20 Oct 10
1 edit
Originally posted by AThousandYoungIt's a question how you define the problem. If I put all the balls I would have first and then remove the same balls I would have put in: {1,2,3,...} and removed all the balls I would have {1,2,3,...} then it's easy to see nothing Same for Achilles. I can breakdown any operation into a sum of infinite ones.
Iteration of what? He is not performing repeated actions over and over again like in the ball swap problem.
I the Achilles case one can breakdown the race such that with each iteration he gets half way to the tortoise. He reaches the tortoise, but not in any finite iteration.
20 Oct 10
Originally posted by PalynkaOne thing that possibly makes a difference is that Achilles' actions get smaller, so that although there are an infinite number of them they converge to a finite result. Whereas with the balls, the "size" of the action does not decrease. Which is what lets some people argue that the infinite number of steps in finite time is not possible (and so the problem is invalid).
In the Achilles case one can breakdown the race such that with each iteration he gets half way to the tortoise. He reaches the tortoise, but not in any finite iteration.
20 Oct 10
Originally posted by mtthwYep, I agree that is a much more defensible position. I mentioned exactly that on my reply to ATY's link on last page. But the quote he mentioned invokes no such convergence condition.
One thing that possibly makes a difference is that Achilles' actions get smaller, so that although there are an infinite number of them they converge to a finite result. Whereas with the balls, the "size" of the action does not decrease. Which is what lets some people argue that the infinite number of steps in finite time is not possible (and so the problem is invalid).
But then again, using set theory axioms, the way we construct sets like the set of natural numbers is by iterating to infinity... So we have to be careful not to jettison that or at least be aware that you are doing so.
20 Oct 10
1 edit
Originally posted by PalynkaIt appears to me that we are left with infinite balls in the sack (balls in the sack tee hee).
Another way to look at it:
What's the set of balls put into the bag?
{1,2,3,4,5...}
What's the set of balls removed from the bag?
{1,2,3,4,5,...}
The sets are the same!
The simplest way to see it is that for every ball removed from the sack two more are added, hence the net result is that one ball is added at each time.
I think that if no mention of numbered balls was made everybody would just say that the answer is infinite because at each step one net ball is being introduced in the sack.
Your first argument only works if the sequence is bounded above and our sequence isn't bounded above since there are infinite halves.
And this last argument seems to fail because even the sets do have the same elements the problem is that don't have it at the same time.
😕
But I do have to say that this apparently simple problem really is head spinner. I'll post on some reputable mathematical blogs and see what the big shots answers will be.
Edit: After having read the rest of the thread I've found out that mathematicians also have teir heads spinning about this "simple" problem...
But the reason of this edit is just to say that I don't feel that the comparisons of this problem to Achiles and the tortoise isn't fair. The thing is that in the Achiles problem the apparent paradox rests on the notion if the distance can ever be 0 or not. And this is neatly solved if we assume that distance is a continuous concept and so one can use real analysis to prove that the distance can in fact be 0. Now in this problem the apparent paradox arises in the question of how many balls are left. And this is a discrete concept.
As a side note: I don't know if all of you know this or not (and I'm willing to take the risk of sounding pedantic - and I'll just do that because I love Zeno paradoxes):
Zeno had more than 20 paradoxes (or maybe Zeno collected more than 20 paradoxes. Questions of originality are really complicated when we go this far back in time) that questioned the nature of space and time.
Not all of them have reached us but we do know four of his paradoxes and these paradoxes are absolutely delicious: two of them assume that time and space are continuous quantities and end up showing that under these assumptions motion (change) is impossible.
The other two start by assuming that time and space are discrete quantities and end up showing that under this new set of assumptions motion (change) is also impossible
So that leaves us with the inescapable fact that motion (change) is impossible and yet we see motion (change) happening all the time.
Sorry for this sideline but Zeno's paradoxes are a very old love affair...
20 Oct 10
Originally posted by adam warlockWhat is my first argument and why does it only work if the sequence is bounded from above?
It appears to me that we are left with infinite balls in the sack (balls in the sack tee hee).
The simplest way to see it is that for every ball removed from the sack two more are added, hence the net result is that one ball is added at each time.
I think that if no mention of numbered balls was made everybody would just say that the answer is infinit ...[text shortened]... he time.
Sorry for this sideline but Zeno's paradoxes are a very old love affair...
20 Oct 10
Originally posted by adam warlockBut I am going to iterate until infinity so the speed of convergence towards the set of natural numbers is irrelevant to the fact that both result in the same set.
And this last argument seems to fail because even the sets do have the same elements the problem is that don't have it at the same time.
20 Oct 10
Originally posted by PalynkaGod Damn you! You're right...
But I am going to iterate until infinity so the speed of convergence towards the set of natural numbers is irrelevant to the fact that both result in the same set.
But if the balls aren't numbered the result is infinite...😲😲😲😲😲😲😲😲😲😲
I'm really pissed at this... 😠😠😠😠😠😠😠😠😠😠😠😠😠😠ðŸ˜
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21 Oct 10
4 edits
I'm pretty bored with this, but I'll just say.
Create a graph.
X axis is step.
Draw the line Y = 2x, at integer X this line shows the number of the ball added to the bag at step X.
Draw the line Y=x, at integer X this line shows the number of the ball removed from the bag at step X.
These lines are provably straight, and provably intersect at step 0, they provably cannot intersect anywhere else, and hence at infinity the balls removed from the bag are not equal to the balls added.
I think the mark of the true answer is that it can be proved with various different approaches, the other answer can only be "proved" with weird juggling of infinite sets, and is dependent on ball numbering and stuff.
21 Oct 10
Originally posted by iamatigerWe've been there before. Using the same argument you could "show" that the natural numbers are infinitely more than the even numbers. Just expand finite sets including in set A all even numbers up to 2*n and in set B all the natural numbers up to 2*n. Do this sequentially and you'd have the same exact graph plotting their cardinalities.
I'm pretty bored with this, but I'll just say.
Create a graph.
X axis is step.
Draw the line Y = 2x, at integer X this line shows the number of the ball added to the bag at step X.
Draw the line Y=x, at integer X this line shows the number of the ball removed from the bag at step X.
These lines are provably straight, and provably intersect at ...[text shortened]... "proved" with weird juggling of infinite sets, and is dependent on ball numbering and stuff.
Yet for the infinite sets the cardinality of natural numbers and even numbers is the same.
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22 Oct 10
Originally posted by Palynkaso the lines intersect at infinity?
We've been there before. Using the same argument you could "show" that the natural numbers are infinitely more than the even numbers. Just expand finite sets including in set A all even numbers up to 2*n and in set B all the natural numbers up to 2*n. Do this sequentially and you'd have the same exact graph plotting their cardinalities.
Yet for the infinite sets the cardinality of natural numbers and even numbers is the same.
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25 Oct 10
Originally posted by PalynkaPossibly the problem lies in the difference between "the same cardinality" and equality?
We've been there before. Using the same argument you could "show" that the natural numbers are infinitely more than the even numbers. Just expand finite sets including in set A all even numbers up to 2*n and in set B all the natural numbers up to 2*n. Do this sequentially and you'd have the same exact graph plotting their cardinalities.
Yet for the infinite sets the cardinality of natural numbers and even numbers is the same.
And the problem with the balls in the bag being zero is that infinity-infinity=infinity?