I really like this question.
The sum of an infinite amount of 1’s (the dots in this case) is undefined.
Except when he finishes. 
infinitely many dots, since he never erases any. I’m guessing it’ll look a lot like his signature–well, that’s what I would do.
Let’s go back to that one chip with the dots. Let’s leave that chip untouched, but add more chips to it. God starts with a one-dot chip. On the first move, he adds a dot to it. On the 2nd move, he adds a dot to it & adds another chip with 4 dots. On the 3rd move, he adds a dot to each of those chips & adds another chip with 6 dots. On the nth move, he adds a dot to each chip that is already there and adds a new chip with 2n dots. As a sequence of sets, this can be written: {2},{3,4},{4,5,6},{5,6,7,8},…,{n+1,…,2n},… , with each element in the nth set representing one chip in the dish after n moves, and the number of the element representing the number of dots on that chip. This is the same as the sequence of sets for the original procedure. In this case, though, we do not end up with 0 chips. We end up with infinitely many chips, each with infinitely many dots on them. Or, if we follow ultrafilter’s statement, the number of chips at the end is undefined, and the number of dots on any given chip is undefined.
The sequence of sets for the 2 procedures are the same. So do the results have to be the same? Let’s tweak this new procedure a little to make that question easier to answer (I hope). Instead of adding 1 dot to every chip that’s already in the dish, on the nth move let’s add n-1 dots to the chip with the smallest number of dots on it (n). Before, when going from {4,5,6} to {5,6,7,8}, old 4 became new 5, old 5 became new 6, old 6 became new 7, and 8 was brand new. Now, old 4 becomes new 7, old 5 & 6 stay 5 & 6, and 8 is brand new. This is a lot closer to the original procedure. And of course the timing remains the same as always.
Now, let’s imagine that God takes chip #4 (or n) and adds 3 (or n-1) dots to it on the 4th (or nth) move. He also happens to have another chip available that is identical to this one - with exactly 7 (or 2n-1) dots on it. He takes his newly made 7 (or 2n-1) and his previously available 7 (or 2n-1) and holds one in each hand, trying to decide which to pick. He inspects them and finds them identical. Maybe he shakes them in his hands to mix them up & asks you which one to put into the dish. They look completely identical to you (though of course He knows their history), and so you pick one. Of course, all of this happens in no time flat.
Now, according to some arguments here, if you always pick the previously available chip, then the dish will end up empty. If you always pick the chip the he just took out and changed, then the dish will have infinitely many (or an undefined number of) chips in it at the end. How can the number of chips in the dish depend on which of 2 seemingly identical chips you choose to put in?
If you’re thinking that the chip’s history is somehow essential, I can make this example more magical and mysterious. God does not need to pick up and move chips - he is all powerful. Instead, he can merely hold the new chip (7 or 2n-1) in one hand, and point to the chip in the dish (4 or n). And ZAP, the next thing you know the chip in the dish has 7 dots on it (or 2n-1) and the one in his hand has only 4 (or n). And you ask God - what were you doing? Did you make the 2 chips switch places? Or, did you just add dots to the chip in the dish & remove dots from the chip in your hand, possibly transferring some of the dots on the chip in your hand to the one in the dish? And God says: “What’s the difference?”
I was tempted to be a smartass and say that there is now a line segment on the coin - except that I know that’s a higher cardinality than integers, so that’s wrong. How about a Cantor dust?
But the dots are finite, not infinitesimal, so it’s OK.
Thank you, knock knock, for creating some interesting variations on the problem.
However, I think that an important change is made when we switch from decimal notation to dot-notation. Decimal notation, by definition, can only represent finite quantities, but dot-notation is much more powerful: it can be used to represent infinite quantities. In fact, dot-notation could be used to represent any infinite cardinal whatsoever. Therefore, this is a very non-trivial change to the parameters of the problem.
With this in mind, here are my answers to your scenarios.
