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• 0 ## Question After reading unreality's problems dealing with infinite numbers, I decided to post my own. To be honest, it's not "my own," since I read it in a book a few years ago (I don't remember which book, but it was really good. You should all read it).

Suppose you have 2 bins, labeled 'A', and 'B'. To start, bin A has an infinite number of ping pong balls, and they are numbered sequentially from 1 to infinity. Each ball has a unique number printed on it. Bins B is empty to start.

You begin an experiment that will last 1 minute, but will include an infinite number of iterations. Each iteration will take exactly half of the time remaining - ie, the first iteration will be completed after 30 seconds, the second after 45, the third after 52.5, etc. At each iteration, you do two things: 1st, take the two lowest numbered balls out of bin A and put them in bin B; 2nd, take the one lowest numbered ball out of bin B, and throw it away.

So, in the first 30 seconds, you take balls 1 and 2 out of A and put them in B, then take ball 1 and throw it away. In the next 15 seconds, you take balls 3 and 4 out of A, put them in B, then throw away ball 2. As you can see, the number of balls in bin A will decrease by 2 with every iteration, and the number of balls in bin B will increase by 1.

The question is this: after the entire minute has elapsed, how many balls will be in bins A and B?

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• 0
There is a flaw in your logic: you are looking at the state after some finite number of operations/iterations, and from there trying to count to infinity.

Let us simplify the problem by first considering only bin A. Bin A doesn't care what happens to the balls once they are removed from it, only that they are no longer contained in A. Removing the lowest, or two lowest, numbers consecutively from bin A is equivalent to counting the natural numbers. Therefore, when I have accomplished my task of removing balls from bin A, I have removed the set of natural numbers from it. But this is precisely what bin A contained initially (and nothing more), so it must now be empty.

Now consider bin B. Everything removed from A is placed directly into B. Therefore, every natural number is contained in B, although not simultaneously. As before, when I discard the lowest numbered ball from B at each iteration, I am counting the natural numbers. Therefore, if I perform this task ad infinitum, I discard the set of natural numbers from B. B must therefore also be empty.

Another way of looking at it is this: because we started only with the natural numbers, any remaining balls in A or B must be subsets of N. There is no subset that satisfies the condition of always removing the lowest numbered ball from B, because this is the same as counting the members of N.

However, you are correct in saying that if you remove the highest numbered ball from B instead of the lowest, you will end up with the odd numbers in it.

[edit: clarification]

1. The number of balls in bin B cannot depend on which ball was removed.
If removing the larger number results in a countably infinite set of numbers [the odd integers] at completion,
then removing the smaller number cannot result in an empty set at completion.

To see this, restate the problem.
Step 1. Move two balls from bin A to bin B.
Step 2. Move one ball from bin B to bin C. [or discard it]
Repeat.

Or equivalently,
Step 1. Move one ball from bin A to bin B.
Step 2. Move one ball from bin A to bin C. [or discard it]
Repeat.

Please describe how in either case bin B eventually becomes empty.

2. If talking about a finite number of operations and then letting that number approach infinity is fallacious [it isn't] your argument is fallacious.

3. A proper subset of a countably infinite set can also be countably infinite [the set of even integers is a countably infinite proper subset of the set of integers].
Thus, just because you have accounted for a countably infinite number of natural numbers [those removed from bin B] it does not follow that you have accounted for all of them.

4. Consider the sequence {1, 2, 3, ..., N, N+1, N+2, N+3, ... 2N}
Remove the first N of the elements.
Let N approach infinity.
You will have removed a countably infinite subset of integers from the sequence.
Does that mean you have removed all the elements of the sequence?
No.

There are as many real numbers in the interval [0, 1) as there are in the interval [0, 2).

But if you remove the uncountably infinite first set from the uncountably infinite second set, you have just as many reals remaining: those in the interval [1, 2).

Removing the countably infinite set of integers [1, N] from the set [1, 2N] as N goes to infinity does not deplete the [1, 2N] set,

any more than removing the countably infinite set of even integers depletes the countably infinite set of integers.

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• 0 You make some interesting points, but I still disagree. See lytefoot's post if you haven't already.

