lomateron 491 Report post Posted September 16, 2014 (edited) I got the numbers {0,1,2,3....infinity} in my table I take half of this numbers and put them in a set(even numbers) and the other half in another set(odd numbers) how many different pairs can I make using a number from each set? How is the answer represented mathematically? Edited September 16, 2014 by lomateron 0 Share this post Link to post Share on other sites
Buckeye 10747 Report post Posted September 16, 2014 As it's been years since I've messed about with sets, this is a guess based on vaguely remembered (or understood) theorems: 1. The pairs are uncountable. 2. The cardinality of that set of pairs is Aleph-One. 0 Share this post Link to post Share on other sites
Samith 2460 Report post Posted September 16, 2014 As it's been years since I've messed about with sets, this is a guess based on vaguely remembered (or understood) theorems: 1. The pairs are uncountable. 2. The cardinality of that set of pairs is Aleph-One. I think the pairs are actually countable. See this link for a proof. So to answer the OP's question, the number of different pairs is countably infinite, so essentially there are as many pairs as there are integers. 2 Share this post Link to post Share on other sites
apatriarca 2365 Report post Posted September 16, 2014 The set is countable and thus have the same cardinality of natural numbers. If you define the following order on the set: (a, b) < (d, e) IFF a + b < d + e OR (a + b = d + e AND b < e) This is well defined and you can use this order to create a sequence of pairs. Given each pair you can determine its position in the sequence (which is finite) and you thus have a surjective, injective and invertible map between the set of the pairs and natural numbers. 0 Share this post Link to post Share on other sites
Buckeye 10747 Report post Posted September 16, 2014 (edited) I think the pairs are actually countable. The set is countable and thus have the same cardinality of natural numbers. You are correct. They are countable - implying Aleph-Null, not Aleph-One. Edited September 16, 2014 by Buckeye 0 Share this post Link to post Share on other sites
lomateron 491 Report post Posted September 16, 2014 (edited) ok, now I have the same set {0,1,2,3...} each number represented using bitarrays, so the length of a number's bitarray is infinite.I imagine a number inside this set, the only thing I know about this number is it starts with two "on" bits, I don't know the values of the next bits in the array, which means it can be one of the 1/4 of the total numbers inside the set. there are two different numbers inside the set "a" and "b", the only things I know about those numbers "a" number starts with one "on" bit "b" number starts with two "on" bits In "a" and "b" I don't know the values of the next bits in the array if the number I am thinking is the same as "a" I win, if it is "b" I lose. I will want an expert in this kinds of mathematics to tell me the probability of winning or losing, I came to the conclusion that the chances of losing is two times the chances of winning Am I right? Edited September 17, 2014 by lomateron 0 Share this post Link to post Share on other sites
alvaro 21273 Report post Posted September 16, 2014 I am sure I know enough mathematics to solve the problem, but you are going to have to pose it correctly first.ok, now I have the same set {0,1,2,3....infinity} each number represented using bitarrays, so the lenght of a number's bitarray is infinite.No, that's not right. For starters, it's not clear to me what set is {0,1,2,3,...infinity}. If you had just said {0,1,2,3,...}, I would imagine you mean the set of natural numbers, but then "infinity" shouldn't be there. Then, if you represent each number using bitarrays, the length of a number's bitarray is finite.So please try again, and be as precise as possible. 0 Share this post Link to post Share on other sites
lomateron 491 Report post Posted September 16, 2014 I didn't know about the first thing you pointed out, thanks. But about the second thing, it is obvious that the number zero can be represented by an infinite bit array where all it's bits are "off" the same for all the other numbers, why are you wronging that? 0 Share this post Link to post Share on other sites
alvaro 21273 Report post Posted September 16, 2014 I am not "wronging" anything. I am just trying to understand what you are saying. 2 Share this post Link to post Share on other sites
lomateron 491 Report post Posted September 16, 2014 so is the problem clear now? 2 Share this post Link to post Share on other sites
