OConnor -- Zeno's Paradox

ZENO’S PARADOX

Derek O’Connor

Started : February 6, 2008Latest : February 15, 2013

1 Introduction

Zeno (or Zenon, or Xeno, Xenon1) was born about 490 BC in Elea, southern Italy. Hewas a pupil of Parmenides at the famous Greek school of philosophy founded by Xeno-phanes. This school denied the existence of the infinite divisibility of space and time.One of the arguments used to support their position was Zeno’s justly-famous paradox.

Zeno’s Paradox. Achilles and a Tortoise are to have a race.

Figure 1. THE MAN HIMSELF

Because Achilles runs 10 times faster than the Tortoise,the Tortoise is given a head start of 10 stadia2 . It is ob-vious that Achilles will always beat the Tortoise over asufficiently long distance. For example, if Achilles coversa distance of 20 stadia in 20 time units then the Tortoisewill have covered 10 + 20

10 = 12 stadia in the same time.

However, Zeno argued as follows : by the time Achillesreaches the 10-stadia mark, the Tortoise is at 10 + 10

10 = 11stadia; by the time Achilles reaches the 11-stadia mark,the Tortoise is at 11 + 1

10 = 11.1 stadia, and so on. Inthis sequence of events, the Tortoise is always ahead ofAchilles. Hence the paradox.

Solutions

We will look at two mathematical solutions to this problem and then two algorithmicsolutions in an effort to eliminate as many distracting mathematical artifacts as possi-ble.

The conclusion reached here will satisfy (or not) both sides in this argument :

Achilles passes the Tortoise but they never meet.

1Hero - Heron, Plato - Platon2One stadium is about 625 ft ≈ 1 furlong

1

Derek O’Connor Zeno’s Paradox

In what follows we assume (1), that we can count, and (2), that we can measure distance(in distance units) and time (in time units).

2 Mathematical Solutions

2.1 First Solution – Algebra

Let DT(t) and DA(t) be the distance of the tortoise and Achilles from the starting point,at time t. Achilles runs at a speed SA distance units in 1 unit of time, while the tortoiseruns at a speed ST distance units in 1 unit of time. The Tortoise is given a head-start of10 units. Let tm be the instant at which they meet. We have DT(0) = 10, ST = 1

10 andDA(0) = 0, SA = 1. Hence the equations of motion are

DT(t) = DT(0) + STt = 10 +110

t (2.1)

DA(t) = DA(0) + SAt = t (2.2)

We plot these equations and see in Figure 2 that the tortoise is ahead of Achilles upto time tm, after which Achilles is ahead of the tortoise. At time tm the tortoise andAchilles are at the same distance from the start, which gives the condition for findingthe value of tm. That is

DT(tm) = DA(tm), or 10 +1

10tm = tm, or tm =

1009

(2.3)

This solution uses simple algebra and obtains an answer that is a finite rational number.

This simple algebraic solution implicitly assumes the infinite divisibility of space andtime, just as the solution of the algebraic equation x2− 2 = 0 assumes these properties.3

3The solution, x =√

2, is irrational and cannot be represented by a finite number of digits.

© Derek O’Connor, February 15, 2013 2


Figure 2. ACHILLES AND THE TORTOISE

2.2 Second Solution – Analysis

A solution more in keeping with Zeno’s argument is this: by the time Achilles hasreached DA(10) = 10, the tortoise has reached the point DT(10) = 10 + 1

1010 = 11;by the time Achilles has reached DA(11) = 11, the tortoise has reached DT(11) =(10 + 1

1010) + 110 = 10 + 1 + 1

10 = 11.1, etc. Repeating these steps we get

DT(tn) = 10 +n

∑k=0

(1

10

)k, (2.4)

where tn is the time the tortoise takes to perform n steps of ever-decreasing length. Inthe limit, as n→ ∞, Achilles catches up with the tortoise at time tm, and we get

DA(tm) = DT(tm) = limn→∞

DT(tn) = 10 + limn→∞

n

∑k=0

(1

10

)k= 10 +

11− 1

10

=1009

. (2.5)

This solution is the same as the first : a finite rational number, but we have used aninfinite series rather than simple algebra to obtain it. The assumption of the infinitedivisibility of space and time is obvious here.