The number of balls in bin B cannot depend on which ball was removed.

If removing the larger number results in a countably infinite set of numbers [the odd integers] at completion,

then removing the smaller number cannot result in an empty set at completion.

Just because, after a finite number of operations, B contains the same number of balls whether you discarded the lowest valued element, or a other element, does not mean the same is true after "infinitely many" operations. This is because the final result of dividing the a set (in this case, N) into subsets is the same if I do so consecutively or simultaneously. If I remove the first ten elements from A, I have ten elements in B. If I remove the first ten odd elements of A, I still have ten elements in B. However, if I remove every element of A, A will be empty, and B will have an infinite number of elements. If I remove every other element, I still have an infinite number of elements in both A and B.

At any point in the sequence, I can say that the union of A, B, and C (or the discarded numbers) is the set of natural numbers, and that the intersection of any two is empty (and therefore the intersection of all three). The initial condition is A=N, and the final condition is C=N. The process is conservative (numbers/balls are only moved, not created or destroyed), so the final condition must also be A=B={}.

To see this, restate the problem.

[etc...]

Please describe how in either case bin B eventually becomes empty.

It depends which ball is moved/discarded. They are not all equivalent, because you can choose from all members of A, or from just a subset of A, as described above.

If talking about a finite number of operations and then letting that number approach infinity is fallacious [it isn't] your argument is fallacious.

You can only define the limit of a sequence if it converges to a finite value as its argument approaches infinity. In this case the sequence is simply (n), which does not converge. My logic does not require the limit to be defined.

A proper subset of a countably infinite set can also be countably infinite [the set of even integers is a countably infinite proper subset of the set of integers].

Thus, just because you have accounted for a countably infinite number of natural numbers [those removed from bin B] it does not follow that you have accounted for all of them.

I am not claiming that A and B are empty because an infinite number of elements are discarded. You could, for example, devise a sequence by which you end up with the prime numbers in A, all other odd numbers in B, and all even numbers greater than 2 discarded. All of these are countably infinite subsets of N. It is because we are always discarding the lowest valued existing element from the remaining natural numbers in A and B that the discarded numbers must comprise the entire set of natural numbers, and are therefore all accounted for. If you can provide an example of just one element of A that does not get discarded, then I will accept my argument as false.

Consider the sequence {1, 2, 3, ..., N, N+1, N+2, N+3, ... 2N}

Remove the first N of the elements.

Let N approach infinity.

You will have removed a countably infinite subset of integers from the sequence.

Does that mean you have removed all the elements of the sequence?

[etc...]

You cannot let N->infinity because the concept of infinity+1 or 2*infinity is invalid.

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• 0

There's no difference between removing half of the balls by taking every other ball or taking the first half of the balls.

The number of balls that remain is the same.

But since you're still wriggling around in your quicksand, try this:

Wipe all the numbers off the balls.

1. Move two balls from A to B.
2. Move one ball from B to C.
3. Repeat.
Note this is equivalent to
1. Move one ball from A to B.
2. Move one ball from A to C.
3. Repeat.
I'll either enjoy - or be embarrassed for you - if you try to get bin B empty this time.

Have at it. ##### Share on other sites
• 0 Bin A has infinite number of balls.

After the first move Bin A has (infinite -2) balls and we all know the result is infinite and Bin B has 1 ball, since 2 goes in and one goes out.

After the 2nd move Bin A has (infinite -2) balls = infinite balls and Bin B has 2 balls.

After (infinite - 1) move Bin A has (infinite -2) balls = infinite balls and Bin B has [1 + (infinite - 1)] = infinite balls.

Therefore after infinite moves Bin A has infinite balls and Bin B also has infinite balls.

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• 0 The original task as stated is not possible -

1. Bin A would have to be infinitely large to contain an infinite number of ping pong balls (unless the balls were infinitely small!)

2. If each ball has a unique number on it and have to be transferred in order, it would take an infinitely long time to find balls 1 and 2 in Bin A so you could never transfer any.

