>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.
<deletedEssentially 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.
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.
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.
 
>That is, I *said* f is coprime to 3 but in considering this
>possibility 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?
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.
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 now
you yourself bring out that with f=3, there's no way to determine
their values, or do you claim there is?
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?
 
>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.
It's *your* position that the w's are functions of m, so that you can
claim that m=0 is a special case.
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, but that's not a lot different from what was there before
when 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 respect
to 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 that
m=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 relevant
mathematics.
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.
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, would
force 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 are
EXTREMELY dedicated to the notion that I'm wrong, but the problem is
that 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 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 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.
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 follow
mathematical 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 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.
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 a
function of something else. It might help if you made it clear at the
beginning that P is a function of f, m, u and x. And that the b_i and
w_i are functions of f and m. You might then be able to show,
subsequently, that something that looked at though it depended on
something 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 that
P(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 be
chosen 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 have
P(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))
etc
Math 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 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. 
> <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.
  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). 
  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.  
> 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, 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}.
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 indeterminate
as 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 this
>possibility 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 
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.
> 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?
> 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 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.
> 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.
> 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.
>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.
  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.
  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 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.
> 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.
> 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.
> 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 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.
>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.
>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?
> 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.
> 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 m
that 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 dedicated
to defending what I believe to be true.  Is that a character flaw in
your 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 else
in 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 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!
> 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.
> 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 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.
> 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. 
> 
> 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 was
clipped 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 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* 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), 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)
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.
 
>   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.
A function is a function.  It doesn't check coprimeness of some
variable 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 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.
>   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 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.
 
> 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.
> 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.
<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.
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 to
try 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 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?
> 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 
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 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?
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 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.
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.
> 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.
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).
Now trapped by mathematical logic, you are squirming trying to find
some 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 why
think 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^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.
>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 I
could 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 correct
mathematics 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 a
conclusion, 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 the
truth 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, in
order 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 energy
and 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 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.
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.
> 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 logical
inconsistencies 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 is
something that can be worked out mathematically.
I'm glad you caught that mistake as I was posting it all over the
place!
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 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.
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.
I've determined that I can destroy your case using f=3, which is what
I'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 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.
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.
Well possibly it's because you realize that then I can use m=1(mod
sqrt(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 any
place or time.
It seems to me that you demonstrate belief in something other than
mathematics.
 
> 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?
>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?
> 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 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.
In your world, do people never change their minds?
If you are someone who refuses to change their minds then that might
explain 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 which
goes 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 about
f^2 dividing off from P(m) in some particular way.
So you're trapped both by what you said, and by the mathematical
logic.
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 out
that it was false, when I checked the math.
Now you have the option to follow the math, or stick with false
beliefs.
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 recent
presentation. But it ->seems<- to me that he is falling back to an old
mistake yet again.
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.
======================================================================
Why do you take so much trouble to expose such a reasoner as
 Mr. Smith? I answer as a deceased friend of mine used to answer
 on like occasions - A man's capacity is no measure of his power
 to do mischief. Mr. Smith has untiring energy, which does 
 something; self-evident honesty of conviction, which does more;
 and a long purse, which does most of all. He has made at least
 ten publications, full of figures few readers can critize. A great
 many people are staggered to this extend, that they imagine there
 must be the indefinite something in the mysterious all this.
 They are brought to the point of suspicion that the mathematicians
 ought not to treat all this with such undisguised contempt,
 at least.
   -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
> 
>    [.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 which
apparently are meant to persuade.  My interest is in what the actual
mathematical truth is, though it is of secondary interest that I
convince others.
For Arturo Magidin, it appears that persuasion is the primary
interest, which can be seen from his comments in contrast with the
facts 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 be
believed.
authoritative or looks mathematical, but he rarely actually proves
anything relevant to his central objections.
Here the pertinent expression, which he doesn't even bother to give
having 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 the
f^2 divides off dependent on m, so that they can challenge that with
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)
f^2 divides off through only two of the a's, which is a result that
can 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 dispute
how 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^2
divides off from the expression.
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 Harris
====
<same old shit 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 Harris
Interesting.  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 these
newsgroups, believe the second of these statements is true.  Arturo (and
many other talented people) have given you way more time than you deserve.
