3 ms·
I guess the following won't work then. 0 / 0.1, 0.2, 0.3...0.9 / 0.01, 0.02, 0.03...0.09, 0.11, 0.12, 0.13...0.19, 0.21...0.99 / 0.001, 0.002, 0.003...0.009, 0
by danbruc 9y ago
I guess the following won't work then.
0 / 0.1, 0.2, 0.3...0.9 / 0.01, 0.02, 0.03...0.09, 0.11, 0.12, 0.13...0.19, 0.21...0.99 / 0.001, 0.002, 0.003...0.009, 0.011, 0.012...
And when ever you list x, you also list -x, 1/x, and -1/x. Except for 0 of course. And you have to throw in +1 and -1 somewhere. Somehow it feels like you should cover all the reals eventually but that can't be true because that would imply there are only countable infinite many reals. It actually looks more like a convoluted version of the standard way to enumerate the rationals by visiting diagonals in order of increasing distance from the origin on a two dimensional integer lattice.
- giomasce 9y agoIf you could provide some way to construct the WO we could ditch AC. I guess it is not happening soon...
- Retra 9y agoYeah, that's obviously just an ordering of the rationals.
- danbruc 9y agoI don't think it's that obvious, actually I think it's not actually an ordering of the rationals. If this were an ordering of the rationals, then it better included rationals with an infinite number of non-zero decimal digits, for example one third. But if this would include one third, then I am unable to see why it wouldn't also include square root two. Why would this be able to include repeating decimal expansions but not non-repeating ones?
- Retra 9y agoYou're right, I misread what you wrote. You're actually only covering a small portion of the rationals. Either way, to order the reals, you have to include the non-computable ones, which cannot be written down in any systematic way. This also means that ordering and equality relations cannot be systematically defined either.