You should be able to solve this.

Create a method to choose a random integer between 0 and infinity such that no integer is more likely to be chosen than any other.

Yes it is possible, I don't care what your probability book says (Notice I didn't even use the word "probability").  You may assume the axiom of choice.

2?

x = anything, where x is the random integer

Another one

x/0=y, where y is the random integer

anything divided by 0 can be anything because any number of 0s fits into 0

>>3
>>4
missed the point completely. OP said <i>choose</i>. you haven't come up with an actual method, just different ways of expressing a random number.

>>5
<i>choose</i>

aww.. fuck me :P

Let F be the set of all continuous functions from [0,inf] to [0,1]. Let f be an biyective function from c to F.
Consider g a (continuous) function from [0,inf] to [0,1] defined by:
g(x) = sum from i=i to w of (1/2^i)f(i)(x). Let y = inf { x in R : g(x) is maximum }.

Choose n = min { m in N : m > y}

Is that what you wanted OP?

Flip a coin forever so that heads is 1 and tails is 0.  You will generate your number in binary.  You won't ever finish, but that's ok because you math weirdos defined real numbers as limits of inifinite sums anyway.

>>8
>You won't ever finish

The axiom of choice says you can ignore that little detail; it just says it is possible to get infinitely many coin flips, or flipping an infinite number of coins. (You can see now why the axiom is controversial.) Other than that you have the solution.

>>1
Isn't this basically a straightforward application of the axiom of choice? You have an infinite set of bins, each containing the digits 0-9. Select one from each bin. Concatenated together, the digits form a number. More formally, value = sum{n=0}{inf}(b_n * 10^n) where b_n is the digit selected out of the nth bin. This is the same solution as >>8 with a different radix.

This is a possible algorithm, though I realized I haven't really proven that the probability is equal. My set theory is a little rusty.

>>7
I don't know if I understand.  In the third line, what's w or "f(i)(x)".  F is going to have higher cardinality than the reals, so if the "c" in the first line is the complex numbers, there is no such bijection.

>>8
>>9
You can't do the coin flip thing, since you're not guaranteed to ever get a finite number.  At some point, you'd have to have chosen a finite number of digits, and get nothing but zeros from then on.



I'll post the answer later today.  If you want a hint, look at this: http://en.wikipedia.org/wiki/Vitali_set

>>9
lol quote failure.

w is the first infinite ordinal (The natural numbers). c is the first ordinal with the same cardinality as the real numbers. f is a function defined from c to F, so f(i) is a continuous function. F has the same cardinality as the reals, there is a easy injection from F to Q^Q, where Q is the set of rational numbers.

>>12

>>10

>>12
>F has the same cardinality as the reals
I'm not sure about that, but I'll go with it for now.

So let me see if I get this.  Of the uncountably infinite set of functions from [0,\infty) \rightarrow [0,1], you're selecting a sequence f_i := f(i) (how? just any sequence?), and setting g(x) := \sum_{i=1}^{\infty} f_i(x).  Then g is a uniform limit of continuous functions, and so is continuous, and the output is the next integer above the smallest local maximum.

How do you know g has a local max at all?  Also, where's the "random" part of it?  Are you selecting the f's at random?  How are you doing it?

>>10
well, it's later today.  how do you manage to finish choosing a number, then, if each is truly equally likely to be chosen.

>>15
By using the standard construction of unmeasurable sets as in the wikipedia article (or, better yet, this version: http://everything2.com/title/unmeasurable%2520set), you can partition the unit interval I=[0,1] into a countable union of disjoint subsets, each of which is identical to all the others, only translated left or right by some real number, so that they're all the same size (even though they don't really have a "size", strictly speaking, since they're unmeasurable).  Number these sets V_1, V_2, ....

Now just select a random number x \in [0,1] (which can be done with the coinflip thing, since an infinite number of non-zero digits *after* the decimal point is OK), and let your random integer be the index i of the set V_i that x happens to be in.

Since you can't have a probability space that contains all the V_i, you can't talk about the "probability" of choosing 7567865 (or whatever), but since the chance of choosing 7567865 only depends on the size and shape of V_{7567865}, which is the same as all the other V_i's, no number is more likely to be chosen than any other.

