Commentary Aristotle Physics
first, he sets forth the doubt;
second, he solves it, at “therefore, we must not hold” (262b15; [1114]).
1113. He therefore says first (262b8) that what was said in order to prove that a reflected motion is not continuous may be applied to solving a certain doubt, which is this. Assume two equal magnitudes, one called E and the other called Z. Let A and D be two equally swift mobiles, such that A is continuously moved from the beginning of the magnitude E to C, and D along Z to I. And let us assume that, in the magnitude E, there is an intermediate point B, which is as far from C as a like point on Z is distant from I. Let us further assume that, at the same time that A, in its continuous motion, is approaching B, D is, in its continuous motion, receding from Z and going to I. Now, since these motions are regular and equally swift, it will follow that D will arrive at I before A arrives at C, because the one that starts first will first arrive to the end of an equal distance. But D left Z before A left B, because D left Z when A was arriving at B. Therefore, according to this, A did not simultaneously arrive at B and leave B, and it consequently follows that it departed after it arrived, for if it arrives and departs at the same time, it will not have begun to move later. And, so, it is necessary that A, while being carried along, rests in B. Therefore, a continuous motion will be composed of periods of rest, as Zeno claimed in book 6.
1114. Then, at “therefore, we must not hold” (262b15), he resolves this doubt in the light of the foregoing. For the objection supposed that A, in its continuous motion, arrives at a point, B, in the magnitude and that, at the same time that A arrived at B, D left the point Z—which is against what was had above. For it was said above that, when something is being moved continuously, it neither arrived at nor departed from any intermediate point. Therefore, what the objection assumes must not be assumed, namely, that, when A was at (that is, approached) B, D was departing from Z. For if it be granted that A arrived at B, then, for the same reason, it should be granted that it left B, and that this did not occur simultaneously, but in two instants, such that it was at rest in the intermediate time between the two instants.
But, as was said previously, when something was being continuously moved, it was neither departing from nor approaching a given point, but was simply there—and this not for a time, because then it would have been resting, but in a division of time, that is, in some now, which divides time.
Therefore, what the objection assumed—namely, that A arrived at and that D left some intermediate point—is impossible to state in a continuous motion.
But, in a reflected motion, this must be stated. For if a mobile, I, is moved to the point D and is then rebounded, it is plain that the mobile uses the ultimate, which is D, as a beginning and as an end; that is, the point is used for two things. Hence, it had to be at rest there.
Nor can it be said that it simultaneously arrived at and left D, for then it would have been there and not have been there in the same instant. For whatever has been moved exists in the terminus to which it was being moved, and whatever begins to be moved is not in the terminus from which it begins to be moved.
But, when we use the expression “to be at” or “to approach,” we mean that a motion is being terminated at that point, and when we say “to be away from” or “to depart,” we mean that the motion is beginning. Hence, it is necessary that whatever arrives at (or is at) a point be in it, while what is leaving it or is departing from it be not in it. Since, therefore, it is impossible to be and not to be in a given point at the same time, it is consequently impossible to be at once at and away from the same, as the objection more than once assumed.
It should be noted that he here uses different letters from those used above. Here I is the mobile and D the terminus; above, it was the opposite.
But the solution given for continuous motion is not to be used with respect to a reflected motion. For it cannot be said that the mobile I is in the terminus D, from which it began to be reflected, only in the division of time—that is, only during the now—and that the mobile neither arrived at nor departed from the same, as was said with respect to a continuous motion.
For in a reflected motion, an end must be reached that is an actual end, and not merely a potential one, as the intermediate point in a continuous motion was only potentially a beginning and an end. What is an intermediate point of a continuous motion is only potentially a beginning and an end, but the point from which a reflected motion begins is actually a beginning and an end. For example, in the case of a stone falling to earth and bouncing upward, the same thing is the end of the downward motion of a stone and the beginning of its upward motion.
Therefore, just as, in the magnitude in which a motion is occurring, a point from which the motion is reflected is actually both a beginning and end, so also in the motions themselves, there is actually an end of one and a beginning of the other. And this would not be so unless an interval of rest occurred.
Therefore, it is necessary that what is reflected in a straight line be at rest. And, so, it follows that, on a straight magnitude, there cannot be a continuous and perpetual motion, because no straight magnitude is infinite. And, so, there could not be perpetual, continuous, rectilinear motion unless reflection is involved.
1115. Then, at the same method (263a4), he presents the second doubt.
About this, he does three things:
first, he mentions the doubt;
second, he rejects a solution given in book 6, at “now, when we first” (263a11; [1116]);
third, he gives the true solution, at “and we must then apply” (263a22; [1118]).
He says first (263a3) that, by the same method, using the things shown above, one can block those who give the objection of Zeno and wish to argue in the following manner: whatever is being moved must first cross what is intermediate before arriving at the end; but, between any two termini, there are infinite intermediates on account of a magnitude’s infinite divisibility; and so it is impossible to traverse the intermediates, because infinites cannot be traversed; therefore, nothing can arrive at any terminus by motion.
Again, the same difficulty can be presented under another form, as some do in fact propose it. Whatever traverses a whole must previously traverse the half; and, since the half is again divided in half, half of the half must be first traversed. And, thus, whatever is being moved counts off every half as it reaches it. But such halves can be multiplied to infinity. Therefore, it follows that, if anything traverses an entire magnitude, it has counted off an infinite number, which is plainly impossible.
1116. Then, at “now, when we first” (263a11), he rejects the solution he had presented above in book 6.
First, he says what it is;
second, he sets it aside, at “but, although this solution” (263a15; [1117]).
He says therefore that the foregoing objection was answered in book 6 when motion in general was being discussed, on the ground that, just as a magnitude is divided infinitely, so also is time. Consequently, time possesses infinities in itself in the same way as a magnitude. And, so, it is not unfitting if the infinites in a magnitude be traversed in the infinites that are in time. For it is not inconsistent for an infinite magnitude to be traversed in an infinite time. But, as shown in book 6, the infinite is found in magnitude and in time in the same way.
1117. Then, at “but, although this solution” (263a15), he sets aside this solution. And he says that this solution is sufficient to answer the questioner who asked whether it was possible in a finite time to traverse and count off infinites. This question was dismissed by saying that a finite time possesses infinities in which the magnitudinal infinites can be traversed. But that solution does not reach the truth of the matter, because, if someone should omit to ask about the magnitude and fails to ask whether it is possible to traverse infinities in finite time, but asked rather this same question about time, namely, whether the infinites that are in time can be traversed—since time is divided to infinity—then the previous solution would not answer this question. Consequently, another solution must be sought.
1118. Then, at “and we must then apply” (263a22), he gives the true solution in the light of his premises above. And he says that the true solution of the present doubt requires us to repeat what was stated in the immediately foregoing arguments: when someone divides a continuum into two halves, he uses the one point at which the continuum is divided as two, because he is making it serve both as the beginning of one part and as the end of the other. He does this by numbering and by dividing into two halves.
But when a continuum has been divided in this manner, it is no longer a continuum, whether it be a magnitude (such as a line) or a motion that is divided, for a motion cannot be continuous unless of something continuous, namely, the motion of the subject, the time, and the magnitude traversed. Therefore, the divided counts, and by counting, it breaks the continuity.