Let's say we decide to race on a track $1000$ km long. You are a $100$ times faster than me, meaning if we both start at the beginning, you obviously win. To make things more fair you give me a head start of $1$m. The distance is still very small, meaning you will obviously win.

A few premises:

-For you to win the race you need to overtake me

-To overtake me you need to reach a point of equivalence

-If there is no point of equivalence you can't beat me

Let's assume it takes you $1$ second to reach 1m. However in that $1$ second, I would have travelled a distance forward-lets say I am now at $1.01$m. You haven't caught up to me- I'm still $0.01$m ahead of you. It takes you $0.01$s to travel that $0.01$m. But in that $0.01$ s I would have travelled $0.0001$m, meaning I'm still ahead of you. Therefore you can never catch up to me- the distance between us will get infinitesimally small, but never $0$. Therefore since you can't catch up to me, you can never win.

This obvious paradox has been resolved through the fact that an infinitesimal series adds up to one- however, doesn't thid simply prove both people will finish the race? How does it prove the faster person will win?

Please don't simply give me a linear solution. I do not want to know when the faster person catches up - I want to know the mathematical flaw in the paradox's logic.

  • $\begingroup$ There are two very different questions here (namely how a mathematical model of this situation behaves, and what this says about the actual physics of motion, which was Zeno's original point). The second question belongs on physics.stackexchange.com. Which question are you asking? $\endgroup$ – Qiaochu Yuan May 9 '12 at 4:24
  • $\begingroup$ I am taking about the mathematical model of the system, not the physics of it. $\endgroup$ – user26649 May 9 '12 at 4:25
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    $\begingroup$ What "flaw"? "Both people finish the race" does not contradict that one person will finish first. The series does not "add up to $1$", the corresponding series for your particular scenario adds up to $\frac{10}{9}$. And that just means 10/9 meters from where we started. So what? What exactly is it you think you are contradicting? $\endgroup$ – Arturo Magidin May 9 '12 at 4:26
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    $\begingroup$ I also don't understand what you mean by "don't simply give me a linear solution." Computing the intersection of two lines tells you when the faster person catches up and that conclusion is not affected by the paradox in any way. $\endgroup$ – Qiaochu Yuan May 9 '12 at 4:28
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    $\begingroup$ What you've done is exhibited an infinite sequence of times at which you are ahead of me. But since that sequence doesn't extend all the way into the future, your logic says nothing about whether you remain ahead of me at any time after all the times in your sequence. $\endgroup$ – Rahul May 9 '12 at 4:40

If you start 1 meter ahead of me, and it takes me 1s to reach your current position (apparently I run at 1 meter per second, and you run at .01 meters per second), $\frac{1}{100}$th of a second to reach your position at $t=1$, etc. I take $$1 + \frac{1}{100}+\frac{1}{100^2} + \cdots = \sum_{i=0}^{\infty}\frac{1}{100^i}$$ seconds to overtake you. Since $$\sum_{i=0}^{\infty}\frac{1}{100^i} = \frac{1}{1-\frac{1}{100}} = \frac{100}{99},$$ then after $\frac{100}{99}$ seconds, I will have overtaken you. This will occur well before we reach the finish line; we've only advanced $\frac{100}{99}$ meters (since apparently I run at 1 meter per second), and the finish line is more than $\frac{100}{99}$ meters from where we started. After I've overtaken you, I will be ahead at any further time.

The implicit error in the original claim that I cannot overtake you is the assumption that an infinite sum of positive quantities will necessarily be infinite. This has long been dealt with, and does not even require the use of infinitesimals.

Of course, it's possible for you to start so far ahead of me that I will only catch up to you when we get to the finish line; but that's hardly a paradox! Nor do I understand what your complaint is with "both people finish the race". Is there some problem with the slower person finishing after the faster one has?

  • $\begingroup$ This answer is correct, of course, although I think some further clarification could help interpret it better. The meaning of the series as you stated the reasoning could be misunderstood if one were to relate the index of the series to the value of $t$. Instead, I am thinking of it as follows: after a certain moment it will take me an infinitely small amount of time to reach your current position and you will advance an infinitely small distance. That moment, of course, comes after $\frac{100}{99}$ seconds. $\endgroup$ – broncoAbierto Mar 1 '15 at 18:54

Below are some pointers to the literature for your philosophical concerns, taken from old sci.math posts of mine. Thus, this is really in the nature of a comment, not an answer, but because of comment length constraints I'm posting this as an answer.

First, of possible interest are the google searches infinity machines and supertasks, as well as the Wikipedia article Supertask.

I've listed the references that follow in order of how helpful/interesting I think they'd be for your concerns.

