The result doesn’t sound so silly if the criteria for removing chips in the OP is changed.
“Place an infinite amount of chips in. Remove odd chips. How many remain?”
vs
“Place an infinite amount of chips in. Remove all chips. How many remain?”
The trick is like the “where’s the missing dollar”… that is, in the presentation.
First, three equations. I’ll assume these summations are carried out in 2 hours in the fashion of the OP, although that really isn’t relevent for most of this post.
Eq. (1):
S[sub]1[/sub] = a + b - a + c + d - b + e + f - c + …
Eq. (2):
S[sub]2[/sub] = a - a + b - b + c - c + d - d + …
where Eq. (1) is the summation we’re interested in and Eq. (2) is obtained be rearranging the elements of Eq. (1). S[sub]2[/sub] = S[sub]1[/sub] if the summation in Eq. (1) is absolutely convergent. If the summation in Eq. (1) is not absolutely convergent, S[sub]2[/sub] is not in general equal to S[sub]1[/sub]. Eq. (1) is absolutely convergent if the sum of the absolute values of all the terms converges, i.e. the following summation is convergent:
Eq. (3):
S[sub]3[/sub] = |a| + |b| + |-a| + |c| + |d| + |-b| + |e| + |f| + |-c| + …
Now, four examples:
Example 1:
a = 1, b = 1/2, c = 1/4, d = 1/8, and so forth. a, b, c, etc. could represent chip thicknesses in the OP, or the distance an infinitesimally small fly travels, beginning at X = 0. In this example, Eq. (1) is absolutely convergent, and Eq. (2) can be used. You can also sum the terms in Eq. (1), and you get the same answer. The fly ends up at X = 0, where it started.
Example 2:
a = 1, b = 1/2, c = 1/3, d = 1/4, and so forth. a, b, c, etc. can again represent chip thicknesses in the OP (as in one of my previous posts), or the distance an infinitesimally small fly travels, beginning at X = 0. In this example, Eq. (1) is Conditionally Convergent (and note Eq. (1) on this page), and Eq. (2) can not be used. You can, however sum the terms in Eq. (1) and the answer converges. The fly ends up at X = log(2).
Example 3:
a = 1, b = 1, c = 1, d = 1, and so forth. a, b, c, etc. can again represent adding chips in the OP (or removing them for the terms with the minus sign), or again the distance an infinitesimally small fly travels, beginning at X = 0. In this example, Eq. (1) is divergent, and again Eq. (2) can not be used. Summing up the terms, the fly heads off in a jerky fashion towards infinity.
Example 4:
a = Chip 1, b = Chip 2, c = Chip 3, d = Chip 4, and so forth. For “Chip 1”, “Chip 2”, etc., you can use sets, or unit vectors in chip space, or however you want to express it. The partial sums give the chips in the dish after each step in the OP. In this example, Eq. (1) is again divergent, and again Eq. (2) can not be used.
When the argument is made (and I’ll paraphrase, since it’s been made many times) that there are zero chips in the dish, or that every chip in the dish has been removed, the writer is implicitly using Eq. (2) to solve Example 3 or 4. By using words instead of equations, it isn’t obvious, but the pairing of a with -a, b with -b, and so forth, as in Eq. (2), is always made. For examples 2 through 4, this is invalid. You simply can’t validly rearrange the terms to pair off each term with its negation if the series is not absolutely convergent.
Everyone should be able to convince themselves of this by applying the same argument to example 2: For every step where the fly flies in the plus direction, there is a corresponding flight in the -x direction. There is no flight in the +x direction not canceled by a similar flight in the -x direction. Hence, the fly ends up at the origin. This argument is clearly false for example 2; the distance between the fly and log(2) is bounded closer and closer to zero as t --> 2 hours. The fly ends up at X = log(2).
I’ll also address the “name one” argument in two ways. First, I think everyone agrees that the limit, as t --> 2 hours, of the number of chips in the dish is infinite, even if we disagree on the number at t = 2 hours. Even so, in the limit t --> 2 hours, no chip can be named that is in the dish.
Secondly, consider example 2 again. For the fly to end up at log(2), instead of the origin, some flight in the +X direction must not have been canceled by a flight in the -X direction. No such flight can be named, but there’s the fly at X = log(2).