Case IA: “Suppose god gets tired of dealing with an infinite number of chips. He just has one chip. He decides that this chip must always be labeled with one of the positive integers. At first he puts a “1” on it. After an hour, he erases the 1 and makes it a 2. A half hour later, he changes it to a 3. 15 minutes later, he changes it to a four. And so on. At any time before 2 hours, he has 1 chip with a positive integer on it, and as t-> 2 hours (from the left) that integer approaches infinity. Now, what does god have at the 2 hour mark?”
My Answer: At the 2 hour mark, God has a blank chip. When Cabbage gave this answer, you responded, “How could the chip be blank? Every time god erases a number from it, he simultaneously inscribes a new number on the other side.” Ah, but God can write an integer on the chip only if there are any integers left to write. But at the two hour mark, he’s run out of integers, and so can write nothing. Moreover, he’s erased everything he ever wrote. Thus the chip is blank.
Case IB: “Or, imagine a number system where each number n is represented by n dots. So 1 is . 2 is : 3 is :. and so on. For each move, God writes a new number on it - that is, he adds a dot. After infinitely many moves, there are infinitely many dots, right? Or are there somehow 0 dots?”
My Answer: There are infinitely many dots on the chip. Unlike in the case were God was restricted to writing integers, He now has something to write at the two-hour mark: infinitely many dots.
Case II: “He doesn’t even have to pick up the piece he removes. While leaving [a coin] in its place in the stack, he reaches around it with his hand, and uses his powers to transform it into the 2 newly labeled pieces. I don’t know if he breaks it in half and relabels it, or if he melts it down and reforms it - it doesn’t really matter. But he uses the same gold that was there. . . . God starts with a solid 1-kg cylinder of gold, with a label on it (2). On each move, God leaves the cylinder where it is, but he breaks it in pieces and relabels the pieces.”
My Answer: God has 1-kg of gold, broken into infinitely many peices, each of which is 0-kg. In fact, if I’m not mistaken, he has uncountably many pieces. Can God write an integer on such insubstantial films of nothingness? If we allow that he can, then I’d say that they are all blank because he ran out of integers to write on them. Certainly, almost all of them are blank, because there are uncountably many pieces and only countably many integers.
Case IIIA: “Let’s go back to that one chip with the dots. Let’s leave that chip untouched, but add more chips to it. God starts with a one-dot chip. On the first move, he adds a dot to it. On the 2nd move, he adds a dot to it & adds another chip with 4 dots. On the 3rd move, he adds a dot to each of those chips & adds another chip with 6 dots. On the nth move, he adds a dot to each chip that is already there and adds a new chip with 2n dots. As a sequence of sets, this can be written: {2},{3,4},{4,5,6},{5,6,7,8},…,{n+1,…,2n},… , with each element in the nth set representing one chip in the dish after n moves, and the number of the element representing the number of dots on that chip.”
My Answer: I agree with your answer. We end up with infinitely many chips, each with infinitely many dots on it.
Case IIIB: “Let’s tweak this new procedure a little to make that question easier to answer (I hope). Instead of adding 1 dot to every chip that’s already in the dish, on the nth move let’s add n-1 dots to the chip with the smallest number of dots on it (n). Before, when going from {4,5,6} to {5,6,7,8}, old 4 became new 5, old 5 became new 6, old 6 became new 7, and 8 was brand new. Now, old 4 becomes new 7, old 5 & 6 stay 5 & 6, and 8 is brand new.”
My Answer: Again, we end up with infinitely many chips, each with infinitely many dots on it.
Cases IIIC and IIID: “Now, let’s imagine that God takes chip #4 (or n) and adds 3 (or n-1) dots to it on the 4th (or nth) move. He also happens to have another chip available that is identical to this one - with exactly 7 (or 2n-1) dots on it. He takes his newly made 7 (or 2n-1) and his previously available 7 (or 2n-1) and holds one in each hand, trying to decide which to pick. He inspects them and finds them identical. Maybe he shakes them in his hands to mix them up & asks you which one to put into the dish. They look completely identical to you (though of course He knows their history), and so you pick one. Of course, all of this happens in no time flat.”
