Infinite Balls and Jugs [solution]
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Re: Infinite Balls and Jugs [solution]
I think you're looking at this problem the wrong way. There ARE an infinite number of natural numbers. You can't run out of numbers. If you don't beleive me, name me a number, I'll give you a natural number that's bigger. Another example is fractions. They are written a/b, where a and b are both natural numbers. However, natural numbers can still be used to represent ANY fraction.
The point of all this is that there are as many balls as we need, we'll never run out. Therefore, there is no reason that the number of balls fall to zero at midnight. Whether you relabel the balls or not doesn't matter.
The point of all this is that there are as many balls as we need, we'll never run out. Therefore, there is no reason that the number of balls fall to zero at midnight. Whether you relabel the balls or not doesn't matter.
Re: Infinite Balls and Jugs [solution]
Dutchflyboy123 wrote:I think you're looking at this problem the wrong way. There ARE an infinite number of natural numbers. You can't run out of numbers. If you don't beleive me, name me a number, I'll give you a natural number that's bigger. Another example is fractions. They are written a/b, where a and b are both natural numbers. However, natural numbers can still be used to represent ANY fraction.
The point of all this is that there are as many balls as we need, we'll never run out. Therefore, there is no reason that the number of balls fall to zero at midnight. Whether you relabel the balls or not doesn't matter.
It's not a question of running out. The only person in the entire thread who ever described running out was WarDraft, and you'll see that the people who responded to him made exactly your point.
The urn is empty at midnight because every ball that goes in is removed. If you don't believe me, name me a number, I'll tell you exactly when it was removed (20/2^n minutes before midnight).
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Re: Infinite Balls and Jugs [solution]
Sorry, I had misunderstood part of the conversation. While reading it again, I saw I was preaching to the converted.

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Re: Infinite Balls and Jugs [solution]
But to answer your question: infinity1! I know that infinity isn't a number. But that doesn't change the fact that there ARE an infinite number of steps. And if you plot the function you'll see it's a line going up. Why would the line change direction, why would all balls vanish? As I said before, you can't shift numbers around in infinite series.
But just a neutral question. Do you agree to the following?
At step n:
The lowest ball in urn: n+1
Highest ball in urn: 10*n
Number of balls in the jar: 10*n(n+1)
But just a neutral question. Do you agree to the following?
At step n:
The lowest ball in urn: n+1
Highest ball in urn: 10*n
Number of balls in the jar: 10*n(n+1)
Re: Infinite Balls and Jugs [solution]
Our set of them is countable of course... just as you can make a countable set of transcendental numbers, however, those numbers are still trancendental... just as these are still from the set of uncountably large numbers, and there are uncountably many numbers possible between each that we actually have.mikel wrote:WarDaft wrote:Aren't numbers with infinitely many digits equally as valid for diagonal proof of uncountability as the real numbers? That's not to say we have uncountably many of them.
You're right that the set of all infinite strings of digits is uncountable.
I was not very specific in my last post, but you'll never have all infinite strings, just particular ones. For example, if your relabeling consists of adding a 1 at the end of the ball, then you'll have all finite strings of digits followed by infinitely many 1's.
It's trivially empty if we permit a step at midnight but only admit natural numbers for the balls. So how would it not be empty without that step, considering that step would not remove any natural numbers?But to answer your question: infinity1! I know that infinity isn't a number. But that doesn't change the fact that there ARE an infinite number of steps. And if you plot the function you'll see it's a line going up. Why would the line change direction, why would all balls vanish? As I said before, you can't shift numbers around in infinite series.
But just a neutral question. Do you agree to the following?
At step n:
The lowest ball in urn: n+1
Highest ball in urn: 10*n
Number of balls in the jar: 10*n(n+1)
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Re: Infinite Balls and Jugs [solution]
But still, do you accept the formula for finding the number of balls at step n?
Re: Infinite Balls and Jugs [solution]
Yes, for natural n.
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Re: Infinite Balls and Jugs [solution]
Dutchflyboy123 wrote:But still, do you accept the formula for finding the number of balls at step n?
