Zeno’s paradox – by Joel David Hamkins
The Greek thinker Zeno of Elea (c. 490–430 BC) argued in antiquity that each one movement is unattainable. It’s merely unattainable to stroll via city and even throughout the room, to go from right here to there. What? We all know, after all, that this is attainable—we stroll from right here to there daily. And but, Zeno affords us his proof that that is an phantasm—we merely can’t do it.
Zeno argued like this. Suppose it had been attainable so that you can transfer from some level A to a different distinct level B.
Earlier than you full the transfer from A to B , nevertheless, you will need to after all have gotten half means there.
However earlier than you get to this halfway level, after all, you will need to get half solution to the halfway level! And earlier than you get to that place, you will need to get half means there.
And so forth, advert infinitum.
Thus, to maneuver from A to B , or certainly wherever in any respect, one will need to have accomplished an infinite variety of duties—a supertask. It follows, in line with Zeno, that you would be able to by no means begin shifting—you can not transfer any quantity in any respect, since earlier than doing that you will need to have already got moved half as a lot. And so, opposite to appearances, you’re frozen immobile, unable to start. All movement is unattainable.
Is the argument convincing? On what grounds would you object to it? Do you suppose, opposite to Zeno, that we will really full infinitely many duties? How would that be attainable?
It will likely be no good, after all, to criticize Zeno’s argument on the grounds that we all know that movement is attainable, for we transfer from one level to a different daily. That’s, to argue merely that the conclusion is fake doesn’t really inform you what’s flawed with the argument—it doesn’t establish any specific flaw in Zeno’s reasoning. In any case, if it had been in truth an phantasm that we expertise movement, then your objection can be groundless.
Reasonably, to actually criticize Zeno’s argument, one should interact extra straight with it. What precisely is perhaps the issue along with his argument?
Achilles and the Tortoise
Contemplate the allegory of Achilles and the Tortoise. The Tortoise has challenged Achilles to a race across the stadium. How laughable! How might the gradual Tortoise win towards the swift warrior? However, the assured Tortoise insists he can win, if given solely a modest head begin, say, one quarter of the best way round. Achilles accepts, and the race begins. They line up at the beginning, and the Tortoise begins off working at a brisk tempo (for a tortoise), whereas Achilles waits patiently for him to achieve the quarterwayaround mark at level A , at which period Achilles takes off at a quick clip from the beginning.
Why does the Tortoise suppose he’ll win? Between steps, panting closely, the Tortoise explains his reasoning like this. He argues that Achilles shall by no means really overtake him. Why? Nicely, when Achilles had began working, the Tortoise had already had his head begin, 1 / 4 means across the stadium to level A. However by the point Achilles will get to that time A, the Tortoise will naturally have moved on to some additional level B. So so as to catch him, Achilles additionally might want to transfer to this additional level B. However once more by the point Achilles will get to level B, the Tortoise may have moved on to some additional level C. And so forth, advert infinitum. Each time Achilles catches as much as the place the Tortoise had been, the Tortoise has moved on to a brand new additional location nonetheless past Achilles. And so, the Tortoise reasoned, Achilles shall by no means catch up!
Does it make sense ever so as to add up infinitely many numbers? Contemplate the next sum:
(frac12+frac14+frac18+frac1{16}+cdots)
Since we’re including up infinitely many optimistic numbers, ought to one anticipate the sum to be infinite?
No, in truth this infinite sum provides up altogether to a finite worth. What I declare is that this summation sums to 1.
(frac12+frac14+frac18+frac1{16}+cdotsquad =quad 1)
To see this, think about a line phase of unit size 1 , as beneath:
Allow us to take the primary half, like so:
And subsequent we take additionally half of what stays, which is able to add 1 / 4 extra:
After which an eight and sixteenth extra:
At every stage, we take half of what stays, which shall be half as a lot as we had simply added on the earlier step, and on this means, we exhaust increasingly of the unique unit interval.
