Re: Logarithm of transfinite numbers

Tony Orlow wrote:
David R Tribble said:
Tony Orlow wrote:
1. 0 is a real number
2. If x is a real number, then 2^x and 2^-x are real numbers.
David R Tribble said:
You omitted the part where you claim that all reals can be derived by
recursively applying rule 2.

They can't. An uncountable number of reals are never derived.
For example, applying rule 2 repeatedly will never result in even a
simple number like 3.

But please, oh please, prove me wrong.
Tony Orlow wrote:
I wish I could give you a good answer. Hopefully, at some point, I can get to
implementing the H-riffics on the computer, and make it do my survey for me so
I can derive rules once the intuition is established, but these gradal-type
numbers are difficult. Right now, I cannot exactly tell you the bit string
corresponding to 3.
I'll give you a hint. It is represented by an H-riffic bitstring of
infinite length. Which means that
a) it cannot be derived from (is not the successor of) any finite
H-riffic; and
b) it has no H-riffic successor (which would required flipping the
"last" bit in the string).



a) is based on your inability to deal with hyperinegers or see that they are the natural outgrowth of the successor function over an actually infinite sequence of increments.
b) is based on the notion that an infinite set cannot have more "after" it, a notion which is flatly counterexemplified by the reals in [0,1]. If it takes an infinite number of iterations to reach 3, then there may be an infinite number after 3 as well. Wherever 3 is, 8 is its successor. I'll have to write up a real proof that the H-Riffics cover all the reals, but I'm working on other things right now.


If it takes an infinite number of iterations to reach something, how do you
know when you've reached it? (And how do you stop at it?)

Matt
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