Back in 1973 when I was an undergraduate studying group theory, I stumbled
onto a paradoxical result.  I spent the better part of a weekend going over
my logic and could not find a mistake, so when Monday rolled around, I went
to my professor's office and asked him to look over my work,  He 
immediately
saw that I had to be wrong, and after a bit, explained that I was making a
bad assumption at one point.  I do not recall the details, because this was
so long ago, but the gist of it was that I was making a mistake about
properties of a group due to my lack of knowledge and experience.  I 
trusted
my teacher, and devoted myself to bettering my understanding of some
fundamentals.    A month or so later I reviewed my work and with the
improved state of my knowledge, could clearly see my errors.  See the
difference?  I didn't see a conspiracy to deny my results, but recognized
that my knowledge was deficient, and I corrected that.
James, you use invalid terminology, you continue to use a colloquial rather
than a formal exposition, you skip important steps (my high school algebra
teacher made that issue very clear to me...), you repeatedly use
intermediate results that are not proven.  You have been given excellent
feedback from some very knowledgeable people - the basic gist of which is:
get yourself some education in fundamentals.  Nobody is going to take you
seriously if it's obvious you haven't been doing your homework...
====
> 
>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 was
> clipped 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 
  ... 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 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.
> 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.
  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 what
you 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 technique
could simplify a great many difficult proofs, but it is not
considered 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 conclusion
without doing any hard mathematics.  Trying to pretend that my argument
requires that equations are intelligent beings or whatever is silly, and
you 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 in
both the leading coefficient (hidden in v^3) and in the
middle coefficient (hidden in v itself).  Certainly there
is no doubt that if you change m, you change the roots of 
the equation.  Now if you factor the equation in the
form
          (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 functions
of m.  If the equation happens to be reducible into
linear terms with rational roots, then you can assume
that c1, c2, and c3 are ordinary integers, and that they
factor the number (u*f)^3.  In general you are not going
to be able to tell HOW they factor (u*f)^3.  That will
depend on the MIDDLE term.  
  Here are two quadratic examples that may clarify
this:
         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 the
factors!  In the first example, the roots follow
the factorization 2*5 = 10.  In the second
example, 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 you
are interested, the term (u*f)^3 stays the same.
Again: it was changing the MIDDLE term which 
resulted in different values of the roots and
a 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 check
anything about their arguments, they must always be constant
with respect to those arguments ???
  Again, THIS IS NOT MATHEMATICS.  This is an attempt to 
ridicule my hard, rigorous mathematical conclusion by implying
that I impute some kind of intelligence to functions, etc..  
It doesn't wash.  IT ISN'T A MATHEMATICAL ARGUMENT.  I give
it 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 your
purposes 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 wanted
to 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, bullshit.  This constant stupid attempt to imply that 
I assume equations choose etc., is getting tiresome.  Plus
IT 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 represents
my 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 idea
that 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 prove
otherwise has been shot down: in my previous post, by
showing that the case f = 3 leaves the w's entirely
indeterminate.  And in the present post, by showing that
you 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 sound
like a dimwitted yokel.  You might consider varying it
from 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 a
lot of mistakes.  If they were just misprints I would
not mind, and I could probably figure out what you 
intended.  But as in this case, they are far more than
misprints.  I had no idea what you intended.  You
presented it as part of a serious argument.  All I could
tell was, it didn't make any sense.  
  Also I pointed this out in one or more previous posts
which you must not have read, or read too quickly to
understand.  You kept repeating it.  I note you have
had 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 act
as 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 you
have 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 weight
on it as justifying your side of the argument.  I was
OBLIGATED to point out that it was wrong.
  The other thing that I must note here is, you often
admit *minor* mistakes like this with reasonable grace.
In such cases our objections are easy to understand, and
impossible to deny.  But when we present rigorous but 
deeper and more difficult objections - equally valid,
in my view - you apparently refuse to even THINK about
them.  Why?  I think, because you don't know enough
math to understand them.  When I object to some statement
of yours and I invoke, say, a simple theorem from 
Galois theory to show it is wrong, to ME it seems just
as wrong as the little algebra mistake you made here.
