My favourite subset of irrational numbers are the uncomputable numbers. Those are the numbers for which no set of instructions can be written down as to how to compute them. By contrast, computable irrational numbers like pi seem very concrete (at least they have formulas!). Most real numbers are uncomputable. This remarkable fact can be seen simply by noting that the entire set of formulas you could imagine writing down is countable. Therefore, there must be uncountably many uncomputable numbers! Some of them can be defined, however (though not computed). I seem to recall from a book of Scott Aaronson's that one example is 'the probability that a turing machine will stop given its own code as input'. From memory, if that probability could be written down it would solve the halting problem, therefore it must be uncomputable!
Very nice post! One minor nitpick for footnote 2: I think calling it 'uncountable' does not explain so much. It is maybe worth pointing to how one would prove that they are uncountable, that is, every possible counting scheme for these numbers can be proven to be incomplete. Therefore, there is no counting scheme for them.
Thanks Ben! Yes, that's fair enough. That footnote was intended just to serve as a justification of the claim in the relevant sentence in the post, but I've added a link readers can follow to learn about the diagonalisation argument in case they are curious.
My favourite subset of irrational numbers are the uncomputable numbers. Those are the numbers for which no set of instructions can be written down as to how to compute them. By contrast, computable irrational numbers like pi seem very concrete (at least they have formulas!). Most real numbers are uncomputable. This remarkable fact can be seen simply by noting that the entire set of formulas you could imagine writing down is countable. Therefore, there must be uncountably many uncomputable numbers! Some of them can be defined, however (though not computed). I seem to recall from a book of Scott Aaronson's that one example is 'the probability that a turing machine will stop given its own code as input'. From memory, if that probability could be written down it would solve the halting problem, therefore it must be uncomputable!
Very nice post! One minor nitpick for footnote 2: I think calling it 'uncountable' does not explain so much. It is maybe worth pointing to how one would prove that they are uncountable, that is, every possible counting scheme for these numbers can be proven to be incomplete. Therefore, there is no counting scheme for them.
Thanks Ben! Yes, that's fair enough. That footnote was intended just to serve as a justification of the claim in the relevant sentence in the post, but I've added a link readers can follow to learn about the diagonalisation argument in case they are curious.