mm-33

===

A mathematical proof begins with a truth, and
proceeds by logical>steps to a conclusion which then must be
true.>>I've pulled a detailed exposition of a short
argument that quickly>shows a problem with algebraic
integers. It starts after the>reference.Now here's
a math proof. Those who doubt that fact can believe it's aclaim of proof, but it's verified to be a proof by tracing
the>argument out.>> Not! See below.I've clipped sections
where there was no claim of error by the posterNora
Baron.<deleted>>Essentially objections to how f^2 divides
off now come down to>claiming that the w's are functions of
m, but consider that w_1 w_2 =>1, when m=0, if f is coprime
to 3.>>Now I'm focusing on what has been revealed to be an
area of confusion.> Apparently some people believe that when
I divide off f^2 that it can>divide off as a *function* of
m, so that m=0 might be a special case. >I'm now starting
the argument to address that belief by noting again>that w_1
w_2 = 1, when m=0, if f is coprime to 3. That is, when fdoesn't have 3 as a factor.>>But that was an arbitrary
choice, so let f=3.>> f = 3 is irrelevant to what you
want. There is> no reason to consider it. As you will see
below, it> is a red herring and it does not show what you
want. Which would imply that you are claiming that there is a
misstep.However, f=3 shows that the w_1 w_2 = 3^{2/3} for ALL
m, when f=3.However, w_1 w_2 is coprime to 3 when f is coprime
to 3, at m=0.Therefore, the w's are not functions of m, and in
fact w_1 w_2 iscoprime to f, for all m, and m=0 is not a
special case.That is the conclusion that follows
logically.Well having read down, the poster does have a point
which in fact isthat the 3's divide off in an indeterminate
way, so it's not so easyto just say that w_1 w_2 = 3^{2/3}
because that could be the case, butthen again, maybe only w_3
has a factor of 3, so w_1 w_2 could stillbe coprime to 3. That is, I *said* f is coprime to 3 but in considering thispossibility it's worth it to relax that restriction and now
consider>what would happen if it equals 3.>>Now w_1 w_2
= 3^{2/3} WITHOUT REGARD TO m. >>Seeing that is as simple
as looking at>> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>> 3(-1+mf^2 )x u^2 + u^3 f>>with f=3 as then you have> P(m)/3^2 = (m^3 3^4 - 3m^2 3^2 + 3m) x^3 - >>
3(-1+m3^2 )x u^2 + 3u^3>>so *every* coefficient has a
factor that is 3, as you can tell by>looking.>> Let's
look at this in detail when m = 1 and u = 1. Then>
P(m)/3^2 = (81 - 27 + 3)*x^3 - 3*8*x + 3 > = 3*(19*x^3 - 8*x
+ 1).Ok.>So with>> P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u
w_2)(b_3 x + u w_3)>>each of the b's and each of the w's
has a factor that is 3^{1/3},>while the b's can have
additional factors in common with 3, the w's>cannot, as when
3 is separated out, notice you have>> P(m)/3^2 = 3((m^3 3^3
- 3m^2 3 + m) x^3 - >> (-1+m3^2 )x u^2 + u^3).>> As
above when m = 1, u = 1, this is> 3*(19*x^3 - 8*x + 1).>
The polynomial inside the parentheses can be factored> in the
form> [1] (19*x^3 - 8*x + 1) = (b1*x + 1)*(b2*x + 1)*(b3*x +
1),> where b1, b2, and b3 are the negatives of the roots> of
the associated polynomial> u^3 + 8*u^2 - 19.> Since the
roots of the latter polynomial are algebraic> integers, one
concludes b1, b2, and b3 are algebraic > integers also.>
Now: how might you distribute the 3 in the expression>
3*(b1*x + 1)*(b2*x + 1)*(b3*x + 1) ?> Answer: LOTS of ways!
There is no unique way. Here are> several:> 1. (b1*x +
1)*(b2*x + 1)*(3*b3*x + 3)> 2. (sqrt(3)*b1*x +
sqrt(3))*(sqrt(3)*b2*x + sqrt(3)*(b3*x + 1)> 3.
(3^{1/3}*b1*x + 3^{1/3})*(3^{2/3}*b2*x + 3^{2/3})*(b3*x + 1) 4. (3^{1/5}*b1*x + 3^{1/5})*(3^{3/5}*b2*x + 3^{3/5})>
*(3^{1/5}*b3*x + 3{1/5})> In fact INFINITELY many ways.
Any way you want to split 3 as a> product of three numbers
gives a factorization. And in all the> examples just given
(and in infinitely many others) note that (1) the>
coefficients of the x's are algebraic integers, and (2) the
w terms> are also algebraic integers.> Proving what,
exactly, you ask?> #### Proving that the f = 3 case tells
you NOTHING useful about the> necessary values of the w's.
There is no unique way to write them> down. This is a special,
exceptional case in which too many of the> of the f terms
can be factored out.It is correct that they are
indeterminate,and that's a VERY good pointthat you brought
up.But now tell me, do you *still* believe the w's are then
functions ofm?That is, once you divide off 3, it's gone, and
the w's don't vary withm, so you can't claim m=0 is a special
case where the w's arefunctions of m.Or do you think you still
can?>But before at m=0, they were coprime to f, now they are
not when f=3,>as they are constant. Clearly, they are
constant in both cases with>respect to m, without regard to
the value of f. Which makes sense as>f^2 is not a function
of m, and it is what is being divided off.>> But in this
case you can divide off *another* factor of f,> and that is
what leads to the nonuniqueness shown above, which> wrecks
your argument. This does not happen when f is a prime > bigger
than 3 and m is coprime to f. f = 3 is a special case > of no
interest or relevance to your main argument. It is a >
nuisance distraction. It proves nothing.Really? Your position
is that the w's are *functions* of m, but nowyou yourself
bring out that with f=3, there's no way to determinetheir
values, or do you claim there is?If you were correct, then
even with f=3, the w's would STILL vary withm, so that m=0
would be a *special* case, and then they'd vary as mvaried.Are
you claiming they do? >That is, if they were functions of m,
so that w_1 w_2 = 1 at m=0 as a>*function* of m, then it
wouldn't matter if f had a factor of 3 or>not, you'd STILL
get w_1 w_2 = 1 at m=0, without regard to the value>of f.>But in fact, if f=3, you have w_1 w_2 = 3^{2/3} at m=0,
which only>works if the w's are independent of m, which they
are.>> See above. In the special case with f = 3, the w's>
are not uniquely determined. You cannot draw conclusions> from
it about how the f terms are distributed among > the w's or
the linear factors. This case is a red > herring, and it is of
no interest in your general > argument, where you require that
f is a prime > 3. Worst> of all, it does not imply what you
want.You seem to be very determined to convince yourself that
themathematics says one thing, when it does not.It's *your*
position that the w's are functions of m, so that you canclaim
that m=0 is a special case.You need that claim so that you can
then claim that the w's havevarying factors of f, as m
varies.Now you need the w's to vary when the polynomial has a
factor of 3,when f=3, but that's not a lot different from what
was there beforewhen the factor was f^2.But here you try to
dismiss one constant factor--3--as a unimportant,but try to
hang on to f^2 which is also a constant factor with respectto
m.Doesn't that bother you just a little bit?Does it seem
rational to you? >It makes sense that they are anyway, as
f^2 isn't a function of > m, but>I've seen that for some
people the idea can take hold after seeing m=0>highlighted.>But if the w's were functions of m, then w_1 w_2 would
equal 1,>without regard to the value of f, but it does not.> w1 = w2 = 1 and w3 = 3 is just one of the infinite range>
of possibilities. It is Example 1 above.Well then, what makes
you think the w's are *functions* of m, so thatm=0 would be a
special case?Where's the mathematical logic in your thinking?Therefore, the factorization is>> P(m)/f^2 = (m^3 f^4 -
3m^2 f^2 + 3m) x^3 - >> 3(-1+mf^2 )x u^2 + u^3 f =>>
(b_1 x + u)(b_2 x + u)(b_3 x + uf)>>where you'll notice
that the b's are algebraic integers with m=1,>f=sqrt(2), but
that's a special case as generally they are not, which>shows
a problem with the ring of algebraic integers.>> In this
special case, P(x) does not even have> rational coefficients.
It too is of no interest> or value for your main
argument.Hmmm...that's a fascinating attempt at dodging the
relevantmathematics.As I pointed out in another thread, with
m=1, f=sqrt(2), y=uf, you get 2x^3 - 3xy^2 + y^3which is
reducible as it factors as 2x^2 - 3xy^2 + y^3 = (x-y)(2x^2 +
2xy - y^2)and yes it is true that since y=uf, where f=sqrt(2),
the y's areirrational, but it doesn't change the fact that the
polynomial isreducible.Now that makes the *second* time I've
pointed that out to you.Possibly you try to dismiss it because
now using m=1(mod sqrt(2))destroys your claim that reducibility
matters, as for instance,m=3=1(mod sqrt(2)) but now the
resulting polynomial is *irreducible*and your claims, where
you believe the w's are functions of m, wouldforce factors of
f to shift.>And here I've packed in a lot of information as
well.>> Not enough, clearly.Not if you refuse to follow
mathematical logic.I think you've shown through your posting
behavior that you areEXTREMELY dedicated to the notion that
I'm wrong, but the problem isthat the mathematical logic is
against you.So far, you've decided to go with your beliefs
against the math.>First, with f coprime to 3, I now know
that the factorization is>> P(m)/f^2 = (b_1 x + u)(b_2 x +
u)(b_3 x + uf)>>as the w's are constant with respect to m,
so I can just check at m=0,>which revealed that w_1 w_2 = 1.
Now that doesn't necessarily force>w_1 and w_2 to each equal
1, but even if they were factors of 1, i.e.>unit factors,
that would only change b_1 and b_2.>>So I have my
factorization without regard to m in terms of where the fgoes, and then I point out that you can actually check my
work using>m=1, f=sqrt(2), as then you get a polynomial
which you can factor>rather simply. So you can actually get
the values for the b's and>check them, and see that they are
all algebraic integers, and all are>coprime to 2.>However, usually, for f values that are coprime to 3, you
don't get>b's that are algebraic integers, which shows a
problem with the ring>of algebraic integers.>> Wrong!
You *can* get algebraic integers, but *** not with> the
properties you want ***, and there is no problem with> the
ring of algebraic integers. Here is how things work when> f =
5, m = 1, u = 1, v = -1 + m*f^2, and> P(x) = (v^3 + 1)*x^3 -
3*v*x*(u*f)^2 + (u*f)^3:> P(x)/f^2 = P(x)/25 = 553*x^3 -
72*x + 5.> By the Magidin-McKinnon theorem (essentially
proved earlier by > someone else [P. M. Cohn?]), this can be
factored in the form> 553*x^3 - 72*x + 5 = (b1*x + w1)*(b2*x
+ w2)*(b3*x + w3),> where b1, b2, b3, and w1, w2, w3 are
algebraic integers. > You can show using elementary Galois
theory that EACH of> w1, w2, and w3 is not coprime to f = 5.  Thus: a factorization of the desired form DOES exist, but it>
does NOT have one of the properties that you desperately want.>
No problem with the ring of algebraic integers, and no valid>
proof for you. You lose on two counts. Too bad!And your
*belief* depends on the w's being functions of m, which
isnonsensical, as follows from your own statements, like where
younoticed that with f=3, the w's are indeterminate, as 3 can
divide offan infinite number of ways.Yet now here you are
claiming that the w's are functions of m.That is NOT following
the math. >Now the nice thing about a mathematical proof is
that if someone>disagrees they have to find some misstep.> See above at #### ! The misstep in the current argument
has been > found. No, but you have shown your own bias and
refusal to followmathematical logic. >Unfortunately, people
can *say* that proof is not a proof, even when>it is, just
like if you tried to say you were human, and not a dog,someone might dispute any proof you might give, claiming it
false.>> In this case, you have tried to use an irrelevant
> red-herring argument to show what you want. Unfortunately,>
in the special case you selected, the number f (= 3) does>
factors *** non-uniquely *** through the linear terms of your>
polynomial factorization, and you end up being able to>
conclude:> *** N O T H I N G ***> about the cases in which
you are interested.> But you have made progress. Do you
realize how long it> took us to get through to you that there
is actually a> nontrivial problem with generalizing from m = 0
to> m <> 0 ? Do you realize how many incorrect arguments> you
have already burned through (including the present> one) in
trying to handle that problem? Do you realize> that all of
this is a waste of time, because your > main claims have
already been shown to be false and> cannot be fixed by
twiddling with the details?> Nora B.And readers should
note that Nora Baron here arrogant and presumablyconfident,
is still fighting to claim that f^2 divides off of apolynomial
as a function of m, though it is itself constant withregard to
m, when the poster has spent a lot of effort trying to
claimthat 3 divides off in an inconsequential way, when
f=3.The point is that the poster Nora Baron is not making
sense.James Harris

===

> A mathematical proof begins with a
truth, and proceeds by logical> steps to a conclusion which
then must be true.> I've pulled a detailed exposition of a
short argument that quickly> shows a problem with algebraic
integers. It starts after the> reference.> Now here's a
math proof. Those who doubt that fact can believe it's a>
claim of proof, but it's verified to be a proof by tracing
the> argument out.> In this case, I begin with an
expression. The expression exists, so> that is the truth from
which you start.> Consider, in the ring of algebraic
integers,> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - >
3(-1+mf^2 )x u^2 + u^3 f).> That is, I have the identity
which defines P(m) in terms of various> symbols, and it's all
in the ring of algebraic integers, which means> that the
symbols can only represent numbers that are algebraic>
integers.> Now using b_1, b_2, b_3, w_1, w_2, and w_3, I
have the factorization> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u
w_2)(b_3 x + u w_3)James,You sometimes get into an argument as
to whether something is afunction of something else. It might
help if you made it clear at thebeginning that P is a function
of f, m, u and x. And that the b_i andw_i are functions of f
and m. You might then be able to show,subsequently, that
something that looked at though it depended onsomething else,
in fact does not.For instance:Let f, m, u and x be algebraic
integers.Consider the function:P(f,m,u,x)=f^2((m^3 f^4 - 3m^2
f^2 + 3m)x^3 - 3(-1+mf^2)x u^2 + u^3 f)Since the set of
algebraic integers is an integral domain, P(f,m,u,x)is
necessarily an algebraic integer.Aside: Is P(f,m,u,x)/f^2
necessarily an algebraic integer?It is possible to find
functions (of f and m) b_1, b_2, b_3, w_1, w_2,and w_3 such
thatP(f,m,u,x)/f^2 =(b_1(f,m) x + u w_1(f,m))*(b_2(f,m) x + u
w_2(f,m))*(b_3(f,m) x + u w_3(f,m))Clearly these functions are
not uniquely determined.Aside: Can the functions b_1, b_2, b_3,
w_1, w_2, and w_3 all bechosen such that, if f and m are
algebraic integers, b_1(f,m) b_2(f,m)b_3(f,m) w_1(f,m)
w_2(f,m) and w_3(f,m) are all algebraic integers?If m=0, we
haveP(f,0,u,x)/f^2 =(b_1(f,0) x + u w_1(f,0))*(b_2(f,0) x + u
w_2(f,0))*(b_3(f,0) x + u w_3(f,0))etcMath Fan

===

>A
mathematical proof begins with a truth, and proceeds by
logical>steps to a conclusion which then must be true.>I've pulled a detailed exposition of a short argument
that quickly>shows a problem with algebraic integers. It
starts after the>reference.Now here's a
math proof. Those who doubt that fact can believe it's aclaim of proof, but it's verified to be a proof by tracing
the>argument out.>> Not! See below.> I've
clipped sections where there was no claim of error by the
poster> Nora Baron.> You also clipped parts where I made
valid claims that disagreewith your conclusions. I hope you
will read ALL of this post, andread it carefully. > <deleted>Essentially objections to how f^2 divides off now come
down to>claiming that the w's are functions of m, but
consider that w_1 w_2 =>1, when m=0, if f is coprime to
3.>>Now I'm focusing on what has been revealed to be
an area of confusion.> Apparently some people believe that
when I divide off f^2 that it can>divide off as a
*function* of m, so that m=0 might be a special case. >I'm
now starting the argument to address that belief by noting
again>that w_1 w_2 = 1, when m=0, if f is coprime to 3.
That is, when f>doesn't have 3 as a factor.>>But
that was an arbitrary choice, so let f=3.>> f = 3 is
irrelevant to what you want. There is> no reason to consider
it. As you will see below, it> is a red herring and it does
not show what you want.> Which would imply that you are
claiming that there is a misstep.> However, f=3 shows that
the w_1 w_2 = 3^{2/3} for ALL m, when f=3.> The expression you
are working with is P(x)/f^2. What I showed in my post is that
w1*w2 can be ANY two algebraic integerfactors of f = 3. It is
NOT necessarily the case that w1*w2 = 3^{2/3}in that case. The
case f = 3 does not give you any information aboutthe w's. What
I have said is that in general, the factorization (including
thevalues of the w's) is dependent on the value of m. That
statementis correct when it is restricted to the cases of
interest to you:the polynomial P(x) is irreducible and
primitive. These conditionswill hold when f is a prime > 3 and
m is coprime to f. They doNOT hold when f = 3 (P(x) is not
primitive). The conditions are not artificial or arbitrary. I
apply a theoremfrom algebraic number theory which requires
irreducibility and(as noted by Arturo Magidin) primitivity.
These conditions alsohappen to prevail in all the cases of
interest to you: f prime > 3and m coprime to f. I do NOT make
statements about theform of the factorization when f = 3, and
certainly not whenm = 0. > However, w_1 w_2 is coprime to 3
when f is coprime to 3, at m=0.> Therefore, the w's are not
functions of m, and in fact w_1 w_2 is> coprime to f, for all
m, and m=0 is not a special case.> No. In all the cases of
interest, the w's ARE dependent on m. If the w's are not
dependent on m, then they must be constant.Right? As you have
noted, when m = 0, you get P(0)/f^2 = w1*w2*u^2*(b3*x +
u*w3)[1] = u^2*(b3*w1*w2*x + u*f). Right?But P(0)/f^2 also
equals (you note) P(0)/f^2 = 3x*u^2 + u^3*f[2] = u^2*(3*x +
u*f). Right?And looking at [1] and [2], you conclude w1*w2 =
1. Right?And that happens to be true regardless of the value
of f. Right?*Including* f = 3.Now you say w1 and w2 do not
depend on m. To me, thatwould mean that they are constant with
respect to m. Right?But now when f = 3 and m <> 0, you say you
have w1*w2 = 3^{2/3}.And when f = 3 and m = 0, from the above,
you have w1*w2 = 1.You call that constant? You call that not
dependent on m?[The fact is, when m = 0 and f = 3, the w's are
indeterminateas discussed below.]> That is the conclusion that
follows logically.> Well having read down, the poster does
have a point which in fact is> that the 3's divide off in an
indeterminate way, so it's not so easy> to just say that w_1
w_2 = 3^{2/3} because that could be the case, but> then again,
maybe only w_3 has a factor of 3, so w_1 w_2 could still> be
coprime to 3.> Aha. So you did get the point. Yay!>That
is, I *said* f is coprime to 3 but in considering thispossibility it's worth it to relax that restriction and now
consider>what would happen if it equals 3.>>Now
w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. >>Seeing that
is as simple as looking at>> P(m)/f^2 = (m^3 f^4 -
3m^2 f^2 + 3m) x^3 - >> 3(-1+mf^2 )x u^2 + u^3 f>with f=3 as then you have>> P(m)/3^2 = (m^3 3^4 -
3m^2 3^2 + 3m) x^3 - >> 3(-1+m3^2 )x u^2 + 3u^3>so *every* coefficient has a factor that is 3, as you can
tell by>looking.>> Let's look at this in detail
when m = 1 and u = 1. Then> P(m)/3^2 = (81 - 27 +
3)*x^3 - 3*8*x + 3 > = 3*(19*x^3 - 8*x + 1).> Ok.So with>> P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u
w_2)(b_3 x + u w_3)>>each of the b's and each of the
w's has a factor that is 3^{1/3},>while the b's can have
additional factors in common with 3, the w's>cannot, as
when 3 is separated out, notice you have>> P(m)/3^2 =
3((m^3 3^3 - 3m^2 3 + m) x^3 - >> (-1+m3^2 )x u^2 +
u^3).>> As above when m = 1, u = 1, this is>
3*(19*x^3 - 8*x + 1).> The polynomial inside the
parentheses can be factored> in the form> [1] (19*x^3
- 8*x + 1) = (b1*x + 1)*(b2*x + 1)*(b3*x + 1),> where
b1, b2, and b3 are the negatives of the roots> of the
associated polynomial> u^3 + 8*u^2 - 19.> Since
the roots of the latter polynomial are algebraic> integers,
one concludes b1, b2, and b3 are algebraic > integers also. Now: how might you distribute the 3 in the expression 3*(b1*x + 1)*(b2*x + 1)*(b3*x + 1) ?> Answer: LOTS
of ways! There is no unique way. Here are> several:> 1.
(b1*x + 1)*(b2*x + 1)*(3*b3*x + 3)> 2. (sqrt(3)*b1*x +
sqrt(3))*(sqrt(3)*b2*x + sqrt(3)*(b3*x + 1)> 3.
(3^{1/3}*b1*x + 3^{1/3})*(3^{2/3}*b2*x + 3^{2/3})*(b3*x + 1) 4. (3^{1/5}*b1*x + 3^{1/5})*(3^{3/5}*b2*x + 3^{3/5})>
*(3^{1/5}*b3*x + 3{1/5})> In fact INFINITELY many
ways. Any way you want to split 3 as a> product of three
numbers gives a factorization. And in all the> examples just
given (and in infinitely many others) note that (1) the>
coefficients of the x's are algebraic integers, and (2) the
w terms> are also algebraic integers.> Proving what,
exactly, you ask?> #### Proving that the f = 3 case tells
you NOTHING useful about the> necessary values of the w's.
There is no unique way to write them> down. This is a
special, exceptional case in which too many of the> of the
f terms can be factored out.> It is correct that they are
indeterminate,and that's a VERY good point> that you brought
up.> But now tell me, do you *still* believe the w's are
then functions of> m?> I believe they are dependent on m when
you restrict to the cases that happen to be of interest to
you: f a prime > 3, m relatively prime to f. my point of
view, these conditions (f = prime > 3, m coprime to f) are
quitenatural also. I base my conclusions on a theorem in
algebraic number theory which requires primitivity and
irreducibility of a polynomial. When f = 3, you do not get
primitivity and irreducibility. So indeed, in this special
case, I do not pretend to say anything about the w's.> That
is, once you divide off 3, it's gone, and the w's don't vary
with> m, so you can't claim m=0 is a special case where the
w's are> functions of m.> You have already divided off f^2. In
this special case you candivide off another f. However, your
factorization statement wasabout P(m)/f^2, not P(m)/f^3. If
you want to consider thelatter, then you get w1 = w2 = w3 = 1,
since the polynomialis then 19*x^3 - 8*x + 1.So I don't think
you want to go there, eh?> Or do you think you still can?But before at m=0, they were coprime to f, now they are not
when f=3,>as they are constant. Clearly, they are constant
in both cases with>respect to m, without regard to the
value of f. Which makes sense as>f^2 is not a function of
m, and it is what is being divided off.>> But in
this case you can divide off *another* factor of f,> and
that is what leads to the nonuniqueness shown above, which>
wrecks your argument. This does not happen when f is a prime  bigger than 3 and m is coprime to f. f = 3 is a special case
> of no interest or relevance to your main argument. It is a
> nuisance distraction. It proves nothing.> Really? Your
position is that the w's are *functions* of m, My position is
that they are dependent on m, which is notquite the same as
being functions of m. And in any case,my proofs regarding the
factorization do not include thespecial cases f = 3 or m = 0.
That does not mean myproofs are incorrect. It does mean that
your entire discussionof f = 3 is irrelevant.> but now> you
yourself bring out that with f=3, there's no way to determine>
their values, or do you claim there is?> I certainly don't. You
however did. You said the values wereindependent of m, i.e.,
constant, and you used the case f = 3to show it. But when f =
3, far from being a constant, they cannoteven be determined at
all. They can be anything that divides 3.> If you were correct,
then even with f=3, the w's would STILL vary with> m, so that
m=0 would be a *special* case, and then they'd vary as m>
varied.> Are you claiming they do?> Yes, when f is a prime >
3 and m is coprime to f. Whichhappens to be exactly the
conditions that you want to consider also in your proof.That is, if they were functions of m, so that w_1 w_2 = 1 at
m=0 as a>*function* of m, then it wouldn't matter if f had
a factor of 3 or>not, you'd STILL get w_1 w_2 = 1 at m=0,
without regard to the value>of f.>>But in fact, if
f=3, you have w_1 w_2 = 3^{2/3} at m=0, which only>works if
the w's are independent of m, which they are.>> See
above. In the special case with f = 3, the w's> are not
uniquely determined. You cannot draw conclusions> from it
about how the f terms are distributed among > the w's or
the linear factors. This case is a red > herring, and it is
of no interest in your general > argument, where you require
that f is a prime > 3. Worst> of all, it does not imply what
you want.> You seem to be very determined to convince
yourself that the> mathematics says one thing, when it does
not.> Not at all! You have noted that I am right about the
indeterminacy of the w's when f = 3. You were using thatcase
in an attempt to prove that w1 and w2 are constantwith respect
to m. I merely noted that in fact that case doesn't give you
any information at all regarding the w's.In that case, the w's
are indeterminate. Your conclusionwas not justified. It seems
to me that it is YOU, not I,who is determined to convince
yourself that the mathsays one thing, when in this case it
doesn't actually sayanything. I thought it was a little bit of
a subtle point. Clearly youhad not considered it previously,
and clearly from whatyou have said here, you understand it.
This in itself is thus a rather rare event. It wrecks this
particular part of your argument. That is the point. > It's
*your* position that the w's are functions of m, so that you
can> claim that m=0 is a special case.> Dependent on m, not
necessarily functions of m. And I canclaim it only when f =
prime > 3 and m is coprime to f. Butthat is a very big set of
values, and is plenty good enough for both our purposes.> You
need that claim so that you can then claim that the w's have>
varying factors of f, as m varies.> Now you need the w's to
vary when the polynomial has a factor of 3,> when f=3, No, I
don't care what happens in that special case. I don't
claimthat my proofs hold in that case. The polynomial in
question isnot irreducible and primitive.> but that's not a
lot different from what was there before> when the factor was
f^2.> When f is prime > 3, it is a LOT different.> But here
you try to dismiss one constant factor--3--as a unimportant,>
but try to hang on to f^2 which is also a constant factor with
respect> to m.> It's not that it is unimportant (though in fact
it is - the case f = 3is of no interest to you in your
arguments either). It is that I don'tmake any claims about
what happens when f = 3. The theorem that I use in that case
does not apply anyway.> Doesn't that bother you just a little
bit?> Nope.> Does it seem rational to you?> Yes. What seems
irrational to me is to claim that when f = 3, w1*w2 = 3^{2/3},
when in fact w1 andw2 are, as you yourself have noted above,
indeterminatewhen f = 3. The f = 3 case adds no information to
yourknowledge of the relationship between m and w1 and w2.You
claimed that it confirmed that w1 and w2 were constant.It did
no such thing.>It makes sense that they are anyway, as f^2
isn't a function of > m, but>I've seen that for some people
the idea can take hold after seeing m=0>highlighted.>But if the w's were functions of m, then w_1 w_2 would
equal 1,>without regard to the value of f, but it does
not.>> w1 = w2 = 1 and w3 = 3 is just one of the
infinite range> of possibilities. It is Example 1 above.>
Well then, what makes you think the w's are *functions* of m,
so that> m=0 would be a special case?> Where's the
mathematical logic in your thinking?> See above. Unlike you, I
do not make claims aboutthe special case f = 3. I say that when
f = prime > 3 andm is coprime to f, the w's are dependent on m.
Proveme wrong in THAT case and I will start to take notice.Therefore, the factorization is>> P(m)/f^2 = (m^3 f^4
- 3m^2 f^2 + 3m) x^3 - >> 3(-1+mf^2 )x u^2 + u^3 f => (b_1 x + u)(b_2 x + u)(b_3 x + uf)>>where you'll
notice that the b's are algebraic integers with m=1,f=sqrt(2), but that's a special case as generally they are
not, which>shows a problem with the ring of algebraic
integers.>> In this special case, P(x) does not even
have> rational coefficients. It too is of no interest> or
value for your main argument.> Hmmm...that's a fascinating
attempt at dodging the relevant> mathematics.> This is 180
degrees wrong! The RELEVANT mathematics arenot when f = 3. The
cases of interest to YOU are f = prime > 3,m coprime to f.
Doesn't it seem a little odd for you to spend allthis effort
in a subset where your own proofs are not concerned?> As I
pointed out in another thread, with m=1, f=sqrt(2), y=uf, you
get> 2x^3 - 3xy^2 + y^3> which is reducible as it factors
as> 2x^2 - 3xy^2 + y^3 = (x-y)(2x^2 + 2xy - y^2)> and yes
it is true that since y=uf, where f=sqrt(2), the y's are>
irrational, but it doesn't change the fact that the polynomial
is> reducible.> There are two things going on here. When you
write 2*x^3 - 3*x*y^2 + y^3you are considering a polynomial in
two variables. Indeedthat does factor in the obvious way. But
when you subsitute y = sqrt(2), you are now considering a
polynomial in ONE variable (x). You areback to the case in
which you are interested. But thispolynomial in ONE variable
has a constant term whichis irrational. You know perfectly
well that it does notfactor with rational coefficients. In
your factorizations, you are invariably factoring apolynomial
in one indeterminate, x. That is why you consider linear
factors of the form (b1*x + u*f), etc.. It is from that point
of view that you should be consideringyour polynomial, not as
a polynomial in two variables.> Now that makes the *second*
time I've pointed that out to you.> Right, thanks. But sqrt(2)
is still irrational, eh? Or did that change recently in
Georgia?> Possibly you try to dismiss it because now using
m=1(mod sqrt(2))> destroys your claim that reducibility
matters, as for instance,> m=3=1(mod sqrt(2)) but now the
resulting polynomial is *irreducible*> and your claims, where
you believe the w's are functions of m, would> force factors
of f to shift.> You have lost me here. Are you saying that two
values of mthat agree mod sqrt(2) are necessarily actually the
same?>And here I've packed in a lot of information as
well.>> Not enough, clearly.> Not if you refuse to
follow mathematical logic.> I think you've shown through your
posting behavior that you are> EXTREMELY dedicated to the
notion that I'm wrong, but the problem is> that the
mathematical logic is against you.> You are correct that I
think you are wrong. Saying I am dedicated to that notion is
not different from saying I am dedicatedto defending what I
believe to be true. Is that a character flaw inyour eyes?
Should I not do that?> So far, you've decided to go with your
beliefs against the math.> My beliefs are based entirely on
the math and nothing elsein this discussion.>First, with
f coprime to 3, I now know that the factorization is>>
P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf)>>as the
w's are constant with respect to m, so I can just check at
m=0,>which revealed that w_1 w_2 = 1. Now that doesn't
necessarily force>w_1 and w_2 to each equal 1, but even if
they were factors of 1, i.e.>unit factors, that would only
change b_1 and b_2.>>So I have my factorization
without regard to m in terms of where the f>goes, and then
I point out that you can actually check my work using>m=1,
f=sqrt(2), as then you get a polynomial which you can factorrather simply. So you can actually get the values for the
b's and>check them, and see that they are all algebraic
integers, and all are>coprime to 2.>>However,
usually, for f values that are coprime to 3, you don't getb's that are algebraic integers, which shows a problem with
the ring>of algebraic integers.>> Wrong! You
*can* get algebraic integers, but *** not with> the
properties you want ***, and there is no problem with> the
ring of algebraic integers. Here is how things work when> f
= 5, m = 1, u = 1, v = -1 + m*f^2, and> P(x) = (v^3 +
1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3:> P(x)/f^2 =
P(x)/25 = 553*x^3 - 72*x + 5.> By the Magidin-McKinnon
theorem (essentially proved earlier by > someone else [P. M.
Cohn?]), this can be factored in the form> 553*x^3 - 72*x
+ 5 = (b1*x + w1)*(b2*x + w2)*(b3*x + w3),> where b1, b2,
b3, and w1, w2, w3 are algebraic integers. > You can show
using elementary Galois theory that EACH of> w1, w2, and w3
is not coprime to f = 5. > Thus: a factorization of the
desired form DOES exist, but it> does NOT have one of the
properties that you desperately want.> No problem with the
ring of algebraic integers, and no valid> proof for you. You
lose on two counts. Too bad!> And your *belief* depends on
the w's being functions of m, which is> nonsensical, as
follows from your own statements, like where you> noticed that
with f=3, the w's are indeterminate, as 3 can divide off> an
infinite number of ways.> It's amazing how you got this turned
around. You claimed thatthe case f = 3 showed that the w's were
constant. In fact in thecase f = 3, the w's are indeterminate.
It clear from what you justsaid that you actually understand
this. The case f = 3 added nothing to your knowledge of the
behavior of the w's. Yet you claim that *** I *** am being
nonsensical !!! Incredible!> Yet now here you are claiming
that the w's are functions of m.> I claim that the w's are
dependent on m under the conditionsthat f = prime > 3, and m
coprime to f. What you have donenot only does not support your
own conclusions in the specialno-interest case where f = 3, it
doesn't even touch the caseswhich are of genuine interest to
both of us.> That is NOT following the math.> Exactly the
opposite is true.>Now the nice thing about a mathematical
proof is that if someone>disagrees they have to find some
misstep.>> See above at #### ! The misstep in the
current argument has been > found.> No, but you have shown
your own bias and refusal to follow> mathematical logic.>
Really, there was nothing in my post but mathematical
logic.Sorry it didn't support your beliefs.>Unfortunately,
people can *say* that proof is not a proof, even when>it
is, just like if you tried to say you were human, and not a
dog,>someone might dispute any proof you might give,
claiming it false.>> In this case, you have tried to
use an irrelevant > red-herring argument to show what you
want. Unfortunately,> in the special case you selected, the
number f (= 3) does> factors *** non-uniquely *** through
the linear terms of your> polynomial factorization, and you
end up being able to> conclude:> *** N O T H I N G
***> about the cases in which you are interested. But you have made progress. Do you realize how long it>
took us to get through to you that there is actually a>
nontrivial problem with generalizing from m = 0 to> m <> 0 ?
Do you realize how many incorrect arguments> you have already
burned through (including the present> one) in trying to
handle that problem? Do you realize> that all of this is a
waste of time, because your > main claims have already been
shown to be false and> cannot be fixed by twiddling with the
details?> Nora B.> And readers should note that
Nora Baron here arrogant and presumably> confident, is still
fighting to claim that f^2 divides off of a> polynomial as a
function of m, though it is itself constant with> regard to m,
when the poster has spent a lot of effort trying to claim> that
3 divides off in an inconsequential way, when f=3.> Well, duh.
Yes, that is what I was saying. I agree that I spent alot of
time on an inconsequential case. I was simply trying to
showthat your conclusions *even in that case* were not
justified. At leastyou got the point this time, so I don't
think it was a complete waste.> The point is that the poster
Nora Baron is not making sense.> It's clear from what you
have said here that you understood what I wassaying, and even,
I daresay, that you were surprised by it. So clearly, it is not
that what I said did not make sense. It is that you did not
like the conclusion. Nora B. > James Harris

