Re: Logarithm of transfinite numbers

In article <MPG.1e80bc50bd90201a98aad1@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:

The 2^x=aleph_0 and aleph_0 is finite. Is it aleph_0?

Yes.

Not if that makes 2^aleph_0 strings.

No. You don't need ALL of the length aleph_0 strings to
make the natural numbers. Just some of them.

Ahem! How many is "some"? (sigh) That's not very mathematically precise.

Actually, to make the naturals, you don't need any infinite strings.

Is zero precise enough for you, TO?

You just need that there is no finite bound on lengths.
.

Relevant Pages

  • Re: Logarithm of transfinite numbers
    ... Tony Orlow wrote: ... But any countably infinite number of bits you have left will still ... produce an uncountable se tof strings. ... Naturals are natural, ...
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  • Re: Logarithm of transfinite numbers
    ... Tony Orlow wrote: ... bit strings. ... How do you cull the set from uncountably infinite to a countably infinite set ... Naturals are natural, ...
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  • Re: Logarithm of transfinite numbers
    ... Tony Orlow wrote: ... Aleph_0 finite naturals, each of which can be represented ... have 2^aleph_0 strings ... It is only if one includes strings of countably infinite ...
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  • Re: Logarithm of transfinite numbers
    ... Tony Orlow wrote: ... Since in OUR strings there are no such things as "infinite bit ... The number of bits required to represent all the naturals in any basal ...
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  • Re: Logarithm of transfinite numbers
    ... What is that constraint? ... Depends on your exact usage of bit strings to represent naturals. ... The second is as "normal" as any other representation scheme. ...
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