We are asked for just the final number in each bin. Therefore the task could be altered so that you just take any two balls at random from bin A and transfer them to bin B. However, with each transfer you have to transfer them faster and faster from A to B, to transfer them in less time. Well before the minute is up, you will be moving balls from bin A to B at the speed of light which cannot be exceeded so the task cannot be completed any way!

Alice

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• 0 There's no difference between removing half of the balls by taking every other ball or taking the first half of the balls.

The number of balls that remain is the same.

I believe I (and others) have already provided ample evidence that it makes all the difference. You, however, still have have not provided any evidence of there being any member or subset of N remaining in either bin A or B. I ask again: if bins A and B are not empty, please identify at least one object in either.

But since you're still wriggling around in your quicksand, try this:

Wipe all the numbers off the balls.

1. [etc...]

Once you wipe the numbers off, the balls are no longer distinct, and therefore do not constitute a set. They may represent some random variable now, but I have already contended that you cannot select randomly from an infinite number of objects. Have at it. En garde! ;-)

I think where our thoughts differ is that you consider the balls to represent a function of sorts, whereas I consider them to be a set. In the former case, the function, or sequence, does not converge to any value as its argument approaches infinity, and its behavior, especially any summation, is undefined. My approach does not require a limit to exist. If I can paint unique sequential numbers on an infinite number of balls and place them in a bin, I must just as easily be able to reverse the operation and empty it. If you adhere to your logic, you should be arguing that you could never fill the bin in the first place, not that it cannot be emptied. Once you allow for the existence of the bin (or the set of natural numbers), you must also allow that it can be exhausted.

D.

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• 0

I stand in awe [i think it's awe ...] of your thinking. I'll close my thoughts by relaying a dream I had last night.

I know it was a dream, because in real life I don't have a butler.

Enjoy.

One night I had a group of friends to dinner.

Because they had not all previously met, I provided them with name tags.

At one point, hoping to determine whether they had all arrived, I sent my butler to the drawing room to count them.

Forty-two guests are present, he informed me, we are waiting on the final six to arrive.

After the string quartet had finished their last minuet, I asked the butler to count them once again.

When he returned he said:

I'm sorry sir, but I'm afraid the guests have removed their name tags, so I have no idea how many are here.

I fired the butler.

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• 0

nice hehe

question: was that actually a real dream? Edited by unreality

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• 0 I was just discussing a different topic with a friend of mine, who brought to my attention the fact that a function or sequence need not converge to a finite value for the limit to be defined, it need only converge monotonically to a value. I don't agree that infinity is a value, but apparently mathematicians do consider it to be one in the case of a limit. I therefore retract my statements in several posts on this topic attempting to invalidate Bonanova's argument based on the limit not being defined. Doing so, however, does not change any of my arguments, or my position on this subject.

I'm off to play baseball now.

D.

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• 0

Here is a simple re-mapping procedure:

1. Every time you remove 2 balls from bin A, chose the next two numbers not divisible by 3. E.g., (1,2), (4,5), (7,8)... etc.

2. Upon placing removed balls into the bin B, wipe out the numbers and initialize the balls with new numbers sequentially. (1,2), (3,4), (5,6), ... etc.

3. Throw away odd numbered ball from each pair.

Those operations are equivalent to the OP in terms of the number of balls moved and thrown away. After one minute has elapsed and you have performed an infinite number of operations you'll have an infinite number of even numbered balls in bin B; and infinite number of balls with numbers divisible by 3 plus some more in bin A.

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• 0 I was just discussing a different topic with a friend of mine, who brought to my attention the fact that a function or sequence need not converge to a finite value for the limit to be defined, it need only converge monotonically to a value. I don't agree that infinity is a value, but apparently mathematicians do consider it to be one in the case of a limit. I therefore retract my statements in several posts on this topic attempting to invalidate Bonanova's argument based on the limit not being defined. Doing so, however, does not change any of my arguments, or my position on this subject.

I'm off to play baseball now.

D.

This is not true. Infinity is not value when considered as part of a limit (and most other places, but I'm not prepared to say "all"). When a mathematician says the limit of something "equals infinity" what they are using is a shorthand method of saying it increases without bound. This means it doesn't converge, and definitely is not defined. This has been drilled into me by many professors and teachers, so I will have to side with them on this issue.