===

A mathematical proof begins with a truth, and proceeds by
logical>steps to a conclusion which then must be true.>I've pulled a detailed exposition of a short
argument that quickly>shows a problem with algebraic
integers. It starts after the>reference.
>Now here's a math proof. Those who doubt that fact can
believe it's a>claim of proof, but it's verified to be
a proof by tracing the>argument out.>>
Not! See below.> I've clipped sections where there was
no claim of error by the poster> Nora Baron.> You
also clipped parts where I made valid claims that disagree>
with your conclusions. I hope you will read ALL of this post,
and> read it carefully. I read all of the previous post. If
you feel something material wasclipped you should have said
what, rather than use an insinuation. > <deleted>Essentially objections to how f^2 divides off now come down
to>claiming that the w's are functions of m, but
consider that w_1 w_2 =>1, when m=0, if f is coprime to
3.>>Now I'm focusing on what has been revealed to
be an area of confusion.> Apparently some people
believe that when I divide off f^2 that it can>divide
off as a *function* of m, so that m=0 might be a special case.
>I'm now starting the argument to address that belief by
noting again>that w_1 w_2 = 1, when m=0, if f is coprime
to 3. That is, when f>doesn't have 3 as a factor.>But that was an arbitrary choice, so let f=3.> f = 3 is irrelevant to what you want. There is>
no reason to consider it. As you will see below, it> is a
red herring and it does not show what you want.> Which
would imply that you are claiming that there is a misstep. However, f=3 shows that the w_1 w_2 = 3^{2/3} for ALL m,
when f=3.>> The expression you are working with is>
P(x)/f^2.> What I showed in my post is that w1*w2 can be ANY
two algebraic integer> factors of f = 3. It is NOT necessarily
the case that w1*w2 = 3^{2/3}> in that case. The case f = 3
does not give you any information about> the w's.Well it turns
out I was in error as the answer is given by consideringthe
factorization P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x +
uf)as with f=3, it's clear that *each* has 3 as a factor,
which gives you P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - (-1+m
3^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u).Now
using the b's and w's gives P(m)/3^2 = 3((m^3 3^3 - 3^2 m^2 +
m) x^3 - (-1+m 3^2 )x u^2 + u^3) = (b_1 x + u w_1)(b_2 x + u
w_2)(b_3 x + u w_3)but at m=0, I have P(0)/3^2 = u^2(3x - 3u)
= u^2(b_3 w_1 w_2 - 3u), soand dividing off the last 3 gives
P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - (-1+m 3^2 )x u^2 +
u^3 = (b_1 x + u)(b_2 x + u)(b_3/3 x + u)so it's NOT true that
it's completely arbitrary what the w's are.Notice that using
m=0 and the constant term P(0) forces the case, asthe only
thing that can vary is the indices as my choices there
werearbitrary. > What I have said is that in general, the
factorization (including the> values of the w's) is dependent
on the value of m. That statement> is correct when it is
restricted to the cases of interest to you:> the polynomial
P(x) is irreducible and primitive. These conditions> will hold
when f is a prime > 3 and m is coprime to f. They do> NOT hold
when f = 3 (P(x) is not primitive). You're saying that
equations *check* to see if P(x) is irreducibleover rationals,
which you claim happens if f is a prime greater than 3and m is
coprime to f.A function is a function. It doesn't check
coprimeness of somevariable to figure out if it's a function.>
The conditions are not artificial or arbitrary. I apply a
theorem> from algebraic number theory which requires
irreducibility and> (as noted by Arturo Magidin) primitivity.
These conditions also> happen to prevail in all the cases of
interest to you: f prime > 3> and m coprime to f. I do NOT
make statements about the> form of the factorization when f =
3, and certainly not when> m = 0. It's *algebra* so you can
use f and if there are functions of m,then they will be
functions of m, whether f is coprime to 3 or not.> However,
w_1 w_2 is coprime to 3 when f is coprime to 3, at m=0.>
Therefore, the w's are not functions of m, and in fact w_1 w_2
is> coprime to f, for all m, and m=0 is not a special case.> No. In all the cases of interest, the w's ARE dependent on
m.That is wrong as I've shown with P(m)/3^3 = (m^3 3^3 - 3^2
m^2 + m) x^3 - (-1+m 3^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x
+ u)(a_3/3 x + u)as a 3 must divide off through *each* of the
a's.Your claims would require that the a's have a factor of f
that variesas a function of m, except you wish to dodge your
assertion herebecause f=3, and make claims that would force
equations to *choose* tobe functions or constants with regard
to m based on whether or not fis coprime to 3.> If the w's are
not dependent on m, then they must be constant.> Right? They
are. > As you have noted, when m = 0, you get> P(0)/f^2 =
w1*w2*u^2*(b3*x + u*w3)> [1] = u^2*(b3*w1*w2*x + u*f).
Right?Yup.> But P(0)/f^2 also equals (you note)> P(0)/f^2 =
3x*u^2 + u^3*f> [2] = u^2*(3*x + u*f). Right?> And looking
at [1] and [2], you conclude w1*w2 = 1. Right?Yup.> And that
happens to be true regardless of the value of f. Right?>
*Including* f = 3.Yup.> Now you say w1 and w2 do not depend on
m. To me, that> would mean that they are constant with respect
to m. Right?> But now when f = 3 and m <> 0, you say you
have w1*w2 = 3^{2/3}.That was a mistake. Now if you were
following the math, then youwould simply note it as a mistake,
but instead you're trying to use*my* mistake to convince that
something else that is not true, istrue. > And when f = 3 and
m = 0, from the above, you have w1*w2 = 1.> You call that
constant? You call that not dependent on m?I call my
statements where I said w_1 w_2 = 3^{2/3} mistaken.> [The fact
is, when m = 0 and f = 3, the w's are indeterminate> as
discussed below.]I did believe that myself for a while, which
is why I posted whatfollows, later I realized it didn't follow
mathematically.> That is the conclusion that follows
logically.> Well having read down, the poster does have
a point which in fact is> that the 3's divide off in an
indeterminate way, so it's not so easy> to just say that w_1
w_2 = 3^{2/3} because that could be the case, but> then
again, maybe only w_3 has a factor of 3, so w_1 w_2 could
still> be coprime to 3.> Aha. So you did get the
point. Yay!Well yes, I did understand what you thought was an
issue, and at firstblush agreed that in fact the w's are
indeterminate when f=3.<deleted>> In fact INFINITELY many
ways. Any way you want to split 3 as a> product of three
numbers gives a factorization. And in all the> examples
just given (and in infinitely many others) note that (1) the coefficients of the x's are algebraic integers, and (2)
the w terms> are also algebraic integers.>
Proving what, exactly, you ask?> #### Proving that
the f = 3 case tells you NOTHING useful about the>
necessary values of the w's. There is no unique way to write
them> down. This is a special, exceptional case in which
too many of the> of the f terms can be factored out. It is correct that they are indeterminate,and that's a
VERY good point> that you brought up.> It is a very
important point, but I made a mistake saying they
areindeterminate, but I'm glad you brought up the issue.In
fact, your *same* points would actually apply in general with
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 +
u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf),where just like
before with 3, you now have f^2 as a factor.Do you see that
now?You might want to avoid seeing it now as you went to so
much effort totry and show that f=3 makes a special case, but
can you see it now?> But now tell me, do you *still* believe
the w's are then functions of> m?> I believe they are
dependent on m when you restrict to the cases > that happen to
be of interest to you: f a prime > 3, m relatively prime to f.
> my point of view, these conditions (f = prime > 3, m coprime
to f) are quite> natural also. I base my conclusions on a
theorem in algebraic number theory > which requires
primitivity and irreducibility of a polynomial. > When f = 3,
you do not get primitivity and irreducibility. So indeed, in
> this special case, I do not pretend to say anything about
the w's. Well forget about the w's for now, and look at P(m) =
f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)and consider that you're
now trying to say that f divides off as somekind of function,
as long as f is coprime to 3.Doesn't that bother you just a
*little* bit, as it goes against theobjections you yourself
raised with f=3?> That is, once you divide off 3, it's gone,
and the w's don't vary with> m, so you can't claim m=0 is a
special case where the w's are> functions of m.> You
have already divided off f^2. In this special case you can>
divide off another f. However, your factorization statement
was> about P(m)/f^2, not P(m)/f^3. If you want to consider
the> latter, then you get w1 = w2 = w3 = 1, since the
polynomial> is then> 19*x^3 - 8*x + 1.> So I don't think
you want to go there, eh? Why not? > Or do you think you
still can?>But before at m=0, they were coprime to
f, now they are not when f=3,>as they are constant.
Clearly, they are constant in both cases with>respect to
m, without regard to the value of f. Which makes sense asf^2 is not a function of m, and it is what is being divided
off.>> But in this case you can divide off
*another* factor of f,> and that is what leads to the
nonuniqueness shown above, which> wrecks your argument.
This does not happen when f is a prime > bigger than 3 and
m is coprime to f. f = 3 is a special case > of no interest
or relevance to your main argument. It is a > nuisance
distraction. It proves nothing.> Really? Your position
is that the w's are *functions* of m, > My position is that
they are dependent on m, which is not> quite the same as being
functions of m. And in any case,> my proofs regarding the
factorization do not include the> special cases f = 3 or m =
0. That does not mean my> proofs are incorrect. It does mean
that your entire discussion> of f = 3 is irrelevant.Now you've
been trying to say that m=0, is a special case, if f=3, sothat
you can claim that certain variables have different factors of
f,depending on m, but now you're saying it's not quite the same
as thembeing functions of m?Why don't you try and explain that
mathematically?And as for f=3 being irrelevant, how is f then
relevant with P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x +
uf)?Your *own* objections would indicate that you must believe
that how fdivides off is actually *indeterminate*, which I can
then show to bewrong, as I have by noting that with f=3, an f
has to divide off from*each* of the a's.Do you think maybe
you've been trying to make something simple *more*complicated
in order to hold a pet belief, against mathematical logic?If
you do not believe so, please explain carefully why *you*
think f^2divides off in some determined way here, after
arguing in another casethat f, when it equaled 3, did not, to
the extent that you now calldiscussion of it irrelevant.>
but now> you yourself bring out that with f=3, there's no
way to determine> their values, or do you claim there is? I certainly don't. You however did. You said the values
were> independent of m, i.e., constant, and you used the case
f = 3> to show it. But when f = 3, far from being a constant,
they cannot> even be determined at all. They can be anything
that divides 3. I made two errors. First I *said* that w_1 w_2
= 3^{2/3}, when f=3,which has been shown to be an error. You,
however, claimed they wereindeterminate, and I agreed with
you, which was a second error.Now you're holding on to your
own error, which would actually shootdown your other
objections anyway, as f=3 can't be distinguished inthe way you
wish, as all that changes is that instead of just f^2dividing
off, f^3 divides off of P(m), where P(m) = f^2((m^3 f^4 - 3m^2
f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x
+ uf)(a_3 x + uf).Now trapped by mathematical logic, you are
squirming trying to findsome way to hold on to your pet
beliefs, against the math.> If you were correct, then even
with f=3, the w's would STILL vary with> m, so that m=0
would be a *special* case, and then they'd vary as m>
varied.> Are you claiming they do?> Yes, when f is
a prime > 3 and m is coprime to f. Which> happens to be exactly
the conditions that you want to > consider also in your
proof.Apparently you have your heart set on my claims being
wrong, but whythink of them as my claims? It's
mathematics.Mathematics isn't true in order to upset you.It's
just true.Now then, can you see that you have no basis for
your opinion that f^2divides off in certain ways, just because
f is coprime to 3, when youyourself have challenged that
position when f=3?Why do you wish to believe that f^2 or f^3
when f=3, divides off ofP(m) as a variable dependent on
m?Isn't it true that you have beliefs you're trying to
protect?But it's *mathematics*, so why would you want to hold
on to beliefsthat are wrong? After all, in math, you don't
have to *protect* beliefs, asmathematical logic is there for
you. It doesn't need your help.>That is, if they were
functions of m, so that w_1 w_2 = 1 at m=0 as a*function* of m, then it wouldn't matter if f had a factor of
3 or>not, you'd STILL get w_1 w_2 = 1 at m=0, without
regard to the value>of f.>>But in fact, if
f=3, you have w_1 w_2 = 3^{2/3} at m=0, which only>works
if the w's are independent of m, which they are.> See above. In the special case with f = 3, the w's>
are not uniquely determined. You cannot draw conclusions>
from it about how the f terms are distributed among >
the w's or the linear factors. This case is a red >
herring, and it is of no interest in your general >
argument, where you require that f is a prime > 3. Worst>
of all, it does not imply what you want.> You seem to be
very determined to convince yourself that the> mathematics
says one thing, when it does not.> Not at all! You have
noted that I am right about the > indeterminacy of the w's
when f = 3. You were using that> case in an attempt to prove
that w1 and w2 are constant> with respect to m. I merely noted
that in fact that case > doesn't give you any information at
all regarding the w's.> In that case, the w's are
indeterminate. Your conclusion> was not justified. It seems to
me that it is YOU, not I,> who is determined to convince
yourself that the math> says one thing, when in this case it
doesn't actually say> anything.Well the evidence is to the
contrary as I'm flexible.I've learned that it's useless to
fight the math.So I made a mistake about the w's. I make
mistakes. But at least Icould entertain the possibility.You
however seem set on a certain belief and damn mathematical
logic. > I thought it was a little bit of a subtle point.
Clearly you> had not considered it previously, and clearly
from what> you have said here, you understand it. This in
itself is thus > a rather rare event.And here you are being
insulting. Why do you think correctmathematics needs
defense?Mathematics is not a child that can be harmed.If you
are right, then I assure you that if you make a logical
case,beginning with a truth and proceeding by logical steps to
aconclusion, then that conclusion is true.> It wrecks this
particular part of your argument. That is the point.That may
be your point, but your emotional needs do not affect thetruth
or falsity of a mathematical argument.The math argument just
does not care.> It's *your* position that the w's are
functions of m, so that you can> claim that m=0 is a special
case.> Dependent on m, not necessarily functions of m.
And I can> claim it only when f = prime > 3 and m is coprime
to f. But> that is a very big set of values, and is plenty
good enough for > both our purposes.However that goes against
the facts. The fact is that P(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x +
uf)(a_3 x + uf)and if f=3, you get *each* of the a's having a
factor of 3, for ALL m,without regard to its value.Do you
understand?Now then, your attempt at distinguishing f=3 from f
coprime to 3, inorder to force a functional dependency falls
flat at that point.So what happens when f is coprime to 3?Then
f^2 divides off through only two of the a's as I've shown.It's
that mathematical result which you are fighting with such
energyand emotion.You're fighting the math, not me.> You
need that claim so that you can then claim that the w's have varying factors of f, as m varies.> Now you need the
w's to vary when the polynomial has a factor of 3,> when
f=3, > No, I don't care what happens in that special case. I
don't claim> that my proofs hold in that case. The polynomial
in question is> not irreducible and primitive.For readers what
the poster Nora Baron is talking about here isreducibility
over rationals otherwise known as reducibility over Q.For an
example, consider P(x)= x^2 + 2x + 1 = (x+1)(x+1), as
apolynomial reducible over Q, whereas P(x) = x^2 + 5x + 2, is
anexample of a polynomial irreducible over Q.The position
Nora Baron is fighting for, is the claim that dependingon
whether or not you can reduce a polynomial over rationals, it
willfactor in certain ways, to where if it is irreducible, ALL
of itsroots MUST not be coprime to any prime INTEGER factor of
its constantterm.So by the poster's beliefs now because x^2 +
5x + 2 is irreducibleover rationals, both its roots must have
some non unit factor incommon with 2 based on that
irreducibility alone.> but that's not a lot different from
what was there before> when the factor was f^2.> When
f is prime > 3, it is a LOT different.> But here you try
to dismiss one constant factor--3--as a unimportant,> but
try to hang on to f^2 which is also a constant factor with
respect> to m.> It's not that it is unimportant
(though in fact it is - the case f = 3> is of no interest to
you in your arguments either). It is that I don't> make any
claims about what happens when f = 3. The > theorem that I use
in that case does not apply anyway.Hmmm...if that's correct,
then how would it apply to P(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)?On what basis have you
been arguing then?> Doesn't that bother you just a little
bit?> Nope.That is you feel confident you are right,
despite logicalinconsistencies in your position?> Does it
seem rational to you?> Yes. > What seems irrational to
me is to claim that > when f = 3, w1*w2 = 3^{2/3}, when in fact
w1 and> w2 are, as you yourself have noted above,
indeterminate> when f = 3. The f = 3 case adds no information
to your> knowledge of the relationship between m and w1 and
w2.> You claimed that it confirmed that w1 and w2 were
constant.> It did no such thing.I made a mistake. It's that
simple.Yup, I was saying w_1 w_2 = 3^{2/3} and that was wrong,
which issomething that can be worked out mathematically.I'm
glad you caught that mistake as I was posting it all over
theplace!There is an answer, which is found by following the
math. >It makes sense that they are anyway, as f^2 isn't
a function of > m, but>I've seen that for some people the
idea can take hold after seeing m=0>highlighted.>But if the w's were functions of m, then w_1 w_2 would
equal 1,>without regard to the value of f, but it does
not.>> w1 = w2 = 1 and w3 = 3 is just one of
the infinite range> of possibilities. It is Example 1
above.> Well then, what makes you think the w's are
*functions* of m, so that> m=0 would be a special case? Where's the mathematical logic in your thinking?> See
above. Unlike you, I do not make claims about> the special
case f = 3. I say that when f = prime > 3 and> m is coprime to
f, the w's are dependent on m. Prove> me wrong in THAT case and
I will start to take notice.Ok, so now f=3 is a special case.
Before it was m=0 as your specialcase and now f=3 is one as
well.It looks to me that when the math shows your beliefs to
be wrong, youfind a way to call it a special case!Therefore, the factorization is>> P(m)/f^2 = (m^3
f^4 - 3m^2 f^2 + 3m) x^3 - >> 3(-1+mf^2 )x u^2 +
u^3 f =>> (b_1 x + u)(b_2 x + u)(b_3 x + uf)>where you'll notice that the b's are algebraic
integers with m=1,>f=sqrt(2), but that's a special case
as generally they are not, which>shows a problem with
the ring of algebraic integers.>> In this
special case, P(x) does not even have> rational
coefficients. It too is of no interest> or value for your
main argument.> Hmmm...that's a fascinating attempt at
dodging the relevant> mathematics.> This is 180
degrees wrong! The RELEVANT mathematics are> not when f = 3.
The cases of interest to YOU are f = prime > 3,> m coprime to
f. Doesn't it seem a little odd for you to spend all> this
effort in a subset where your own proofs are not concerned?I
go where necessity and the mathematical logic lead.Here I'm
working to refute your popular position that reducibilityover
Q actually forces factors to move around in a
factorization,which depends on claims that require that f^2,
constant with regard tom, and a factor of P(m) divides off
from P(m) as a variable dependenton m.I've determined that I
can destroy your case using f=3, which is whatI'm doing.> As
I pointed out in another thread, with m=1, f=sqrt(2), y=uf, you
get> 2x^3 - 3xy^2 + y^3> which is reducible as it
factors as> 2x^2 - 3xy^2 + y^3 = (x-y)(2x^2 + 2xy -
y^2)> and yes it is true that since y=uf, where
f=sqrt(2), the y's are> irrational, but it doesn't change
the fact that the polynomial is> reducible.> There are
two things going on here. When you write> 2*x^3 - 3*x*y^2 +
y^3> you are considering a polynomial in two variables.
Indeed> that does factor in the obvious way.> But when you
subsitute y = sqrt(2), you are now > considering a polynomial
in ONE variable (x). You are> back to the case in which you
are interested. But this> polynomial in ONE variable has a
constant term which> is irrational. You know perfectly well
that it does not> factor with rational coefficients. > In
your factorizations, you are invariably factoring a>
polynomial in one indeterminate, x. That is why you > consider
linear factors of the form (b1*x + u*f), etc.. > It is from
that point of view that you should be considering> your
polynomial, not as a polynomial in two variables.What you said
is not true. In fact, the factorizations I show arethose like
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 +
u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)which is in more
than one variable.I can *relate* such factorizations to
simpler polynomials, where youand others have argued.I find it
interesting that you continue to defend your claims about 2x^2
- 3xy^2 + y^3 = (x-y)(2x^2 + 2xy - y^2)as you demonstrate a
troubling inflexibility.Well possibly it's because you realize
that then I can use m=1(modsqrt(2)) to destroy your claims. >
Now that makes the *second* time I've pointed that out to you. Right, thanks. But sqrt(2) is still irrational, eh? Or
did > that change recently in Georgia?That's what's grand
about mathematics, it's not particular to anyplace or time.It
seems to me that you demonstrate belief in something other
thanmathematics. > Possibly you try to dismiss it because
now using m=1(mod sqrt(2))> destroys your claim that
reducibility matters, as for instance,> m=3=1(mod sqrt(2))
but now the resulting polynomial is *irreducible*> and your
claims, where you believe the w's are functions of m, would>
force factors of f to shift.> You have lost me here. Are
you saying that two values of m> that agree mod sqrt(2) are
necessarily actually the same?Your claim is to how factors in
common with f will distribute withinthe factorization where
you have called m=0 a special case.But with f=sqrt(2), m=1, I
can demonstrate another case.You've tried to challenge that
because f=sqrt(2), and others mightclaim that maybe m=1 is a
special case, so I've pointed out that usingm=1(mod sqrt(2))
destroys your claims as then factors of f are fixed.Think
about it like this, if f=sqrt(2), then I can take the m=1
case,and use m=1(mod sqrt(2)) to make factors of 2 stay in
their place.Do you understand?>And here I've packed in a
lot of information as well.>> Not enough,
clearly.> Not if you refuse to follow mathematical
logic.> I think you've shown through your posting
behavior that you are> EXTREMELY dedicated to the notion
that I'm wrong, but the problem is> that the mathematical
logic is against you.>> You are correct that I think you
are wrong. Saying I am > dedicated to that notion is not
different from saying I am dedicated> to defending what I
believe to be true. Is that a character flaw in> your eyes?
Should I not do that?You can believe any number of things,
even when they're false.But in mathematics, you are supposed
to adjust your beliefs ifnecessary to the math.Your beliefs
are NOT more important than mathematical truth.Can you
understand?> So far, you've decided to go with your beliefs
against the math.> My beliefs are based entirely on
the math and nothing else> in this discussion.I've proven
that statement false.>First, with f coprime to 3, I now
know that the factorization is>> P(m)/f^2 = (b_1 x
+ u)(b_2 x + u)(b_3 x + uf)>>as the w's are
constant with respect to m, so I can just check at m=0,which revealed that w_1 w_2 = 1. Now that doesn't necessarily
force>w_1 and w_2 to each equal 1, but even if they were
factors of 1, i.e.>unit factors, that would only change
b_1 and b_2.>>So I have my factorization without
regard to m in terms of where the f>goes, and then I
point out that you can actually check my work using>m=1,
f=sqrt(2), as then you get a polynomial which you can factorrather simply. So you can actually get the values for the
b's and>check them, and see that they are all algebraic
integers, and all are>coprime to 2.>However, usually, for f values that are coprime to 3, you
don't get>b's that are algebraic integers, which shows a
problem with the ring>of algebraic integers.> Wrong! You *can* get algebraic integers, but *** not
with> the properties you want ***, and there is no problem
with> the ring of algebraic integers. Here is how things
work when> f = 5, m = 1, u = 1, v = -1 + m*f^2, and P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3: P(x)/f^2 = P(x)/25 = 553*x^3 - 72*x + 5.> By
the Magidin-McKinnon theorem (essentially proved earlier by  someone else [P. M. Cohn?]), this can be factored in the
form> 553*x^3 - 72*x + 5 = (b1*x + w1)*(b2*x +
w2)*(b3*x + w3),> where b1, b2, b3, and w1, w2, w3
are algebraic integers. > You can show using
elementary Galois theory that EACH of> w1, w2, and w3 is
not coprime to f = 5. > Thus: a factorization of the
desired form DOES exist, but it> does NOT have one of the
properties that you desperately want.> No problem with the
ring of algebraic integers, and no valid> proof for you.
You lose on two counts. Too bad!> And your *belief*
depends on the w's being functions of m, which is>
nonsensical, as follows from your own statements, like where
you> noticed that with f=3, the w's are indeterminate, as 3
can divide off> an infinite number of ways.> It's
amazing how you got this turned around. You claimed that> the
case f = 3 showed that the w's were constant. In fact in the>
case f = 3, the w's are indeterminate. It clear from what you
just> said that you actually understand this. The case f = 3
added > nothing to your knowledge of the behavior of the w's.
Yet you > claim that *** I *** am being nonsensical !!!
Incredible!I was convinced by you that the w's are
indeterminate, after I had thewrong position that w_1 w_2 =
3^{2/3}, and later found out you werewrong by checking the
mathematical logic.In your world, do people never change their
minds?If you are someone who refuses to change their minds then
that mightexplain things. > Yet now here you are claiming
that the w's are functions of m.> I claim that the w's
are dependent on m under the conditions> that f = prime > 3,
and m coprime to f. What you have done> not only does not
support your own conclusions in the special> no-interest case
where f = 3, it doesn't even touch the cases> which are of
genuine interest to both of us.Here you've declared that f=3
is a special no-interest case whichgoes against the math.>
That is NOT following the math.> Exactly the opposite is
true.Then prove it, mathematically.Don't *claim* one thing and
do another.Prove your claims.>Now the nice thing about a
mathematical proof is that if someone>disagrees they
have to find some misstep.>> See above at ####
! The misstep in the current argument has been > found. No, but you have shown your own bias and refusal to
follow> mathematical logic.> Really, there was nothing
in my post but mathematical logic.> Sorry it didn't support
your beliefs.Then discussions should be quick.You show the
logical chain in your reasoning and that will be it.Show the
mathematical logic that you believe supports your position,if
you can.>Unfortunately, people can *say* that proof is
not a proof, even when>it is, just like if you tried to
say you were human, and not a dog,>someone might dispute
any proof you might give, claiming it false.>>
In this case, you have tried to use an irrelevant >
red-herring argument to show what you want. Unfortunately, in the special case you selected, the number f (= 3) does factors *** non-uniquely *** through the linear terms of
your> polynomial factorization, and you end up being able
to> conclude:> *** N O T H I N G ***>
about the cases in which you are interested.>
But you have made progress. Do you realize how long it>
took us to get through to you that there is actually a>
nontrivial problem with generalizing from m = 0 to> m <> 0
? Do you realize how many incorrect arguments> you have
already burned through (including the present> one) in
trying to handle that problem? Do you realize> that all of
this is a waste of time, because your > main claims have
already been shown to be false and> cannot be fixed by
twiddling with the details?> Nora B.>
And readers should note that Nora Baron here arrogant and
presumably> confident, is still fighting to claim that f^2
divides off of a> polynomial as a function of m, though it
is itself constant with> regard to m, when the poster has
spent a lot of effort trying to claim> that 3 divides off in
an inconsequential way, when f=3.>> Well, duh. Yes, that is
what I was saying. I agree that I spent a> lot of time on an
inconsequential case. I was simply trying to show> that your
conclusions *even in that case* were not justified. At least>
you got the point this time, so I don't think it was a
complete waste.However, the arguments you used would destroy
your own claims aboutf^2 dividing off from P(m) in some
particular way.So you're trapped both by what you said, and by
the mathematicallogic.You can, of course, restate your
position, if you wish.> The point is that the poster Nora
Baron is not making sense.>> It's clear from what you
have said here that you understood what I was> saying, and
even, I daresay, that you were surprised by it. So clearly, >
it is not that what I said did not make sense. It is that you
did not > like the conclusion.> Nora B. That is not
correct. I *accepted* your conclusion and then found outthat
it was false, when I checked the math.Now you have the option
to follow the math, or stick with falsebeliefs.James
Harris