[1] Wesley C. Salmon (editor), Zeno's Paradoxes, Bobbs-Merrill, 1970, 309 pages. [Reprinted by Hackett Publishing Company in 2001; 320 pages.]

[2] Adolf Grünbaum, Modern Science and Zeno's Paradoxes, Wesleyan University Press, 1967, 148 pages. [Reprinted by Allen and Unwin in 1968; 153 pages.]

[3] Florian Cajori, The history of Zeno's arguments on motion (in 9 parts), American Mathematical Monthly 22 (1915), 1-6, 39-47, 77-82, 109-115, 143-149, 179-186, 215-220, 253-258, 292-297. Jstor: 1 2 4 5 6 7 8 9 10

[4] Florian Cajori, The purpose of Zeno's arguments on motion, Isis 3 (1920), 7-20. jstor

[5] Clive W. Kilmister, Zeno, Aristotle, Weyl and Shuard: two-and-a-half millenia of worries over number, Mathematical Gazette 64 #429 (October 1980), 149-158. jstor

  • $\begingroup$ @Martin Sleziak: I just noticed your compression of my URLs into the [item](item's URL) form, a format I only recently figured out by selecting "edit" on other people's posts (but not actually performing an edit on their posts) and examining the underlying code for doing this. Anyway, I've included the JSTOR link for Isis, which you overlooked. I don't have access to JSTOR, but I just learned that even without access one can obtain the URLs for their papers. $\endgroup$ – Dave L. Renfro May 21 '12 at 12:16
  • $\begingroup$ Such formats for links is mentioned e.g. in the following meta thread: How do you cite a reference in a comment?. It's particularly useful in comments. $\endgroup$ – Martin Sleziak May 21 '12 at 13:56
  • $\begingroup$ @Martin Sleziak: Thanks for the link, which led me to editing devices, which I've bookmarked for later use. And thanks for the clean-up on my post. I've been reluctant to do this to my old posts, as this will top-list them, thereby possibly suggesting to some that I was doing it just to pick up new votes! $\endgroup$ – Dave L. Renfro May 21 '12 at 14:42
  • $\begingroup$ @Martin Sleziak: By the way, a week ago I completed most of a very extensive addition to my comments at Is there a $\sigma$-algebra on $\mathbb R$ strictly between the Borel and Lebesgue algebras?, but suddenly something come up at work that required me to devote all my time there (early morning, weekends, etc.), and so I won't get around to finishing my additions until sometime this weekend or early next week. FWIW, I've been incorporating the [item](item's URL) form very extensively there, even for things like on-line Zbl and JFM reviews. $\endgroup$ – Dave L. Renfro May 21 '12 at 14:49

I think the mathematical "explanation" of the Zeno's paradox (convergence of infinite series) is quite unsatisfying. Assuming that each term in the series corresponds to one step of Achilles's and considering that he indeed overtakes the turtle in finite time, which of Achilles feet is forward at the moment when he reaches the turtle?

Or a slightly different, but equivalent, presentation of the paradox: assume that the turtle changes direction at each discrete instant of time Achilles reaches her previous position, alternatively moving NE and SE. Achilles just follows her path. What direction is the turtle facing the moment Achilles reaches her?

Achilles's and the turtle is no paradox at all, but a refutation of the hypotheses that the space is continuous. Zeno's arrow paradox is a refutation of the hypothesis that the space is discrete. Together they form a paradox and an explanation is probably not easy. For Zeno the explanation was that what we perceive as motion is an illusion. In any case, I don't think that convergent infinite series have anything to do with the heart of Zeno's paradoxes.

EDIT: The same argument can be made point-like particles, only assuming that physical reality is continuous and infinitely divisible. Imagine a photon travelling between an infinite sequence of mirrors placed in a zig-zag shape with distance between mirrors decreasing at a geometric rate. So the photon bounces from a NE to SE direction and back, with the distance travelled decreasing "fast". Since the length of the total path is finite (sum of a geometric series), the photon will emerge from the sequence of mirrors in finite time. What direction will it travel? The heart of the Zeno's argument is that there is no logical way to decide that. You may argue that it is impossible to build such a sequence of mirrors, however this is just conceding Zeno's point that physical reality is NOT continuous and infinitely divisible.

I think the mathematical model of the Zeno's paradox is a great pedagogical tool in first year calculus, probably could be made even earlier in high school, but it misses an important aspect of Zeno's argument. Granted, this argument lies at the boundary of math, physics and perhaps philosophy.

