Zeno's book is gone
Zeno's own book is lost. We know his paradoxes from other writers' accounts, above all Aristotle, who took them apart in his "Physics" in order to refute them.
Zeno of Elea argued that Achilles can never catch the tortoise: by the time he reaches where it was, it has crawled a little further, and so on without end. Let's test the idea on three walkers. Each takes infinitely many steps, and each step is shorter than the one before:
A: 1 km, then ½, ¼, ⅛ km…
B: 1 km, then ¼, 1/9, 1/16 km…
C: 1 km, then ½, ⅓, ¼ km…
A milestone stands 10 km from the start. Which of them will ever pass it? Lock in your answer, then check it.
Achilles catches the tortoise in 11.11 seconds: infinitely many stages fit into a finite time. With the walkers, neither simple answer works: not "infinitely many steps, so infinitely far", and not "the steps shrink to zero, so it stops". A never gets even to 2 km, B not even to 1.645 km. But C, whose steps also shrink to zero, passes the milestone at step 12,367 and goes as far as you like. What decides is not how many steps there are but how fast they shrink.
Zeno lived in the 5th century BC and composed several paradoxes, arguments about motion that look impossible to answer. The most famous is about Achilles. We start with it, then ask a question where intuition goes wrong in a more modern way.
Achilles runs 10 m/s, the tortoise crawls 1 m/s and starts 100 m ahead. To catch it, Achilles first has to reach where it was: 100 m in 10 s. But in that time the tortoise has crawled 10 m. He runs them in 1 s, and it is already 1 m further on. He runs that metre too, and it is still 10 cm ahead.
There are infinitely many stages, and after each one the tortoise is still ahead. So, says Zeno, Achilles never catches it. It sounds like a joke, but pinning down exactly where the argument fails took people more than two thousand years.
Each one walks without stopping and takes infinitely many steps. Each step is shorter than the one before, and the steps shrink towards zero.
A milestone stands 10 km from the start. Will each of the three walkers pass it, if they walk forever?
"There are infinitely many steps, so the walker gets as far as you like"
versus"The steps shrink to zero, so sooner or later he stops"
Which one is right, and for whom? Notice that A and C start the same: 1 km, then ½ km.
Both are wrong as a general rule. A and B never get even to 2 km despite infinitely many steps, while C goes as far as you like even though his steps shrink to zero.
The first eight steps of each, on a scale from 0 to 3 km. So far all three are shuffling near the start, and the milestone is far away.
For each walker, pick whether he ever passes the milestone. Lock in your answer before reading on.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Look at walker A from the 2 km mark. What part of the remaining distance to it does each of his steps cover?
For B, try bounding each step 1/n² from above by the difference 1/(n − 1) − 1/n. What do such differences add up to?
Group C's steps: ⅓ + ¼, then ⅕ to ⅛, then 1/9 to 1/16. At least how much does each group give?
The next rounds show where each walker ends up: a game with Achilles, millions of steps on a computer and the explanation. Answer round 1 first, or go on?
You chose to always show all rounds.
First go through a few of Zeno's stages with Achilles, then let ordinary time run. Next, move the step slider for the three walkers all the way to ten million.
In stage mode the picture zooms in tenfold each time, so every stage looks the same. Only the labels change: the gap to the tortoise is 10 times smaller, and the clock adds 10 times less.
The slider is logarithmic: each seventh of its travel multiplies the number of steps by 10. The dashed lines are the limits that A and B approach. The ticks on the lanes mark the first steps, and you can see them getting finer.
There is no randomness here, so instead of a simulation the computer simply adds the steps one by one, 10,000,000 steps for each walker, and compares the sums with formulas. Dots are what it counted, lines are the formulas.
