2519 = 363,797,880,709,171,295,166,015,625.
We write this number out explicitly to re-echo the vastness of the numbers woven through the Library. Simply adding 19 orthographic symbols on the spine magnifies the Library more than 300 septillion times. For comparison, this number is roughly the number of microscopic plant cells comprising a grove of 364 oak trees.3 So if the Library of 251,312,000 books is considered as one imperceptible plant cell, accounting for differing symbols on the spine multiplies the Library into a grove of 364 giant oak trees.
However, since we cannot be sure of either the maximum number of symbols on the spine of each book or of Borges' intent, we restrict ourselves to 251,312,000 books. This number, so easy to write, is, in a powerful sense, utterly unimaginable. To see that we can't see it, let's begin by converting this number to a power of 10, which puts it in a more familiar context.
251,312,000 is just a little bit larger than 101,834,097;
which is, of course, a 1 followed by one million, eight hundred thirty-four thousand, and ninety-seven 0s. We accomplish this conversion to a power of 10 notation using the logarithmic function and discuss the mechanics in the Math Aftermath portion of this chapter.
Could our universe possibly contain the Library? Current research approximates the size of the universe as being about 1.5 x 1026 meters across. Let's simplify calculations and create an upper bound to the universe by overestimating its size and supposing that our universe is shaped like a cube, each side of which measures 1027 meters (figure 3).
So, we'll say that our cubic universe consists of approximately 1027 ∙ 1027 ∙ 1027 = 1081 cubic meters. If we assume that we may fit 1,000 = 103 Library books in a cubic meter—and this is an exceedingly generous assumption—then our universe, if it consisted of nothing except books, would contain
1081 103 = 1084 books.
This doesn't make the slightest dent in the Library; it would take
universes the size of ours to hold just the books of the Library. What if the books were each as small as a grain of sand?
Using a ruler shows that an average grain of sand is approximately one millimeter across. If we assume a cubical shape combined with a perfect packing, then we could fit approximately
grain-of-sand books in a cubic meter. Multiplying by the size of the universe, we find that the universe holds only
such books.
That is, if the universe consisted of nothing but sand, it would hold at most about 1090 grains of sand. As we promised at the beginning of the chapter, using exponential notation allows us to estimate the number of grains of sand considerably faster than Archimedes.
Once again, though, this hardly impacts the Library's collection. As a final illustration of this point, suppose that each book is shrunk to the size of a proton; that is, shrunk to about 10-15 meters across. Given that each book is 10-15 meters across, we could pack 1015 of them in a very narrow one-meter-long strip. Thus, packing a cubic meter with proton-sized books yields
Our universe holds merely
of these subatomic books.
Let's adopt one more viewpoint in our efforts to conceptualize the enormity and complexity of the Library. Perhaps the simplest books to imagine, of which there are exactly 25, are those that consist of nothing except one symbol, repeated for the entire book. For example, one such book would consist of its 1,312,000 slots filled by the letter g.4 The first two lines of that book would read
gggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg
gggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg
and so on for another 38 lines on the first page, followed by 40 more lines on each of the remaining 409 pages: a veritable rhapsody in g.
Now allow a slight variation. The next set of books we consider are those that consist of entirely the orthographic symbol g except for one h. That is, exactly 1,311,999 slots will be filled with the letter g, while exactly one slot will contain the letter h. One such book will begin
gggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg
gggggggggggggggggggggggggggggggggggggggghggggggggggggggggggggggggggggggggggggggg
and, as above, all of the rest of the symbols in the book are the letter g.
How many books like this are there? Well, there are exactly 1,312,000 different slots that the single h can occupy, and every other slot must be filled with a g. Thus, there are exactly 1,312,000 such books.
Now, we allow ourselves to imagine a book that consists of 1,311,998 slots filled with the symbol g and two slots—not necessarily adjacent— filled with an h. There are precisely
such books. (At the end of this chapter, see the second Math Aftermath, "An Example of the Ars Combinatoria," for an explanation of this and the next two formulas.) Put into human terms, assuming the world population is currently somewhat less than seven billion people, this translates to every one of us enjoying a personal library of about 123 of these books.
If we next consider books that, excepting three instances of the letter h, are all g, all the time, we perform a similar calculation to find that there are exactly
such books. This number, although perhaps not appearing much larger than the preceding one, expands these monotonous libraries to about 53 million distinct books for each person currently alive.