Yes I accept your forumla for the number of balls in the urn at step n. Your concern has been addressed numerous times in the last 5 pages. Look at the pencil/eraser on a ruler example. Or look at my post about how the '1' terms don't belong to the '+10' terms in the same step, they belong to the +10 from way long ago when they were put in the jug. You are right that you can't shift the numbers around in an infinite series, you have to put them where they belong in the first place.
Infinity  1 was not written on any of the balls I started with, nor any ball I put in. I'm only doing things at time 20/2^n minutes before midnight, for natural number n. During that time, I deposit balls 10n9..10n, and I remove ball n. Please tell me the number of a ball that is in the urn at midnight, and what time it was added at. This shouldn't be hard to do if there are infinitely many balls in the urn at midnight, (though you'll find it quite difficult to do, since it's actually empty!)
WarDaft wrote:Our set of them is countable of course... just as you can make a countable set of transcendental numbers, however, those numbers are still trancendental... just as these are still from the set of uncountably large numbers, and there are uncountably many numbers possible between each that we actually have.
Uncountable is a property of a set, not a property of a string of digits. So while it's true that the set of all infinite strings of digits is uncountable, I've never heard anyone refer to such strings as uncountably large numbers before. You'll have to forgive me for interpretting 'the balls in the urn are uncoutable' as 'the set of balls in the urn is uncoutable'.
It's trivially empty if we permit a step at midnight but only admit natural numbers for the balls. So how would it not be empty without that step, considering that step would not remove any natural numbers?
The urn is empty without doing ANYTHING at midnight. This is why I was pressing before for what exactly you planned to do at midnight. Before midnight, I've only added natural numbers, and I've removed every single one, so it's empty when you get to midnight.
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Re: Infinite Balls and Jugs [solution]
And I've already said, if we constrain the balls to natural numbers, it is necessarily empty. However, I fail to see how we can constrain the states of the jar to natural numbers, and then ask what is in it after midnight, as all states the jar took on then occurred before midnight.The urn is empty without doing ANYTHING at midnight. This is why I was pressing before for what exactly you planned to do at midnight. Before midnight, I've only added natural numbers, and I've removed every single one, so it's empty when you get to midnight.
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Re: Infinite Balls and Jugs [solution]
WarDaft wrote:And I've already said, if we constrain the balls to natural numbers, it is necessarily empty. However, I fail to see how we can constrain the states of the jar to natural numbers, and then ask what is in it after midnight, as all states the jar took on then occurred before midnight.The urn is empty without doing ANYTHING at midnight. This is why I was pressing before for what exactly you planned to do at midnight. Before midnight, I've only added natural numbers, and I've removed every single one, so it's empty when you get to midnight.
You just said that you needed the step at midnight to empty the urn. Please tell me what ball is in the urn before this step is performed, and when it was put in the urn.
I'm perfectly fine with you calling the state of the jar at midnight [imath]\omega[/imath], but calling it such doesn't mean you did a step there.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
Re: Infinite Balls and Jugs [solution]
No, I said if the balls have to have finite numbers that it's trivially easy for it to be empty with a step at midnight. Then I asked that, given such a step would not itself remove any finite numbers, why it would be necessary to construct a situation which gives us an empty jar, if that was the goal.
My intent was never to imply that it was impossible to have an empty jar, but that it is possible for it to be nonempty, depending on our exact description.
Can I ask what conditions you would consider sufficient for the presence of an ω ball?
My intent was never to imply that it was impossible to have an empty jar, but that it is possible for it to be nonempty, depending on our exact description.
Can I ask what conditions you would consider sufficient for the presence of an ω ball?
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Re: Infinite Balls and Jugs [solution]
WarDaft wrote:No, I said if the balls have to have finite numbers that it's trivially easy for it to be empty with a step at midnight. Then I asked that, given such a step would not itself remove any finite numbers, why it would be necessary to construct a situation which gives us an empty jar, if that was the goal.
My intent was never to imply that it was impossible to have an empty jar, but that it is possible for it to be nonempty, depending on our exact description.
Can I ask what conditions you would consider sufficient for the presence of an ω ball?
The insertion and nonremoval of said ball.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
Re: Infinite Balls and Jugs [solution]
What starting conditions?mikel wrote:The insertion and nonremoval of said ball.