Left: the step axis is logarithmic, each division multiplies n by 10. A and B quickly reach their ceilings, 2 and π²/6 ≈ 1.645. On this axis C climbs almost in a straight line: every tenfold increase in steps adds another ln 10 ≈ 2.3 km. Near n = 1 the formulas for B and C are still rough, but further on the dots sit exactly on the lines. Right: for C to cover one more kilometre, he needs about e ≈ 2.72 times as many steps. That is why 10 km takes him 12,367 steps, while 20 km would take ≈ 2.7·10⁸.
Once you answer, the simulation code appears here so you can run it yourself.
The same simulation, short and with no libraries. Run the JavaScript in your browser console (F12) or in Node; the Python needs only the standard library.
// Three walkers: steps of 1/2ⁿ⁻¹, 1/n² and 1/n km. Where do they get in 10 million steps?
const N = 1e7, GAMMA = 0.5772156649015329;
let a = 0, b = 0, c = 0, step = 1, cross = 0;
for (let n = 1; n <= N; n++) {
a += step; step /= 2; // A: each step half the one before
b += 1 / (n * n); // B: a step of 1/n²
c += 1 / n; // C: a step of 1/n
if (!cross && c > 10) cross = n; // the first step past the 10 km milestone
}
console.log('A', a.toFixed(10), 'limit 2');
console.log('B', b.toFixed(10), 'limit π²/6 =', (Math.PI ** 2 / 6).toFixed(10));
console.log('C', c.toFixed(10), 'ln n + γ =', (Math.log(N) + GAMMA).toFixed(10), 'and still growing');
console.log('C passes 10 km at step', cross);
# Three walkers: steps of 1/2ⁿ⁻¹, 1/n² and 1/n km. Where do they get in a million steps?
import math
N = 1_000_000
GAMMA = 0.5772156649015329
a = b = c = 0.0
step, cross = 1.0, 0
for n in range(1, N + 1):
a += step; step /= 2 # A: each step half the one before
b += 1 / n**2 # B: a step of 1/n²
c += 1 / n # C: a step of 1/n
if not cross and c > 10: # the first step past the 10 km milestone
cross = n
print("A", f"{a:.10f}", "limit 2")
print("B", f"{b:.10f}", "limit π²/6 =", f"{math.pi**2 / 6:.10f}")
print("C", f"{c:.10f}", "ln n + γ =", f"{math.log(N) + GAMMA:.10f}", "and still growing")
print("C passes 10 km at step", cross)
Zeno counts stages; we count distance. A sum of infinitely many terms can be finite, and it can also be infinite even when the terms shrink to zero. It all depends on how fast they shrink.
“In plain words” is hidden on all puzzles.
Zeno's stages last 10 s, 1 s, 0.1 s, 0.01 s and so on. Together that is 11.111… s, which is 11 and 1/9 seconds. There are infinitely many stages, but they crowd ever closer to one instant and never go past it.
Zeno only describes the time up to that instant and says nothing about what comes after. But the clock keeps going: at 12 seconds Achilles is already 8 m ahead. Infinitely many stages fit into a finite time. This card explains Achilles; for the walkers it is not enough on its own, because C also has infinitely many shrinking steps.
Look at walker A from the 2 km mark. After the first step 1 km is left to go, and the next step is ½ km, exactly half of what is left. After that ½ km is left, and the step is again half, ¼. Every step covers half of what is left to 2 km.
Half of what is left never exceeds what is left. So A gets as close to 2 km as you like but never reaches it, let alone the 10 km milestone. This is Zeno's second paradox, the "dichotomy", told the other way round: there, to cover a distance you must first cover half of it, then half of the rest, and so on.
Group C's steps. The step ½ gives ½. The steps ⅓ and ¼ together are more than ¼ + ¼ = ½. The next four steps, ⅕ to ⅛, are more than 4 · ⅛ = ½. The next eight, 1/9 to 1/16, are more than 8 · 1/16 = ½. And so on without end.
There are infinitely many groups, and each adds at least half a kilometre. Twenty groups already give more than 10 km. C's steps shrink, but so slowly that there is always enough of them for another half. Nicole Oresme found this proof around 1350. The same series saves the ant on a rubber rope.