===

>[...]>>Well it turns out I was in error [...]>That
was a mistake. [...]>>It is a very important point, but I made
a mistake [...]>>Apparently you have your heart set on my
claims being wrong, but why>think of them as my claims? It's
mathematics.************************David C. Ullrich

===

[.snip.]>>Essentially objections to how f^2 divides off now
come down to>>claiming that the w's are functions of m, but
consider that w_1 w_2 =>>1, when m=0, if f is coprime to
3.>>Now I'm focusing on what has been revealed to be an area
of confusion.>> Apparently some people believe that when I
divide off f^2 that it can>>divide off as a *function* of m,
so that m=0 might be a special case. >>I'm now starting the
argument to address that belief by noting again>>that w_1 w_2
= 1, when m=0, if f is coprime to 3. That is, when f>>doesn't
have 3 as a factor.>>But that was an arbitrary choice, so
let f=3.>> f = 3 is irrelevant to what you want. There is>no
reason to consider it. As you will see below, it>is a red
herring and it does not show what you want.I have not been
following all the details of James's most recentpresentation.
But it ->seems<- to me that he is falling back to an
oldmistake yet again.It is of course true that any polynomial
identity will become a truestatement about specific numbers
when evaluated (to be more precise,if you have an identity in
the polynomial ring R[x], where R is acommutative ring, then
under the evaluation homomorphism x|->a it willyield a true
statement in R, for any a in R). However, it is not thecase
that any identity obtained under evaluation can be lifted to
apolynomial identity.Here James develops certain formulas in
general; he is being sloppy,in so far as he is not being
careful to remain in his polynomial ring,not all terms that
appear lie there. It so happens that when aparticular
evaluation is made, certain identities hold, and Jamestries to
deduce that these are polynomial identities. That's why
whenconfronted, he always falls back on claiming that
sincex^2+2x+1=(x+1)^2, then this holds for any value of x.
I.e., he isaffirming the
consequent.

===


===

Why do you take so much trouble to expose
such a reasoner as Mr. Smith? I answer as a deceased friend of
mine used to answer on like occasions - A man's capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does most
of all. He has made at least ten publications, full of figures
few readers can critize. A great many people are staggered to
this extend, that they imagine there must be the indefinite
something in the mysterious all this. They are brought to
the point of suspicion that the mathematicians ought not to
treat all this with such undisguised contempt, at least. --
A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de
Morgan

===
>
[.snip.]>>Essentially objections to how f^2 divides off
now come down to>>claiming that the w's are functions of m,
but consider that w_1 w_2 =>>1, when m=0, if f is coprime to
3.>>Now I'm focusing on what has been revealed to be an
area of confusion.>> Apparently some people believe that
when I divide off f^2 that it can>>divide off as a
*function* of m, so that m=0 might be a special case. >>I'm
now starting the argument to address that belief by noting
again>>that w_1 w_2 = 1, when m=0, if f is coprime to 3.
That is, when f>>doesn't have 3 as a factor.>>But that
was an arbitrary choice, so let f=3.>> f = 3 is
irrelevant to what you want. There is>no reason to consider
it. As you will see below, it>is a red herring and it does
not show what you want.> I have not been following all the
details of James's most recent> presentation. But it ->seems<-
to me that he is falling back to an old> mistake yet again.What
I've seen is that Arturo Magidin engages in making posts
whichapparently are meant to persuade. My interest is in what
the actualmathematical truth is, though it is of secondary
interest that Iconvince others.For Arturo Magidin, it appears
that persuasion is the primaryinterest, which can be seen from
his comments in contrast with thefacts which I present below. >
It is of course true that any polynomial identity will become a
true> statement about specific numbers when evaluated (to be
more precise,> if you have an identity in the polynomial ring
R[x], where R is a> commutative ring, then under the
evaluation homomorphism x|->a it will> yield a true statement
in R, for any a in R). However, it is not the> case that any
identity obtained under evaluation can be lifted to a>
polynomial identity.> Here James develops certain formulas
in general; he is being sloppy,> in so far as he is not being
careful to remain in his polynomial ring,> not all terms that
appear lie there. It so happens that when a> particular
evaluation is made, certain identities hold, and James> tries
to deduce that these are polynomial identities. That's why
when> confronted, he always falls back on claiming that since>
x^2+2x+1=(x+1)^2, then this holds for any value of x. I.e., he
is> affirming the consequent.What I've noticed is that Arturo
Magidin works VERY hard to bebelieved.authoritative or looks
mathematical, but he rarely actually provesanything relevant
to his central objections.Here the pertinent expression, which
he doesn't even bother to givehaving deleted it out of the post
to which he is replying, is P(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f).Notice the factor f^2,
which is, of course, constant with regard to m,but Arturo
Magidin and his diligent supporters try to argue that thef^2
divides off dependent on m, so that they can challenge that
withthe factorization P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3
- 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x +
uf)f^2 divides off through only two of the a's, which is a
result thatcan be looked at with m=1, f=sqrt(2), as then you
have P(m) = 2x^3 -3xy^2 + y^3, using y=sqrt(2)u,and that
factors as P(m) = (x-y)(2x^2 + 2xy - 1).However, Arturo
Magidin and his supporters try to ignore that case,while
continuing to claim that m=0 is a special case.Their focus on
m=0 comes from the fact that P(0)/f^2 = u^2(3x + uf)which I
use to show how f^2 divides off from P(m).Usually the claim by
Arturo Magidin and others like him who disputehow f^2 divides
off, trying to claim it divides off as a factor of m,is to say
that the reducibilty over *rationals* of f^2((m^3 f^4 - 3m^2
f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)considered to be a
polynomial with respect to x, determines how f^2divides off
from the expression.What I've found is that despite repeated
demonstrations to thecontrary posters like Arturo Magidin
refuse to back down, but insteadengage in various tactics to
convince people that they are right.James Harris

===

<same old
snipped>> What I've found is that despite repeated
demonstrations to the> contrary posters like Arturo Magidin
refuse to back down, but instead> engage in various tactics to
convince people that they are right.>> James HarrisInteresting.
Contrast this with:What I've found is that despite repeated
demonstrations to the contrary James Harris refuses to back
down, but instead engages in various tactics to convince
people that he is right.Be assured that the vast majority (if
not all) of the readers of thesenewsgroups, believe the second
of these statements is true. Arturo (andmany other talented
people) have given you way more time than you deserve.Back in
1973 when I was an undergraduate studying group theory, I
stumbledonto a paradoxical result. I spent the better part of
a weekend going overmy logic and could not find a mistake, so
when Monday rolled around, I wentto my professor's office and
asked him to look over my work, He immediatelysaw that I had
to be wrong, and after a bit, explained that I was making abad
assumption at one point. I do not recall the details, because
this wasso long ago, but the gist of it was that I was making
a mistake aboutproperties of a group due to my lack of
knowledge and experience. I trustedmy teacher, and devoted
myself to bettering my understanding of somefundamentals. A
month or so later I reviewed my work and with theimproved
state of my knowledge, could clearly see my errors. See
thedifference? I didn't see a conspiracy to deny my results,
but recognizedthat my knowledge was deficient, and I corrected
that.James, you use invalid terminology, you continue to use a
colloquial ratherthan a formal exposition, you skip important
steps (my high school algebrateacher made that issue very
clear to me...), you repeatedly useintermediate results that
are not proven. You have been given excellentfeedback from
some very knowledgeable people - the basic gist of which
is:get yourself some education in fundamentals. Nobody is
going to take youseriously if it's obvious you haven't been
doing your homework...

