pretty sure I did not make any mistakes, but if
anyone spots
>> any, let me know).
>> So, what happens when m=0?
>Well, for one thing, you show whatÕs *independent* of m,
which is the
>entire point.
No, the entire point is that the conclusion you are claiming
does not
follow. The conclusion here is not m=0 gives the Ōconstant
termÕ,
and the conclusion is not what is *independent* of m. The
conclusion
that does not follow is your apparent claim that a1(m) - a(0)
must be
a multiple of f; that does not follow.
>For those who wonder, what I do is find whatÕs 
independent of
m for
>P(m), and then later look at whatÕs independent of m for
P(m)/f^2.
>It just so happens that to isolate whatÕs independent of m,
I set m=0,
>which isnÕt such an impossible thing to figure 
out, now is it?
And having isolated what is independent of m, why do you
conclude
that it must be a multiple of f?
>> When m=0, you get C=D=0, so
>> a1 = m*f^{2}-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
>> = -1 + (1/2^{1/3})*[cuberoot(-2) + cuberoot(-2)].
>> = -1 + 2^{-1/3}*(-2^{1/3}-2^{1/3})
>> = -1 + (-1) + (-1) = 3.
>>
>> a2 = m*f^{2}-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
>> = -1 + [(-1+sqrt(-3))/2]*2^{-1/3}(-2^{1/3}) +
[(-1-sqrt(-3))/2]*2^{-1/3}(-2^{1/3})
>> = -1 - (-1+sqrt(-3))/2 - (-1-sqrt(-3))/2
>> = -1 +(1/2) - sqrt(-3)/2 + (1/2) + sqrt(-3)/2
>> = 0.
>> a3 = m*f^{2}-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
>> = -1 + [(-1+sqrt(-3))/2]*2^{-1/3}(-2^{1/3}) +
[(-1-sqrt(-3))/2]*2^{-1/3}(-2^{1/3})
>> = -1 - (-1+sqrt(-3))/2 - (-1-sqrt(-3))/2
>> = -1 +(1/2) - sqrt(-3)/2 + (1/2) + sqrt(-3)/2
>> = 0.
>> So, if we write
>> g_1(m) = [g_1(m) - g_1(0)] + g_1(0)
>> = [g_1(m) - (3x+1)] + (3x+1)
>> this is supposed to somehow tells us something about
g_1(m) in
>> general. Specifically, about how it is divisible by f.
>I check for P(m) and P(m)/f^2 for their factors g_1, g_2 and
g_3, and
>note that in one case you have u^2 f^2 (3x + uf) for P(0),
and in the
>other you have u^2 (3x + uf) for P(0)/f^2, which necessarily
means
>that the factors that were g_1, g_2, and g_3 now donÕt have
f as a
>factor.
>Given that at m=0, g_1 = uf, g_2 = uf, but g_3 = 3x + uf, it
stands to
>reason that the terms independent of m, in g_1 and g_2 are
the ones
>that have to lose the f,
No, it does NOT stand to reason. That is precisely what you
are
claiming to prove, and you cannot assume it.
Look at the EXPLICIT formulas I gave you and try to follow
your own
reasoning. You will see that is simply does not apply.
> which requires that g_1 and g_2 have f as a
>factor, as otherwise, if f divides off dependent on m,
But it does. Because your expressions are expressions in mf^2.
> then it canÕt
>be true that uf is independent of m, when it must be as I
get it by
>setting m=0.
This claim is false.
[.snip.]
>> Can we get that
>> a1 = m*f^{2}-1 +
(1/2^{1/3})*(cuberoot(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
>> C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
>> + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^{2}.
>> D = m^3*f^{6} - 3*m^2*f^{4} + 3*m*f^{2}
>> is necessarily a multiple of f whenever m is not equal to
zero? C and
>> D are multiples of f, but look at the expression.
>> C is a multiple of f^2 but, if f is coprime to 12m (which
cannot occur
>> when m is zero, but can happen for other values), then
C/f^2 is
>> coprime to f. D is a multiple of f^2, and if it is coprime
to 3m then
>> D/f^2 is coprime to f. So sqrt(C) is a multiple of f, but
sqrt(C)/f is
>> coprime to f. So D+sqrt(C) would be divisible by f, and we
would
>> have (D+sqrt(C))/f = (D/f) + (sqrt(C)/f), which would be
coprime to
>> f. If 2 is coprime to f, then D + sqrt(C)-2 is coprime to
f, in which
>> case cuberoot(D-2+sqrt(C)) is coprime to f, as is
>> cuberoot(D-2-sqrt(C)).
>> So then a1 is: a multiple of f, minus 1, plus two things
which are
>> coprime to f added together. So you have a multiple of f
plus three
>> things which are coprime to f. That could be coprime to f
or not
>> coprime to f, and there is no reason to believe that it
would
>> necessarily be divisible to f in the latter case.
>> Why do you conclude that this is ALWAYS coprime to f? The
sum of two
>> things coprime to f could be non-coprime to f, as the
trivial example
>> of 3 and 5 show with respect to 2.
>I look at terms that are *independent* of m, and it makes
things easy.
Non responsive.
>> And say instead that f is equal to 2, and m is odd. Then
D-2+sqrt(C)
>> is a multiple of 2; D is a multiple of 4, and D/4 is
coprime to 2. C
>> is a multiple of 16, and C/16 is coprime to 2. So sqrt(C)
is a
>> multiple of 4, and sqrt(C)/4 is coprime to 2. So
D-2+sqrt(C) is a
>> multiple of 2, and (D-2+sqrt(C))/2 = (D/2) + (sqrt(C)/2) -
1, which is
>> coprime to 2. So cuberoot(D-2+sqrt(C)) is a multiple of
2^{1/3}, and
>> cuberoot(D-2+sqrt(C))/2^{1/3} is coprime to 2. So after
dividing by
>> 2^{1/3} you get something which is coprime to 2, so you
end up with a
>> multple of f minus 1, plus something which is coprime to
2, plus
>> something which is coprime to 2 (same argument holds for
D-2-sqrt(C)).
>> If they were all integers, you would have something even,
minus 1,
>> plus something odd plus something odd, so you would indeed
conclude
>> that it is odd (coprime to 2). But here you could have the
sum of two
>> things which are coprime to 2 once again being coprime to
2, or the
>> sum of two things coprime to 2 being not coprime to 2 but
not a
>> multiple of 2. Yet you think we can always conclude that
it is coprime
>> to 2. Why?
>> And why can you conclude that
>> a2 = m*f^{2}-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
>> will be divisible by f, just because when m=0 you got 0?
Just because,
out the -1?
>> They cannot be expressed as (something)*f + a2(0), as you
seem to
>> think they can. If f=2, again we have that
>> cuberoot(D-2+sqrt(C))/2^{1/3} should be coprime to 2, so
now you have
>> the sum of m*f^2 and THREE things coprime to 2. If they
were integers
>> you would conclude that the resulting number is an even
minus 1 plus
>> two odds, which would make it odd again, yet here you are
concluding
>> that they will be even. Why?
>You can conclude that those terms which are independent of m,
Non responsive, irrelevant, and immaterial
> are
>indeed independent of m, and with that conclusion there is no
>mathematical objection.
Non responsive, irrelevant, and immaterial.
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 manÕs 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
magidin@math.berkeley.edu
===
Subject: Re: JSH: About time
> [.snip.]
>I point out to reader that the simple principle IÕm using 
is
that
>setting m=0 shows terms *without* a dependency on m, which
is not
>complicated.
> And people point out that the problem is what you attempt
to CONCLUDE
> from this. You seem to be claiming that g(m)-g(0) is always
a multiple
> of f, but you have never proven this.
> [.snip.]
ItÕs simple. You have P(0) = u^2 f^2 (3x + uf) while P(m) 
has
f^2 as
a factor as well. Now setting m=0 takes m out of the picture,
that
is, it gives me terms independent of m.
That is verified by looking as
P(0) = u^2 f^2 (3x + uf).
Now then, I also have P(m) = g_1 g_2 g_3, where
at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
and the requirement is given that f is coprime to 3, x and u.
Notice that lack of symmetry between whatÕs *independent* of
m with
the gÕs.
Now then, giveen that dividing off f^2 from P(0), gives you
P(0)/f^2 = u^2(3x + uf)
it stands to reason that both g_1 and g_2 each contribute f
as a
factor, without regard to mÕs value, as you can focus on
whatÕs
independent of m, and in this case itÕs uf for both g_1 and
g_2.
That math is so easy. However the *social* consequences are
huge.
Readers should note that if Arturo Magidin accepts that 
whatÕs
independent of m is in fact independent of m, there is no
basis for
argument.
>> In the specific case at issue, where the values of a1, a2,
a3 are just
>> the roots of a cubic, I once worked out the explicit
formulas of a1,
>> a2, and a3 in terms of m and f. Maybe that will help to fix
the
>> discussion?
>>
>> I got the formulas from the Cardano formulas, applied to
>>
>> x^3 -3vx^2 + (v^3+1)
>>
>> where v = -1+mf^2, which I believe is the case James is
dealing
>> with. If it is not, then of course my calculations are
useless for the
>> purposes of giving something to fix on.
>>
>>
>> a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
>>
>> a2 = m*f^2-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
>>
>> a3 = m*f^2-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
>>
>>
>> where
>>
>> C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
>> + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2.
>>
>> D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2
>>
>> (IÕm pretty sure I did not make any mistakes, but if
anyone spots
>> any, let me know).
>>
>>
>>
>> So, what happens when m=0?
>Well, for one thing, you show whatÕs *independent* of m,
which is the
>entire point.
> No, the entire point is that the conclusion you are
claiming does not
> follow. The conclusion here is not m=0 gives the Ōconstant
termÕ,
> and the conclusion is not what is *independent* of m. The
conclusion
> that does not follow is your apparent claim that a1(m) -
a(0) must be
> a multiple of f; that does not follow.
That paragraph was not logical.
As this poster is illogical IÕll stop here.
Possibly the poster would like to try again, and produce a
logical
statement.
James Harris
===
Subject: Re: JSH: About time
>> [.snip.]
>I point out to reader that the simple principle IÕm using 
is
that
>setting m=0 shows terms *without* a dependency on m, which
is not
>complicated.
>>And people point out that the problem is what you attempt
to CONCLUDE
>>from this. You seem to be claiming that g(m)-g(0) is always
a multiple
>>of f, but you have never proven this.
>> [.snip.]
> ItÕs simple. You have P(0) = u^2 f^2 (3x + uf) while P(m)
has f^2 as
> a factor as well. Now setting m=0 takes m out of the
picture, that
> is, it gives me terms independent of m.
> That is verified by looking as
> P(0) = u^2 f^2 (3x + uf).
> Now then, I also have P(m) = g_1 g_2 g_3, where
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
> and the requirement is given that f is coprime to 3, x and
u.
> Notice that lack of symmetry between whatÕs *independent*
of m with
> the gÕs.
Perhaps it would help to write:
at m=0, g_1 = 0x+uf, g_2 = 0x+uf, g_3 = 3x+uf
Then it is easier to see that g_1(m)-g_1(0) = a_1(m)x.
Now, is a_1(m)x divisble by f for all m?
The issue is not with whatÕs independent of m, but with 
whatÕs
*dependent* on m. You have not addressed this.
> Now then, giveen that dividing off f^2 from P(0), gives you
> P(0)/f^2 = u^2(3x + uf)
> it stands to reason that both g_1 and g_2 each contribute f
as a
> factor, without regard to mÕs value, as you can focus on
whatÕs
> independent of m, and in this case itÕs uf for both g_1 
and
g_2.
It doesnÕt stand to reason to me. Please clarify. Perhaps my
hangup
with the notion that g_1(m) = a_1(m)x+uf (and similar for
g_2) is at
fault. If you see that as my problem as well, perhaps working
explicitly with the a_1(m)x+uf form will allow you to make
the situation
clearer.
> That math is so easy. However the *social* consequences are
huge.
There are no social consequences, only potential mathematical
consequences. As you have observed, itÕs about math, not
society.
> Readers should note that if Arturo Magidin accepts that
whatÕs
> independent of m is in fact independent of m, there is no
basis for
> argument.
His concern is with the dependent parts, which you fail to
address.
Either that or you believe that P(m) is independent of m,
which seems
unlikely.
>>In the specific case at issue, where the values of a1, a2,
a3 are just
>>the roots of a cubic, I once worked out the explicit
formulas of a1,
>>a2, and a3 in terms of m and f. Maybe that will help to fix
the
>>discussion?
>I got the formulas from the Cardano formulas, applied to
>x^3 -3vx^2 + (v^3+1)
>where v = -1+mf^2, which I believe is the case James is
dealing
>>with. If it is not, then of course my calculations are
useless for the
>>purposes of giving something to fix on.
>>a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
>a2 = m*f^2-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> + [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
>a3 = m*f^2-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> + [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
>>where
>C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
>> + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2.
>D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2
>(IÕm pretty sure I did not make any mistakes, but if anyone
spots
>> any, let me know).
>So, what happens when m=0?
>Well, for one thing, you show whatÕs *independent* of m,
which is the
>entire point.
>>No, the entire point is that the conclusion you are
claiming does not
>>follow. The conclusion here is not m=0 gives the Ōconstant
termÕ,
>>and the conclusion is not what is *independent* of m. The
conclusion
>>that does not follow is your apparent claim that a1(m) -
a(0) must be
>>a multiple of f; that does not follow.
> That paragraph was not logical.
> As this poster is illogical IÕll stop here.
> Possibly the poster would like to try again, and produce a
logical
> statement.
