Let's discuss Mathematics
- Thread starter ParadigmShifter
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ParadigmShifter
Random Nonsense Generator
Right on, comrade.
On countably infinite:
The normal defition is: a set A is countably infinite if there is a bijection between A and N.
What you're probably looking at is: a set A is countably infinite if and only if there is a successor function S:A->A which satisfies the Peano axioms (http://en.wikipedia.org/wiki/Peano_axioms#Models). Proving that these are equivalent should be pretty easy.
The normal defition is: a set A is countably infinite if there is a bijection between A and N.
What you're probably looking at is: a set A is countably infinite if and only if there is a successor function S:A->A which satisfies the Peano axioms (http://en.wikipedia.org/wiki/Peano_axioms#Models). Proving that these are equivalent should be pretty easy.
ParadigmShifter
Random Nonsense Generator
Yeah, but I was attempting to explain it in layman's terms.
Thanks. But, you sure that every well-ordered set gives rise to an induction principle? I ask, because according to Zermelo's well-ordering theorem, every set can get well-ordered. So, R can get well-ordered. Thus, there would have to exist (in the hypothetical set of all proofs) induction proofs on R. Know of any? Can that really happen?
Weird fact #4 If the rationals come as the universe of discourse, there exist *no* reals between any rational numbers, since all reals get considered as nonsense.
ParadigmShifter
Random Nonsense Generator
This link (from your link) seems to be directly contradictory to that though
http://planetmath.org/encyclopedia/WellOrdering.html
EDIT: OK, not with the "usual order".
We can always order a vector space Rn by doing a lexicographical sort (I think). I rely on that all the time in computer programming.
http://planetmath.org/encyclopedia/WellOrdering.html
EDIT: OK, not with the "usual order".
We can always order a vector space Rn by doing a lexicographical sort (I think). I rely on that all the time in computer programming.
ParadigmShifter
Random Nonsense Generator
Or two headaches, if you use Banach-Tarski.
There is a mathematician with maximal headache.
ParadigmShifter
Random Nonsense Generator
Because he ate far too much cake?
ParadigmShifter
Random Nonsense Generator
And forgot his own norm...
Because infinities come in different sizes. You can't actually have a bus with infinite people, but you can talk about it. You can assign a natural number. But you cannot assign each person a real number, without having numbers left over.
Ergo you cannot have more than a countably infinite number of people.
On the other hand, you can have more than a countable number of groups of people, so long as each person can belong to multiple groups.
ThinkTank
RL Addict
Yes, me sure. See here http://planetmath.org/encyclopedia/WellFoundedInduction.html for the satement of the principle and here http://planetmath.org/encyclopedia/ProofOfTheWellFoundedInductionPrinciple.html for the proof.
As noted by dutchfire, this requires the axiom of choice. So then you know there is a well-order of R; but actually specifying one gets you into fundamental problems on what you can express in set theory, see here is-there-a-known-well-ordering-of-the-reals?
Yeah ok. Well there isn't such a function on the reals anyway.
Weird fact #1: Between every 2 rational numbers there is an irrational number
Weird fact #2: Between every 2 irrational numbers there is a rational number
Not that weird those facts
But weird fact #3: There are infinitely many more irrationals than rationals
EDIT: Ok, add "distinct" to the above. Beer #4
Admittedly, my knowledge of infinities is based only on what I can logically fathom for myself, but surely #3 contradicts #1 and #2? I mean, can't you just pair them up? For example take the first rational number, and pair it up with the irrational that comes in between it and the next rational number, and so on...?
What's the first rational number greater than zero?
IdiotsOpposite
Boom, headshot.
Is there a way to create a function that generates the natural numbers that are NOT a sum of two Fibonacci numbers (or 1)? I believe the first one is 10, if that helps.
Obviously, if a number is Fibonacci and not 1, it is also a sum of two Fibonacci numbers.
Obviously, if a number is Fibonacci and not 1, it is also a sum of two Fibonacci numbers.
0 is a rational number, so we can use that for the first one... I fail to see your point?
Well 10 is 2+8...?
Mise
isle of lucy
First one's 12. You won't be able to generate numbers between F(n)+F(n-1) and F(n)+F(n-2) just by adding 2 (different) fibonacci numbers. So for example you won't be able to generate any numbers between 144+55 and 144+89.
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