2.2: TMS VII #2

This text is rather intricate. Who finds it too opaque may skip it and eventually return to it once familiarized with the Babylonian mode of thought.

17 The fourth of the width to the length I have joined, its seventh

18 until 11 I have gone, over the heap

19 of length and width \(5^{\prime}\) it went beyond. You, 4 posit;

20 7 posit; 11 posit; and \(5^{\prime}\) posit.

21 \(5^{\prime}\) to 7 raise, \(35^{\prime}\) you see.

22 \(30^{\prime}\) and \(5^{\prime}\) posit. \(5^{\prime}\) to 11 raise, \(55^{\prime}\) you see.

23 \(30^{\prime}\), \(20^{\prime}\), and \(5^{\prime}\), to tear out, posit. \(5^{\prime}\) to 4

24 raise, \(20^{\prime}\) you see, 20 the width. \(30^{\prime}\) to 4 raise:

25 2 you see, 2, lengths. \(20^{\prime}\) from \(20^{\prime}\) tear out.

26 \(30^{\prime}\) from 2 tear out, \(1^{\circ} 30^{\prime}\) posit, and \(5^{\prime}\) to ^¿50′, the heap of length and width, join^?

27 7 to 4, of the fourth, raise, 28 you see.

28 11, the heaps, from 28 tear out, 17 you see.

29 From 4, of the fourth, 1 tear out, 3 you see.

30 igi 3 detach, \(20^{\prime}\) you see. \(20^{\prime}\) to 17 raise,

31 \(5^{\circ} 40^{\prime}\) you see, \(5^{\circ} 40^{\prime}\), (for) the length. \(20^{\prime}\) to \(5^{\prime}\), the going-beyond, raise,

32 \(1^{\prime} 40^{\prime \prime}\) you see, \(1^{\prime} 40^{\prime \prime}\), the to-be-joined of the length. \(5^{\circ} 40^{\prime}\), (for) the length,

33 from 11, heaps, tear out, \(5^{\circ} 20^{\prime}\) you see.

34 \(1^{\prime} 40^{\prime \prime}\) to \(5^{\prime}\), the going-beyond, join, \(6^{\prime} 40^{\prime \prime}\) you see.

35 \(6^{\prime} 40^{\prime \prime}\), the to-be-torn-out of the width. \(5^{\prime}\), the step,

36 to \(5^{\circ} 40^{\prime}\), lengths, raise, \(28^{\prime} 20^{\prime \prime}\) you see.

37 \(1^{\prime} 40^{\prime \prime}\), the to-be-joined of the length, to \(28^{\prime} 20^{\prime \prime}\) join,

38 \(30^{\prime}\) you see, \(30^{\prime}\) the length. \(5^{\prime}\) to \(5^{\circ} 20^{\prime}\)

39 raise: \(26^{\prime} 40^{\prime \prime}\) you see. \(6^{\prime} 40^{\prime \prime}\),

40 the to-be-torn-out of the width, from \(26^{\prime} 40^{\prime \prime}\) tear out,

41 \(20^{\prime}\) you see, \(20^{\prime}\) the width.

This is the second, difficult problem from a tablet. The first, easy one (found on page 118 in English translation) can be expressed in symbols in this way:

\(10 \cdot\left(\frac{1}{7}\left[\ell+\frac{1}{4} w\right]\right)=\ell+w\).

After reduction, this gives the equation

\(\ell \cdot 10=6 \cdot(\ell+w)\).

This is an “indeterminate” equation, and has an infinity of solutions. If we have found one of them (\(\ell_{o}, w_{o}\)), all the others can be written (\(k \cdot \ell_{o}, k \cdot w_{o}\)). The text finds one by taking the first factor to the left to be equal to the first factor to the right (thus \(\ell=6\)), and the second factor to the right to be equal to the second factor to the right (thus \(\ell+w=10\), whence \(w=4\)). Afterwards the solution that has been tacitly aimed at from the beginning is obtained through “raising” to \(5^{\prime}\) (the “step” \(\frac{1}{7}\left[\ell+\frac{1}{4} w\right]\) that has been “gone” 10 times). Indeed, if \(\ell=6\), \(w=4\), then the “step” is 1; if we want it to be \(5^{\prime}\) (which corresponds to the normal dimensions of a “school rectangle,” \(\ell=30^{\prime}, w=20^{\prime}\)), then the solution must be multiplied by this value. All of this—which is not obvious—is useful for understanding the second problem.

The first problem is “homogeneous”—all its terms are in the first degree in \(\ell\) and \(w\). The second, the one translated above, is inhomogeneous, and can be expressed in symbols in this way:

\(11 \cdot\left(\frac{1}{7}\left[\ell+\frac{1}{4} w\right]\right)=[\ell+w]+5^{\prime}\).

Figure \(2.7\): Interpretation of TMS XII, lines 21–23.

We take note that \(\frac{1}{4} w\) is "joined" to the length; that we take \(\frac{1}{7}\) of the outcome; and that afterwards we "go" this segment 11 times. What results "goes beyond" the "heap" of length and width by \(5^{\prime}\). The "heap" is thus no part of what results from the repetition of the step—if it were it could have been "torn out.”

The solution begins with a pedagogical explanation in the style of TMS XVI #1, the preceding quasi-problem. Reading well we see that the \(5^{\prime}\) which is "raised" to 7 in line 21 must be the "step" \(\frac{1}{7}\left[\ell+\frac{1}{4} w\right]\)—the raising is a verification that it is really the 7th—and not the "going-beyond" reffered to in line 20. Once again the student is supposed to understand that the text is based on the rectangle (\(30^{\prime}\), \(20^{\prime}\)). Having this configuration in mind we will be able to follow the explanation of lines 21 to 23 on Figure 2.7: when the "step" \(5^{\prime}\) is "raised" to 7, we get \(35^{\prime}\) (A), which can be decomposed as \(\ell\) and \(\frac{1}{4} w\) (B). When it is "raised" to 11 we find \(55^{\prime}\) (C), which can be decomposed as \(\ell\), \(w\), and \(5^{\prime}\) (D).