Why is illogical for him to bring the discussion back to the
aÕs that
you are making claims about? Your argument above neatly fails
to look
at them at all. Your claim is this: a_1(m) is divisible by f
for all m
in the algebraic integers. a_2(m) is divisible by f for all m
in the
algebraic integers. The key phrase is *for all m*. These are
the same
aÕs that are not independent of m.
If I have misrepresented your claim, please restate it
clearly.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: JSH: About time
>
> [.snip.]
>I point out to reader that the simple principle IÕm using 
is
that
>setting m=0 shows terms *without* a dependency on m, which
is not
>complicated.
>And people point out that the problem is what you attempt to
CONCLUDE
>>from this. You seem to be claiming that g(m)-g(0) is always
a multiple
>>of f, but you have never proven this.
> [.snip.]
>
>
> ItÕs simple. You have P(0) = u^2 f^2 (3x + uf) while P(m)
has f^2 as
> a factor as well. Now setting m=0 takes m out of the
picture, that
> is, it gives me terms independent of m.
>
> That is verified by looking as
>
> P(0) = u^2 f^2 (3x + uf).
>
> Now then, I also have P(m) = g_1 g_2 g_3, where
>
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
>
> and the requirement is given that f is coprime to 3, x and
u.
>
> Notice that lack of symmetry between whatÕs *independent*
of m with
> the gÕs.
> Perhaps it would help to write:
> at m=0, g_1 = 0x+uf, g_2 = 0x+uf, g_3 = 3x+uf
> Then it is easier to see that g_1(m)-g_1(0) = a_1(m)x.
> Now, is a_1(m)x divisble by f for all m?
> The issue is not with whatÕs independent of m, but with
whatÕs
> *dependent* on m. You have not addressed this.
Yes I have.
Now I notice that posters keep putting just a *piece* of the
argument,
as if they canÕt read and see all those times I put in
P(0)/f^2 =
u^2(3x + uf).
I see it as a tactic.
They ignore the actual argument, but then claim IÕm not
putting in all
the details.
>
> Now then, giveen that dividing off f^2 from P(0), gives you
>
> P(0)/f^2 = u^2(3x + uf)
>
> it stands to reason that both g_1 and g_2 each contribute f
as a
> factor, without regard to mÕs value, as you can focus on
whatÕs
> independent of m, and in this case itÕs uf for both g_1 
and
g_2.
> It doesnÕt stand to reason to me. Please clarify. Perhaps
my hangup
> with the notion that g_1(m) = a_1(m)x+uf (and similar for
g_2) is at
> fault. If you see that as my problem as well, perhaps
working
> explicitly with the a_1(m)x+uf form will allow you to make
the situation
> clearer.
It should help *you* as you should notice that uf is
INDEPENDENT of m.
Given that P(0)/f^2 = u^2(3x + uf) how do you suppose the
factor f
disappears from a_1(m) x + uf, so that you get that result?
I look at terms INDEPENDENT of m for a reason.
>
> That math is so easy. However the *social* consequences are
huge.
> There are no social consequences, only potential
mathematical
> consequences. As you have observed, itÕs about math, not
society.
Mathematicians are running away in droves, and thatÕs about
social
consequences.
I based that assessment not only from Usenet, but from
contacts with
mathematicians in other areas, like that email from Barry
Mazur, or
the emails from Granville, or my visit with McKenzie.
Mathematicians are clearly terrified, and I think 
itÕs about
social
consequences.
>
> Readers should note that if Arturo Magidin accepts that
whatÕs
> independent of m is in fact independent of m, there is no
basis for
> argument.
> His concern is with the dependent parts, which you fail to
address.
> Either that or you believe that P(m) is independent of m,
which seems
> unlikely.
The independent terms by being independent decide things.
ItÕs basic and simple.
>>In the specific case at issue, where the values of a1, a2,
a3 are
just
>>the roots of a cubic, I once worked out the explicit
formulas of a1,
>>a2, and a3 in terms of m and f. Maybe that will help to fix
the
>>discussion?
>I got the formulas from the Cardano formulas, applied to
>x^3 -3vx^2 + (v^3+1)
>where v = -1+mf^2, which I believe is the case James is
dealing
>>with. If it is not, then of course my calculations are
useless for
the
>>purposes of giving something to fix on.
>>a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
>a2 = m*f^2-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
>a3 = m*f^2-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
>>where
>C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
>> + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2.
>D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2
>(IÕm pretty sure I did not make any mistakes, but if anyone
spots
>> any, let me know).
>So, what happens when m=0?
>Well, for one thing, you show whatÕs *independent* of m,
which is the
>entire point.
>No, the entire point is that the conclusion you are claiming
does not
>>follow. The conclusion here is not m=0 gives the Ōconstant
termÕ,
>>and the conclusion is not what is *independent* of m. The
conclusion
>>that does not follow is your apparent claim that a1(m) -
a(0) must be
>>a multiple of f; that does not follow.
>
> That paragraph was not logical.
>
> As this poster is illogical IÕll stop here.
>
> Possibly the poster would like to try again, and produce a
logical
> statement.
> Why is illogical for him to bring the discussion back to
the aÕs that
> you are making claims about? Your argument above neatly
fails to look
> at them at all. Your claim is this: a_1(m) is divisible by
f for all m
> in the algebraic integers. a_2(m) is divisible by f for all
m in the
> algebraic integers. The key phrase is *for all m*. These
are the same
> aÕs that are not independent of m.
> If I have misrepresented your claim, please restate it
clearly.
That only two of the aÕs have a factor that is f, is almost
accidental
in a way, as itÕs the terms INDEPENDENT of m that determine
things.
The aÕs, by being dependent on m, are constrained in a way
that the
INDEPENDENT terms are NOT.
So the INDEPENDENT terms decide things, not the aÕs.
TheyÕre just along for the ride.
James Harris
===
Subject: Re: JSH: About time
> [.snip.]
>I point out to reader that the simple principle IÕm using 
is
that
>setting m=0 shows terms *without* a dependency on m, which
is not
>complicated.
>And people point out that the problem is what you attempt to
CONCLUDE
>>from this. You seem to be claiming that g(m)-g(0) is always
a
multiple
>>of f, but you have never proven this.
> [.snip.]
>
> ItÕs simple. You have P(0) = u^2 f^2 (3x + uf) while P(m)
has f^2 as
> a factor as well. Now setting m=0 takes m out of the
picture, that
> is, it gives me terms independent of m.
> That is verified by looking as
> P(0) = u^2 f^2 (3x + uf).
> Now then, I also have P(m) = g_1 g_2 g_3, where
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
> and the requirement is given that f is coprime to 3, x and
u.
> Notice that lack of symmetry between whatÕs *independent*
of m with
> the gÕs.
> Perhaps it would help to write:
> at m=0, g_1 = 0x+uf, g_2 = 0x+uf, g_3 = 3x+uf
> Then it is easier to see that g_1(m)-g_1(0) = a_1(m)x.
> Now, is a_1(m)x divisble by f for all m?
> The issue is not with whatÕs independent of m, but with
whatÕs
> *dependent* on m. You have not addressed this.
> Yes I have.
> Now I notice that posters keep putting just a *piece* of
the argument,
> as if they canÕt read and see all those times I put in
P(0)/f^2 =
> u^2(3x + uf).
> I see it as a tactic.
> They ignore the actual argument, but then claim IÕm not
putting in all
> the details.
> Now then, giveen that dividing off f^2 from P(0), gives you
> P(0)/f^2 = u^2(3x + uf)
> it stands to reason that both g_1 and g_2 each contribute f
as a
> factor, without regard to mÕs value, as you can focus on
whatÕs
> independent of m, and in this case itÕs uf for both g_1 
and
g_2.
> It doesnÕt stand to reason to me. Please clarify. Perhaps
my hangup
> with the notion that g_1(m) = a_1(m)x+uf (and similar for
g_2) is at
> fault. If you see that as my problem as well, perhaps
working
> explicitly with the a_1(m)x+uf form will allow you to make
the
situation
> clearer.
> It should help *you* as you should notice that uf is
INDEPENDENT of m.
> Given that P(0)/f^2 = u^2(3x + uf) how do you suppose the
factor f
> disappears from a_1(m) x + uf, so that you get that result?
> I look at terms INDEPENDENT of m for a reason.
> That math is so easy. However the *social* consequences are
huge.
> There are no social consequences, only potential
mathematical
> consequences. As you have observed, itÕs about math, not
society.
> Mathematicians are running away in droves, and thatÕs 
about
social
> consequences.
> I based that assessment not only from Usenet, but from
contacts with
> mathematicians in other areas, like that email from Barry
Mazur, or
> the emails from Granville, or my visit with McKenzie.
> Mathematicians are clearly terrified, and I think 
itÕs about
social
> consequences.
Bullcrap. You wouldnÕt get this type of treatment if you 
were
rational. By
rational, I mean you claim something, someone rebuts it, you
come back and
either accept it or say Well, what about...... Instead, if
someone
rebuts
you their, automatically deemed a liar. You cannot blame
anyone else but
yourself for how others have treated you. Seems to me that
you can dish it
out, but you canÕt take it.
David Moran
> Readers should note that if Arturo Magidin accepts that
whatÕs
> independent of m is in fact independent of m, there is no
basis for
> argument.
> His concern is with the dependent parts, which you fail to
address.
> Either that or you believe that P(m) is independent of m,
which seems
> unlikely.
> The independent terms by being independent decide things.
> ItÕs basic and simple.
>>In the specific case at issue, where the values of a1, a2,
a3 are
just
>>the roots of a cubic, I once worked out the explicit
formulas of
a1,
>>a2, and a3 in terms of m and f. Maybe that will help to fix
the
>>discussion?
>I got the formulas from the Cardano formulas, applied to
>x^3 -3vx^2 + (v^3+1)
>where v = -1+mf^2, which I believe is the case James is
dealing
>>with. If it is not, then of course my calculations are
useless for
the
>>purposes of giving something to fix on.
>>a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
>a2 = m*f^2-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
>a3 = m*f^2-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
>>where
>C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
>> + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2.
>D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2
>(IÕm pretty sure I did not make any mistakes, but if anyone
spots
>> any, let me know).
>So, what happens when m=0?
>Well, for one thing, you show whatÕs *independent* of m,
which is
the
>entire point.
>No, the entire point is that the conclusion you are claiming
does not
>>follow. The conclusion here is not m=0 gives the Ōconstant
termÕ,
>>and the conclusion is not what is *independent* of m. The
conclusion
>>that does not follow is your apparent claim that a1(m) -
a(0) must be
>>a multiple of f; that does not follow.
> That paragraph was not logical.
> As this poster is illogical IÕll stop here.
> Possibly the poster would like to try again, and produce a
logical
> statement.
> Why is illogical for him to bring the discussion back to
the aÕs that
> you are making claims about? Your argument above neatly
fails to look
> at them at all. Your claim is this: a_1(m) is divisible by
f for all m
> in the algebraic integers. a_2(m) is divisible by f for all
m in the
> algebraic integers. The key phrase is *for all m*. These
are the same
> aÕs that are not independent of m.
> If I have misrepresented your claim, please restate it
clearly.
> That only two of the aÕs have a factor that is f, is 
almost
accidental
> in a way, as itÕs the terms INDEPENDENT of m that 
determine
things.
> The aÕs, by being dependent on m, are constrained in a way
that the
> INDEPENDENT terms are NOT.
> So the INDEPENDENT terms decide things, not the aÕs.
> TheyÕre just along for the ride.
> James Harris
===
Subject: Re: JSH: About time
>
> [.snip.]
>I point out to reader that the simple principle IÕm using 
is
that
>setting m=0 shows terms *without* a dependency on m, which
is not
>complicated.
>And people point out that the problem is what you attempt to
CONCLUDE
>>from this. You seem to be claiming that g(m)-g(0) is always
a
multiple
>>of f, but you have never proven this.
> [.snip.]
>
> ItÕs simple. You have P(0) = u^2 f^2 (3x + uf) while P(m)
has f^2
as
> a factor as well. Now setting m=0 takes m out of the
picture, that
> is, it gives me terms independent of m.
> That is verified by looking as
> P(0) = u^2 f^2 (3x + uf).
> Now then, I also have P(m) = g_1 g_2 g_3, where
> at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
> and the requirement is given that f is coprime to 3, x and
u.
> Notice that lack of symmetry between whatÕs *independent*
of m with
> the gÕs.
> Perhaps it would help to write:
> at m=0, g_1 = 0x+uf, g_2 = 0x+uf, g_3 = 3x+uf
> Then it is easier to see that g_1(m)-g_1(0) = a_1(m)x.
> Now, is a_1(m)x divisble by f for all m?
> The issue is not with whatÕs independent of m, but with
whatÕs
> *dependent* on m. You have not addressed this.
> Yes I have.
> Now I notice that posters keep putting just a *piece* of
the argument,
> as if they canÕt read and see all those times I put in
P(0)/f^2 =
> u^2(3x + uf).
> I see it as a tactic.
> They ignore the actual argument, but then claim IÕm not
putting in all
> the details.
> Now then, giveen that dividing off f^2 from P(0), gives you
> P(0)/f^2 = u^2(3x + uf)
> it stands to reason that both g_1 and g_2 each contribute f
as a
> factor, without regard to mÕs value, as you can focus on
whatÕs
> independent of m, and in this case itÕs uf for both g_1 
and
g_2.
> It doesnÕt stand to reason to me. Please clarify. Perhaps
my hangup
> with the notion that g_1(m) = a_1(m)x+uf (and similar for
g_2) is at
> fault. If you see that as my problem as well, perhaps
working
> explicitly with the a_1(m)x+uf form will allow you to make
the
situation
> clearer.