But you act as if my Galois theory argument cannot be
trusted and that I must be lying.  All you have to do
is go learn a little Galois theory or algebraic number
theory 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 was
right.  Just as right as in the present little-algebra-
mistake case.  The only theory I used was an algebraic
number theory theorem which you have seen before and
actually accepted.  Even so, I was taking a huge
chance.  I am not perfect.  I make mistakes.  I could
have had an error in that post, and you could have
picked it up.  Still could, for all I know.  I was,
however, extremely confident of my argument.  No one
else raised the slightest objection to it.  
  I think it completely clinched the case against
your proof in Advance Polynomial Factorization.
  So what happened?  You totally ignored it as far
as I can tell.
  The point?  I have paid great attention to your
arguments.  I have dug into them in great detail.
I have presented proofs that in my view blast your
claims 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 using
any theory that you haven't bothered to learn, and
(2) we must also find where your mistakes are that
led you to your erroneous claims.  
  The latter is what the present thread is all
about.  It shouldn't even be necessary.  We are
being MORE THAN FAIR to you.  For most people,
a rigorous disproof of their claims is enough
to 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, to
boot, (2) find exactly where the child screwed
up, a task that should be left to the child
himself.
  The real point here: from my point of view,
my big objections to your bogus math are just
as valid as my little objections to simple
algebra errors.  You should at least try to
understand 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 you
any 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 that
even 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 wrong
with my argument that they are indeterminate.  How can you
deny 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 are
divisible by 5.  
> Do you see that now?
> 
  No.  In the case where f = 3, again with m = 1 and u = 1, you
have
  P(x)/f^2 = 57*x^3 - 24*x + 3,
and I CAN factor another 3 out of that.  That makes an enormous
difference 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 5
does 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 visible
justification 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.  The
M-M theorem does not guarantee more than that.  Certainly
it 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 CAN
assume 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 divisible
by any factor of 3.  That too does is not in agreement
with 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 
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 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 if
m is divisible by 13, n is an even integer.  That does NOT
mean that n is a function of m.  It does mean that if m1
and m2 are two different values for n, you do not expect
that the corresponding n's,  n1 and n2, are necessarily
the 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 simply
not true in general, and it is not stated as a part of
the Magidin-McKinnon theorem.
  Secondly, if here you are talking about f = 3, then
another f DOES can be divided off, and it leads to nonunique
b's and w's as I noted in my previous post.
  See my previous discussion of this also.  My previous post
showing 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 of
trying 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 as
APF, you claim certain things are true.  They demonstrably 
are not.  You think you have proved them.  Clearly your
proof has an error.  That is what we are talking about
here.
> Mathematics isn't true in order to upset you.
> 
  What ???
> It's just true.
> 
  A tautology obviously.  And implicitly, another claim that
you yourself are infallible.  You are really saying MY
mathematics is invariably true.  I cannot swallow that.  
By now a little humility should have crept in.  You DO
make mistakes, after all.  Until you find an error
in things like Challenge of the Century, I will have
to 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 is
beginning 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 the
only 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 be
attacked!
> Mathematics is not a child that can be harmed.
> 
  You come up with the most bizarre statements.  In one
breath, 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't
you 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 little
people, making decisions about how they factor 
depending on the value of m, etc..  And now YOU
seem to somehow have inside information on the
emotional 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, very
explicitly, 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 are
constant on the basis of what happens when f = 3 is what
falls 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 the
form
        (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 the
roots themselves is an algebraic integer, and the product of the roots
is 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, for
example, f = 5, m = 1, and u = 1, the polynomial inside the
parentheses 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 go
look 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=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 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 go
where 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 the
forest by nibbling on a little of the foliage out at the
edge.
> 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 respect
to the polynomial indeterminate x.  The other 
variables are not like the y in the bivariate 
polynomial above.  In general you think of them as
fixed constants.  Unlike y, they do not appear explicitly in
the final factorization, i.e., in the linear
terms 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 to
have it both ways: on the one hand, you wanted to say
that 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, and
make the same claim.  IT SIMPLY DOESN'T WORK.  A
polynomial with irrational coefficients cannot possibly
be considered reducible over the rationals.  It makes
no sense to even consider it.  It's not a matter of
flexibility.  
