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For extra craziness: you can extend the list argument to list all (finite) programs that produce increasing but bounded sequences. We'll call this the computab
by moomin 13y ago
For extra craziness: you can extend the list argument to list all (finite) programs that produce increasing but bounded sequences. We'll call this the computable numbers. The transcendental number described in the article is computable.
But as I've just mentioned, the computable numbers are still countable. This really hurts my head. There's an uncountable set of numbers that aren't expressible in any closed form whatsoever.
Continuous numbers get weirder the harder you look at them.
- ColinWright 13y agoProblem there is you have to list all finite terminating programs that produce increasing bu bounded sequences. Then you start to have to worry about being really, really careful in your definitions. See here: https://news.ycombinator.com/item?id=6967954 https://news.ycombinator.com/item?id=6967954 That's a classic "paradox" being created by having insufficient care in the definitions. But yes, most (in a technical sense) reals can't be described or computed, and yes, the reals get more and more weird the harder you look at them. In some sense you never really understand them, you just get used to them. Mostly.
- poizan42 13y agoI don't see the problem here. It shouldn't be a surprise that listing the computable numbers is itself uncomputable as it is equivalent to the halting problem - to determine whether an algorithm will either keep on producing new decimals or halt at some point is clearly undecidable. It's when you introduce things as constructible/defineable numbers that things gets hairy.