> It should help *you* as you should notice that uf is
INDEPENDENT of m.
> Given that P(0)/f^2 = u^2(3x + uf) how do you suppose the
factor f
> disappears from a_1(m) x + uf, so that you get that result?
> I look at terms INDEPENDENT of m for a reason.
> That math is so easy. However the *social* consequences are
huge.
> There are no social consequences, only potential
mathematical
> consequences. As you have observed, itÕs about math, not
society.
> Mathematicians are running away in droves, and thatÕs 
about
social
> consequences.
> I based that assessment not only from Usenet, but from
contacts with
> mathematicians in other areas, like that email from Barry
Mazur, or
> the emails from Granville, or my visit with McKenzie.
> Mathematicians are clearly terrified, and I think 
itÕs about
social
> consequences.
> Bullcrap. You wouldnÕt get this type of treatment if you
were rational.
By
> rational, I mean you claim something, someone rebuts it,
you come back
and
> either accept it or say Well, what about...... Instead, if
someone
rebuts
> you their, automatically deemed a liar. You cannot blame
anyone else but
> yourself for how others have treated you. Seems to me that
you can dish
it
> out, but you canÕt take it.
> David Moran
** SHOULDÕVE BEEN Instead, if someone rebuts
you, theyÕre automatically deemed a liar. Late night :).
David Moran
> Readers should note that if Arturo Magidin accepts that
whatÕs
> independent of m is in fact independent of m, there is no
basis for
> argument.
> His concern is with the dependent parts, which you fail to
address.
> Either that or you believe that P(m) is independent of m,
which seems
> unlikely.
> The independent terms by being independent decide things.
> ItÕs basic and simple.
>>In the specific case at issue, where the values of a1, a2,
a3 are
> just
>>the roots of a cubic, I once worked out the explicit
formulas of
a1,
>>a2, and a3 in terms of m and f. Maybe that will help to fix
the
>>discussion?
>I got the formulas from the Cardano formulas, applied to
>x^3 -3vx^2 + (v^3+1)
>where v = -1+mf^2, which I believe is the case James is
dealing
>>with. If it is not, then of course my calculations are
useless
for
> the
>>purposes of giving something to fix on.
>>a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot
> (D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
>a2 = m*f^2-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> +
> [(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
>a3 = m*f^2-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
>> +
> [(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
>>where
>C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
>> + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2.
>D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2
>(IÕm pretty sure I did not make any mistakes, but if anyone
spots
>> any, let me know).
>So, what happens when m=0?
>Well, for one thing, you show whatÕs *independent* of m,
which is
the
>entire point.
>No, the entire point is that the conclusion you are claiming
does
not
>>follow. The conclusion here is not m=0 gives the Ōconstant
termÕ,
>>and the conclusion is not what is *independent* of m. The
conclusion
>>that does not follow is your apparent claim that a1(m) -
a(0) must
be
>>a multiple of f; that does not follow.
>
> That paragraph was not logical.
> As this poster is illogical IÕll stop here.
> Possibly the poster would like to try again, and produce a
logical
> statement.
> Why is illogical for him to bring the discussion back to
the aÕs that
> you are making claims about? Your argument above neatly
fails to
look
> at them at all. Your claim is this: a_1(m) is divisible by
f for all
m
> in the algebraic integers. a_2(m) is divisible by f for all
m in the
> algebraic integers. The key phrase is *for all m*. These
are the
same
> aÕs that are not independent of m.
> If I have misrepresented your claim, please restate it
clearly.
> That only two of the aÕs have a factor that is f, is 
almost
accidental
> in a way, as itÕs the terms INDEPENDENT of m that 
determine
things.
> The aÕs, by being dependent on m, are constrained in a way
that the
> INDEPENDENT terms are NOT.
> So the INDEPENDENT terms decide things, not the aÕs.
> TheyÕre just along for the ride.
> James Harris
===
Subject: Re: JSH: About time
Visiting Assistant Professor at the University of Montana.
>> [.snip.]
>>I point out to reader that the simple principle IÕm using
is that
>>setting m=0 shows terms *without* a dependency on m, which
is not
>>complicated.
>> And people point out that the problem is what you attempt
to CONCLUDE
>> from this. You seem to be claiming that g(m)-g(0) is
always a multiple
>> of f, but you have never proven this.
>> [.snip.]
>ItÕs simple. You have P(0) = u^2 f^2 (3x + uf) while P(m)
has f^2 as
>a factor as well. Now setting m=0 takes m out of the
picture, that
>is, it gives me terms independent of m.
So ->what<-?
P(x) = x^2 + 3x + 2 is always a multiple of 2. P(x) =
(x+1)(x+2). P(0)
is a multiple of 2, and takes x out of the picture, giving
the terms
independent of m. If g_1(x) = x+1, then g_1(0) = 1, which
does not
have any factors of 2. If g_2(x)=x+2, then g_2(0)=2, which is
a
multiple of 2. So the 2 from P(0)=2 is factored with a 1 in
g_1 and
a 2 in g_2.
But from this I CANNOT conclude that the 2 from P(x) for
arbitrary x
will always come from g_2(x), and never from g_1(x), even
when P(x)
is a multiple of 2 and not of 4. For instance, when x=5, P(5)
=
g_1(5)*g_2(5), but now the factor of 2 comes from g_1, and
not from
g_2.
But that is ->exactly<- what you are attempting to conclude
in your
argument: that because the f^2 from P(0) factors as one f
from g_1,
one f from g_2, and nothing from g_3, then we will always
have one f
from g_1, one f from g_2, and nothing from g_3.
All your talk about P(0) is nothing but a red herring. It
does not
give you sufficient information about what happens when m is
not
zero. This is OBVIOUS when you look at the actual values of
a_1, a_2,
and a_3 in terms of m and f which I provided.
>That is verified by looking as
>P(0) = u^2 f^2 (3x + uf).
Yes, nobody disputes that when m=0, a1=a2=0 and a3=3; the
argument is
on why you think that this ->implies<- that a1 and a2 are
ALWAYS
multiples of f and a3 is NEVER a multiple of f.
To conclude that, you seem to be arguing that the nonconstant
part
of a_1(m) (and so the nonconstant part of g_1(m)) is always a
multiple of f; but that is the conclusion you ->want<-, so
you cannot
use it as an assumption.
In other wrods, you agree that g_1, g_2, g_3 are functions of
m. So
you want to write them as
g_1(m) = (g_1(m) - g_1(0)) + g_1(0).
Then you claim that since g_1(0) is a multiple of f, then
g_1(m) is
ALWAYS a multiple of f. But that is only true if you can
prove that
g_1(m)-g_1(0) is always a multiple of f, and that is
->equivalent<- to
the conclusion that g_1(m) is always a multiple of f. And you
have not
proven either, you have just gone round and round about
constant
terms.
You are not factoring the constant term of P(m), you are
factoring the
VALUE of P(m).
>Now then, I also have P(m) = g_1 g_2 g_3, where
>at m=0, g_1 = uf, g_2 = uf, g_3 = 3x + uf
>and the requirement is given that f is coprime to 3, x and u.
>Notice that lack of symmetry between whatÕs *independent* 
of
m with
>the gÕs.
The lack of symmetry arises because when m=0, your polynomial
is not
irreducible. The polynomial defining the aÕs 
becomes a product
of
three linear terms over Q; the symmetry appears when the
polynomial is
->IRREDUCIBLE<-, as we have told you time and time again.
>Now then, giveen that dividing off f^2 from P(0), gives you
>P(0)/f^2 = u^2(3x + uf)
>it stands to reason that both g_1 and g_2 each contribute f
as a
>factor, without regard to mÕs value, as you can focus on
whatÕs
>independent of m, and in this case itÕs uf for both g_1 and
g_2.
No, it does ->NOT<=- stand to reason that the contribution is
independent of the value of m. THAT is your leap. There is no
reason
for that leap, no matter how many times you claim that it is
obvious
and clear.
In my example above, P(x) = x^2 + 3x + 2, g_1(x) = x+1,
g_2(x)=x+2, we
have that P(0)/2 = 1, which is coprime to 2, but it does NOT
stand to
reason that g_1 contributes no 2 and g_2 contributes one 2
without
regard to xÕs value.
>That math is so easy. However the *social* consequences are
huge.
>Readers should note that if Arturo Magidin accepts that
whatÕs
>independent of m is in fact independent of m, there is no
basis for
>argument.
Red herring, strawman, and appeal to the gallery. Your
specialties.
Being a multiple of f is NOT independent of m. If you look at
the
formulas that define a1, a2, a3 and hence g1, g2, g3, you can
plainly
see that being coprime to f and being a multiple of Mis in
fact
NOT independent of m. Your colossal error lies in thinking
that it
->is<- independent of m, when it is not, just as being
coprime to 2
and being divisible by 2 is not independent of x in the
example I
gave.
> In the specific case at issue, where the values of a1, a2,
a3 are
just
> the roots of a cubic, I once worked out the explicit
formulas of a1,
> a2, and a3 in terms of m and f. Maybe that will help to fix
the
> discussion?
>
> I got the formulas from the Cardano formulas, applied to
>
> x^3 -3vx^2 + (v^3+1)
>
> where v = -1+mf^2, which I believe is the case James is
dealing
> with. If it is not, then of course my calculations are
useless for
the
> purposes of giving something to fix on.
>
>
> a1 = m*f^2-1 + (1/2^{1/3})*(cuberoot
(D-2+sqrt(C))+cuberoot(D-2-sqrt(C)))
>
> a2 = m*f^2-1 +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
> +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C)))
>
> a3 = m*f^2-1 +
[(-1-sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2+sqrt(C)))
> +
[(-1+sqrt(-3))/2]*(1/2^{1/3})(cuberoot(D-2-sqrt(C))).
>
>
> where
>
> C = -3*m^6*f^{12} + 18*m^5*f^{10} - 45*m^4*f^{8}
> + 58*m^3*f^{6} - 39*m^2*f^{4} + 12*m*f^2.
>
> D = m^3*f^6 - 3*m^2*f^4 + 3*m*f^2
>
> (IÕm pretty sure I did not make any mistakes, but if 
anyone
spots
> any, let me know).
>
>
>
> So, what happens when m=0?
>Well, for one thing, you show whatÕs *independent* of m,
which is the
>>entire point.
>> No, the entire point is that the conclusion you are
claiming does not
>> follow. The conclusion here is not m=0 gives the Ōconstant
termÕ,
>> and the conclusion is not what is *independent* of m. The
conclusion
>> that does not follow is your apparent claim that a1(m) -
a(0) must be
>> a multiple of f; that does not follow.
>That paragraph was not logical.
ThatÕs false.
It is completely logical.
You talk about something that is independent of m. Nobody is
saying
that P(0) is not a constant, or that g_1(0) is not a
constant, or that
a_1(0) is not a constant.
But you are, apparently, using that divisible by f ->is<-
independent of m. But that is ->equivalent<- to the claim that
a_1(m)-a_1(0) is a multiple of f.
See, you argue that if a_1(0) is a multiple of f, then a_1(m)
is a
multiple of f. You are arguing that, because you are arguing
that
since
g_1(m) = a_1(m)x + uf, and g_1(0) = uf, a_1(0)=0, that
g_1(m)-uf = [a_1(m)x]; so if f is coprime to x, then saying
that
g_1(m)-uf is a multiple of f is the same as saying that
a_1(m) is a
multiple of f.
So you are saying that a_1(m) is always a multiple of f,
because
a_1(0) is a multiple of f. So you are saying that is a
multiple of f
is independent of m. But that is your CONCLUSION, so you
cannot use it
as an assumption.
Likewise, you have g_3(m) = a_3(m)x + uf, a_3(0)=3, so g_3(0)
= 3x+uf;
if f is coprime to 3 and x, then this is coprime to f. When
you claim
that coprime to f is independent of m, you are claiming that
g_3(m) = a_3(m)x + uf is coprime to f; that means claiming
that
a_3(m)x is coprime to f; that means claiming that a_3(m) is
coprime to
f (since x is already coprime to f).
g_3(m) - g_3(0) = a_3(m)x + uf - g_3(0)
= a_3(m)x + uf - (3x+uf)
= a_3(m)x + uf -3x - uf
= [a_3(m)-3] x
= [a_3(m) - a_3(0)] x.
So you are claiming that a_3(m) - a_3(0) are is coprime to f;
so you
are claiming that a_3(m) is never congruent to 3 modulo f. So
you are
claiming that the congruence modulo f of the function a_3(m)
DOES NOT
DEPEND ON m. But that is your desired conclusion, you cannot
assume it
as well.
The question is ->what<- is it that depends on m. You are
claiming
that divisibility by and coprimeness to f of the FUNCTIONS
a_1(m),
a_2(m), and a_3(m), are independent of m. But that is plainly
false,
simply by looking at the formulas that define them. And by
looking at
the explicit examples where we have shown that they are NOT
either
always divisible or always coprime to f.
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 manÕs 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
magidin@math.berkeley.edu
===
Subject: Re: JSH: About time
>Now that IÕve revealed the odd and you could say esoteric
error in
>core mathematics
Also totally ignoring and/or misunderstanding perfectly clear
and simple explanations of why youÕre all wrong. 
DonÕt forget
about that. (And as long as weÕre listing your recent
accomplishments,
you shouldnÕt omit the fact that youÕve 
managed to assume
people
are being serious when theyÕre making fun of you...)
>with such a short, and rather simple argument, the
>issue now is how long until mathematicians decide that
theyÕd rather
>have correct mathematics versus the *belief* that they had
been
>perfect in keeping error out of the collected body of work
that is
>called mathematics.