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Re: Infinite Balls and Jugs [solution]
I don't know about sufficient, but certainly a necessary condition would be a step in which an ω ball is added.
All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.
Re: Infinite Balls and Jugs [solution]
WarDaft wrote:What starting conditions?mikel wrote:The insertion and nonremoval of said ball.
Starting conditions that made the provisions of such a ball.
For example, place 10 balls in the urn at each 20/2^n minutes before midnight, then at midnight, insert ball [imath]\omega[/imath]. In the absence of such a statement, I assume it is not there. Much like I also assume there is no horse in the urn. Nobody said anything about doing anything at midnight, so I assumed we weren't. Nobody said anything about ordinal numbers, so I assumed we weren't using them. (And by nobody, I of course mean me, since I posted the problem). You should note that the natural numbers are perfectly sufficient to do all of the operations before midnight as well, so there's no 'argument by necessity' here. There's no reason in the statement of the question that there should be a ball labeled [imath]\omega[/imath], much like there is no reason there is a horse, a Jello shot, or a ICBM. (but maybe there are puppies, because everyone likes puppies)
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Re: Infinite Balls and Jugs [solution]
I still don't follow (or at least I do not accept the logic behind the argument). Would the following not be the same problem?
Add 10 balls and remove 1 every day for an infinite amount of time. How many remain at time infinity?
It would seem that the answer to that problem is infinity, its just an infinite sum that does not converge.
In addition, how can the labeling matter? What happens if the balls go in without numbers, and are identical. Does the sum then become infinity?
Add 10 balls and remove 1 every day for an infinite amount of time. How many remain at time infinity?
It would seem that the answer to that problem is infinity, its just an infinite sum that does not converge.
In addition, how can the labeling matter? What happens if the balls go in without numbers, and are identical. Does the sum then become infinity?
Re: Infinite Balls and Jugs [solution]
The exact rules are important. It's not the labeling that matters so much as the implication of the labeling.
If we start off with some natural number, and iterate through a process where we select a new number based solely on the condition that it is greater than the previous number, is it possible to have a number less than ω as the result of the process iterated ω times? (Without using real numbers, obviously)I don't know about sufficient, but certainly a necessary condition would be a step in which an ω ball is added.
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Re: Infinite Balls and Jugs [solution]
Muffins3141 wrote:I still don't follow (or at least I do not accept the logic behind the argument). Would the following not be the same problem?
Add 10 balls and remove 1 every day for an infinite amount of time. How many remain at time infinity?
If I've understood your "time infinity" notion, I don't see how this changes anything except that the identity of the balls is unspecified.
Muffins3141 wrote:It would seem that the answer to that problem is infinity, its just an infinite sum that does not converge.
I reject the idea that the problem is about an infinite sum. Its about an infinite set, the set of balls, and how it is partitioned into balls in the urn and balls out of the urn at different stages. The important idea for analysing the problem as far as I am concerned is the idea of the limit of sets, described here. You can also see some of my earlier confusion there.
Muffins3141 wrote:In addition, how can the labeling matter? What happens if the balls go in without numbers, and are identical. Does the sum then become infinity?
I would say that the state of the urn is indeterminate if we do not label the balls. To show that the labeling might matter, consider what would occur if we had one blue ball and an infinity of red balls. If we add and remove the blue ball at the first step, and never return it to the urn, then there is definitely no blue ball in the urn. If, on the other hand, the blue ball goes in and never comes out, then the urn must contain at least a blue ball at the end of the process. The labeling allows us to track the fate of every individual ball in the system, and by doing that we can determine the disposition of the balls and the urn.
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Re: Infinite Balls and Jugs [solution]
jestingrabbit wrote:No, I would say the urn ends empty. Consider that every ball enters the urn and leaves the urn, and afterwards is untouched. The fact that the labeling is messing around changes nothing, except the numbers on the balls removed. You end up with an empty urn and huge pile of balls labeled 1.
But there will always be a ball in the jar with the number 1 on it, and every time you remove it the jar conspires that there's another one there.
