B, for instance, traverses two miles (c d) in an hour, A in the same time, one mile (d e); if they are two miles (c d) removed from one another, B has in one hour come to where A was at the beginning of the hour. While B, in the next half hour, goes over the distance crossed by A of one mile (d e), A has got half a mile (e f) further, and so on into infinity. Quicker motion does not help the second body at all in passing over the interval of space by which he is behind: the time which he requires, the slower body always has at its avail in order to accomplish some, although an ever shorter advance; and this, because of the continual division, never quite disappears. Aristotle, in speaking of this, puts it shortly thus. “This proof asserts the same endless divisibility, but it is untrue, for the quick will overtake the slow body if the limits to be traversed be granted to it.” This answer is correct and contains all that can be said; that is, there are in this representation two periods of time and two distances, which are separated from one another, i.e. they are limited in relation to one another; when, on the contrary, we admit that time and space are continuous, so that two periods of time or points of space are related to one another as continuous, they are, while being two, not two, but identical. In ordinary language we solve the matter in the easiest way, for we say: “Because the second is quicker, it covers a greater distance in the same time as the slow; it can therefore come to the place from which the first started and get further still.” After B, at the end of the first hour, arrives at d and A at e, A in one and the same period, that is, in the second hour, goes over the distance e g, and B the distance d g. But this period of time which should be one, is divisible into that in which B accomplishes d e and that in which B passes through e g. A has a start of the first, by which it gets over the distance e f, so that A is at f at the same period as B is at e. The limitation which, according to Aristotle, is to be overcome, which must be penetrated, is thus that of time; since it is continuous, it must, for the solution of the difficulty, be said that what is divisible into two spaces of time is to be conceived of as one, in which B gets from d to e and from e to g, while A passes over the distance e g. In motion two periods, as well as two points in space, are indeed one.