"I can make this example more magical and mysterious. God does not need to pick up and move chips - he is all powerful. Instead, he can merely hold the new chip (7 or 2n-1) in one hand, and point to the chip in the dish (4 or n). And ZAP, the next thing you know the chip in the dish has 7 dots on it (or 2n-1) and the one in his hand has only 4 (or n). And you ask God - what were you doing? Did you make the 2 chips switch places? Or, did you just add dots to the chip in the dish & remove dots from the chip in your hand, possibly transferring some of the dots on the chip in your hand to the one in the dish? And God says: “What’s the difference?”
My (non)Answer: I’m going to have to think more about these when I have the time. Right now, my inclination is to say that, God’s protestations to the contrary, it really does make a difference which it is that God is doing. See this post of mine from the previous page where I gave some of my thoughts on a related phenomenon.
I really like knock knock’s analysis of the cylinder of gold, because we don’t need to be God to pull that one off. Let’s say that, instead of a physical cylinder of gold, we just use the line segment [0,1), and we’re dividing it into smaller and smaller line segments (for sake of definiteness, let’s consider all of the subsegments to be half-open, so they can exactly add up to the original segment). I can define a labelling scheme as a function of time which exactly simulates the gold cylinder situation, and just say that that’s the labelling scheme. But if I do write down that labelling scheme, that doesn’t mean that two hours later, the number line will all of a sudden have to skip directly from 0 to 1. The line segment is still there. The manner in which it’s divided up is not well-defined after two hours, but the segment is there nonetheless.
With due respect to Tyrell McAllister’s excellent analysis so far, I’ll add my comments on this part :
Calling them ‘seemingly identical’ is only circular reasoning, though. There is a distinction between the chips – those that have been in the bowl and those that haven’t been in the bowl.
If IIIC = use the previous (nth) chip and IIID = use the new (2n-1) chip, in IIIC a chip put in the bowl is never removed; in IIID it is. Suppose we paint the #4 chip black. Now looking at the state of the bowl after each step, we have :
IIIC IIID
1 1
2 2
3 **4** 3 **4**
**4** 5 6 **4** 5 6
5 6 **7** 8 5 6 7 8
6 **7** 8 9 10 6 7 8 9 10
**7** 8 9 10 11 12 7 8 9 10 11
8 9 10 11 12 **13** 14 8 9 10 11 12 13 14
etc. etc.
In IIIC, at no time does the black chip ever leave the bowl. It gets marked with some higher (odd) number. In fact, had we painted all the (originally even) chips black, we end up with a bowl full of black chips in IIIC. In IIID, the chips actually leave the bowl. So those cases are not identical.
Now, if God’s just ‘zapping’ the chips, everything’s fine up until t=2. But what actually happens at infinity? Following the ‘blank coin’ example, there cannot be a ‘infinity’ coin to zap with. I’ll admit this doesn’t explain much, but I have difficulty trying to explain how one ‘accomplishes’ an infinite number of tasks.
ZenBeam, I’m sorry, it’s the end of the semester, and all of a sudden I’ve got a big pile of work to do. I’ll try and make a response to you now, but it’s going to be much briefer than I’d like.
Actually, I haven’t addressed your infinite series example specifically because I’m not sure you wrote what you intended to write. We know
log(2) = 1 - 1/2 + 1/3 - 1/4 +…
but that’s not what I’m seeing in your second example, which reads to me as
1 + 1/2 - 1 + 1/3 + 1/4 - 1/2 + 1/5 + 1/6 - 1/3 + …
I don’t know offhand what that series converges to (if it converges at all).
I figure that’s probably irrelevant to your point, so I was trying to address your point specifically, which seems (to me) to be summed up in two statements:
- The position of the fly always (even at the end) corresponds to the number of chips in the box (or maybe the height of the chips).
- The movement of the fly is continuous.
None of this changes my position. If I grant you 1., I simply can’t grant you 2. If I grant you 2., I simply can’t grant you 1. The fly is merely an additional element to the experiment–maybe it moves continuously, maybe its position always corresponds to the height of the chips, but I claim it can’t do both.
