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<jstevh@xxxxxxx> wrote in message
news:1159644259.872275.191160@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> 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
>
James, let me ask you this. Have you ever thought "Hmm...All these people
disagree with me and/or tell me to go read certain books. Maybe I need to do
so."? I can tell you as a professional mathematician, I've read more math
books than I did when I was working on my degree. Not because I forgot what
I learned so much, but I've come across problems that I admittedly needed to
get some more information on certain topics that I knew little to nothing
about before I tackled the problem at hand. I think that's what any good
mathematician does.
Dave
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