How's that any better than the flipping the coin thing?  We imagine that all infinity digits are chosen at once anyhow, not simply from left to right as though we were reading them.

>>17
It's obviously better, because it always converges, the other way doesn't.

In fact the other diverges with probability 1.

>>18
I don't see that it does, for any accepted meaning of convergence.

it's impossible to do

>>19
You flip a coin an infinite amount of times to decide a, for the sake of simplicity, binary decimal expansion.

So You have numbers X_i

X_1 = 0.a_1
X_2 = 0.a_1a_2
X_3 = 0.a_1a_2a_3

etc

We define the number we pick to be X = Lim X_i as i-> inf

Now obviously X_i is a cauchy sequence and |X_i - X_j| is at most 1/2^i for all j > i

So the sequence converges. This is all fine with the axiom of choice.


The other way though, if we pick digits going to the right, you can easily see that the sequence you get is not cauchy and doesn't converges.


As a very simple example what value do you give to the number if all your coin flips are heads?

>>19
There's nothing that needs to converge in the first place.  You just pick a random number in [0,1], and then pick the set that number is in.

throw a ball out into space, and count how long it takes to get back to you in seconds.
i guarantee you, every number from 0 to infinity is equally as unlikely to be the number of seconds it takes for the ball to get back to you.

>>23
oh right, uh, just to cover a technicality, round to the nearest second

>>23
<i guarantee you, every number from 0 to infinity is equally as unlikely to be the number of seconds it takes for the ball to get back to you.
<i guarantee you,

Prove it.

>>14
You're right, g doesn't necessary have a max; instead restrict F to the sets of functions with at least one maximum and just take f(0) as g. The random part is at the moment of taking f, which i dont have a flipcoin/throw a ball in the universe argument (And maybe will never have it, as c doesn't have a physicall representation)

PD: F being of size c is a consecuence of the fact that two continuous functions are equal iff they are equal in a dense set, in this case, the rational numbers. So c <= |F| <= |R^Q| = c^w = c

>>21
Where does Axiom of Choice actually come in? AC implies "there exists" such an infinite sequence, but you're already defining one hence don't need it.

>>21
That part obviously converges, but that's just an intermediate result. There's no convergence after doing the mapping.

>>22
Right, so no convergence. It's all or nothing.

>>16
Assuming that "countable union of disjoint subsets" refers to the method on the everything2 page, you can't say that they have the same size and shape after you've cut them to the [0, 1] range.

I wondered if you could choose the elements in X so as to all have a difference of more than 1, but then you'd have covered a real interval with no more than one element from each of an allegedly countable amount of sets, which can't be good.

Meh, too hot right now, will think more about it later.

>>28 All the W_i have the same external measure (In fact, they are constructed as translations modulo 1 IIRC). In that way you have the 'same probability' of being chosen, just as the OP said.

>>29

*translations by a rational number modulo 1 of some set

>>9 here...

>>10
You can't do the coin flip thing, since you're not guaranteed to ever get a finite number.  At some point, you'd have to have chosen a finite number of digits, and get nothing but zeros from then on.

Your second statement is correct here; after a certain point, you'd need an infinite number of zeros. This is not a problem. This is in fact what the axiom of choice allows us to do.

>>17
>>18
What does this have to do about convergence?

>"method...random"
wut

>>31
>This is not a problem. This is in fact what the axiom of choice allows us to do.

Uhhh, no.

>>27
Might not, I'm never sure, just use it as a cover all.

But I'd say picking the a_i uses choice, cause you're basically defining a choice function, but every A_i is just the set {0,1}, so maybe you're ok.

>>33
Wtf? Of course it does. Are you going to provide any rational explanation for why you think it doesn't?

>>35

Why do you think it does??  You're looking at the (countable) set of all infinite sequences of coinflips which are eventually always zero, and just claiming that AC lets you pick a random element of that set, which is basically the same problem we started with.

http://en.wikipedia.org/wiki/Axiom_of_choice

>>36
You're looking at the (countable) set of all infinite sequences of coinflips which are eventually always zero
Yeah...

and just claiming that AC lets you pick a random element of that set
It does...

which is basically the same problem we started with.
What the fuck? No it's not, it a valid answer to OP's question. "Create a method to..." well there's the fucking method!

