Re: Logarithm of transfinite numbers

Randy Poe said:

Tony Orlow wrote:
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.

The point is that because your analysis of the second one
does not ever consider the first one, your conclusions about
the second one do not bear on the first.

No more than analyzing dogs exclusively allows you to
draw conclusions about cats.

The second is sfficient to draw a concluson about the elements which has
bearing on the set size, as I showed.


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,

I'm paying plenty of attention, that's why I can keep identifying
the same pattern.

That's why you keep thinking something happens in the set of elements when
there is no element in the set at which it can possibly happen. Very attentive.


and stop accusing me of being illogical on
points you can't refute, or even grasp.

My refutation remains that #1 and #2 do not imply #3.

Your 1, 2 and 3 are yours. Those are not my statements, but an obnoxious
attempt of yours to discredit my point. I am not discussing your bastardization
of my argument.


I grasp that you think you can draw conclusions of
the form #3, but I continue to insist you can't and that
it's just wishful thinking rather than logic that gets you there.

There is no finite bit position where the string up to that point can represent
an infinite value.

Statement #2.

It requires an infinite bit position to achieve that,

Unjustified assertion, and in fact untrue.

If it does not happen at any finite position in the set, but happens in the
set, then it happens at an infinite position, so the statement is perfectly
well justified and undeniably true.


All we know from your first statement is that no terminating
sum can represent an infinite value. It does not follow that
the sum over all finite bits is finite.

Forget temrinating sums. You are hung up on the notion that you can prove
anything you want based on the lack of a lrgest finite. Grow up. If there is no
finite natural at which point the set is infinite, and finite naturals are the
only points in your set, the set can't become infinite at any point in the set.


which I demonstrated,

(wished for over and over and over and over...)

and to which you've agreed.

I have never agreed that "it requires an infinite bit position
to achieve that".

Yes, you did. You agreed that every bit in a finite position represents a
finite value, that all bits to the right can only sum to 1 less than that
finite value, and so the entire string up to and including any finite bit
position can only have a finite value. Since you only have finite bit positions
in the set, you can only have finite values, because this fact is true for
EVERY SINGLE finite bit position in your countably "infinite" string.


You have no such infinite bit
positions in any countable string, whether you consider it finite or infinite.

Right.

So, no countable string can represent anything but a finite value. Exactly
which statement is wrong?

The one that says "it requires an infinite position to
achieve that". Also the one that says I agreed with that
absurd claim.

Then explain how something occurs in a set of finite bit positions which cannot
occur at a finite bit position. Stop dancing around the question.


- Randy



--
Smiles,

Tony
.

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