Re: Logarithm of transfinite numbers


Tony Orlow wrote:
Randy Poe said:

matt271829-n...@xxxxxxxxxxx wrote:
What value, if any, can be ascribed to the logarithm of aleph_n?

If you define it, we can see if anything fits that definition.


Take log_2(aleph_0) as an example.

log_2(aleph_0) can't be finite, and it can't be bigger than aleph_0, so
it has to equal aleph_0.

You're drawing conclusions about something you haven't yet
defined.


But log_2(aleph_0) = aleph_0 implies 2^aleph_0 = aleph_0, whereas in
fact 2^aleph_0 = aleph_1.

Ah, so by "log_2(aleph_0)" you mean "the cardinality of a set
S such that |P(S)|, the cardinality of the powerset of S, is aleph_0".

No such S exists.

Where did it go wrong?

You haven't come up with a definition such that anything
meets that definition. There is no such thing as "log2(aleph_0)".

- Randy


That seems kinda weak, Randy, no offense. "Undefined". Hmph! Sorry.

We have the set of natural numbers, of size aleph_0. Let's represent each one
in binary. How many bits will we need to represent aleph_0 naturals? Well,
given N=S^L

And L is the thing that is undefined here. No L satisfies that.

, where N=aleph_0 and S=2, aleph_0=2^L, so L=log2(aleph_0), and that
is the length of binary strings you need to represent all the naturals.

No finite quantity is enough to represent all the naturals, so L is not
any finite value.

Is L infinite? Well, no, because the set of infinite-length binary
strings has more than aleph_0 elements.

So since L is neither finite nor infinite, it doesn't exist.

It's valid.

Hint: Declaring nonsense as "valid" does not actually make it
valid, any more than declaring a paper boat seaworthy actually
makes it so.

Is the example concrete enough for you? How can you say this number does
not exist, given this example?

Because no number has the properties needed for your example.

- Randy

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