At the nth repetition, the n+1th ball will never have been removed, though it will always have been placed in the jar. So as far as I understand, attempting to prove that all balls are removed by probability will leave you with an equation of an indefinite form. Your other method of proving that all the balls were gone doesn't work, and more importantly I can say for sure that there will always be a ball numbered 1 in the urn, regardless how many times you perform that step.
doogly wrote:Oh yea, obviously they wouldn't know Griffiths from Sakurai if I were throwing them at them.
Re: Infinite Balls and Jugs [solution]
dedalus wrote:jestingrabbit wrote:No, I would say the urn ends empty. Consider that every ball enters the urn and leaves the urn, and afterwards is untouched. The fact that the labeling is messing around changes nothing, except the numbers on the balls removed. You end up with an empty urn and huge pile of balls labeled 1.
But there will always be a ball in the jar with the number 1 on it, and every time you remove it the jar conspires that there's another one there.
At the nth repetition, the n+1th ball will never have been removed, though it will always have been placed in the jar. So as far as I understand, attempting to prove that all balls are removed by probability will leave you with an equation of an indefinite form. Your other method of proving that all the balls were gone doesn't work, and more importantly I can say for sure that there will always be a ball numbered 1 in the urn, regardless how many times you perform that step.
That doesn't actually change the situation so long as we always know which ball has been in the jar the longest time.
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Re: Infinite Balls and Jugs [solution]
We can keep track of different balls, whether they are labeled or not, by keeping track of a balls position as a function of time. The labels are merely a convenience that let us dispense with this tracking, so whether the balls are labeled differently, unlabeled, or all labeled identically makes no difference, so long as the first ball in is removed, then the second ball in is removed, then the third ball in is removed, and so on.
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Re: Infinite Balls and Jugs [solution]
um... say we have a jug with a ball in it(labeled 1). we insert one ball(labeled 2) and take the previous one out. if we do this an infinite amount of times, it is obvious that every ball we put in will be taken out, but it is also obvious that there is never a case where there is no ball in the jug.
the ball that is in the jug changes infinitely many times, so any ball (say ball n) will be taken out, so the probability of there being any ball in there is 0, but the probability of there being no balls, according to the "rules" is also none.
there is always at least one ball in there, but it is impossible to name a ball that is in there.
the ball that is in the jug changes infinitely many times, so any ball (say ball n) will be taken out, so the probability of there being any ball in there is 0, but the probability of there being no balls, according to the "rules" is also none.
there is always at least one ball in there, but it is impossible to name a ball that is in there.
I actually posted the real final version right after your post. It's so deep that the post containing it doesn't exist.
I'm not confirming or denying whether I posted an even more final version after your post, but leave it as a hypothetical possibility.
I'm not confirming or denying whether I posted an even more final version after your post, but leave it as a hypothetical possibility.
Re: Infinite Balls and Jugs [solution]
Well I guess the case is that if we need to name a ball in the jug at t=midnight, we're considering the limit as n approaches infinity of the nth repetition of the process. But you've never removed the n+1th ball, so in retrospect, that should be the ball that's always still in the urn.
doogly wrote:Oh yea, obviously they wouldn't know Griffiths from Sakurai if I were throwing them at them.
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Re: Infinite Balls and Jugs [solution]
You can't say that's "the" ball, since that's a sequence of balls, not any one ball.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
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Re: Infinite Balls and Jugs [solution]
yes. the point is, people are talking about the end of infinity, but it is a continous process. the argument against zero balls falls apart when talking about "in the end, any ball will be taken out" because you are ignoring the fact that it never ends. thus, the conclusion of zero balls is fallacious.
I actually posted the real final version right after your post. It's so deep that the post containing it doesn't exist.
I'm not confirming or denying whether I posted an even more final version after your post, but leave it as a hypothetical possibility.
I'm not confirming or denying whether I posted an even more final version after your post, but leave it as a hypothetical possibility.