Arguments have been made that the number of chips in the dish is not necessarily continuous at t = 2 hours, but without the above arguments for why the number of chips in the dish must be zero, this argument doesn’t lead anywhere.
The “name one” argument, and the arguments pairing each chip with its removal are, I believe, the only arguments in this thread for there being zero chips in the dish at t >= 2 hours (although it is a long thread). Since these argments are invalid, there is no justification for saying there are zero chips in the dish. Now, if someone can show, using valid equations and valid manipulations of those equations that the sums in examples 3 and 4 are zero, I’d like to see that. I haven’t yet.
Equations aren’t what matters here. The order in which you remove chips is what matters.
A different example is in order. Suppose I’m playing at a slot machine, with a countably infinite number of quarters. Each time I play, I win a countably infinite number of quarters. As above, let’s assume that the kth play takes 1/k minutes.
If I just keep playing the quarters I started with, I’ll always have a countably infinite number of quarters. But if I dovetail, I can be out of quarters at the end of 2 minutes.
I play the same number of quarters in each scenario. The only difference is the order in which I play them.
Equations are how you ensure you are removing chips in the right order.
No. What ensures you’re removing chips in the right order is the sequence of sets of chips in the bowl after each step. All the equations can do is track the cardinality, which isn’t enough information to determine the behavior of the function at infinity.
I hate to use that term when it has such potential for confusion, but I really can’t think of anything else.
ZenBeam, please see my post from the previous page, where I point out that we (those of us who argue that the dish is empty at two hours) are not evaluating a series, implicitly or otherwise, to arrive at our conclusion. Nothing so sophisticated as infinite series are necessary. The problem can be analysed with simple set threory (and if set theoretic reasoning is invalid, then the entire ediface of mathematics, including the mathematics of series evaluations, is worthless).
I will try to make the set theoretic reasoning more explicit.
Let N be the set of positive integers. Given any subset S of N, denote by S[sup]c[/sup] the complement of S in N; that is, S[sup]c[/sup] is the set of all those numbers in N that are not in S.
Lemma:Given any subset S of N, if, for each number n in N, we have that n is in S[sup]c[/sup], then S is empty.
Proof: This is a trivial proof by contradiction. Suppose that S is not empty. Then S contains some positive integer, say m. By hypothesis, every positive integer, including m, is in S[sup]c[/sup]. Therefore, by the definition of S[sup]c[/sup], m is not in S, a contradiction.
In the theorem that follows, I’m going to identify each chip with the number written on it. So now N may be thought of as the set of all the chips.
Theorem:The set D of chips in the dish at two hours is empty.
Proof: We show that each chip is in D[sup]c[/sup]; the conclusion will then follow from the Lemma. In principle, D[sup]c[/sup] contains two kinds of chips:
ulthose chips that aren’t placed in the dish before the two hour mark, and
(ii)those chips that are removed from the dish at a time before the two hour mark.[/ul]I think that everyone here agrees that the set of chips satisfying condition (i) is empty. That is, every chip is eventually placed in the dish at some time before the 2 hour mark. So the problem reduces to showing that every chip in N is of type (ii).
To prove this, suppose that n is a chip in N. Then n is removed from the bowl at time t[sub]n[/sub] = 2 - (1/2)[sup]n-1[/sup]. Note that, regardless of the value of n, we have that t[sub]n[/sub] < 2. Therefore, chip n is removed from the dish before the two hour mark, so chip n is in D[sup]c[/sup].
Since we showed this for an arbitrary chip n in N, we’ve shown that every chip in N is in D[sup]c[/sup]. Therefore, by the Lemma, D, the set of chips in the dish, is empty.
At 2 hours and 1 second, where will the odd chips be? The 2 hour mark does not necessarily end the experiment, it’s just that, before, nothing interesting happened after the 2 hour mark. If the majority are right, and there are 0 chips in the dish in the situation in the OP, then I think all of the odd chips must be instantaneously moved from not yet in the dish to inside the dish to taken out of the dish at the moment when t=2 hr.
I think what is happening at the 2 hr mark precisely is somewhat ill-defined, but before, not all the evens have been in the dish, and after, all the evens are already out.