At every stage, the present sum stays lower than 1 , however the distinction between it and 1 decays to zero. The entire sum, which suggests the restrict worth of the finite partial sums, will subsequently be precisely 1. The size of the unique interval is exhausted by the infinite sum, which subsequently sums to 1.
(frac12+frac14+frac18+frac1{16}+cdotsquad =quad 1.)
So we have now an infinite sum with a finite worth.
A unique perspective on Zeno
Maybe this evaluation offers an alternate perspective on Zeno’s paradox. Allow us to think about a forwardoriented model of his argument. That’s, allow us to argue like Zeno that so as to get from one level to a different, we should first traverse half the space, after which after that traverse half the remaining distance, after which half the nonetheless remaining distance, and so forth advert infinitum. That is precisely the method we had used within the figures above to research the 1/2 + 1/4 + 1/8 + ··· and so forth. We are able to think about an alternate Zeno arguing that it’s unattainable to reach on the vacation spot, since one should traverse individually all these infinitely many disjoint intervals. Such a forwardoriented model of Zeno’s argument appears to have a lot in frequent with Zeno’s unique concepts. And but, in the meantime, the paradox appears to be a minimum of partly addressed by our evaluation that though there are infinitely many distinct segments, however the entire size is finite, and so one can certainly generally traverse infinitely many segments in a finite time. This remark appears to pour some chilly water on the concept it’s inherently unattainable to traverse infinitely many distinct intervals. Does this paved the way to a decision of Zeno’s paradox?
Another illustration
Right here is an alternate solution to analyze the identical sum. Contemplate the unit sq. proven right here, with space 1 . We have now successively divided it into smaller items: a rectangle of space 1/2, a sq. of space 1/4 , a rectangle of space 1/8, and so forth. Every rectangle is adopted by a sq. of half the dimensions, and every sq. is adopted by a rectangle of half the dimensions. So the sum of all of the items is
(frac 12+frac 14+frac 18+frac1{16}+cdots, )
and since they exhaust the unit sq. within the restrict, the sum is 1, as claimed.
Such a sum is named a geometric sequence—every successive time period reveals the identical ratio with the earlier time period, getting half as massive at every step.
Which equation has spurred numerous hours of contested argumentation and debate in our center colleges and much past? I refer, after all, to the id:
(0.999999999cdotsqquad =qquad 1.)
It’s true? Is 0.99999··· totally equal to 1, or is there some minute or infinitesimal distinction between them?
Sure, certainly, it’s true. Allow us to show it. Right here is one frequent argument. Let x = 0.999999···, no matter worth which may be. It follows that 10x = 9.999999··· and we might compute the distinction 10x – x just by lining up the decimal and performing the subtraction:
On the left we have now 9x, and on the precise, all of the 9s within the decimal elements cancel, leaving simply 9. So 9x = 9, which suggests x = 1. And so we appear to have proved that certainly 0.9999999999··· is the same as 1.
Right here is one other proof. Many people know the decimal enlargement of 1 third to be level three repeating, like this:
( frac13 quad=quad 0.33333cdots)
This may be simply seen if one merely performs lengthy division. However on this case,
(0.9999999cdots= 3times0.3333333cdots=3timesfrac13=1)
which exhibits the specified id.
Maybe each of those proofs will be criticized, nevertheless, on the grounds that they assume 0.99999··· and 0.3333··· are significant expressions in our actual quantity system. The primary argument was actually of the shape: if 0.9999··· is a significant expression, then its worth is 1.
However is it a significant mathematical expression? Sure, it seems that these expressions are each situations of the geometric sequence. So allow us to talk about that earlier than returning to this subject.
Let me clarify. Our acquainted placedbased decimal notation for the pure numbers offers a method of describing each quantity by way of the variety of ones, tens, a whole bunch, hundreds, and so forth. Thus, the quantity 7547 is written that means exactly as a result of it’s the worth of:
7547 = 7 hundreds + 5 a whole bunch + 4 tens + 7 ones.