===

>A mathematical
proof begins with a truth, and proceeds by logicalsteps to a conclusion which then must be true.>I've pulled a detailed exposition of a short argument that
quickly>shows a problem with algebraic integers. It
starts after the>reference.
>Now here's a math proof. Those who doubt that fact can
believe it's a>claim of proof, but it's verified to
be a proof by tracing the>argument out.> Not! See below.> I've clipped sections
where there was no claim of error by the poster> Nora
Baron.> You also clipped parts where I made valid
claims that disagree> with your conclusions. I hope you will
read ALL of this post, and> read it carefully. > I read
all of the previous post. If you feel something material was>
clipped you should have said what, rather than use an
insinuation.> <deleted>>Essentially
objections to how f^2 divides off now come down toclaiming that the w's are functions of m, but consider that
w_1 w_2 =>1, when m=0, if f is coprime to 3.>Now I'm focusing on what has been revealed to be an
area of confusion.> Apparently some people believe
that when I divide off f^2 that it can>divide off as
a *function* of m, so that m=0 might be a special case. I'm now starting the argument to address that belief by
noting again>that w_1 w_2 = 1, when m=0, if f is
coprime to 3. That is, when f>doesn't have 3 as a
factor.>>But that was an arbitrary choice, so
let f=3.>> f = 3 is irrelevant to what
you want. There is> no reason to consider it. As you
will see below, it> is a red herring and it does not
show what you want.> Which would imply that you are
claiming that there is a misstep.> However, f=3
shows that the w_1 w_2 = 3^{2/3} for ALL m, when f=3.> The expression you are working with is> P(x)/f^2. What I showed in my post is that w1*w2 can be ANY two
algebraic integer> factors of f = 3. It is NOT necessarily
the case that w1*w2 = 3^{2/3}> in that case. The case f = 3
does not give you any information about> the w's.> Well it
turns out I was in error ... not for the first time, or, I
suspect, the last.as the answer is given by considering> the
factorization> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -  3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3
x + uf)> as with f=3, it's clear that *each* each what?> has
3 as a factor, which gives you> P(m)/3^3 = (m^3 3^3 - 3^2
m^2 + m) x^3 - > (-1+m 3^2 )x u^2 + u^3 = > (a_1/3 x +
u)(a_2/3 x + u)(a_3/3 x + u).> What I showed in my previous
post implies that ifa1, a2, and a3 are the negatives of the
roots of the polynomial u^3 - 8*u^2 - 19,then a perfectly
legit factorization is:[****] (a1*x + 1)*(a2*x + 1)*(27*a3*x +
27)which happens to equal 27*(a1*x + 1)*(a2*x + 1)*(a3*x + 1).I
think you can work out the reasoningfor yourself on that. If
not, let me know.> Now using the b's and w's gives> P(m)/3^2
= 3((m^3 3^3 - 3^2 m^2 + m) x^3 - > (-1+m 3^2 )x u^2 + u^3) =
> (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)> but at
m=0, I have> P(0)/3^2 = u^2(3x - 3u) = u^2(b_3 w_1 w_2 -
3u), so> and dividing off the last 3 gives> P(m)/3^3 =
(m^3 3^3 - 3^2 m^2 + m) x^3 - > (-1+m 3^2 )x u^2 + u^3 = >
(b_1 x + u)(b_2 x + u)(b_3/3 x + u)> Nope, this is NOT OK.
Why? Because you are assuming what you want to prove. You want
to show that thefactorization when m <> 0 is similar to that
when m = 0.Here m = 3. In case you have not noticed, 3 is not
equalto 0. Plus, if you look at what I said above at [****},
you will see a counterexample. Thus I guess there is no point
in going through the rest of whatyou have here. I will do so
anyway just to be fair.> so it's NOT true that it's completely
arbitrary what the w's are.> Notice that using m=0 and the
constant term P(0) forces the case, as> the only thing that
can vary is the indices as my choices there were> arbitrary.>
No, sorry. You are using what you want to prove. This
techniquecould simplify a great many difficult proofs, but it
is notconsidered a fair tactic in mathematics. In politics,
possibly. > What I have said is that in general, the
factorization (including the> values of the w's) is
dependent on the value of m. That statement> is correct when
it is restricted to the cases of interest to you:> the
polynomial P(x) is irreducible and primitive. These
conditions> will hold when f is a prime > 3 and m is coprime
to f. They do> NOT hold when f = 3 (P(x) is not primitive).  You're saying that equations *check* to see if P(x) is
irreducible> over rationals, which you claim happens if f is a
prime greater than 3> and m is coprime to f.> I am of course
not saying any such thing. Equations don't check
anything.Your putting it in that way is an attempt to deny
the conclusionwithout doing any hard mathematics. Trying to
pretend that my argumentrequires that equations are
intelligent beings or whatever is silly, andyou know damn well
it is not mathematics. In your typical equation, P(x) = (v^3 +
1)*x^x - 3*v*x*(u*f)^2 + (u*f)^3,where v = -1 + m*f^2, note
that the variable m occurs inboth the leading coefficient
(hidden in v^3) and in themiddle coefficient (hidden in v
itself). Certainly thereis no doubt that if you change m, you
change the roots of the equation. Now if you factor the
equation in theform (a1*x + c1)*(a2*x + c2)*(a3*x + c3),where
a1, a2, a3, and c1, c2, c3 are algebraic integers,the roots
are r1 = -c1/a1, r2 = -c2/a2, and r3 = -c3/a3.We have agreed
here that a1, a2, and a3 are functionsof m. If the equation
happens to be reducible intolinear terms with rational roots,
then you can assumethat c1, c2, and c3 are ordinary integers,
and that theyfactor the number (u*f)^3. In general you are not
goingto be able to tell HOW they factor (u*f)^3. That
willdepend on the MIDDLE term. Here are two quadratic examples
that may clarifythis: x^2 - 7*x + 10 = (x - 2)*(x - 5). x^2 -
11*x + 10 = (x - 1)*(x - 10).Note that changing the MIDDLE
term changed thefactors! In the first example, the roots
followthe factorization 2*5 = 10. In the secondexample, the
roots follow the factorization 1*10 = 10.The constant term,
10, stayed the same in both examples, just as, in the case in
which youare interested, the term (u*f)^3 stays the
same.Again: it was changing the MIDDLE term which resulted in
different values of the roots anda different form of
factorization.And AGAIN: in your polynomial, if you change m,
you will change the coefficient of the MIDDLE term,
-3*v*(u*f)^2.because you will have changed v.> A function is a
function. It doesn't check coprimeness of some> variable to
figure out if it's a function.> Silly. Are you saying that
because functions cannot checkanything about their
arguments, they must always be constantwith respect to those
arguments ??? Again, THIS IS NOT MATHEMATICS. This is an
attempt to ridicule my hard, rigorous mathematical conclusion
by implyingthat I impute some kind of intelligence to
functions, etc.. It doesn't wash. IT ISN'T A MATHEMATICAL
ARGUMENT. I giveit no weight whatsoever, nor should anyone
else.> The conditions are not artificial or arbitrary. I
apply a theorem> from algebraic number theory which requires
irreducibility and> (as noted by Arturo Magidin) primitivity.
These conditions also> happen to prevail in all the cases of
interest to you: f prime > 3> and m coprime to f. I do NOT
make statements about the> form of the factorization when f
= 3, and certainly not when> m = 0. > It's *algebra* so
you can use f and if there are functions of m,> then they
will be functions of m, whether f is coprime to 3 or not.>
Nope. My examples showed very clearly that 3 is a special
case.Plus I don't care about it anyway. And it is irrelevant
for yourpurposes too.> However, w_1 w_2 is coprime to 3
when f is coprime to 3, at m=0.> Therefore, the w's
are not functions of m, and in fact w_1 w_2 is> coprime to
f, for all m, and m=0 is not a special case.>> No. In
all the cases of interest, the w's ARE dependent on m.>
That is wrong as I've shown with> P(m)/3^3 = (m^3 3^3 - 3^2
m^2 + m) x^3 - > (-1+m 3^2 )x u^2 + u^3 = > (a_1/3 x +
u)(a_2/3 x + u)(a_3/3 x + u)> as a 3 must divide off through
*each* of the a's.> Your argument is incorrect. You assumed
what you wantedto prove. You generalized from m = 0 to m <>
0.> Your claims would require that the a's have a factor of f
that varies> as a function of m, except you wish to dodge your
assertion here> because f=3, and make claims that would force
equations to *choose* to> be functions or constants with
regard to m based on whether or not f> is coprime to 3.> Oh,
bull. This constant stupid attempt to imply that I assume
equations choose etc., is getting tiresome. PlusIT ISN'T
MATHEMATICS. Here are two ways to state the situation: (1) The
values of the w coefficients depend on the coefficients of the
polynomial, which in turn depend on m. (2) The w's choose
what their values are after they see what values m has. They
are conscious beings who look around and make decisions about
things that they see, and change their behavior accordingly.
Which of those do you think more accurately representsmy
position? Which is a (NON-MATHEMATICAL) attempt to ridicule
it? Don't you think the default assumption would be (1) ?Don't
you think the natural assumption would be to assume (1) until
somebody proved otherwise? The ideathat elements of a
factorization might be expected to depend on the coefficients
of the polynomial, especially the MIDDLE coefficients, is not
exactly rocket science, eh? > If the w's are not dependent
on m, then they must be constant.> Right? > They are.>
Not. Definitely not. And again, your attempt to proveotherwise
has been shot down: in my previous post, byshowing that the
case f = 3 leaves the w's entirelyindeterminate. And in the
present post, by showing thatyou tried to assume what you
wanted to prove.> As you have noted, when m = 0, you get P(0)/f^2 = w1*w2*u^2*(b3*x + u*w3)> [1] =
u^2*(b3*w1*w2*x + u*f). Right?> Yup.> You know, always
saying Yup instead of yes makes you soundlike a dimwitted
yokel. You might consider varying itfrom time to time.> But
P(0)/f^2 also equals (you note)> P(0)/f^2 = 3x*u^2 +
u^3*f> [2] = u^2*(3*x + u*f). Right?> And looking
at [1] and [2], you conclude w1*w2 = 1. Right?> Yup.> See
above.> And that happens to be true regardless of the value
of f. Right?> *Including* f = 3.> Yup.> And see above
again, Abner.> Now you say w1 and w2 do not depend on m. To
me, that> would mean that they are constant with respect to
m. Right?> But now when f = 3 and m <> 0, you say you
have w1*w2 = 3^{2/3}.> That was a mistake. Now if you were
following the math, then you> would simply note it as a
mistake, but instead you're trying to use> *my* mistake to
convince that something else that is not true, is> true.> Now
you are asking me to read your mind and correct your mistakes
for you. That is too much. You make alot of mistakes. If they
were just misprints I wouldnot mind, and I could probably
figure out what you intended. But as in this case, they are
far more thanmisprints. I had no idea what you intended.
Youpresented it as part of a serious argument. All I couldtell
was, it didn't make any sense. Also I pointed this out in one
or more previous postswhich you must not have read, or read
too quickly tounderstand. You kept repeating it. I note you
havehad to retract it also in posts in other threads.Certainly
not my fault!> And when f = 3 and m = 0, from the above, you
have w1*w2 = 1.> You call that constant? You call that
not dependent on m?> I call my statements where I said w_1
w_2 = 3^{2/3} mistaken.> [The fact is, when m = 0 and f =
3, the w's are indeterminate> as discussed below.]> I did
believe that myself for a while, which is why I posted what>
follows, later I realized it didn't follow mathematically.> In
which case, I am VERY ANNOYED that you continue to actas if you
are infallible, and frequently accuse the rest of us of lying.
I could easily have said you were lying here, or trying to
confuse people, or cover up the truth. That is what youhave
done repeatedly EVEN WHEN I PRESENT CORRECT MATH. It is *not*
what I did in your case. I assumed you had made a mistake of
some kind, but I didn't know whether it was a superficial
error, and you really had not noticed it, or what. But you
seemed to be putting a lot of weighton it as justifying your
side of the argument. I wasOBLIGATED to point out that it was
wrong. The other thing that I must note here is, you
oftenadmit *minor* mistakes like this with reasonable grace.In
such cases our objections are easy to understand, andimpossible
to deny. But when we present rigorous but deeper and more
difficult objections - equally valid,in my view - you
apparently refuse to even THINK aboutthem. Why? I think,
because you don't know enoughmath to understand them. When I
object to some statementof yours and I invoke, say, a simple
theorem from Galois theory to show it is wrong, to ME it seems
justas wrong as the little algebra mistake you made here.But
you act as if my Galois theory argument cannot betrusted and
that I must be lying. All you have to dois go learn a little
Galois theory or algebraic numbertheory for yourself. This too
you refuse to do. You recall my thread, Challenge of the
Century ?I was (and am) pretty sure in that case that I
wasright. Just as right as in the present
little-algebra-mistake case. The only theory I used was an
algebraicnumber theory theorem which you have seen before
andactually accepted. Even so, I was taking a hugechance. I am
not perfect. I make mistakes. I couldhave had an error in that
post, and you could havepicked it up. Still could, for all I
know. I was,however, extremely confident of my argument. No
oneelse raised the slightest objection to it. I think it
completely clinched the case againstyour proof in Advance
Polynomial Factorization. So what happened? You totally
ignored it as faras I can tell. The point? I have paid great
attention to yourarguments. I have dug into them in great
detail.I have presented proofs that in my view blast
yourclaims out of the water. You have ignored them.You insist
that we not only disprove your claims,but we must also (1)
disprove them without usingany theory that you haven't
bothered to learn, and(2) we must also find where your
mistakes are thatled you to your erroneous claims. The latter
is what the present thread is allabout. It shouldn't even be
necessary. We arebeing MORE THAN FAIR to you. For most
people,a rigorous disproof of their claims is enoughto send
them scurrying back to the drawing board.For you, we have to
(1) simplify our disproofs so that a child could understand
them, and, toboot, (2) find exactly where the child screwedup,
a task that should be left to the childhimself. The real point
here: from my point of view,my big objections to your bogus
math are justas valid as my little objections to
simplealgebra errors. You should at least try tounderstand
them. If you cannot refute them,it is just like refusing to
acknowledge and change the kind of error you noted above. And
finally: do NOT expect me to cut youany slack for making
algebra mistakes. I honestly cannot read your mind.> That
is the conclusion that follows logically.> Well
having read down, the poster does have a point which in fact
is> that the 3's divide off in an indeterminate way, so
it's not so easy> to just say that w_1 w_2 = 3^{2/3}
because that could be the case, but> then again, maybe
only w_3 has a factor of 3, so w_1 w_2 could still> be
coprime to 3.> Aha. So you did get the point.
Yay!> Well yes, I did understand what you thought was an
issue, and at first> blush agreed that in fact the w's are
indeterminate when f=3.> And at second blush, you have tried
to pull back again,*assuming the very thing you were trying to
prove*. Note thateven in that case, you did not actually
identify anything wrong with what I said. You went off on
another (unjustified)tangent. > <deleted>> In fact
INFINITELY many ways. Any way you want to split 3 as a>
product of three numbers gives a factorization. And in all
the> examples just given (and in infinitely many others)
note that (1) the> coefficients of the x's are algebraic
integers, and (2) the w terms> are also algebraic
integers.> Proving what, exactly, you ask? #### Proving that the f = 3 case tells you NOTHING
useful about the> necessary values of the w's. There is
no unique way to write them> down. This is a special,
exceptional case in which too many of the> of the f
terms can be factored out.> It is correct that they
are indeterminate,and that's a VERY good point> that you
brought up.> It is a very important point, but I
made a mistake saying they are> indeterminate, but I'm glad
you brought up the issue.> No, you were right the first time.
There is absolutely nothing wrongwith my argument that they
are indeterminate. How can youdeny explicit examples?> In
fact, your *same* points would actually apply in general with P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x
u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf),>
where just like before with 3, you now have f^2 as a factor.>
No, I don't see that at all. In the case of f = 3,you know
that another 3 factors out of the left side.In the case where
f <> 3, that is not true. For example,when f = 5, m = 1, u =
1, you have P(x)/f^2 = 553*x^3 - 72*x + 5,and you cannot
factor any more 5's through that polynomial:f^2 = 25 is the
most you can do. Neither 553 nor 72 aredivisible by 5. > Do
you see that now?> No. In the case where f = 3, again with m =
1 and u = 1, youhave P(x)/f^2 = 57*x^3 - 24*x + 3,and I CAN
factor another 3 out of that. That makes an enormousdifference
as my previous post showed.> You might want to avoid seeing it
now as you went to so much effort to> try and show that f=3
makes a special case, but can you see it now?> See above. It
looks to me like 3 divides 57 and 24, but 5does NOT divide 553
and 72. Cold, hard arithmetic.> But now tell me, do you
*still* believe the w's are then functions of> m? I believe they are dependent on m when you restrict to the
cases > that happen to be of interest to you: f a prime > 3,
m relatively prime to f. > my point of view, these
conditions (f = prime > 3, m coprime to f) are quite>
natural also. I base my conclusions on a theorem in algebraic
number theory > which requires primitivity and
irreducibility of a polynomial. > When f = 3, you do not get
primitivity and irreducibility. So indeed, in > this special
case, I do not pretend to say anything about the w's.> Well
forget about the w's for now, and look at> P(m) = f^2((m^3
f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) = >
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)> and consider that
you're now trying to say that f divides off as some> kind of
function, as long as f is coprime to 3.> Doesn't that bother
you just a *little* bit, as it goes against the> objections you
yourself raised with f=3?> Actually, no. First of all, you are
assuming without visiblejustification that the factorization
is of the form (a1*x + uf)*(a2*x + uf)*(a3*x + uf),when you
should be assuming it is (a1*x + c1)*(a2*x + c2)*(a3*x +
c3),where, as guaranteed by the Magidin-McKinnon theorem, a1,
a2, a3, and c1, c2, c3 are algebraic integers. TheM-M theorem
does not guarantee more than that. Certainlyit does not
guarantee that c1 = uf, c2 = uf, c3 = uf.What makes you think
that ???> That is, once you divide off 3, it's gone, and
the w's don't vary with> m, so you can't claim m=0 is a
special case where the w's are> functions of m. You have already divided off f^2. In this special case you
can> divide off another f. However, your factorization
statement was> about P(m)/f^2, not P(m)/f^3. If you want to
consider the> latter, then you get w1 = w2 = w3 = 1, since
the polynomial> is then> 19*x^3 - 8*x + 1.> So I
don't think you want to go there, eh?> Why not?> Because the
constant term is 1, and IN THIS CASE, you CANassume that the
factorization is of the form (a1*x + 1)*(a2*x + 1)*(a3*x +
1).[Would you like to see a proof of that? Probably not.]That
is, the w's are all equal to 1. None are divisibleby any
factor of 3. That too does is not in agreementwith your
previous claims about the w's. > Or do you think you still
can?>But before at m=0, they were coprime to f,
now they are not when f=3,>as they are constant.
Clearly, they are constant in both cases with>respect
to m, without regard to the value of f. Which makes sense asf^2 is not a function of m, and it is what is being
divided off.>> But in this case you can
divide off *another* factor of f,> and that is what
leads to the nonuniqueness shown above, which> wrecks
your argument. This does not happen when f is a prime >
bigger than 3 and m is coprime to f. f = 3 is a special case  of no interest or relevance to your main argument. It is
a > nuisance distraction. It proves nothing.>
Really? Your position is that the w's are *functions* of m,  My position is that they are dependent on m, which is
not> quite the same as being functions of m. And in any
case,> my proofs regarding the factorization do not include
the> special cases f = 3 or m = 0. That does not mean my>
proofs are incorrect. It does mean that your entire
discussion> of f = 3 is irrelevant.> Now you've been
trying to say that m=0, is a special case, if f=3, so> that
you can claim that certain variables have different factors of
f,> depending on m, but now you're saying it's not quite the
same as them> being functions of m?> Dependent on m and
functions of m are not the same.For example, suppose n
depends on m in the sense that ifm is divisible by 13, n is an
even integer. That does NOTmean that n is a function of m. It
does mean that if m1and m2 are two different values for n, you
do not expectthat the corresponding n's, n1 and n2, are
necessarilythe same.> Why don't you try and explain that
mathematically?> See above. > And as for f=3 being irrelevant,
how is f then relevant with> P(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2
x + uf)(a_3 x + uf)?> Your *own* objections would indicate
that you must believe that how f> divides off is actually
*indeterminate*, which I can then show to be> wrong, as I have
by noting that with f=3, an f has to divide off from> *each* of
the a's.> First of all, as I noted above, when f<> 3, I do not
agree that you can assume the factorization has the form (a1*x
+ uf)*(a2*x + uf)*(a3*x + uf),and I don't see why you assume
that either. It is simplynot true in general, and it is not
stated as a part ofthe Magidin-McKinnon theorem. Secondly, if
here you are talking about f = 3, thenanother f DOES can be
divided off, and it leads to nonuniqueb's and w's as I noted
in my previous post. See my previous discussion of this also.
My previous postshowing nonuniqueness still stands. You cannot
deny explicit examples with bogus theory.> Do you think maybe
you've been trying to make something simple *more*>
complicated in order to hold a pet belief, against
mathematical logic?> If you do not believe so, please
explain carefully why *you* think f^2> divides off in some
determined way here, after arguing in another case> that f,
when it equaled 3, did not, to the extent that you now call>
discussion of it irrelevant.> Jesus. See above. I have given
more than ample explanations of this.*Read them*.> but now you yourself bring out that with f=3, there's no way to
determine> their values, or do you claim there is? I certainly don't. You however did. You said the values
were> independent of m, i.e., constant, and you used the
case f = 3> to show it. But when f = 3, far from being a
constant, they cannot> even be determined at all. They can
be anything that divides 3.> I made two errors. First I
*said* that w_1 w_2 = 3^{2/3}, when f=3,> which has been shown
to be an error. You, however, claimed they were> indeterminate,
and I agreed with you, which was a second error.> Plus in this
post, you made the really egregious error oftrying to assume
what you wanted to prove.> Now you're holding on to your own
error, which would actually shoot> down your other objections
anyway, as f=3 can't be distinguished in> the way you wish, as
all that changes is that instead of just f^2> dividing off, f^3
divides off of P(m), where> P(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2
x + uf)(a_3 x + uf).> See above.> Now trapped by math> If
you were correct, then even with f=3, the w's would STILL
vary with> m, so that m=0 would be a *special* case, and
then they'd vary as m> varied.> Are you claiming
they do?> Yes, when f is a prime > 3 and m is
coprime to f. Which> happens to be exactly the conditions
that you want to > consider also in your proof.>
Apparently you have your heart set on my claims being wrong,
but why> think of them as my claims? It's mathematics.> Has
nothing to do with my heart. In various places, such asAPF,
you claim certain things are true. They demonstrably are not.
You think you have proved them. Clearly yourproof has an
error. That is what we are talking abouthere.> Mathematics
isn't true in order to upset you.> What ???> It's just true.>
A tautology obviously. And implicitly, another claim thatyou
yourself are infallible. You are really saying MYmathematics
is invariably true. I cannot swallow that. By now a little
humility should have crept in. You DOmake mistakes, after all.
Until you find an errorin things like Challenge of the
Century, I will haveto regard your claims as just that:
claims.> Now then, can you see that you have no basis for your
opinion that f^2> divides off in certain ways, just because f
is coprime to 3, when you> yourself have challenged that
position when f=3?> Why do you wish to believe that f^2 or
f^3 when f=3, divides off of> P(m) as a variable dependent on
m?> Isn't it true that you have beliefs you're trying to
protect?> But it's *mathematics*, so why would you want to
hold on to beliefs> that are wrong?> After all, in math, you
don't have to *protect* beliefs, as> mathematical logic is
there for you. It doesn't need your help.> See above. You are
beating an extremely dead horse. It isbeginning to reek.That is, if they were functions of m, so that w_1 w_2 = 1
at m=0 as a>*function* of m, then it wouldn't matter
if f had a factor of 3 or>not, you'd STILL get w_1 w_2
= 1 at m=0, without regard to the value>of f.>But in fact, if f=3, you have w_1 w_2 = 3^{2/3} at
m=0, which only>works if the w's are independent of m,
which they are.>> See above. In the
special case with f = 3, the w's> are not uniquely
determined. You cannot draw conclusions> from it about
how the f terms are distributed among > the w's or the
linear factors. This case is a red > herring, and it is
of no interest in your general > argument, where you
require that f is a prime > 3. Worst> of all, it does
not imply what you want.> You seem to be very
determined to convince yourself that the> mathematics says
one thing, when it does not.> Not at all! You have
noted that I am right about the > indeterminacy of the w's
when f = 3. You were using that> case in an attempt to prove
that w1 and w2 are constant> with respect to m. I merely
noted that in fact that case > doesn't give you any
information at all regarding the w's.> In that case, the w's
are indeterminate. Your conclusion> was not justified. It
seems to me that it is YOU, not I,> who is determined to
convince yourself that the math> says one thing, when in
this case it doesn't actually say> anything.> Well the
evidence is to the contrary as I'm flexible.> I've learned
that it's useless to fight the math.> I see little evidence of
such learning.> So I made a mistake about the w's. I make
mistakes. But at least I> could entertain the possibility.>
You however seem set on a certain belief and damn mathematical
logic.> On the contrary. That damned mathematical logic is
theonly reason I have for what I think. > I thought it was a
little bit of a subtle point. Clearly you> had not considered
it previously, and clearly from what> you have said here, you
understand it. This in itself is thus > a rather rare event. And here you are being insulting. I apologize. However I have
seen worse.> Why do you think correct> mathematics needs
defense?> I would be the last to think that. Incorrect
mathematics, on the other hand, needs to beattacked!>
Mathematics is not a child that can be harmed.> You come up
with the most bizarre statements. In onebreath, you say you
make mistakes. In the next breath,you are implying that what
you have done is *unassailable*.Incredible !!! Plus you imply
I am a would-be child-beater!> If you are right, then I assure
you that if you make a logical case,> beginning with a truth
and proceeding by logical steps to a> conclusion, then that
conclusion is true.> Good. Glad we agree on that. In view of
that, why don'tyou have a look at the argument in Challenge
of the Century,and tell me which step is not logical ?> It
wrecks this particular part of your argument. That is the
point.> That may be your point, but your emotional needs do
not affect the> truth or falsity of a mathematical argument.>
Forget my emotional needs. You have nothing to do with them
and don't know what they are.> The math argument just does not
care.> Ah, yes. You falsely ridicule my argument on the grounds
that I must believe the w's act like littlepeople, making
decisions about how they factor depending on the value of m,
etc.. And now YOUseem to somehow have inside information on
theemotional state of the math. Are you familiar with the
word hubris ?If not, are you familiar with the word irony
?> It's *your* position that the w's are functions of m,
so that you can> claim that m=0 is a special case. Dependent on m, not necessarily functions of m. And I
can> claim it only when f = prime > 3 and m is coprime to f.
But> that is a very big set of values, and is plenty good
enough for > both our purposes.> However that goes against
the facts. The fact is that> P(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2
x + uf)(a_3 x + uf)> and if f=3, you get *each* of the a's
having a factor of 3, for ALL m,> without regard to its
value.> No, little man. My previous post showed very clearly,
veryexplicitly, the contrary. A possible factorization could
be (b1*x + 1)*(b2*x + 1)*(27*b3*x + 27),where b1, b2, and b3
are the negatives of the roots of u^3 - 8*u^2 - 19.Your
blather above does not change the fact that*such examples
exist*.> Do you understand?> Assuredly. See just above.> Now
then, your attempt at distinguishing f=3 from f coprime to 3,
in> order to force a functional dependency falls flat at that
point.> On the contrary. Your attempt to show that the w's
areconstant on the basis of what happens when f = 3 is
whatfalls flat. See above.> So what happens when f is coprime
to 3?> Then f^2 divides off through only two of the a's as
I've shown.> When f <> 3, you cannot assume the factorization
takes theform (a1*x + uf)*(a2*x + uf)*(a3*x + uf). The most
you can say is that it can be factored as (a1*x + c1)*(a2*x +
c2)*(a3*x + c3).> It's that mathematical result which you are
fighting with such energy> and emotion.> Oh please. > You're
fighting the math, not me.> Again, you are 180 degrees off.
The math is my friend.> You need that claim so that you
can then claim that the w's have> varying factors of f, as
m varies.> Now you need the w's to vary when the
polynomial has a factor of 3,> when f=3, > No, I
don't care what happens in that special case. I don't claim>
that my proofs hold in that case. The polynomial in question
is> not irreducible and primitive.> For readers what the
poster Nora Baron is talking about here is> reducibility
over rationals otherwise known as reducibility over Q.> For
an example, consider P(x)= x^2 + 2x + 1 = (x+1)(x+1), as a>
polynomial reducible over Q, whereas P(x) = x^2 + 5x + 2, is
an> example of a polynomial irreducible over Q.> The
position Nora Baron is fighting for, is the claim that
depending> on whether or not you can reduce a polynomial over
rationals, it will> factor in certain ways, to where if it is
irreducible, ALL of its> roots MUST not be coprime to any
prime INTEGER factor of its constant> term.> Yes. You have
stated that correctly.> So by the poster's beliefs now because
x^2 + 5x + 2 is irreducible> over rationals, both its roots
must have some non unit factor in> common with 2 based on that
irreducibility alone.> Definitely true. Don't you agree? After
all, each of theroots themselves is an algebraic integer, and
the product of the rootsis 2. AND neither of the roots is a
unit. It's pretty obvious, isn't it ???> but that's not a
lot different from what was there before> when the factor
was f^2.> When f is prime > 3, it is a LOT
different.> But here you try to dismiss one constant
factor--3--as a unimportant,> but try to hang on to f^2
which is also a constant factor with respect> to m. It's not that it is unimportant (though in fact it is
- the case f = 3> is of no interest to you in your
arguments either). It is that I don't> make any claims about
what happens when f = 3. The > theorem that I use in that
case does not apply anyway.> Hmmm...if that's correct, then
how would it apply to> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m)
x^3 - > 3(-1+mf^2 )x u^2 + u^3 f)?> When f = 3, this
polynomial is not primitive. When, forexample, f = 5, m = 1,
and u = 1, the polynomial inside theparentheses is
irreducible. Do you now really want to see the rest of the
argument?If so, why have you ignored it previously? Why don't
you golook at Challenge of the Century ?> On what basis have
you been arguing then?ematical logic, you are squirming trying
to find> some way to hold on to your pet beliefs, against the
math.> The math is the only reason I have for my pet
beliefs.> Doesn't that bother you just a little bit? Nope.> That is you feel confident you are right,
despite logical> inconsistencies in your position?> Have you
stopped beating your wife?> Does it seem rational to you? Yes. > What seems irrational to me is to
claim that > when f = 3, w1*w2 = 3^{2/3}, when in fact w1
and> w2 are, as you yourself have noted above,
indeterminate> when f = 3. The f = 3 case adds no
information to your> knowledge of the relationship between m
and w1 and w2.> You claimed that it confirmed that w1 and w2
were constant.> It did no such thing.> I made a mistake.
It's that simple.> Yup, I was saying w_1 w_2 = 3^{2/3} and
that was wrong, which is> something that can be worked out
mathematically.> I'm glad you caught that mistake as I was
posting it all over the> place!> No problem! Glad to
oblige!> There is an answer, which is found by following the
math.> I totally agree.>It makes sense that they are
anyway, as f^2 isn't a function of > m, but>I've seen
that for some people the idea can take hold after seeing m=0highlighted.>>But if the w's were
functions of m, then w_1 w_2 would equal 1,>without
regard to the value of f, but it does not.> w1 = w2 = 1 and w3 = 3 is just one of the infinite range of possibilities. It is Example 1 above.> Well
then, what makes you think the w's are *functions* of m, so
that> m=0 would be a special case?> Where's the
mathematical logic in your thinking?> See above.
Unlike you, I do not make claims about> the special case f =
3. I say that when f = prime > 3 and> m is coprime to f, the
w's are dependent on m. Prove> me wrong in THAT case and I
will start to take notice.> Ok, so now f=3 is a special
case. Before it was m=0 as your special> case and now f=3 is
one as well.> It looks to me that when the math shows your
beliefs to be wrong, you> find a way to call it a special
case!>Therefore, the factorization is> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f =>> (b_1 x +
u)(b_2 x + u)(b_3 x + uf)>>where you'll notice
that the b's are algebraic integers with m=1,f=sqrt(2), but that's a special case as generally they are
not, which>shows a problem with the ring of algebraic
integers.>> In this special case, P(x)
does not even have> rational coefficients. It too is of
no interest> or value for your main argument.>
Hmmm...that's a fascinating attempt at dodging the relevant mathematics.> This is 180 degrees wrong! The
RELEVANT mathematics are> not when f = 3. The cases of
interest to YOU are f = prime > 3,> m coprime to f. Doesn't
it seem a little odd for you to spend all> this effort in a
subset where your own proofs are not concerned?> I go
where necessity and the mathematical logic lead.> The point of
my question was, there is no necessity to gowhere you did, and
it doesn't do you any good anyway. You want to come to certain
conclusions when f = 5, 7, 11, 13,etc.. You cannot get there by
restricting yourself only to f = 3. It is like trying to deduce
what is inside theforest by nibbling on a little of the foliage
out at theedge.> Here I'm working to refute your popular
position that reducibility> over Q actually forces factors to
move around in a factorization,> which depends on claims that
require that f^2, constant with regard to> m, and a factor of
P(m) divides off from P(m) as a variable dependent> on m.>
Yes, I am aware of what you are TRYING to do.> I've determined
that I can destroy your case using f=3, which is what> I'm
doing.> No. You have determined no such thing. Your argument
here assumes exactly what you want to prove, that you can
generalize from m = 0 to m <> 0. > As I pointed out in
another thread, with m=1, f=sqrt(2), y=uf, you get>
2x^3 - 3xy^2 + y^3> which is reducible as it factors
as> 2x^2 - 3xy^2 + y^3 = (x-y)(2x^2 + 2xy - y^2) and yes it is true that since y=uf, where f=sqrt(2),
the y's are> irrational, but it doesn't change the fact
that the polynomial is> reducible.> There are
two things going on here. When you write> 2*x^3 -
3*x*y^2 + y^3> you are considering a polynomial in two
variables. Indeed> that does factor in the obvious way. But when you subsitute y = sqrt(2), you are now >
considering a polynomial in ONE variable (x). You are> back
to the case in which you are interested. But this>
polynomial in ONE variable has a constant term which> is
irrational. You know perfectly well that it does not> factor
with rational coefficients. > In your factorizations, you
are invariably factoring a> polynomial in one indeterminate,
x. That is why you > consider linear factors of the form
(b1*x + u*f), etc.. > It is from that point of view that you
should be considering> your polynomial, not as a polynomial
in two variables.> What you said is not true. In fact, the
factorizations I show are> those like> P(m) = f^2((m^3 f^4 -
3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x
+ uf)(a_2 x + uf)(a_3 x + uf)> which is in more than one
variable.> You are considering only factorizations with
respectto the polynomial indeterminate x. The other
variables are not like the y in the bivariate polynomial
above. In general you think of them asfixed constants. Unlike
y, they do not appear explicitly inthe final factorization,
i.e., in the linearterms of the general form (ai*x + ci).> I
can *relate* such factorizations to simpler polynomials, where
you> and others have argued.> ???> I find it interesting that
you continue to defend your claims about> 2x^2 - 3xy^2 + y^3
= (x-y)(2x^2 + 2xy - y^2)> as you demonstrate a troubling
inflexibility.> You applied this when y = sqrt(2). You wanted
tohave it both ways: on the one hand, you wanted to saythat as
a polynomial in TWO variables, it factored over the rationals.
Then you wanted to let y = sqrt(2)and consider it as a
polynomial in ONE variable, andmake the same claim. IT SIMPLY
DOESN'T WORK. Apolynomial with irrational coefficients cannot
possiblybe considered reducible over the rationals. It makesno
sense to even consider it. It's not a matter offlexibility. >
Well possibly it's because you realize that then I can use
m=1(mod> sqrt(2)) to destroy your claims.> Nope. See above.
You cannot have it both ways.> Now that makes the *second*
time I've pointed that out to you.> Right, thanks.
But sqrt(2) is still irrational, eh? Or did > that change
recently in Georgia?> That's what's grand about mathematics,
it's not particular to any> place or time.> Agreed. But answer
the question about Georgia.> It seems to me that you
demonstrate belief in something other than> mathematics.> The
old mind-reader at work again.> Possibly you try to
dismiss it because now using m=1(mod sqrt(2))> destroys
your claim that reducibility matters, as for instance,>
m=3=1(mod sqrt(2)) but now the resulting polynomial is
*irreducible*> and your claims, where you believe the w's
are functions of m, would> force factors of f to shift. You have lost me here. Are you saying that two values
of m> that agree mod sqrt(2) are necessarily actually the
same?> Your claim is to how factors in common with f will
distribute within> the factorization where you have called m=0
a special case.> But with f=sqrt(2), m=1, I can demonstrate
another case.> You've tried to challenge that because
f=sqrt(2), and others might> claim that maybe m=1 is a special
case, so I've pointed out that using> m=1(mod sqrt(2)) destroys
your claims as then factors of f are fixed.> Think about it
like this, if f=sqrt(2), then I can take the m=1 case,> and
use m=1(mod sqrt(2)) to make factors of 2 stay in their
place.> Do you understand?> You know, I just don't care
about this example. It is notrelevant no matter how you cut
it.>And here I've packed in a lot of information as
well.>> Not enough, clearly.>
Not if you refuse to follow mathematical logic.> I
think you've shown through your posting behavior that you are EXTREMELY dedicated to the notion that I'm wrong, but the
problem is> that the mathematical logic is against you.> You are correct that I think you are wrong. Saying I
am > dedicated to that notion is not different from saying
I am dedicated> to defending what I believe to be true. Is
that a character flaw in> your eyes? Should I not do that? You can believe any number of things, even when they're
false.> But in mathematics, you are supposed to adjust your
beliefs if> necessary to the math.> Your beliefs are NOT
more important than mathematical truth.> Can you
understand?> Here again you are saying: I am infallible. I am
uniquelyin possession of mathematical truth. If you disagree
with me you are automatically wrong. Right? You are already
back to saying you never make mistakes, right? Why in the HELL
do you make that assumption? Given your history, the opposite
assumption (that you areinvariably wrong) would be more
justified. Have youbeen hearing voices from God telling you
you are The One,or something? Have a little humility, man! So far, you've decided to go with your beliefs against the
math.> My beliefs are based entirely on the
math and nothing else> in this discussion.> I've proven
that statement false.> Nosirree bob. You have assumed what you
wanted to prove.In math, that's a no-no. Try to avoid it.First, with f coprime to 3, I now know that the
factorization is>> P(m)/f^2 = (b_1 x + u)(b_2
x + u)(b_3 x + uf)>>as the w's are constant
with respect to m, so I can just check at m=0,>which
revealed that w_1 w_2 = 1. Now that doesn't necessarily
force>w_1 and w_2 to each equal 1, but even if they
were factors of 1, i.e.>unit factors, that would only
change b_1 and b_2.>>So I have my factorization
without regard to m in terms of where the f>goes, and
then I point out that you can actually check my work usingm=1, f=sqrt(2), as then you get a polynomial which you can
factor>rather simply. So you can actually get the
values for the b's and>check them, and see that they
are all algebraic integers, and all are>coprime to
2.>>However, usually, for f values that are
coprime to 3, you don't get>b's that are algebraic
integers, which shows a problem with the ring>of
algebraic integers.>> Wrong! You *can*
get algebraic integers, but *** not with> the properties
you want ***, and there is no problem with> the ring of
algebraic integers. Here is how things work when> f = 5,
m = 1, u = 1, v = -1 + m*f^2, and> P(x) = (v^3 +
1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3:>
P(x)/f^2 = P(x)/25 = 553*x^3 - 72*x + 5.> By the
Magidin-McKinnon theorem (essentially proved earlier by >
someone else [P. M. Cohn?]), this can be factored in the form 553*x^3 - 72*x + 5 = (b1*x + w1)*(b2*x +
w2)*(b3*x + w3),> where b1, b2, b3, and w1, w2,
w3 are algebraic integers. > You can show using
elementary Galois theory that EACH of> w1, w2, and w3 is
not coprime to f = 5. > Thus: a factorization of
the desired form DOES exist, but it> does NOT have one
of the properties that you desperately want.> No problem
with the ring of algebraic integers, and no valid> proof
for you. You lose on two counts. Too bad!> And your
*belief* depends on the w's being functions of m, which is nonsensical, as follows from your own statements, like where
you> noticed that with f=3, the w's are indeterminate, as 3
can divide off> an infinite number of ways.>
It's amazing how you got this turned around. You claimed that the case f = 3 showed that the w's were constant. In fact in
the> case f = 3, the w's are indeterminate. It clear from
what you just> said that you actually understand this. The
case f = 3 added > nothing to your knowledge of the behavior
of the w's. Yet you > claim that *** I *** am being
nonsensical !!! Incredible!> I was convinced by you that the
w's are indeterminate, after I had the> wrong position that w_1
w_2 = 3^{2/3}, and later found out you were> wrong by checking
the mathematical logic.> No. You just replaced one error with
another. I must say, you are incredibly inventive and
incrediblystubborn. You keep getting shot down, like Yul
Brynner inthe old movie Westworld, and you just keep coming
back with yet another bogus argument.> In your world, do
people never change their minds?> Yes. I mean, Yup.> If you
are someone who refuses to change their minds then that might>
explain things.> I change my mind several times a day, just
like myunderwear.> Yet now here you are claiming that the
w's are functions of m.> I claim that the w's are
dependent on m under the conditions> that f = prime > 3, and
m coprime to f. What you have done> not only does not support
your own conclusions in the special> no-interest case where f
= 3, it doesn't even touch the cases> which are of genuine
interest to both of us.> Here you've declared that f=3 is a
special no-interest case which> goes against the math. That is NOT following the math.> Exactly the
opposite is true.> Then prove it, mathematically.> Did.
Previous post.> Don't *claim* one thing and do another.> Did.
My conscience is clear on that one.> Prove your claims.>
Wasn't once enough?>Now the nice thing about a
mathematical proof is that if someone>disagrees they
have to find some misstep.>> See above
at #### ! The misstep in the current argument has been  found.> No, but you have shown your own bias and
refusal to follow> mathematical logic.>
Really, there was nothing in my post but mathematical logic. Sorry it didn't support your beliefs.> Then discussions
should be quick.> I wish they were quicker than this one!> You
show the logical chain in your reasoning and that will be it.>
Right. See all of above. Or go have a look at Challenge of
theCentury.> Show the mathematical logic that you believe
supports your position,> if you can.> Done. See all of above,
plus previous post, plus several dozenothers.Unfortunately, people can *say* that proof is not a proof,
even when>it is, just like if you tried to say you
were human, and not a dog,>someone might dispute any
proof you might give, claiming it false.> In this case, you have tried to use an irrelevant >
red-herring argument to show what you want. Unfortunately, in the special case you selected, the number f (= 3) does factors *** non-uniquely *** through the linear terms of
your> polynomial factorization, and you end up being
able to> conclude:> *** N O T H I N G *** about the cases in which you are interested. But you have made progress. Do you realize
how long it> took us to get through to you that there is
actually a> nontrivial problem with generalizing from m =
0 to> m <> 0 ? Do you realize how many incorrect
arguments> you have already burned through (including
the present> one) in trying to handle that problem? Do
you realize> that all of this is a waste of time,
because your > main claims have already been shown to be
false and> cannot be fixed by twiddling with the
details?> Nora B.> And readers
should note that Nora Baron here arrogant and presumably confident, is still fighting to claim that f^2 divides off
of a> polynomial as a function of m, though it is itself
constant with> regard to m, when the poster has spent a
lot of effort trying to claim> that 3 divides off in an
inconsequential way, when f=3.>> Well, duh. Yes,
that is what I was saying. I agree that I spent a> lot of
time on an inconsequential case. I was simply trying to show that your conclusions *even in that case* were not
justified. At least> you got the point this time, so I don't
think it was a complete waste.> However, the arguments you
used would destroy your own claims about> f^2 dividing off
from P(m) in some particular way.> No, I am not going to give
an inch on that. See above.> So you're trapped both by what
you said, and by the mathematical> logic.> Oddly, I don't feel
trapped at all.> You can, of course, restate your position, if
you wish.> I don't wish.> The point is that the poster
Nora Baron is not making sense.>> It's clear from
what you have said here that you understood what I was>
saying, and even, I daresay, that you were surprised by it. So
clearly, > it is not that what I said did not make sense. It
is that you did not > like the conclusion.> Nora B.
> That is not correct. I *accepted* your conclusion and then
found out> that it was false, when I checked the math.> No -
you were right when you accepted it. Then yourelapsed.> Now
you have the option to follow the math, or stick with false>
beliefs.> I invariably go with the former. Nora B.> James
Harris