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Re: Infinite Balls and Jugs [solution]
The process described in the OP ends at midnight. If a set is infinite then it doesn't preclude the possibility that it can be exhausted by a process. For instance, if you consider all the points on the number line between 0 and 1 as the set, and the process being a point moving from 0 to 1 at a speed of one unit of length per unit of time, then the point arrives at 1 after one unit of time and has exhausted the set. There are an infinity of points, but they have been used up. Likewise, in the process that the OP describes, the nth ball is used up at time midnight minus 20/2^{n} minutes. All the balls get used up and the set of balls is exhausted. The process ends.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Infinite Balls and Jugs [solution]
Here's another one:
If I put a machine in the jar that switches the numbers of the highest numbered ball and the lowest numbered ball in between the act of placing the ten in the jar and removing one, then you're performing task a (from the problem thread; removing the highest numbered ball); any ball that isn't the largest of its group (so a multiple of 10) will never be removed, but you have exactly the same effect as task b. So, what happens?
Oh, and insert the condition of inserting a new number rather then deleting and then rewriting; so the number of the ball is stored upon multiple flipflops, and you insert the new number as needed.
If I put a machine in the jar that switches the numbers of the highest numbered ball and the lowest numbered ball in between the act of placing the ten in the jar and removing one, then you're performing task a (from the problem thread; removing the highest numbered ball); any ball that isn't the largest of its group (so a multiple of 10) will never be removed, but you have exactly the same effect as task b. So, what happens?
Oh, and insert the condition of inserting a new number rather then deleting and then rewriting; so the number of the ball is stored upon multiple flipflops, and you insert the new number as needed.
doogly wrote:Oh yea, obviously they wouldn't know Griffiths from Sakurai if I were throwing them at them.
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Re: Infinite Balls and Jugs [solution]
If there is some entity monkeying around with labeling inside the jar which we are unaware of, then that naturally implies that we might be mistaken regarding the fate of particular balls. However, if we know what happens, with progressive labeling preserved in some way, then we can see what happens.
Here, after you deposit a new batch of balls, you remove the last after putting a label on it (which is obtained by removing the final 0 from the previous label). Every other ball will be given an infinite sequence of labels, each with another appended 0. So, balls removed have two labels, the first divisible by ten, the second what you get when you divide it by 10. Every other ball has an infinity of labels, the first a number not divisible by 10, and every subsequent label obtained by appending another 0 to its label.
Again, trace the fate of each individual ball to determine what happens.
Here, after you deposit a new batch of balls, you remove the last after putting a label on it (which is obtained by removing the final 0 from the previous label). Every other ball will be given an infinite sequence of labels, each with another appended 0. So, balls removed have two labels, the first divisible by ten, the second what you get when you divide it by 10. Every other ball has an infinity of labels, the first a number not divisible by 10, and every subsequent label obtained by appending another 0 to its label.
Again, trace the fate of each individual ball to determine what happens.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Infinite Balls and Jugs [solution]
But by inserting the new label the old one is immediately removed (flipflops work such that they hold a binary yes or binary no until you insert a different number, in which case they hold the new one). The point is though, that at any time n this urn holds the exact same configuration of balls as the case where you remove the smallest numbered ball, so surely that wouldn't change at midnight, except that at midnight (because this follows the same removal rules as the case where you remove the largest numbered ball) this urn will be full whereas the other one will be empty. And yes, at midnight there will be an arbitrarily infinite and meaningless number on each of the balls left in the urn, but won't that still be the state of the other?
doogly wrote:Oh yea, obviously they wouldn't know Griffiths from Sakurai if I were throwing them at them.
Re: Infinite Balls and Jugs [solution]
dedalus wrote: The point is though, that at any time n this urn holds the exact same configuration of balls as the case where you remove the smallest numbered ball, so surely that wouldn't change at midnight, except that at midnight (because this follows the same removal rules as the case where you remove the largest numbered ball) this urn will be full whereas the other one will be empty.
I can't really bring myself to believe (as jestingrabbit seems to) that you can change the result of an experiment by swapping the positions of identical things. Nor do I want to believe that you can change the final position of one of the balls simply by changing how you move around the others. I can't prove that these desired properties are contradictory, but I can't find a way of answering these types of problems while maintaining both of them either.
