3.3: YBC 6967

Obv.

1 The igibûm over the igûm, 7 it goes beyond

2 igûm and igibûm what?

3 You, 7 which the igibûm

4 over the igûm goes beyond

5 to two break: \(3^{\circ} 30^{\prime}\);

6 \(3^{\circ} 30^{\prime}\) together with \(3^{\circ} 30^{\prime}\)

7 make hold: \(12^{\circ} 15^{\prime}\).

8 To \(12^{\circ} 15^{\prime}\) which comes up for you

9 \(1^{\prime}\) the surface join: \(1^{\prime} 12^{\circ} 15^{\prime}\).

10 The equal of \(1^{\prime} 12^{\circ} 15^{\prime}\) what? \(8^{\circ} 30^{\prime}\).

11 \(8^{\circ} 30^{\prime}\) and \(8^{\circ} 30^{\prime}\), its counterpart, lay down.

Rev.

1 \(3^{\circ} 30^{\prime}\), the made-hold,

2 from one tear out,

3 to one join.

4 The first is 12, the second is 5.

5 12 is the igibûm, 5 is the igûm.

Second-degree problems dealing with rectangles are more copious than those about squares. Two problem types belong to this category; others, more complex, can be reduced to these basic types. In one of these, the area and the sum of the sides are known; in the other, the area and their difference are given.

The above exercise belongs to the latter type—if we neglect the fact that it does not deal with a rectangle at all but with a pair of numbers belonging together in the table of reciprocals (see page 20 and Figure 1.2). Igûm is the Babylonian pronunciation of Sumerian igi, and igibûm that of igi.bi, “its igi” (the relation between the two is indeed symmetric: if \(10^{\prime}\) is igi 6, then 6 is igi \(10^{\prime}\)).

One might expect the product of igûm and igibûm to be 1; in the present problem, however, this is not the case, here the product is supposed to be \(1^{\prime}\), that is, 60. The two numbers are represented by the sides of a rectangle of area \(1^{\prime}\) (see line Obv. 9); the situation is depicted in Figure 3.4, A. Once more we thus have to do with a rectangle with known area and known difference between the length and the width, respectively \(1^{\prime}\) and 7.

Figure \(3.4\): The procedure of YBC 6967

It is important to notice that here the "fundamental representation" (the measurable geometric quantities) serves to represent magnitudes of a different kind: the two numbers igûm and igibûm. In our algebra, the situation is the inverse: our fundamental representation is provided by the realm of abstract numbers, which serves to represent magnitudes of other kinds: prices, weights, speeds, distances, etc. (see page 16).

As in the two analogous cases that precede, the rectangle is transformed into a gnomon, and as usually the gnomon is completed as a square “held” by the two “moieties” of the excess (lines Obv. 3–10). The procedure can be followed on the Figures 3.4, B and 3.4, C.

The next steps are remarkable. The “moiety” that was detached and moved around (the “made-hold,” that is, that which was “made hold” the complementary square) in the formation of the gnomon is put back into place. Since it is the same piece which is concerned it must in principle be available before it can be “joined.” That has two consequences. Firstly, the "equal" \(8^{\circ} 30^{\prime}\) must be "laid down”³ twice, as we see in Figure 3.4, D: in this way, the piece can be “torn out” from one (leaving the width igûm) and “joined” to the other (giving the length igibûm). Secondly, “tearing-out” must precede “joining” (lines Rev. 1–3), even though the Babylonians (as we) would normally prefer to add before subtracting—cf. BM 13901 #1–2: the first problem adds the side, the second subtracts: \(3^{\circ} 30^{\prime}\), the made-hold, from one tear out, to one join.

In BM 13901 #1 and #2, the complement was “joined” to the gnomon, here it is the gnomon that is “joined.” Since both remain in place, either is possible. When \(3^{\circ} 30^{\prime}\) is joined to \(8^{\circ} 30^{\prime}\) in the construction of the igibûm, this is not the case: if one magnitude stays in place and the other is displaced it is always the latter that is “joined.” Differently from our addition and the “heaping” of the Babylonians, “joining” is no symmetric operation.