2.1: TMS XVI #1

1 The 4th of the width, from the length and the width I have torn out, \(45^{\prime}\). You, \(45^{\prime}\)

2 to 4 raise, 3 you see. 3, what is that? 4 and 1 posit,

3 \(50^{\prime}\) and \(5^{\prime}\), to tear out, posit. \(5^{\prime}\) to 4 raise, 1 width. \(20^{\prime}\) to 4 raise,

4 \(1^{\circ} 20^{\prime}\) you see ,² 4 widths. \(30^{\prime}\) to 4 raise, 2 you see , 4 lengths. \(20^{\prime}\), 1 width, to tear out,

5 from \(1^{\circ} 20^{\prime}\), 4 widths, tear out, 1 you see. 2, the lengths, and 1, 3 widths, heap, 3 you see.

6 igi 4 detach, \(15^{\prime}\) you see. \(15^{\prime}\) to 2, lengths, raise, \(30^{\prime}\) you see , \(30^{\prime}\) the length.

7 \(15^{\prime}\) to 1 raise, \(15^{\prime}\) the contribution of the width. \(30^{\prime}\) and \(15^{\prime}\) hold.

8 Since “The 4th of the width, to tear out,” it is said to you, from 4, 1 tear out, 3 you see.

9 igi 4 de tach , \(15^{\prime}\) you see, \(15^{\prime}\) to 3 raise, \(45^{\prime}\) you see , \(45^{\prime}\) as much as (there is) of widths.

10 1 as much as (there is) of lengths posit. 20, the true width take, 20 to \(1^{\prime}\) raise, \(20^{\prime}\) you see.

11 \(20^{\prime}\) to \(45^{\prime}\) raise, \(15^{\prime}\) you see. \(15^{\prime}\) from \(30_{15^{\prime}}\) tear out,

12 \(30^{\prime}\) you see, \(30^{\prime}\) the length.

This text differs in character from the immense majority of Old Babylonian mathematical texts: it does not state a problem, and it solves none. Instead, it gives a didactic explanation of the concepts and procedures that serve to understand and reduce a certain often occurring equation type.

Figure \(2.1\): The geometry of TMS XVI #1.

Even though many of the terms that appear in the translation were already explained in the section “A new interpretation,” it may be useful to go through the text word for word.

Line 1 formulates an equation: The 4th of the width, from the length and the width I have torn out, \(45^{\prime}\).

The equation thus concerns a length and a width. That tells us that the object is a rectangle—from the Old Babylonian point of view, the rectangle is the simplest figure determined by a length and a width alone.³ Concerning the number notation, see the box “The sexagesimal system,” page 14. If \(\ell\) is the length and \(w\) the width, we may express the equation in symbols in this way:

\((\ell+w)-\frac{1}{4} w=45^{\prime}\).

Something, however, is lost in this translation. Indeed, the length and the width is a condensed expression for a “heaping,” the symmetric addition of two magnitudes (or their measuring numbers; see page 18). The length is thus not prolonged by the width, the two magnitudes are combined on an equal footing, independently of the rectangle. The sole role of the rectangle is to put its dimensions at disposal as unknown magnitudes (Figure 2.1).

Figure \(2.2\): "The equation" of TMS XVI #1.

Once the length and the width have been "heaped," it is possible to "tear out" \(\frac{1}{4} w\), since this entity is a part of the width and hence also of the total. To "tear out" as we remember, is the inverse operation of "joining," and thus the removal of a magnitude from another one of which it is a part (Figure 2.2).

Line 1 shows the nature of a Babylonian equation: a combination of measurable magnitudes (often, as here, geometric magnitudes), for which the total is given. Alternatively the text states that the measure of one combination is equal to that of another one, or by how much one exceeds the other. That is not exactly the type of equation which is taught in present-day school mathematics, which normally deals with pure number—but it is quite similar to the equations manipulated by engineers, physicists or economists. To speak of “equations” in the Babylonian context is thus not at all anachronistic.

Next, lines 1 and 2 ask the student to multiply the \(45^{\prime}\) (on the right-hand side of the version in symbols) by 4: You, \(45^{\prime}\) to 4 raise, 3 you see. To "raise," we remember from page 13, stands for multiplying a concrete magnitude—here the number which represents a composite line segment. The outcome of this multiplication is 3, and the text asks a rhetorical questions: 3, what is that?

Figure \(2.3\): Interpretation of TMS XVI, lines 1–3.