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Re: Infinite Balls and Jugs [solution]
lets have an example. if i start a pencil from zero on a ruler(infinitely long), and an eraser is busy erasing the pencil line(starting from zero as well). the pencil is moving ten times as fast as the eraser. so here we have the same case as with the balls. a pencil marks ten integers, and the eraser removes the lowest. over an infinite amount of time, the eraser will erase any point on the ruler, but there will still be marks left in front of it.
i think part of the problem is with degrees of infinity. the eraser goes to infinity, but the pencil goes to ten times infinity. this may seem strange, but for those who know the formula for e or (1 + 1/n)^n when n goes to infinity. if we take e^10 and substitute, you get (1+1/n)^10n. so 10*infinity is different than infinity.
i think part of the problem is with degrees of infinity. the eraser goes to infinity, but the pencil goes to ten times infinity. this may seem strange, but for those who know the formula for e or (1 + 1/n)^n when n goes to infinity. if we take e^10 and substitute, you get (1+1/n)^10n. so 10*infinity is different than infinity.
I actually posted the real final version right after your post. It's so deep that the post containing it doesn't exist.
I'm not confirming or denying whether I posted an even more final version after your post, but leave it as a hypothetical possibility.
I'm not confirming or denying whether I posted an even more final version after your post, but leave it as a hypothetical possibility.

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Re: Infinite Balls and Jugs [solution]
Although there are different kinds of infinity, the two present in this problem as well as the ones in your (fairly nonillustrating example) are actually the same, so that's not the issue. Just as there are in a naive sense 10 times more powers of 10 than there are natural numbers, the amount of balls put in is in a naive sense 10 times bigger than the amount of balls taken out. Never the less, the cardinality of all these sets is exactly the same. Infinity can be a weird concept, and you come to strange conclusions if you tackle it with just a naive concept. It has been studied for a while, and arguably the most meaningful way to deal with it is the approach introduced by Cantor.
And as a slight note, when you write "when n goes to infinity", no actual infinity is involved.
And as a slight note, when you write "when n goes to infinity", no actual infinity is involved.
Re: Infinite Balls and Jugs [solution]
They aren't identical if you know which was added first. Changing the labels doesn't make it functionally equivalent to simply doing the operations on the balls that used to bear those labels.Macbi wrote:I can't really bring myself to believe (as jestingrabbit seems to) that you can change the result of an experiment by swapping the positions of identical things. Nor do I want to believe that you can change the final position of one of the balls simply by changing how you move around the others. I can't prove that these desired properties are contradictory, but I can't find a way of answering these types of problems while maintaining both of them either.
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Re: Infinite Balls and Jugs [solution]
first off, thanks to LLCoolDave for setting me straight. I don't know too much about limits, obviously.
second, my first post still holds. if someone can prove me wrong, please do. until then, there are balls in the jar.
also, infinity may produce strange results, but it will not work miracles. if you add ten balls to a jar and take away one over and over, i do not care how many times you do this, it will never become zero. it gets bigger.
second, my first post still holds. if someone can prove me wrong, please do. until then, there are balls in the jar.
also, infinity may produce strange results, but it will not work miracles. if you add ten balls to a jar and take away one over and over, i do not care how many times you do this, it will never become zero. it gets bigger.
I actually posted the real final version right after your post. It's so deep that the post containing it doesn't exist.
I'm not confirming or denying whether I posted an even more final version after your post, but leave it as a hypothetical possibility.
I'm not confirming or denying whether I posted an even more final version after your post, but leave it as a hypothetical possibility.
Re: Infinite Balls and Jugs [solution]
Octavian Starr wrote:first off, thanks to LLCoolDave for setting me straight. I don't know too much about limits, obviously.
second, my first post still holds. if someone can prove me wrong, please do.
In general, the limit of a function as its parameter goes to a limit does NOT equal the result of the function evaluated at the limit of its parameter,
[math]\large\lim_{x\longrightarrow a}f\left(x\right)\neq f\left(\lim_{x\longrightarrow a}x\right)[/math]
This is also true for the limit as [imath]x\longrightarrow\infty[/imath]. Equality only holds for continuous functions.