My apologies if I’ve actually missed your point entirely.
knock knock, I applaud you for your interesting variations on the problem.
Tyrrell McAllister, I applaud you for your analysis of knock knock’s variations.
I wish I could respond more, but I gotta get back to work. Hopefully in a day or two I’ll be able to participate more.
Try this:
log(2) = 1 + (1/2 - 1) + 1/3 + (1/4 - 1/2) + 1/5 + (1/6 - 1/3) + …
Anyway, my point remains that I’m trying to use the same methods Tyrrell McAllister used to determine the dish is empty on a different problem, that of the fly in example 2. Forget about the chip thicknesses. I’m using Tyrrell McAllister’s proof because he’s the only one who attempted to actually show it with better than handwaving arguments.
Can anyone (and I’m especially looking at the “dish is empty” crowd, although those of you with outside lives get a bit of a pass) confirm or deny whether I’ve correctly modified Tyrrell McAllister’s proof in this post to show that the fly is at zero for t=2? Is the fly at 0 or at log(2) at t=2? Surely, if everyone can answer the OP, whcih doesn’t even converge, someone can figure this one out.
While we’re at it, what about Tyrrell McAllister’s original proof showing the dish is empty, about three posts prior to the linked one. Is this a valid proof in the eyes of the rest of the DIE gang?
In the case where we remove the largest, the sum of all the numbers on the chips in the bowl—whatever those numbers are—at time k is
[symbol]S[/symbol][sub]n=0[/sub][sup]k[/sup] (2n+1) = (k+1)[sup]2[/sup]
In the case where we remove the smallest, the sum of all the numbers on the chips in the bowl—whatever those numbers are—at time k is… well, I can’t develop a series for it as the terms change each time n changes, but there is a recursive sequence of partial sums, which is
k[sub]n+1[/sub] = k[sub]n[/sub] + 5 + 3(n-1); k[sub]0[/sub]=2
If the sum of the numbers on the chips grows without bound, then I don’t think one can claim the bowl is ever empty, whether we can name a single number or not.
I’m so confused.
But if we cannot create a series representation of the second case, I’m not sure I can say my argument holds any water.
Thanks everyone for your compliments, and for your interesting arguments. I’m mostly going to be responding to Tyrrell McAllister’s post here, and, since I don’t want to make my post longer than it already is with an unwieldy amount of cutting and pasting, I’ll just refer to the cases as they have been nicely labeled above and give a few important quotes.
Case IA: “Ah, but God can write an integer on the chip only if there are any integers left to write. But at the two hour mark, he’s run out of integers, and so can write nothing. Moreover, he’s erased everything he ever wrote. Thus the chip is blank.” I don’t see how this “running out” of things to write can happen any sooner than a running out of things to erase. Every time God is in the process of erasing an integer n, he can always write another integer, namely n+1. Imagine that God had this quirk - he abhors a blank chip more than nature abhors a vacuum. So every time he erases a number from a chip, he is always sure to write another number on the chip simultaneously, if not sooner. Now, would this keep God from completing the task I described? I think He’d still be able to do it just fine, because this is precisely what he does on every move - he replaces a number with a higher one. And there’s always exactly one number written on the chip.
Case II: (the cylinder) “God has 1-kg of gold, broken into infinitely many peices, each of which is 0-kg. In fact, if I’m not mistaken, he has uncountably many pieces.” In this case I believe you are mistaken, possibly because I wasn’t quite clear to begin with. On each move, God only breaks one of the pieces of the cylinder in half (and it’s always one of the largest remaining pieces of the cylinder). I was trying to keep this problem isomorphic with the original. The number of pieces of cylinder after n moves = number of dots on coin after n moves = number of chips in dish after n moves. The number of pieces increases by 1 each move, so it stays countable.
“I’d say that they are all blank because he ran out of integers to write on them.” I’m not sure if this comment is based on the belief that there are uncountably many pieces. If he used dot notation, would they still be blank?
