Re: infinity
- From: "David R Tribble" <david@xxxxxxxxxxx>
- Date: 4 Oct 2005 14:15:16 -0700
David R Tribble said:
>> 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.
>>
>> 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.
>
Tony Orlow wrote:
> 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.
Yes, and every one of them is finite. I proved it above, but it
bears repeating: Every '9' digit corresponds to a member in set S
of 9x10^p; and for each finite member 9x10^p there is another larger
member 9x10^(p+1), which is also finite.
You don't ever get anything but finite members in S, because each
finite member (starting with 9) is followed by another member that
is just the previous multiplied by 10. Multiplying a finite value
by 10 never produces anything but another finite value, right?
>> 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.
But I just did (above). If it's not true, please tell us which digit
in N-1, or which finite member of S, is followed by an infinite value.
>> 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.
It's based on your theory that any set of only finite elements must
contain only a finite number of elements. You've said repeatedly
that such a set is "unbounded but finite", and that it must have an
"unidentifiable" largest member.
Thus, by your own rules, set S must have only a finite number of
members, and it must have a largest (unidentifiable) member. And
since S was constructed from the digits of N-1, then N-1 must have
only a finite number of digits.
(The "Cantorian spin" says that set S is infinite, not finite, which
is exactly the opposite of what your rules say.)
>> 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.
Just show us which digit of N-1, or which member in S, is infinite.
I thought all the digits were just '9', but apparently that's not
correct?
Or, show us how to construct set S properly, including all the finite
and infinite digits of N-1.
.
- Follow-Ups:
- Re: infinity
- From: Tony Orlow
- Re: infinity
- References:
- Re: infinity
- From: Tony Orlow
- Re: infinity
- Prev by Date: Re: set proof valid?
- Next by Date: Re: area of a triangle determinant
- Previous by thread: Re: infinity
- Next by thread: Re: infinity
- Index(es):
Relevant Pages
|
