ZENO’S PARADOX RENEWED Miguel Á. León Untiveros Universidad Nacional Mayor de San Marcos
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1.
The Zeno’s version, according to Aristotle.
Zeno is more famous for his defense of Parmenides’ claim that there is no motion. Plato does not mention any of these arguments. We learn about them principally through Aristotle. The best known of these puzzles is the bisection paradox. Can you walk across a room? To reach the opposite side, you must first walk halfway across. After that, you must walk half of the remaining distance. And then half the new remainder. There are infinitely many of these halfway points. No one can perform infinitely many acts in a finite amount of time. Zeno’s second paradox of motion pits Achilles against a tortoise. Since Achilles is the faster runner, we give the tortoise a head start. Can Achilles overtake the tortoise? To pass the tortoise, Achilles must first make up for the head start. But by the time he has covered that distance, the tortoise has moved ahead further. Achilles must therefore make up for that distance. But once Achilles has done that, the tortoise has moved again. Although this new distance is shorter, Achilles must still make up for it. But the enterprise of making up this endless sequence of distance debts is futile. Achilles cannot pass the tortoise because he cannot catch up infinitely many times. A known solution to it is: let S be a sum like this 1
1
1
1
1
S= 2 + 4 + 8 + 16 + 32 …
(1)
According to infinitesimal calculus, a mathematic theory raised in XVII century (Isaak Newton and Gottfried Leibniz), S has a limit, that is Multiplying (1) by 2, 1
1
1
1
1
2S= 1+2 + 4 + 8 + 16 + 32 …
(2)
From (1) and (2) 2S=1+S S=1 So S tends to 1 as a limit, using a notion of potential infinite. This means that Achilles reaches the tortoise. In this context, there is no more paradox.
2.
A renewed version of Achilles and the tortoise paradox.
Roy T. Cook argues that the last solution takes a divergent concept of infinite despite of Zeno proposed actual infinite concept (Cook, 2013, p. 23). In that sense, I will give a new version of this argument, in order to show that yet this is paradoxical. This new version is:
Let Achilles be a computer, assume he is the most fast one in Earth. Let us take a real interval [0, 1], and put a tortoise on point 1. The tortoise do not move, is quit. Now assign Achilles to reach the tortoise running through all the points from 0 to 1, inclusive. Does Achilles reach the tortoise? According to the fundamental theorem of set theory (Zermelo-Fraenkel+Choice), a
denumerable set N of all integers is no equivalent to the continuum but only to a proper subset of the continuum. In other words, the continuum is more-thatdenumerable set. A proof is1. Assume the continuum is the interval [0, 1], and denote it by C. First we shall represent the elements of C in a convenient way, i.e. decimals. For this we must consider that every positive real number can be expanded in one, and only one, way into and “infinite” decimal, i.e., into a decimal that after any digit contains yet any other digit differing from 0. In this presentation we shall not include the number 0. So we can write the elements of the continuum C as a sequence: 0, n11 n12 n13 n14… 0, n21 n22 n23 n24… 0, n31 n32 n33 n34… 0, n41 n42 n43 n44… . . .
Fig. 1. Infinite Square where all nik (i,k=1, 2,…) are digits, i.e., numerals between 0 and 9, inclusive. If we disregard the common initial 0, we have here an infinite square of digits with the vertex n11, which spread to the righty and downward,
n11 n12 n13 n14… n21 n22 n23 n24… n31 n32 n33 n34… n41 n42 n43 n44… . . .
In particular, no “row” of the square can finally consist of zeros only, since all decimal are infinite. For any nik, the first index, i, stand for the ith row of the square, and the second index, k, for its kth column; nik is then the kth digit of the ith decimal of C. For instance, n75 is the fifth digit of the seventh decimal.
Essentially, this proof in taken from (Fraenkel, 1966, pp. 20-22) with some adaptations. This is the famous proof of the diagonal. 1
We shall now specify an infinite (uniquely defined) decimal d beginning with 0, hence member of C, which is not contained in the infinite square; that is to say, there exists no integer i such that d is the ith decimal of C. n11 n12 n13 n14… n21 n22 n23 n24… n31 n32 n33 n34… n41 n42 n43 n44… . . .
To construct d, we consider the diagonal of the square which passes through the digits n11,n22,n33,n44,…, nii, …(cf. the arrow inserted in the square). d is the infinite decimal
d=d1d2… di… whose digits are defined as follows. For any i for which nii is different from the numeral 1, di=1; for those i for which nii=1, di=2. Hence, always di≠nii; that is to say, for every i, di is different from the ith digit of the ith decimal. The member d of C is therefore not contained in our square, which was assumed to contain all infinite decimals beginning with 0, i.e., all positive real numbers up to the number 1. We have thus obtained a contradiction which shows that our assumption that the continuum C is equivalent to the set N of integers is untenable. Now, regarding the Achilles and the tortoise paradox, the fundamental theorem means the new Achilles (a computer) cannot run through all decimals of C in any case, because C is greater than N. Zeno’s paradox is still alive! Referencias Cook, R. T. (2013). Paradoxes. Malden: Polity. Fraenkel, A. A. (1966). Set Theory and Logic. Massachusetts et al.: Addison-Wesley Publishing. Sorensen, R. (2003). A Brief History of the Paradox. Philosophy and the Labyrinths of the Mind. New York: Oxford University Press.