You agree that the cylinder is there, and that its mass is still 1 kg., though it’s composed of infinitely many “insubstantial films of nothingness.” Now, topple the cylinder over. Knock it into an amorphous pile (but don’t break or combine anything). Now, maybe you can’t see it, but, if the whole thing is sitting on a scale, the scale should still read 1 kg.
And this would still happen, presumably, if each move consisted of taking out a coin (i.e. piece of cylinder), melting it down, and smelting 2 new coins. Although I guess you would say these coins wouldn’t really be “new”. I gather that you would say that the crucial difference between the different conditions I stated in that post is whether the coin going into the dish is made out of the material that was taken out, or if it was made out of new material.
Case IIIC,D: And here’s the fun one. I’m not sure, panamajack, if you’re allowed to paint a chip black. The 2 chips you choose between are supposed to be indistinguishable except for their history. Tyrrell McAllister’s historical view is intriguing, but here’s an argument for why history doesn’t matter:
Say God is on an arbitrary one of his moves (call it the nth) and He has kindly stopped time to ask you the question: which chip should go in the dish? Now, the chips look identical to you, and you are feeling very nervous and concerned about your choice. You have a bet with your friend who says that there won’t be any chips in the dish at the end of this. You bet that there will be chips, and you’re suspecting infinitely many of them. You’re feeling a lot of pressure because you are familiar with arguments that show that the result depends entirely on your choices (and also because you’re telling God what to do, but you’re already kinda getting used to that). But then you realize: after this move, there will be n chips in the dish, whatever choice I make. So I can start from there, with the problem beginning with n chips. Those n chips can be the special, marked chips that I worry about. My choice here is irrelevant.
And you’re right. You can choose whichever chip on the first n moves, and then choose right (i.e. as in Case IIIC) the rest of the way, and the number of chips will end up infinite.
So, on any move n, your choice is irrelevant. Since n was an arbitrary move, your choice on any move is irrelevant. So all of your choices are irrelevant.
And I just realized that my argument began with the statement “here’s an argument for why history doesn’t matter:” and ended “So all of your choices are irrelevant.” I could change that, but instead I’ll just bring it to everyone’s attention.
I was painting it black only so we could follow it; to see the difference between a chip always chosen to remain in the bowl and the one that gets removed. I hope that shows that if you do care about which chip is where, it makes a difference. I’ll admit to not having a completely solid answer to the ‘zapping’ or choosing case.
But it did make me think of an interesting and potentially annoying variation :
Starting similarly to the original problem statements …
Suppose that after adding two chips to the bowl, God now flips a perfect coin (or rolls a three-sided die and ignores the 3’s) to decide whether he will remove the largest or the smallest chip from the pot. What is the expected number of chips left in the bowl after 2 hours?
It’s too late for me to think about the answer to that one, so I’ll flip a coin and say it’s infinite.
I have a modification of example 2 with the fly, I’ll call it example 2b. (In example 2, ignore anything about chips, I think that just confused people.) I’m going to use this to show a contradiction in Tyrrell McAllister’s method of proof.
For both this example and example 2, I need to further specify that the fly only travel in the last, say, 1/60th of the time period. So in the first step, the fly rests 59 minutes, then travels in the final minute before t=1hour, in the next step it rests 29.5 minutes and travels in the final 30 seconds before t=1.5 hours, and so forth. Further, the fly travels at a uniform speed during the travel period. The fly’s motion in both examples is now fully specified for t<2.
I need an additional equation for reference.
Eq. (4):
S[sub]4[/sub] = alpha + b[sub]0[/sub] - c[sub]0[/sub] - d[sub]0[/sub] - e[sub]1[/sub] + a[sub]1[/sub] + b[sub]1[/sub] - c[sub]1[/sub] - d[sub]1[/sub] - e[sub]2[/sub] + a[sub]2[/sub] + …
Example 2b.
In this example, the fly travels in the X direction in the following amounts:
1 + 1/2 - 1/2 - 1/6 - 1/3 + 1/3 + 1/4 - 1/4 - 1/20 - 1/5 + 1/5 + 1/6 + …
so alpha and all of a[sub]i[/sub], b[sub]i[/sub], …, e[sub]i[/sub] are positive.
