Zeno of Elea · c. 450 BCE

Achilles and the Tortoise

The swiftest runner can never overtake the slowest. Give a tortoise a head start and — Zeno argued — Achilles is doomed: to pass it he must first reach where it was, but in that time it has crept a little further on; reach that spot and it has crept again; and so on, forever. Infinitely many stages, so he never arrives. Below, step through the stages yourself — and watch the infinite collapse onto a single, finite point.

Set up the race

Full track — Achilles ●, tortoise ▮, the catch ┃. Ticks mark each stage's end.

Zoom on the gap — re-magnified each stage by the speed ratio, so every stage looks the same size. It never runs out of stages. Yet the catch line never moves.

Press Take the next stage to begin Zeno's chase. Stage 0 — Achilles at the start line, the tortoise 100 m ahead.

Achilles' distance run (sum of stages so far) 0%
Stage number n0
Achilles has run0 m
Tortoise still ahead by100 m

Stages remaining

infinitely many

the chase truly never ends

…yet they fit before

1000⁄9 m

≈ 111.111111 m · at t ≈ 11.111111 s

Zeno's logic is airtight up to one buried assumption: that infinitely many stages must add up to an infinite amount — of distance, and of time. They don't. The stage lengths shrink by the same factor every time (here, ÷10), so they form a convergent geometric series. Add all of them — every last one of the infinitely many — and you get a finite total. Achilles passes the tortoise at exactly that total, at exactly the finite moment the stage-times sum to. The "never" was hiding a false sum.

Why the sum is finite

distance = H + H·r + H·r² + … = H / (1 − r)   with r = 1/k < 1

With k = 10 and H = 100 m, that is 100 / (1 − 1/10) = 1000⁄9 ≈ 111.11 m — the catch point you can read off the cards above. The time works identically: each stage takes ten times less than the last, so the durations sum to H/(k−1) = 100/9 ≈ 11.11 s. Both totals are recomputed in exact whole-number ratios as you move the sliders — never rounded into agreement.

The check

Every number on this page is exact rational arithmetic (BigInt numerator⁄denominator), so nothing is rounded into looking right. The catch point is L = H·k / (k − 1); after n stages Achilles has run Sn = H·(kⁿ − 1)·k / ((k−1)·kⁿ), strictly below L at every finite n, and the tortoise's lead is exactly H / kⁿ — which shrinks toward zero but is never zero at any finite stage. Step far enough and you'll see a lead like 1/10⁴⁸ m: tiny, positive, real. The offline verifier research/zeno-paradox/verify.mjs runs the same identities (catch point = H·A/(A−T), the stage-times summing to H/(A−T), the Dichotomy ½+¼+⅛+… = 1) across a batch of races — all passing.

What this does not settle

It is often said, glibly, that "calculus solved Zeno." Be careful. The convergent series dissolves the quantitative paradox: the distance and the time are finite, so "Achilles needs forever" is simply false. But it does not, on its own, answer the deeper question Zeno's stages raise — whether completing infinitely many distinct acts (a supertask) is even coherent. That is a live problem in philosophy (Thomson's lamp; Benacerraf, 1962): a series of times can converge to a limit while the thing being done at each step has no well-defined limiting state. The maths tells you where the stages are heading; it does not by itself prove the completion is unproblematic. We show the sum and name the gap — we don't paper over it.