Zeno's Achilles and the Tortoise, math that surprises you lesson

Achilles gives a tortoise a head start of 10 meters and runs ten times as fast. Before he can pass it, he has to reach the point where it started, and by then it has crawled a little farther. Then he has to reach that new point, and so on without end, which gives infinitely many stages with the tortoise still ahead at the end of each one. How can a runner finish infinitely many shrinking stages in a finite time? This page adds up the times of the stages, finds the moment Achilles draws level, and shows that counting stages and reading the clock tell the same story.

Key terms

Zeno’s paradoxes

Definition Zeno’s paradoxes are arguments attributed to Zeno of Elea, a Greek philosopher of the fifth century BCE, that seem to show that motion is impossible. They are known mainly through Aristotle’s Physics, and the best known are Achilles and the tortoise, the dichotomy, the arrow and the stadium. In Achilles and the tortoise, two runners move along the same straight track in the same direction at constant positive speeds and start at the same moment. The tortoise starts a distance $d > 0$ ahead and runs at speed $v_T$ , and Achilles runs at a greater speed $v_A > v_T$ . The argument splits the chase into stages. In the first stage Achilles runs to the tortoise’s starting point, and in each later stage he runs to where the tortoise was at the end of the previous stage. At the end of every stage the tortoise has moved on and is still ahead, and there is no last stage, so the argument concludes that Achilles never catches up. The resolution in the mathematical model is that the stage times form a geometric series with ratio $r = v_T / v_A$ , which is less than $1$ , and this convergent series has the finite sum

\frac{d / v_A}{1 - r} = \frac{d}{v_A - v_T}.

Every stage ends before this time, and at this time Achilles draws level with the tortoise, after which he is ahead. The stages cut the time before the meeting into infinitely many pieces, but that time is finite, and the argument never looks at the meeting itself. This settles the chase in a model where time and distance are measured by real numbers and motion is continuous. Whether that model describes physical space and time, and what it would mean to complete infinitely many separate tasks, are further questions that philosophers still discuss.

In plain words The stages get shorter fast enough that all of them together fit into a fixed, finite stretch of time. The argument looks at the race at more and more moments before the meeting, but never at the meeting or after it.

Example Take $d = 10$ meters, $v_A = 10$ meters per second and $v_T = 1$ meter per second, with times in seconds from the start and positions in meters from the starting line. Each stage takes one tenth as long as the one before, because the gap that Achilles has to close shrinks to one tenth of its size during each stage.

Stage	Duration	Ends at time	Achilles at	Tortoise at
1	$1$	$1$	$10$	$11$
2	$0.1$	$1.1$	$11$	$11.1$
3	$0.01$	$1.11$	$11.1$	$11.11$
4	$0.001$	$1.111$	$11.11$	$11.111$

The end times approach $10/9 = 1.111\ldots$ seconds, and the durations of all the stages add up to

\begin{aligned} &1 + 0.1 + 0.01 + \cdots \\[0.4em] &\quad = \frac{1}{1 - 0.1} = \frac{10}{9}. \end{aligned}

At the time $t = 10/9$ Achilles has run $10 \cdot 10/9 = 100/9$ meters, and the tortoise is at $10 + 10/9 = 100/9$ meters, the same point, about $11.11$ meters from the starting line.

Geometric series

Definition A geometric series is a sum of the form

a + ar + ar^2 + ar^3 + \cdots

in which each term is the previous one times a fixed number $r$ , called the common ratio, and $a$ is the first term. When $r \ne 1$ , the sum of its first $n$ terms is

S_n = a \cdot \frac{1 - r^n}{1 - r}.

If $-1 < r < 1$ , then $r^n$ approaches $0$ as $n$ grows, and the series converges to

a + ar + ar^2 + \cdots = \frac{a}{1 - r}.

If $r \ge 1$ or $r \le -1$ , and $a \ne 0$ , the terms do not approach $0$ , and the partial sums do not approach any number.

In plain words Start with a number, multiply by the same factor again and again, and add up the results. If the factor lies strictly between $-1$ and $1$ , the terms shrink so fast that the running total settles on one finite value.

Example The stage times of the race form the geometric series with $a = 1$ and $r = 1/10$ . The sum of the first $n$ stage times is

\begin{aligned} S_n &= \frac{1 - 10^{-n}}{1 - \frac{1}{10}} \\[0.4em] &= \frac{10}{9}\left(1 - 10^{-n}\right), \end{aligned}

so $S_2 = 1.1$ and $S_3 = 1.11$ , and the whole series sums to $10/9$ . In any chase of this kind the first stage takes $d / v_A$ and the ratio is $v_T / v_A$ , so the stages add up to $d/(v_A - v_T)$ . The dichotomy, another of Zeno’s paradoxes, says that a walker must first cover half the distance, then half of what is left, and so on. For a distance of $1$ it uses the series $\tfrac12 + \tfrac14 + \tfrac18 + \cdots$ with $a = r = 1/2$ , whose sum is $1$ .

Convergent series

Definition An infinite series $a_1 + a_2 + a_3 + \cdots$ has the partial sums $S_n = a_1 + a_2 + \cdots + a_n$ . The series converges to a real number $S$ , called its sum, if the partial sums approach $S$ . This means that for every positive number $\varepsilon$ there is a whole number $N$ such that $\lvert S_n - S \rvert < \varepsilon$ for every $n \ge N$ . A series that does not converge is said to diverge. A convergent series can have infinitely many positive terms. Its terms must approach $0$ , but terms that approach $0$ do not guarantee that the series converges.

