Adding the successive halves of a track gives exactly 1, even though there are infinitely many terms. That is the mathematical answer to Zeno’s dichotomy: each stretch is half the previous one, and together they cover the whole distance. The adding fractions calculator checks partial sums with an exact result. What the sum does not settle is the philosophical debate, and the two are kept apart below.
Two paradoxes that often get mixed up
Zeno of Elea, a Greek philosopher of the fifth century BC and a pupil of Parmenides, is known for arguments against motion. The statements of his four arguments about motion (the dichotomy, Achilles and the tortoise, the arrow and the stadium) survive mainly through Aristotle’s Physics (VI.9), with additions from Simplicius and Plato.
In the dichotomy, to cover a distance you must first reach the halfway point, then half of what is left, and so on without end. In Achilles and the tortoise, the fast runner gives the slow one a head start and, when he reaches the point where it stood, it has moved a little farther. Aristotle treats them almost as one, although some authors argue their structure differs. Both lead to a sum of infinitely many ever-shorter stretches, which is why fractions help in studying them.
The dichotomy with fractions
Each stretch is a unit fraction of the total distance. A partial sum is how much you have covered after n stretches.
Worked example: six stretches of the dichotomy
A runner covers half of the track, then half of what is left, and repeats this six times in all. What part of the track has the runner covered, and how much is left?
- Write each stretch as a fraction. They are 1/2, 1/4, 1/8, 1/16, 1/32 and 1/64 of the track.
- Choose the common denominator. The largest denominator, 64, is a multiple of the others, so the stretches become 32/64, 16/64, 8/64, 4/64, 2/64 and 1/64.
- Add the numerators. 32 + 16 + 8 + 4 + 2 + 1 = 63.
- Find what is left. 1 − 63/64 = 1/64.
Result: the runner has covered 63/64 of the track, and 1/64 is left.
Check: what is left equals the last stretch covered, and each new stretch covers exactly half of what was left.
The rule that follows is that after n stretches the partial sum is (2ⁿ − 1)/2ⁿ and the gap is 1 over 2ⁿ.
| n | Partial sum | Gap to 1 | Gap as a decimal |
|---|---|---|---|
| 3 | 7/8 | 1/8 | 0.125 |
| 4 | 15/16 | 1/16 | 0.0625 |
| 6 | 63/64 | 1/64 | 0.015625 |
| 10 | 1023/1024 | 1/1024 | 0.0009765625 |
The gap is never zero after a finite number of steps, but it becomes smaller than any amount you choose. On the number line, the points in the middle column crowd up against 1 without passing it.
Why infinitely many terms can have a finite total
An infinite sum is defined as the limit of its partial sums. Here the gap approaches 0 as n grows, so the partial sums approach 1 and the limit is 1. Infinitely many terms do not mean an infinite amount when each term is a fixed fraction of the one before.
This pattern is called a geometric series: a first term a and a ratio r that multiplies each step. If −1 < r < 1, the total is a/(1 − r). Zeno’s series has a equal to one half and r equal to one half, and gives 1. With a = 1/3 and r = 1/3 you get (1/3)/(2/3), which is one half, and the partial sums 1/3, 4/9 and 13/27 approach that value. If r is 1 or more, the terms do not shrink and the sum grows without limit.
Achilles and the tortoise with numbers
Achilles’ version is also a geometric series. Suppose the tortoise starts 100 m ahead, Achilles runs at 10 m/s and the tortoise at 1 m/s. Achilles first covers 100 m while the tortoise moves 10 m. Then he covers those 10 m while the tortoise moves 1 m, and so on.
Achilles’ stretches are 100 m, 10 m, 1 m, 0.1 m and so on, with a = 100 and r = 1/10. The sum is 100 ÷ (1 − 1/10) = 1000/9 m, about 111.1 m (the rounding error is less than 0.05 m). The time works the same way: 10 s + 1 s + 0.1 s + … adds up to 100/9 s, about 11.1 s (error less than 0.05 s). A check without series confirms it. Achilles gains 9 m every second and the head start is 100 m, so it takes him 100/9 s, and in that time he runs 10 · 100/9 = 1000/9 m.
What math settles and what stays open
Summing the series shows that the total distance and time are finite, and that Achilles does catch the tortoise. The calculation goes that far. The philosophical objection is a separate matter: Zeno could reply that the sum is built stretch by stretch and that this count never finishes, or that it is built from the wrong end. The MacTutor biography notes that the infinite sum alone does not answer the dichotomy for this reason, and Wikipedia notes that some philosophers still regard the paradoxes as open problems. This page only states the mathematical result.
A sister series: 0.999… = 1
With a = 9/10 and r = 1/10, the series 9/10 + 9/100 + 9/1000 + … equals (9/10) ÷ (1 − 1/10) = 1. Its partial sums are 0.9, 0.99 and 0.999, and the gap is 1/10ⁿ. That is why the repeating decimal 0.9̅ is another way of writing 1; the guide on exact and repeating decimals explains it and the repeating decimal to fraction calculator turns it into a fraction. Another case of approaching a limit, this time from both sides, is the ratios of Fibonacci numbers and the golden ratio.
Practice
- Find 1/6 + 1/36 + 1/216 + … with the formula a/(1 − r).
- The tortoise starts 50 m ahead, Achilles runs at 10 m/s and it runs at 2 m/s. How many meters does Achilles run before he catches it?
Show the answers
- 1/5, because a = 1/6, r = 1/6 and (1/6) ÷ (5/6) = 1/5. Check: 1/6 + 1/36 = 7/36 already exceeds 0.19, and 1/5 = 0.2.
- 62.5 m. The stretches are 50 + 10 + 2 + …, with ratio one fifth, and 50 ÷ (1 − 1/5) = 125/2. Check: he gains 8 m every second, takes 50/8 = 6.25 s and runs 10 · 6.25 = 62.5 m.
Sources
- Wikipedia, Zeno’s paradoxes: statement of the dichotomy and of Achilles and the tortoise, quotation from Aristotle’s Physics VI.9, resolution with limits and the philosophical debate that continues (accessed October 17, 2026).
- J. J. O’Connor and E. F. Robertson, Zeno of Elea, MacTutor History of Mathematics, University of St Andrews: Parmenides, Aristotle’s four arguments in his Physics, and why the sum of the halves does not settle the dichotomy (accessed October 17, 2026).