===

> <histrionics deleted> In which case, I
am VERY ANNOYED that you continue to act> as if you are
infallible, and frequently accuse the rest of > us of lying.
Nora Baron, you have relinquished any right to be very
annoyed atthe behaviour of James Harris. Nora Baron, you
have made *HUNDREDS*of replies to his posts, going back over
two and a half years now.There is no *possible* way it took
until now for the clue train tosmack you broadside. James
Harris displays a sickness, and is clearly*NOT* amenable to
change. The role he has settled into on Usenet, isthe one he
is comfortable with, and one he will likely keep until theday
he dies.You, and some other posters, have displayed signs of
codependancy. Asif James' sickness draws you to him, like
flies on .James will *NOT* change. Whether you change is
up to you. Arguingwith him is like arguing with a rock, or
your bathroom wall. If youcontinue to define your online life
by replying to his trolling, youare most certainly a diseased
person also.This is my last post to you on this subject, Nora
Baron, because asfor myself I *refuse* to be drawn into James'
sickness for years andyears on end.<more histrionics deleted> Nora B.> James HarrisFlatrings

===

[snip]> You, and
some other posters, have displayed signs of codependancy. As>
if James' sickness draws you to him, like flies on .Say,
aren't you the same guy who posts all the James Harris
parodies? (Talk about codependancy!)--There are two things
you must never attempt to prove: the unprovable -- and the
obvious.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com

===

> What I've noticed is
that Arturo Magidin works VERY hard to be> believed.He doesn't
have to work hard to be believed. The things he says areeasily
believable and are backed up with solid math.> authoritative
or looks mathematical, but he rarely actually proves> anything
relevant to his central objections.Your failure to understand
his arguments and their relevance doesn'tmake them
irrelevant.-- Wayne Brown | When your tail's in a crack, you
improvisefwbrown@bellsouth.net | if you're good enough.
Otherwise you give | your pelt to the trapper.e^(i*pi) = -1
-- Euler | -- John Myers Myers, Silverlock

===

<deleted>>
What I showed in my post is that w1*w2 can be ANY two algebraic
integer> factors of f = 3. It is NOT necessarily the case
that w1*w2 = 3^{2/3}> in that case. The case f = 3 does
not give you any information about> the w's.> Well
it turns out I was in error > ... not for the first time,
or, I suspect, the last.Oh, so now you want to attack me for
admitting a mistake.That tells a lot about you Nora
Baron.Why don't you tell the newsgroups who you actually are
and quit hidingbehind the pseudonym? > as the answer is given
by considering> the factorization> P(m) = f^2((m^3 f^4
- 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) =  (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)> as with f=3,
it's clear that *each* > each what?Each of the a's has 3
as a factor when f=3, and that is without thedependency on m
that you seem so dedicated on claiming exists when fis coprime
to 3.> has 3 as a factor, which gives you> P(m)/3^3 =
(m^3 3^3 - 3^2 m^2 + m) x^3 - > (-1+m 3^2 )x u^2 + u^3 =
> (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u).> What
I showed in my previous post implies that if> a1, a2, and a3
are the negatives of the roots of > the polynomial > u^3 -
8*u^2 - 19,> then a perfectly legit factorization is:>
[****] (a1*x + 1)*(a2*x + 1)*(27*a3*x + 27)> which happens
to equal> 27*(a1*x + 1)*(a2*x + 1)*(a3*x + 1).> I think
you can work out the reasoning> for yourself on that. If not,
let me know.But Nora Baron the a's are NOT the negatives of
the roots of u^3 - 8*u^2 - 19as in fact, P(m) = f^2((m^3 f^4 -
3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x +
uf)(a_2 x + uf)(a_3 x + uf)so the a's are the roots of a^3 +
3(-1+9m)a^2 - ((-1+9m)^3+1).> Now using the b's and w's
gives> P(m)/3^2 = 3((m^3 3^3 - 3^2 m^2 + m) x^3 -  (-1+m 3^2 )x u^2 + u^3) = > (b_1 x + u w_1)(b_2 x + u
w_2)(b_3 x + u w_3)> but at m=0, I have> P(0)/3^2
= u^2(3x - 3u) = u^2(b_3 w_1 w_2 - 3u), so> and dividing
off the last 3 gives> P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m)
x^3 - > (-1+m 3^2 )x u^2 + u^3 = > (b_1 x + u)(b_2
x + u)(b_3/3 x + u)> Nope, this is NOT OK. Why?
Because you are assuming > what you want to prove. You want to
show that the> factorization when m <> 0 is similar to that
when m = 0.> Here m = 3. In case you have not noticed, 3 is
not equal> to 0.But m does not equal 3, as readers can verify
by looking above at theexpressions where they see m all over
the place, as well as 3, andthere is no claim that m=3. >
Plus, if you look at what I said above at [****}, you > will
see a counterexample.<deleted remainder to focus on strange
statements>I'm curious about how you'd like you to explain the
weird statementsyou've already made, and I'll finish out by
focusing on the last oneI've included.Here's what you had
above:> What I showed in my previous post implies that if>
a1, a2, and a3 are the negatives of the roots of > the
polynomial > u^3 - 8*u^2 - 19,> then a perfectly legit
factorization is:> [****] (a1*x + 1)*(a2*x + 1)*(27*a3*x +
27)But in fact, the a's are the roots of a^3 + 3(-1+9m)a^2 -
((-1+9m)^3+1).Yet after you said if in one place, further
down in your post yousaid:> Plus, if you look at what I said
above at [****}, you > will see a counterexample.Why don't you
try and explain yourself to the newsgroups?James
Harris

===

[snip]> Why don't you tell the newsgroups who you
actually are and quit hiding> behind the pseudonym?That's not
a math issue. Perhaps James Harris is an acronym for Mr. J.
Harie Ass. So what? Argumentum ad hominemis a fallacy.A
better question is: Why don't you post one of the numbers you
claim to have discovered which *should be* analgebraic
integer, but which is *left out* of the ring of algebraic
integers? Those numbers are the basis for yourrepeated
accusations that math professors are teaching lies to students
and form the basis of your threats to sue theuniversities.
Here's a great opportunity to post *one* such number for the
enlightenment of students andmathematicians alike.I'll make
it easy for you.I, James Harris, have found the following
number (numeric quantity) which should be an algebraic
integer (because itis the root of a monic polynomial with
integer coefficients), but which is not:____________This is
a much simpler challenge than the only posed by Nora Baron,
because her challenge requires you to think andunderstand.
Mine only requires that you post a number you claim to have
already found.--There are two things you must never attempt
to prove: the unprovable -- and the obvious.--Democracy: The
triumph of popularity over
principle.--http://www.crbond.com

===

http://www.geocities.com/
dharwadker/Has this proof been peer reviewed? Is it correct or
is it just garbage?Charles

===

>
http://www.geocities.com/dharwadker/> Has this proof been
peer reviewed? Is it correct or is it just garbage?I found it
impossible to read on my browser (Mozilla).The text was fine;
but the equations appeared to be in some strange format.Could
anyone decipher it?-- Timothy Murphy tel: +353-86-233
6090

===

Is (sqrt(sqrt(5)+3/2)-cubrt(sqrt(14/22)-9))^2/5 an
'algebreic' number?It's composed completely of a combination
of rational numbers raisedto rational exponents, so shouldn't
it be algebreic? As in,nontranscendental?(...Starblade Riven
Darksquall...)

===

> Is
(sqrt(sqrt(5)+3/2)-cubrt(sqrt(14/22)-9))^2/5 an 'algebreic'
number?> It's composed completely of a combination of rational
numbers raised> to rational exponents, so shouldn't it be
algebreic? As in,> nontranscendental?Yes, any such number is
algebraic,and so -- by definition -- not
transcendental.(However, the converse is not true;not every
algebraic number can be expressed in such a form.)-- Timothy
Murphy tel: +353-86-233 6090

===

>Is
(sqrt(sqrt(5)+3/2)-cubrt(sqrt(14/22)-9))^2/5 an 'algebreic'
number?Yes.>It's composed completely of a combination of
rational numbers raised>to rational exponents, so shouldn't it
be algebreic? Well, there's a little more to the proof than
that. But in factthe algebraic numbers form an algebraically
closed field,from which it follows that the number above is
algebraic.>As in,>nontranscendental?>>(...Starblade Riven
Darksquall...)************************David C. Ullrich

===

Is (sqrt(sqrt(5)+3/2)-cubrt(sqrt(14/22)-9))^2/5 an
'algebreic' number?> Yes.>It's composed completely of a
combination of rational numbers raised>to rational
exponents, so shouldn't it be algebreic? > Well, there's a
little more to the proof than that. But in fact> the algebraic
numbers form an algebraically closed field,> from which it
follows that the number above is algebraic.>As in,nontranscendental?>>(...Starblade Riven Darksquall...)>
************************> David C. UllrichOkay, I understand.
But is the number of such algebreic numberscountably infinite
or uncountably infinite?(...Starblade Riven
Darksquall...)

===

Is
(sqrt(sqrt(5)+3/2)-cubrt(sqrt(14/22)-9))^2/5 an 'algebreic'
number? Yes.It's composed completely of a
combination of rational numbers raised>>to rational
exponents, so shouldn't it be algebreic?  Well, there's a
little more to the proof than that. But in fact>> the algebraic
numbers form an algebraically closed field,>> from which it
follows that the number above is algebraic.As in,>nontranscendental?>>(...Starblade Riven
Darksquall...) ************************ David C.
Ullrich>>Okay, I understand. But is the number of such
algebreic numbers>countably infinite or uncountably
infinite?The algebr_a_ic numbers are countable.>(...Starblade
Riven Darksquall...)************************David C.
Ullrich

===

>Is
(sqrt(sqrt(5)+3/2)-cubrt(sqrt(14/22)-9))^2/5 an 'algebreic'
number?> Yes.>It's composed completely of a
combination of rational numbers raised>to rational
exponents, so shouldn't it be algebreic? > Well, there's
a little more to the proof than that. But in fact> the
algebraic numbers form an algebraically closed field,> from
which it follows that the number above is algebraic.As in,>nontranscendental?>>(...Starblade Riven
Darksquall...)> ************************> David C.
Ullrich> Okay, I understand. But is the number of such
algebreic numbers> countably infinite or uncountably
infinite?Countably. That follows from noting that there are
countablymany polynomials with integer coefficients. - Randy
Suppose p,q, and r are any three consecutive primes. Is there
a constant Ksuch that there always exist nonzero a,b,c with
a*p+b*q+c*r=0 and 0<=a+b+c<K? I doubt the answer is yes, but
it is kind of curious to note that one canoften find a,b,c
s.t a+b+c is quite small. For example,
ifp=2^307-255,q=2^307-187 and r=2^307-147 then there exist
a,b,c with a+b+c=4. Rich Burge

===

> Suppose p,q, and r are
any three consecutive primes. Is there a constant K> such
that there always exist nonzero a,b,c with a*p+b*q+c*r=0 and
0<=a+b+c<K? Let the three consecutive primes be p, p + s, p +
t, s < t. Let a = -t + s, b = t, c = -s. Then ap + bq + cr =
(-t + s)p + t(p + s) + -s(p + t) = 0 and a + b + c = 0.--

===

>Let the three consecutive primes be p, p + s, p + t, s <
t. >Let a = -t + s, b = t, c = -s. >Then ap + bq + cr = (-t +
s)p + t(p + s) + -s(p + t) = 0 >and a + b + c = 0.>Yes.
Clearly I should have thought about this a little
more.Rich

===

Maybe more interesting to use the magnitudes of
a,b,c. Thus:Question: Does there exist a constant K so that
for any threeconsecutive primes p,q,r there exist integers
a,b,c with the propertythat ap+bq+cr=0 and |a|+|b|+|c| < K?The
solution to your original question obviously won't work,
sinceusing r3769's notation, either s or t can become
arbitrarily large.(There are arbitrarily long stretches of
composite numbers.)So a related question (possibly easy?) is
the following:Question: Given a positive integer K, is it
always possible to findthree consecutive primes p,q,r so that
q > p+K and r > q+K? (If so,there's an obvious generalization
to more than three consecutiveprimes.)>Let the three
consecutive primes be p, p + s, p + t, s < t. >Let a = -t +
s, b = t, c = -s. >Then ap + bq + cr = (-t + s)p + t(p + s)
+ -s(p + t) = 0 >and a + b + c = 0.>> Yes. Clearly I
should have thought about this a little more.>
Rich

===

suppose that you have 2 different sets, S1 and S2.
there are many,many sample points in each one. S1 has a mean
of x1 and a standarddeviation, sd, of sd1. moreover, S2 has a
mean and sd of x2 and sd2. moreover, S1 and S2 are slightly
correllated with each other with R =-0.4.suppose that i
created a new set called S* with 45% of its samplepoints from
S1 and 55% from S2, what would the new standard deviationand
mean be? is there an algorithm/calculation for this?finally,
suppose that you have 3 different sets, S1, S2, and S3. there
are many, many sample points in each one. S1 has a mean of
x1and a standard deviation, sd, of sd1. moreover, S2 has a
mean and sdof x2 and sd2, and s3 has a mean of x3 and sd of
sd3. S1, S2, and S3are all correllated with each other like
this: S1 S2 S3S1 1.00 -0.55 0.22S2 -0.55 1.00 0.02S3 0.22 0.02
1.00and suppose that i create a new set with a mixture of k1%
of S1, k2%of S2, and k3% of S3 (k1 + k2 + k3 = 1.00). what
would be the newmean and sd of this? does a math exist for
these kinds of problems? finally, what measurement of
correllation do i use: R^2, or R?

