Re: Logarithm of transfinite numbers

matt271829-news@xxxxxxxxxxx wrote:
Shmuel (Seymour J.) Metz wrote:
In <1142422359.828461.187230@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, on
03/15/2006
at 03:32 AM, matt271829-news@xxxxxxxxxxx said:

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.

And, indeed, it isn't right, because only those strings in which all
but finitely many bits are zero represent naturals. There are only
Aleph_null such strings.


Several people have said similar things, but unfortunately I still do
not see why only those strings with finitely many non-zero bits
represent naturals. I do not see why the number of non-zero bits in
strings that map to natural numbers cannot increase without bound. For
example, in the sequence 2^1-1, 2^2-1, 2^3-1,...

Can the number of non-zero bits increase without bound? If so, how does
this square with the statement that only those with finitely many
non-zero bits represent naturals? If not, what happens with the
sequence 2^1-1, 2^2-1, 2^3-1,...?

What do you think happens with that sequence? It looks like
...0000001
...0000011
...0000111
...0001111
etc.

Every element in that sequence begins with a infinite number
of 0's, and ends with a finite number of 1's. There are an
infinite number of such sequences.

You seem to have a problem with the fact that there are
an infinite number of finite natural numbers. Perhaps
you should try to clarify what 'infinite' and 'finite'
mean to you, and see if it is consistent with the
mathematical definitions.

Stephen


.

Relevant Pages

  • Re: infinity
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  • Re: Logarithm of transfinite numbers
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