The answer to this question is found in lines 2–5. 4 and 1 posit: First, the student should "posit" 4 and 1. To "posit" means to give a material representation; here, the numbers should probably be written in the appropriate place in a diagram (Figure 2.3 is a possible interpretation). The number «1» corresponds to the fact that the number \(45^{\prime}\) to the right in the initial equation as well as the magnitudes to the left are all used a single time. The number «4» is "posited" because we are to explain what happens when \(45^{\prime}\) and the corresponding magnitudes are taken 4 times.

\(50^{\prime}\) and \(5^{\prime}\), to tear out, posit: the numbers \(50^{\prime}\) and \(5^{\prime}\) are placed on level «1» of the diagram. This should surprise us: it shows that the student is supposed to know already that the width is \(20^{\prime}\) and the length is \(30^{\prime}\). If he did not, he would not understand that \(\ell+w=50^{\prime}\) and that \(\frac{1}{4} w\) (that which is to be torn out) is \(5^{\prime}\). For the sake of clarity not only the numbers \(50^{\prime}\) and \(5^{\prime}\) but also \(30^{\prime}\) and \(20^{\prime}\) are indicated at the level «1» in our diagram even though the text does not speak about them.

Lines 3–5 prove even more convincingly that the student is supposed to know already the solution to the problem (which is thus only a quasi-problem). The aim of the text is thus not to find a solution. As already stated, it is to explain the concepts and procedures that serve to understand and reduce the equation.

These lines explain how and why the initial equation

\((\ell+w)-\frac{1}{4} w=45^{\prime}\)

is transformed into

\(4 \ell+(4-1) w=3\)

through multiplication by 4.

Figure \(2.4\): Interpretation of TMS XVI, lines 3–5.

This calculation can be followed in Figure 2.4, where the numbers on level «1» are multiplied by 4, giving thereby rise to those of level «4»:

\(5^{\prime}\)to 4 raise, 1 width: \(5^{\prime}\), that is, the \(\frac{1}{4}\) of the width, is multiplied by 4, from which results \(20^{\prime}\), that is, one width.

\(20^{\prime}\) to 4 raise, \(1^{\circ} 20^{\prime}\) you see , 4 widths: \(20^{\prime}\), that is, 1 width, is multiplied by 4, from which comes \(1^{\circ} 20^{\prime}\), thus 4 widths.

\(\(30^{\prime}\)\) to 4 raise, 2 you see , 4 lengths: \(\(30^{\prime}\)\), that is 1 length, is multiplied by 4. This gives 2, 4 lengths.

After having multiplied all the numbers of level «1» by 4, and finding thus their counterparts on level «4», the text indicates (lines 4 and 5) what remains when 1 width is eliminated from 4 widths: \(\(20^{\prime}\)\), 1 width, to tear out, from \(1^{\circ} 20^{\prime}\), 4 widths, tear out, 1 you see.

Finally, the individual constituents of the sum are identified, as shown in Figure 2.5 2, the lengths, and 1, 3 widths, heap, 3 you see: 2, that is, 4 lengths, and 1, that is, widths, are added. This gives the number 3. We have now found the answer to the question of line 2, 3 you see. 3, what is that?.

Figure \(2.5\): Interpretation of TMS XVI, line 5.

But the lesson does not stop here. While lines 1–5 explain how the equation \((\ell+w)-\frac{1}{4} w=45^{\prime}\) can be transformed into \(4 \cdot \ell+(4-1) \cdot w=3\), what follows in lines 6–10 leads, through division by 4, to a transformation of this equation into

\(1 \cdot \ell+\frac{3}{4} \cdot w=45^{\prime}\).

For the Babylonians, division by 4 is indeed effectuated as a multiplication by \(\frac{1}{4}\). Therefore, line 6 states that \(\frac{1}{4}=15^{\prime}\): igi 4 detach, \(15^{\prime}\) you see. igi 4 can be found in the table of igi, that is, of reciprocals (see page 20).

Figure 2.6 shows that this corresponds to a return to level «1»:

\(15^{\prime}\) to 2, lengths, raise, \(30^{\prime}\) you see , \(30^{\prime}\) the length: 2, that is, 4 lengths, when multiplied by \(\frac{1}{4}\) gives \(30^{\prime}\), that is, 1 length.

Figure \(2.6\): Interpretation of TMS XVI, lines 6–12.

\(15^{\prime}\) to 1 raise, \(15^{\prime}\) the contribution of the width. (line 7): 1, that is, 3 widths, is multiplied by \(\frac{1}{4}\), which gives \(15^{\prime}\), the contribution of the width to the sum \(45^{\prime}\). The quantity of idths to which this contribution corresponds is determined in line 8 and 9. In the meantime, the contributions of the length and the width are memorized: \(30^{\prime}\) and \(15^{\prime}\) hold —a shorter expression for may you head hold, the formulation used in other texts. We notice the contrast to the material takin note of the numbers 1, 4, \(50^{\prime}\) and \(5^{\prime}\) by "positing" in the beginning.