Consider the function
[math]\text{sign}(x) = \begin{cases}1&\text{if }x>0\\0&\text{if }x=0\\1&\text{if }x<0\end{cases}[/math]
Now the sequence [imath]\large\{a_n = \frac{1}{n}\}[/imath] for positive integers [imath]n[/imath] clearly has limit 0,
[math]\large\lim_{n\longrightarrow\infty}a_n = \lim_{n\longrightarrow\infty}\left(\frac{1}{n}\right) = 0[/math]
So we know that
[math]\large\text{sign}\left(\lim_{n\longrightarrow\infty}a_n\right) = \text{sign}(0)= 0[/math]
But for each positive integer [imath]n[/imath] we have [imath]\large a_n = \frac{1}{n} > 0[/imath], so
[math]\large\text{sign}(a_n) = 1[/math]
Therefore the sequence of [imath]\{\text{sign}(a_n)\}[/imath] is just [imath]\{1,\;1,\;1,\;\dots\}[/imath] and we immediately see it has limit 1,
[math]\large\lim_{n\longrightarrow\infty}\text{sign}(a_n) = 1 \neq\text{sign}\left(\lim_{n\longrightarrow\infty}a_n\right) = 0[/math]
In the particular case at hand, the set of balls in the jar is a function of time, [imath]S = f(t)[/imath].
The number of balls in the jar is a function of the set of balls in the jar [imath]N = g(S)[/imath]. Thus we have [imath]N = g\left(f(t)\right)[/imath] and we see the number of balls in the jar is a function of time. So we can write [imath]N = h(t)[/imath].
This function is discontinuous because the set changes instantly between distinct discrete compositions: at each time step, the set of balls goes from state [imath]S_t[/imath] to state [imath]S_{t+1}[/imath]. Thus the cardinality of the set, which is the number of balls in the jug, has a discontinuous jump at every time step. So we already know the limit of the number of balls in the jug as time approaches a step where we add balls does not equal the number of balls in the jug at that time step.
Furthermore, between any time before midnight (no matter how close to midnight) and midnight itself, there is a time step where we add balls to the jug. Therefore we know that the number of balls in the jug is discontinuous somewhere in every neighborhood of midnight. That is guaranteed by the statement of the problem. So there is no reason to expect that the number of balls should be continuous at midnight.
As a result, any argument involving a limit of the number of balls in the jug is definitely false. Let me reiterate: to claim that the limit equals the value on principle, you must first show the function is continuous. In order to show the function is continuous, you must by definition show that the value equals the limit. This is called begging the question. For your argument to be valid you must have continuity. But to show continuity you must prove a statement that is equivalent to the conclusion of your argument. You cannot use your unproved conclusion to prove itself. Therefore the argument from continuity is fallacious.
We know precisely what [imath]f[/imath] and [imath]g[/imath] are for all times before midnight, so we know [imath]h[/imath] as well. Indeed,
Dutchflyboy123 wrote:But just a neutral question. Do you agree to the following?
At step n:
The lowest ball in urn: n+1
Highest ball in urn: 10*n
Number of balls in the jar: 10*n(n+1)
WarDaft wrote:Yes, for natural n.
mikel wrote:Yes I accept your forumla for the number of balls in the urn at step n.
Uh, guys, you know that formula is wrong don't you? I mean, just plug in for step 1. We know there are 9 balls. But the formula gives 10*1(1+1) = 102 = 8. The correct formula is 10*nn = 9n. Remember that a total of 9 balls are added at each step, starting from zero.
Octavian Starr wrote:second, my first post still holds. if someone can prove me wrong, please do. until then, there are balls in the jar.
No. Absolutely not. This is not how mathematics works. The answer exists independently of any of our abilities to comprehend or prove it. The answer does not change in time based on anything any of us says.
As it happens, several people in this thread have comprehended and proved the answer. In the mind of anyone who does not yet understand the proofs, the question is either still unresolved or is resolved in a false and incorrect manner. That does not change the fact that there is a unique, unchanging, correct solution.
Whether your first post holds is also a statement with a single truth value that will not change unless you edit the post (or others edit their posts in this thread to the point where the context around your post renders it to have a different meaning.) The rest of us will believe your post if and only if your claims are proved with mathematical rigor.
