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by nawitus
4690 days ago
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Your proof seems to only prove that a finite number of programs (described by you) which can produce a finite number of irrational numbers, while there are infinite number of irrational numbers. But we're surely not talking about a finite number of programs. I'm not even sure what the Kolmogorov complexity of a "complex irrational number" means. If you need the sequence of digits and you cannot use an algorithm to produce the digits, then the complexity is infinite? |
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The "countable" part means we can put the set into one-to-one correspondence with the natural numbers. The correspondence starts with 0 mapping to the empty program, then 1-256 mapping to programs of a single byte, then 257-65793 mapping to the programs of two bytes, and so on. This mapping will hit each program exactly once, and it will hit every program because every individual program has a finite length.
Different types of infinities are not intuitive so don't feel bad if the concepts are confusing. These issues troubled lots of very smart mathematicians for decades. The existence of irrational numbers was hugely troubling to Pythagoreans. The existence of uncountable infinities discovered by Cantor was shocking [1].
[1]: http://en.wikipedia.org/wiki/Controversy_over_Cantor%27s_the...