In plain words An infinite sum means the number that the running totals settle on. Having infinitely many pieces does not make a total infinite, and having shrinking pieces does not make it finite. What matters is how fast the pieces shrink. The same idea explains Why 0.999… Equals 1.

Example After $n$ stages of the race, the time still left before the meeting is

\frac{10}{9} - S_n = \frac{10}{9 \cdot 10^n}

seconds, which is $1/9$ , $1/90$ and $1/900$ of a second after one, two and three stages. For $\varepsilon = 0.001$ the choice $N = 4$ works, since $1/9000 < 0.001$ and the later remainders are smaller still, while $N = 3$ fails, since $1/900$ is more than $0.001$ . Every positive $\varepsilon$ , however small, is beaten in the same way by a large enough $N$ , so the stage times converge to $10/9$ . By contrast, the harmonic series

1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots

has terms that approach $0$ , yet it diverges. The terms from $\tfrac13$ to $\tfrac14$ add up to more than $\tfrac12$ , and so do the terms from $\tfrac15$ to $\tfrac18$ , the terms from $\tfrac19$ to $\tfrac1{16}$ , and so on, so the partial sums grow past every number.

Relative speed

Definition When two objects move along the same straight line at constant speeds, the rate at which the distance between them changes is called their relative speed. If they move in the same direction at speeds $v_A > v_T$ , the faster one gains on the slower one at the rate $v_A - v_T$ . If they move toward each other, the distance between them shrinks at the rate $v_A + v_T$ . A faster object that starts a distance $d$ behind a slower one moving the same way is still $d - (v_A - v_T)t$ behind at time $t$ , so it draws level at the single time

t^* = \frac{d}{v_A - v_T}

and is ahead at every later time. If $v_A = v_T$ the gap never changes, and if $v_A < v_T$ it grows.

In plain words Picture the race as the tortoise sees it. Achilles comes up from behind at the difference of the two speeds, so the head start is closed at that rate in one step, with no stages at all.

Example With positions $x_A = 10t$ and $x_T = 10 + t$ in meters, the gap is $x_T - x_A = 10 - 9t$ . Achilles gains $9$ meters each second, and the gap is $0$ at $t = 10/9$ seconds, the same time as the sum of Zeno’s stages. Two seconds after the start, Achilles is at $20$ meters and the tortoise at $12$ , so Achilles leads by $8$ meters.

Check yourself

Give the tortoise a head start of $12$ meters and a speed of $2$ meters per second, and let Achilles run at $8$ meters per second. Find when Achilles catches up, first from the relative speed and then by adding Zeno’s stages.

Show the answer

Achilles gains $8 - 2 = 6$ meters each second, so he closes the gap of $12$ meters in $12/6 = 2$ seconds. They meet at the $16$ meter mark, since $8 \cdot 2 = 16$ and $12 + 2 \cdot 2 = 16$ .

In Zeno’s stages, the first one takes $12/8 = 1.5$ seconds, and during it the tortoise moves $3$ meters, to the $15$ meter mark. The second stage takes $3/8 = 0.375$ seconds, and the gap shrinks to $0.75$ meters. Each stage takes one quarter as long as the one before, because the ratio of the speeds is $2/8 = 1/4$ . The stage times form a geometric series with $a = 1.5$ and $r = 1/4$ , so they add up to

\frac{1.5}{1 - \frac{1}{4}} = \frac{1.5}{0.75} = 2

seconds, the same answer.

Change the rule so that Achilles runs exactly as fast as the tortoise, $5$ meters per second each, with the tortoise $10$ meters ahead. What happens to Zeno’s stages, and does Achilles catch up?

Show the answer

Each stage now takes $10/5 = 2$ seconds, and during it the tortoise also moves $10$ meters, so the gap is still $10$ meters after every stage. The stage times form the geometric series $2 + 2 + 2 + \cdots$ with ratio $r = 1$ , whose partial sums $2n$ grow without bound, so the series diverges. The relative speed is $5 - 5 = 0$ , so the gap never changes and Achilles never catches up.

Here the conclusion of Zeno’s argument is true. The number of stages is the same as in the original race, so the infinitely many stages are not what decides the outcome. What decides it is whether the stage times add up to a finite total, and that happens exactly when Achilles is faster, so that the ratio $v_T / v_A$ is less than $1$ .

In the original race, with a head start of $10$ meters and speeds of $10$ and $1$ meters per second, how many stages must pass before less than one millionth of a second remains until Achilles draws level with the tortoise?

Show the answer

After $n$ stages the remaining time is $10/(9 \cdot 10^n)$ seconds. After $6$ stages this is

\frac{10}{9 \cdot 10^6} \approx 1.11 \times 10^{-6}

seconds, still more than one millionth, and after $7$ stages it is about $1.11 \times 10^{-7}$ seconds, less than one millionth. So $7$ stages are needed.

The meeting itself is never one of the stages. Every stage ends before $10/9$ seconds, and the end times of the stages come as close to $10/9$ as we like, which is what it means for them to have the limit $10/9$ .

Zeno's Achilles and the Tortoise

Video 1, chapter 17

Key terms

Zeno’s paradoxes

Geometric series

Convergent series

Relative speed

Check yourself