## I. The Dichotomy

Therefore,

### (3) Nothing moves.

Is the argument valid? Is it sound (i. e., are all the premises true)?

Aristotle rejects premise (2) on the grounds that Zeno has failed to distinguish between infinite divisibility and infinite extension. How might one justify (2)? Aristotle suggests the following argument:

Therefore,

### (A4) Nothing can perform infinitely many tasks.

Aristotle rejects this argument in terms of the distinction between infinite divisibility and infinite extension. What then, he asks, is the justification for (2)?

But, Aristotle's argument is not quite right, since the infinite divisibility of the motion is reduced to the infinite divisibility of the temporal interval of the motion. If this interval is infinitely divisible then it has an infinite number of parts and to get from one moment to the next requires an infinite series of steps. Aristotle introduces a new distinction between potential and actual infinities. Infinitely divisible intervals are only potentially infinite.

But, the fact remains that the interval (either of motion or time or space) contains an infinite sequence of non-zero intervals.

The following Zenoesque argument for (2) can be constructed:

(Z1) Every interval can be subdivided into an infinite number of non-zero parts.

(Z2) To reconstruct the interval we would have to "stitch together" all the parts.

(Z3) The sum of an infinite number of non-zero parts is infinite, i. e., can never be completed.

Therefore,

(Z4)No intervals can be constructed.

Mathematical techniques developed in the 19th century (2500 years or so after Zeno first put forth the paradoxes) suggest that (Z3) is false. This argument for (2) fails. But, we have no argument that (2) is false. What, then, if anything is wrong with the argument?

### II. The Achilles

(1) For every i, if Achilles is at T(i), then the tortoise is at T(i+1).

Therefore,

(C) For every point P on the race course, if Achilles is at P, the tortoise is at some point P" such that P" is farther from A(0) than P.

In order to derive (C) from (1), Zeno needs another premise

(2)For all i, the distance from T(0) to T(i) is shorter than the distance from T(0) to T(i+1).

Zeno's Arrow

1.At every moment during its flight, the arrow occupies a space equal to itself.

2.When the arrow occupies a space equal to itself, it is not moving.

3.If the arrow is not moving when it occupies a space equal to itself, it must be at rest.

Therefore,

4.At every moment in its flight, the arrow is at rest.

Therefore,

5.The arrow never moves.

This is a modern restatement of the paradox as it has been passed down to us from Aristotle.

Aristotle challenges premise (3) and what he takes to be the atomistic assumption underlying the argument. For Aristotle, an object, to be at rest, must remain in the same position for a certain non -zero interval of time. For Aristotle, there is no "rest at an instant." Aristotle's objection to premise 3 then, is that, at an instant, no object is either at rest or moving. Both these predicates are only applicable to objects considered over finite intervals of elapsed time.

Even if you do not accept Aristotle's argument (from "ordinary Greek," as it were) the argument, as formulated is invalid. The conclusion (5) does not follow from (4). This inferential step commits what is called the fallacy of composition (whatever is true of all the parts of a thing is true of the whole).

But Zeno is not down and out yet. The fact that one way of formulating a line of reasoning is invalid does not mean that all formulations of the line of reasoning are invalid. Zeno, were he alive today, might well respond by offering the following reformulation of his view.

A Valid Reconstruction of the Arrow.

6.If the arrow is moving at t, then (at t) the arrow occupies a space equal to its own volume. [If P then Q]

7.If the arrow occupies (at t) a space equal to its own volume, then it is not moving at t. [If Q then not-P]

Therefore,

8.If the arrow is moving at t, then the arrow is not moving at t. [If P then not P: Hypothetical Syllogism from 6 and 7]

Therefore,

9.The arrow is not moving at t. [not-P: this follows from 8 (check this by showing that there is no way to make the premise (=8) come out TRUE when the conclusion (=9) is FALSE]

This reconstruction of the argument IS VALID. Since the conclusion, presumably, is unacceptable either premise 6 or premise 7 must be false. But, which one and why?

Note that this formulation implicitly rejects Aristotle's argument that one cannot meaningfully talk of rest (or motion) at an instant.

SOLUTION(Uncover only after you have given up or think you have the answer)

The resolution depends on the concept of instantaneous velocity. At a given instant, it is true that the arrow does not move from one point to another BUT it is not at rest if by "at rest" we mean "has a velocity = 0. The instantaneous velocity of moving objects is > 0. The power of Zeno's paradox is that this "solution" requires the notion of a "limit" and an understanding of how to compute infinite sums. The mathematics of these procedures was only put on a solid foundation in the 1800's - 2300 years after Zeno originally formulated his puzzle!