>>37
>It does...
You don't need AC to pick an element from one infinite set.  Besides, AC doesn't say anything about the distribution of the element chosen.

>What the fuck? No it's not, it a valid answer to OP's question. "Create a method to..." well there's the fucking method!

The problem was to pick an element uniformly at random from the integers (a countable set) which is the same thing as picking an element at random from the set of sequences of coinflips that are eventually all zero, since the two sets are in bijection.

This is a cute thread, and I like the Vitali set solution, but isn't this cheating?  The notion of "likelihood" just seems like an alternate phrasing of "probability" and if it is indeed supposed to be a different animal, then it hasn't been defined.

I don't feel we can speak about one choice being "more likely" than another in any definite way without resorting to some notion of measure.  The solution OP proposed seems to rely on saying that sets that are congruent modulo translation have the same measure - even though they are nonmeasurable!  For this to work we need to assert that a number randomly selected from the unit interval is as 'likely' to appear in one element in the partition as in any other - but again, this seems to resort to the idea of measure.

imo, this is a clever way of dodging the question rather than a solution to the question as asked.  I'd love for someone to convince me otherwise, though!

>>39 here again, with a more serious objection to the alleged "solution" in
>>16
You assert that if a subset of [0,1] is obtained from another by partitioning it into finitely many pieces and translating them, then if we choose a random number in [0,1] it is "as likely" to appear in either one.  If we apply the same reasoning to [0,1]^3 we run into problems with Banach-Tarski.

Namely, if we have three disjoint balls in [0,1]^3 of the same radius, and we choose a random point in the unit cube, it should be "equally likely" for this point to lie in balls 1, 2, or 3.  But with AC, balls 2 and 3 can be obtained from 1 by partitioning 1 into finitely many pieces, and translating them; so by your reasoning it should be "as likely" for this point to be in ball 1 as in the union of balls 2 and 3.

Anyway, I sort of doubt what OP is suggesting is possible without an utterly trivial notion of "likelihood", but this is still a fun question to think about.

>>39
>cheating

OP here: A little bit, maybe. ^^ (notice I avoided the word "probability")  It makes intuitive sense, though, imo.  I think maybe if you threw Vitali sets in as extra generators of your sigma field, and used outer measure instead of lebesgue measure for your probability, you could still get a finitely additive probability space such that the probability of any single integer is zero. I don't know much about how finitely additive probability works (does your sigma field still have to be closed under infinite unions?) so maybe I'm wrong.

I know it's possible to get a uniform distribution on the natural numbers with finitely additive probability (if someone wants, I can try to find the paper I read about it in).  What I was trying to get at though is some kind of algorithm to actually pick one.  This doesn't really work for that, though, since you can't explicitly describe Vitali sets or figure out which one some given real number is in.

>>40
B-T doesn't work in dimension 1, though.

(Alright fine then, maybe I was wrong. :P )

Here are the papers anyway, just to prove it can be done, even if possibly not by that method:

Using Finitely Additive Probability: Uniform Distributions on the Natural Numbers
Joseph B. Kadane, Anthony O'Hagan; Journal of the American Statistical Association, Vol. 90, 1995

Uniform Distributions on the Natural Numbers
Oliver Schirokauer, Joseph B. Kadane; Journal of Theoretical Probability Volume 20, Number 3, September, 2007

Any solution using the Axiom of Choice fails OPs condition. He said "Create a method to...", not "Show that a method exists to...". Since none of the proofs shown using the Axiom of Choice show how to find the element they says exists, the method as a whole fails to show how to find a random natural number.

>>43
Roughly what I was saying in >>27

AC is pretty irrelevent here. It doesn't magically allow you to flip a coin an infinite number of times.

>>43
Well, if a usable method *did* exist, then with probability 1 the number chosen would be too large to actually write down using every bit of matter in the observable universe.

>>41
\sigma-algebras are by definition closed under countable unions, hence the "\sigma".  Otherwise it's just an algebra.  And I'm not positive this is what you're saying, but an outer measure must satisfy

so that if the Vitali sets are in your \sigma-algebra and each have measure 0, then [0,1] has measure 0, which you probably don't want.
>>42
cool, these look interesting and accessible.  Unfortunately I can't read them because I don't have a springerlink subscription :(

This is what mathematicians waste their time thinking about.

>>47
yeah, and we get paid to do it. ain't life great?

