Ant On A Rubber Rope Paradox – How To Solve It?

The Ant on a rubber rope paradox is a veridical paradox that exploits a deviation between intuition and mathematical logic. The paradox starts with an ant crawling on a one-dimensional rubber rope.

This rubber rope is perfectly and infinitely elastic, such that it can stretch infinitely along its length-wise direction. Now, to start, let us say that the rope is 1 Kilometre (Km) long and the ant starts at the far end of the rope.

As the ant starts to crawl at a speed of 1 centimetre/second (cm/s), the rope simultaneously starts to stretch (uniformly throughout its length) at a rate of 1 Kilometre/second (Km/s). The question posed by this problem is as follows:

Will the ant reach the end of the rope?

Our intuition says that there is no way that the ant can ever reach the end of the rope, provided that it is stretching at a much faster rate than the ant’s crawling speed. However, mathematical logic says that it WILL!

Are you intrigued? I know I was, when I first learned of this paradox. In this essay, I will cover a simplified mathematical approach to treat this problem in the discrete domain. Later on, I will briefly touch upon an approach in the continuous domain as well. Let us begin.

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Treating the Problem in the Discrete Domain

The reason why I chose the discrete domain first is that this problem is a bit easier to solve for in such a domain. To make the problem discrete, let us assume that the rope stretches suddenly and instantaneously just before each second completes.

We make this assumption to put the ant at a slight disadvantage compared to the original problem statement. In other words, if the ant can make it to the end of the rope in this variation of the problem, then it can do so without a doubt in the original version as well.

I am also going to go ahead and generalise the problem; don’t worry, we will be returning to the specific numbers later. Let ‘t’ represent the time.

In the beginning, that is at t = 0 seconds (s), let the initial length of the rope be ‘l’ units and let the ant be at a displacement of x = 0 units (left-hand far end of the rope).

As soon as the clock starts ticking, the ant starts crawling at a rate of, say, ‘a’ units/s with respect to the rope at is current location. At the same time, the rope stretches discretely at a rate of ‘v’ units/s. Given this premise, let us see how we can solve this problem further.

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The Discrete Math Behind the Ant on a Rubber Rope Paradox

The target for the ant is to reach the end of the rope. The rope-end is originally at a length of ‘l’ units. However, each second, this target point moves along the x-direction at a rate of ‘v’ units/s.

Therefore, the x-position of the target point at any given time ‘t’ is given by its original length plus the rope’s stretch-rate multiplied by the time ‘t’:

Note that even though we consider only the equation for the target point, the rope is uniformly stretching throughout its length. At the end of the first second, the ant would have covered ‘a’ units, whereas the target would be at x = l + (v*1) = (l + v) units.

At the end of two seconds, the ant would have covered another ‘a’ units, while the target would be at x = l + (v*2) = (l + 2v) units. Consequently, at any given time ‘t’, the proportion of the distance of rope covered by the ant, say F(t), is as follows:

For the ant to achieve its target, the value of this F(t) should at least reach ‘1’. Now, let us say that we are interested in knowing what happens at t = k seconds. The corresponding series is represented using the following equation:

Next, let us modify F(k) a bit further by multiplying the ‘l’ term in the denominator with ‘k’ as well and call the new proportion G(k). Since we require the value of ‘k’ to be greater than zero, it follows that F(k) ≥ G(k) for k > 0.

The Discrete Solution

From the above relation, we could say for sure that if G(k) reaches a value of at least 1 for some ‘k’, F(k) must at least be equal to or greater than that value. Now, since we have ‘k’ as a common factor among the denominator terms, we could factor it out as follows:

As you can see, the resulting series component is nothing but the famous harmonic series. This just happens to be a divergent series. To understand the context of convergence and divergence, check out my essay on the concept of a limit.

Since we are dealing with a divergent series here, there is no upper bound. So, it follows there exists some value of ‘k’ for which G(k) ≥ 1:

Therefore, the ant is bound to reach the end of the rope at some point. All we need to do is give it enough time. While we are on the topic of time, it is worth noting that this solution does not give us a precise value for ‘k’. To obtain that, we will have to look elsewhere.

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Solving the Problem in the Continuous Domain

It is possible to solve the same problem in the continuous domain using calculus. However, since the method involves slightly advanced calculus, I plan to cover it later as part of my calculus series.

For now, you can take it from me that given a = 1 cm/s, v = 1 km/s, and l = 1km, the solution comes out as follows:

k = (e¹⁰⁰⁰⁰⁰ — 1) = 2.8 * 10⁴³⁴²⁹ seconds (approximately).

Just to give you a benchmark, our known universe is estimated to be 4*10¹⁷ seconds old. So, it is safe to say that we have to wait quite a while before the ant reaches the end of the long rope (approximately 2.8 * 10⁴³⁴²⁹ Km).

Intuition for the Ant on a Rubber Rope Paradox

It just so happens to be the case that when we play around with the numbers and reduce the scales low enough, we can graphically “see” what happens.

In the below image, you can clearly see how the rope expands at a rate of 2 cm/s from an initial length of 4 cm. The red dot (representing the ant) crawls at rate of 1 cm/s and reaches the end of the rope at 12.8 seconds.

The astute reader might have already noted this. This paradox is a nice proxy for light from distant galaxies travelling across our expanding universe. Our universe is akin to the expanding rope, whereas the light ray represents the ant.

It would be nice if light could “finally” reach the end of our universe. But alas! Given our current knowledge, our universe’s expansion rate is increasing! The problem we just solved deals with constant rates.

If we accelerated the stretch enough, the ant might never reach the end of the rope, just like that lonely light ray from that distant galaxy!

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Postscript

Observation from reader Dan Foley post publishing:

I think you should stress in your explanation that the rope is always stretching uniformly all along its length, as opposed to the additional length simply being appended to the far end of the rope each second.

I realize this should be obvious to the reader, given the description of the rope as “elastic” and “stretching”, but still. If the additional length WAS being simply added to the far end of the rope, then the ant would promptly be left in the dust forever.

But because the additional length is applied uniformly along the rope, the rope growth that occurs BEHIND the ant presents no obstacle to his progress, and this rear portion is an increasing fraction of the total (approaching one) as the ant marches along.

Once the fraction of rope growth behind the ant becomes large enough (>0.99999 in the problem as given, which admittedly could take a really, really long time), the rope growth in front of the ant becomes < 1 cm/sec.

It’s important to note that until that moment, the distance from the ant to his goal has been increasing each second, despite him now being >0.99999 of the way there (measured along the now much, much longer rope).

But thereafter, the ant gets closer to the far end of the rope each second, and ultimately reaches his goal.

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Reference and credit: Martin Gardner.

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Further reading that might interest you:

• How To Really Solve This Packing Spheres Puzzle?
• How To Perfectly Predict Impossible Events?
• How To Actually Solve The Königsberg Bridge Problem?

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