Re: Logarithm of transfinite numbers
- From: Tony Orlow <aeo6@xxxxxxxxxxx>
- Date: Mon, 20 Mar 2006 12:02:05 -0500
Virgil said:
In article <MPG.1e84b7c66271a2f898ab0f@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
Randy Poe said:
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?
YES!
Can any of them
represent an infinite value
NO!
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.
What is the allegedly finite value that TO assigns to the string with
1's in all infinitely many postions?
There aren't infinitely many positions. There is a finite but unbounded
sequence of bits, which you call countably infinite. That's no matter. The
string you talk about, with 1's in all finite positions, in binary, would be
your largest finite. So, YOU tell me what that is.
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?
That appears to have happened well before TO's appearance in sci.math,
and is apparently uncurable.
You didn't answer my question above. How can you have an infinite value in a
string with only finite positions? You can't. Up to any finite position, the
string still has a finite value. You don't have infinite positions. Therefore,
you don't have infinite values in the set of possible values for your countably
infinite bit string.
--
Smiles,
Tony
.
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