The problem is you cannot come up with a point in time where the odds have started moving through the dish! The description of the problem requires that the odd coins themselves have a point in time that they are removed from the dish. Saying the odd coins are taken from the dish after the 2 hour mark does no good, because the time at which a coin would be removed from the dish has no meaning (requires taking the logarithm of a negative number to solve). Before the 2 hour mark, only the evens are leaving the dish.
So as you say, they’d all have to leave the dish exactly at the two hour mark which is ill defined because the problem supposes every coin is moved at a different time (that is, after some time t since the last coin, a new coin is moved).
I claim this is enough to show that the odd coins never enter or leave the dish.
Tyrrell McAllister, thanks for your response. It gives me a chance to show the dilemma I’m seeing in something someone else wrote. I’m going to rewrite much of your post, applying it to my example 2, with the fly:
In the theorem that follows, I’m going to identify each move in the +X direction (hereafter “move” for brevity) with a number. So now N may be thought of as the set of all the moves.
Theorem:The set D of moves to the right not compensated for by an equal move in the -X direction (hereafter “antimove”) at two hours is empty.
Proof: We show that each move is in D[sup]c[/sup]; the conclusion will then follow from the Lemma. In principle, D[sup]c[/sup] contains two kinds of moves:
(i)those moves that were never made before the two hour mark, and
(ii)those moves that are compensated for by an antimove before the two hour mark.
I think that everyone here agrees that the set of moves satisfying condition (i) is empty. That is, every move is eventually made at some time before the 2 hour mark. So the problem reduces to showing that every move in N is of type (ii).
To prove this, suppose that n is a move in N. Then n is compensated for by an antimove at time tn = 2 - (1/2)n-1(*****). Note that, regardless of the value of n, we have that tn < 2. Therefore, move n is compensated for before the two hour mark, so move n is in D[sup]c[/sup].
Since we showed this for an arbitrary move n in N, we’ve shown that every move in N is in D[sup]c[/sup]. Therefore, by the Lemma, D, the set of moves not compensated for by a corresponding antimove, is empty.
(ZenBeam again) We’ve just proved that in example 2, the fly ends up at the origin. How do you reconcile this with the fly approaching log(2) when the series in example 2 is summed? I reconcile this by asserting that this proof implicitly converts Eq. (1) to Eq. (2). This happens when each move is paired with its antimove in the sentence marked with (). The sentence in your original post corresponding to the one with the () is likewise suspect. Certainly, if this example fails for a conditionally converging series, there is no reason to believe it will not also fail for a diverging series.
Now, you referred me to a previous post where you assert that “the more natural definition of H(infinity)” is zero, but that value of zero is based on reasoning like above, which gives an incorrect value for a case where we do know the value. At any rate, saying H(infinity) = 0 or the number of chips in the dish = 0 because I’m defining them to be zero is highly unsatisfying, and no better than saying the number of chips in the dish is infinite because I’m defining it to be infinite.
knock knock, the experiment you’re proposing can’t be done with the standard division of time into “steps” that we’ve been using.
For example, in the original division, before each step occurs, only finitely many steps have already been taken. For your new experiment, some of the steps (the steps involving the odds) don’t take place until infinitely many steps have already taken place.
Not that there’s a problem with this–we only have to devise a new plan for the steps. The most obvious would be to simply repeat, in the following two hours, the same division into steps that we utilized in the first two hours. So essentially we have a two hour period involving the evens, followed by a two hour period involving the odds. It’s essentially like doing the original experiment twice in succession, and now, at the end of the four full hours, the box will still be empty.
You may be interested in reading some on the ordinal numbers. For example, in the original experiment, the steps were orderd by the ordinal omega. In our new experiment, the steps are ordered by the ordinal omega + omega.
ZenBeam, I think you’re missing Tyrrell McAllister’s point–that we shouldn’t be using limits in the first place.
Starting from the beginning, we’re considering a time interval [0,2] (time expressed in hours).
We have a function f:[0,2] -> {0,1,2,3,…,omega (or infinite, if you prefer)}, that, when you plug in a given time, will spit out the number of chips in the box at that time.
The question we’ve all been considering is, what is f(2)?
Certainly, no one denies that (lim f(t) as t->2) = omega.
However, as any first year calculus student will tell you (provided they’ve been paying attention), you can’t assume that
f(2) = lim f(t) as t -> 2.