In different phrases,
(7547quad=quad 7cdot 10^3+5cdot 10^2+4cdot 10+7)
The decimal notation for the actual numbers works equally, besides that fractions of a unit are thought of. So the quantity 5.385 means:
( 5.385quad =quad 5+frac3{10}+frac8{100}+frac5{1000})
However now the purpose is that this similar concept applies to infinitary decimal representations, whether or not they’re repeating or not. Each infinite decimal expression represents an infinite sum. The decimal quantity 5.171717··· is solely a means of writing the infinite sum:
( 5.17171cdotsquad=quad 5+frac1{10}+frac7{100}+frac1{1000}+frac7{10000}+frac1{10000}+cdots
)
The quantity π = 3.14159265358979323846··· is a notation for sure sum:
(3.14159265358979323846cdots=3+frac1{10}+frac4{100}+frac1{1000}+frac5{10000}+cdots)
To have digit d within the okayth place after the decimal level implies that we’re including d/10^okay to the sum.
From this perspective, the expression 0.99999··· refers back to the infinite sum:
( 0.99999cdots = frac9{10}+frac9{100}+frac9{1000}+cdots+frac9{10^okay}+cdots)
That is an occasion of what’s known as a geometric sequence.
Geometric sequence
A geometrical sequence is an infinite summation whose successive phrases all stand in the identical ratio to the earlier time period. That’s, you get the subsequent time period at all times by multiplying by the identical issue. For instance, the sequence we thought of earlier is a geometrical sequence
( frac12+frac14+frac18+frac1{16}+cdots)
as a result of every time period is half as massive because the earlier time period. Usually, a geometrical sequence has the shape
( a+ar+ar^2+ar^3+cdots)
the place a is the beginning worth and r is the fixed ratio of successive phrases. Every time period is obtained by multiplying the earlier time period by r .
Allow us to think about the easy case a = 1, which is however common sufficient, since each geometric sequence is obtained from this case by multiplying via each time period by the issue a. So we think about
( 1+r+r^2+r^3+r^4+r^5+r^6+cdots)
Such an infinite sequence is alleged to converge to a restrict worth L, if the sequence of finite partial sums
will get as shut as desired to L as n will increase. That’s, for any optimistic diploma of accuracy ε > 0, there’s a quantity N of phrases within the sequence, such that for all bigger n ≥ N the finite partial sum is inside ε of L. Such a type of limitvalue evaluation is a core conception of the calculus, and a few of my readers could also be aware of that.
It’s clear that if r ≥ 1, then the sequence can’t converge, because the particular person phrases are a minimum of 1 and the sum shall be infinite. The same evaluation works if r is unfavourable, however with magnitude r ≥ 1.
So allow us to assume that r is smaller than 1 in magnitude, that means r < 1. We will calculate the precise worth of the finite partial sums. Let x be the finite partial sum
( x=1+r+r^2+r^3+cdots+r^n,)
taking the phrases as much as r^n for some fastened n. Observe that
And so x + r^{n+1} = 1 + rx, which will be simply solved for x to acquire
( x=frac{1r^{n+1}}{1r})
So that is the worth of the phrases within the geometric sequence as much as r^n. The factor to note about this expression is that as a result of r < 1, it follows that the r^{n+1} time period turns into more and more small towards zero. And so the restrict of the expression is strictly 1/(1r). Thus, we have now discovered the worth of the geometric sequence!
( 1+r+r^2+r^3+cdotsquad=quad frac1{1r},qquad textual content{offered }r<1.)
Within the common case, we have now
( a+ar+ar^2+ar^3+cdotsquad=quad frac a{1r},qquad textual content{offered }r<1.)