>>46
But the point is that in a finitely additive space, you can't add up all of them at once, so you don't run into the problem of [0,1] having measure 0.  The first tricky bit I can see would be proving that the outer measure is actually additive instead of subadditive in this case, which definitely would still be required for a finitely additive probability.

>>47
I prefer stuff that's a bit more concrete myself usually, but yeah.

>>48
>ain't life great?

Yeah, until that moment when you look out the window at the ripped, suntanned dude with the perfect blonde wrapped around him, then you look down at your stack of papers and books and have one of those "What the fuck am I doing wasting my life on this crap?" moments.

rand() * (1/0)

/thread

You should be able to solve this.
lie

>>38
You don't need AC to pick an element from one infinite set.
The problem was to pick an element uniformly at random from the integers (a countable set) which is the same thing as picking an element at random from the set of sequences of coinflips that are eventually all zero, since the two sets are in bijection.
God damnit. YOU brought up this representation of the question fuckhat, not me. I didn't say shit about sequences. YOU rephrased the question, then accused me of having done it, and then said it wasn't useful, and then accused me of using AC to wave it away.

I don't disagree that YOUR sequence representation is not useful. THAT'S WHY I DIDN'T FUCKING SAY IT!

Fucking listen. MY statement is that picking an element uniformly at random from the integers is the same as picking a random integer from a finite set an infinite number of times by construction via decimal representation. In other words I have an INFINITE number of bins (countably many numbered 0 and up), each containing a FINITE set. AC allows me to choose an element from EACH BIN, even though there are infinitely many. I can certainly choose an element from ONE BIN uniformly randomly since it has finite contents. Therefore it is possible to choose uniformly randomly one element from each bin. Using this as the decimal representation of the number, I get a uniformly randomly generated finite number between zero and infinity.

Notice I didn't say fuck all about sequences? It's not a fucking goddamn sequence. THIS HAS NOTHING TO DO WITH SEQUENCES.

Notice how >>16 and >>40 both agree that infinite coinflips are implicitly valid by the assumption of AC. The reason you're thinking of it as a sequence is because you can't wrap your head around the concept of 'infinite bins' or 'infinite coinflips'. Well guess what, you've successfully stumbled onto the controversial aspect of AC. Congratulations for demonstrating to the rest of the class.

>>52
facepalm.jpg

lrn2 math. A decimal representation is a sequence, and with probability 1, your method doesn't select a finite number, since there will be infinitely many non-zero digits.

Tard.

>>52
I'm just going to chime in and agree that you have made it abundantly clear that you have absolutely no idea what you're talking about, congratulations.

>>53
no it isn't, the reason the math dinks keep whining that 0.999... means 1 is because the math dinks defined real numbers as the sum of the sequence rather than the sequence itself.  It's their ball, and they can take it from us and go home no matter how we protest.  Or we can just quietly ignore them because really who gives a shit.

>>54
play your bells in fag town

>>55
Get it right, it's the limit of the sequence, not the sum of the sequence.
.1, .5, .9, .99, .999, ... and
.9, 1.01, .999, 1.0001, .99999, ...
are very different sequences with the same limit.  It seems rather nonsensical to define real numbers as sequences, as you seem to be suggesting.

>>57
It's not nonsensical to represent a real number as the sequence of its decimal expansion. You can certainly represent every real number this way. It's just rarely useful because it's not unique.

>>53
A decimal representation is a sequence, and with probability 1, your method doesn't select a finite number, since there will be infinitely many non-zero digits.
I'm not convinced. It is certainly possible for a finite number to be generated with this method. A finite number is generated if every digit after a certain bound B is zero. Therefore you can fix this by running the algorithm, and seeing if B exists; if not, try again. Eventually you'll stumble on a finite number. It doesn't matter that the probability of ever generating a finite number this way is zero, just as it doesn't matter that the probability of generating any given number is zero.

>>54
I can easily say the same about you; if you're >>38 you've done nothing but come up with strawman arguments, either acting smug when you disprove things totally unrelated to what I've actually said, or just making snide remarks like this without even attempting to contradict anyone. That frustrates the fuck out of me. If you'd like to try your own hand at answering the question then be my guest.