Since your claims are directly contradictory to a result which has been proved with mathematical rigor, there is no possible way for your claims to be proved unless the entire foundation of our mathematical system contains mutually imcompatible axioms. If there were such internal contradictions in mathematics, then every statement, whether true or false, could be proved, and thus a "proof" would not guarantee truth of the statement. Luckily, our mathematical systems are constructed so as to be internally consistent, and thus by Gödel incomplete in the sense that there are true statements we cannot prove. However in the situation we are dealing with, the correct answer can be and has been proved, and it has even been presented in straightforward, intuitive ways.
Also, for the cases where labels get changed, just make the labels that change or are changeable blue, and add a red label to each ball that lists its "true" or "original" ID number, which does not change. Then the path of every red label can be tracked, and hence each ball accounted for. If each red label performs a simple task, then the final position of the ball is known. If the blue labels change infinitely often or otherwise perform a supertask, their final positions may not be welldefined. In those cases the state of the balls at midnight will be known but the state of the blue labels may not be.
Last edited by Qaanol on Tue Nov 17, 2009 3:31 am UTC, edited 1 time in total.
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Re: Infinite Balls and Jugs [solution]
Qaanol wrote:Dutchflyboy123 wrote:But just a neutral question. Do you agree to the following?
At step n:
The lowest ball in urn: n+1
Highest ball in urn: 10*n
Number of balls in the jar: 10*n(n+1)WarDaft wrote:Yes, for natural n.mikel wrote:Yes I accept your forumla for the number of balls in the urn at step n.
Uh, guys, you know that formula is wrong don't you? I mean, just plug in for step 1. We know there are 9 balls. But the formula gives 10*1(1+1) = 102 = 8. The correct formula is 10*nn = 9n. Remember that a total of 9 balls are added at each step, starting from zero.
Owch, but in my defense, the point was this didn't matter (in the part you snipped from my quote) Besides, being off by a constant doesn't matter that much even if the formula did matter.
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Re: Infinite Balls and Jugs [solution]
i'm sorry for not being rigorous, but i simply pointed out the fact that just because any ball put in the jar will be removed eventually(which was the original argument) that does not mean that there are no balls in the jar in the end. since this is just pointing out the possibility, there should be no need for me to prove it rigorously. i am just pointing out that you have not proved your point yet, and thus, i am hesitant to accept this.
also, i understand your point about continuity, but i am just wondering at which point the number of balls begins to become smaller. as far as i can see, it gets bigger and bigger until suddenly it becomes zero. infinity is strange, yes, but in this case infinity is describing doing a process over and over. the process seems to be one that makes the number of balls in the jug larger, and this does not change over time. so why does the number of balls get smaller?
also, i understand your point about continuity, but i am just wondering at which point the number of balls begins to become smaller. as far as i can see, it gets bigger and bigger until suddenly it becomes zero. infinity is strange, yes, but in this case infinity is describing doing a process over and over. the process seems to be one that makes the number of balls in the jug larger, and this does not change over time. so why does the number of balls get smaller?
I actually posted the real final version right after your post. It's so deep that the post containing it doesn't exist.
I'm not confirming or denying whether I posted an even more final version after your post, but leave it as a hypothetical possibility.
I'm not confirming or denying whether I posted an even more final version after your post, but leave it as a hypothetical possibility.
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Re: Infinite Balls and Jugs [solution]
Octavian Starr wrote:i'm sorry for not being rigorous, but i simply pointed out the fact that just because any ball put in the jar will be removed eventually(which was the original argument) that does not mean that there are no balls in the jar in the end.
If there are any balls in the jar at the end, then there must be at least one example of a ball in the jug at the end. Name one.
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Re: Infinite Balls and Jugs [solution]
as far as i can see, it gets bigger and bigger until suddenly it becomes zero.
Integer overflow.
Re: Infinite Balls and Jugs [solution]
If you are the only person performing the steps (adding and removing balls from the jug) is your hand inside or outside the jug at midnight?
If infinitely many people are involved so person n only adds and removes balls at step n, where is each person's hand at midnight (inside or outside the jug)?
If infinitely many people are involved so person n only adds and removes balls at step n, where is each person's hand at midnight (inside or outside the jug)?
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