Re: Logarithm of transfinite numbers

In article <MPG.1e82410e145bdb4b98aaf1@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:

Randy Poe said:

Tony Orlow wrote:
Shmuel (Seymour J.) Metz said:
In <1142386560.427503.17760@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, on
03/14/2006
at 05:36 PM, matt271829-news@xxxxxxxxxxx said:

To represent all the reals you need aleph_0 bits.

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

Aleph_null. Of course, there is a constraint on which
combinations represent integers.



What is that constraint?

Depends on your exact usage of bit strings to represent naturals.

I can think of two obvious representations. In one, the constraint
is that only finitely many bits are nonzero. In another, the
constraint is that only ONE bit is nonzero.

The first is the same as there being a finite number of bits overall,
assuming all bit positions are finite, and so doesn't help get any
better answer than the self-contradictory aleph_0 as a count of bits.

Is TO saying that it takes more that a countable set of strings to
represent the countable set of naturals?
Is TO saying that there is aANY natural that requires aleph_0 bits?
Unless TO is claiming one of these, there is no problem to solve.



The second is not the normal binary numbering scheme, and so doesn't
address the question.

The second is as "normal" as any other representation scheme. All
representation schemes are artificial, and have no intrinsic
significance. They are used merely for convenience.


Wouldn't any countably infinite number of bits produce an
uncountably infinite set of strings?

If you used all possible combinations of 1's and 0's. But remember
a couple of sentences ago that we're restricting the combinations
we use.

Yes, I answered that silly suggestion twice

And was twice sillily wrong.


I asked exactly what constraint you were suggesting, and one
suggestion is the same as having finite strings, while the other is
to change the number system entirely and not address the question at
all. The second is rejected as irrelevant and a diversion. The first
doesn't solve anything, because it still leads one to claim there are
aleph_0 bits, since it is the set of all finite bit positions, and
2^aleph_0 strings.

Then TO must be claiming that the English language contains an infinite
number of words, since there is no limit on the number of letters a word
can have. A silly sugstion, but no more so than TO's that every string
must be considered when not every one is needed.





What number of bits produces all the natural binary numbers?

What's the difference between a duck?

The question is as relevant as TO's and its answer is equally relevant.

So, despite the fact that we have a very specifically defined binary
representation system for the natural numbers, where each unique
combination of bits represent a unique value, you would like to
pretend that there are some finite naturals that use strings of
length aleph_0, but that not all unique strings of length aleph_0
represent a unique natural number?

Name any natural number whose binary representation uses more 1's than
its value. In fact, except for 0, 1 and 2, name any require use as many
binary DIGITS as its value.




Sorry, you have to justify that
statement, which you can't.

It is trivially true outside of TOmania that no natural requires an
infintie string to represent it in any base, including base 1.


If you want to cull out some vast
majority of an uncountable set of strings to leave a countable set,
then you are going to have to do a lot better than pretend they don't
all use the "last" bit.

They all, except 0, have in their representations a first, or most
significant, non-zero bit, followed by a finite number, possibly 0, of
bits in other bit positions..





The number of bits required to list the set of naturals is aleph_0.

Great, then there's 2^aleph_0 strings, and the naturals are
uncountable. Fantabulous!

The TO is implying that he can construct a bijection between the
naturals and some uncountable set. Let him present it here!

On the other hand, it may only mean that there are injections from the
set of naturals to the set of infinite strings without ever there being
any surjections.

So that TO must provide us with his alleged bijection or be shown wrong.


You really can't tell the difference between that statement and the
statement "every bit string of length aleph_0 represents a
natural"? The second statement is false. You rightly point out that
the number of bit strings of length aleph_0 is larger than aleph_0.

But what's so unfathomable about the idea that from this set larger
than N, there's a subset of size aleph_0?

Of course there's a subset of size aleph_0, your set of all binary
strings of length aleph_0 with exactly one non-zero bit. So what?

There is also a subset of size aleph-0 of all strings with only finitely
many non-zero digits, and a natural bijection between this countable set
of strings and the countable set of naturals.

TO unnaturally denies the existence of this natural bijection.
.

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