Re: Basic argument, algebraic integers
From: Nora Baron (norabaron_at_hotmail.com)
Date: 10/05/04
- Next message: The Last Danish Pastry: "Re: How to calculate pi in binary ?"
- Previous message: krosty: "An integration problem"
- In reply to: James Harris: "Re: Basic argument, algebraic integers"
- Messages sorted by: [ date ] [ thread ]
Date: 5 Oct 2004 08:18:34 -0700
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0410050242.7aab0913@posting.google.com>...
> CORRECTED ARGUMENT:
>
> So it's easy.
>
It's still wrong, and it's not so easy. See below.
> I. First section
>
> Start with
>
> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f)
>
> with the factorization
>
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> and note that at
>
> m=0, P(0) = u^2 f^2(3x + uf),
>
> which gives you terms that do not vary as m varies.
>
> So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)?
>
> Well, at m=0, two of the a's have to equal 0, and it's convenient to
> just arbitrarily select a_1 and a_2 as those two.
>
> Then you have uf for the first, uf for the second and 3x + uf for the
> third as terms that do not vary when m varies.
>
> Now then, if m=1, what are the *constant* terms?
>
> They are uf, for the first, uf for the second, and 3x + uf for the
> third.
>
> That's logical because they do not vary with m, so if m=1003909273,
> what are the constant terms?
>
> They are uf, for the first, uf for the second, and 3x + uf for the
> third.
>
> Now divide f^2 from both sides, which gives
>
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f
>
> P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2
>
> and you note that P(0)/f^2 = u^2(3x + uf), which means that now your
> constant terms are u for the first, u for the second and 3x + uf for
> the third.
>
> Now then, if m=1, what are the constant terms now?
>
> They are u for the first, u for the second, and 3x + uf for the third.
>
> If m = 2938479378, what are the constant terms now?
>
> They are u for the first, u for the second, and 3x + uf for the third.
>
> How can the constant terms of the first two go from uf to u?
>
> They must be divided by f.
>
> Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided
> from P(m), it is u; therefore, a_1 x + uf is divided by f, and you
> have
>
> a_1 x/f + u
>
> and the constant term of a_2 x + uf is uf, but when f^2 is divided
> from P(m), it is u; therefore, a_2 x + uf is divided by f, and you
> have
>
> a_2 x/f + u
>
> while the constant term of a_3 x + uf is 3x + uf, and after f^2 is
> divided off, it is 3x + uf, so you have
>
> a_3 x + uf
>
> so, dividing P(m) by f^2 gives
>
> P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf).
>
> There is no way to mathematically argue with the result.
>
> II. Second section
>
> Now take
>
> P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf)
>
> and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside
> by f^2(a_1 a_2 a_3) and you have
>
> P(m)/f^2 = ((a_1 a_2 a_3)/f^2)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3)
>
> and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is
>
> P(m)/f^2 =
>
> (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3).
>
> So I have the ratios of algebraic integers:
>
> uf/a_1, uf/a_2, and uf/a_3,
>
> and now let
>
> v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_2
>
> where the v's and w's are algebraic integers in each case coprime to
> each other.
>
> Making the substitutions I have
>
> P(m)/f^2 =
>
> (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3).
>
> And I have from before that
>
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f
>
> so
>
> (m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = f
>
What happened to u^3 ?
> as that is the last coefficient, which proves that
>
> (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let
>
No, you cannot conclude that. It is true that v_1 and w_1
are coprime, but you do not know that v_2 and w_1 are coprime,
for example. Therefore you cannot conclude that w_1 divides
(m^3 f^4 - 3m^2 f^2 + 3m). You still need a nontrivial step.
> (m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3
>
> then I have
>
> P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3)
>
> but I still have that
>
> P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf).
>
> So finding algebraic integers w_1, w_2, w_3, and v_1, v_2 and v_3 are
> irrelevant to that result, as they could always be found as proven
> using some basic results where the only theorem used without naming or
> giving proof covers the existence of coprime algebraic integers v and
> w, given a ratio of algebraic integers.
>
You are right that w_1, w_2, w_3, v_1, v_2, v_3 can be
found to factor P(m)/f^2 as you describe, but your method to
find them is inadequate as noted above. You need a more
sophisticated approach.
You are wrong that finding such a factorization is "irrelevant
to the result". What this factorization implies (after a little
additional work) is that P(m)/f^2 factors in exactly the way you
say it cannot. The algebraic integers are not incomplete as
you have claimed: there SHOULD be a factorization of this form,
and you have come fairly close to finding a valid proof that
there IS. That is, you have yourself undermined APF and all the
related B.S. that you have been trying to promote.
Good work!
Nora B.
>
> James Harris
- Next message: The Last Danish Pastry: "Re: How to calculate pi in binary ?"
- Previous message: krosty: "An integration problem"
- In reply to: James Harris: "Re: Basic argument, algebraic integers"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