lomateron 491 Report post Posted September 16, 2014 (edited) ohhh, I did a confucious, my correct conclusions were If the number I imagine starts with one "on" bit then the probability of winning or losing is the same. If the number I imagine starts with two "on" bits then the probability of losing is two times the probability of winning. Edited September 16, 2014 by lomateron 0 Share this post Link to post Share on other sites
alvaro 21273 Report post Posted September 16, 2014 No, I still don't understand your problem. 0 Share this post Link to post Share on other sites
lomateron 491 Report post Posted September 17, 2014 (edited) do you understand why "which means it can be one of the 1/4 of the total numbers inside the set"? lets say I have a number represented by an infinite length bitarray, if the only thing I know about this number is that the first bit is "off" I can say this number is even, which means this number can one of the 1/2 of the total numbers inside the set. Now If I know two bits inside array I know this number can one of the 1/4 of the total numbers inside the set If I know three bits inside array I know this number can one of the 1/8 of the total numbers inside the set If I know four bits inside array I know this number can one of the 1/16 of the total numbers inside the set ... etc here the chances of winning or losing are measured in x/y of the total numbers inside the set... Edited September 17, 2014 by lomateron 0 Share this post Link to post Share on other sites
apatriarca 2365 Report post Posted September 17, 2014 Each bit has a probability of 1/2 of being on and 1/2 off regardless of the state of the other bits and its position in the binary representation. You problem is thus (if I understand it correctly) equivalent to the following: Choose a number between 0 and 3. You win if the number is 1 or 3, and you lose if the number is 3. It doesn't really make sense as a problem since the winning event and the losing events are not disjoint. Moreover, what happen in all other cases? 0 Share this post Link to post Share on other sites
lomateron 491 Report post Posted September 17, 2014 (edited) I wrote " I imagine a number inside this set, it starts with two "on" bits " after reading this people will think the number I am thinking is "3" and this is the mistake, so I edited and added more words to the main problem to clarify... some explanation... lets change this... " I imagine a number inside this set, it starts with ONE "on" bit " <--- I CHANGE TWO TO ONE Here the number I am thinking is not "1", it is an odd number... Edited September 17, 2014 by lomateron 0 Share this post Link to post Share on other sites
Bacterius 13165 Report post Posted September 17, 2014 (edited) What exactly does it mean to "think of a number in {0, 1, 2, 3, ...}"? Please explain how you intend to select an arbitrary integer from an unbounded set! Like others have hinted to above, this is not well-defined without specifying the probability distribution you use to select the number, because there exists no uniform probability distribution over an unbounded set! Be specific, your question is meaningless at this time! Adding more examples won't help if the problem is not well-defined or is based on a faulty premise. And you can't just shake it off as "obvious" or "self-evident", rigor is essential in mathematics, and in the study of the infinite more than anywhere else. Edited September 17, 2014 by Bacterius 0 Share this post Link to post Share on other sites
lomateron 491 Report post Posted September 17, 2014 " I imagine a number inside this set, it starts with ONE "on" bit " that is the same I saying " I imagine a number inside this set, and it is an odd number " So how I select the odd number?...whut?....what is the problem with selecting a random odd number? the probability of selecting a random odd number is the same for all odd numbers in the set 0 Share this post Link to post Share on other sites
Bacterius 13165 Report post Posted September 17, 2014 (edited) So how I select the odd number?...whut?....what is the problem with selecting a random odd number? the probability of selecting a random odd number is the same for all odd numbers in the setRight, and there infinitely many of them, so they can't have the same probability of being selected. Please, read this: http://math.stackexchange.com/questions/500545/why-does-the-probability-of-a-random-natural-number-being-prime-make-no-sense (first paragraph of the first answer) and think again about what you just wrote. Edited September 17, 2014 by Bacterius 1 Share this post Link to post Share on other sites