> 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 not
relevant 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 uniquely
in 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 are
invariably wrong) would be more justified.  Have you
been 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 
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 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 incredibly
stubborn.  You keep getting shot down, like Yul Brynner in
the 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 my
underwear.
> 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 the
Century.
> Show the mathematical logic that you believe supports your position,
> if you can.
> 
  Done.  See all of above, plus previous post, plus several dozen
others.
>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 you
relapsed.
> 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 at
the 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 to
smack you broadside.  James Harris displays a sickness, and is clearly
*NOT* amenable to change.  The role he has settled into on Usenet, is
the one he is comfortable with, and one he will likely keep until the
day he dies.
You, and some other posters, have displayed signs of codependancy.  As
if James' sickness draws you to him, like flies on shit.
James will *NOT* change.  Whether you change is up to you.  Arguing
with him is like arguing with a rock, or your bathroom wall.  If you
continue to define your online life by replying to his trolling, you
are most certainly a diseased person also.
This is my last post to you on this subject, Nora Baron, because as
for myself I *refuse* to be drawn into James' sickness for years and
years on end.
<more histrionics deleted 
>   Nora B.
> 
> 
> James Harris
Flatrings
====
[snip]
> You, and some other posters, have displayed signs of codependancy.  As
> if James' sickness draws you to him, like flies on shit.
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 are
easily 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't
make them irrelevant.
-- 
Wayne Brown                | When your tail's in a crack, you improvise
fwbrown@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 hiding
behind 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 the
dependency on m that you seem so dedicated on claiming exists when f
is 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 - 19
as 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 the
expressions where they see m all over the place, as well as 3, and
there 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 statements
you've already made, and I'll finish out by focusing on the last one
I'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 you
said:
>   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 hominem
is a fallacy.
A better question is: Why don't you post one of the numbers you claim to 
have discovered which *should be* an
algebraic integer, but which is *left out* of the ring of algebraic 
integers? Those numbers are the basis for your
repeated accusations that math professors are teaching lies to students and 
form the basis of your threats to sue the
universities. Here's a great opportunity to post *one* such number for the 
enlightenment of students and
mathematicians 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 it
is 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 and
understand. 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 raised
to 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 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
====
> 
>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?
(...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 countably
many polynomials with integer coefficients.
       - Randy
   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? 
   I doubt the answer is yes, but it is kind of curious to note that one 
can
often find a,b,c s.t a+b+c is quite small.  For example, if
p=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 three
consecutive primes p,q,r there exist integers a,b,c with the property
that
  ap+bq+cr=0  and |a|+|b|+|c| < K?
The solution to your original question obviously won't work, since
using 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 find
three 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 consecutive
primes.)
>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 standard
deviation, 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 sample
points from S1 and 55% from S2, what would the new standard deviation
and 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 x1
and a standard deviation, sd, of sd1.  moreover, S2 has a mean and sd
of x2 and sd2, and s3 has a mean of x3 and sd of sd3.  S1, S2, and S3
are all correllated with each other like this:
    S1     S2     S3
S1  1.00  -0.55   0.22
S2 -0.55   1.00   0.02
S3  0.22   0.02   1.00
and 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 new
mean 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?  
> josh
rational - rational'  Now I get it.
So many wonderful ways of putting it.  I am beginning to see how this
all 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 irrational
Assume p/q + n = r/s
n = r/s - p/q
Now 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 irrational
Assume p/q + n = r/s
n = r/s - p/q
Now 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?  
> josh
Assume 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 = 14
2)
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.
Tapio
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

>    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?
> 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 dollar
if private university, 2000~2500 dollar
your 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 
I
>>    
>>care 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 years
than 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 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?
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 Tooth
http://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 is
proved 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?An
um=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.de
http://www.frank-buss.de, http://www.it4-systems.de
====
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???
For brief info see
http://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 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 when 
interpreted
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).
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).
--- 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) we
consider???