"Infinitely many steps, so he goes infinitely far" is false: A and B walk forever and still face an invisible wall. "The steps shrink to zero, so he stops" is false too: C's steps get smaller than any number you name, and he still passes any milestone.
Steps shrinking to zero is only a necessary condition for a finite sum, not a sufficient one. What matters is how fast they shrink. Halving at every step (A) is enough. Like 1/n² (B) is enough too. Like 1/n (C) is not.
“Rigorous” is hidden on all puzzles.
Each of B's steps from the second on is smaller than a difference of two fractions: 1/n² < 1/((n − 1)n) = 1/(n − 1) − 1/n.
The right-hand sides fold up like a telescope: each fraction appears once with a plus and once with a minus, and 1 − 1/n is left. So all the steps after the first add up to less than 1 km, and with the first one B never covers 2 km. The exact limit is smaller: π²/6 ≈ 1.645 km. Leonhard Euler found it in 1734.
| n | A | B | C |
|---|---|---|---|
| 10 | 1.99805 | 1.54977 | 2.92897 |
| 100 | 2 − 2−99 | 1.63498 | 5.18738 |
| 1000 | 2 − 2−999 | 1.64393 | 7.48547 |
| 106 | 2 − 2−999999 | 1.644933 | 14.39273 |
| 107 | 2 − 2−9999999 | 1.6449340 − 10−7 | 16.69531 |
| limit | 2 | π²/6 ≈ 1.644934 | none |
After a hundred steps A and B are already almost at their limits, and millions of steps after that add only crumbs. C has covered only 5.2 km after 100 steps and 14.4 km after a million, but he never stops: every tenfold increase in steps adds another 2.3 km.
The geometric series: each term is r times the one before. It is finite when |r| < 1. For A, a = 1 and r = ½; for Achilles, a = 100 m and r = 1/10.
The p-series: finite if and only if p > 1. For B, p = 2 and the sum is π²/6 (Euler, 1734). For C, p = 1 and the series diverges.
Harmonic numbers grow like a logarithm. γ ≈ 0.5772 is the Euler–Mascheroni constant. Hence C covers k km in about ek−γ steps: to get the exact step number, subtract ½ and round up.
The sum of an infinite series means a limit: the number that the sums of the first n terms get as close to as you like as n grows. For A that limit is 2, for B π²/6. For C there is no such number: the sums outgrow any of them. This clarification of what "the sum of infinitely many terms" means is what dissolves the Achilles paradox: he catches the tortoise at 1000/9 m, at the moment 100/9 s, after all of Zeno's stages.
“Deeper” is hidden on all puzzles.
Zeno of Elea lived in the 5th century BC. His book is lost, and we know the paradoxes mostly from Aristotle's "Physics". Four of them are about motion: Achilles, the "dichotomy" (to get anywhere you must first go half the way, and before that a quarter, without end), the "arrow" (at every instant a flying arrow is at rest) and the "stadium" (about rows of bodies moving past each other).
Around 1350 Nicole Oresme proved that the harmonic series diverges. In 1650 Pietro Mengoli asked what 1 + ¼ + 1/9 + … equals, a question that became known as the Basel problem. Euler gave the answer π²/6 in 1734. The rigorous definition of a limit, on which the modern answer to Zeno rests, came in the 19th century, above all in the work of Augustin-Louis Cauchy.
In 1954 the philosopher James F. Thomson proposed this lamp in the journal Analysis. It is switched on at minute 1, off at minute 1.5, on at minute 1.75 and so on, each time twice as fast. Before minute 2 there are infinitely many switches. Is the lamp on or off at 2:00?
The time here is finite, just as for Achilles: 1 + ½ + ¼ + … = 2 minutes. But there is no last switch, so the description says nothing about the moment 2:00. In 1962 Paul Benacerraf showed that both answers are consistent with the setup. The limit tells us where Achilles will be, because his position changes continuously. But not every question about the "end" of an infinite process has an answer just because the process fits into a finite time.
