|
|
jstevh@xxxxxxx wrote:
> Some recent threads simply explaining yet another way to see the
> coverage problem of the ring of algebraic integers have floundered on
> the issue of convergence and what it means with rings like the ring of
> algebraic integers, where I introduce the concept of an exclusionary
> ring.
>
> The concepts are simple, luckily, as I can mostly use something basic:
>
> S = 1 + x + x^2 + x^3 +...
>
> where the issue is convergence, so that if S converges in whatever ring
> you're in--notice none given yet--you can go to
>
> S = 1 + x*S
>
> and solve to get
>
> S = 1/(1-x)
>
> where to get convergence in the ring of complex numbers, which of
> course is also the field of complex numbers, let's choose x = sqrt(5) -
> 2, considering only the positive solution, then in complex numbers, I
> have
>
> S = 1/(1 - sqrt(5))
>
> which is a complex number, and it's also an algebraic number, but it's
> NOT an algebraic integer.
>
> But if I go back to
>
> S = 1 + x + x^2 + x^3 +...
>
> and plug in x=sqrt(5) - 2, I'll have
>
> S = 1 + (sqrt(5) - 2) + (sqrt(5) - 2)^2 + (sqrt(5) - 2)^3 +...
>
> and if I stop at some point like just look at
>
> 1 + (sqrt(5) - 2) + (sqrt(5) - 2)^2
>
> I have an algebraic integer, and in fact, I can have an arbitrarily
> large number of terms added together, and stop--ever closer to
> 1/(1-sqrt(5)) and STILL have an algebraic integer.
>
> BUT in the ring of algebraic integers, you cannot reach 1/(1-sqrt(5))
> because it is NOT an algebraic integer!
>
> Yet you can approach it out to infinity, so you have an asymptotic
> approach to that value.
>
> A corollary to this is 1/x where you can let x go out to infinity and
> that approaches 0 but never reaches it, or you can let x approach 0,
> but never reach it, but the difference with the infinite sum example is
> that the asymptotic nature is created by the exclusionary nature of the
> ring of algebraici integers!!!
>
> That is, the reason you can never reach 1/(1-sqrt(5)) with the series
> is that 1/(1-sqrt(5)) is NOT the root of any monic polynomial with
> integer coefficients, which is the rule that defines algebraic integers
> and excludes that value!
>
> So the mathematics holds on a definition, as logically, if
> 1/(1-sqrt(5)) is not an algebraic integer, then there is no way to
> reach that value in the ring, so
>
> S = 1 + (sqrt(5) - 2) + (sqrt(5) - 2)^2 + (sqrt(5) - 2)^3 +...
>
> approaches it asymptotically, in the ring of algebraic integers.
The problem is that you previously said that if all the terms
in a series were in a ring the sum must be in the ring
and this must hold for any ring.
Have you changed your mind?
- William Hughes
|
|