3 Algorithmic Solutions

3.1 Third Solution – Successive Approximation

We recast the problem as

Solve DA(tm) = DT(tm) for tm, (3.1)

where tm is the time at which Achilles and the Tortoise meet.

If Achilles moves a distance DA(t) = t in time t, then the tortoise has moved a distanceDT(t) = 10 + 1

10 DA(t), in the same time t. We re-write (3.1) to get it into fixed point orsuccessive approximation form:

Solve the equation t = D−1A (DT(t)) = 10 +

110

t, for t. (3.2)

This is a fixed point equation, t = f (t), which can be solved by the successive approx-imation sequence {tk+1 := f (tk) = 10 + 1

10 tk}, starting with t0 = 0. The algorithmZenoSA generates the successive approximation sequence, where k is a count of the iter-ations.

algorithm ZenoSA

DT := 10; DA := 0k := 0while DA < DT do

k := k + 1DA := DT Achilles gets to where the Tortoise was,

DT := 10 + 110 DA but the Tortoise has moved on.

endwhile

We can see from Table 1 that this is the algorithmic equivalent of the infinite series

Table 1. SUCCESSIVE RATIONAL APPROXIMATIONS TO tm

k 0 1 2 3 4 5 6 7 · · · ∞

DT 10 11 11110

1111100

111111000

11111110000

1111111100000

111111111000000 · · · 100

9

DA 0 10 11 11110

1111100

111111000

11111110000

1111111100000 · · · 100

9

solution and the sequence is converging to 1009 . It is quite obvious that the while -loop

never stops and that k→ ∞.



Figure 3. ACHILLES AND THE TORTOISE APPROACH THE FIXED POINT 32

In Figure 3 we have plotted the the positions of the Tortoise and Achilles with DT =16 + 1

2 t and DA = t, as the ZenoSA algorithm approaches the fixed point 32 which isrepresentable as a base-2 floating point number. If ZenoSA is run using MATLAB thenit fortuitously converges to 32 after 50 iterations. This is because MATLAB uses finiteprecision floating point arithmetic with 16-digit precision.If we run ZenoSA using MAXIMA with exact rational arithmetic we see that the Tortoiseis still ahead of Achilles at k = 50:

DT =502, 251, 799, 813, 685, 247

70, 368, 744, 177, 664= 31.9999999999999857891452847979962825775146484375

> DA =1, 125, 899, 906, 842, 623

35, 184, 372, 088, 832= 31.999999999999971578290569595992565155029296875

Note that in IEEE double precision arithmetic, fl(DT) = fl(DA) = 32, and so we getconvergence in a finite number of steps because of rounding.

3.2 Fourth Solution – Simulation

This algorithmic solution has the advantage that it uses nothing more complicated thanthe addition of the natural numbers – no analysis, algebra, or rational or floating-point



numbers are here to distract us.4

To ensure that all numbers used are natural we have re-stated the problem: the Tortoisetravels 1 distance unit in 1 time unit and Achilles travels 10 distance units in 1 timeunit. The Tortoise is given a 50 distance-unit head-start.

The state of this system is the distance vector D(t) = [DA(t), DT(t)]>, with D(0) =[0, 50]>. This vector difference equation describes the ‘dynamics’ of the system:[

DA(t + 1)

DT(t + 1)

]=

[DA(t)

DT(t)

]+

[10

1

],

[DA(0)

DT(0)

]=

[0

50

]. (3.3)

The algorithm for ‘simulating’ this difference equation is trivial:

algorithm ZenoSim

DA := 0; DT := 50t := 0while DA 6= DT do

t := t + 1DA := DA + 10DT := DT + 1

endwhile

In ZenoSim DA and DT are measured in distance units – these may be yards, furlongs,stadia, or nano-meters. Time is recorded by the variable t in time units – these maybe years, micro-seconds, or æons. The clock is incremented by 1 time unit for eachiteration of the while -loop.

When we run this algorithm we get the following result:

Table 2. ACHILLES PASSES THE TORTOISE BETWEEN t = 5 AND t = 6

t 0 1 2 3 4 5 6 7 8 9 10

DT 50 51 52 53 54 55 56 57 58 59 60

DA 0 10 20 30 40 50 60 70 80 90 100

Conclusion: Achilles passes the Tortoise but they nevermeet.

4Kronecker, I hope, would approve.


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OConnor -- Zeno's Paradox