The contribution of the width is thus \(15^{\prime}\). The end of line 9 indicates that the number of widths to which that corresponds—the coefficient of the width, in our language—is \(\frac{3}{4}\) (= \(45^{\prime}\)): \(45^{\prime}\) as much as (there is) of widths. the argument leading to this is of a type known as "simple false position.”⁴

Line 8 quotes the statement of the quasi-problem as a justification of what is done (such justifications by quotation are standard): Since "The 4th of the width, to tear out", it is said to you. We must therefore find out how much remains of the width when \(\frac{1}{4}\) has been removed.

For the sake of convenience, it is “posited” that the quantity of widths is 4 (this is the "false position"). \(\frac{1}{4}\) of 4 equals 1 (the text gives this number without calculation). When it is eliminated, 3 remains: from 4, 1 tear out, 3 you see.

In order to see to which part of the falsely posited 4 this 3 corresponds, we multiply by \(\frac{1}{4}\). Even though this was already said in line 6, it is repeated in line 9 that \(\frac{1}{4}\) corresponds to \(15^{\prime}\): igi 4 de tach , \(15^{\prime}\) you see.

Still in line 9, multiplication by 3 gives the coefficient of the width as \(45^{\prime}\) (= \(\frac{3}{4}\)): \(15^{\prime}\) to 3 raise, \(45^{\prime}\) you see , \(45^{\prime}\) as much as (there is) of widths.

Without calculating it line 10 announces that the coefficient of the length is 1. We know indeed from line 1 that a sole length enters into the \(45^{\prime}\), without addition nor subtraction. We have thus explained how the equation \(4 \cdot \ell+(4-1) \cdot w=3\) is transformed into

\(1 \cdot \ell+\frac{3}{4} \cdot w=45^{\prime}\).

The end of line 10 presents us with a small riddle: what is the relation between the “true width” and the width which figures in the equations?

The explanation could be the following: a true field might measure 30 [\(\mathrm{NINDAN}\)] by 20 [ \(\mathrm{NINDAN}\)] (c. 180 m by 120 m, that is, \(\frac{1}{3}\) bùr), but certainly not \(30^{\prime}\) by \(20^{\prime}\) (3 m by 2 m). On the other hand it would be impossible to draw a field with the dimensions \(30 \times 20\) in the courtyard of the schoolmaster's house (or any other school; actually, a sandstrewn courtyard is the most plausible support for the diagrams used in teaching). But \(30^{\prime}\) by \(20^{\prime}\) would fit perfectly (we know from excavated houses), and this order of magnitude is the one that normally appears in mathematical problems. Since there is no difference in writing between 20 and \(20^{\prime}\), this is nothing but a possible explanation—but a plausible one, since no alternative seems to be available.

In any case, in line 11 it is found again that the width contributes with \(15^{\prime}\), namely by multiplying \(20^{\prime}\) (1 width) by the coefficient \(45^{\prime}\): \(20^{\prime}\) to \(45^{\prime}\) raise, \(15^{\prime}\) you see.

In the end, the contribution of the width is eliminated from \(45^{\prime}\) (already written 30^{15}, that is, as the sum of \(30^{\prime}\) and \(15^{\prime}\), in agreement with the partition memorized in the end of line 7). \(30^{\prime}\) remains, that is, the length: \(15^{\prime}\) from ³⁰_15′ tear out, \(30^{\prime}\) you see, \(30^{\prime}\) the length.

All in all, a nice pedagogical explanation, which guides the student by the hand crisscross through the subject “how to transform a first-degree equation, and how to understand what goes on.”

Before leaving the text, we may linger on the actors that appear, and which recur in most of those texts that state a problem together with the procedure leading to its solution.⁵ Firstly, a “voice” speaking in the first person singular describes the situation which he has established, and formulates the question. Next a different voice addresses the student, giving orders in the imperative or in the second person singular, present tense; this voice cannot be identical with the one that stated the problem, since it often quotes it in the third person, “since he has said.”

In a school context, one may imagine that the voice that states the problem is that of the school master, and that the one which addresses the student is an assistant or instructor—“edubba texts,”⁶ literary texts about the school and about school life, often refer to an “older brother” whose task it is to give instructions. However, the origin of the scheme appears to be different. Certain texts from the early eighteenth century begin “If somebody asks you thus , ‘I have …’.” In these texts the one who asks is a hypothetical person not belonging to the didactical situation—a pretext for a mathematical riddle. The anonymous guide is then the master, originally probably to be identified with a master-surveyor explaining the methods of the trade to his apprentice.