4.2: AO 8862 #2
- Page ID
- 47610
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30 Length, width. Length and width
31 I have made hold: A surface I have built.
32 I turned around (it). The half of the length
33 and the third of the width
34 to the inside of my surface
35 I have joined: 15.
36 I turned back. Length and width
37 I have heaped: 7.
II
1 Length and width what?
2 You, by your proceeding,
3 2 (as) inscription of the half
4 and 3 (as) inscription
5 of the third you inscribe:
6 igi 2, \(30^{\prime}\, you detach:
7 \(30^{\prime}\ steps of 7, \(3^{\circ} 30^{\prime}\); to 7,
8 the things heaped, length and width,
9 I bring:
10 \(3^{\circ} 30^{\prime}\) from 15, my things heaped,
11 cut off:
12 \(11^{\circ} 30^{\prime}\) the remainder.
13 Do not go beyond. 2 and 3 make hold:
14 3 steps of 2, 6.
15 igi 6, \(10^{\prime}\ it gives you.
16 \(10^{\prime}\ from 7, your things heaped,
17 length and width, I tear out:
18 \(6^{\circ} 50^{\prime}\) the remainder.
19 Its moiety, that of \(6^{\circ} 50^{\prime}\), I break:
20 \(3^{\circ} 25^{\prime}\) it gives you.
21 \(3^{\circ} 25^{\prime}\) until twice
22 you inscribe; \(3^{\circ} 25^{\prime}\) steps of \(3^{\circ} 25^{\prime}\),
23 \(11^{\circ} 40^{\prime} 25^{\prime \prime}\); from the inside
24 \(11^{\circ} 30^{\prime}\) I tear out:
25 \(10^{\prime} 25^{\prime \prime}\) the remainder. 
By \(10^{\prime} 25^{\prime \prime}\), \(25^{prime}\) is equal 
.
26 To the first \(3^{\circ} 25^{\prime}\)
27 \(25^{prime}\) you join: \(3^{\circ} 50^{\prime}\),
28 and (that) which from the things heaped of
29 length and width I have torn out
30 to \(3^{\circ} 50^{\prime}\) you join:
31 4 the length. From the second \(3^{\circ} 25^{\prime}\)
32 \(25^{prime}\) I tear out: 3 the width.
33 7 the things heaped.
34 4, the length; 3, the width; 12, the surface.
The first two words of the first line (I.30) tell us that we are dealing with a figure that is fully characterized by its length and its width, that is, with a rectangle (cf. page 28)—or rather with a rectangular field: references to surveyors’ practice can be found in the text (for instance, I turned around it in line I.32 probably means that the surveyor, after having laid out a field, has walked around it; in I.36 he turned back).
Before studying the procedure, we may concentrate on certain aspects of the formulation of the text. In line I.31 we see that the operation “to make hold” does not immediately produce a numerical result—since the measures of the sides are still unknown, that would indeed be difficult. The text only says that a “surface” has been “built”; we are probably meant to understand that it has been laid out in the terrain. Later, when two known segments are to “hold” (lines II.13–14, and perhaps II.21–22), the numerical determination of the area appears as a distinct operation, described with the words of the table of multiplication. Finally, we observe that the text defines the outcome of a “heaping” addition as a plural, translated “the things heaped,” and that the normal alternating pattern of grammatical person is not respected.
The text, almost certainly from Larsa, seems to be from c. 1750 bce and thus to belong to the early phase of the adoption of algebra by the southern scribe school (see page 109). These particularities may therefore give us information about the ideas on which it was based—such ideas were to become less visible once the language and format became standardized.
The topic of the problem is thus a rectangle. Lines I.36-37 tell us that the "heap" of its length and width is 7, while the lines I.32-35 state that "joining" half of the length and one third of the width to the "surface" produces 15:2
\((\ell, w)+\frac{1}{2} \ell+\frac{1}{3} w=15 \quad, \quad \ell+w=7\).The solution could have followed the pattern of TMS IX #3 (page 57). By introducing an "extended length" \(\lambda=\ell+\frac{1}{3}\) and an "extended width" \(\phi=w+\frac{1}{2}\), and adding (according to the "Akkadian method") the rectangle \(\left(\frac{1}{2}, \frac{1}{3}\right)\) which is lacking in the corner where 2 and 3 are "inscribed," we would have reduced the problem to
\((\lambda, \phi)=15+\mathrm{C} \sqsupset\left(\frac{1}{2}, \frac{1}{3}\right)=15^{\circ} 10^{\prime}, \lambda+\phi=7+\frac{1}{2}+\frac{1}{3}=7^{\circ} 50^{\prime}\).
In this operation, it is obvious that the (shaded) half of the length that had been "joined" according to the statement has been eliminated. However, more than the (equally shaded) third of the width has disappeared. How much more precisely?
\((3,2)\) is constructed (perhaps one should imagine it in the corner where 2 and 3 are "inscribed" in Figure 4.2; in any case Figure 4.3 shows the situation). Without further argument it is seen that the half (three small squares) exceeds the third (two small squares) by one of six small squares, that is, by a sixth—another case of reasoning by “false position.” Exceptionally, igi 6 is not “detached” but “given” (namely by the table of reciprocals).Figure \(4.3\)
We thus know that, in addition to the third of the width, we have eliminated a piece \(\left(w, 10^{\prime}\right)\) (drawn in black); if \(\lambda=\ell-10^{\prime}\), we therefore have
\((\lambda, w)=11^{\circ} 30^{\prime}\).Figure \(4.4\)
Once more we therefore have a rectangle of which we know the area and the sum of length and width. The procedure is the same as in the final part of TMS IX #3—Figure 4.4; the area that is to be displaced is shown again in light shading in the position from where it is to be take and in heavy shading where it has to be placed. The only difference is terminological: in TMS IX #3, the two "moieties" are "made hold," here they are "inscribed"—but since a multiplication of a number by a number follows immediately, the usual construction of a rectangle (here a square) must be intedned (lines II.13-14).5
In the end, the final addition of the side of the square precedes the subtraction, as in TMS IX #3. Once more, indeed, it is not hte same piece that is involved in the two operations; there is therefore no need to make it available before it is added.