>My work is out there and rather easy to go over as can be
seen at the
>Hong Konk math site:
>See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
>And I send people there because their allowal of the use of
LaTeX
>makes for a *much* better presentation, and given the
*social* issues
>IÕm facing, I need all the help I can get.
>Some of you are now facing the reality of the human brain
versus any
>fantasy you might have had about being completely rational.
Human
>beings are NOT rational creatures but necessarily rely on
social
>forces to determine what they believe.
>You are creatures of society.
>You may have believed that your mathematical knowledge was
based
>completely on logic and rationality, but human beings donÕt
work that
>way; itÕs built-in to your wiring NOT to work that way.
>Some of you must learn to be more than human.
>You must learn to be truly rational, for the first times in
your
>lives.
>So itÕs about time, as I wait, and wonder, how many of you
can handle
>the truth.
>And how many of you prefer the fantasy which was the world
you
>believed in, which actually never existed, except in your
>imaginations; your wishes for a nicer world, where your
wishes matter.
Some day when you buy a mirror all this will make sense.
>James Harris
************************
===
Subject: Algebra problem
Hi..
Pardon any translation misses and the fact that i dont know
how to
write certain things in ascii..
IÕm cramming for an algebra test and cannot find 
the solution
for the
following to similar problems, the chapter is on congruenses
so I
suppose that what I should use.
1/ Prove that (17^47 + 2^12)^14 - 4 is divisible by 13.
In other words that (17^47 + 2^12)^14 is congurent with 4 mod
13.
2/ Let n be a natural number.
Prove that 11^2n + 5^(2n+1) - 6 is divisible by 24 for any n
Again, 11^2n + 5^(2n+1) is congruent with 6 mod 24
--
Sigblock empty. By choice.
===
Subject: Re: Algebra problem
>Hi..
> Pardon any translation misses and the fact that i dont know
how to
> write certain things in ascii..
>IÕm cramming for an algebra test and cannot 
find the solution
for the
>following to similar problems, the chapter is on congruenses
so I
>suppose that what I should use.
>1/ Prove that (17^47 + 2^12)^14 - 4 is divisible by 13.
> In other words that (17^47 + 2^12)^14 is congurent with 4
mod 13.
Show that x^{12} = 1 (mod 13) for x=1,2,3,4,5,6,7,8,9,10,11,
and 12.
Then note that since 17 = 4 (mod 14), we have
(17^47 + 2^12)^14 = (4^{47} + 2^12)^14 (mod 13)
= (4^{47} + 1)^14 (mod 13)
= (4^{36}*4^{9} + 1)^14 (mod 13);
if the number inside the parenthesis is not 0 (mod 13), then
its 14th
power should equal its square (why?), so just show that the
square is
congruent to 4 mod 13.
>2/ Let n be a natural number.
> Prove that 11^2n + 5^(2n+1) - 6 is divisible by 24 for any n
> Again, 11^2n + 5^(2n+1) is congruent with 6 mod 24
Think about what happens to the powers of 5 modulo 24: 5^1 =
5, 5^2=1,
so what is 5^3, 5^4, 5^6,....?
What is 5^{2n+1} for any natural number n?
What about the powers of 11? 11^2 = 121 = 120 + 1 = 5(24) + 1
= 1 (mod
24). So what about 11^2n for an arbitrary n?
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Algebra problem
>>1/ Prove that (17^47 + 2^12)^14 - 4 is divisible by 13.
>> In other words that (17^47 + 2^12)^14 is congurent with 4
mod 13.
> Show that x^{12} = 1 (mod 13) for
x=1,2,3,4,5,6,7,8,9,10,11, and 12.
> Then note that since 17 = 4 (mod 14), we have
> (17^47 + 2^12)^14 = (4^{47} + 2^12)^14 (mod 13)
> = (4^{47} + 1)^14 (mod 13)
> = (4^{36}*4^{9} + 1)^14 (mod 13);
> if the number inside the parenthesis is not 0 (mod 13),
then its 14th
> power should equal its square (why?), so just show that the
square is
> congruent to 4 mod 13.
though. What does the proving x^12 tell me? Sorry if IÕm
daft, I just
dont get it.
--
Sigblock empty. By choice.
===
Subject: Re: Algebra problem
>1/ Prove that (17^47 + 2^12)^14 - 4 is divisible by 13.
> In other words that (17^47 + 2^12)^14 is congurent with 4
mod 13.
>> Show that x^{12} = 1 (mod 13) for
x=1,2,3,4,5,6,7,8,9,10,11, and 12.
>> Then note that since 17 = 4 (mod 14), we have
>> (17^47 + 2^12)^14 = (4^{47} + 2^12)^14 (mod 13)
>> = (4^{47} + 1)^14 (mod 13)
>> = (4^{36}*4^{9} + 1)^14 (mod 13);
>> if the number inside the parenthesis is not 0 (mod 13),
then its 14th
>> power should equal its square (why?), so just show that
the square is
>> congruent to 4 mod 13.
>though. What does the proving x^12 tell me? Sorry if IÕm
daft, I just
>dont get it.
If x^12 = 1 (mod 13) for all values of x not divisible by 13,
then,
first of all, what happens to x^{14}?
Well, x^{14} = x^{12+2} = x^{12}*x^2. If you know that
x^{12}=1 (mod
13), then it must follow that
x^{14} = x^{12}*x^2 = 1*x^2 = x^2 (mod 13)
so instead of having to calculate (17^{47} + 2^{12})^14, you
only need
to calculate (17^{47} + 2^{12})^2.
Same argument gives you that 4^{36} = 1 (mod 13), so that the
entire
thing is really just a much simpler calculation that you can
do
directly. (You might want to write 4^9 as (2^2)^9 = 2^{18} =
2^{12}*2^{6}).
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Line formula questions
Is the following information correct?
Slope Intercept ~ y = mx+b
Point Slope ~ y2-y1=m(x2-x1)
Standard ~ Ax + By = C
Slope = Y2 - Y1
-------
X2 - X1
Dependent - X
Independent - Y
-----
Find the equation of the line described in slope-intercept
form.
Through(3,5) and (4, -2)
I would find the slope via the slope formula and then I put it
in Point
Slope form:
-2 - 5 -7
------ = -
4 - 3 1
y-5 = -7 (x-3)
-
1
Right?
Has an x-intercept of 4 and a y-intercept of -3.
Ok... what do I do here? IÕm lost :(
Slope of 0 and through (2, -3)
IÕm lost again :(, answer I have down is y+3 = 0(x-2)
---
Vertical Lines are y =
Horizontal lines are x =
Right?
---
The price you pay for 4 pieces of pastry is $3.15
for 10 pieces of pastery is $7.65
Write the particular equation expressing price in terms of
pastry.
Ok.. please save me!
How do I determin which is the x (dependent or independent?)
and which is
the y (dependent or independent?)
The price depends on the number of pasterys so let x be the
number of
pasterys and let y be the price
$7.65 - $3.15 $4.5
----------- = --
10 - 4 6 pasteries
y - 3.15 = (4.5/6) (x - 4)
Did I do that correctly?
I could really use some advice =
---
FYI: Yes this was posted in both alt.algebra.help and
alt.math.undergrad in
an attempt to recive the most amount of replys in the
shortest amount of
time. Forgive me?
---
For your time.
===
Subject: Re: Line formula questions
>Is the following information correct?
>Dependent - X
>Independent - Y
As I teach that material, itÕs the other way round. y is the
dependent variable because its value depends on the value of
x.
>Find the equation of the line described in slope-intercept
form.
>Through(3,5) and (4, -2)
>I would find the slope via the slope formula and then I put
it in Point
>Slope form:
>-2 - 5 -7
>------ = -
>4 - 3 1
>y-5 = -7 (x-3)
> -
> 1
>Right?
Right so far, but you have more to do. First, 7/1 = 7:
y - 5 = -7(x-3)
Next, you were asked for slope-intercept form. Always read a
problem
when you think youÕve finished, to make sure 
youÕre answering
the
actual problem and not your own version. :-)
y - 5 = -7x + 21
y = -7x + 26
>Has an x-intercept of 4 and a y-intercept of -3.
>Ok... what do I do here? IÕm lost :(
Hint: Can you express an x intercept of 4 as a point in
(__,__)
form? And the same for the y intercept? Then you have two
points and
can solve the problem as above. (There are easier methods,
but I
find most beginning students would rather learn a few methods
thoroughly and rely on them.)
>Slope of 0 and through (2, -3)
>IÕm lost again :(, answer I have down is y+3 = 0(x-2)
Hmm... You have the slope, and you have a point. You have
used the
point-slope form correctly, but now again you must finish your
answer.
>Vertical Lines are y =
>Horizontal lines are x =
>Right?
No. Try plotting x = 4, for example, by points: (4,-3), (4,0),
(4,5). What is the orientation of that line?
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Re: Line formula questions
>Is the following information correct?
>Dependent - X
>Independent - Y
> As I teach that material, itÕs the other way round. y is 
the
> dependent variable because its value depends on the value
of x.
so
y dependent
x independent
>Find the equation of the line described in slope-intercept
form.
>Through(3,5) and (4, -2)
>I would find the slope via the slope formula and then I put
it in Point
>Slope form:
>-2 - 5 -7
>------ = -
>4 - 3 1
>y-5 = -7 (x-3)
> -
> 1
>Right?
> Right so far, but you have more to do. First, 7/1 = 7:
> y - 5 = -7(x-3)
> Next, you were asked for slope-intercept form. Always read
a problem
> when you think youÕve finished, to make sure 
youÕre
answering the
> actual problem and not your own version. :-)
> y - 5 = -7x + 21
> y = -7x + 26
Ok, I was asking if it was correct up to that point. Thats
what I get for
not being clear = The advice is great though.
>Has an x-intercept of 4 and a y-intercept of -3.
>Ok... what do I do here? IÕm lost :(
> Hint: Can you express an x intercept of 4 as a point in
(__,__)
> form? And the same for the y intercept? Then you have two
points and
> can solve the problem as above. (There are easier methods,
but I
> find most beginning students would rather learn a few 
methods
> thoroughly and rely on them.)
4,0
0,-3
^_^, this was not covered in class for some reason.
>Slope of 0 and through (2, -3)
>IÕm lost again :(, answer I have down is y+3 = 0(x-2)
> Hmm... You have the slope, and you have a point. You have
used the
> point-slope form correctly, but now again you must finish
your
> answer.
Ok
y = -3
>Vertical Lines are y =
>Horizontal lines are x =
>Right?
> No. Try plotting x = 4, for example, by points: (4,-3),
(4,0),
> (4,5). What is the orientation of that line?
Vertical, so xÕs are vertical and yÕs are 
horizontal.
> --
> Stan Brown, Oak Road Systems, Cortland County, New York, USA
> http://OakRoadSystems.com
surely
> reduces the number of useful answers you get.
Address munging is an attempt to prevent bots from harvesting
your e-mail
number of
useful
answers one recives is dependent upon the person replying.
Not everyone
will
killfile a user for not runing around screaming private
information.
> http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Re: Line formula questions
> Is the following information correct?
> Slope Intercept ~ y = mx+b
> Point Slope ~ y2-y1=m(x2-x1)
> Standard ~ Ax + By = C
> Slope = Y2 - Y1
> -------
> X2 - X1
> Dependent - X
> Independent - Y
> -----
> Find the equation of the line described in slope-intercept
form.
> Through(3,5) and (4, -2)
> I would find the slope via the slope formula and then I put
it in Point
> Slope form:
> -2 - 5 -7
> ------ = -
> 4 - 3 1
> y-5 = -7 (x-3)
> -
> 1
> Right?
Well, thatÕs not slope-intercept. You know the slope is -7
and you know
y = -7*x + b and you know 5 = -7*3 + b (or -2 = -7*4 + b) so
b = 26 and
y = -7*x + 26.
> Has an x-intercept of 4 and a y-intercept of -3.
> Ok... what do I do here? IÕm lost :(
y-intercept is 26, x-intercept is 26/7.
> Slope of 0 and through (2, -3)
> IÕm lost again :(, answer I have down is y+3 = 0(x-2)
Which is correct. y = -3
y = 0*x + b, y = -3 when x = 2, -3 = 0*2 + b, b = -3, y = -3.
> ---
> Vertical Lines are y =
_Horizontal_ lines are y = const.
> Horizontal lines are x =
_Vertical_ lines are x = const.
> Right?
> ---
> The price you pay for 4 pieces of pastry is $3.15
> for 10 pieces of pastery is $7.65
> Write the particular equation expressing price in terms of
pastry.
> Ok.. please save me!
> How do I determin which is the x (dependent or
independent?) and which is
> the y (dependent or independent?)
Price (dependent) in terms of pastry (independent).
> The price depends on the number of pasterys so let x be the
number of
> pasterys and let y be the price
Good start.
> $7.65 - $3.15 $4.5
> ----------- = --
> 10 - 4 6 pasteries
> y - 3.15 = (4.5/6) (x - 4)
> Did I do that correctly?
Well, letÕs see. The slope is 4.5/6 = 3/4 so y = (3/4)*x + 
b.
y = 3.15
when x = 4 (or 7.65 when x = 10) so 3.15 = (3/4)*4 + b so b =
0.15 and
y = (3/4)*x + 0.15.
You have y = (3/4)*(x - 4) + 3.15 = (3/4)*x -3 + 3.15. So we
agree.
> I could really use some advice =
My best advice is to forget everything except y = m*x + b
(except, of
course,for vertical lines which are x = const).
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: Line formula questions
> Price (dependent) in terms of pastry (independent).
so Price (dependent) = y and Pastry (independent) = x ?
Y is dependent upon the value of X ?
