Re: Logarithm of transfinite numbers


matt271829-news@xxxxxxxxxxx wrote:
matthias@xxxxxxxxxxx wrote:
matt271829-news@xxxxxxxxxxx wrote:
I can ask the same question in a different way, if that helps...

To represent all the reals you need aleph_0 bits.

To represent all the natural numbers you need how many bits?

In what way do you claim you can represent the entire set of real
numbers with only countably many bits? Countably many bits makes one
real number. Are you saying that somehow all the real numbers can be
coded into a single real number? That seems... unlikely.

No, rather poor wording on my part I'm afraid. I mean you can represent
any real number with aleph_0 bits. Therefore the number of different
real numbers, c, is, in some sense, equal to 2^aleph_0 (I could have
used any base instead of 2).

Now do the same with the natural numbers. To be sure you can represent
any natural number you also need aleph_0 bits. Therefore the number of
natural numbers, aleph_0, is also equal to 2^aleph_0, which doesn't
seem right.

This logic is fallacious, for the same reason I tried to explain
to Tony.

Informally, you can see the fallacy with a finite analogy.

"Now do the same with the natural numbers 1, 2, ..., 20. To
be sure you can represent any of these numbers you need
5 bits. Therefore the number of values in 1, 2, ..., 20 is
also equal to 2^5 = 32, which doesn't seem right."

It doesn't seem right because it isn't right. The bit string
representations of natural numbers, whatever representation
you choose, consist of only a countable subset of the
uncountable bit strings of length aleph_0. For instance
you might choose to represent the natural numbers as
....001, ...010, ...100, ...1000, i.e, infinite bit strings where only
a single bit is nonzero.

Do you have any problem seeing that all though the total
number of countably-long bit strings is 2^aleph_0, the
number of countably-long bit strings where only one
bit is set is aleph_0?

- Randy

.

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