This is true if and only if f is continuous at 2.
My (and others) approach to finding f(2) is simple. Simply analyze what chips are in the box at time 2, then count them. This is, after all, what f(2) is defined to be–how many chips there are in the box at time 2.
Your approach is more involved. You count the number chips before time 2, notice an upward trend, then assume this upward trend will be preserved, by a limiting process, at t=2.
This is your error. Can you demonstrate that f is continuous at t=2? If you can’t, this approach must be discarded.
My “approach” has been to call into question your approach, and to ask you to show that your approach is valid. The reason I doubt your approach is that when I attempt to apply it to a problem where we do know the answer (my example 2 above), your approach gives the wrong answer. What you are doing is equivalent to starting with Eq. (1), and rearranging to get Eq. (2), and you haven’t given any justification for doing so. Basically for the reason “we say so”, the fly ends up at 0, when it had been approaching log(2).
Do you believe in example 2, that the fly is at X = 0 at t = 2? Does this make more sense to you than that the fly is at log(2)?
Well, in the first example, the equation forbids us from removing odd numbered chips. Thus, an infinite amount remain. In the second example, the equation compells us to remove every chip. Thus, none remain.
What am I missing?
Involving the fly seems to be a clever way of concealing the fact that you’re still dealing with the limit of an infinite series. I haven’t used any infinite series in concluding that the box is empty, so I fail to see how this calls my approach into question.
You keep avoiding answering those questions…
I explained my reasoning in the paragraph above the one you quoted:
Now I’m assuming here that your and Tyrrell McAllister’s approaches are essentially the same.
Now, either you believe the fly is at zero at t=2 or you don’t. The fly being at zero at t = 2 is, to me, counterintuitive. If you tell me you believe the fly is at zero at t=2, I’m going to ask the other mathmeticians who hang out here if they agree with that, and more to the point, if that is the consensus of current mathematical thinking. If you tell me you don’t, I’m going to ask you why my rewrite of Tyrrell McAllister’s proof fails. To emphasize, this is important because Tyrrell McAllister used essentially that same proof to show that the number of chips in the dish is empty, and your subsequent responses imply you agree with that proof.
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? Everything from Tyrrell McAllister’s proof seems to apply here, in slightly revised form. The chip always has a positive integer on it. After 2 hours, it cannot have a positive integer on it, because, for any integer n, I can tell you the moment when n was erased from it, and once it is erased, it never comes back. So there can be no positive integer on the chip. As long as the chip exists, it always has a positive integer on it (and can never stop having a positive integer on it, because whenever one is erased another replaces it instantaneously), it must not exist. He doesn’t really have a chip.
That is just like the reasoning of those who say that there are no chips left after 2 hours in the case described in the OP. But how could the chip disappear just from writing and erasing numbers on it? Where would it go? The one chip must still be there, just like it always has been. It’s just hard (or impossible) to say what it’s labeled.
I think that writing the sets out in some mathematical notation does help the problem a bit.
In owl’s original problem:
The first set is “all integers” (except those divisible by 2), or:
S[sub]1[/sub] = {All n in N such that there is no m in N where n = 2m}
This will consist of positive integers n where {n = 2m + 1, and m is in N}. This will have the same cardinality as N, which is an infinite set (specifically, aleph null).
The second set is “all integers” (except those that have an integer higher than them), or:
S[sub]2[/sub] = {All n in N such that there is no m in N where m > n}
This will consist of positive integers where there is no higher integer - which is none of them, there is no highest integer. It’s an empty set.
The problem comes when you try to count the chips by looking at what happens to a series of similarly constructed finite sets. That is what we are doing, if you think about it - after turn 1, you have one finite set; after turn 2, you have a different finite set; and so forth. It’s tempting to expect the total “size” of S[sub]2[/sub] to be the limit of the finite sizes of these sets, but there is no mathematical guarantee that it will be. When the limit does not converge, we may say it “goes to infinity”, but that does not mean that it converges at the cardinal number aleph null; it means that it does not converge, and we cannot extend the function in that way. I think that Cabbage and ZenBeam have explained the limits and their limits (snort) pretty well.
knock knock, I would argue that the chip is blank at the end (the chip still exists, of course). Everything that was ever written on it has been erased, so it’s blank.
