4.5: BM 13901 #12
- Page ID
- 47613
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27 The surfaces of my two confrontations I have heaped: \(21^{\prime} 40^{\prime \prime}\).
28 My confrontations I have made hold: \(10^{\prime}\).
29 The moiety of \(21^{\prime} 40^{\prime \prime}\) you break: \(10^{\prime} 50^{\prime \prime}\) and \(10^{\prime} 50^{\prime \prime}\) you make hold,
30 \(1^{\prime} 57^{\prime \prime} 21+25^{\prime \prime \prime} 40^{\prime \prime \prime \prime}\)8 is it. \(10^{\prime}\) and \(10^{\prime}\) you make hold, \(10^{\prime} 40^{\prime \prime}\)
31 inside \(1^{\prime} 57^{\prime \prime} 21\{+25\}^{\prime \prime \prime} 40^{\prime \prime \prime \prime}\) you tear out: by \(17^{\prime} 21\{+25\}^{\prime \prime \prime} 40^{\prime \prime \prime \prime}\), \(4^{\prime} 10^{\prime \prime}\) is equal.
32 \(4^{\prime} 10^{\prime \prime}\) to one \(10^{\prime} 50^{\prime \prime}\) you join: by \(15^{\prime}\), \(30^{\prime}\) is equal.
33 \(30^{\prime}\) the first confrontation.
34 \(4^{\prime} 10^{\prime \prime}\) inside the second \(10^{\prime} 50^{\prime \prime}\) you tear out: by \(6^{\prime} 50^{\prime \prime}\), \(20^{\prime}\) is equal.
35 \(20^{\prime}\) the second confrontation.
With this problem we leave the domain of fake practical life and return to the geometry of measured geometrical magnitudes. However, the problem we are going to approach may confront us with a possibly even more striking case of representation.
This problem comes from the collection of problems about squares which we have already drawn upon a number of times. The actual problem deals with two squares; the sum of their areas is given, and so is that of the rectangle “held” by the two “confrontations” \(c_{1}\) and \(c_{2}\) (Figure 4.9):
\(\left(c_{1}, c_{2}\right)=10^{\prime}\).Figure \(4.9\): The two squares and the rectangle of BM 13901 #12.
The problem could have been solved by means of the diagram shown in Figure 4.10, apparently already used to solve problem #8 of the same tablet, which can be expressed symbolically as follows:
\(\square\left(c_{1}\right)+\square\left(c_{2}\right)=21^{\prime} 40^{\prime \prime} \quad, \quad c_{1}+c_{2}=50^{\prime}\).
However, the author chooses a different method, showing thus the flexibility of the algebraic technique. He takes the two areas \(\square\left(c_{1}\right)\) and \(\square\left(c_{2}\right)\) as sides of a rectangle, whose area can be found by making \(10^{\prime}\) and \(10^{\prime}\) "hold (Figure 4.10):
\(\square\left(c_{1}\right)+\square\left(c_{2}\right)=21^{\prime} 40^{\prime \prime}\), \(\left(\square\left(c_{1}\right), \square\left(c_{2}\right)\right)=10^{\prime} \times 10^{\prime}=1^{\prime} 40^{\prime \prime}\).
\(\left(c_{1}, c_{2}\right)\)—which corresponds to our arithmetical rule \(p^{2} \cdot q^{2}=(p q)^{2}\).
What is to be taken note of in this problem is hence that it represents areas by line segments and the square of an area by an area. Together with the other instances of representation we have encountered, the present example will allow us to characterize the Old Babylonian technique as a genuine algebra on page 99.