Re: Logarithm of transfinite numbers

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.

Looking at it as logs, we can say log_2(c) = aleph_0. We also seem to
have, from this argument, log_2(aleph_0) = aleph_0. But I think the
latter should somehow be infinitely large but less than aleph_0.

Do you see what I mean?

.

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