From Aristotle:

The fourth is an argument concerning two rows with an equal number of bodies all of equal length, the rows extending from the opposite ends of the stadium to the midpoint and moving in opposite directions with the same speed; and the conclusion in this argument, so Zeno thinks, is that the half of an interval of time is equal to its double . . . For example, let A1A2A3A4 be a set of stationary bodies all of equal length, B1B2B3B4 another equal set of moving bodies starting on the right from the middle of the A's and having lengths equal to the A's, and C1C2C3C4 a third equal set with speed equal to and contrary to that of the B's, also of lengths equal to those of the A's and ending on the right with the end of the stadium

Now as the B's and the C's pass over one another, B1 will be over C4 at the same time that C1 will be over B4

Thus, (1) C1 will have passed all the B's but only half of the A's, and, as C1 takes an equal time to go through each B as through each A, its time in covering half the A's will be half that in covering all the B's. Also (2) during this same time the B's will have passed all the C's; for since C1 takes an equal time to pass each A as each B, C1 and B1 will reach the contrary ends of the courses at the same time because each of them takes an equal time to pass each A. (240a, 5-15)

The argument can be more formally set out as follows:

(1)At t1, the A's, B's, and C's, each one unit long, are arranged as in the first figure.

(2)The B's move to the right at some fixed speed which we may take, for convenience, to be one unit per unit time.

(3)The C's move to the left at the same speed of one unit per unit time.

(4)The A's are stationary.

(5)Let t2 be the time at which the front of the C1 has passed every A.

(6)The front of C1 has passed 2 A's in the interval T= t2 - t1.

(7)The front of C1 has passed 4 B's in the interval T= t2 - t1.

(8)The time it takes the front of C1 to pass each A is equal to the time it takes the front of C1 to pass each B.

From (6), (7) and (8), we conclude

(9)T = 2T.

But, this is impossible. Therefore,

(10)Motion is impossible.

Aristotle rightly focusses on (8) as suspect and he argues that it is false. But, he does not pick up on the significance of relative motion. No one does until the 17th century AD.

Nonetheless, the argument can be reconstrued as an argument against the view that spatial and temporal magnitudes can be decomposed into atoms. The first seven premises are the same. The argument proceeds

(8')In the time it takes C1 to pas one A [A1], i. e., one atomic unit of time T, C1 passes 2 B's. [This follows from a consideration of premisses (6) and (7)].

Therefore,

(9')In that time T, C1 must have first passed B1 and then passed B2.

(10)The time interval T can be subdivided into TB1 (the time during which C1 is passing B1) and TB2 (the time during which C1 is passing B2) such that T = TB1 + TB2, and TB1 comes before TB2.

But, (10') contradicts the assumption that T is an atomic (i. e., indivisible)
unit of time. Therefore,

(11')Motion in opposite directions is impossible.

Does this argument rule out indivisible units of space and time? If so, then it seems to conflict with modern quantum theory which suggests that space and time might be quantized - i. e., be composed of minimal indivisible units.

### III The Paradox of Plurality

1)The magnitude of any finite extended interval is  0.

2)The interval can be infinitely subdivided.

3)The ultimate parts produced by this division, once completed, will

either

a)be an infinite number of small atomic parts of some positive (  0) magnitude,

or

b)be an infinite number of parts of zero ( = 0) magnitude.

4)If 3a, then the sum of the parts will be an interval of infinite magnitude.

5)If 3b, then the sum of the parts will be an interval of zero magnitude.

6)The interval has either zero magnitude or is infinitely long.
Measure Theory and the Plurality Paradox

The bottom line is that

(1)Summing up ¿0 "0's" = 0

(2)Summing up c 0's can lead to non-0 length intervals

The key is to generalize the concept of "length" to allow us to assign "lengths" to a variety of sets of different cardinalities. This generalization of "length" is called the "measure" of a set. There is a precise mathematical theory called measure theory which deals with these issues. For our purposes, we can ignore the details and focus on the qualitative implications of this approach.

The measure of a set

(1)The measure of a set is a generalization of the ordinary concept of length. We can assign measures to finite sets of points as well as continuous intervals.

(2)KEY POINT: The measure of an interval is

(a)not determined the cardinality of the interval [ i. e., different intervals can contain the same number of points [ have cardinality c] and yet have different lengths or measures. For example, the interval (0,2) is twice as long as (0,1) but the number of points in each interval is the same!