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 when
interpreted
>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 each
for all values of the variables.  If all such evaluations produce TRUE
then output it.  To generate all wffs, one simple way is to use the
fact that P, Q, R, . . . are wffs, and if X and Y are wffs, so are ~X
and X^Y and XvY etc, (for whatever logical connectives you choose to
use).  I have had to throw together a subroutine to do this a number
of 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 Volkstorf
Cambridge, 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 which
can be determined in closed form.  Then try a few nearby integers to
get 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 whether
it should be M or M+1 without having to evaluate f(N).
Robert Israel                                israel@math.ubc.ca
Department 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 views
are 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 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_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_k
Suppose 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]]);if
iszero(evalm(G^4)) then print('G'^4=0,`so`,T=1+G+T*T,`has
solution`,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 whether
you believe the continuum hypothesis.
You might as well propose belief in the lord our god.
Roland
--
Roland and Lisa Paterson-Jones
Forest Lodge, Stirrup Lane, Hout Bay
http://www.rolandpj.com/forest-lodge
mobile: +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 you
mean 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 that
picks a component from a representation of a point?
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?
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.
-- 
A good plan executed right now is far better than a perfect plan
executed 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 from
R^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 Alfred
Bester 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 a
tensor is, but I'm wonder if a simple layman's explaination of what
one is would help to better appreciate this song. Do tensors have
something to do with tension? With reducing eight to one? Is tenser
also a pun on ten, sir!? (Except the countdown begins at eight and
not ten, but maybe the nature of tensor analysis has something to do
with 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 a
informal expository discussion you can look at
http://www.math.ucr.edu/home/baez/gr/gr.html.
They are used in general relativity (hence the gr), which was
Einstein's theory of gravitation, still the best available theory
of gravitation.
|Do tensors have something to do with tension?
Yes.
Stress (and tension) is described using the stress tensor. Perhaps
it will give an idea of how they generalize vectors. Forces are given
by vectors. But stress isn't generally just a matter of forces being
applied along a single axis. If we place an imaginary plane within
a solid, the forces exerted by the solid on one side on the solid on
the other side (either tension or compression, possibly with shear)
depend upon the orientation of the plane you are putting there. But
this dependence is well-behaved, and can be determined from the
forces exerted across just three planes, parallel to the xy, xz, and
yz 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 coordinate
system, the components transform in a certain way, so it's more
of a geometric type of thing than just 9 quantities lumped together,
just like vectors are more geometric than just 3 quantities that
belong together.
Part of the magic of general relativity is that three space coordinates
plus time can be represented mathematically as a four-dimensional
continuum. In that context, instead of just 9 components for the
stress, there's a similar kind of tensor with 16 = 4*4 components,
called the energy-momentum-stress tensor because the additional
components describe the energy and momentum as well as the
stress.
There's a second tensor describing a kind of curvature of space, and
what GR says is that the curvature tensor is proportional to the
energy-stress-momentum tensor.
So there are gravitational effects relating to momentum and tension
or compression as well as just energy; they are just usually too
weak to be noticed. Two long rods arranged parallel to each other
will 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's
an 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 better
than the usual trash that is forced upon undergraduates.
For instance Springer sells calculus books by Serge Lang in the range
of 50-60 EUR. At least for US and Canada, Dover offers books
in 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 because
those who pay the price are not those who make the decisions about
which 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 1000
pages etc., I would at least suspect that all this buzz is meant as a blind
for shortcomings of the main text.
During a course, there are instructors and T.A.'s who can make the material
digestible 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 a
book 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 he
shows that the continuum is a countable union of countable sets, he shows
that aleph-1 is the sup of an increasing countable sequence of countable
ordinals.
|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 subsequence
b_1 < b_2 < ... where for each k, b_k is the first term in the a's
such 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 has
sup aleph-1 then my argument works again.
So it seems the model in Cohen's book is exactly the kind needed
to 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, titled
Set Theory without choice: not everything on cofinality is possible.
at http://arxiv.org/pdf/math.LO/9512227, and tell us what the result
of Gitik on cofinality is. I was only able to fetch Google's cached
copy in which one reads 'By Gitik [Gi] ZF + every aleph has cofinality
{splotch} 0 is consistent.' It sounds entertaining, but I'd like to be
sure 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 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 the 
geometric
objects without resorting to any set speak.
