Re: infinity

David R Tribble said:
> David R Tribble said:
> >> However, your 0:999...999 and 1:000...000 are naturals, so it's
> >> entirely reasonable to ask if they are odd or even, prime or
> >> composite, or any number of other questions about whether they
> >> behave like naturals.
> >
>
> Tony Orlow wrote:
> > let's answer this now. If this is decimal N, then 1:000...000 is even, as a
> > power of 10, so 999...999 is odd. That's obvious anyway, since it ends in a 9.
> > Does this cause a contradiction?
>
> Funny you should ask...
>
> Let's have a closer look at Tony's 'N'. Let x = 999...999, an infinite
> natural number. In Tony's number theory, this is equal to N-1, or
> 1:000...000 - 1, where 'N' is Tony's arbitrary unit infinity. (This
> may imply that N-1 is the largest natural number, but we'll ignore that
> for now.)
It does not imply any such thing. The unit infinity is not a specific number,
and is not the smallest infinity, so it has nothing to do with a largest
finite. Nice try.
>
> It is obvious that:
> x = N-1
> x = 999...999 (base 10)
> x = 9 + 90 + 900 + 9000 + ...
> x = 9x10^0 + 9x10^1 + 9x10^2 + 9x10^3 + ...
>
> We can also represent x as the set of its digit values, like this:
> Sx = {9, 90, 900, 9000, ...}
>
> All of the digits and their corresponding 9x10^p values are, of course,
> finite, because each one represents a single decimal digit and a
> single power of 10; obviously for each member 9x10^p, the next member
> is simply 9x10^(p+1). Which means that Sx is a set containing only
> finite members.
I disagree. If p can take on infinite values, the the terms 9x^p become
infinite. There is no reason to restrict your digits to finite positions.
999...999 has an infinite number of 9's.
>
> According to Tony's set theory, a set of only finite elements can
> contain only a finite number of elements, which means that set Sx has
> only a finite number of elements.
Only if you restrict your digits to finite positions as you just did. You can't
prove all numbers are finite, based on that very assumption. Sorry.
>
> Which in turn means that x = N-1 has a finite number of digits; exactly
> ceil(log(x)) digits, in fact, according to Tony's number theory.
No, that is according to your Cantorian spin.
>
> But this means that N-1 cannot be an infinite number, because it does
> not have an infinite number of digits.
>
> Or it means that sets of finite elements can, in fact, have an infinite
> number of elements.
>
> Which is it, Tony?
It is that there is no restriction to finite digits. The problem you point out
is in your own misassumptions, not in my system.
>
>

--
Smiles,

Tony
.

Relevant Pages

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