Pursuing this notion to its conclusion, by considering the number of books consisting of a mere 16 occurrences of the letter h in an otherwise uniform desert of the letter g, we find there are
3,683,681,259,485,362,310,918,865,543,989,208,654,728,931,149,486,911,733,618,072,454,576,141,229,488,660,718,000
distinct books—about 3.7 x 1084 books—more than enough to fill three cubic universes. These books, droning wearily of g with a little respite provided only by the scant 16 instances of h, are not typographical phantasmagoria to inflame the imagination or addle the senses, and yet if they were all collected into a subsection of the Library, they would occupy a space greater than three times our known universe.
Finally, it would be a tedious, uninspired, but straightforward calculation to determine how big the Library needs be to hold the books in the hexagonal configurations described by Borges. Given the work we've just done, it should be clear that however the Library is constructed, any sort of ambulatory circumnavigation would be utterly impossible for a human being: a vigorous, long-lived librarian who managed to walk a little over 60 miles—about 100 kilometers—every day for 100 years would cover somewhat less distance than light travels in two minutes. To cross our universe, which is incomprehensibly dwarfed by the Library, light would need to travel for at least 15 billion years.
The number of books in the Library, although easily notated, is unimaginable.
Math Aftermath I: The Logos of Logarithms
There are those who dance to the rhythm that is played to them, those who only dance to their own rhythm, and those who don't dance at all.
—Jose Bergamin, The Rocket and the Star
This Aftermath is included for two purposes, one explanatory and one hortatory. The expository side is to provide a basis for those who wish to understand the details of how certain approximations and calculations are made in this chapter, as well as the chapter "Real Analysis." The public relations portion is to reconceive of the logarithm as a function imbued with a friendly collection of useful, easily manipulated properties.
For the purposes of this book, we'll say that a function is a rule such that for each legitimate number the rule is applied to, it returns back exactly one number. The output number might be the same or different from the input number; however, the important thing is that given a specific input number, the output number for it is always the same. (There are many interesting generalizations of this idea, including that of studying spaces whose elements are themselves functions.) One of the functions most misunderstood and maligned by generations of students is that of the logarithm.
The logarithm (bas
e 10) is typically notated log; frequently it is written log(x) to emphasize it is a function: given an input of one number, x, it outputs another number, log(x). The modern notation is quite evocative:
x → log(x).
We could, at this juncture, include a graph of the logarithmic function; after all, a picture is useful for nurturing our visual awareness. However, we deliberately exclude such an illustration to hammer home a point: the logarithm, as it turns out, is a function that may be defined by a number of truly remarkable properties. Since really we only need to use one of the properties, let's jump right in: if x is any positive number, and n is any number, then
log(xn) = n log(x).
That is, the logarithm, remarkably, "lowers" the exponential, thereby reducing it to a much more familiar operation—multiplication. There are many marvelous implications of this property, but for our purposes, the property alone will give us what we need.
Earlier in the chapter, using exponential notation, we found that there are 251,312,000 distinct volumes in the Library. We'd like to contextualize the number of books by putting that number into a somewhat more familiar form. We choose to convert it to the power of 10 notation, 10n, because we may think of that as a single 1 followed by n 0s. Therefore, we set up the following equation and endeavor to solve it for n.
251,312,000 = 10n
When we solve this equation for n, we thus gain a greater intuition for the number of books in the Library.
Here's the key point: even though 251,312,000 and 10n are written differently and look different, if we choose some n such that the two numbers are equal, then they are, in fact, the same number. Since they are the same number, by the definition of a function, it must the case that using both representations of the number as inputs to the function entails that both of the outputs must continue to be equal to each other. So we apply the logarithm to both sides of the equation and get
log (251,312,000) = log (10n).
Now, the remarkable property of the logarithm "brings the exponential down" and gives
1,312,000 log (25) = n log (10).
(In fact, things are even better than they appear, for log(10) is equal to 1, but that need not concern us here.) Divide both sides by log(10) to solve for n:
By using a calculator, a computer, or even Henry Brigg's log tables from 1617, we find
Therefore,
Math Aftermath II: An Example of the Ars Combinatoria
Drawing is a struggle between nature and the artist, in which the better the artist understands the intentions of nature, the more easily he will triumph over it. For him it is not a question of copying, but of interpreting in a simpler and more luminous language.
—Charles Baudelaire, The Salon of 1846, VII. "On the Ideal and the Model"
In the final analysis, a drawing simply is no longer a drawing, no matter how self-sufficient its execution may be. It is a symbol, and the more profoundly the imaginary lines ofprojection meet higher dimensions, the better.
—Paul Klee, The Diaries of Paul Klee 1898-1918, no. 681, entry for July 1905
Here, we endeavor to explain the origins of the (possibly) mysterious formulas appearing earlier in the chapter. The first one arises in the context of trying to determine the number of distinct books in the Library consisting of 1,311,998 occurrences of the letter g and two instances of the letter h .
