3.1: BM 13901 #1

Obv. I

1 The surface and my confrontation I have heaped: \(45^{\prime}\) is it. 1, the projection,

2 you posit. The moiety of 1 you break, \(30^{\prime}\) and \(30^{\prime}\) you make hold.

3 \(15^{\prime}\) to \(45^{\prime}\) you join: by 1, 1 is equal. \(30^{\prime}\) which you have made hold

4 from the inside of 1 you tear out: \(30^{\prime}\) the confrontation.

This is the problem that was quoted on page 9 in the Assyriologists’ “transliteration” and on page 13 in a traditional translation. A translation into modern mathematical symbolism is found on page 12.

Even though we know it well from this point of view, we shall once again examine the text and terminology in detail so as to be able to deal with it in the perspective of its author.

Line 1 states the problem: it deals with a surface, here a square, and with its corresponding confrontation, that is, the square configuration parametrized by its side, see page 22. It is the appearance of the “confrontation” that tells us that the “surface” is that of a square.

“Surface” and “confrontation” are heaped. This addition is the one that must be used when dissimilar magnitudes are involved, here an area (two dimensions) and a side (one dimension). The text tells the sum of the two magnitudes—that is, of their measuring numbers: \(45^{\prime}\). If \(c\) stands for the side of the square and (\(c\)) for its area, the problem can thus be expressed in symbols in this way:

\(\square(c)+c=45^{\prime}\left(=\frac{3}{4}\right)\).

Figure \(3.1\): The procedure of BM 13901 #1, in slightly distorted proportions.

Figure 3.1 shows the steps of the procedure leading to the solution as they are explained in the text:

A: 1, the projection, you posit. That means that a rectangle (\(c\),1) is drawn alongside the square \(\square\)(\(c\)). Thereby the sum of a length and an area, absurd in itself, is made geometrically meaningful, namely as a rectangular area \((c, c+1)=\frac{3}{4}=45^{\prime}\). This geometric interpretation explains the appearance of the "projection," since the rectangle (\(c\),1) "projects" from the square as a projection protruding from a building. We remember (see page 15) that the word was originally translated as "unity" or "coefficient" simply because the translators did not understand how a number 1 could "project.”

B: The moiety of 1 you break. The “projection” with adjacent rectangle (\(c\),1) is “broken” into two “natural” halves.

C: \(30^{\prime}\) and \(30^{\prime}\) you make hold. The outer half of the projection (shaded in grey) is moved around in such a way that its two parts (each of length \(30^{\prime}\)) “hold” the square with dotted border below to the left. This cut-and-paste procedure has thus allowed us to transform the rectangle (\(c\),\(c+1\)) into a “gnomon,” a square from which a smaller square is lacking in a corner.

D: \(15^{\prime}\) to \(45^{\prime}\) you join: 1. \(15^{\prime}\) is the area of the square held by the two halves (\(30^{\prime}\) and \(30^{\prime}\)), and \(45^{\prime}\) that of the gnomon. As we remember from page 18, to “join” one magnitude to another one is an enlargement of the latter and only possible if both are concrete and of the same kind, for instance areas. We thus “join” the missing square, completing in this way the gnomon in order to get a new square. The area of the completed square will be \(45^{\prime}+15^{\prime}=1\).

by 1, 1 is equal. In general, the phrase “by \(Q\), \(s\) is equal” means (see page 23) that the area \(Q\) laid out as a square has \(s\) as one of its equal sides (in arithmetical language, \(s=\sqrt{Q}\)). In the present case, the text thus tells us that the side of the completed square is 1, as indicated in D immediately to the left of the square.

\(30^{\prime}\) which you have made hold from the inside of 1 you tear out. In order to find the side \(c\) of the original square we must now remove that piece of length \(\frac{1}{2}=30^{\prime}\) which was added to it below. To “tear out” \(a\) from \(H\), as we have seen on page 18, is the inverse operation of a “joining,” a concrete elimination which presupposes that \(a\) is actually a part of \(H\). As observed above (page 15), the phrase “from the inside” was omitted from the early translations, being meaningless as long as everything was supposed to deal with abstract numbers. If instead the number 1 represents a segment, the phrase does make sense.

\(30^{\prime}\) the confrontation. Removing from 1 the segment \(\frac{1}{2}=30^{\prime}\) which was added, we get the initial side \(c\), the “confrontation,” which is hence equal to \(1-30^{\prime}=30^{\prime}=\frac{1}{2}\) (extreme left in D).

That solves the problem. In this geometric interpretation, not only the numbers are explained but also the words and explanations used in the text.

The new translation calls for some observation. We take note that no explicit argument is given that the cut-and-paste procedure leads to a correct result. On the other hand it is intuitively clear that it must be so. We may speak of a “naive” approach—while keeping in mind that our normal way to operate on equations, for instance in the example solving the same problem on page 12, is no less naive. Just as the Old Babylonian calculator we proceed from step to step without giving any explicit proof that the operations we make are justified, “seeing” merely that they are appropriate.

Figure \(3.2\)

The essential stratagem of the Old Babylonian method is the completion of the gnomon as shown in Figure 3.2. This stratagem is called a “quadratic completion”; the same term is used about the corresponding step in our solution by means of symbols

\(\begin{aligned}
x^{2}+1 \cdot x=\frac{3}{4} & \Leftrightarrow x^{2}+1 \cdot x+\left(\frac{1}{2}\right)^{2}=\frac{3}{4}+\left(\frac{1}{2}\right)^{2} \\
& \Leftrightarrow x^{2}+1 \cdot x+\left(\frac{1}{2}\right)^{2}=\frac{3}{4}+\frac{1}{4}=1 \\
& \Leftrightarrow\left(x+\frac{1}{2}\right)^{2}=1
\end{aligned}\).

However, the name seems to apply even better to the geometric procedure.

It is obvious that a negative solution would make no sense in this concrete interpretation. Old Babylonian algebra was based on tangible quantities even in cases where its problems were not really practical. No length (nor surface, volume or weight) could be negative. The only idea found in the Old Babylonian texts that approaches negativity is that a magnitude can be subtractive, that is, pre-determined to be torn out. We have encountered such magnitudes in the text TMS XVI #1 (lines 3 and 4—see page 27) as well as TMS VII #2 (line 35, the "to-be-torn-out of the width"—see page 34). In line 25 of the latter text we also observe that the Babylonians did not consider the outcome of a subtraction of \(20^{\prime}\) from \(20^{\prime}\) as a number but, literally, as something not worth speaking of.

Certain general expositions of the history of mathematics claim that the Babylonians did know of negative numbers. This is a legend based on sloppy reading. As mentioned, some texts state for reasons of style not that a magnitude \(A\) exceeds another one by the amount \(d\) but that \(B\) falls short of \(A\) by \(d\); we shall encounter an example in BM 13901 #10, see note 4, page 46. In his mathematical commentaries Neugebauer expressed these as respectively \(A-B=d\) and \(B-A=-d\) (\(A=B+d\) and \(B=A-d\) would have been closer to the ancient texts, but even Neugebauer had his reasons of style). In this way, mathematicians who only read the translations into formulas and not the explanations of the meaning of these (and certainly not the translated texts) found their “Babylonian” negative numbers.

As the French Orientalist Léon Rodet wrote in 1881 when criticizing modernizing interpretations of an ancient Egyptian mathematical papyrus:

For studying the history of a science, just as when one wants to obtain something, ‘it is better to have business with God than with his saints’.¹