I think when you suggest I think about sets in disguised form you are
letting your bias for sets show.  I would say that when I talk about sets
that I am talking about geometric objects in disguised form.  So I 
think
you 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 out
that I assume properties of sets.  Sets are useful to me only if they have
properties that approximate (in some sense) the behavior of geometric
objects.  So I am really assuming lots of properties about them, where
them does not refer to sets really.  
As an example, you seem to be chastising me for thinking of sets as fixed
totalities.  Putting aside the fact that I am not really thinking of
sets as you may think I am, let me address this issue of fixed
totalities.  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 set
with a fixed totality, is a perfectly good way to talk about it because it
captures what I'm actually thinking (the second way).  To only be able to
talk about finite sets would be silly in this context.  It would not
capture my idea of sphere.
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.  
> 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 that
objects I talk about have some properties.  I don't see that because I'm
not interested in explaining my assumptions in the language of some
axiomatic theory this is a failing of some kind.  The fact that you think
it is a failing of mathematicians shows your own bias.
====

> [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 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 
the
geometric
> 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 fixed
totalities.
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 in
explicitizing
the assumptions you actually use / need in this kind of geometry.
But are they? And are geometers themselves interested in making these
assumptions
more explicit?
I know most mathematicians are not; and when they teach their subject to
students,
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 that
distracts
young people from a study or a career in math. And it is also not a good
sign,
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?
> 
> Kevin
York 1985, ISBN 0-393-01921-7).
      <d2c23ee6.0308060523.4e9c498f@posting.google.com> 
      <3f314280$7$fuzhry+tra$mr2ice@news.patriot.net> 
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      <bgu1eh$ctj$1@panix1.panix.com> 
      <d2c23ee6.0308090857.5935b0cd@posting.google.com>
X-Cise: tanbanso@iinet.net.au
X-CompuServe-Customer: Yes
X-Coriate: admin@interspeed.co.nz
X-Ecrate: tanandtanlawyers.com
X-Punge: Micro$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 may
not converge.
-- 
     Shmuel (Seymour J.) Metz, SysProg and JOAT
Reply to domain Patriot dot net user shmuel+news to contact me.  Do not 
reply
to 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 made
directly 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 Jesus
were supposedly made by God then it would follow that they would
both 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 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.
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/weight.htm
There is another type of measure that is also very 
ancient and that is the measure of mass. . . .
  As time passed, each nation and region developed
its own standard masses against which unknown masses
could be compared.  The chief such unit is called
pound 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 the
acceleration of free fall; that is constant, and a measure of mass that is
_constant_; because the acceleration of free fall varies with time, being
dependent on such things as the relative positions of the sun and moon wrt
earth.
====
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.  Then again, it's clear SI is far more coherent on this
particular point anyway, although I have identified an inaccuracy
which is probably so minute it's meaningless.  However, has
anyone 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 curved
grav field, which varies.  While the variation is extremely
small, there is one, in theory -- and I'm wondering if
it's detectable at all on a lab bench (Hafele and Keating
took a trip around the world with cesium-beam clocks and
did detect it, so it's doable in theory).
It's also far less than the 0.4% g variation anyway so I
am probably worrying too much.... :-)
[.sigsnip]
-- 
#191, ewill3@earthlink.net
It'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 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.
The most common values used for that g_n, the standard acceleration of
free fall, include 9.80665 m/sÓ (which is official for 
defining grams
force and kilograms force), 32.16 ft/sÓ, 386.1 
in/sÓ, and others
practically indistinguishable, in almost all applications in the real
world, from the first such as 32.174 ft/sÓ and 1 lbf/1 lb = 
4.44822
N/0.45359237 kg or approximately 9.806646... m/sÓ.  Many 
textbooks
also use 32.2 ft/sÓ and even 32 ft/sÓ, though 
I haven't encountered
any 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 of
units is, with a poundal defined as exactly 1 
lbáft/sÓ.  It is
something totally immaterial in either of these systems.  In neither
of these systems does a standard acceleration of free fall serve any
purpose whatsoever.