===
Subject: Re: Line formula questions
> Price (dependent) in terms of pastry (independent).
> so Price (dependent) = y and Pastry (independent) = x ?
> Y is dependent upon the value of X ?
Not really. The price is dependent on the number of pastries.
The
letters are entirely up to you as long as you say what they
represent.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: Line formula questions
> Price (dependent) in terms of pastry (independent).
> so Price (dependent) = y and Pastry (independent) = x ?
> Y is dependent upon the value of X ?
I believe that the names were originally chosen to cover the
case of
uniqueness.
For instance, if y = x^2 then the value of Y is UNIQUELY
determined by the
choice of x, but not vice versa.
( this is a necessary condition that a formula describe a
Function)
Bob Pease
===
Subject: Re: Line formula questions
> Is the following information correct?
> Slope Intercept ~ y = mx+b
> Point Slope ~ y2-y1=m(x2-x1)
> Standard ~ Ax + By = C
Yes.
> Has an x-intercept of 4 and a y-intercept of -3.
y-intercept of -3 means the point (0, -3).
> Ok... what do I do here? IÕm lost :(
Now find the line through those two points.
> Slope of 0 and through (2, -3)
> IÕm lost again :(, answer I have down is y+3 = 0(x-2)
ThatÕs correct, though not in simplest form.
Multiply it out.
y + 3 = 0(x - 2)
y + 3 = 0
y = -3
> ---
> Vertical Lines are y =
> Horizontal lines are x =
> Right?
Yes.
> The price you pay for 4 pieces of pastry is $3.15
> for 10 pieces of pastery is $7.65
> Write the particular equation expressing price in terms of
pastry.
> Ok.. please save me!
> How do I determin which is the x (dependent or
independent?) and which is
> the y (dependent or independent?)
The problem tells you. It says to find price in terms of
pieces of
pastry. Dependent is price, independent is pieces of pastry.
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Re: Line formula questions
> Vertical Lines are y =
> Horizontal lines are x =
> Right?
> Yes.
Whoops! ThatÕs backwards, of course.
*vertical* is x = c
*horizontal* is y = c
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Limit Proof
The question is Suppose there exists a that is in all Reals
such that x>a
implies f(x)>0. Prove: lim as x
-> infinity = infinity if and only if lim as 
x->1/infinity = 0.
IÕm first assuming true that lim as x-> 
infinity =infinity, and
IÕm trying
to show lim x->1/infinity = 0.
I have used the definition of the limit, or more 
specifically
the cluster
point definition of limit, that says for all cluster points
(a), such that
itÕs in the domain of function f, lim as x->a =L if and only
if for all E>0
there exists D>0 such that x is in domain and 0 < abs(x-a) <
D implies
abs( f (x) - L )<E.
Here is where IÕm stuck...I know that if lim as x-> 
infinity
implies that
for all E>0 there exists D>0 s.t. x is in domain and 0<
abs(x-infinity) < D
implies that abs (f (a)-L )<E... How do I follow up with
this...
===
Subject: Re: Limit Proof
> The question is Suppose there exists a that is in all Reals
such that
x>a
> implies f(x)>0.
Do I understand your question?
Suppose thereÕs an a such that for all real x, x > a implies
f(x) > 0.
> Prove: lim as x -> infinity = infinity if and 
only if lim as
> x->1/infinity = 0.
lim as x -> oo of what? Yes, lim(x->oo) x = oo = infinity.
Do you mean lim(x->oo) f(x), lim x->oo of f(x)? 1/oo has no
meaning.
===
Subject: Re: Limit Proof
> The question is Suppose there exists a that is in all Reals
such that
> implies f(x)>0.
> Do I understand your question?
> Suppose thereÕs an a such that for all real x, x > a
implies f(x) > 0.
ThatÕs what I meant.
> Prove: lim as x -> infinity = infinity if and 
only if lim as
> x->1/infinity = 0.
> lim as x -> oo of what? Yes, lim(x->oo) x = oo = infinity.
lim as x-> Infinity of f(x) = Infinity if and only 
if lim as
of
1/f(x) = 0.
> Do you mean lim(x->oo) f(x), lim x->oo of f(x)? 1/oo has no
meaning.
===
Subject: Re: Limit Proof
<%5dhb.700$_f.39@news1.central.cox.net> Suppose thereÕs an a
such that for all real x, x > a implies f(x) > 0.
> ThatÕs what I meant.
> lim as x-> Infinity of f(x) = Infinity if and 
only if lim as
of
> 1/f(x) = 0.
Assume lim(x->oo) f(x) = oo. Thus
for all n > 0, some N with for all x > N, f(x) > n
if eps > 0, then 1/eps > 0, some N with all for x > N, f(x) >
1/eps
also for all x > a, 0 < f(x). Thus
for all x > max(a,N), 0 < 1/f(x) < eps, |1/f(x)| < eps
Hence conclude
lim(x->oo) 1/f(x) = 0
The converse I leave for you.
===
Subject: Complex taylor expansion
How to expand the following function using Taylor series,
f(w) = (1+|w|^2)^{-1}, where w is a complex number?
Zedtoe
===
Subject: Re: Complex taylor expansion
> How to expand the following function using Taylor series,
> f(w) = (1+|w|^2)^{-1}, where w is a complex number?
> Zedtoe
The Taylor series is no more complicated using complex
numbers than it is
using real numbers. So, we make a = 1.
f(w) = f(a) + (w-a)fÕ(a) + 
((x-a)^2/2!)*fÕÕ(a) + . . .
You can work out that f(1) = 0.5, f Ō(1) = -0.5 and f 
ŌÕ(1) =
-1. Keep
going with that until you have enough values to zoom in on
your answer.
You
COULD use a complex number in place of a, but the idea is to
make the maths
easier, so use a real number. The complex part will come when
you plug it
back into w.
So keep that for future reference. But you have modulus
brackets around
the
w. That removes the fact that itÕs a complex number and 
makes
it real.
You
are only dealing with the magnitude, so there are no jÕs or
iÕs involved.
Just do it as you would normally with a Taylor Series.
MadJock
===
Subject: Re: Complex taylor expansion
I should also mention that f(w) = (1+|w|^2)^{-1) is a fairly
standard
expansion (or can be forced to fit it anyway)
if we substitude |w|^2 = -r, then we have (1-r)^(-1) This can
be written
as
a sum as follows
Sigma (k=0 to infinity) r^k
We can then substitute r back in:
Sigma (k=0 to infinity) (-|w|^2)^k
or more simply
Sigma (k=0 to infinity) (-|w|^(2k))
So itÕs -|w|^0 -|w|^2 - |w|^4 . . .
MadJock
===
Subject: Re: Complex taylor expansion
in your explanation, the derivative is actually delta f over
delta r
(r is the modulus of w).
If the function is f(w) = w/(1+|w|^2)
and since |w|^2 = w w^*,
Am I right to do the following:
delta f(w)/ delta w = (1+ww^*)^(-1) - w(w^*)(1+ww^*)^(-2),
by treating w and w^* indepent,
etc. to find higher order derivatives,
Zedtoe
> I should also mention that f(w) = (1+|w|^2)^{-1) is a
fairly standard
> expansion (or can be forced to fit it anyway)
> if we substitude |w|^2 = -r, then we have (1-r)^(-1) This
can be written
as
> a sum as follows
> Sigma (k=0 to infinity) r^k
> We can then substitute r back in:
> Sigma (k=0 to infinity) (-|w|^2)^k
> or more simply
> Sigma (k=0 to infinity) (-|w|^(2k))
> So itÕs -|w|^0 -|w|^2 - |w|^4 . . .
> MadJock
===
Subject: Re: Complex taylor expansion
No. Sorry I used f Ō(w). When youÕre expanding 
this, note
that the
outcome
will never be complex. It will always be real. So instead of
differentiating with respect to w, differentiate with respect
to |w|.
df/d|w|
MadJock
> in your explanation, the derivative is actually delta f
over delta r
> (r is the modulus of w).
> If the function is f(w) = w/(1+|w|^2)
> and since |w|^2 = w w^*,
> Am I right to do the following:
> delta f(w)/ delta w = (1+ww^*)^(-1) - w(w^*)(1+ww^*)^(-2),
> by treating w and w^* indepent,
> etc. to find higher order derivatives,
> Zedtoe
> I should also mention that f(w) = (1+|w|^2)^{-1) is a
fairly standard
> expansion (or can be forced to fit it anyway)
> if we substitude |w|^2 = -r, then we have (1-r)^(-1) This
can be
written as
> a sum as follows
> Sigma (k=0 to infinity) r^k
> We can then substitute r back in:
> Sigma (k=0 to infinity) (-|w|^2)^k
> or more simply
> Sigma (k=0 to infinity) (-|w|^(2k))
> So itÕs -|w|^0 -|w|^2 - |w|^4 . . .
> MadJock
===
Subject: Fourier Transform of sine or cosine
Hi
IÕm looking for the proof of the Fourier transform of a sine
or cosine
wave.
The answer is a delta function positioned at f = +/- f0,
where f0 = the
frequency of the sine/cosine. Here is my working so far - am
I far out?
If g(t) = sin(wt)
F{g(t)} = S sin(wt)exp(-jwt) dt
=(1/2) S (exp(jwt)-exp(-jwt))exp(-jwt) dt
=(1/2) S (1-exp(-j2wt)) dt
Notes: S - integral from -infinity to +infinity.
j = square root of -1, also known as i.
w = 2*pi*f
sin(wt) = exp(jwt)-exp(-jwt)
I am stuck on where to go from here. I donÕt know how to
translate this
expression into the dirac delta function. IÕd appreciate a
proof if anyone
can help!
MadJock
===
Subject: Re: Fourier Transform of sine or cosine
> Hi
> IÕm looking for the proof of the Fourier transform of a
sine or cosine
wave.
> The answer is a delta function positioned at f = +/- f0,
where f0 = the
> frequency of the sine/cosine. Here is my working so far -
am I far out?
> If g(t) = sin(wt)
> F{g(t)} = S sin(wt)exp(-jwt) dt
> =(1/2) S (exp(jwt)-exp(-jwt))exp(-jwt) dt
> =(1/2) S (1-exp(-j2wt)) dt
1 - e^-2jwt = 1 - cos -2wt - j.sin -2wt
= 1 - cos 2wt + j.sin 2wt
= 1 - cos^2 wt + j.sin^2 wt
= cos^2 wt + sin^2 wt - cos^2 wt + j.sin^2 wt
= sin^2 wt + j.sin^2 wt
= (1+j)sin^2 wt
So the integral = 0 when w = n.pi and oo otherwise
Have you missed a condition on g(t)? My reference book
requires absolute convergence of S g(t) dt. It also gives
the constant as 1/sqr 2.pi instead of 1/2.
> Notes: S - integral from -infinity to 
+infinity.
> j = square root of -1, also known as i.
> w = 2*pi*f
> sin(wt) = exp(jwt)-exp(-jwt)
> I am stuck on where to go from here. I donÕt know how to
translate this
> expression into the dirac delta function. IÕd appreciate a
proof if
anyone
> can help!
> MadJock
===
Subject: Tricks
I wonder if there are any more relations between the digits
in a number
written base 10 and simple operations on the line of the
following:
A natural number a = a_n * 10^n a_(n-1)* 10^(n-1) ... * a_0 *
10^0. Where every term a_j, is less then 10. (A number
written in base
10)
The number a is divisible by 11 if and only if the
alternating sum of
the terms is divisible by 11.
_n
11|a <=> 11 | (-1)^j * a_j
/_
j=0
The same number a is divisible by three if and only if three
divides the
sum of the terms
_n
3|a <=>3 | a_j
/_
j=0
Any help or references to litterature is welcome!
--
Sigblock empty. By choice.
===
Subject: Re: Tricks
> I wonder if there are any more relations between the digits
in a number
> written base 10 and simple operations on the line of the
following:
> A natural number a = a_n * 10^n a_(n-1)* 10^(n-1) ... * a_0
*
WhereÕs the plus signs?
> 10^0. Where every term a_j, is less then 10. (A number
written in base
> 10)
and an integer >= 0.
> The number a is divisible by 11 if and only if the
alternating sum of
> the terms is divisible by 11.
> _n
> 11|a <=> 11 | (-1)^j * a_j
> /_
> j=0
> The same number a is divisible by three if and only if
three divides the
> sum of the terms
> _n
> 3|a <=>3 | a_j
> /_
> j=0
Do you know about modular arithematic?
As 10 = 1 (mod 3)
sum a_j 10^j = sum a_j (mod 3)
As theyÕre equal modulus 3, their divisibility by 3 is the
same.
Same can be said about 9 as 10 = 1 (mod 9)
which gives us the process of casting out 9Õs.
As for 11Õs, note 10 = -1 (mod 11). Thus
sum a_j 10^j = sum (-1)^j a_j (mod 11)
As theyÕre equal modulus 11, their divisibility by 11 is the
same.
===
Subject: Re: Tricks
>> I wonder if there are any more relations between the
digits in a number
>> written base 10 and simple operations on the line of the
following:
>> A natural number a = a_n * 10^n a_(n-1)* 10^(n-1) ... *
a_0 *
> WhereÕs the plus signs?
Woops, sorry :)
> Do you know about modular arithematic?
> As 10 = 1 (mod 3)
> sum a_j 10^j = sum a_j (mod 3)
> As theyÕre equal modulus 3, their divisibility by 3 is the
same.
> Same can be said about 9 as 10 = 1 (mod 9)
> which gives us the process of casting out 9Õs.
> As for 11Õs, note 10 = -1 (mod 11). Thus
> sum a_j 10^j = sum (-1)^j a_j (mod 11)
> As theyÕre equal modulus 11, their divisibility by 11 is
the same.