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• 0
... This means it doesn't converge, and definitely is not defined. ...

When an infinite series does not converge, it does not mean that it's not defined. If it was not defined, we wouldn't be able to discuss it intelligently.

Mathematicians have been tackling with understanding of infinity for centuries and came to a certain concensus in some respects and not so much in others.

The OP gives an illustration of a typical mapping problem. With such problems the mathematical concensus has been reached long time ago.

Regard famous Galileo's paradox, where two infinite sets, one of all natural numbers, anoher -- of all their squares are mapped one to one.

1 --> 1

2 --> 4

3 --> 9

4 --> 16

and so on. On one hand, there is one to one correspondence between the two sets, on the other hand, second set does not seem to have some numbers that the first set has. Which is bigger? The thing is, assigning numbers to the memebers of an infinite set, does not change the nature of that set. You may or may not be able to do so (assign numbers). Since some infinite sets are, indeed, larger than others. A typical example is an infinite set of all natural numbers versus infinite set of all real numbers between zero and 1. The latter is larger than the former. That is, you cannot make one to one correspondence between the set of real numbers (0<=r<1) and set of all natural numbers from 1 to infinity, like we did in Galileo's paradox above. Some real numbers will always be left unpaired with natural number. But tat's more advanced topic.

With respect to the OP, we are dealing wih infinite sets of the same order here, where one to one correspondence can always be found. One property of infinite set, which mathematicians had agreed upon long time ago: you can divide infinity by any finite number and the result would be an infinity. Furthermore, the resulting infinite set is not "smaller" than the original set.

I'll refer you to another well known example -- Hilbert's Hotel. Famous mathematician David Hilbert used it as an illustration in his lectures (a century or so ago). The variation that I like goes like this:

In a small town N, there is an hotel with infinite number of rooms, all of which are occupied. There happens to be a superball game in that town, and an infinite number of fans come to watch the game. As a manager of that hotel, how would you place the newcomers?

Answer: Announce on your infinite intercome for all room occupants to multiply their room number by 2 and move there. This way you'll have infinite number of odd-numbered rooms vacant.

In conclusion, I'd recommend to reading "Infinity and the Mind" by Rudy Rucker to all who is interested in the subject. It's a popular, yet in-depth well-written book.

Edited by Prime

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• 0
After reading unreality's problems dealing with infinite numbers, I decided to post my own. To be honest, it's not "my own," since I read it in a book a few years ago (I don't remember which book, but it was really good. You should all read it).

Suppose you have 2 bins, labeled 'A', and 'B'. To start, bin A has an infinite number of ping pong balls, and they are numbered sequentially from 1 to infinity. Each ball has a unique number printed on it. Bins B is empty to start.

You begin an experiment that will last 1 minute, but will include an infinite number of iterations. Each iteration will take exactly half of the time remaining - ie, the first iteration will be completed after 30 seconds, the second after 45, the third after 52.5, etc. At each iteration, you do two things: 1st, take the two lowest numbered balls out of bin A and put them in bin B; 2nd, take the one lowest numbered ball out of bin B, and throw it away.

So, in the first 30 seconds, you take balls 1 and 2 out of A and put them in B, then take ball 1 and throw it away. In the next 15 seconds, you take balls 3 and 4 out of A, put them in B, then throw away ball 2. As you can see, the number of balls in bin A will decrease by 2 with every iteration, and the number of balls in bin B will increase by 1.

The question is this: after the entire minute has elapsed, how many balls will be in bins A and B?

I don't know

that one minute will never elapse (reminds me of Xenon's race paradox)because you will never finish repeating those actions!!!

So... ##### Share on other sites
• 0 After reading unreality's problems dealing with infinite numbers, I decided to post my own. To be honest, it's not "my own," since I read it in a book a few years ago (I don't remember which book, but it was really good. You should all read it).

Suppose you have 2 bins, labeled 'A', and 'B'. To start, bin A has an infinite number of ping pong balls, and they are numbered sequentially from 1 to infinity. Each ball has a unique number printed on it. Bins B is empty to start.