We abstract the books and hexagons away from the problem by noting that what we are really interested in can be characterized as the question "How many distinct ways exist to pick two objects from 1,312,000?" The two objects, of course, correspond to the two slots that we will fill with the letter h. So, the number of distinct ways to choose two objects from 1,312,000 corresponds precisely to the number of distinct books under discussion.
As it turns out, for millennia combinatorialists have known a formula for this and related questions; in the most general terms, the number of different ways to choose a subset of k objects from a set of n objects is equal to
One way to uncover the derivation of this formula is to break the analysis into two parts, first explaining the terms appearing in the numerator, and then understanding the term in the denominator. (Joe Roberts, the professor who introduced me to combinatorial analysis, helpfully said "attic" and "basement" instead of "numerator" and "denominator.")
We wish to choose one object from n distinct objects. Thus, we have n choices for our first object and then we are left with n — 1 objects. So, when we choose the second object, we have n — 1 distinct objects to choose from. This means that choosing two objects is tantamount to having ways to pick them: n ways to choose the first object multiplied by ways to choose the second.
If we pick a third object, we are choosing from distinct objects, and so the numerator grows accordingly. Notice that when we pick a fourth object, we choose from distinct objects; thus, extending the developing pattern, when we pick the kth object, we are selecting it from the remaining distinct objects. Multiplying, in succession, all of the choices yields the numerator (attic):
At this juncture, it's reasonable to wonder why there needs to be a denominator (basement). Why can't we simply stop at the numerator, or, put another way, what is wrong with what we've derived? The answer is devilishly simple: there are a number of different ways to pick the exact same subset of size k, and we don't care in what order the k objects are chosen. We just want to know which are the chosen ones.
Let's illustrate this with an easy example. We have a set, a collection, of three distinct objects, {A, B, C}. Let's choose all subsets consisting of two distinct objects:
{A, B} {B, A}
{A, C} {C, A}
{B, C} {C, B}
If the order in which the objects are picked is important, then we have a complete list of all subsets of size two. However, if order is unimportant, then {A, B} and {B, A} are both names for the same subset. Really, then, we would be happy with, say, this list.
{A, B}
{A, C}
{C, B}
Since all we care about is the number of ways to choose two things, and we don't care about the order, we need to divide out by the number of repetitions, which in this case, is two. We thus arrive at the complete formula for this example,
Another way to think about repetitions is as the number of distinct orderings of, for instance, a set of k objects. There is a critical difference between this and the exponential work from earlier in the chapter; when calculating the number of books, we allowed an orthographic symbol to be used over and over and over again—possibly 1,312,000 times. Or, conversely, a symbol didn't need to appear at all. In an ordering, every object needs to appear exactly once. In the beginning of the chapter "More Combinatorics," we show how to calculate the number of distinct orderings of a set of k objects: it's a product of k integers, notated k! and pronounced "k factorial." For now, suffice it to note that
This explains the denominator of the formula: we divide out by all the repetitions given by all the different orderings of the k chosen objects and thus achieve a masterpiece of the ars combinatorial:
Applying this formula to the situation of choosing subsets of size 16 from a set of size 1,312,000 yields the expression
= 3,683,681,259,485,362,310,918,865,543,989,208,654,728,931,149,486,911,733,618,072,454,576,141,229,488,660,718,000
.
· 16765220490415253625065478163110488777596070684631829708120311409986396665091758868942316900907777384574090574408577882732061772110931659947399568714591497545824796138075835421197279779754323576490572256786468422800398414001130840404432159220567873647879819752992180116091963070003460102877057133859986460838201334698105992713225457349777667823840107714018295679082043307285550872688827887567010456660198813317308577461625092980751975955442225426797719393203367532575001211842556594519778330069767047797344180140352992420259949470026323167037321871020156554080028628985372035016289304847323104057902026971342243620895518683161620610971532819079644261674019733075609639725425948141117929760571410501529175736939057142480970571052799
5642620280697196621430275793093225927800376559882994925327612689196008920829563638966405968151079193703516798977935410410870485480473180206692696460141319574750537162302401458151912894683905017520492915492610250607658200820459233579973871624581533039027827192594822076477326080909994846009681777529003361408645173508147190013663404830519365501647324846666372695454023369419855605974124635054913613707789078539963199486512143281891270633487234820460978516962245945218404332537360951568826338781616551557083534695665518111841590380729315478105653632803123719714062985622464780873761799170525539055256885813059255491913245295763000439144465356103197575576731159299217928919322435311018790938012446381695777636352
The Unimaginable Mathematics of Borges' Library of Babel Page 4