B and C are two points on one scale: steps of 1/np with p = 2 and p = 1. Turn p and see exactly where the sum stops being finite. Then change A's geometric steps.
For p ≤ 1 the sum grows without limit: with p = ½ the walker passes the milestone already at step 33, with p = 0.8 at step 201, with p = 1 at step 12,367. For p > 1 the sum has a limit: ζ(1.5) ≈ 2.612, ζ(2) ≈ 1.645, ζ(3) ≈ 1.202. Here ζ(p) is the Riemann zeta function, just a name for this sum.
The most interesting case is p = 1.1. The limit ζ(1.1) ≈ 10.584 km is more than 10: the walker passes the milestone and still stops before 10.6 km. Only it takes ≈ 2.15·10¹² steps. After 10 million steps he has covered just 8.6 km, and on the chart you cannot tell him from C. That is why no computer can tell "by eye" whether a sum is finite: you need a proof, like Oresme's or the telescope for B.
Zeno's mistake and the intuition's mistake about C are the same: both judge a sum from a single fact, that there are infinitely many terms, or that the terms shrink. The right question is different: how fast they shrink. Geometric steps with a fixed ratio r < 1 always give a finite sum, even if r = 0.99 and the limit is as far as 100 km.
A physical example is a bouncing ball. It drops from 1 m and keeps 0.8 of its speed after every bounce, so the bounce height is multiplied by 0.64 each time. There are infinitely many bounces, but they all fit into ≈ 4.06 s, and the ball travels ≈ 4.56 m (g = 9.81 m/s²). Engineers who model such systems call this "Zeno behaviour": a program that counts bounces one by one never gets to the moment the ball comes to rest. Another ball with other numbers is in the fly between two trains.
Whenever something repeats without end and shrinks, ask: by a fixed factor (then the sum is finite) or more slowly, like 1/n (then it can grow without limit)? The numbers in the table are easy to check on a calculator.
| Situation | Series | What comes out |
|---|---|---|
| A perpetuity: 100 a year forever at 5% | Geometric, r = 1/1.05 | Infinitely many payments are worth a finite sum today: 100/0.05 = 2,000. More in the fly between two trains. |
| Binary search: halve the list each time | Dichotomy, r = ½ | A million records narrow down to one in 20 steps, because 2²⁰ = 1,048,576. Zeno's dichotomy, put to work. |
| A stack of books leaning over a table edge | Harmonic, Hn/2 | n books can overhang by Hn/2 book lengths. Just 4 books give 1.04, so the top one is entirely past the edge, 31 books give more than 2 lengths, and in principle the overhang is unlimited (Paul B. Johnson, 1955). More in the ant on a rubber rope. |
| A bouncing ball | Geometric, r = 0.8 for time | Infinitely many bounces in ≈ 4.06 s. A model that counts them one at a time gets stuck. |
Zeno's own book is lost. We know his paradoxes from other writers' accounts, above all Aristotle, who took them apart in his "Physics" in order to refute them.
Oresme's proof that the harmonic series diverges was forgotten for a long time. In the 17th century it had to be found again, by Pietro Mengoli and Johann Bernoulli.
Mengoli asked about the sum 1 + ¼ + 1/9 + … in 1650. Euler found the answer π²/6 in 1734, 84 years later. What surprised people was the π itself: there are no circles anywhere in the sum.
Walker C needs 12,367 steps to cover 10 km, and as many as ≈ 2.7·10⁸ steps to cover 20 km. At one step a second, the second 10 km would take more than eight years.
A stack of 31 books can overhang a table edge by more than two book lengths. Paul B. Johnson described it in 1955 in the American Journal of Physics in a note called "Leaning Tower of Lire".
A quantum system that is observed often enough barely changes, rather like Zeno's "arrow". Misra and Sudarshan named the effect in 1977, and in 1990 Wayne Itano and colleagues demonstrated it with trapped beryllium ions.
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