Utility to our preliminary examples
Within the particular case of our preliminary geometric sequence, we have now a = 1/2 and r = 1/2, resulting in
( frac12+frac14+frac18+frac1{16}+cdotsquad=frac{frac12}{1frac12}=1,)
which agrees with our earlier evaluation of the worth of that geometric sequence.
The case of 0.9999··· is the geometric sequence with a = 9/10 and r = 1/10, and so
( 0.999cdots quad=quad frac 9{10}+frac9{100}+frac9{1000}+cdotsquad = quad frac a{1r} quad=quad frac{frac 9{10}}{1frac 1{10}}quad =quad 1)
which once more agrees with our earlier evaluation.
And equally, the case of 0.3333··· is the geometric sequence with a = 3/10 and r = 1/10, and so
(0.333cdots quad=quad frac 3{10}+frac3{100}+frac3{1000}+cdotsquad = quad frac a{1r} = frac{frac 3{10}}{1frac 1{10}} = frac39 = frac13,)
as anticipated.
Harmonic sequence diverges
And what concerning the following sequence, often called the harmonic sequence:
( 1+frac12+frac13+frac14+frac15+frac16+cdots)
Does it converge? The person phrases 1/n are vanishing to zero, which is a needed requirement for convergence of the sequence. However is there a finite restrict worth for this sequence?
No, in truth there’s not. The harmonic sequence diverges to infinity. To see this, suppose that we have now thought of the phrases of the sequence as much as 1/n, like this:
( 1+frac12+frac13+cdots+frac1n.)
Allow us to think about doubling the variety of phrases, like this:
(left(1+frac12+frac13+cdots+frac1nright)+left(frac1{n+1}+frac1{n+2}+cdots+frac1{2n}proper).
)
Discover that we have now added n additional phrases in that second grouped expression, and every of them is a minimum of as massive as the ultimate time period 1/2n. So the additional quantity added by the second group is a minimum of n · (1/2n) = 1/2.
In different phrases, we will at all times add worth 1/2 extra to a finite partial sum just by doubling the variety of phrases showing in it. By doing this twice (taking 4 occasions as many phrases), we will add 1 extra. By doing that twice, we will add 2 extra to the worth. And so forth. We are able to make the worth as huge as desired, just by doubling the variety of phrases we take a adequate variety of occasions. This sequence subsequently diverges to infinity.
Alternating harmonic sequence
A curious variation on the divergent harmonic sequence is the alternating harmonic sequence, whose phrases are identical to within the harmonic sequence, besides that they alternate in signal from optimistic to unfavourable to optimistic once more.
( 1frac12+frac13frac14+frac15frac16+cdots)
Does it converge? Sure, certainly it does. This may be seen by contemplating the character of the graph of the finite partial sums because the sequence progresses.
The factors on this graph characterize the worth of the working tally as we add successive phrases to the sum as indicated. Discover how the sum jumps up when we have now simply added a optimistic time period and down when we have now simply added a unfavourable time period. And since these jumps additionally change into smaller because the sequence progresses, it implies that the higher sums are progressively descending and the decrease sums are progressively ascending. Moreover, these approximations are getting nearer to one another, as shut as desired, as a result of the magnitude of the jumps is 1/n, which tends to zero. And so the sequence will converge to some restrict worth, which is solely the infimum of the higher approximations, which is similar because the supremum of the decrease approximations. Actually one can present utilizing strategies from calculus that the restrict worth is exactly ln 2, the pure logarithm of two, which is about 0.693 . So the alternating harmonic sequence converges to ln 2.
The alternating harmonic sequence is an instance of a conditionally convergent sequence. Specifically, it’s convergent, with restrict worth ln 2 as we have now talked about, but when one makes all of the phrases optimistic, then it turns into the harmonic sequence, which diverges to infinity.