===

> rational number +
rational number = rational number> and> rational number +
irrational = irrational> This is what I have:> 1) since all rational numbers can be expressed as a ratio of two
integers, n=p/q> then p1/q1 + p2/q2 = (p1q2+p2q1)/q1q2 which
is still a ratio of integers> i.e. 1*4+2*3/4*3=10/12 2) using the above, with irrationals, there is no ratio
such that n=p/q>> Suppose that rational + irrational =
rational, > then rational - rational = irrational> which
contracdicts what you proved in (1), so it must be the case
that> rational + irrational = irrational.>> i.e
n=sqrt(x)> so that p/q + sqrt(x) ...> that is where I
begin to get really informal. > anyone wanna let me know
how I could improve on this?> would induction or an indirect
proof be of any use? > joshrational - rational' Now I get
it.So many wonderful ways of putting it. I am beginning to see
how thisall works.Josh

===

> what would be the best way to
prove>> rational number + rational number = rational number>
and> rational number + irrational = irrational>> This is what
I have:>> 1)> since all rational numbers can be expressed as a
ratio of two integers,n=p/q> then p1/q1 + p2/q2 =
(p1q2+p2q1)/q1q2 which is still a ratio of integers>> i.e.
1*4+2*3/4*3=10/12>> 2) using the above, with irrationals,
there is no ratio such that n=p/q> i.e n=sqrt(x)> so that p/q
+ sqrt(x) ...> that is where I begin to get really
informal.>Its just a redo of (1)Let n be irrationalAssume p/q
+ n = r/sn = r/s - p/qNow finish up the right hand side and
look for a contradiction-Tralfaz

===

> what would be the best
way to prove>> rational number + rational number = rational
number> and> rational number + irrational = irrational>> This
is what I have:>> 1)> since all rational numbers can be
expressed as a ratio of two integers,n=p/q> then p1/q1 + p2/q2
= (p1q2+p2q1)/q1q2 which is still a ratio of integers>> i.e.
1*4+2*3/4*3=10/12>> 2) using the above, with irrationals,
there is no ratio such that n=p/q> i.e n=sqrt(x)> so that p/q
+ sqrt(x) ...> that is where I begin to get really
informal.>Its just a redo of (1)Let n be irrationalAssume p/q
+ n = r/sn = r/s - p/qNow finish up the right hand side and
look for a contradiction-Tralfaz

===

> what would be the best
way to prove> rational number + rational number = rational
number> and> rational number + irrational = irrational> This
is what I have:> 1)> since all rational numbers can be
expressed as a ratio of two integers, n=p/q> then p1/q1 +
p2/q2 = (p1q2+p2q1)/q1q2 which is still a ratio of integers>
i.e. 1*4+2*3/4*3=10/12> 2) using the above, with irrationals,
there is no ratio such that n=p/q> i.e n=sqrt(x)> so that p/q +
sqrt(x) ...> that is where I begin to get really informal. >
anyone wanna let me know how I could improve on this?> would
induction or an indirect proof be of any use? > joshAssume
that x + y = z, x rational, y irrational, z rational. Then y =
z - x. But rational - rational is rational, so the assumption
is wrong.

===

1)a= sqrt(3+b) , where b = sqrt(17)c=
sqrt(5+3*b)d =sqrt(2)Sum = (-d*a) - (2*b) + (b*d*a) - (2*c*d)
+ 2*c*a = 142)The sum of two irrational: The most common
example to remember: sin^2(A)+cos^2(A)=1,where for some A both
sin^(A) and cos^2(B) are two irrationals.Tapiowould be the best
way to prove>> rational number + rational number =
rational number> and> rational number + irrational =
irrational>> This is what I have:>> 1)>
since all rational numbers can be expressed as a ratio of
twointegers, n=p/q> then p1/q1 + p2/q2 = (p1q2+p2q1)/q1q2
which is still a ratio ofintegers>> i.e.
1*4+2*3/4*3=10/12>> 2) using the above, with
irrationals, there is no ratio such thatn=p/q
Suppose that rational + irrational = rational,> then
rational - rational = irrational>> which contracdicts what
you proved in (1), so it must be the casethat>> rational +
irrational = irrational.> i.e n=sqrt(x)> so
that p/q + sqrt(x) ...> that is where I begin to get really
informal.>> anyone wanna let me know how I could
improve on this?> would induction or an indirect proof be
of any use?> josh>> rational - rational' Now I get it.> So
many wonderful ways of putting it. I am beginning to see how
this> all works.> Josh

===

I am anxious to university
registration fee for a term at your nation.in my case(korea)if
national university, usually 850~1300 dollarif private
university, 2000~2500 dollaryour nation ?? USA, Britain,
Switzerland, Germany....etc...please, tell me what
price~

===

>>Is this homework? <<I kinda wish
it was, but I've been out of school for many more years than
Icare to admit and my algebra/trig/geometry is quite
rusty.>>am trying to figure out if 'A' and 'w' are enough
info to enable one to find'b' and 'c'.>>Ok.>>A =
pi(c^2-b^2)>w = c-b [not b-a as you give above]>>So,>So, since
it's a programming project and not homework, do you get a share
of the sales?As a rough estimate, the flat rate is .1 hr *
$60/hr. = $6. Close? (Depends on what your professional hourly
rate is, and I don't know your profession.)Jon Miller

===


>>Is this homework? <<>>I kinda wish it was, but
I've been out of school for many more yearsthan I>>care
to admit and my algebra/trig/geometry is quite rusty.>>I>am trying to figure out if 'A' and 'w' are enough info to
enable one tofind>>'b' and 'c'.>Ok.>>A =
pi(c^2-b^2)>w = c-b [not b-a as you give above]>>So,>>
So, since it's a programming project and not homework, do you
get a> share of the sales?Sadly, I forgot to negotiate a
contract before blurting out the answer.Darn.> As a rough
estimate, the flat rate is .1 hr * $60/hr. = $6. Close?>
(Depends on what your professional hourly rate is, and I don't
know> your profession.)-- Clive
Toothhttp://www.clivetooth.dk

===

> so evidently something's
going on here. you sure about that 87382,> though? the
encyclopedia predicts 2^18.Now I've implemented a symbolic Xor
(see below) and the sequence isproved until length 62: 1, 3, 7,
7, 31, 63, 15, 15, 511, 63, 2047, 1023, 511, 16383, 31,
31,4095, 87381, 4095, 1023, 127, 4095, 8388607, 2097151, 255,
67108863,1048575, 511, 536870911, 1073741823, 63 You can find
it at
OEIS:http://www.research.att.com/cgi-bin/access.cgi/as/njas/
sequences/eisA.cgi?Anum=A086839 (I've send a comment to add
the new elements).The program with symbolic Xor
calculating:public class Rule90 { public static void
main(String args[]) { long mul2[] = new long[64]; long start =
1; for (int i = 0; i < 64; i++, start *= 2) mul2[i] = start;
long pattern[] = new long[64]; long next[] = new long[64];
long reverse[] = new long[64]; for (int len = 2; len < 64; len
+= 2) { /* init pattern and reverse check */ for (int i = 0; i
< len; i++) { pattern[i] = mul2[i]; reverse[i] = mul2[len - i
- 1]; } /* calculate steps needed for reversing */ long steps
= 0; while (true) { /* calculate next step */ next[0] =
pattern[1]; for (int i = 1; i < len - 1; i++) { next[i] =
pattern[i - 1] ^ pattern[i + 1]; } next[len - 1] = pattern[len
- 2]; /* check for reverse */ boolean found = true; for (int i
= 0; i < len; i++) { if (pattern[i] != reverse[i]) { found =
false; break; } } if (found) break; /* swap pattern and
next */ long tmp[] = next; next = pattern; pattern = tmp; /*
increment step count */ steps++; } /* print number of steps */
System.out.println(steps + , ); } }}-- Frank Bu,
fb@frank-buss.dehttp://www.frank-buss.de,
http://www.it4-systems.de

===

Let A be a relational system and
let L be a f.o.language for predicatecalculus. How we prove
that the formula((Vx)f(x)) -> f(y)is satisfable in A for every
A (s.t. L is appropriate for A) we consider???For brief info
seehttp://mathworld.wolfram.com/Satisfaction.html .

===

>Let A
be a relational system and let L be a f.o.language for
predicate>calculus. How we prove that the formula>>((Vx)f(x))
-> f(y)>>is satisfable in A for every A (s.t. L is appropriate
for A) we consider???Let me change notation a bit (A becomes
M, f(x) becomes P(x), /x becomesEx).Let M be any L-structure
where L contains a 1-place predicate P. Let P^M bethe extension
of P in M. You want to show that There is an assignment v such
that (M,v) |= ExP(x) -> P(y).(v is an assignment of values to
variables. I.e., v : var(L) -> dom(M). Thenotation (M, v) |=
phi means that the formula phi is true when interpretedby
the structure M, relative to assignment v.)Case 1. Suppose P^M
is empty. Then, ExP(x) is false in M. Hence, M |=~ExP(x).
Recall that (by truth tables), if ~A is true, then the
materialconditional A->B is true, for any formula B. So, for
any assignment v, (A,v)|= ExP(x) -> P(y).Case 2. Suppose P^M
has an element, say a. Then M |= ExP(x). Define anassignment v
by letting v(y) = a (i.e.,. v assigns the object a to
thevariable y). Then, (M, v) |= P(y). So, by the truth table
for ->, (A, v) |=ExP(x) -> P(y).--- Jeff

===

>>Let A be a
relational system and let L be a f.o.language for
predicate>>calculus. How we prove that the
formula>>((Vx)f(x)) -> f(y)>>is satisfable in A for every
A (s.t. L is appropriate for A) weconsider???>>Let me change
notation a bit (A becomes M, f(x) becomes P(x), /x
becomes>Ex).>>Let M be any L-structure where L contains a
1-place predicate P. Let P^M be>the extension of P in M. You
want to show that>> There is an assignment v such that (M,v)
|= ExP(x) -> P(y).>>(v is an assignment of values to
variables. I.e., v : var(L) -> dom(M). The>notation (M, v)
|= phi means that the formula phi is true wheninterpreted>by
the structure M, relative to assignment v.)>>Case 1. Suppose
P^M is empty. Then, ExP(x) is false in M. Hence, M |=>~ExP(x).
Recall that (by truth tables), if ~A is true, then the
material>conditional A->B is true, for any formula B. So, for
any assignment v,(A,v)>|= ExP(x) -> P(y).Whoops. Last bit
should read: (M, v) |= ExP(x) -> P(y).>Case 2. Suppose P^M has
an element, say a. Then M |= ExP(x). Define an>assignment v by
letting v(y) = a (i.e.,. v assigns the object a to
the>variable y). Then, (M, v) |= P(y). So, by the truth table
for ->, (A, v) |=>ExP(x) -> P(y).Again, should read: (M, v)
|= ExP(x) -> P(y).--- Jeff

===

> I'm searching for a list of
tautology formula of symbolic logic.> If anybody knows,please
let me know where on the net!!!It's fairly trivial to write a
program to list them. For example:Generate all wffs (formulas
- up to a certain size) and evaluate eachfor all values of the
variables. If all such evaluations produce TRUEthen output it.
To generate all wffs, one simple way is to use thefact that P,
Q, R, . . . are wffs, and if X and Y are wffs, so are ~Xand X^Y
and XvY etc, (for whatever logical connectives you choose
touse). I have had to throw together a subroutine to do this a
numberof times in my research.If there is interest, I will post
the first few hundred tautologies. Which connectives (^, v, ~,
=>, <=>, etc.) and variables (p, Q, R,etc.) shall I
use?Charlie VolkstorfCambridge, MA

===

Dear NG,Heres a
quickie.. Given epsilon (small real number). What is the
smallest integer N, such that sum_(k=N to infinity) 1/k^2 <=
epsilon.Sincerely,Jose Capco

===

> Dear NG,> Heres a
quickie.. > Given epsilon (small real number). What is the
smallest integer N, such that > sum_(k=N to infinity) 1/k^2
<= epsilon.> Sincerely,> Jose Capco> Probably solving for
that N is not possible in elementary form.But the sum is close
to the integral x=N to infinity of 1/x^2 whichcan be determined
in closed form. Then try a few nearby integers toget your
answer.

===

>> Given epsilon (small real number). What is the
smallest integer N, such that >> sum_(k=N to infinity) 1/k^2
<= epsilon. >Probably solving for that N is not possible in
elementary form.>But the sum is close to the integral x=N to
infinity of 1/x^2 which>can be determined in closed form. Then
try a few nearby integers to>get your answer.Well, there are
just two possibilities: if f(N) is your sum,1/N < f(N) <
1/(N-1) so if 1/M <= epsilon < 1/(M-1), you want either N = M
or N = M+1. More precise estimates are available,e.g.
1/(N-1/2) < f(N) < 1/(N-1/2 + 1/(12 N) + 1/(24 N^2))where the
difference between the left and right sides is O(1/N^3),so for
most values of epsilon it will be possible to tell whetherit
should be M or M+1 without having to evaluate f(N).Robert
Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2

===

>> Dear NG,>> Heres a
quickie.. >> Given epsilon (small real number). What is the
smallest integer N, such that >> sum_(k=N to infinity) 1/k^2
<= epsilon.>> Sincerely,>> Jose Capco>Probably solving for
that N is not possible in elementary form.>But the sum is
close to the integral x=N to infinity of 1/x^2 which>can be
determined in closed form. Then try a few nearby integers
to>get your answer.It can be reduced even further; the sum is
between 1/N and 1/(N-.5),with 1/(N-.5) being closer, and
relatively accurate for large N.-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue University

===

on a
rock and hollered:>or would you just as soon I shut-up whether
I'm right or not>because you're just sick of
me,Bingo.(followups set: sci.skeptic)-- No collection of
individuals is less vindictive than an audience at amateur
theatricals. - P. G. Wodehouse, _The Intrusion of
Jimmy_

===

Can't touch this, it's HAMMER TIME
===
Can't touch
this, it's HAMMER TIME

===

Can't touch this, it's HAMMER
TIME

===

Can't touch this, it's HAMMER TIME
===
Please help
me figure out how to solve this problem. This is with
relationto ratio and root tests.If {a_k} is any sequence of
positive numbers, thenlim inf _{k->infty} a_{k+1}/a_k
leq lim inf _{k -> infty} a_k^{1/k}leqlim sup
_{k->infty} a_k^{1/k} leq lim sup _{k->infty}
a_{k+1}/a_k

===

>Please help me figure out how to solve this
problem. This is with relation>to ratio and root tests.>>If
{a_k} is any sequence of positive numbers, then>>lim inf
_{k->infty} a_{k+1}/a_k leq lim inf _{k -> infty}
a_k^{1/k}Then there exists N so that a_{k+1}/a_k > L' for all
k >= N.Think about this - it follows that a_k > ________ for
all k > N,and then that implies that lim inf _{k -> infty}
a_k^{1/k} > L'.Since that holds for all L' < L, in fact lim
inf _{k -> infty} a_k^{1/k} >= L.>leq>>lim sup
_{k->infty} a_k^{1/k} leq lim sup _{k->infty}
a_{k+1}/a_k>>************************David C. Ullrich

===

>
Please help me figure out how to solve this problem. This is
with relation> to ratio and root tests.> If {a_k} is any
sequence of positive numbers, then> lim inf _{k->infty}
a_{k+1}/a_k leq lim inf _{k -> infty} a_k^{1/k}>
leq> lim sup _{k->infty} a_k^{1/k} leq lim sup
_{k->infty} a_{k+1}/a_kSuppose liminf a_{k+1}/a_k = L, and
suppose M < L.Then a_{k+1}/a_k > M for all k >= n.Hence a_k >
a_N M^{k-n} for k > n.Thus a_k > C M^k for some constant C and
all k > n.So a_k^{1/k} > C^{1/k} M for all k > n.Since C^{1/k}
-> 1 as k -> infty, ...-- Timothy Murphy tel: +353-86-233
6090

===

> Given a square matrix G, how do I solve the
equation> X = I + G + X*X> ? Any references to the
literature? Can computer algebra systems do that?> (I know
the solution of X = I + G + G*X, it's X=(I-G)^(-1), but
blissfully> unaware of the method;-)Some Maple code for
matrix G
...with(linalg):G:=matrix(6,6,[[0,0,0,0,0,0],[1,0,0,0,0,0],[
1,0,0,0,0,0],[1,0,0,0,0,0],[0,0,1,1,0,0],[0,1,0,0,1,0]]);
ifiszero(evalm(G^4)) then
print('G'^4=0,`so`,T=1+G+T*T,`hassolution`,T=collect(convert(
series(RootOf(T=1+G+T*T,T),G,4),polynom),RootOf));T=collect(
convert(series(RootOf(T=1+G+T*T,T),G,4),polynom),RootOf,evalm)
; fi;Chris

===

> I'm looking for an (possibly efficient)
algorithm to solve systems of> quadratic equalities and
inequalities where the number of quadratic> terms (i.e. x*y or
x^y) is small.> Losely speaking, I have a lot of linear
constraints, but sometimes> there could be some products.x^y
is quadratic?--Ron Bruck

===

> Let z = sum ( i = 0,1,2,3,... )
1/(2^(f(i)))>> where>> f(x) = x iff the Continuum Hypothesis
is true> f(x) = x+1 iff the Continuum Hypothesis is false>> Is
z well defined? Is z a number? Is z some sort of weird>
Shroedinger's Cat (sp.)? Is z complete gibberish?Well, clearly
z is either sum 1/2^i or sum 1/2^(i+1) depending on whetheryou
believe the continuum hypothesis.You might as well propose
belief in the lord our god.Roland--Roland and Lisa
Paterson-JonesForest Lodge, Stirrup Lane, Hout
Bayhttp://www.rolandpj.com/forest-lodgemobile: +27 72 386
8045

===

> * Doug B.>> Let z = sum ( i = 0,1,2,3,... )
1/(2^(f(i))) where f(x) = x iff the Continuum
Hypothesis is true>> f(x) = x+1 iff the Continuum Hypothesis
is false Is z well defined? Is z a number? Is z some sort
of weird>> Shroedinger's Cat (sp.)? Is z complete gibberish?>>
It depends on your mathematical philosophical position:>> An
intuitionist will say it is not a number.> A platonist and a
formalist think it is a number.Why would a formalist think
that z is a number?How would a formalist interpret true
here?(If you mean: a formalist would interpret true to mean
provable,then certainly z is a number -- a known quantity in
this case. If youmean something else, then please elaborate.)--
Jesse Hughes Depression hits more people than thought.
--headline in Lexington, KY newspaper, as reported on NPR's
Morning Edition

===

Bishop and Goldberg said that the standard
topology on R^n is d_p(x,y) = [sum |u^i x|^p + |u^i
y|^p]^(1/p)for p >= 1.I don't think they defined the u^i, are
they just some operator thatpicks a component from a
representation of a point?I don't think they explained why
p>=1, although it may have been in aproblem that I solved
incorrectly. Looks to me like they'd all bestrongly equivalent
if p!=0, for instance. And what's wrong with p=1/2?What can we
say about a type of Pythagorean theorem that looked like r =
(sqrt|x2-x1| + sqrt|y2-y1| + sqrt|z2-z1|)^2They sure don't
have a lot of pictures in their book.-- A good plan executed
right now is far better than a perfect planexecuted next
week. -Gen. George S. Patton

===

> Bishop and Goldberg said
that the standard topology on R^n is> d_p(x,y) = [sum |u^i
x|^p + |u^i y|^p]^(1/p)> for p >= 1.They don't.-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen

===

>>Bishop and Goldberg said that the standard
topology on R^n is>> d_p(x,y) = [sum |u^i x|^p + |u^i
y|^p]^(1/p)>>for p >= 1.That's nonsense - how does the formula
actually read in the book?>I don't think they defined the u^i,
are they just some operator that>picks a component from a
representation of a point?Probably.>I don't think they
explained why p>=1, although it may have been in a>problem
that I solved incorrectly. Looks to me like they'd all
be>strongly equivalent if p!=0, for instance. And what's wrong
with p=1/2?If 0 < p < 1 then the correct version of the formula
above does not define a metric. (The fact that d(x,y) <= d(x,z)
+ d(z,y) needs to be proved; for p < 1 it can't be proved
because it's false.)>What can we say about a type of
Pythagorean theorem that looked like>> r = (sqrt|x2-x1| +
sqrt|y2-y1| + sqrt|z2-z1|)^2>>They sure don't have a lot of
pictures in their book.************************David C.
Ullrich

===

>>Bishop and Goldberg said that the standard
topology on R^n is>> d_p(x,y) = [sum |u^i x|^p + |u^i
y|^p]^(1/p)>>for p >= 1.> That's nonsense - how does the
formula actually read in the book?It's a category mistake too.
That's a metric not a topology.The book says something
like[sum_i|u^i x - u^i y|^p]^(1/p),u^i having been previously
defined as the i-th projection fromR^n to R (ugh!). >>They
sure don't have a lot of pictures in their book.What is the
use of a book without pictures or conversations? :-)-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen

===

In the excellent classic scifi novel The
Demolished Man by AlfredBester there is the following
song:Eight, sir; seven, sir;Six, sir; five, sir;Four, sir;
three, sir;Two, sir; one!Tenser, said the Tensor.Tenser, said
the Tensor.Tension, apprehension,And dissension have begun.I
don't know much about what tensor analysis is all about or
what atensor is, but I'm wonder if a simple layman's
explaination of whatone is would help to better appreciate
this song. Do tensors havesomething to do with tension? With
reducing eight to one? Is tenseralso a pun on ten, sir!?
(Except the countdown begins at eight andnot ten, but maybe
the nature of tensor analysis has something to dowith
that?)Well, thank you for any explainations!

===

|I don't know
much about what tensor analysis is all about or what a|tensor
is, but I'm wonder if a simple layman's explaination of
what|one is would help to better appreciate this song.Tensors
are geometric constructions generalizing vectors. For
ainformal expository discussion you can look
athttp://www.math.ucr.edu/home/baez/gr/gr.html.They are used
in general relativity (hence the gr), which wasEinstein's
theory of gravitation, still the best available theoryof
gravitation.|Do tensors have something to do with
tension?Yes.Stress (and tension) is described using the stress
tensor. Perhapsit will give an idea of how they generalize
vectors. Forces are givenby vectors. But stress isn't
generally just a matter of forces beingapplied along a single
axis. If we place an imaginary plane withina solid, the forces
exerted by the solid on one side on the solid onthe other side
(either tension or compression, possibly with shear)depend
upon the orientation of the plane you are putting there.
Butthis dependence is well-behaved, and can be determined from
theforces exerted across just three planes, parallel to the xy,
xz, andyz planes.So in order to describe stress we wind up
using 9 components,rather than just 3 for an ordinary vector.
If we rotate the coordinatesystem, the components transform in
a certain way, so it's moreof a geometric type of thing than
just 9 quantities lumped together,just like vectors are more
geometric than just 3 quantities thatbelong together.Part of
the magic of general relativity is that three space
coordinatesplus time can be represented mathematically as a
four-dimensionalcontinuum. In that context, instead of just 9
components for thestress, there's a similar kind of tensor
with 16 = 4*4 components,called the energy-momentum-stress
tensor because the additionalcomponents describe the energy
and momentum as well as thestress.There's a second tensor
describing a kind of curvature of space, andwhat GR says is
that the curvature tensor is proportional to
theenergy-stress-momentum tensor.So there are gravitational
effects relating to momentum and tensionor compression as well
as just energy; they are just usually tooweak to be noticed.
Two long rods arranged parallel to each otherwill attract each
other less, gravitationally, if under tension, for
example.|With reducing eight to one?Doubtful.|Is tenser
also a pun on ten, sir!?Evidently so! But your guess is as
good as mine whether it'san intentional pun.Keith
Ramsay

===

>[...]> It seems that many some readers of my post
inferred that I was> complaining about the price of the
textbook. My complaint was for not> including the worked out
solutions to the odd numbered questions> either together with
the textbook or on the CDROM which is already> included with
the textbook.> I agree that academic textbooks are expensive
and even more so if you> live in Canada. Since Canadian
universities and colleges use the same> textbooks as in the
US, the price of textbooks are even more expensive> because of
the lower exchange rate for Canadian currency. For a> discrete
mathematics course that I had taken recently, the course used>
Rosen's Discrete Mathematics and Its Applications which costs
$135> CDN plus the solutions manual for $55 and after tax, the
amount was> over $200 (the US cost after currency conversion
would have been even> more). Fortunately, some publishers
charge a lower price for Canadian> purchasers otherwise
textbooks would routinely cost over $150 CDN for> each
textbook. Unfortunately, higher level textbooks are even more>
expensive relative to the size of the books and publishers are
less> likely to offer differential pricing to Canadian
purchasers. A> introductory proof text that I had recently
looked at was a small 400> page hardcover text that cost $133
CDN plus tax. I am not> knowledgeable about the academic book
publishing business so I do not> have an informed opinion as
to whether the textbook costs are> reasonable or not.> It
looks like I will be shopping for secondhand textbooks.Well,
the above prices _are_ to high.You can get reasonable books
around 50 US$. Most of these books are betterthan the usual
trash that is forced upon undergraduates.For instance Springer
sells calculus books by Serge Lang in the rangeof 50-60 EUR. At
least for US and Canada, Dover offers booksin the range of
10-20$. Then there is the AMS etc.I was well aware, that your
complaint was not about the price as such;but still, the
prices indicate that something is wrong.As mentioned before,
publishers can sell these overpriced books becausethose who
pay the price are not those who make the decisions aboutwhich
text to adopt for a course.There is also the question if a
textbook is really suitable for self-study.If I saw a book for
a first year student that comes along with'solution manual',
CD-ROM, where the main text perhaps has more than 1000pages
etc., I would at least suspect that all this buzz is meant as
a blindfor shortcomings of the main text.During a course,
there are instructors and T.A.'s who can make the
materialdigestible in spite of a poor book. During self-study
you are on your own.Therefore it really helps to use the books
in the library until you find abook that suits your
needs.Marc

