**Zeno’s Paradox – Achilles and the Tortoise**

This is a very famous paradox from the Greek philosopher Zeno – who argued that a runner (Achilles) who constantly halved the distance between himself and a tortoise would never actually catch the tortoise. The video above explains the concept.

There are two slightly different versions to this paradox. The first version has the tortoise as stationary, and Achilles as constantly halving the distance, but never reaching the tortoise (technically this is called the dichotomy paradox). The second version is where Achilles always manages to run to the point where the tortoise was previously, but by the time he reaches that point the tortoise has moved a little bit further away.

**Dichotomy Paradox**

The first version we can think of as follows:

Say the tortoise is 2 metres away from Achilles. Initially Achilles halves this distance by travelling 1 metre. He halves this distance again by travelling a further 1/2 metre. Halving again he is now 1/4 metres away. This process is infinite, and so Zeno argued that in a finite length of time you would never actually reach the tortoise. Mathematically we can express this idea as an infinite summation of the distances travelled each time:

1 + 1/2 + 1/4 + 1/8 …

Now, this is actually a geometric series – which has first term a = 1 and common ratio r = 1/2. Therefore we can use the infinite summation formula for a geometric series (which was derived about 2000 years after Zeno!):

sum = a/(1-r)

sum = 1/(1-0.5)

sum = 2

This shows that the summation does in fact converge – and so Achilles would actually reach the tortoise that remained 2 metres away. There is still however something of a sleight of hand being employed here however – given an *infinite* length of time we have shown that Achilles would reach the tortoise, but what about reaching the tortoise in a *finite* length of time? Well, as the distances get ever smaller, the time required to traverse them also gets ever closer to zero, so we can say that as the distance converges to 2 metres, the time taken will also converge to a finite number.

There is an alternative method to showing that this is a convergent series:

S = 1+ 1/2 + 1/4 + 1/8 + 1/16 + …

0.5S = 1/2+ 1/4 + 1/8 + 1/16 + …

S – 0.5S = 1

0.5S = 1

S = 2

Here we notice that in doing S – 0.5S all the terms will cancel out except the first one.

**Achilles and the Tortoise**

The second version also makes use of geometric series. If we say that the tortoise has been given a 10 m head start, and that whilst the tortoise runs at 1 m/s, Achilles runs at 10 m/s, we can try to calculate when Achilles would catch the tortoise. So in the first instance, Achilles runs to where the tortoise was (10 metres away). But because the tortoise runs at 1/10th the speed of Achilles, he is now a further 1m away. So, in the second instance, Achilles now runs to where the tortoise now is (a further 1 metre). But the tortoise has now moved 0.1 metres further away. And so on to infinity.

This is represented by a geometric series:

10 + 1 + 0.1 + 0.01 …

Which has first time a = 10 and common ratio r = 0.1. So using the same formula as before:

sum = a/(1-r)

sum = 10/(1-0.1)

sum = 11.11m

So, again we can show that because this geometric series converges to a finite value (11.11), then after a finite time Achilles will indeed catch the tortoise (11.11m away from where Achilles started from).

We often think of mathematics and philosophy as completely distinct subjects – one based on empirical measurement, the other on thought processes – but back in the day of the Greeks there was no such distinction. The resolution of Zeno’s paradox by use of calculus and limits to infinity some 2000 years after it was first posed is a nice reminder of the power of mathematics in solving problems across a wide range of disciplines.

**The Chess Board Problem**

The chess board problem is nothing to do with Zeno (it was first recorded about 1000 years ago) but is nevertheless another interesting example of the power of geometric series. It is explained in the video above. If I put 1 grain of rice on the first square of a chess board, 2 grains of rice on the second square, 4 grains on the third square, how much rice in total will be on the chess board by the time I finish the 64th square?

The mathematical series will be:

1+ 2 + 4 + 8 + 16 +……

So a = 1 and r = 2

Sum = a(1-r^{64})/(1-r)

Sum = (1-2^{64})/(1-2)

Sum = 2^{64 }-1

Sum = 18, 446,744, 073, 709, 551, 615

This is such a large number that, if stretched from end to end the rice would reach all the way to the star Alpha Centura and back 2 times.

**IB Maths Revision**

I’d strongly recommend starting your revision of topics from Y12 – certainly if you want to target a top grade in Y13. My favourite revision site is Revision Village – which has a huge amount of great resources – questions graded by level, full video solutions, practice tests, and even exam predictions. Standard Level students and Higher Level students have their own revision areas. Have a look!

## 3 comments

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November 11, 2014 at 11:50 pm

TalalHi, i would just like to ask whether or not this would be an appropriate topic for a Math SL exploration?

If yes, how would i approach it?

June 7, 2016 at 2:09 pm

Nipun2^64 -1 is not a prime number

June 7, 2016 at 2:28 pm

Ibmathsresources.comThanks for the correction – I have now amended the post.