The situation is also different for the non-SI grams force and
kilograms force, which are officially defined, since the CGPM adopted
for this purpose in 1901 a standard acceleration of gravity equal to
980.665 cm/sÓ.
Gene Nygaard
http://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 fall
varies at various locations on Earth; for various reasons: Mainly depending
on the strength of gravity at that location and the centrifugal effect due
to Earth's rotation.
_Tables_ have been prepared showing this acceleration; as measured at
numerous locations. There is no standard acceleration of free fall: It is
what it is at those locations because it depends on the strength of gravity
and the centrifugal effect due to Earth's rotation there.
No national or international standards laboratory, no professional
organization, no national legislature, or anybody else can change the
acceleration 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/sÓ (which is official for 
defining grams
> force and kilograms force), 32.16 ft/sÓ, 386.1 
in/sÓ, and others
> practically indistinguishable, in almost all applications in the real
> world, from the first such as 32.174 ft/sÓ and 1 lbf/1 lb = 
4.44822
> N/0.45359237 kg or approximately 9.806646... m/sÓ.  Many 
textbooks
> also use 32.2 ft/sÓ and even 32 ft/sÓ, 
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 convenient
approximations: Close enough for the practical applications to which they
pertain.
National and international standards laboratories and professional
organizations, and national legislatures have banded together and adopted a
_standard weight_; the weight of the kilogram artifact that is stored for
safe keeping at the international archives in Sevres France.
This artifact serves as the standard for weight all over the civilized
world, and accurate copies of it have been distributed to all member
countries - including the U.S. - who use it as the basis for weight
measurements.
The kilogram is called a _unit_ of mass, because originally it was a 
cubic
decimeter of pure water at its maximum density; with a scale-weight that 
was
numerically; _exactly_ equal to the acceleration of free fall at the
location 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 
to
the ratio of the net force [f] to the forced displacement [s/t] that
constitutes inertia; the property of an object that requires the 
application
of an external force in order to change the object's velocity. That a
kilogram 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 to
activity may take even longer(;^)
====
 I must respectfully disagree with the seeming consensus of the
> mathematicians whose analyses I scanned.  The seeming paradox has 
little
to
> 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 a
winning
> strategy would be to switch in every case in which you observe a value
less
> than the maximum value you have previously observed, otherwise to stick
with
> your first choice.
 Of course if you are told a maximum bound ahead of time there is no
problem.
> Always switch unless you see that value.

> knowing the bound IS having knowledge of the distribution!
 Herc
Of course.  But the resolution of the paradox does not depend on knowing a
bound, or on any reasoning about any particular kinds of distribution.   
The
paradox arises because of the conflict between the provable impossibility 
of
overall gain from unconditional switching, and a seemingly valid 
calculation
which says one will indeed gain from unconditional switching.  To defeat 
the
paradox, one need only recognize that the estimation of benefit from
unconditional switching must also include the guaranteed loss at the
(typically unknown) upper bound of the values encountered in any finite
series of trials.
In an infinite series of trials with no finite upper bound, switching might
be said to improve the rate of gain, but your expectation in any case is
either infinite or undefined depending on which aspects of the picture you
want to quibble about, so there is no paradox to be worried about in the
first place...
I mentioned knowing the bound only in context with explicating the
conditions under which one could actually gain by a conditional switching
strategy.
Bill
====
Hey James,
Sorry, I forgot to tell you that it's under renovation now (till
around 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, which
luckily 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 REGARD
TO 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 be
through 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 f
can 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 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 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 Harris
I 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.ca
Department 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 an
infinite 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 eepdoos
I believe, however that Robert Wilensky has argued that we have a perfectly
good *natural* experiment:  The hypothesis is that an infinite number of
monkeys at an infinite number of keyboards will eventually produce the 
complete
works of Shakespeare; our experience of Usenet refutes this hypothesis
rather definitively. 
It's also been observed (I forget by whom) that if an infinite number of
monkeys are given an infinite number of typewriters, then with probability
one at least one monkey will figure out how to beat the monkeys around it
to death with its typewriter.
