Re: Logarithm of transfinite numbers

Randy Poe said:

Tony Orlow wrote:
Randy Poe said:

Tony Orlow wrote:
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?

Consider the string with a 1 in all odd positions.

Its value is 2^1 + 2^3 + 2^5 + ....

That is larger than any finite value. I think you agree that
something larger than any finite value is not given the name
"infinite".

sum(n=1->N: 2^(2n-1)) diverges to an infinite value, for sure, if N is
infinite, but is finite for any finite value of N.

sum(n=1,...) 2^(2n+1) is not the same as

sum(n=1,...M) 2^(2n+1) for any value of M.

Right. The first one doesn't have a specific answer. the second allows for some
analysis.


Do any of your naturals achieve any infinite values?

No. And no sum ending at M represents the sum over the
entire set, or the value of the string I am discussing.

Again we have your usual logic:

1. Here I have a collection of things (strings that
have a leftmost 1) that all have property P (their
value is finite).

2. Here I have an object (the string which has a 1
in all odd positions) that we all agree is not in the
collection.

3. Therefore I conclude my object has property P.

Yes it's true that every finite string with 1's in odd bits
has a finite value. Now can we get back to discussing
the value of the non-terminating string?

I am getting tired of this, Randy, and may stop responding to this nonsense if
you don't pay better attention, and stop accusing me of being illogical on
points you can't refute, or even grasp.

There is no finite bit position where the string up to that point can represent
an infinite value. It requires an infinite bit position to achieve that, which
I demonstrated, and to which you've agreed. You have no such infinite bit
positions in any countable string, whether you consider it finite or infinite.
So, no countable string can represent anything but a finite value. Exactly
which statement is wrong?


- Randy



--
Smiles,

Tony
.

Relevant Pages

  • Re: Logarithm of transfinite numbers
    ... Tony Orlow wrote: ... Can all 1's up to any finite position represent an infinite ... There is no infinite position in the string. ... And no sum ending at M represents the sum over the ...
    (sci.math)
  • Re: Logarithm of transfinite numbers
    ... Tony Orlow wrote: ... and is greater than the sum of the string to the right of it. ... How does not being a member of a collection lead you ... infinite value in the set. ...
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  • Re: An uncountable countable set
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  • Re: infinity
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    ... >>2) The point of reference means how we define the counting results ... the string length and the members in the placeholders of the string. ... Note that supernatural numbers are that I have called earlier infinite ... omega 1 is behind our standard infinity in the next infinity. ...
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