#Sets
16 messages · Page 1 of 1 (latest)
note that all transcendentals are irrational and that algebraic numbers are countable
but the question itself is meaningless
Maybe its meaningless but if we were asked nevertheless to construct the most inclusive set that is countable, it would be Q union {algebraic irrationals} right?
there's no such thing as the most inclusive countable numerical set, you can always add just an element countably many times to get a countable set
Flippo is still right, but there are also the computable numbers.
recursive reals?
if construction here means algorithmical explicit one (with the real condition over the numerical set), then we need to take account of certain other definitions
okay I see where I'm wrong then
what are the computable numbers?
Numbers that can be calculated to arbitrary precision in a finite number of steps
For example, e and pi
oke I see
anyways thanks for clearing my q up
.solved