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    $\begingroup$ "Assuming that each term in the series corresponds to one step of Achilles's..." Why would you assume that? You seem to reject the idea that a finite interval of time can contain infinitely many instants of time. But it is clearly the case that the real numbers do have the property that a finite interval contains infinitely many numbers. If you model distance and time with real numbers, the convergence of infinite series deals with all of your questions; if you wish to reject this model of the universe that's fine, but it's a discussion to have with philosophers, not mathematicians. $\endgroup$ – Michael Joyce May 9 '12 at 5:58
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    $\begingroup$ @MichaelJoyce I am not rejecting this model of the Universe, but rather express the opinion that Zeno's paradox is a limitation of the model. If we set up the problem entirely in a mathematical setting, then it is trivial. Read the setting of the paradox in the second paragraph: the question of which direction is the turtle "facing", makes no mathematical sense, but it makes physical sense. Convergence of serires cannot answer that. I think the true nature of the paradox is much more complex than what its mathematical explanation suggests. $\endgroup$ – Theo May 9 '12 at 6:20
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    $\begingroup$ Does it really make physical sense to speak of a turtle that can turn a right angle in arbitrarily small intervals of time? This is like saying, I put a ball in the box at time $t = 0$, take it out of the box at $t = 0.9$, put it back in at $t = 0.99$, take it out at $t = 0.999$, and so on; is the ball in the box at $t = 1$? The scenario described is impossible in real life. I feel that you have stretched this paradox past its breaking point. $\endgroup$ – Rahul May 9 '12 at 7:00
  • $\begingroup$ @RahulNarain Please see the edits. I think your example is just conceding Zeno's point, that at some level physical reality consists of discrete, indivisible entities. In itself, that's not paradoxical at all. Together with Zeno's arrow argument, which argues against a physical reality consisting of finite indivisible entities, it becomes a paradox. $\endgroup$ – Theo May 9 '12 at 15:08

Another way to look at it without calculus or infinitesimals: Between any two distinct points A and B on a line L, there are infinitely many other points. An object going from A to B along L, will pass through each one of these points in a finite length of time. What may seem paradoxical about this scenario is that an infinite number events (the arrival at each point) will happen in a finite length of time. But there is nothing really paradoxical about it. If the object in question maintains a constant speed $s$ (as both the tortoise and Achilles are assumed to do in your example), you can calculate its distance traveled $d$ after any given time $t$ using a simple formula ($d=st$).

Now calculate the position of both the tortoise and Achilles at $\frac{1}{99}$ seconds after Achilles sets out. They will be at the same point at this time.

For the sake of completeness, here is the "linear solution:"

Achilles (the faster runner): $d_A=100t$

Tortoise (the slower runner): $d_T=1+t$

where $t=0$ when Achilles sets out.

When is $d_A=d_T$? When $t=\frac{1}{99}$ seconds.


The Paradox for Achilles and the Tortoise

Zeno’s “paradox” is that the swift Achilles cannot catch the plodding tortoise. In this paradox Zeno bases his argument on Dichotomy.

A dichotomy is any splitting of a whole into two non-overlapping parts, meaning it is a procedure in which a whole is divided into two parts. It is a partition of a whole (or a set) into two parts. The 2 parts are then treated seperately. In above example, the movement of Achilles and the tortoise are treated independent from each other and consequentlt Achilles will never catch the tortoise.

To put Zeno’s paradox into math, let the speed of Achilles A be denoted by VA and that of the tortoise T by VT.

The time of movement t for both is the same. At the start, there is a distance DO between them. Let’s look at some of the intervals.

DO = VA x t D1 = VT x t

Since t is the same, substitute t = DO / VA into D1 which yields D1 = VT x DO / VA Now Achilles has to cover the distance D1 needing time t = D1 / VA But the tortoise, in the meantime goes ahead D2 = VT x D1 / VA. Now Achilles has to cover the distance D2 needing time t = D2 / VA But the tortoise, in the meantime goes ahead D3 = VT x D2 / VA. As can be seen this is a recursive equation

Di+1 = VT x Di / VA where i = 0, 1, 2, 3, …..infinity. The distances getting smaller and smaller but A will never catch T

Now let’s solve the problem using physics. Achilles and the tortoise will meet at time t. Achilles needs to run the distance DO plus the distance the tortoise moved in the meantime. Thus, the time needed is t = (DO + DT}/VA The tortoise, in the same time covers t = DT / VT Equating the 2 equations leads to

(DO + DT}/VA = DT / VT

Solving for the unknown distance DT the tortoise has to run before being caught yields DT = (DO x VT)/( VA - VT)

Achilles has to run DA = DO + DT

The time needed for intersection is T = ( DO/ VA) x [1 + VT/( VA - VT)]

Verification: If VA = VT then DT approaches infinity that means they never meet. If VT = 0 then they meet at the initial distance DO

Test example DO = 10m, VA = 2m/s, VT = 1m/s Results: DT = 10m, DO = 20m, t = 10s


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