===

|> But it's possible to prove in ZF that the
union of a countable family of|> countable subsets of w_1 is
countable. If S_1, S_2, ... is a sequence of|> countable
subsets of w_1, then they have least upper bounds a_1, a_2,
...|> in w_1, and the sequence a_1, a_2,... in turn has a
least upper bound a.||How do we know that the sup of
a_1,a_2,... is < w_1? For that matter|how do we even know
a_1,a_2... are not equal to w_1?Sorry, my mistake here! In
fact in the same model of Cohen's where heshows that the
continuum is a countable union of countable sets, he showsthat
aleph-1 is the sup of an increasing countable sequence of
countableordinals.|Certainly you can reduce the statement to
showing that a countable set|of countable ordinals is bounded
above by a countable ordinal. In fact|you can even replace
countable set of countable ordinals with|strictly
increasing sequence of countable ordinals without too
much|difficulty but I still don't see how to prove that the
sup is strictly|less than w_1. And I've been told you can't
prove it without AC,|though of course I can't justify
that.Given a countable increasing sequence of ordinals a_1 <
a_2 < a_3 <...with sup aleph-1, define (not needing choice!)
the subsequenceb_1 < b_2 < ... where for each k, b_k is the
first term in the a'ssuch that {u : u<b_k but u>=b_j for j<k}
is infinite. Then the sets{u : u<b_k but u>=b_j for j<k} are
countable and their union is aleph-1.Conversely, if no
increasing sequence of countable ordinals hassup aleph-1 then
my argument works again.So it seems the model in Cohen's book
is exactly the kind neededto show you heard right.|I didn't
actually know for certain that the same held for
the|continuum. That's even worse - it means there's no hope
of|constructing a sensible measure on R in ZF (i.e. a measure
for which|we can prove the singletons are null and R itself
isn't).Isn't dependent choice enough?I'd also like it if
someone would look at the paper of Shelah, titledSet Theory
without choice: not everything on cofinality is possible.at
http://arxiv.org/pdf/math.LO/9512227, and tell us what the
resultof Gitik on cofinality is. I was only able to fetch
Google's cachedcopy in which one reads 'By Gitik [Gi] ZF +
every aleph has cofinality{splotch} 0 is consistent.' It
sounds entertaining, but I'd like to besure of what the
special symbol is supposed to be.Keith Ramsay

===

> [snip]> Well, we do have someone who doesn't believe infinite sets
exist, and goes>> so far as to say sets don't exist at all!
That person is me.>> I don't believe in the Platonic realm
or whatever, but I do assign greater>> degrees of realness
to mathematical objects. Sets have my lowest>> realness
rating.> But you do talk and think about them, right?
Perhaps only in disguised form.> And when you do, you
probably think about them *as if* they are fixed,> completed
totalities? In that case, you are assuming lots of properties>
about them....>Well, I try and think about them as little as
possible. For me, sets arelinguistic constructs that are
useful to me insofar as they give me aframework in which to
talk about actual mathematical objects (actualmeans they
have a high realness rating; I don't claim this is a
precisenotion ;-)). In geometric contexts, most of the time I
can talk about the geometricobjects without resorting to any
set speak.I think when you suggest I think about sets in
disguised form you areletting your bias for sets show. I
would say that when I talk about setsthat I am talking about
geometric objects in disguised form. So I thinkyou and I
have a difference of opinion about what is disguising
what.Continuing in the same vein, I don't see the relevance of
pointing outthat I assume properties of sets. Sets are useful
to me only if they haveproperties that approximate (in some
sense) the behavior of geometricobjects. So I am really
assuming lots of properties about them, wherethem does not
refer to sets really. As an example, you seem to be chastising
me for thinking of sets as fixedtotalities. Putting aside the
fact that I am not really thinking ofsets as you may think I
am, let me address this issue of fixedtotalities. Take the
2-sphere. I like to think of it as a fixed object. Where you
might like to say, let S^2 be a sphere (the set of all
points,etc...) and x be an element in S^2,my perspective,
talking about the sphere in the first manner, as a setwith a
fixed totality, is a perfectly good way to talk about it
because itcaptures what I'm actually thinking (the second
way). To only be able totalk about finite sets would be silly
in this context. It would notcapture my idea of sphere.The
point here is that I'm working in a specific context. I'm fine
withother foundational theories, but not all of them will be
useful to me. I'm not espousing that ZFC (or just ZF or
whatever) is *the* foundation ofmathematics or anything of
that sort. For me, that's irrelevant. > And, indeed, most
mathematicians are not aware that they do exactly that:>
assuming lots of things silently. They don't even care that
they do.> But what can be done about it? Sigh.> I don't see
what the problem is here. Of course I have to assume
thatobjects I talk about have some properties. I don't see
that because I'mnot interested in explaining my assumptions in
the language of someaxiomatic theory this is a failing of some
kind. The fact that you thinkit is a failing of mathematicians
shows your own bias.

===

 [snip] Well, we do have
someone who doesn't believe infinite sets exist, andgoes>>
so far as to say sets don't exist at all! That person is me.> I don't believe in the Platonic realm or whatever, but
I do assigngreater>> degrees of realness to mathematical
objects. Sets have my lowest>> realness rating.>> But
you do talk and think about them, right? Perhaps only in
disguisedform.> And when you do, you probably think about
them *as if* they are fixed,> completed totalities? In that
case, you are assuming lots of properties> about them.... Well, I try and think about them as little as possible.
For me, sets are> linguistic constructs that are useful to me
insofar as they give me a> framework in which to talk about
actual mathematical objects (actual> means they have a high
realness rating; I don't claim this is a precise> notion
;-)). In geometric contexts, most of the time I can talk about
thegeometric> objects without resorting to any set speak.You
got a point. (I smell a sense of irritation on your side; i
apologize:-) )You are right, for geometry, it makes sense to
view, say, 'spaces' as fixedtotalities.And, of course, i don't
know how you talk about your subject to students.[snip]> The
point here is that I'm working in a specific context. I'm fine
with> other foundational theories, but not all of them will be
useful to me.> I'm not espousing that ZFC (or just ZF or
whatever) is *the* foundation of> mathematics or anything of
that sort. For me, that's irrelevant.That is very good, and
very interesting.People doing foundations should be interested
in what you do, and inexplicitizingthe assumptions you actually
use / need in this kind of geometry.But are they? And are
geometers themselves interested in making theseassumptionsmore
explicit?I know most mathematicians are not; and when they
teach their subject tostudents,they simply continue the
tradition: ZFC or an informal version of it,without thinking
further.In that sense, the math culture has stopped
questioning the basics.Doesn't that worry you? It does worry
me. Because i think it is this thatdistractsyoung people from
a study or a career in math. And it is also not a goodsign,in
general, for the future of a science.Herman Jurjus

===

>
[snip] _Pace_ Wigner, what's unreasonable isn't that
correct>> mathematics correctly understood and applied can be
effective>> in understanding the world, it's that Cargo Cult
Mathematics>> seems to actually work sometimes (whereas Cargo
Cult Science>> hardly ever does).  Lee Rudolph> what
does cargo cult mathematics mean?> KevinYork 1985, ISBN
0-393-01921-7).
<d2c23ee6.0308060523.4e9c498f@posting.google.com>
<3f314280$7$fuzhry+tra$mr2ice@news.patriot.net>
<d2c23ee6.0308070822.79d4b86b@posting.google.com>
<bgu1eh$ctj$1@panix1.panix.com>
<d2c23ee6.0308090857.5935b0cd@posting.google.com>X-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
Micro$oft

===

at 09:57 AM, kevin.roberge@umit.maine.edu (Kevin
Roberge) said:>what does cargo cult mathematics mean?Dirac
Delta Function. Renormalization. Formal power series that
maynot converge.-- Shmuel (Seymour J.) Metz, SysProg and
JOATReply to domain Patriot dot net user shmuel+news to
contact me. Do not replyto spamtrap@library.lspace.org

===

>So are you saying that the Y-chromosome that God made
for Jesus would>>somehow be different than the one he made
for Adam?> Why would you ask such a silly
question? There isn't nearly enough> information in
scripture to draw that sort of conclusion. Odds favor> the
conjecture that they were different, of course, unless you
propose> that Jesus and Adam were identical twins.>
Never mind all that. Do you propose that Adam existed?> Well,
assuming it would make the thread even funnier. ;+)> Some
claim that Jesus = God. Which means Mother of Jesus = Mother
of God. > So, Jesus' Y chomosome would be God's. However,
AFAICT, Nobody says Adam> had God's Chromosomes. God likely
made Adam's genes like he made> everything else's -- complete
with a billion-year history.Why do you assume that Jesus and
Adam had different Y-chromosomes?After all, if Jesus was made
directly by God and Adam was madedirectly by God shouldn't
they both have the same Y-chromosome? ;)

===

> So are you
saying that the Y-chromosome that God made for Jesus would>
somehow be different than the one he made for Adam?> Why
would you ask such a silly question? There isn't nearly
enough> information in scripture to draw that sort of
conclusion. Odds favor> the conjecture that they were
different, of course, unless you propose> that Jesus and Adam
were identical twins.> I made the fairly simple statement
that scripture calls Christ the> only _begotten_ son of God,
while distinctly calling others than> Christ sons of God.
This clearly implies a distinction.They don't have to be
identical twins to have the same Y-chromosome.All they need to
have is the same father. Since both Adam and Jesuswere
supposedly made by God then it would follow that they
wouldboth have the same Y-chromosome. ;)

===

> Actually 1 slug
= 32 (lbm), give or take a few ounces. It's> either a
numerical accident or by planning by the appropriate>
Standards Committee(s) long ago that the force exerted by> 1
lbm in the Earth's gravitational field is exactly 1 lbf.No
standards committee has ever officially defined a pound
force. In 1995, the best NIST could do is to give us a
qualified definition,with a big if depending on the choice
of a standard acceleration offree fall (NIST generally uses
the one which is official for definingthe kilogram force,
something that was done not so very long ago, justa little
more than a century ago in 1901, by the CGPM).The force
exerted by 1 lb (the symbol used by the experts such as
NISTand NPL, the U.K. standards agency; it is only the
recentbastardization which needs to distinguish itself by
using the symbollbf) in Earth's gravitational field is
somewhere between 15.93 ozf and16.04 ozf. There is no place at
all on earth where the force exertedby 1 lb is always exactly
1 lbf or even always 1.0000 lbf expressedto that precision,
because the local acceleration of gravity varieswith time,
being dependent on such things as the relative positions ofthe
sun and moon wrt earth.Gene
Nygaardhttp://ourworld.compuserve.com/homepages/Gene_Nygaard/
weight.htmThere is another type of measure that is also very
ancient and that is the measure of mass. . . . As time passed,
each nation and region developedits own standard masses against
which unknown massescould be compared. The chief such unit is
calledpound in English, from a Latin word meaning a weight.
Isaac Asimov Realm of Measure,l960

===

Cut<>> No standards
committee has ever officially defined a pound force.> In
1995, the best NIST could do is to give us a qualified
definition,> with a big if depending on the choice of a
standard acceleration of> free fall (NIST generally uses the
one which is official for defining> the kilogram force,
something that was done not so very long ago, just> a little
more than a century ago in 1901, by the CGPM).>> The force
exerted by 1 lb (the symbol used by the experts such as NIST>
and NPL, the U.K. standards agency; it is only the recent>
bastardization which needs to distinguish itself by using the
symbol> lbf) in Earth's gravitational field is somewhere
between 15.93 ozf and> 16.04 ozf. There is no place at all on
earth where the force exerted> by 1 lb is always exactly 1
lbf or even always 1.0000 lbf expressed> to that precision,
because the local acceleration of gravity varies> with time,
being dependent on such things as the relative positions of>
the sun and moon wrt earth.>That's right Gene: It's the
inertia - the _ratio_ of 1 lbf to theacceleration of free
fall; that is constant, and a measure of mass that
is_constant_; because the acceleration of free fall varies
with time, beingdependent on such things as the relative
positions of the sun and moon wrtearth.

===

In sci.math, Gene
Nygaard<gnygaard@nccray.com><df336888.0308100420.5de6e834@
posting.google.com>:>> Actually 1 slug = 32 (lbm), give or
take a few ounces. It's>> either a numerical accident or by
planning by the appropriate>> Standards Committee(s) long ago
that the force exerted by>> 1 lbm in the Earth's gravitational
field is exactly 1 lbf.> No standards committee has ever
officially defined a pound force. > In 1995, the best NIST
could do is to give us a qualified definition,> with a big
if depending on the choice of a standard acceleration of>
free fall (NIST generally uses the one which is official for
defining> the kilogram force, something that was done not so
very long ago, just> a little more than a century ago in 1901,
by the CGPM).> The force exerted by 1 lb (the symbol used by
the experts such as NIST> and NPL, the U.K. standards agency;
it is only the recent> bastardization which needs to
distinguish itself by using the symbol> lbf) in Earth's
gravitational field is somewhere between 15.93 ozf and> 16.04
ozf. There is no place at all on earth where the force
exerted> by 1 lb is always exactly 1 lbf or even always
1.0000 lbf expressed> to that precision, because the local
acceleration of gravity varies> with time, being dependent on
such things as the relative positions of> the sun and moon wrt
earth.but I for one would have at least thought that 1 lbf
would havebeen exactly the force exerted by 1 lb, assuming a
certain valueof g. Then again, it's clear SI is far more
coherent on thisparticular point anyway, although I have
identified an inaccuracywhich is probably so minute it's
meaningless. However, hasanyone looked into it?Basically, the
acceleration depends on a unit of time:1 m/s/s. This unit of
time is being measured in a curvedgrav field, which varies.
While the variation is extremelysmall, there is one, in theory
-- and I'm wondering ifit's detectable at all on a lab bench
(Hafele and Keatingtook a trip around the world with
cesium-beam clocks anddid detect it, so it's doable in
theory).It's also far less than the 0.4% g variation anyway so
Iam probably worrying too much.... :-)[.sigsnip]-- #191,
ewill3@earthlink.netIt's still legal to go .sigless.

===

>In
sci.math, Gene
Nygaard><gnygaard@nccray.com>><df336888.0308100420.5de6e834@
posting.google.com>:>> Actually 1 slug = 32 (lbm), give or
take a few ounces. It's> either a numerical accident or by
planning by the appropriate> Standards Committee(s) long ago
that the force exerted by> 1 lbm in the Earth's gravitational
field is exactly 1 lbf. No standards committee has ever
officially defined a pound force. >> In 1995, the best NIST
could do is to give us a qualified definition,>> with a big
if depending on the choice of a standard acceleration of>>
free fall (NIST generally uses the one which is official for
defining>> the kilogram force, something that was done not so
very long ago, just>> a little more than a century ago in
1901, by the CGPM). The force exerted by 1 lb (the symbol
used by the experts such as NIST>> and NPL, the U.K. standards
agency; it is only the recent>> bastardization which needs to
distinguish itself by using the symbol>> lbf) in Earth's
gravitational field is somewhere between 15.93 ozf and>> 16.04
ozf. There is no place at all on earth where the force
exerted>> by 1 lb is always exactly 1 lbf or even always
1.0000 lbf expressed>> to that precision, because the local
acceleration of gravity varies>> with time, being dependent on
such things as the relative positions of>> the sun and moon wrt
earth.>>but I for one would have at least thought that 1 lbf
would have>been exactly the force exerted by 1 lb, assuming a
certain value>of g. Yes, that's what it is, of course. It's
just that no standardscommittee has ever bothered to set a
*standard* value to be used bythe people over whom that
standards committee has jurisdiction--nonational or
international standards laboratory, no
professionalorganization, no national legislature, etc.The
most common values used for that g_n, the standard
acceleration offree fall, include 9.80665 m/s (which is
official for defining gramsforce and kilograms force), 32.16
ft/s, 386.1 in/s, and otherspractically indistinguishable, in
almost all applications in the realworld, from the first such
as 32.174 ft/s and 1 lbf/1 lb = 4.44822N/0.45359237 kg or
approximately 9.806646... m/s. Many textbooksalso use 32.2
ft/s and even 32 ft/s, though I haven't encounteredany that
clearly treat this as a precise definition.>Then again, it's
clear SI is far more coherent on this>particular point anyway,
SI is exactly as clear on this point as the absolute fps system
ofunits is, with a poundal defined as exactly 1 lbft/s. It
issomething totally immaterial in either of these systems. In
neitherof these systems does a standard acceleration of free
fall serve anypurpose whatsoever.The situation is also
different for the non-SI grams force andkilograms force, which
are officially defined, since the CGPM adoptedfor this purpose
in 1901 a standard acceleration of gravity equal to980.665
cm/s.Gene
Nygaardhttp://ourworld.compuserve.com/homepages/Gene_Nygaard/=

===

Cut<>> Yes, that's what it is, of course. It's just that no
standards> committee has ever bothered to set a *standard*
value to be used by> the people over whom that standards
committee has jurisdiction--no> national or international
standards laboratory, no professional> organization, no
national legislature, etc.>That would be pretty stupid don't
you think? The acceleration of free fallvaries at various
locations on Earth; for various reasons: Mainly dependingon
the strength of gravity at that location and the centrifugal
effect dueto Earth's rotation._Tables_ have been prepared
showing this acceleration; as measured atnumerous locations.
There is no standard acceleration of free fall: It iswhat it
is at those locations because it depends on the strength of
gravityand the centrifugal effect due to Earth's rotation
there.No national or international standards laboratory, no
professionalorganization, no national legislature, or anybody
else can change theacceleration of free fall: that's up to
mother nature!> The most common values used for that g_n,
the standard acceleration of> free fall, include 9.80665 m/s
(which is official for defining grams> force and kilograms
force), 32.16 ft/s, 386.1 in/s, and others> practically
indistinguishable, in almost all applications in the real>
world, from the first such as 32.174 ft/s and 1 lbf/1 lb =
4.44822> N/0.45359237 kg or approximately 9.806646... m/s.
Many textbooks> also use 32.2 ft/s and even 32 ft/s, though
I haven't encountered> any that clearly treat this as a
precise definition.>Woe be it when they start doing this:
Those are just convenientapproximations: Close enough for the
practical applications to which theypertain.National and
international standards laboratories and
professionalorganizations, and national legislatures have
banded together and adopted a_standard weight_; the weight of
the kilogram artifact that is stored forsafe keeping at the
international archives in Sevres France.This artifact serves
as the standard for weight all over the civilizedworld, and
accurate copies of it have been distributed to all
membercountries - including the U.S. - who use it as the basis
for weightmeasurements.The kilogram is called a _unit_ of
mass, because originally it was a cubicdecimeter of pure
water at its maximum density; with a scale-weight that
wasnumerically; _exactly_ equal to the acceleration of free
fall at thelocation of the scale on which it was weighed; so
that its weight [w]divided by the acceleration at which it
will free fall [g] equals One!It's taking a while for science
to realize that this ratio [w/g] is equal tothe ratio of the
net force [f] to the forced displacement [s/t] thatconstitutes
inertia; the property of an object that requires the
applicationof an external force in order to change the
object's velocity. That akilogram is a unit of inertia, and a
measure of mass may take a while(;^)That a kilogram is a
measure of sluggishness; and/or disinclination toactivity may
take even longer(;^)

===

>> I must respectfully disagree with
the seeming consensus of the> mathematicians whose analyses I
scanned. The seeming paradox has littleto> do with the
specific distributions assumed for the source of the pairs.It is resolved in all cases by inclusion of the boundary cases
in the> calculation of expected gain over any finite
series.>> I haven't thought it through completely, but it
would seem that awinning> strategy would be to switch in
every case in which you observe a valueless> than the
maximum value you have previously observed, otherwise to
stickwith> your first choice.>> Of course if you are
told a maximum bound ahead of time there is noproblem.>
Always switch unless you see that value.> knowing the
bound IS having knowledge of the distribution!>> HercOf
course. But the resolution of the paradox does not depend on
knowing abound, or on any reasoning about any particular kinds
of distribution. Theparadox arises because of the conflict
between the provable impossibility ofoverall gain from
unconditional switching, and a seemingly valid
calculationwhich says one will indeed gain from unconditional
switching. To defeat theparadox, one need only recognize that
the estimation of benefit fromunconditional switching must
also include the guaranteed loss at the(typically unknown)
upper bound of the values encountered in any finiteseries of
trials.In an infinite series of trials with no finite upper
bound, switching mightbe said to improve the rate of gain, but
your expectation in any case iseither infinite or undefined
depending on which aspects of the picture youwant to quibble
about, so there is no paradox to be worried about in thefirst
place...I mentioned knowing the bound only in context with
explicating theconditions under which one could actually gain
by a conditional switchingstrategy.Bill

===

Hey James,Sorry, I
forgot to tell you that it's under renovation now
(tillaround the 11th of August). So perhaps, some entries do
not work.Math Observer

===

> Hey James,> I gave an
HONEST suggestion. I thought that your elementary proof of>
the FLT could perhaps be propagated here in HK as well (with
so many> young math enthusiasts around!).> Math
Observer> PS: I'm an HK Math Observer (and I don't know
the guys you> mentioned!).> PPS: I am simply
interested in making the amateurs here do something>
useful (discuss and scrutinize your elementary proof (?)
using> elementary methods).> You're replying to the
imposter. In any event, it's worth my saying> again that I
went to the link you gave and tried to post.> Here's a copy
of what I said in my previous reply:> How about you do me a
favor and post the following there and post back> if there are
any interesting comments?<deleted>Well there's a problem that
someone pointed out with what I had, whichluckily is not a big
deal. More below...> which is> P(0)/f^2 = u^2 (b_3 w_1 w_2
x + u f) = u^2(3x + uf)> proving that w_1 w_2 must equal 1,
if f is coprime to 3, which leaves> b_3 = 3.Yup, and I should
have remembered that fact.> Essentially objections to how f^2
divides off now come down to> claiming that the w's are
functions of m, but consider that w_1 w_2 => 1, when m=0, if f
is coprime to 3.> But that was an arbitrary choice, so let
f=3.> Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.OOPS!!! Now
w_1 w_2 is still coprime to 3, but w_3 = 3 WITHOUT REGARDTO m,
which follows from P(0)/3^2 = u^2 (b_3 w_1 w_2 x + 3u) = u^2(3x
+ uf)as that 3 factor of 3u has to get out of there someway and
it can't bethrough w_1 w_2.> That is, the w's are now all
constant with regard to m and have the> same value no matter
what the value of m is.It's not a big deal as that is still
true.> Therefore, the factorization is> P(m)/f^2 = (m^3 f^4
- 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f => (b_1
x + u)(b_2 x + u)(b_3 x + uf)which still works with f=3, as
then all that happens is that last fcan divide through> where
you'll notice that the b's are algebraic integers with m=1,>
f=sqrt(2), but that's a special case as generally they are
not, which> shows a problem with the ring of algebraic
integers.> I've found the Ring of Objects which includes the
ring of algebraic> integers, and does not have this problem, as
the b's are all included> in it.> The Ring of Objects is the
set of all numbers where -1 and 1 are the> only members that
are both a unit, i.e. factor of 1, and an integer,> where no
non-unit member is a factor of any two integers that are>
coprime.That's a nifty definition that's extremely powerful.>
That definition and more is linked to from my primary website http://groups.msn.com/AmateurMath> where you can also find
information on my other math research. James Harris

===

>
That's a nifty definition that's extremely powerful.Nifty or
not nifty, please post it there, James.(Math forum works ---
and so is its LateX stuff)Math Observer

===

[snip]> My hope
is that these issues can be resolved to some extent through>
Usenet's strength: Discussion.> Ah, the triumph of hope over
experience.[snip]> James Harris

===

> That's a nifty
definition that's extremely powerful.>> Nifty or not nifty,
please post it there, James.He is not yet finished. All you
see here are just his drafts.

===

> The Ring of Objects is the
set of all numbers where -1 and 1 are the> only members that
are both a unit, i.e. factor of 1, and an integer,> where no
non-unit member is a factor of any two integers that are>
coprime.You forgot to define operations here.Without
operations it's just a set of numbers, not a ring.Thus your
whole FLT proof is invalid.

===

> That's a nifty definition
that's extremely powerful.> Nifty or not nifty, please post
it there, James.> (Math forum works --- and so is its LateX
stuff)> Math ObserverI've posted using the directions given
at the site for the version ofLaTex used by the site.Hopefully
tomorrow it will display properly.In any event, it is readable,
even with the formatting code.James Harris

===

>That's a
nifty definition that's extremely powerful.>>Nifty or not
nifty, please post it there, James.>>(Math forum works --- and
so is its LateX stuff)>>Math Observer> I've posted using
the directions given at the site for the version of> LaTex
used by the site.> Hopefully tomorrow it will display
properly.> In any event, it is readable, even with the
formatting code.> James HarrisI find it interesting that
no one there had a clue what you were talking about.-- Will
Twentyman

===

> If R is a noncomutative integral domain and
M is a R-module. How do you> prove that Tor(M) = { m in M | r
m = 0 for some r in R} is a submodule of> ^^^^^^> nonzero!
:-)> M??> I don't think you do :-(Robin is correct as
usual. But oddly enough, exactly the same error occurs in an
exercise in Godement's Algebra (Hermann 1968), exercise
11.11: Let M be a left module over a ring K. a) For each x
in M, the annihilator of x is defined to be the set of all
lambda in K such that lambda x = 0.... b) Suppose that K
is an integral domain. Show that the elements x of M whose
annihilator is not {0} form a submodule T of M.... I checked -
he does not assume rings or integral domains are
commutative.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2

===

Can anybody
give me a more intuitive description of the total
derivative?Not the definition, please.TIA,Lurch

===

> -- I'm
CanSpice in that thread and the person on the other side of
the > argument is Aylutar) about trying to prove the old adage
that if you had > an infinite number of monkeys typing on an
infinite number of typewriters > for an infinite amount of
time, you would produce the works of > Shakespeare.Actually,
this theory has never been tested due to lack of research
funds.The closest approximation was an experiment with
three-quarters of aninfinite number of monkeys, who typed: To
be or not to be; that is the question. Whether 'tis nobler in
the mind to suffer The slings and arrows of outrageous fortune
Eeepps jksaagj effauyp dsajfjkd eepdoosI believe, however that
Robert Wilensky has argued that we have a perfectlygood
*natural* experiment: The hypothesis is that an infinite
number ofmonkeys at an infinite number of keyboards will
eventually produce the completeworks of Shakespeare; our
experience of Usenet refutes this hypothesisrather
definitively. It's also been observed (I forget by whom) that
if an infinite number ofmonkeys are given an infinite number
of typewriters, then with probabilityone at least one monkey
will figure out how to beat the monkeys around itto death with
its typewriter.Steven E.
Landsburgwww.landsburg.com/about2.html

===

> |>Some countable
ordinals have a finite desciption in English> |>(abbreviated
hereon to FDIE).> The restriction to countability isn't
strictly necessary, of course.> |[...]> |> |>This raises the
question of whether the FIRST countable ordinal which> |>has no
FDIE has a FDIE.> |> |Sounds to me like the proof that each
natural number is an interesting > |number...> I agree. I
think the possibility of describing a mathematical object>
like an ordinal (by way of a well-ordering on the natural
numbers, say)> is intimately related to its holding an
interest for us, enabling it to be> distinguished from among
the closely related objects by means of a> finite amount of
pointing out. I would call it a strengthening of the>
paradox of the first uninteresting natural number, because
while one> can imagine that the natural numbers are all at
least microscopically> interesting, it's not so easy to
imagine from a realist perspective that> ALL ordinals (going
up to ones of arbitrarily large cardinality) continue> to have
some positive interest to them.> The paradox is that these
concepts are slippery enough not to be> able to reflect back
on themselves in such a way.> Keith Ramsay> It sounds to me
like these are simply inconsistent self-referential
definitions. In essence, each of them boils down to a
description phrase like, an object that is not described by
this description. An inconsistent definition isn't a very
interesting paradox; it's just nonsense.