ZenBeam, I’m not ignoring you, I just don’t have time to respond at the moment.
How could the chip be blank? Every time god erases a number from it, he simultaneously inscribes a new number on the other side.
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?
I’m going to go back to the original problem and make a few changes in the procedure. I’ll try not to make any important changes - that is, I’ll try not to change any facts that have been used in the reasoning to support the majority opinion that no chips remain after 2 hours.
First of all, let’s use gold coins instead of chips. 'Cause I like gold.
Now, have a move consist of removing a coin and then adding 2. Since these 2 steps occur together, perhaps instantaneously, we can do it this way instead of adding 2 and then removing 1. Every coin is still getting removed at the same time as before. Except we can’t do this on the first move, so let’s make the moves go as follows: 1) add 1 & 2, take away 1. 2) remove 2, add 3 & 4. 3) remove 3, add 5 & 6. And so on. Or, better yet, we can just start at the 1-hr mark, with only coin 2 in the dish, and begin with step 2).
Now, does it matter that God has this infinite pile of coins beforehand, or could he just mint them right before he puts them in? It doesn’t matter, since we don’t care exactly how he got them before he put them into the dish, or how long they existed beforehand. As long as they come in the order of the positive integers, our interest in the coins begins the moment they get put into the dish. So, we could imagine that God has an infinite vat of liquid gold, and on every move he reaches in and pinches, thereby creating 2 new coins, with the next 2 integers on them. Then he puts these into the dish.
What about coins that have already been taken out? Do they have to all go together into a pile? No, we don’t care. They can be destroyed. God can create a coin right before it goes into the dish, and destroy it right after it comes out, and that has no effect on what’s happening in the dish - each integer goes in, each integer comes out, according to a regular pattern that has a limit at t=2 hr.
Since God cares about the environment, he reduces the creation and destruction by recycling. Each coin that comes out can be destroyed by tossing it into the vat, where it melts, and that may at some point be used in the creation of a new coin. None of this effects the dish - each integer goes in, each integer comes out, even if the coin its on is made, in part, out of recycled gold.
Now, did any argument make use of the size of the coins, or of the fact that they were all the same size? Nope. So we can vary the size. We can make the coins progressively smaller. God doesn’t want to use an infinite vat of gold where a finite will one suffice.
So God could make coin 2 out of 1 kg of gold, coins 3 & 4 out of 1/2 kg, coins 5, 6, 7, & 8 out of 1/4 kg, 9-16 out of 1/8 kg, and, in general, coins (2^n)+1 through 2^(n+1) out of 1/(2^n) kg of gold. This way, there is always 1 kg of gold in the dish. Since the coin that comes out is always the same weight as the 2 coins that go in, God could manage with exactly 1 kg of gold. The nth move would be as follows: God takes out the coin in the dish with the smallest label: n. He melts it down & forms it into 2 coins, half the weight of the original, identical to each other except that they are labeled with consecutive integers 2n-1 and 2n. He puts these new coins into the dish. Total elapsed time: as small as you want.
Now, imagine that the coins are stacked in a cylinder. They are all the same diameter, and only their thickness varies. Of course, they’re all still labeled with integers. The nth move involves taking out a piece of the cylinder, cutting it in half, and removing the label on that piece and replacing it with a new label on each piece.
Since God strives for efficiency, he finds a fast way to do this. He doesn’t even have to pick up the piece he removes. While leaving it 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.
Now, take a step back and look at the process. 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. The whole cylinder is always there - it never moves - so it must be there after 2 hours. Where else would it go? Since the coins are getting thinner and thinner during the 2 hours, they must be infinitesimally thin by the 2 hour mark. Since they still have the same total height as at the beginning, there must be infinitely many of them. The numbers written on the pieces were increasing without bound, so I do not know how they are labeled after 2 hours. But that doesn’t mean that they can’t exist - that the whole cylinder suddenly disappears at the 2-hr mark, just because its pieces were being cut in half & relabeled up until then. So, in this case, there are infinitely many coins after 2 hours.
So, if you said that there were 0 chips after 2 hours in the original question, what happened? Are there still somehow 0 coins after 2 hours? If not, then which change in the process caused the answer to change?