This I do know, and had to prove it in an exercise. My point,
and why I
ask for more similar relations, is that theese two relations
make
checking if a natural number can be divided by 3 and 11 childs
play. That whould be nice to have for more divisors, ideallly
if the
relation in some form existed for say all, of even just most,
primes. I
very well.
--
Sigblock empty. By choice.
===
Subject: Re: Tricks
> As 10 = 1 (mod 3)
> a = sum a_j 10^j = sum a_j (mod 3)
> Same can be said about 9 as 10 = 1 (mod 9)
> which gives us the process of casting out 9Õs.
> As for 11Õs, note 10 = -1 (mod 11). Thus
> a = sum a_j 10^j = sum (-1)^j a_j (mod 11)
> This I do know, and had to prove it in an exercise. My
point, and why I
> ask for more similar relations, is that theese two
relations make
> checking if a natural number can be divided by 3 and 11
childs
> play. That whould be nice to have for more divisors,
ideallly if the
> relation in some form existed for say all, of even just
most, primes. I
> very well.
ThereÕs trival
2|a iff a_0 even
5|a iff 5|a_0
10|a iff a_0 = 0
by similar modular reasonings. Also
20|a iff a_1 even, a_0 = 0
25|a iff 25 | a_1 10 + a_0
50|a iff 5|a_1, a_0 = 0
100|a iff a_1 = a_0 = 0
by different reasonings. These four seem generalizable to any
number of
digets, 200, 250, 500, 1000, 2000, 2500, ...
Some more
4|a iff 4 | a_1 10 + a_0
8|a iff 8 | a_2 100 + a_1 10 + a_0
16|a iff ..
...
also generalizes to 40, 80 etc
===
Subject: Re: Tricks
<cd3afd3d3cd14c9cb1c7351fbae1a9fe@news.teranews.com> As 10 =
1 (mod 3)
>> a = sum a_j 10^j = sum a_j (mod 3)
> Same can be said about 9 as 10 = 1 (mod 9)
>> which gives us the process of casting out 9Õs.
> As for 11Õs, note 10 = -1 (mod 11). Thus
>> a = sum a_j 10^j = sum (-1)^j a_j (mod 11)
>> This I do know, and had to prove it in an exercise. My
point, and why I
>> ask for more similar relations, is that theese two
relations make
>> checking if a natural number can be divided by 3 and 11
childs
>> play. That whould be nice to have for more divisors,
ideallly if the
>> relation in some form existed for say all, of even just
most, primes. I
>> very well.
> ThereÕs trival
> 2|a iff a_0 even
> 5|a iff 5|a_0
> 10|a iff a_0 = 0
> by similar modular reasonings. Also
> 20|a iff a_1 even, a_0 = 0
> 25|a iff 25 | a_1 10 + a_0
> 50|a iff 5|a_1, a_0 = 0
> 100|a iff a_1 = a_0 = 0
> by different reasonings. These four seem generalizable to
any number of
> digets, 200, 250, 500, 1000, 2000, 2500, ...
> Some more
> 4|a iff 4 | a_1 10 + a_0
> 8|a iff 8 | a_2 100 + a_1 10 + a_0
> 16|a iff ..
> ...
> also generalizes to 40, 80 etc
Duh! HadnÕt though about 2 and five, have to 
have been almost
sleeping
when I read it. Anyway, suppose the following(bear with me,
IÕm
practicing my formulation skills, comments welcome):
p is a prime
n is a natural number
set A = { a_0, a_1, , a_j} where n = a_0 * 10^0 + ... + a_j *
10^j
For each p there exists a relation, R_p, between p and one or
more
elements in A so that n = (p R_p A) (mod p).
(btw is this a correct use of a relation ???)
Examples:
p=2:
2|a_0 <=> 2|n
p=3:
3|sum(k=0,j) a_k <=> 3
same for 5 and 11
Now some Qs:
Is it possible give a counterexample and how would one go
about doing
that?
Is it possible to prove that there exists such a relation for
all p and
has anyone attempted it?
The latter of course is what is interesting. Again any
reference to work
on this is appreciated. For all I know this might be very
basic, and
IÕve missed out on it so far in my life :)
--
Sigblock empty. By choice.
===
Subject: Re: Tricks
===
Subject: Re: Tricks
>> a = sum a_j 10^j
>> a = sum a_j 10^j = sum a_j (mod 3)
>> a = sum a_j 10^j = sum a_j (mod 9)
>> a = sum a_j 10^j = sum (-1)^j a_j (mod 11)
>> 2|a iff a_0 even
>> 5|a iff 5|a_0
>> 10|a iff a_0 = 0
>> 20|a iff a_1 even, a_0 = 0
>> 25|a iff 25 | a_1 10 + a_0
>> 50|a iff 5|a_1, a_0 = 0
> 100|a iff a_1 = a_0 = 0
>> generalizes to any number of digets 200, 250, 500, 1000,
2000, ...
>> 4|a iff 4 | a_1 10 + a_0
>> 8|a iff 8 | a_2 100 + a_1 10 + a_0
>> 16|a iff ..
>> ...
>> also generalizes to 40, 80 etc
>p is a prime
>n is a natural number
>set A = { a_0, a_1, , a_j} where n = a_0 * 10^0 + ... + a_j
* 10^j
>For each p there exists a relation, R_p, between p and one
or more
>elements in A so that n = (p R_p A) (mod p).
>(btw is this a correct use of a relation ???)
Makes no sense. (mod p) is a relation between two numbers.
Tho n is a number, p R_p A isnÕt a number, 
itÕs statement.
In addition R_p is sloppy, in need of clearer description.
>Examples:
>p=2: 2|a_0 <=> 2|n
>p=3: 3|sum(k=0,j) a_k <=> 3
>same for 5 and 11
----
===
Subject: Re: Tricks
<cd3afd3d3cd14c9cb1c7351fbae1a9fe@news.teranews.com>
<5cface913fd5dc576acfa5885b5993d0@news.teranews.com>> p is a
prime
>> n is a natural number
>> set A = { a_0, a_1, , a_j} where n = a_0 * 10^0 + ... +
a_j * 10^j
>> For each p there exists a relation, R_p, between p and one
or more
>> elements in A so that n = (p R_p A) (mod p).
>> (btw is this a correct use of a relation ???)
> Makes no sense. (mod p) is a relation between two numbers.
> Tho n is a number, p R_p A isnÕt a number, 
itÕs statement.
> In addition R_p is sloppy, in need of clearer description.
Ok,
well say then instead that there exists a function f from
A^i, where
i <= j, to N so that n = f (mod p). Would that be more
correct?
Or maybe just defining the function to be f: N -> N, and
leaving the a_j
stuff to be defined in the function.
Better??
--
Sigblock empty. By choice.
===
Subject: Cheap U.S. College with good undergrad Math program?
Any suggestions anyone? IÕm interested in getting a
bachelorÕs, but IÕm
looking for a quality program at a bargain price. I
appreciate any
feedback...
--
===
Subject: Re: Use of variable independence, core error
Visiting Assistant Professor at the University of Montana.
[mindless repetition of the same ßawed argument removed]
>Well it turns out that over a hundred years ago algebraic
integers
>were defined as roots of monic polynomials with integer
coefficients.
>It was a good idea for the time, but mathematicians failed
to realize
>that it was slightly ßawed in that some roots of non-monic
>polynomials with integer coefficients need to be included or
you can
>create supposed proofs of two different and opposite
conclusions in
>the ring of algebraic integers.
Once again, this makes absolutely no logical sense
whatsoever. You are
grasping at straws.
There can be no contradiction in introducing a definition.
Defining
the set of algebraic integers to be the set of all complex
roots of
monic polynomials with integer coefficients cannot introduce a
contradiction. The set is well defined.
So, you must mean that SOME property of the set of algebraic
integers
that we are using is false. Specifically, you MUST be saying
that the
algebraic integers do NOT form a commutative ring. So you
must be
saying one of the following:
(a) 0 is not an algebraic integer; or
(b) there are algebraic integers a and b such that a+b is not
an
algebraic integer; or
(c) there is an algebraic integer a such that -a is not an
algebraic
integer; or
(d) 1 is not an algebraic integer; or
(e) there are algebraic integers a and b such that a*b is not
an
algebraic integer; or
(f) the sum of algebraic integers is not associative; or
(g) the sum of algebraic integers is not commutative; or
(h) the multiplication of algebraic integers is not
associative; or
(i) the multiplication of algebraic integers is not
commutative; or
(j) the multiplication of algebraic integers does not
distribute over
the sum.
However, (f)-(j) are clearly false, since the respective
properties
are true of COMPLEX numbers, and the sum and multiplication of
algebraic integers is just the multiplication of complex
numbers. Also, (a) and (d) are clearly false, as 0 is a root
of
p(x)=x, and 1 is a root of P(x)=x-1.
And (c) is false, because if a is a root of
x^n+a_{n-1}x^{n-1}+...+a_1x+a_0, then -a is a root of
x^n-a_{n-1}x^{n-1}+...+a_1*(-1)^(n-1)x + a_0*(-1)^n.
So you must be saying that there are algebraic integers whose
sum is
not an algebraic integer, or else that there are algebraic
integers
whose products are not algebraic integers.
Which one?
The reason you are making this ludicrous claims is plain to
everyone:
you cannot see an error in YOUR argument (despite the fact
that the
exact spot where your leap of logic occurs has been pinpointed
repeatedly). You cannot find an error in the explicit
disproves of
your claim, because there isnÕt one (these disproofs consist
merely of
exhibiting explicit numbers that satisfy your hypothesis but
not your
conclusions). So, since you are unwilling to grant the
possibility
that your argument may be wrong, you conclude that there must
be some
contradiction in Ōcorecreated by the 
definition of algebraic
integer, even though it makes absolutely no sense to make
such a
claim.
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 manÕs 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
magidin@math.berkeley.edu
===
Subject: Re: Use of variable independence, core error
> [mindless repetition of the same ßawed argument removed]
Readers can, of course, see my original post.
But notice that Arturo Magidin clearly is running away from
the math.
>Well it turns out that over a hundred years ago algebraic
integers
>were defined as roots of monic polynomials with integer
coefficients.
>It was a good idea for the time, but mathematicians failed
to realize
>that it was slightly ßawed in that some roots of non-monic
>polynomials with integer coefficients need to be included or
you can
>create supposed proofs of two different and opposite
conclusions in
>the ring of algebraic integers.
> Once again, this makes absolutely no logical sense
whatsoever. You are
> grasping at straws.
Actually it *does* make sense as itÕs a compelling story 
that
fits the
facts.
Algebraic integers were defined as roots of *monic*
polynomials with
integer coefficients.
However, integers are roots of both monic and non-monic
polynomials
with integer coefficients, e.g. (2x+1)(x+2) = 2x^2 + 3x + 2.
So algebraic integers were defined in such a way that an
asymmetry was
introduced. And it turns out that allows you to find proofs
that
contradict each other.
> There can be no contradiction in introducing a definition.
Defining
> the set of algebraic integers to be the set of all complex
roots of
> monic polynomials with integer coefficients cannot introduce
a
> contradiction. The set is well defined.
Well, would you rather dismiss algebra?
Readers should note how simple the algebra is in the argument
which
IÕve used to highlight that there must be an error in core.
See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782
There you see basic *axioms* of mathematics leading to the
conclusion,
while here you have this poster Arturo Magidin holding fast
to an
arbitrary *definition*.
It amazes me that any of you would for a second consider
tossing
*axioms* to keep a definition as arbitrary as saying, ok,
include
roots of monic polynomials with integer coefficients.
IÕve concluded that fear can explain the irrationality.
I doubt any of you would toss axioms out otherwise.
But remember, what you call mathematics is a body of
discoveries by
people like me who were willing to look at the unknown and
see its
true face. They were brave enough to face fear and make
discoveries
for the benefit of humanity.
Unfortunately, clearly, many of you have never had to be
brave.
> So, you must mean that SOME property of the set of
algebraic integers
> that we are using is false. Specifically, you MUST be saying
that the
> algebraic integers do NOT form a commutative ring. So you
must be
> saying one of the following:
Nope. This poster is irrational.
The definition of algebraic integers *does* give a commutative
ring;
however, itÕs ßawed as IÕve explained 
using rather basic
algebra.
> (a) 0 is not an algebraic integer; or
> (b) there are algebraic integers a and b such that a+b is
not an
> algebraic integer; or
> (c) there is an algebraic integer a such that -a is not an
algebraic
> integer; or
> (d) 1 is not an algebraic integer; or
> (e) there are algebraic integers a and b such that a*b is
not an
> algebraic integer; or
> (f) the sum of algebraic integers is not associative; or
> (g) the sum of algebraic integers is not commutative; or
> (h) the multiplication of algebraic integers is not
associative; or
> (i) the multiplication of algebraic integers is not
commutative; or
> (j) the multiplication of algebraic integers does not
distribute over
> the sum.
> However, (f)-(j) are clearly false, since the respective
properties
> are true of COMPLEX numbers, and the sum and multiplication
of
> algebraic integers is just the multiplication of complex
> numbers. Also, (a) and (d) are clearly false, as 0 is a
root of
> p(x)=x, and 1 is a root of P(x)=x-1.
> And (c) is false, because if a is a root of
> x^n+a_{n-1}x^{n-1}+...+a_1x+a_0, then -a is a root of
> x^n-a_{n-1}x^{n-1}+...+a_1*(-1)^(n-1)x + a_0*(-1)^n.
> So you must be saying that there are algebraic integers
whose sum is
> not an algebraic integer, or else that there are algebraic
integers
> whose products are not algebraic integers.
