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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.
I make mistakes, but I care about what is correct.
So I can admit when I'm wrong, as mathematics is beautiful in that
what's right is absolutely right.
And people fighting for their own delusions of worth do not matter to
that correctness.
After all, these arguments will one day be gone, all of you will be
dead and your children, and even your children's children will be dead,
but the correct mathematical arguments will still be correct.
The mathematician looks beyond the moment to any point further when all
the arguing is meaningless, which is why to some mathematicians are
unearthly or almost mystical.
Only true mathematicians care nothing for social crap or the accolades
of the moment as they quest for absolute knowledge knowing that even
God cannot change it.
And in that way, the mathematician stands in the presence of the
divine.
While lesser beings stoop at the feet of social needs, begging for
approval, priding themselves on social acceptance, and when they are
gone--there is nothing left.
James Harris
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