Re: abundance of irrationals!)
- From: Virgil <ITSnetNOTcom#virgil@xxxxxxxxxxx>
- Date: Mon, 16 May 2005 12:43:37 -0600
In article <MPG.1cf2ae05a97c583e989c63@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> Randy Poe said:
> >
> > aeo6 Tony Orlow wrote:
> > > I already said you can have your set of finite counting numbers. Just
> > don't
> > > claim the set has an infinite number of members unless you can come
> > up with
> > > some infinite set of symbols to represent them. S^L is finite for
> > finite S and
> > > L, period.
> >
> > What do you think the value of L is for the base-10
> > representation of the naturals? Do you think there
> > is a value L such that there are no naturals with
> > L+1 digits?
> >
> > - Randy
> >
> >
> I think that if there are an infinite number of naturals, then there must be
> some with infinitely many significant digits.
A finite sequence is one with a last term,
An infinite (non-finite) sequence is one that has no last term.
All finite sequences are bounded.
Some infinite sequences are bounded, but even the unbounded ones are
sequences of all finite elements.
According to TO's argument, at some point one adds 1 to a natural with
finitely many digits and gets something with infinitely many digits, but
TO does not explain either how or why this happens.
Is there a largest finite natural to which adding 1 still gives a finite
natural? I don't think so, but apparently TO does.
.
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