(b)not determined by the order of the points in the interval

(3)All finite sets of points and all countably infinite sets of points have measure = 0. [That's one way of seeing why, after assuming that an infinite division of an interval has been completed, we can't "stitch" the interval back again to get a finite length. The points are all dimensionless [i. e., have measure = 0] and an infinite division only produces a countable infinity of cuts. So, the measure of the set of points produced by the division in the plurality paradox is = 0. The resolution of the paradox depends on the differences between our concept of a continuum [an uncountable collection of points] and Aristotle's concept of a continuum [a countably infinitely divisible continuum].

(4)Sets whose cardinality = c can have

(1)Infinite measure: Example: The interval (- ƒ, + ƒ)

(2)Any finite measure: Examples: intervals (0,1), (0,5)

(3)Zero measure:Example: The Cantor Discontinuum.

The basic idea of the construction [see handout 3 for details] is this: Construct an interval of any size, say the interval (0,1). Remove the middle third of the interval and then the middle third of the remaining parts and so on. The measure of the sum of the removed parts = 1. The points that remain form a discontinuous set with cardinality of c but measure = 0.

The set has measure = 0 because

measure( remainder set) + measure (removed set) = m(0,1) = 1

The set has cardinality = c because there is a one to one correspondence between the elements of the remainder set and the real numbers in the interval (0, 1).

Cantor's Discontinuum

A set of cardinality = c with measure = 0.

Construction:

1. Consider the interval 0 ¾ x ¾ 1.

0 1
[________________________]

2. Remove open middle thirds in successive steps.

After the first cut:

0 1/3 2/3 1
[_________] [________]

After the second cut:

0 1/9 2/9 1/3 2/3 7/9 8/9 1
[______] [______] [______] [______]

3. The removed parts sum up to 1 (i. e., has measure = 1).

4. What remains is the Cantor set.

5.Measure(of the interval from 0 to 1) =

Measure(of removed parts) + Measure(of the Cantor set)

But, the measure(of the interval from 0 to 1) = 1. By the result in (3),

1 = 1 + Measure(of the Cantor set)

i. e., Measure(of the Cantor set) = 0.

6. Cardinality of the Cantor set = c (the cardinality of reals)

Proof:

1.Represent the real numbers from 0 to 1 in ternary notation, i. e., using 0, 1, 2.

2.The first decimal place of each number in the original interval is:
0: for numbers in first 1/3
1: for numbers in second 1/3
2: for numbers in third 1/3

3.The (n+1)th decimal place of each number in the nth cut follows the same pattern.

4.The removed parts contain all numbers with 1's anywhere in their decimal expansion.

5.Therefore, the Cantor set (= all the remaining parts) contains all combinations of infinite sequences of 0's and 2's.

6.Change all the 2's to 1's (relabel the numbers).

7.Therefore, the Cantor set contains all combinations of infinite sequences of 0's and 1's.

8.Every real number between 0 and 1 can be represented as an infinite sequence of 0's and 1's. Similarly, each such sequence corresponds to a real number in the interval from 0 to 1.

9.Therefore, the cardinality of the Cantor set = the cardinality of the reals between 0 and 1.

10.But, there is a 1-1 correspondence between the reals lying in the interval from 0 to 1 and the total set of all the reals.

11.Therefore, the cardinality of the Cantor set = c.
Representation of Numbers in Different Bases

1.For numbers greater than 0, divide through by new base number b until you reach 0. The remainders, in reverse order, are the digits in the number in the base b.

Example: Convert 25910 to base 3.

Quotients Remainders

3/259
86 1
28 2
9 1
3 0
1 0
0 1

The required number (base 3) = 100121

Check: 1001213 =

1x30 + 2x31 + 1x32 + 0x33 + 0x34 + 1x35

= 1 + 6 + 9 + 243 = 259 QED

2.Converting Fractions: The following theorem holds:

Thm.If k/l is a rational number (k, l are relatively prime) then (k/l)b has a finite representation IFF every prime factor of l divides b.

Examples:

1. Represent (1/2)(base 10) in base 2.

1/2 = 1x2-1 = .12

2. Represent 1/9(base 10) in base 3.

1/9 = 9-1 = (32)-1 = 1x3-2 = (.01)3

3. Represent (1/3) and (2/3) (base 10 in base 3.

(1/3)10 = 1x3-1 = 1x(.1)3 = (.1)3

(2/3)10 = 2x3-1 = 2x(.1)3 = (.2)3