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?

Yes. You can choose an arbitarily large finite value M, and there will
be a value in your sequence 2^M-1 whose binary representation
has M one bits.

This is true for arbitrarily large finite M. That's what it means to
"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,...?

No matter how large a finite M you pick, there will be values
in the sequence with that many bits and more. All of which
have finitely many non-zero bits.

- Randy

.

Relevant Pages

  • Re: Logarithm of transfinite numbers
    ... not see why only those strings with finitely many non-zero bits ... strings that map to natural numbers cannot increase without bound. ... non-zero bits represent naturals? ...
    (sci.math)
  • Re: Logarithm of transfinite numbers
    ... but finitely many bits are zero represent naturals. ... Aleph_null such strings. ... Every element in that sequence begins with a infinite number ...
    (sci.math)
  • Re: Logarithm of transfinite numbers
    ... but finitely many bits are zero represent naturals. ... Aleph_null such strings. ... not see why only those strings with finitely many non-zero bits ...
    (sci.math)
  • Re: Logarithm of transfinite numbers
    ... but finitely many bits are zero represent naturals. ... Aleph_null such strings. ... Every element in that sequence begins with a infinite number ...
    (sci.math)
  • Re: Logarithm of transfinite numbers
    ... but finitely many bits are zero represent naturals. ... Aleph_null such strings. ... Suppose we start with one bit, mapping to the numbers 0, 1. ...
    (sci.math)