# La Biblioteca de Babel de Borges

## Some definitions from the story

• Each line has 80 characters

• There are 40 lines on each page and 410 pages on each book

We can calculate the number of characters on each book.

## With the computer

\begin{aligned} pages\_per\_book &= 410\\ char\_per\_book &= 80 \cdot 40 \cdot pages\_per\_book \end{aligned}

## How many books

Each character can be any of 25 symbols

The total number of books is $$25^{1312000}$$ which is not easy to calculate without the proper tools

## Back of the envelope

A quick “back-of-the-envelop” calculation can confirm this result. We have $25^n = 5^{2n} ≈ 4^{2n}=2^{4n}$ Thus we are speaking about a little more than 5.2 million bits

Using the rule of thumb $$2^{10}≈10^3$$ we get $$10^{1.6\text{ million}}$$ books.

## Trying harder

The difference between $$5^{2n}$$ and $$4^{2n}$$ is important, so we will try better. We have $25^n = 5^{2n} = 4^{2n\log_4(5)}$

We have that $$\log_4(5)≈1.16$$ so to change from base 5 to base 4 we need to increase the exponent by 16%. So 1.3 millions become 1.5 millions, and so we have near $$4^{3\text{ million}}=2^{6\text{ million}}$$ books.

## Another way

In fact it may be easier to see that $$\log_2(5)$$ is a little more than 2, in fact near 2.32, thus $25^n = 2^{2n\log_2(5)}≈ 2^{4.64 n}$ When $$n$$ is 1.3 million, then $$4.64 n$$ is about 6 million. We need 6 megabits just to count the number of books in The Library. Now, $$2^n ≈10^{3n/10},$$ therefore there are near 101800000 books.

We need to work with orders-of-magnitude of orders-of-magnitude.

## How many rooms?

Using more data from the story, we can calculate how many books fit on each room, and therefore how many rooms are in the Library.

\begin{aligned} books\_per\_shelves =& 32\\ shelves\_per\_wall=& 5\\ walls\_per\_room =& 4\\ books\_per\_room =& books\_per\_shelves \cdot shelves\_per\_wall\\ & \cdot walls\_per\_room\\ rooms =& books / books\_per\_room\\ \end{aligned}

## Area of each room

The area of a hexagon of side $$s$$ is $\frac{3\sqrt{3}·s²}{2}$ Thus, we need to know the width of each wall

We know that there are 32 books on each shelf, so we need to estimate the width of each book.

## We need to make some assumptions

• Paper thickness is around 0.0001m, that is, 10 pages per millimeter. This is based on the fact that a pack of 500 page takes more or less 5cm.
• Each sheet has 2 pages
• The covers are 1 mm each

\begin{aligned} paper\_width &= 1⋅ 10^{-4} m\\ book\_width &= paper\_width \cdot \frac{pages\_per\_book}{2} + 2e-3 m \end{aligned}

## Therefore, the wall width is

$wall\_width = book\_width \cdot books\_per\_shelves$

## Now we calculate the area of one room

$room\_area = \frac{3 \sqrt 3}{2} \cdot wall\_width^2$

and then the area of the whole library

$library\_area = room\_area \cdot rooms$

Compare this area to the surface of Earth

$earth\_radius = 6371 ⋅ 10^3 m\\ earth\_area = 4 \pi \cdot earth\_radius^2$

## Book size

\begin{aligned} room\_height &= 1.6 m\\ book\_height &= room\_height / shelves\_per\_wall\\ aspect\_ratio &= 3/4\\ page\_area &= aspect\_ratio \cdot book\_height^2 \end{aligned}

## Weight of book

\begin{aligned} paper\_weight &= 50⋅ 10^{-3} kg/m^2\\ page\_weight &= page\_area \cdot paper\_weight\\ book\_weight &= pages\_per\_book \cdot page\_weight \end{aligned}

## Weight of room

$room\_weight = books\_per\_room \cdot book\_weight$