lomateron 491 Report post Posted September 17, 2014 the probability of selecting a random odd number is the same for all odd numbers in the set, it is 1/(total number in the set), so this makes no sense because 1/(total number in the set) = 0, but does it makes sense to say 1/(total number in the set) ? 0 (it is just close but not equal) and continue solving the problem? 0 Share this post Link to post Share on other sites
apatriarca 2365 Report post Posted September 17, 2014 While I agree that the probability of selecting a single number is zero, I think it is possible to make sense of something like "the probability of a natural number to be odd" by using a limit, i.e. P({n in {0, 1, 2, ..} : n is odd}) = lim_{M to infinity} P({n in {0, 1, 2, .., M-1} : n is odd}). This is a well defined limit and its value is 1/2. But even if we use a definition like this one, the rest of the sentences in the problem still does not make any sense to me. When I said the problem was equivalent to the one where we limit ourself to the first two bits I used the definition above. Indeed, if we take M = 2^m, we may represent the numbers from 0 to M-1 as strings of m bits. The set we are sampling from is then equal to {0, 1}^m and the set of odd numbers is {1} x {0, 1}^{m-1}. Since the remaining (m-1) bits can be chosen at random without limitations, this is equivalent to the choice of just the first bit value (the rest does not matter for the probability). For this reason, we have that P({n in {0, 1, 2, ..} : n is odd}) = lim_{M to infinity} P({n in {0, 1, 2, .., M-1} : n is odd}) = lim_{m to infinity} P({n in {0,1, .. 2^(m-1)} : n is odd} = lim_{m to infinity} P({n in {0, 1} : n == 1}) = P({n in {0,1} : n == 1}). 0 Share this post Link to post Share on other sites
alvaro 21273 Report post Posted September 17, 2014 Apatriarca defined a function P that maps some subsets of the natural numbers to real numbers between 0 and 1. This function is not a probability, as it does not satisfy the third axiom described [url="http://en.wikipedia.org/wiki/Probability_axioms"]here[/url]. If you relax the axiom to require only finite sums to work, then it would be fine. Alternatively, we could forget about natural numbers and work with infinite sequences of bits. Or, almost equivalently, we could imagine there is a "0." before the sequence of bits, and what we are doing is picking a random real number between 0 and 1 uniformly (technically we are assigning probabilities to subsets of [0,1] using the Lebesgue measure, for the sigma-algebra of measurable sets). Now "the first bit is set" means "x >= 0.5" and the "first two bits are set" means "x>=0.75". Computing those kinds of probabilities is easy, since the probability of an interval is its length. Now, what was the question again? 0 Share this post Link to post Share on other sites
Norman Barrows 7179 Report post Posted October 3, 2014 if you know the low bit is set, that eliminates half of them. if you know the second from lowest bit is also set, that eliminates half of those remaining. the problem is you have an infinite number, so half or 1/4 of them is still an infinite number. only when you have a finite number (say 1000) does calculating odds make sense. if you have 1000, and you pick one, your chances of picking the right one are 1 in 1000. if you know the low bit is set, you now only have 500 to choose from and the odds drop to 1 in 500. if you know the two low bits are set, you now have 250 to choose from and the odds are 1 in 250. 0 Share this post Link to post Share on other sites
JohnnyCode 1046 Report post Posted October 3, 2014 The OP is quite unable to express him meaningly, But this tricks me too, I will elaborate further. Someone picks any positive whole number (N={0,1,2.....}). - what is the probability I pick the same one? (it cannot be 0 since it is possible to happen, but extremly unlikely to happen ) - if someone says the number is odd, how bigger is probability I will gess what number is picked? We are speaking of a number from N set. (I do not know how those numbers are called in english exactly,I am talking of the set N={0,1,2,...}) Interisting question becouse of wheather disputing meaning of them, the question just makes sence. The probability being a real number : f e R; f e (0,0,1.0) ; f>0.0 ;f<1.0; 0 Share this post Link to post Share on other sites
alvaro 21273 Report post Posted October 3, 2014 There is no such thing as a uniformly random natural number. If someone picks a natural number randomly, he must do it with some probability distribution (the probability of the singletons {n} forms a sequence of non-negative numbers whose partial sum converges to 1). Then you can answer any question you want.Read my previous post (#21) for more details. 0 Share this post Link to post Share on other sites