Once again, I’m going to rewrite much of Tyrrell McAllister’s proof showing the dish is empty, applying it now to my example 2b, with the fly:
In the theorem that follows, I’m going to identify each move in the +X direction (hereafter “move” for brevity) or in the -X direction (hereafter “antimove”) with a number. So now N may be thought of as the set of all the moves and antimoves. alpha is move 1, b[sub]0[/sub] is move 2, c[sub]0[/sub] is antimove 3 and so forth.
Theorem: The set D of moves not compensated for by an equal antimove and antimoves not compensating a move at two hours is the set consisting of move alpha, and all antimoves d[sub]i[/sub].
Proof: We show that each move Or antimove) a[sub]i[/sub], b[sub]i[/sub], c[sub]i[/sub], e[sub]i[/sub] is in D[sup]c[/sup] and that alpha and d[sub]i[/sub] are in D. In principle, D[sup]c[/sup] contains three kinds of moves and antimoves:
(i)those moves or antimoves that were never made before the two hour mark,
(ii)those moves that are compensated for by an antimove before the two hour mark,
(iii)those antimoves that are compensating a move before the two hour mark,
while D contains:
(iv)those moves that were never compensated for by an antimove before the two hour mark, and
(v)those antimoves that are never used to compensate a move before the two hour mark.
I think that everyone here agrees that the set of moves and antimoves satisfying condition (i) is empty. That is, every move or antimove is eventually made at some time before the 2 hour mark. The set of every move b[sub]i[/sub] satisfies condition (ii) (compensated by antimove c[sub]i[/sub]).
To prove this, suppose that n is a move b[sub]i[/sub] in N. Then n is compensated for by the antimove n+1 (c[sub]i[/sub]) immediately following. Note that, regardless of the value of n, we have that t[sub]n[/sub] < 2. Therefore, move n is compensated for before the two hour mark, so move n is in D[sup]c[/sup].
Likewise the set of every move a[sub]i[/sub] also satisfies condition (ii) (compensated by e[sub]i[/sub]). Each of the set of antimoves c[sub]i[/sub] and e[sub]i[/sub] satisfies conditon (iii) (compensating b[sub]i[/sub] and a[sub]i[/sub] respectively).
alpha clearly satisfies condition (iv) (since all antimoves are less than 1). The set of moves d[sub]i[/sub] satisfy condition (v) (all the moves except alpha are already compensated by an antimove, and alpha is too large). Thus the theorme is proved.
The set of moves and antimoves satisfying conditons (i), (ii), and (iii) do not contribute to the fly’s final position (they were either never made, were compensated for, or were compensating a move). The remaining moves and antimoves satisfying (iv) and (v) can now be easily summed, since they form an absolutely convergent sequence:
1 - 1/6 - 1/20 - 1/42 - …
The sum converges (absolutely) to log(2). So for example 2b, the fly ends up at log(2).
I’ve already shown, using the same method, that in example 2, the fly ends up at zero.
The important point here is that the fly in example 2, and the fly in example 2b is the exact same fly. The fly in example 2 cannot. e.g travel 1 in the -X direction (i.e. travel -1) without traveling 1/2 then 1/6 then 1/3, all in the -X direction. Likewise, the fly in example 2b cannot travel 1/2 then 1/6 then 1/3, all in the -X direction without having traveled a total of 1 in the -X direction. Examples 2 and 2b are the same example.
Thus, I’ve used Tyrrell McAllister’s method of showing the dish is empty on a different problem, and obtained two different answers. Tyrrell McAllister’s method of showing that the dish is empty is flawed.
I’m not going to track down every statement that the dish is empty, since they’re all variations on the same theme, and the same approach can be taken onal of them. For example, the tersest “proof” that the dish is empty is this one, for which I’ll give an equally terse refutation:
I can simply analyze what moves are not compensated for in examples 2 and 2b and sum them up, but when I do so I get zero in the one case, and log(2) in the other, even though examples 2 and 2b are the same problem.