Next follows the prescription for solving the equation; is it still formulated in such a way that the solution is supposed to be known. “Raising” to 4 (lines 23 to 25) gives the equivalent of the symbolic equation

\(11 \cdot\left(\frac{1}{7}\left[4 \ell+4 \cdot \frac{1}{4} w\right]\right)=4 \cdot\left([\ell+w]+5^{\prime}\right)\).

Not having access to our symbols, the text speaks of \(\frac{1}{4} w\) as \(5^{\prime}\), finds that \(4 \cdot \frac{1}{4} w\) is equal to \(20^{\prime}\), and identifies that with the width (line 24); then \(4 \ell\) appears as 2, said to represent lengths (line 25).

Now, by means of a ruse which is elegant but not easy to follow, the equation is made homogeneous. The text decomposes \(4 \ell+w\) as

\((4-1) \ell-5^{\prime}+(w-w)+\left(\ell+w+5^{\prime}\right)\)

and “raises” the whole equation to 7. We may follow the calculation in modern symbolic translation:

\(\begin{aligned}
11 \cdot\left([4-1] \ell-5^{\prime}+0+\left[\ell+w+5^{\prime}\right]\right) &=(7 \cdot 4) \cdot\left([\ell+w]+5^{\prime}\right) \\
\Leftrightarrow \quad 11 \cdot\left([4-1] \ell-5^{\prime}\right) &=(28-11) \cdot\left([\ell+w]+5^{\prime}\right) \\
&=17 \cdot\left([\ell+w]+5^{\prime}\right) \\
\Leftrightarrow \quad 11 \cdot\left(\ell-\frac{1}{3} \cdot 5^{\prime}\right) &=\frac{1}{3} \cdot 17 \cdot\left(\ell+w+5^{\prime}\right) \\
\Leftrightarrow \quad\left(\ell-1^{\prime} 40^{\prime \prime}\right) \cdot 11 &=5^{\circ} 40^{\prime} \cdot\left(\ell+w+5^{\prime}\right)
\end{aligned}\).

However, the Babylonians did not operate with such equations; they are likely to have inscribed the numbers along the lines of a diagram (Figure 2.8); that is the reason that the “coefficient” \((4-1)\) does not need to appear before line 29.

As in the first problem of the text, a solution to the homogeneous equation is found by identification of the factors “to the left” with those “to the right” (which is the reason that the factors have been inverted on the left-hand side of the last equation): \(\ell-1^{\prime} 40^{\prime \prime}\) (now called “the length” and therefore designated \(\lambda\) in Figure 2.8 thus corresponds to \(5^{\circ} 40^{\prime}\), while \(\ell+w+5^{\prime}\) (referred to as “the heap” of the new length \(\lambda\) and a new width \(\phi\), that is, \(\lambda+\phi\)) equals 11; \(\phi\) must therefore be \(11-5^{\circ} 40^{\prime}=5^{\circ} 20^{\prime}\). Next the text determines the “to-be-joined” (wāṣbum) of the length, that is, that which must be joined to the length \(\lambda\) in order to produce the original length \(\ell\): it equals \(1^{\prime} 40^{\prime \prime}\), since \(\lambda=t-1^{\prime} 40^{\prime \prime}\). Further it finds “the to-be-torn-out” (nāsum) of the width, that is, that which must be “torn out” from \(\phi\) in order to produce \(w\). Since \(\ell+w+5^{\prime}=11\), \(w\) must equal \(11-\ell-5^{\prime}=11-\left(\lambda+1^{\prime} 40^{\prime \prime}\right)-5^{\prime}=(11-\lambda)-\left(1^{\prime} 40^{\prime \prime}+5^{\prime}\right)=\phi-6^{\prime} 40^{\prime \prime}\); the “to-be-torn-out” is thus \(6^{\prime} 40^{\prime \prime}\).

But “joining” to \(\lambda\) and “tearing out” from \(\phi\) only gives a possible solution, not the one which is intended. In order to have the values for \(\ell\) and \(w\) that are aimed at, the step \(5^{\prime}\) is “raised” (as in the first problem) to \(5^{\circ} 40^{\prime}\) and \(5^{\circ} 20\). This gives, respectively, \(28^{\prime} 20^{\prime \prime}\) and \(26^{\prime} 40^{\prime \prime}\); by “joining” to the former its “to-be-joined” and by “tearing out” from the latter its “to-be-torn-out” we finally get \(\ell=30^{\prime}, w=20^{\prime}\).

Figure \(2.8\): The resolution of TMS VII #2.

We must take note of the mastery with which the author avoids to make use in the procedure of his knowledge of the solution (except in the end, where he needs to know the "step" in order to pick the solution that is aimed at among all the possible solutions). The numerical values that are known without being given serve in the pdegogical explanations; afterwards, their funciton is to provide names—having no symbols like \(\ell\) and \(\lambda\), the Babylonian needs to use identifications like "the length \(30^{\prime}\)" and "the length 5^{\prime} 40^{\prime \prime}" (both are lengths, so the name "length" without any qualifier will not suffice).

Numerical values serve as identifiers in many texts; nontheless, misunderstandings resulting from a mix-up of given and merely known numbers are extremely rare.