===

>> |>Some
countable ordinals have a finite desciption in English>
|>(abbreviated hereon to FDIE).>> |>This raises the question
of whether the FIRST countable ordinal which> |>has no FDIE
has a FDIE.> |> |Sounds to me like the proof that each natural
number is an interesting> |number...>> I would call it a
strengthening of the paradox of the first> uninteresting
natural number, because while one can imagine that the>
natural numbers are all at least microscopically interesting,
it's not> so easy to imagine from a realist perspective that
ALL ordinals (going> up to ones of arbitrarily large
cardinality) continue to have some> positive interest to
them.>> The paradox is that these concepts are slippery enough
not to be> able to reflect back on themselves in such a
way.>Similar to the first non-recursively enumerable
ordinal?Also reminds me of the ordinal number of all
ordinals.

===

>Some countable ordinals have a finite desciption
in English>(abbreviated hereon to FDIE). For example 'the first
countable>ordinal which is greater than any finite ordinal' is
a FDIE of a>countable ordinal. So is 'the first countable
ordinal which is>greater than any finite ordinal raised to the
power of the first>countable ordinal which is greater than any
finite ordinal'.>>However, not all countable ordinals do have
a FDIE. The reason for>this is because the number of FDIEs is
countable (quite>straightforwardly, the set of finite
combinations of a finite set of>symbols is countable); On the
other hand, the set of countable>ordinals is
UNcountable.>>This raises the question of whether the FIRST
countable ordinal which>has no FDIE has a FDIE.>>If it DOES,
this would contradict the fact that it's the first>countable
ordinal which does NOT have a FDIE.>>If it DOESN'T, this would
contradict the fact that I've just give a>FDIE of it, namely:
'The first countable ordinal which does not have a>FDIE'.The
resolution of the paradox is to note that the phraseFDIE is
unacceptably vague. If you give a precisedefinition of what
strings of English words constituteFDIE's of ordinals
(together with an explanation ofexactly which ordinal is
defined by a FDIE) then youfind that smallest ordinal that
has a FDIE does notqualify as a
FDIE.************************David C. Ullrich

===

> The
resolution of the paradox is to note that the phrase> FDIE
is unacceptably vague. If you give a precise> definition of
what strings of English words constitute> FDIE's of ordinals
(together with an explanation of> exactly which ordinal is
defined by a FDIE) then you> find that smallest ordinal that
has a FDIE does not> qualify as a FDIE.>
************************> David C. UllrichOkay, suppose we
have a precise definition D of what strings inEnglish
constitute FDIEs of ordinals.Consider the string:the smallest
ordinal which is not in the set of ordinals which aredescribed
by FDIEs under definition DSince D is a precise definition it
must include the above string.If it DOES include the above
string we're back at the paradox.Richard Clark

===

 The
resolution of the paradox is to note that the phrase>> FDIE
is unacceptably vague. If you give a precise>> definition of
what strings of English words constitute>> FDIE's of ordinals
(together with an explanation of>> exactly which ordinal is
defined by a FDIE) then you>> find that smallest ordinal that
has a FDIE does not>> qualify as a FDIE.
************************ David C. Ullrich>>Okay, suppose
we have a precise definition D of what strings in>English
constitute FDIEs of ordinals.>>Consider the string:>>the
smallest ordinal which is not in the set of ordinals which
are>described by FDIEs under definition D>>Since D is a
precise definition it must include the above string.Huh???
That doesn't follow at all.>If it DOES include the above
string we're back at the paradox.Yes, if the above _is_ an
FDIE of an ordinal. This is no problem, until you _give_ a
precise definition of FDIE of anordinal such that the above
_is_ an FDIE of an ordinal.Exactly what definition of FDIE of
an ordinal are you using?>Richard
Clark************************David C. Ullrich

===

|It sounds to
me like these are simply inconsistent self-referential
|definitions. In essence, each of them boils down to a
description |phrase like, an object that is not described by
this description. An |inconsistent definition isn't a very
interesting paradox; it's just |nonsense.Well, why didn't we
all think of that earlier? :-)Everybody who looks at
paradoxes, involving self-reference like these do,concludes
that there's something wrong with the definitions and that
ithas something to do with their being circular. That's not
what people areafter when they spend more than a little bit of
time trying to resolvethem. That's just the starting point. One
might hope, for example, that anexplanation for why the
definitions are wrong would give some kind ofcriterion for
when circularity of various kinds is okay and when it
isn't.Saying you know precisely why a problem has occurred but
are unable togive a reasonable rule for avoiding the problem in
the future is a littledubious.Impredicative definitions are an
example. There have been people whothought that when you
define a set, it should be a set composed of elementsthat you
could have defined before defining the set itself. Otherwise,
theysay, there's a kind of invalid circularity.The
consequences for mathematics are substantial because the
definition ofthe real numbers is impredicative. By the
Dedekind cut construction, a realnumber is given by a
collection of rational numbers (satisfying someconditions like
being downward closed). Such a set of rationals is given as {r
: r satisfies ...}where we can refer to just about anything in
.... So this kind ofalleged circularity exists, because ...
could talk about the real numberline and its properties.On the
other hand, there have been lots of people with the point of
viewthat definitions like this are okay, because the reals are
all therealready, and by making a definition we are merely
selecting from outof a pre-existing domain.A thorough
unravelling of the paradox should help us to see whether
thiskind of circularity or apparent circularity is valid, or
whether it justmakes the same mistake as the paradox does but
inadvertently avoids runninginto a direct contradiction.
That's the kind of thing that makes suchparadoxes interesting
to some people.Keith Ramsay

===

>> The resolution of the
paradox is to note that the phrase>> FDIE is unacceptably
vague. If you give a precise>> definition of what strings of
English words constitute>> FDIE's of ordinals (together with
an explanation of>> exactly which ordinal is defined by a
FDIE) then you>> find that smallest ordinal that has a
FDIE does not>> qualify as a FDIE.
************************ David C. Ullrich>>Okay,
suppose we have a precise definition D of what strings inEnglish constitute FDIEs of ordinals.>>Consider the
string:>>the smallest ordinal which is not in the set of
ordinals which are>described by FDIEs under definition D>Since D is a precise definition it must include the above
string.> Huh??? That doesn't follow at all.> Well, I guess
that depends on what the precise definition of
'precisedefinition' is ...

===

> Nonsense, just take the
probability of an event to be> its Lebesgue measure. Try
some post-1850 mathematics> for a change. Game over.> Aw,
Wade, you're right but also that's kind of harsh,> considering
that a significant percentage of posts have> no math content
whatsoever, posts in this thread being> among them.Perhaps,
but I hold the mathematical posts to a higher standard.> By
the way, have you found a job yet?This doesn't ring a bell
with me. Can you remind me what has prompted your question
here?

===

> Perhaps, but I hold the mathematical posts to a
higher standard.How silly of me.I thought you were taking a
shot at me,after the fact, in sympathy with your colleague.I
say after the fact, because in this thread,Prof. Israel had
explained, succinctly, the errorof my ways.> By the way,
have you found a job yet?> This doesn't ring a bell with me.
Can you> remind me what has prompted your> question here?You
took your shot. Tit for tat.

===

> Perhaps, but I hold the
mathematical posts to a higher standard.> How silly of me.>
I thought you were taking a shot at me,> after the fact, in
sympathy with your colleague.> I say after the fact, because
in this thread,> Prof. Israel had explained, succinctly, the
error> of my ways.> By the way, have you found a job yet? This doesn't ring a bell with me. Can you> remind me what
has prompted your> question here?> You took your shot. Tit
for tat.My shot (your term) was based on the mathematics. Your
shot is from the bizarro world.

===

>> Suppose that you flip a
coin infinitely many times. What is the>> probability that you
get all heads? Zero. Does that mean that it's ^^^^>>
impossible to get all heads? No,...> Yes.>> you get
_some_sequence, and any given sequence is just as likely as
any>> other.> Although, a priori, any_finite_sequence of a
given length is as likely as> any other, I claim: When
flipping a fair coin, an infinite sequence of> heads is
impossible, theoretically. Such a sequence, violates the
axioms> of probability. If an event has non-zero probability
it must occur in ^^^^^^^^> infinitely-many trials with
probability 1. Granted, though, there may be> arbitrarily long
sequences, beginning HHH...HT... As you write:Pay attention,
Nat.Phil

===

>
http://mr-31238.mr.valuehost.co.uk/assets/Flash/psychic.swf Well... With 40, I decided to do: 40 + 4+0 = 44,
and the symbol> for 44 didn't appear...>> David Bernier>But
with 40, this becomes 40 - (4 + 0) = 36 and the symbol for 36
does infact appear

===

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===

>For
continuous data f(x) on [a,b], we can calculate the mean
or>average value as>A = 1/(a-b) int{a,b} f(x) dx>My question
is can we calculate the variance of a continuous>function
as>>1/(a-b) int{a-b} (f(x) - A)^2 dxI assume you mean 1/(b-a)
int_a^b (f(x) - A)^2 dx.This is the variance of f(X) where X
is a random variable with uniform distribution on [a,b].>For N
discrete data points, we can divide by either N or N-1 to>get
biased or unbiased estimator, but is there this distinction
in>continuous case?I'm not sure what you mean by continuous
data here. I think you mayhave to explain the situation you
have in mind. Where does the functionf come from?Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2

===

> Carol Rees schrieb: I need to solve the sum of 1 + 2^2 +3^2 +4^2+......n^2 as
part of a> statistics problem. It seems simple but it
doesn't seem to fit in with the> usual solutions for
arithmetic or geometric progression.> look at
http://www.research.att.com/~njas/sequences/ and type in a few
members> of your series. You get n(n+1)(2n+1)/6.> hth>
KlausThat's very interesting because I have just been
interested in somethingvery similar. Lets say you have a
series of numbers (lets call it A) 1, 4, 9, 16, 25, 36, 49,
64, 81 ...just as Carol Rees talked about. Now you have a
second series of numbers(call this B) x+1, x+4, x+9, x+16,
x+25, x+36, x+49, x+64, x+81 ... wherex is some positive
integer.These are two totally regular (aren't they) series of
numbers. Now how can Idetermine the set of numbers which are
common to A and B? The intersection ofset A and set B.

===

X
is not a random variable. Better notation would be f(t) where
t istime. If we have N discrete points f1, f2, ..., fN, we can
estimate thevariance of this series by Var = 1/(N-1) sum{i=1,N}
(fi - Mean)^2This is the usual unbiased estimator. Now consider
the continuous functionf(t) on [a,b], and I want to measure how
wavy or how spread out thefunction is so I want to know if we
can calculate the variance as Var = 1/(a-b) int(a,b) (f(t) -
Mean)^2 dt> I'm not sure what you mean by continuous data
here. I think you may> have to explain the situation you have
in mind. Where does the function> f come from?

===

>X is not a
random variable. Better notation would be f(t) where t
is>time. If we have N discrete points f1, f2, ..., fN, we can
estimate the>variance of this series by> Var = 1/(N-1)
sum{i=1,N} (fi - Mean)^2>This is the usual unbiased estimator.
Now consider the continuous function>f(t) on [a,b], and I want
to measure how wavy or how spread out the>function is so I
want to know if we can calculate the variance as> Var =
1/(a-b) int(a,b) (f(t) - Mean)^2 dtThe unbiased estimator of
_what_? You can't have an unbiased estimatorunless there's
some underlying probability model.If you have a collection of
random variables X_1,...,X_n whose probability distribution
depends on some unknown parameters, an unbiased estimator of
some parameter q is Q(X_1,...,X_n)(a function of the given
random variables) such that E[Q(X_1,...,X_n)] = q for all
values of the unknown parameters.In the case of your N points
f1,...,fN, these are assumed to be independent random
variables with the same unknown probability distribution.But
in the case of your continuous function f(t), what kind
ofdistribution is this supposed to be taken from? You
certainlycan't assume that f(t) for different t's are
independent, becausethen (except in the trivial case where
f(t) is almost surely constant)f(t) won't be a measurable
function and the integral won't exist.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2

===

> X is not a random variable.
Better notation would be f(t) where t is> time. If we have N
discrete points f1, f2, ..., fN, we can estimate the> variance
of this series by> Var = 1/(N-1) sum{i=1,N} (fi - Mean)^2>
This is the usual unbiased estimator. Now consider the
continuous function> f(t) on [a,b], and I want to measure how
wavy or how spread out the> function is so I want to know if
we can calculate the variance as> Var = 1/(a-b) int(a,b)
(f(t) - Mean)^2 dtThis is the mean of the function
(f(t)-mu)^2, which is thedefinition of the variance. As with
the discrete case, you canprove that it's equal to E[f(t)^2] -
Mean^2.So long as f(t) is well-behaved I don't think there's
anyproblem with defining this, which is an estimator of a
statisticover the uncountable number of samples {f(t): t in
[a,b]}. - Randy

===

>That's very interesting because I have
just been interested in something>very similar. Lets say you
have a series of numbers (lets call it A) >1, 4, 9, 16, 25,
36, 49, 64, 81 ...>just as Carol Rees talked about. Now you
have a second series of numbers>(call this B) x+1, x+4, x+9,
x+16, x+25, x+36, x+49, x+64, x+81 ... where>x is some
positive integer.>These are two totally regular (aren't they)
series of numbers. Now how can I>determine the set of numbers
which are common to A and B? The intersection of>set A and set
B.I don't know how this got into the variance thread. Anyway,
it really has very little to do with Rees's question. You just
want to solve theDiophantine equation a^2 = x + b^2 for a and
b. Write this as(a-b)(a+b) = x, look for ways to factor x such
that both factors areodd or both are even, and you have your
answer.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israelUniversity of
British ColumbiaVancouver, BC, Canada V6T 1Z2

===

> In the case
of your N points f1,...,fN, these are assumed to be>
independent random variables with the same unknown
probability> distribution.Yes, that was precisely what I had
in mind: E(fi) = mu and Var(fi) = s^2 forall i.For estimating
the mean, we compute 1/N sum(i=1,N) fi for N samples f1 to
fN.I had thought this was related to the notion of average
value of a continuousfunction given by 1/(a-b) int{a,b} f(t)
dt (that everybody learns in calculusI), but maybe not since
f(t) here is deterministic??Actually what I have in mind for
f(t) is a continuous time stochastic process(a function of
possibly correlated Wiener processes), and so f(t) denotes
onesample path. I'm trying to find a way of measuring how
spread out the _path_f(t), a<t<b is, but not the variance
of f at a particular time t.

===

>> In the case of your N
points f1,...,fN, these are assumed to be>> independent random
variables with the same unknown probability>>
distribution.>Yes, that was precisely what I had in mind:
E(fi) = mu and Var(fi) = s^2 for>all i.>For estimating the
mean, we compute 1/N sum(i=1,N) fi for N samples f1 to fN.>I
had thought this was related to the notion of average value of
a continuous>function given by 1/(a-b) int{a,b} f(t) dt (that
everybody learns in calculus>I), but maybe not since f(t)
here is deterministic??It's related in the sense that both are
integrals, but the situation issomewhat different.>Actually
what I have in mind for f(t) is a continuous time stochastic
process>(a function of possibly correlated Wiener processes),
and so f(t) denotes one>sample path. I'm trying to find a way
of measuring how spread out the _path_>f(t), a<t<b is, but
not the variance of f at a particular time t.OK then, you can
call your integral the variance of the path if you want.What
its expectation is in terms of the parameters of the process,
I don't know. In certain cases, the Ergodic Theorems (Birkhoff
and von Neumann) might tell you something about what happens as
b -> infinity.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2

===

Under AC, of
course, does there exist an algortihm which develops a
well-ordering of the reals (not necessarily in finite time)
and which has the property that, given any two real numbers,
it determines in finite time which of the two is no greater
than the other under this ordering?My guess is no. Such an
algorithm would need to take countably many steps. Infinitely
many of the steps must order an uncountable subset of R in
finite time. If one could do that, one could well-order R
without choice [not sure about this step]. Have I just
answered my own question?-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu

===

> Under AC, of course, does
there exist an algortihm which develops a> well-ordering of
the reals (not necessarily in finite time) and which> has the
property that, given any two real numbers, it determines in>
finite time which of the two is no greater than the other
under this> ordering?Exactly what do you mean by given any
two real numbers?On the face of it, it seems like only
equality would be a problem, no?

===

* Stephen J. Herschkorn>
Under AC, of course, does there exist an algortihm which
develops a> well-ordering of the reals (not necessarily in
finite time) and which> has the property that, given any two
real numbers, it determines in> finite time which of the two
is no greater than the other under this> ordering?I am not
sure if I understand the question. How do you
definealgortithm? Most definitions include the condition
that it has toterminate in finite time. What does it mean for
an algorithm, or acomputer program, to terminate in infinite
time? (How do youdistinguish such a program from, say WHILE
(TRUE) DO NOTHING.-- Jon Haugsand

===

JH:>* Stephen J.
Herschkorn>> Under AC, of course, does there exist an
algortihm which develops a> well-ordering of the reals (not
necessarily in finite time) and which> has the property
that, given any two real numbers, it determines in> finite
time which of the two is no greater than the other under
this> ordering?I am not sure if I understand the
question. How do you define>algortithm? Most definitions
include the condition that it has to>terminate in finite time.
What does it mean for an algorithm, or a>computer program, to
terminate in infinite time? (How do you>distinguish such a
program from, say WHILE (TRUE) DO NOTHING.>An algorithm is a
finitely-specified list of steps. Such an algorithm would
define a well-ordering S on the reals. The algorithm would
take as input two reals x and y (specified as you wish) and,
in finite time, return as an output either (x,y) in S or (y,x)
in S.I see that the specification of x and y may present
problems. For example, if we wish to specify a real by a
decimal expansion, we may not even be able to specify it in
finite time.What I am trying to do is construct a
well-ordering of reals, whose existence the axiom of choice
implies.NS:>Exactly what do you mean by given any two real
numbers?>On the face of it, it seems like only equality would
be a problem, no?>I am talking about a *well*-ordering here,
not the usual ordering of the reals.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu

===

> Under AC, of course, does
there exist an algortihm which develops a>> well-ordering of
the reals (not necessarily in finite time) and which>> has the
property that, given any two real numbers, it determines in>>
finite time which of the two is no greater than the other
under this>> ordering?>>Exactly what do you mean by given any
two real numbers?>On the face of it, it seems like only
equality would be a problem, no???? My best guess here is that
you're unaware of what the phrasewell-ordering means, or
possibly you missed the fact that we'retalking about _a_
(well-)ordering, not talking about the standardorder on the
reals.************************David C. Ullrich

===

Forget about
algorithms. Here is a simpler question:Does there exist an
explicit well-ordering of 2^N (the set of all binary
sequences)?Again, I suspect not, fora) it probably would be
well-known;b) it would yield a well-ordering of R and there
wouldn't have been so much controversy about the axiom of
choice, andc) it seems to be somehow equivalent to exhibiting
an explicit Hamel basis, which I gather is also impossible.--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu

===

Stephen
J. Herschkorn> Forget about algorithms.I will, with pleasure
:)> Here is a simpler question:>> Does there exist an explicit
well-ordering of 2^N (the set of all> binary sequences)?I
expect so, although I haven't seen any. In Conway's theory of
surrealnumbers,R gets two different order relations:-- the
usual total order; call that T, a subset of RxR.-- a
well-founded forest order which we can describe as x is an
ancestor ofy;call that relation F.Write W = F cup (F-T).
It's not altogether trivial, but W is a well-order.I'll try
to come up with something more explicit in terms of binary
orternary sequences.Larry

===

>Forget about algorithms. Here is
a simpler question:>>Does there exist an explicit well-ordering
of 2^N (the set of all >binary sequences)?No.>Again, I suspect
not, for>a) it probably would be well-known;>b) it would yield
a well-ordering of R Yes.>and there wouldn't have been so >much
controversy about the axiom of choice, and>c) it seems to be
somehow equivalent to exhibiting an explicit Hamel >basis,
which I gather is also impossible.Hmm, never noticed that.
It's clear that a well-ordering of R allowsone to construct
a Hamel basis for R over Q; come to think ofit a Hamel basis
also allows one to construct a
well-ordering.************************David C. Ullrich

===

>
Does there exist an explicit well-ordering of 2^N (the set of
all > binary sequences)?There are an explicit bijections
between R and 2^N and N^N, sologicians sometimes consider the
most convenient one.Goedel defined an explicit well-ordering
of a certain subset of R, andshowed that ZF cannot disprove
the assertion that the subset is all ofR (provided ZF is
consistent).-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/

===

I've been seeing
this term a lot in James Harris's proofs. Can someoneplease
give me the definition, so that I can check those proofs
formyself?Jack Rudd

===

> I've been seeing this term a lot in
James Harris's proofs. Can someone> please give me the
definition, so that I can check those proofs for>
myself?>Algebraic integers are solutions to polynomials x^n +
a_(n-1) x^(n-1) +..+ a_1 x + a_0 = 0with integer coefficients
and a_n = 1.For example sqr 3, 1 + sqr 2, (1 + sqr 5)/2.Don't
feed the troll, maybe it'll go away.

===

>> I've been seeing
this term a lot in James Harris's proofs.<Snipped stuff died
here>> Don't feed the troll, maybe it'll go away.Dear William,
I have a bridge to sell.

===

> I've been seeing this term a lot
in James Harris's proofs. Can someone> please give me the
definition, so that I can check those proofs for> myself?>
Jack RuddFor a complex number to be an algebraic integer, it
is necessary and sufficient that it be a zero of an
irriducible monic polynomial in one variable with integer
coefficients.These form a subset, and subring, of the set, and
field, of algebraic numbers, also zeros of polynomials with
integer coeficients, but for which the polynomials need not be
monic.

===

In sci.math, Jack
Rudd<b64489b2.0308100338.2fb0d2a3@posting.google.com>:> I've
been seeing this term a lot in James Harris's proofs. Can
someone> please give me the definition, so that I can check
those proofs for> myself?> Jack RuddFactorization problems
are tricky in algebraic integers. Othershave already defined
the field, or one can look
athttp://planetmath.org/encyclopedia/
AlgebraicInteger.htmlwhich is as good as any although
extremely terse (wolfram.comis being fussy today for some
reason).However, certain assumptions break down. In
particular,1 has an infinite number of factorizations:1 = 1 *
1= (2 - sqrt(3)) * (2 + sqrt(3))= (3 - 2*sqrt(2)) * (3 +
2*sqrt(2))= (4 - sqrt(15)) * (4 + sqrt(15))= (5 - 2*sqrt(6)) *
(5 + 2*sqrt(6))= ...= (n - 2*sqrt(n^2-1)) * (n +
2*sqrt(n^2-1))= ...All of these are degree-2 algebraic
integers, solutionsof the equation x^2 - 2nx + 1 = 0, for any
integer n > 1.(n=1 yields a reducible quadratic; n=0 gives us
no realsbut is another factorization: 1 = i * (-i).n = -1
gives us (-1) * (-1); n < -1 gives us the negativesof most of
the roots we've found already.)Since Mr. Harris likes to have
things dependent on coprimality of 3to a root, it's clear that
we have some minor difficulties here.The equation x^2 - (2n+1)x
+ 1 = 0 yields even more factorizations,which have a 2 in the
denominator:1 = ( (-(2n+1) + sqrt(4n^2+4n-3))/2 ) * ( (-(2n+1)
- sqrt(4n^2+4n-3))/2 )All these are not integers (except for 1
and -1),but they are algebraic integers.It's also clear that
there is no smallest positivealgebraic integer. Consider the
equation y^n - 2, n > 1.This equation has n roots, exactly one
of which is realand positive (if n is even we get a negative
one as well).Subtract 1 from this root (or, if one prefers,
solve theequation (x+1)^n-2=0 instead) and one sees that one
cangenerate an infinite sequence of algebraic integers inan
ordering, each smaller than the last one but all ofthem
positive.One can use the above quadratic equation as a
generatoras well; the smaller root will tend towards 0 as n
getslarger and larger; the bigger root tends towards
positiveinfinity.Funny things, these algebraic integers. :-)
It's clearMr. Harris is making a strange assumption or two in
hisproof.-- #191, ewill3@earthlink.netIt's still legal to go
.sigless.

===

> The square terms in 2^n - 1 make a pretty
pattern:> 2^n - 1 divides p if n divides (p - 1)>
miniproof:> m = (p - 1) to abbreviate> 2^mn - 1 = (2^m - 1)
(2^(n-1)m + 2^(n-2)m + ... + 1)> 2^m = 1 (mod p) of course.>
But, apparently, 2^n - 1 divides p^2 iff n divides p(p - 1).
Why is that?First, you certainly mean p divides N, i.e. N
is a multiple of p,whenever you write N divides p.
Secondly, the if part of your apparent iff statement is
due to Euler, as phi(p^2)=p(p-1);the only if part is not
true, as e.g. 7^2 divides 2^21-1.Miro Chlebik

===

Fred Galvin
<galvin@math.ukans.edu> escreveu na mensagem> Search the google
archives for a posting to this newsgroup by Robin> Chapman with
the title Bertrand's Postulate.I did find it, at

===