Steven E. Landsburg
www.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 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
====
> 
> 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.
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 an
ordinal 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 it
has something to do with their being circular. That's not what people are
after when they spend more than a little bit of time trying to resolve
them. That's just the starting point. One might hope, for example, that an
explanation for why the definitions are wrong would give some kind of
criterion 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 to
give a reasonable rule for avoiding the problem in the future is a little
dubious.
Impredicative definitions are an example. There have been people who
thought that when you define a set, it should be a set composed of elements
that you could have defined before defining the set itself. Otherwise, they
say, there's a kind of invalid circularity.
The consequences for mathematics are substantial because the definition of
the real numbers is impredicative. By the Dedekind cut construction, a real
number is given by a collection of rational numbers (satisfying some
conditions 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 of
alleged circularity exists, because ... could talk about the real 
number
line and its properties.
On the other hand, there have been lots of people with the point of view
that definitions like this are okay, because the reals are all there
already, and by making a definition we are merely selecting from out
of a pre-existing domain.
A thorough unravelling of the paradox should help us to see whether this
kind of circularity or apparent circularity is valid, or whether it just
makes the same mistake as the paradox does but inadvertently avoids running
into a direct contradiction. That's the kind of thing that makes such
paradoxes 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 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.
> 
Well, I guess that depends on what the precise definition of 'precise
definition' 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 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.
====
> 
> 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 in
fact 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 dx
I 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 may
have to explain the situation you have in mind.  Where does the function
f come from?
Robert Israel                                israel@math.ubc.ca
Department 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
> Klaus
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.
====
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 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 dt
The unbiased estimator of _what_?  You can't have an unbiased estimator
unless 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 of
distribution is this supposed to be taken from?  You certainly
can't assume that f(t) for different t's are independent, because
then (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.ca
Department 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 dt
This is the mean of the function (f(t)-mu)^2, which is the
definition of the variance. As with the discrete case, you can
prove 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 any
problem with defining this, which is an estimator of a statistic
over 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 the
Diophantine 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 are
odd or both are even, and you have your answer.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, 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 
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??
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.
====
>> 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 is
somewhat 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.ca
Department 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 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.
-- 
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 phrase
well-ordering means, or possibly you missed the fact that we're
talking about _a_ (well-)ordering, not talking about the standard
order 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, for
a) 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, and
c) 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 surreal
numbers,
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 
of
y;
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 or
ternary 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 allows
one to construct a Hamel basis for R over Q; come to think of
it 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, so
logicians sometimes consider the most convenient one.
Goedel defined an explicit well-ordering of a certain subset of R, and
showed that ZF cannot disprove the assertion that the subset is all of
R (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 someone
please give me the definition, so that I can check those proofs for
myself?
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 = 0
with 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 Rudd
For 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 Rudd
Factorization problems are tricky in algebraic integers.  Others
have already defined the field, or one can look at
http://planetmath.org/encyclopedia/AlgebraicInteger.html
which is as good as any although extremely terse (wolfram.com
is 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, solutions
of the equation x^2 - 2nx + 1 = 0, for any integer n > 1.
(n=1 yields a reducible quadratic; n=0 gives us no reals
but is another factorization: 1 = i * (-i).
n = -1 gives us (-1) * (-1); n < -1 gives us the negatives
of most of the roots we've found already.)
Since Mr. Harris likes to have things dependent on coprimality of 3
to 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 positive
algebraic integer.  Consider the equation y^n - 2, n > 1.
This equation has n roots, exactly one of which is real
and positive (if n is even we get a negative one as well).
Subtract 1 from this root (or, if one prefers, solve the
equation (x+1)^n-2=0 instead) and one sees that one can
generate an infinite sequence of algebraic integers in
an ordering, each smaller than the last one but all of
them positive.
One can use the above quadratic equation as a generator
as well; the smaller root will tend towards 0 as n gets
larger and larger; the bigger root tends towards positive
infinity.
Funny things, these algebraic integers. :-)  It's clear
Mr. Harris is making a strange assumption or two in his
proof.
-- 
#191, ewill3@earthlink.net
It'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
====