So, proof by induction works in N because there is a clear and defined "next" element. Induction doesn't work for real numbers, since that property is gone.
However, both Q and Z are countable, there are bijections between N and Q/Z respectively, you can give each rational number a corresponding natural number.
Thus, in theory, induction should work both for the whole numbers and for the rational numbers, right?
So, proof by induction works in N because there is a clear and defined "next" element...
Ian Martin
Other urls found in this thread:
Jonathan Myers
How is Q countable?
Dominic Perez
Every set can be well-order with AoC, so induction "works" even for the reals - however while a "next" relation is existing, you don't get the nice structure of +1 of the natural numbers and it would be useless in proofs (and you can't give a formula for it)
You're a retard
Evan Cook
>How is Q countable?
You can map Q to ZxN, and because the cartesian product of countable sets is countable, Q is countable.
Liam Morris
...
David Perez
not OP but how the fuck can you order irrational numbers. Since they're part of the real numbers and there's an uncountable number of them seems like a big roadblock
Jonathan Harris
The lemma you use is de facto a reformulation of the problem
Benjamin Thompson
This. Yes, you can do induction on integers or rational numbers and even on real numbers if you want, but you'd be hard pressed to find a situation where that actually helps you anything.
Jason Price
See a construction of R, there you see how the usual order on R is defined.
The post you quoted is about a well-order on R though, which is a different thing. The existence of such a well-order depends on choice, so we will never be able to explicitly construct one.
Easton Edwards
there's no starting point for q,z,q/z hence why it doesn't work