> Which one?
Nope.
> The reason you are making this ludicrous claims is plain to
everyone:
> you cannot see an error in YOUR argument (despite the fact
that the
> exact spot where your leap of logic occurs has been
pinpointed
> repeatedly). You cannot find an error in the explicit
disproves of
> your claim, because there isnÕt one (these disproofs
consist merely of
> exhibiting explicit numbers that satisfy your hypothesis
but not your
> conclusions). So, since you are unwilling to grant the
possibility
> that your argument may be wrong, you conclude that there
must be some
> contradiction in Ōcorecreated by the 
definition of
algebraic
> integer, even though it makes absolutely no sense to make
such a
> claim.
The poster is lying about finding a leap of logic.
I can go step-by-step to any level of detail necessary, and
it should
be possible to use proof checkers on the argument as well.
But youÕre terrified, so math society runs 
away.
LISTEN, I said you should be able to use proof checkers and I
can go
step-by-step to ANY level of detail necessary.
James Harris
===
Subject: Re: Use of variable independence, core error
>> [mindless repetition of the same ßawed argument removed]
>Readers can, of course, see my original post.
>But notice that Arturo Magidin clearly is running away from
the math.
IÕve addressed it elsewhere. Whenever I post anything that
you donÕt
understand, you either run away or accuse me of lying.
[.snip.]
>> The reason you are making this ludicrous claims is plain
to everyone:
>> you cannot see an error in YOUR argument (despite the fact
that the
>> exact spot where your leap of logic occurs has been
pinpointed
>> repeatedly). You cannot find an error in the explicit
disproves of
>> your claim, because there isnÕt one (these disproofs
consist merely of
>> exhibiting explicit numbers that satisfy your hypothesis
but not your
>> conclusions). So, since you are unwilling to grant the
possibility
>> that your argument may be wrong, you conclude that there
must be some
>> contradiction in Ōcorecreated by the 
definition of
algebraic
>> integer, even though it makes absolutely no sense to make
such a
>> claim.
>The poster is lying about finding a leap of logic.
No, James.
At ->best<- you could say that you disagree that what I have
pointed
out is a leap of logic, but whenever you claim I am lying
about
pointing to something in your proof, it is ->you<- who are
engaging in
malicious character assassination, libel, and dishonesty.
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 manÕs 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
magidin@math.berkeley.edu
===
Subject: Re: Use of variable independence, core error
...
> Algebraic integers were defined as roots of *monic*
polynomials with
> integer coefficients.
> However, integers are roots of both monic and non-monic
polynomials
> with integer coefficients, e.g. (2x+1)(x+2) = 2x^2 + 3x + 2.
Yes, and so what? Algebraic integers are also roots of both
monic and
non-monic polynomials. The definition of algebraic integers
only states
that an algebraic number is an algebraic integer only if it
is a root of
at least one monic polynomial.
> So algebraic integers were defined in such a way that an
asymmetry was
> introduced.
What asymmetry? Note that the distinction between integers
and true
rationals is that rationals are *not* a root of a monic
polynomial with
integer coefficients. The same distinction we have between
algebraic
integers and other algebraic numbers.
> And it turns out that allows you to find proofs that
> contradict each other.
This is nonsense. The definition of algebraic integers only
label a
set of numbers with the name algebraic integers. How can such
naming
lead to a contradiction? The name could also have been
numbers that
are roots of monic polynomials. Would that name lead to a
contradiction?
The only possibility is that in your contradiction you use a
result
on algebraic integers that is wrong. Which is either a
theorem you
use or your own logic. So you should identify the theorem you
are
using that is wrong. The definition in and of itself is *not*
wrong,
it just gives a labeling.
One such theorem you use is that the algebraic integers form
a ring.
Yes, that *is* a theorem that must be proven. To prove it is
a ring
is to show the following (a, b and c algebraic integers):
R1: given a and b, a + b is also an algebraic integer (closed
under
addition).
R2: given a and b, a * b is also an algebraic integer (closed
under
multiplication)
the remaining requirements are trivial (a + b = b + a, etc)
and follow
because algebraic integers are complex numbers. So to show
that the
algebraic integers do not form a ring you have to show that
either
R1 or R2 is false. The simplest proof of R1 and R2 is one
where
given monic polynomials Pa(x) and Pb(x) of which a resp. b
are roots,
two new monic polynomials Pc(x) and Pd(x) are constructed of
which
a+b resp. a*b are roots. So either the construction is wrong,
or
indeed the algebraic integers do form a ring.
So what do you think? Is the construction wrong or do the
algebraic
integers indeed form a ring?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Use of variable independence, core error
> ...
> > Algebraic integers were defined as roots of *monic*
polynomials with
> > integer coefficients.
> >
> > However, integers are roots of both monic and non-monic
polynomials
> > with integer coefficients, e.g. (2x+1)(x+2) = 2x^2 + 3x +
2.
> Yes, and so what? Algebraic integers are also roots of both
monic and
> non-monic polynomials. The definition of algebraic integers
only states
> that an algebraic number is an algebraic integer only if it
is a root of
> at least one monic polynomial.
However, as repeatedly shown by various posters on this board,
algebraic integers cannot be roots of non-monic polynomials
with
integer coefficients that are irreducible over Q.
ThatÕs the exclusion that leads to the ability to *appear* 
to
prove
two different and contradictory things, in the ring of
algebraic
integers.
> > So algebraic integers were defined in such a way that an
asymmetry was
> > introduced.
> What asymmetry? Note that the distinction between integers
and true
> rationals is that rationals are *not* a root of a monic
polynomial with
> integer coefficients. The same distinction we have between
algebraic
> integers and other algebraic numbers.
However, as repeatedly shown by various posters on this board,
algebraic integers cannot be roots of non-monic polynomials
with
integer coefficients that are irreducible over Q.
> > And it turns out that allows you to find proofs that
> > contradict each other.
> This is nonsense. The definition of algebraic integers only
label a
> set of numbers with the name algebraic integers. How can
such naming
> lead to a contradiction? The name could also have been
numbers that
> are roots of monic polynomials. Would that name lead to a
> contradiction?
ItÕs not the name; itÕs the exclusion.
> The only possibility is that in your contradiction you use
a result
> on algebraic integers that is wrong. Which is either a
theorem you
> use or your own logic. So you should identify the theorem
you are
> using that is wrong. The definition in and of itself is
*not* wrong,
> it just gives a labeling.
But my proof is very short, which means it is machine
checkable.
It relies primarily on the notion that factors of P(m) have
terms that
are independent of m, found by setting m=0, and that these
terms are
in fact independent of m.
If the factors were polynomials, thereÕd be no argument, as
who out
there believes that you can have a polynomial P(x), a
polynomial
factor f(x) of P(x), where P(x) has 7 as a factor, for all
algebraic
integers x, and f(0)=7, that 7 can be divided off such that
P(0) is
coprime to 7, without f(x) having 7 as a factor?
It looks complicated, but *writing* it out with actual
polynomials
makes it easy.
However, with non-polynomial factors you have to trust the
mathematical logic.
ItÕs like ßying a plane by its instruments at night or 
during
a
storm.
Mathematicians are acting like they have to physically *see*
when the
mathematical logic suffices.
> One such theorem you use is that the algebraic integers
form a ring.
> Yes, that *is* a theorem that must be proven. To prove it
is a ring
> is to show the following (a, b and c algebraic integers):
> R1: given a and b, a + b is also an algebraic integer
(closed under
> addition).
> R2: given a and b, a * b is also an algebraic integer
(closed under
> multiplication)
> the remaining requirements are trivial (a + b = b + a, etc)
and follow
> because algebraic integers are complex numbers. So to show
that the
> algebraic integers do not form a ring you have to show that
either
> R1 or R2 is false. The simplest proof of R1 and R2 is one
where
> given monic polynomials Pa(x) and Pb(x) of which a resp. b
are roots,
> two new monic polynomials Pc(x) and Pd(x) are constructed
of which
> a+b resp. a*b are roots. So either the construction is
wrong, or
> indeed the algebraic integers do form a ring.
> So what do you think? Is the construction wrong or do the
algebraic
> integers indeed form a ring?
IÕve never said that algebraic integers donÕt 
form a ring.
What they form is a ßawed ring, which doesnÕt include all 
the
numbers
it must to prevent the possibility of appearing to prove two
different
but opposite things.
James Harris
===
Subject: help with velocity
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h944pol01702;
Needing help with velocity word problem can anyone help me
come up
with the equasion? HereÕs the question:
A ball is thrown upward from the top of a 196 ft building
with an
initial velocity of 64 ft per second. Approximate the maximum
height
of the ball.
What height of the ball am I looking for? I donÕt think I 
get
the
question and thatÕd why the equasion is so hard for me. Can
anyone
please help.
samanthe79
===
Subject: Re: help with velocity
>Needing help with velocity word problem can anyone help me
come up
>with the equasion? HereÕs the question:
>A ball is thrown upward from the top of a 196 ft building
with an
>initial velocity of 64 ft per second. Approximate the
maximum height
>of the ball.
>What height of the ball am I looking for?
YouÕre looking for the maximum height. That is so obvious
that I
fear I have misunderstood your question.
general advice: Write the equation of height for this ball.
Your
textbook will have shown you how the height of any
free-moving body
varies with time because of gravity.
Then use that equation to find the maximum height. If 
youÕre in
calculus, use derivatives; if youÕre in algebra, 
find the
vertex of
the parabola.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Help with velocity
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h94KMFd28338;
This is a standard free fall problem.
The equation for the height is given by:
y = Ho + Vo(t) - 16t^2
where Ho is the initial height, and Vo is the initial
velocity.
For your problem: y = 196 + 64t - 16t^2
Using Calculus, find yÕ, set it equal to 0, and 
solve.
y= 64 - 32t = 0 . . . hence, t = 2 seconds.
Then maximum height is: y = 196 + 64(2) - 16(2^2) = 260 feet.
[I assume they want the height above the ground.
Otherwise, they wouldnÕt have mentioned the height of the
building.)
If youÕre NOT in a Calculus course, we can still solve it.
The height function is a quadratic; its graph is a parabola.
A parabola has its vertex at x = (-b)/{2a}
In your equation: a = -16, b = 64, c = 196
The vertex is at: x = (-64)/(2[-16]) = 2
Then y = 196 + 64(2) - 16(2^2) = 260
The vertex is at (2,260).
Since the parabola opens DOWNward (a = -16),
the vertex is the maximum (heighest) point of the parabola.
Therefore, the maximum height is 260 feet.
===
Subject: Help with inequality
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h95LO7531058;
Can someone help me solve this inequality?
(n / 2 ^ i) <= 2
Read : n divided by 2 to the power i is less than or equal to
2
Solve for i
-Sam
===
Subject: Re: Help with inequality
>Can someone help me solve this inequality?
> (n / 2 ^ i) <= 2
>Read : n divided by 2 to the power i is less than or equal
to 2
>Solve for i
Are we to assume that n is a positive number? (If n is
negative or
zero, the statement is true for every i.)
Assuming n is positive, take the base-2 log of both sides:
log_2(n) - i <= 1
and itÕs pretty straightforward from there.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Help with inequality
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h97E9eS23737;
We have: n/(2^i) <= 2
Since 2^i is positive (for real i),
we can multiply both sides: n <= 2(2^1) = 2^(i+1)
We have: 2^(i+1) >= n
Take logs (any base): log[2^(i+1)] = log(n)
Then we have: (i+1)log(2) = log(n)
Hence: i + 1 = (log n)/(log 2)
Therefore: i = (log n)/(log 2) - 1
===
Subject: Re: Help with inequality
For some reason you like to not include context of problem,
nor quote post
thatÕs being replied to. That is not good. I do same, enjoy.
The OP asked for i when <=, not for =.
You solved for i, but didnÕt answer the < part.
===
Subject: DE
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h95LO5L31044;
does anyone know how the method of undetermined coefficients
(educated
guess for a particular solution depending on the form of g(x)
in non
homogenous DE ayÕ+ by+cy = 
g(x) ) can be used to find a
particular
solution of yÕ+ y = sinxcos2x. ?
===
Subject: Reply to DE
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h962hgE19755;
On 5 Oct 03 16:24:14 -0400 (EDT) JC
<jodelcharles@hotmail.com> asked:
>does anyone know how the method of undetermined coefficients
>(educated
>guess for a particular solution depending on the form of
g(x) in non
>homogenous DE ayÕ+ by+cy = 
g(x) ) can be used to find a
>particular
>solution of yÕ+ y = sinxcos2x. ?
sin(x + 2x) = sinx cos(2x) + sin(2x) cosx ... (1)
sin(x - 2x) = sinx cos(2x) - sin(2x) cosx ... (2)
Add equations (1) and (2) together:
sin(3x) + sin(-x) = 2 sinx cos(2x)
=> sinx cos(2x) = 1/2 sin(3x) - 1/2 sinx.
Make this replacement and then use the method of undetermined
coefficients in the usual way.
===
Subject: Addition to the problem for a math genius.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h97LQCW23496;
For x^a and a^x, a is a real number, nonnegative.
Algebra needed. No calculators.
===
Subject: Limits
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h97JmLf15830;
I havenÕt taken an analysis class so IÕm not 
exactly sure how
to prove
statements formally. I understand the statement but how would
I show
that this example is valid?
Let f(z) be a continuous function for all z. Show that if
f(zo) does
not equal 0, then there must be a neighborhood of zo in which
f(z)
does not equal 0.
===
Subject: Re: Limits
>I havenÕt taken an analysis class so IÕm not 
exactly sure
how to prove
>statements formally. I understand the statement but how
would I show
>that this example is valid?