>>57
In set theory and logic real numbers are traditionally defined as equivalence classes of cauchy sequences of real numbers (dedekind cuts are also used).

>>59
I suppose, but they're the same thing.  Two items are in the same equivalence class if they have the same limit, which is equivalent to the "definition" of real numbers I stated.
From a logic point of view, I can see how one would be more fundamental/easier to show from fundamental stuff than the other.
I take it you mean "cauchy sequences of rational numbers".

>>60
Yes, sorry, that was an error.

I'm not OP here, but you're all being fucking retard (Including OP)

As far as I can tell here's OP's argument

I take sets A_i = {0,1,2,3,4,5,6,7,8,9} for all natural numbers i.

I then define a real number in [0,1] by picking a_i a representative from each A_i and forming the number

a=0.a_1a_2a_3a_4......

>>53 I think here you misunderstood what he was saying

I've also partitioned [0,1] into a countable number of sets, labelled V_1, V_2..... using the vitali sets

Then I look at which V_i a is in.

This argument is fine, it all works, not certain it works in the way he wants it to, the very point of vitali sets is all ideas of "probability" is useless for them.


Now, one actual criticism here is that it doesn't use the axiom of choice.

Sure we have an infinite number of sets, and we have to choose an element of each, but I can explicity construct a choice function f(i)=0, ta-fucking-da.

Also the axiom of choice, even if it were applicable here, wouldn't make your choices evenly distributed along [0,1] for any reason.

Luckily you can just do this with normal probability, so we're all fine, picking a random number in [0,1] is just define random variables X_i, all unformly distributed on {0,1,2....9}


and define X = X_1/10 + X_2/10^2 + X_3/10^3.....

nothing to do with the axiom of choice, the only place you use it here is in constructing the vitali sets

>>62
>"probability" is useless for them.

OP (i.e. me) is relying on a fuzzy definition of "probability" that makes some intuitive sense but can't be expressed in the usual probability space way.  Whether it actually works or not is up for debate.

>it doesn't use the axiom of choice.
>...
>the only place you use it here is in constructing the vitali sets

¬_¬

>...define X = X_1/10 + X_2/10^2 + X_3/10^3.....

I think that's been mentioned at least 5 times so far slready, except with a binary expansion.

>>63

No, everyone's been getting in a fucking tizzy about flipping coins, and whether constructing a random number in [0,1] uses AC (It doesn't)

It's been hinted at informally, which is why everyone's been getting confused, I thought it best to state it properly.

Me saying it doesn't use the axiom of choice is referring to that problem specifically, if you're OP didn't you make that silly post about bins in which you claimed to pick an element out of each bin uses the axiom of choice?

Because it doesn't. You don't need AC if the sets are well-ordered, cause you can just specify the choice function, and also the existence of a choice function doesn't guarantee you a set of ones uniformly distributed. Those are specific points people are being ridiculous about.

>>64
AC is needed for the existence of the vitali sets.  That's all.

>>65
That's exactly what I said!

We don't need vitali sets even mentioned in the first place.

>>67
orly?

I know a way you can finish! Take an infinite number of robots that can each flip one 10 sided die. They all flip at once and display their digit. Done.

>>69
Wtf? You flip a coin. You roll a die.

>>16
I'm confused as to how you know the vitali sets are of the same "size and shape", since they are unmeasurable and pretty much undefinable.  Forgive my ignorance, I'm only in undergrad.

Well, could you not simply roll infinite ten-sided dice in order? If I'm not mistaken, the odds of choosing any one number between zero and infinity are exactly zero, which would seem to correspond to the never-ending "rolling". Perhaps the only problem is with orders of infinity; specifically between zero and which order of infinity? (For the purposes of font, x=aleph null) Were the upper bound set at x, then you would need to perform at least x steps in order to not leave any numbers out, so to speak. This, however, would mean that after x steps, you could have a number equal to 10^x. I believe you would run into the same problem were you to set the upper bound at 10^x; you would get a number after 10^x steps equal to 10^10^x. I'm not sure what exactly this means, but perhaps it has no solution?

I think the problem with the dice-rolling ideas is that your result has to be finite, so (rolling lower digits first), at some point your rolls would have to be eventually constant at zero, and with an infinite number of rolls this does not happen.  I could be wrong though.

>>71
Sorry that I'm not english but I think you got it.