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jst...@xxxxxxx wrote:
> William Hughes wrote:
> > 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
>
> Well that's wrong, so yes, I had to, following what is mathematically
> correct.
>
So is the following true or false?
Let:
S = 2 + 2 a_1 + 2 a_2 + 2 a_3 +...
where S and a_1,a_2,a_3 ... are objects. (So we have an example
of a series that does converge).
Then: S is divisible by 2 in the ring of objects (i.e.
S/2 is an object).
- William Hughes
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