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'''Solution A'''

One traditional answer to this problem is that in fact we only looking a constant amount of time ahead. To see this let us assume that the tortoise advances at unit speed and Achilles at speed x, starting y units behind the tortoise (say, at point 0). Then it takes Achilles <math> \frac{y}{x} </math> units in time (say seconds) to reach the starting point of the tortoise. In the meantime the tortoise has advanced to the point <math> y + \frac{y}{x} </math> being now <math> \frac{y}{x} </math> units adhead. Since x > 1 the distance has decreased. To reach the second point of the tortoise it takes Achilles the time of <math> \frac{y/x}{x} = \frac{y}{x^2} </math>. By induction we have, that it takes Achilles the time of <math> y\sum_{i=1}^n{x^{-i}} </math> to reach the point the tortoise reached after her <math>n</math>th advance.

For <math> n \to \infty </math> this expression converges to <math> \frac{y}{1 - \frac{1}{x}} =:T </math> which is clearly finite. Therefore in whole our race we are not considering points in time greater or equal than <math>T</math>. In fact at <math>T</math> Achilles has reached the tortoise and is going to overtake it at any infinitesimal time thereafter.

However, the whole argument assumes that time is continuous. Is this true in turn?


All of these problems are flawed because at each 'step' they slow down time, and if you slow down time enough, obviously you don't achieve anything. If you move between the line A and B and take the time measurements each time you reduce the distance by half, there are an infinite measurements each being taken faster and faster. This does not mean Achilles will not catch the tortious, it just means they are taking an infinite number of measurements until he does.

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