>Let f(z) be a continuous function for all z. Show that if
f(zo) does
>not equal 0, then there must be a neighborhood of zo in
which f(z)
>does not equal 0.
Start with the definition of continuous. It is usually
expressed in
terms of delta and epsilon but instead of the greek
characters I will
use d and e.
f(z) is continuous at z0 if for any e>0 there exists a d>0
such that
|f(z)-f(z0)|<e for all z where |z-z0|<d
Expressed colloquially, you can force f(z) to become
arbitrarily close
to f(z0) by making z come sufficiently close to z0.
You are given that f(z0) is not 0. Let e = |f(z0)|. e
satisfies the
condition stated above. Therefore, there exists a d such that
when
|z-z0)|<d, |f(z)-f(z0)|<e. If f(z) were 0, then the previous
relation
would be |0-f(z0)|<e => |-f(z0)|<e => |f(z0)|<e which
contradicts the
fact that |f(z0)|=e.
Therefore, d is a neighborhood satisfying the request.
<<Remove the del for email>>
===
Subject: Re
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h98Ftxe31780;
Simply say that f(z)=a where a>0 without loss of generality;
then by
definition of continuity there is a neighborhood of z, say the
open
interval (z-d, z+d) where d>0, s.t. a-a/2<f(z)<a+a/2, i.e.
for every x
in this neighborhood f(z)>a/2>0. The other case a<0 is proved
in an
analogous way.
===
Subject: A problem for a math genius.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h97LQCM23490;
Prove algebraically that curves x^a (x to the a) and a^x (a
to the
x)have a number of crossings (intersects) dependent on a,
and, if a=e
(~2.71) then the curves are tangential, i.e. they cross
exactly once.
Hint: Observe functions z=x^a-a^x and zÕ=ax^(a-1)-a^xln(a)
===
Subject: Analyzing Random Selections
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h97Lrgo25449;
I need help with the following problem, I will try to explain
it as
simple as I can.
Im doing a test on processors. There are different tools the
test is
conducted on. There are 50 tests (Lots 1 thru 50) performed.
There are
4 sets of Tools, each ranging from 2 to 5 tools each.
Set1: Tool 1 Tool 2 Tool 3
Set2: Tool 1 Tool 2 Tool 3 Tool 4 Tool 5
Set3: Tool 1 Tool 2
Set4: Tool 1 Tool 2 Tool 3
The tests are conducted in a random order using random tools.
So the result table would look something like.
--------------------------------------------------------------
Lot1 Lot2 Lot3 Lot4 Lot5 ......... Lot49 Lot50
--------------------------------------------------------------
Tool1 Tool3 Tool1 Tool2 - ......... Tool3 Tool2
Tool4 Tool4 - Tool5 Tool3 ......... - Tool4
- Tool2 Tool2 - Tool1 ......... Tool1 -
Tool1 Tool3 Tool1 Tool2 Tool3 ......... Tool2 Tool1
--------------------------------------------------------------
60% 40% 90% 77% 100% 40% 100%
--------------------------------------------------------------
So Lot1 means that the processor passed through (Tool1|Set1),
(Tool4|Set2) and (Tool1|Set4) with a success ratio of 60%.
Lot2 means that the processor passed through (Tool3|Set1),
(Tool4|Set2), (Tool2|Set3) and (Tool3|Set4) with a success
ratio of
40%.
Lot3 means the processor passed through (Tool1|Set1),
(Tool2|Set3)and
(Tool1|Set4) with a success ratio of 90%.
Lot4 means the processor passed through (Tool2|Set1),
(Tool5|Set2)
and (Tool2|Set4) with a success ratio of 77%.
.
.
.
Lot50 means the processor passed through (Tool2|Set1),
(Tool4|Set2),
(Tool1|Set4) with a success ratio of 100%.
I am writing an algorithm to find the tool which is not 
working
properly or which has the lowest overall yeild. So say that
the reult
of the processor passing from Lot 1 is 60%, Lot 2 is 40%
there are
chances that there is a problem with the common tool in both
i.e. Tool
4 from Set 2.
Now all the tests are random that means there can be any
number of
Tools selected from each Set.
Any ideas/input will be appreciated.
VK
===
Subject: funny boards
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h97MsI129660;
doing grad studies (math) at ga tech, and stumbled across
this forum.
funny stuff, for sure. people think that artists and
mathematicians
are so different, but we have one thing in common: fairly
large egos
:)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
ed
webmaster: <a href=http://www.edu-directory.org/>Continuing
Education</a>
===
Subject: need help with calculus problems
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h991Os108084;
F(x)=cosx-sinx
Find the local extrema of f and the intervals on which f is
increasing or is decreasing, and sketch the graph of f.
===
Subject: Re: need help with calculus problems
>F(x)=cosx-sinx
>Find the local extrema of f and the intervals on which f is
>increasing or is decreasing, and sketch the graph of f.
This is a completely routine problem. Please explain
_specifically_
where you are getting stuck.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
surely
reduces the number of useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Re: need help with calculus problems
> F(x)=cosx-sinx
> Find the local extrema of f and the intervals on which f is
> increasing or is decreasing, and sketch the graph of f.
Step one, whatÕs fÕ(x).
Step two, solve fÕ(x) = 0.
Step three, for what intervals is
fÕ(x) > 0 and for what intervals is fÕ(x) < 
0?
===
Subject: interesting
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h99Fi2d01038;
It is very interesting to see the life in the usernet. I have
learned
a great deal following this thread because I was interested
in High IQ
but I also now achnoledge that virute is also improtant in a
person.
roddy
===
Subject: RE: Seating problem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h98Fv5s31914;
What a lovely problem. I have never seen it before. I will
use it.
Lets take the string aabbccdd to be the four sets of
identical
twins. We can rearrange this string in (8!)/(2)^4 ways. Use
the
inclusion-exclusion principle to find the number of ways to
rearrange
the string in which at least two twins (same letters) are
together:
SUM{(-1)^(k+1)*combin(4,k)*[(8-k)!/2^(4-k)], k=1 to 4}. Now
the
difference in those two numbers is 864, no two are standing
together.
===
Subject: science and must religion agree or religion is wrong
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h99FiFt01055;
http://www.mazepath.com/uncleal/eotvos.htm
visited Uncle Al web site and was very impressed and emailed
it to two
of my friends to look over for comment. Mega foundation
postings with
of the words brought me to this site and enjoyed reading the
pair
reveiws for should I say banana spits in stead with axes or
asses
being the weapon of choice. but peer or banana reviews asside
I like
that you are able to have others come along and read what you
have
talking of. Only thought the class of differing opinions does
the
spark of truth come. www.bcca.org
Roddy young
===
Subject: mega foundation
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h99Fi5501043;
I find it interesting to see that a test on IQ that unites a
group of
people and then asks for 25 dollars to join in and view the
work of
this group is different from circuses from years ago. But
high IQ is
what is genetically being selected for in the Human Gnome
which was
not that case for circus acts. So we have an interesting
dilema here.
High IQ is going to cost more to access and life on the
planet is
going to be dominated by High IQ people and processes.
Roddy
===
Subject: f(x)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9AJI2822789;
I wanna know a history about Function( in simplified form,
f(x))
please....
===
Subject: Group Theory
Please help me with this horrible question.
For n>0, Rn is defined to be the multiplicative group {x is an
interger such that 1=< X < n, (n,x)=1} taken modulo n. Find
the orders
Rn for n= 5, 8, 10, 12, 16, 210. For those cases where the
order of
Rn=4, decide whether Rn is isomorphic to a cyclic group or to
a Klein
Four Group.
===
Subject: Re: Group Theory
>Please help me with this horrible question.
>For n>0, Rn is defined to be the multiplicative group {x is 
an
>interger such that 1=< X < n, (n,x)=1} taken modulo n. Find
the orders
>Rn for n= 5, 8, 10, 12, 16, 210.
Suppose that n = 5; you need to find how many positive 
integers
less than 5 are relatively prime to 5. Since 5 is prime,
thatÕs
easy: all four of them. Thus, |R_5| = 4. Except for n = 210,
the others are almost as easy; if all else fails, the numbers
are
small enough so that you can simply count the elements of R_n.
That you ask suggests that either youÕre missing something
pretty
basic, or you havenÕt tried very hard yet. The case n = 210,
on
the other hand, is a bit harder, unless youÕve learned about
the
Euler phi function. (If you have, you shouldnÕt need to ask
about this part of the problem!)
210 = 2 * 3 * 5 * 7, so the positive integers less than 210
that
are relatively prime to 210 are the ones that are not
multiples
of 2, 3, 5, or 7. Look at the set {1, 2, 3, ..., 210}; exactly
half of its members, or 105 of them, are even. A third, or 70,
are multiples of 3. A fifth, or 42, are multiples of 5. And a
seventh, or 30, are multiples of 7. Thus, 105 + 70 + 42 + 30 =
247 of them are multiples of 2, 3, 5, or 7 and therefore 
ŌbadÕ
(not in R_210). But wait: since there are only 210 numbers in
the set to begin with, how can 247 of them be 
ŌbadÕ? The
problem, of course, is that we counted some 
Ōbadnumbers more
than once: 6, for instance, is both even and a multiple of 3.
To
compensate for this multiple counting we have to subtract from
the Ōbadtotal everything that got 
double-counted, which
would
be the 210/6 = 35 multiples of 6, the 210/10 = 21 multiples of
10, the 210/14 = 15 multiples of 14, the 210/15 = 14
multiples of
15, the 210/21 = 10 multiples of 21, and the 210/35 = 6
multiples
of 35; that leaves 247 - 101 = 146 Ōbad
numbers, if my quick
mental arithmetic is correct. But we still have a problem.
Consider the number 2 * 3 * 5 = 30: itÕs a multiple of 2, of
3,
and of 5, so it was counted three times in the original total
of
247. But itÕs also a multiple of 6, of 10, and of 15, so 
weÕve
subtracted it three times in the adjustment. The net effect is
that we havenÕt counted it at all in the set of 
Ōbadnumbers.
The same is true of every other multiple of 30 and of the
multiples of 2 * 3 * 7 = 42, 2 * 5 * 7 = 70, and 3 * 5 * 7 =
105
as well. There are 210/30 = 7 multiples of 30, 5 multiples of
42, 3 multiples of 70, and 2 multiples of 105 in the set, so
we
have to add in 7 + 5 + 3 + 2 = 17 Ōbadnumbers 
to get 146 +
17 =
163. And now we find that 210, the sole multiple of all four 
of
2, 3, 5, and 7, has been (1) counted Ōbad4 
times in the
original calculation, (2) Ōuncounted6 times 
(as a multiple
of
6, 10, 14, 15, 21, and 35) in the first adjustment, and (3)
counted back in 4 times in the second adjustment; the net
effect
is that itÕs been counted twice and has to be 
ŌuncountedÕ
once,
leaving a grand total of 163 - 1 = 162 Ōbad
numbers. This
means
that the set must contain 210 - 162 = 48 Ōgood
numbers, i.e.,
numbers relatively prime to 210.
The technique that I used here is a basic counting technique,
often called the inclusion-exclusion principle or the like;
itÕs
well worth knowing.
>For those cases where the order of
>Rn=4, decide whether Rn is isomorphic to a cyclic group or
to a Klein
>Four Group.
These cases are small enough to work out completely by hand:
just
find the four elements of R_n, and see whether or not they
form a
cyclic group (by seeing whether one of them generates the
whole
group).
Brian
===
Subject: Group Theory
Please help me with this question
Let G be a group. Prove that the order of g^-1 = order of g
for all g
in G. and that the order of gh = the order of hg, for all g,h
in G.
===
Subject: Re: Group Theory
> Let G be a group. Prove that the order of g^-1 = order of g
for all g
Show
a^n = e iff (a^-1) = e
and from that conclude
o(a) = o(a^-1)
> in G. and that the order of gh = the order of hg, for all
g,h in G.
Show
b^n = e iff (aba^-1)^n = e
and from that conclude
o(b) = o(aba^-1)
Thus
o(ba) = o(abaa^-1) = o(ab)
questions iÕve posted.
===
Subject: Group Theory
Please help me with this question,
For n>0, let Cn denote a cyclic group of order n. Prove that
Cn x Cm
is isomorphic to Cmn iff m and n are relatively prime.
===
Subject: Re: Group Theory
>Please help me with this question,
>For n>0, let Cn denote a cyclic group of order n. Prove that
Cn x Cm
>is isomorphic to Cmn iff m and n are relatively prime.
You have two things to show:
(1) If n and m are relatively prime, then C_n x C_m is
isomorphic to C_{mn}; and
(2) If C_n x C_m is isomorphic to C_{mn}, then n and m
are relatively prime.
(1) is fairly easy. If C_n is cyclic, then it has a generator
g,
i.e., an element such that {g, g^2, g^3, ..., g^n} = C_n.
Similarly, C_m has a generator h. Prove that (g, h) is a
generator of C_{mn}. To do this you have only to prove that if
1 <= k < mn, then (g, h)^k is not the identity of C_n x C_m.
Hint: (g, h)^k = (g^k, h^k), and mn is the least common
multiple
of m and n.
(2) is almost as easy once youÕve dont (1), but you want to
prove
the contrapositive: if m and n are *not* relatively prime,
then
C_n x C_m is *not* isomorphic to C_{mn}. Since C_n x C_m has
mn
elements, the only way it can fail to be isomorphic to C_{mn}
is
to be non-cyclic, so you want to show that if m and n are
*not*
relatively prime, then C_n x C_m is not cyclic. Hint: if k is
the least common multiple of m and n, then x^k is the identity
for any x in C_n x C_m.