Riemann proved a outstanding theorem about such conditionally convergent sequence. Specifically, for any conditionally convergent sequence, we will rearrange the phrases of the sequence in order to make the restrict sum no matter worth we need. Sure, you learn that accurately—rearranging the phrases of a conditionally convergent infinite sequence may cause the ultimate sum to have a distinct worth, and certainly, appropriate rearrangements can understand any desired sum. On this respect, the essential arithmetic of infinite sums will be fairly totally different from what we used to with finite sums, that are after all invariant below rearrangement.
For instance of the rearrangement theorem, we will rearrangement the phrases of the alternating harmonic sequence to make the restrict worth no matter we would like. Within the ordinary order, we have now talked about that the sum is ln 2, which is about 0.693. Suppose we take a brand new goal worth of 0.9 for a rearranged model. Can we understand it?
Sure, certainly we will. The thought of the proof is that we construct the rearrangement by taking optimistic phrases solely till they meet or exceed the goal, after which unfavourable phrases till it drops beneath the goal, after which we return to optimistic phrases till it reaches or exceeds the goal once more.
Discover how a number of occasions we wanted to take optimistic values twice in succession so as to attain the brand new greater goal worth. The graph exhibits that for this goal worth we should always use the rearranged alternating harmonic sequence starting with
( 1frac12+frac13+frac15frac14+frac17frac16+frac19+frac1{11}frac18+cdots)
That is not strictly an alternating sequence—it doesn’t merely alternate from optimistic phrases to unfavourable to optimistic once more, however reasonably makes use of optimistic phrases at slightly higher frequency than the unfavourable phrases. There’ll afterward generally even be unfavourable values twice or extra in succession.
The primary level is that the rearrangement process will produce a rearrangement realizing precisely the goal worth as the specified sum. Because the optimistic elements sum to infinity, and the unfavourable elements to minus infinity, we are going to at all times be capable to zig zag backwards and forwards throughout the goal as the development proceeds. And because the particular person phrases within the sequence have gotten very small, they supply a sure for a way far the finite partial sums shall be from the goal. So the rearranged sum converges precisely the goal, as desired.
The argument is totally common and works with any conditionally convergent sequence and any desired goal worth, whether or not optimistic or unfavourable. Just by taking extra optimistic values when the present partial sum is beneath the goal and unfavourable values when the present sum is above the goal, one can on this means attain any desired goal worth because the restrict of the rearranged sequence. We are able to additionally understand an infinite sum, or unfavourable infinity, by the identical methodology, in addition to a sequence whose partial sums oscillate wildly. It’s actually outstanding.
Put up your solutions, feedback, and questions beneath!

Present an evaluation of Zeno’s paradoxes of movement. What’s the flaw, if any, in Zeno’s argument that each one movement is unattainable? Will Achilles overtake the Tortoise? In that case, what’s the flaw within the Tortoise’s argument?

Is it attainable for us really to finish a supertask, a process involving infinitely many separate steps?

Do the restrict conceptions of calculus, particularly, the concepts surrounding the convergence and divergence of sequence, present an enough decision of Zeno’s paradox? Why or why not?

Can you utilize the concepts of Riemann’s theorem to show that each optimistic actual quantity r is the worth of an acceptable subseries of the harmonic sequence? That’s, for each r > 0 there’s a set A ⊆ ℕ for which
(r=sum_{nin A}frac 1n)
Is A distinctive for a given r? If not, what number of A are there?

Equally, are you able to show a generalization of Riemann’s rearrangement theorem, exhibiting that if a sequence is conditionally convergent, then for any goal worth r there are numerous totally different rearrangements of the sequence having sum r?

Huggett, Nick. 2019. Zeno’s Paradoxes, Winter 2019 edn. In The Stanford encyclopedia of philosophy, ed. Edward N. Zalta. Metaphysics Analysis Lab, Stanford College. A abstract philosophical account of Zeno’s paradoxes of movement.
The redorange determine exhibiting the sum of the geometric sequence 1/2 + 1/4 + ··· was tailored from my ebook, Proof and the Art of Mathematics, MIT Press, 2020.