## Thursday, October 11, 2007

### Achilles and the Tortoise: A False Paradox

According to Aristotle (restating Zeno):
In a race, the quickest runner [Achilles] can never overtake the slowest [Tortoise], since the pursuer must first reach the point whence the pursued started [i.e. the pursued has a head start], so that the slower must always hold a lead.
Can anyone really believe that Achilles fails to catch the Tortoise? Yes. See this, this, this, and this, for example.

To show what's wrong with Aristotle's analysis, I begin with an example:
• Achilles (A), a quasi-god with a tricky tendon, runs at a mortal speed of 15 miles an hour (a 4-minute miler, he).
• Tortoise (T) "runs" at a speed of 1 mile an hour. (I exaggerate for simplicity of illustration.)
• If A gives T a 15-mile lead, A reaches T's starting point in 1 hour. T has, in that hour, moved ahead by 1 mile.
• A covers that mile in 1/15 of an hour, in which time T has moved ahead by 1/15 of a mile.
• A runs the 1/15 of a mile in 16 seconds, in which time T has moved ahead by another 23.47 feet.
• And so on.
• Therefore, A can never catch T.
What's the catch? It's verbal sleight-of-hand, much like the "proof" that 1 + 1 = 3. We know that A must be able to catch T, but we are trapped in a fallacious argument which seems to prove that A can't catch T.

The trick lies in the presentation of A's and T's movements as occurring alternately instead of simultaneously. A is always described as going to where T was, not to where T will be when A catches him.

Such reasoning contradicts what we know of reality. Fast runners often catch (relatively) slow runners on the football field. The best sprinter on any high-school track team could give me a 25-yard head start and beat me to the finish line of a 100-yard dash. And so on.

When a real A catches a real T, the real A does so by covering a greater distance than the real T, but the real A covers that distance by running for exactly the same length of time as the real T runs. Along the way, A will pass points already passed by T, but A will not pause at any of those points and allow T to move a bit farther ahead -- which is the trick that lies behind the Aristotelian "proof" of A's inability to catch T.

Going back to the example (A runs 15 miles an hour, etc.), we can determine when and where A catches T simply by describing the race correctly:
• A's time (in hours) x A's velocity (in miles per hour) = A's distance (in miles).
• If A catches T, T's time = A's time; T's distance (including his head start) = A's distance.
• Therefore, when A has run for 15/14 hours at 15 miles an hour he has covered a distance of 225/14 miles (16 and 1/14 miles).
• In that same 15/14 hours, T (moving at 1 mile an hour) has covered a distance of 1-1/14 mile.
• Adding the distance T has traveled in 15/14 hours to T's head start of 15 miles, we see that T is, at the end of those 15/14 hours, exactly 16-1/4 miles from A's starting point.
• In sum, A catches T after both have been running for 15/14 hours, and at a distance of 16-1/4 miles from A's starting point.
• A, having caught T, then moves farther ahead of him with each stride because A is running at 15 miles an hour, whereas T is moving at only 1 mile and hour.
• Therefore, it is true that T, having been caught and surpassed by A, can never catch A as long as A and T continue to run at their respective velocities of 15 miles an hour and 1 mile an hour.
Generally, if A is able to overtake T at time t and distance d (from A's starting point):
ta = tt = t, where ta is A's time since the start of the race and tt is T's time since the start of the race
da = dt = d, where da is A's distance from his starting point and dt is T's distance from A's starting point, which include's T's head start: h
da = (va)(t), where va is A's speed
dt = h + (vt)(t), where vt is T's speed
If da = dt = d:
(va)(t) = h + (vt)(t)
(va)(t) - (vt)(t) = h
(va - vt)(t) = h
t = h/(va - vt)
Having solved this equation for t -- given h, va, and vt -- it is then trivial to solve the equations for da and dt, and to show that both yield the same result, which is d.

Meaningful values can be found for t and d only if va > vt. Otherwise, t and d take meaningless negative values. This null possibility means that the answer "A catches T" is not assumed in the algebraic statement of the problem; that is, the algebraic formulation given here is not a circular proof of A's ability to catch T.