Re: Logarithm of transfinite numbers

Randy Poe said:

matt271829-news@xxxxxxxxxxx wrote:
Tony Orlow wrote:
Randy Poe said:

Tony Orlow wrote:
Randy Poe said:


That's the same as saying all strings of finite length, meaning bit positions
in N, which would be, according to your theory, aleph_0 bit positions, and 2
^aleph_0 strings. That's not an additional constraint at all.

It leaves out infinitely many strings. How is something that leaves
out some of the possible strings not a constraint? You are
not allowed the string with all 1's, nor the string with a 1
in every prime position, nor the string with a 1 in every
multiple of 17, nor ....

Those are all infinitely long strings,

They're infinitely-long strings that have infinitely-many 1's.

so you have not reduced the number of
finitely long strings

I didn't say I'm reducing the number of finitely-long strings.
Where did you get that from?

The argument is this: There are 2^aleph_0 infinitely-long
strings, and we're not using all of those infinitely-long strings.
Therefore we're using less than 2^aleph_0 strings.

There are aleph_0 finite naturals, each of which indexes a bit in a string, so
you have aleph_0 bits, each in a finite position. Any combination of bits in
finite positions yields a finite value, and every finite value is a combination
of bits in finite positions. In other words, the set of finite bit strings,
which is the set of all combinations of 0's and 1's in finite positions, is the
set of representations of finite naturals. Given aleph_0 finite bit positions,
each corresponding to a finite natural, a string of this length has 2^aleph_0
possible values, each finite. But, the set of finite naturals has aleph_0
elements, not 2^aleph_0, and you have a contradiction.

Now, you are going to claim that there are an infinite number of finite bit
positions, but even if that were true, it's irrelevant. Consider a 1 in any
finite bit position. The value of this 1 is the sum of the values of all 1's in
previous bit positions, plus 1. If that position is n, the largest value you
can get in n bits is 2^n-1. So, any string up to any finite bit position, which
is what you have, is a finite value.

There are no infinite bit strings under discussion.

I'm quite interested in this stuff, so I'll bite again. By "infinite
bit strings" do you mean bit strings that never terminate?

Tony uses words vaguely and in his own way, so I don't
know what *he* means. The rest of us in this discussion
mean that the natural numbers N form an index to the
string: there's a bit in the n-th position for every n in N.

Are all of those bits, then, in finite positions? Can any of them represent an
infinite value, if the interpretation of the nth bit is 2^n? Can all 1's up to
any finite position represent an infinite value? Nope. Your countably infinite
bit string can only represent finite values. There is no infinite position in
the string.


If so, then
isn't the whole point that we *are* talking about bit strings that
never terminate? Each bit string is a string of "length aleph_0"; i.e.
it never terminates. Amongst those strings there are infinitely many
that have "infinitely many 1's" - in other words, if you took out all
the 1's and wrote them one after another then you would never finish
writing them.

That's what I mean by those terms.

And none of those strings can map to a natural number?

No, not in the standard binary representation of natural numbers,
all of which are finite.

It's possible I am viewing this at a more naive level than you.

I doubt it. You've already shown the ability to follow a
mathematical argument about this stuff.

Funny, you think he's following you, and yet, he asked me the question, as far
as I can tell, and hasn't objected to what I've said. He says, "the penny
dropped", and you declare victory. Whatever shakes your tree, Randy. Try
responding directly to my statement above. Where did I go wrong? Is there any
bit in the string which can terminate an infinite value? Not if all positions
are finite.


- Randy



--
Smiles,

Tony
.

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