You begin an experiment that will last 1 minute, but will include an infinite number of iterations. Each iteration will take exactly half of the time remaining - ie, the first iteration will be completed after 30 seconds, the second after 45, the third after 52.5, etc. At each iteration, you do two things: 1st, take the two lowest numbered balls out of bin A and put them in bin B; 2nd, take the one lowest numbered ball out of bin B, and throw it away.

So, in the first 30 seconds, you take balls 1 and 2 out of A and put them in B, then take ball 1 and throw it away. In the next 15 seconds, you take balls 3 and 4 out of A, put them in B, then throw away ball 2. As you can see, the number of balls in bin A will decrease by 2 with every iteration, and the number of balls in bin B will increase by 1.

The question is this: after the entire minute has elapsed, how many balls will be in bins A and B?

Let F1 and F2 represent functions of natural numbers that yield natural numbers.

If,

limit(N->infinity): F1(N)=infinity

limit(N->infinity): F2(N)=infinity

then limit(N->infinity): F1(N)-F2(N) is not any of:

* Negative infinity

* Zero

* Infinity

It's merely undefined.

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• 0 What color are the balls? heh, that reminded me back in highschool in a physics class.. We were given a question that had something to do with balls and my answer was "I need to know the color"... It wasnt the right answer Edited by killaklown

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• 0

Resurrecting a discussion ...

Once you wipe the numbers off, the balls are no longer distinct, and therefore do not constitute a set. They may represent some random variable now, but I have already contended that you cannot select randomly from an infinite number of objects. Regarding wiping the numbers from the balls..

You see immediately that each operation puts a ball into B and another into C [or discards it]

If you want the identity of a ball in B after it's all over, simply place a mark on the first ball to get into B.

You contend two things, without proof.

1. If you wipe the numbers off the balls they do not constitute a set.
2. You cannot select randomly from an infinite number of objects.
Let's see if they make sense.
1. I have 15 balls. 5 are red, 5 are blue, 5 have no identifying marks.
The 15 balls do not constitute a set because they are not numbered.
Or perhaps the 5 red balls are a set because they are red and the 5 blue balls are a set because they are blue.
But the 5 unmarked balls are not a set because they are unmarked.
Interesting.

If the numbers are erased consider that bin A is an infinitely long trough, and the balls are ordered, left to right.
Now they are distinguishable, so even your criterion makes them a set.
Take the first two and put one of them into B [and put a dot on it] and the other into C [or discard it].
Repeat.
That ball with a dot is the identity of one ball that is in bin B, as requested.

2. You cant choose randomly from an infinite set.
From the countably infinite set of rational numbers on the interval [3,4) i choose the number 3.14
Prove that that choice was not random.

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• 0 From the countably infinite set of rational numbers on the interval [3,4) i choose the number 3.14

Prove that that choice was not random.

Do you mean uniformly random over the interval [3,4)? If so, I can prove that the number I just read on this forum was not. Proving that your actual choice was not random is more difficult.

There is only a certain amount of computer memory in the world. More limiting, brainden has a much smaller amount of memory available, as does my PC. While that is rarely if ever a problem for any practical number, the fact is that there are some (in fact, an infinite number of) rational numbers between 3 and 4 which cannot possibly represented on a computer. The probability of me seeing your choice of these numbers on this forum is then 0, and the distribution from which you chose 3.14 is not uniform on [3,4). QED

I could make a similar argument about your actual choice of numbers, since there are an infinite number of numbers which you could not possibly express as a choice in any manner before your own death, or the death of the universe for that matter.

Going back to your original statement, if you were not restraining yourself to uniform randomness, then it is very likely that your choice was random. You certainly could have written 3.1416, or 3.1415926, or 3.5, and the fact that you chose 3.14 is (or can at least be described as) random.

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• 0 wouldnt they both be empty?      <_<  ##### Share on other sites
• 0 Resurrecting a discussion ...

I disappear for a couple of days, and look what happens... thought you could slip one past ol' D, did you?

Just so I know though, are you really not seeing where I'm coming from, or are you just having a laugh now? I don't mind either way...

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