Counting the Continued Fractions, Which Count the Real Numbers
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Re: Counting the Continued Fractions, Which Count the Real Numbers
#2I realize this isn't an expected possibility, but I've reached a point where I am not able to uncover where it might have gone wrong.
I've tried to keep things direct and in point form, to (attempt to) make it easier to spot any assumptions made.
This might not make it as readable as it could be, and specifically avoids the formalization required for a proof, but I have more to add if it holds up.
Thank you in advance for any feedback provided or questions to clarify on.
Re: Counting the Continued Fractions, Which Count the Real Numbers
#3There's a much better introduction to write here, so I'll jump right into avoiding one. I realize this isn't an expected possibility, but I've reached a point where I am not able to uncover where it might have gone wrong. I've tried to keep things direct and in point form, to (attempt to) make it easier to spot any assumptions made. This might not make it as readable as it could be, and specifically avoids the formal…
Re: Counting the Continued Fractions, Which Count the Real Numbers
#4There's a much better introduction to write here, so I'll jump right into avoiding one. I realize this isn't an expected possibility, but I've reached a point where I am not able to uncover where it might have gone wrong. I've tried to keep things direct and in point form, to (attempt to) make it easier to spot any assumptions made. This might not make it as readable as it could be, and specifically avoids the formal…
are you sure you're counting reals? not rationals?
If that holds true without missing any possibilities, then this maps to the continued fractions. I've gone through several such mappings that missed at least one before, so looking to see if this one has any such as well.
https://math.libretexts.org/Bookshelves/Combinatorics_and_Di... A bit before and after section (6.1.13) covers how the continued fractions then provide a counting of the reals, provided any finite representations are > 2 for the last number.
So well I'm not sure it all holds up, it does appear that way.