I therefore feel justified in stating that every proof offered in this thread that the dish is empty is flawed. Therefore no valid justification has been given for asserting that the dish is empty. I’ll further assert that any future “proof” offered that the dish is empty needs to be applied to examples 2 and 2b, showing that different answers to these two examples are not possible. This has been a lot of work, and I’m not going to repeat it for every different phrasing that comes along.
I’ve still got a lot of work to do, so I’ll have to be brief, but:
The problem is, I’m dealing with a set, while you’re dealing with an infinite series.
With an infinite series, if I rearrange the terms of the series, I can end up with it converging to different values. With a set, if I rearrange the elements of the set, I still have the same set. You can’t find fault with my argument by considering series, because my argument has nothing to do with series in the first place.
Also, you’re still making the assumption of continuity, by implicitly assuming that the fly moves continuously (i.e., that the fly converging on the limit of the series of its moves).
As far as I’m concerned, you are free to assume the fly moves continuously, but then you’ll have a hard time convincing me that it’s in any way analagous to the original problem involving the number of chips in the dish.
Oh, and why won’t anyone respond to my variation, mentioned earlier?:
In the original problem, God throws a few positive integers in at a time.
In my variation, God decides to just throw them all in at the beginning, rather than merely two at a time. In other words, the dish already contains all the positive integers at the start. Then he begins removing them, removing the smallest one at each step.
How many chips are left in the dish when he’s finished?
In both my proofs, I’m dealing with a set of moves. Only at the end, when the set has been reduced to a set of moves (possibly empty) which form an absolutely convergent series do I sum the series.
I specifed that the fly moves continuously and still got two different answers. And neither you nor anyone else has ever answered where the fly is at t=2. If the number of disks at t=2 is answerable, surely the fly’s location is also.
Yeah, that’s why I liked to rephrase it as “God removed all even numbers” and “god removes all numbers”. When it is phrased like that it does seem to be obvious.
I find both arguments totally convincing. And I like recursive sequences.
ZenBeam, let me check that I’m understanding you. Consider one of your earlier series:
1 + 1/2 - 1 + 1/3 + 1/4 - 1/2 + 1/5 + 1/6 - 1/3 + …
As I understand it, you’re offering, in some sense, two different interpretations of the terms in the above series:
-
+x and -x correspond to the fly moving x units to the right (move) and x units to the left (antimove), respectively.
-
(Relating it to our original example) +1/n and -1/n correspond to adding chip n to the box, and removing chip n from the box, respectively.
Assume the fly moves continuously, so that the location of the fly at t=2 will be the sum of the infinite series.
My original argument was that, according to the series, every chip is eventually removed, since -1/n occurs somewhere in the series.
Your argument is that this doesn’t make sense, since if we simply rearrange the series, we can get the fly’s motion to converge at various points.
I claim that one has nothing to do with the other. Sure, we can rearrange the series and get the fly going to different places, but, no matter how we rearrange the series, I still claim the dish is empty–I see no contradiction here. This is, of course, assuming that our rearrangement makes sense. For example, suppose our rearrangement begins:
-1/2 + 1/2 +…
This is not a proper rearrangement; it doesn’t make sense to remove chip 2 before we place chip 2 in the dish, so we have to discard arrangments such as these (where -1/n comes before +1/n) on the grounds that it doesn’t preserve the nature of the original problem.
Also, we must not allow discard combining terms. For example, the original series
1 + 1/2 - 1 + 1/3 + 1/4 - 1/2 + 1/5 + 1/6 - 1/3 + …
is significantly different from
1 + 1/2 - 1 + 1/3 - 1/4 + 1/5 + 1/6 - 1/3 + …
since clearly, when it comes to discussing the dish, “adding chip 4 and removing chip 2” is not equivalent to simply “removing chip 4”.
Again, my claim is that, regardless of what the fly may do, any rearrangement (subject to the above restrictions) will leave the dish empty of chips.
Let me stop it here and ask if I’ve understood you so far; please correct me if I haven’t, and we’ll take it from there…