Achilles is asked to race against the tortoise, with the condition that the tortoise is given a ten meter head start. Achilles knows that he is ten times faster than the tortoise, so he figures he can catch up without any problems.

The race begins, and when the tortoise reaches the 10 meter mark, as agreed, Achilles takes off. He soon reaches the 10 meter mark, but by then the tortoise, whose speed is one tenth that of Achilles, is at the 11 meter mark. By the time Achilles runs another meter and reaches the 11 meter mark, the tortoise has moved another tenth of a meter, to the 11.1 meter mark, and so on. Every time Achilles reaches the point the tortoise was, the tortoise has moved ahead again. Therefore, Achilles can never catch up to the tortoise, much less pass him to win the race.

Zeno was an ancient Greek philosopher who lived from about 490-425 BC.

Explanation:

This problem somehow uses a logical argument to contradict common knowledge. Before hearing the race described in this way, everyone would probably answer that Achilles wins the race, unless of course the length of the race is shorter than eleven meters or so. When the race is imagined, the athlete speeds past the tortoise with the greatest of ease. However, Zeno’s explanation may indeed cause us to ask why poor Achilles ever gave that conniving turtle ten meters and lost the race.

The truth of the matter is that Achilles does win the race. He passes the tortoise at some point between the eleven and twelve meter marks (see graph).

Why is this a confusing problem? I think that we hear the situation described and translate it into the statement that Achilles has an infinite number of distance intervals to travel in order to pass the tortoise, and the tortoise has already traveled those intervals plus an extra one. The error lies in assuming that an infinite number of distances will add up to an infinite distance. However, we can show that the infinite distances between Achilles and the tortoise add up to a finite sum, and that sum lies somewhere between eleven and twelve meters.

Let us first consider a similar problem. In order to walk across a room, you must first walk halfway. Then, to walk the remaining distance, you must first walk half of that, or one fourth the total distance. This goes on and on until you realize that you are always walking only halfway to the wall, and will therefore never reach it. This problem is confusing because we tell ourselves that an infinite number of positive distances will combine to form an infinite distance, which will take an infinite amount of time to travel. However, the sum cannot be infinite because it never exceeds the length of the room. This sum of distances is equal to:

We know that the this sum is one because by adding enough terms, we can approach one as closely as we please, but we will never reach it or exceed it. It immediately follows that travelling across the room by halves takes exactly as much time as walking across the room in one shot, as we may have suspected.

And now we return to Achilles. His problem is that he must catch a tortoise with a ten meter head start. At first he is 10 meters behind. After catching up 10 meters, he finds himself 1 meter behind, since the tortoise proceeds at one-tenth the pace of our hero. After running that meter, he is 1/10 of a meter behind, then 1/100, then 1/1000 of a meter.

This tells us that Achilles will catch the tortoise 11 1/9 meters into the race. Another way to find this result is to solve the two equations for position of Achilles and the Tortoise. Let y be the position in meters, and x be time, expressed in units equal to the time it takes the tortoise to move one meter. Then the position of the tortoise is described by x, so y=x. Achilles travels at 10 times the pace, but does not begin until later, so y=10x+b. We can solve for b using a point which we know is on the line, such as (x, y)=(11,10), since Achilles is at the ten meter mark after the tortoise travels eleven meters. Then y=10x+b becomes 10=110+b, so the y-intercept b=-100. We set the two lines equal to find the intersection: