Mathematical justifi cation: the Babylonian example 373
statement but identifying the area as 10 ́. In line 3, this is told to be equiva-
lent to adding ‘a base’ 1 to the width, as shown in Figure 11.3 – in symbols,
( l , w ) + l = ( l , w ) + ( l ,1) = ( l , w + 1); the ‘base’ evidently fulfi ls
the same role as the ‘projection’ of BM 13901. Line 4 tells us that this means
that we get a (new) width 1°20 ́, and line 5 checks that the rectangle con-
tained by this new width and the original length 30 ́ is indeed 40 ́, as it
should be. Lines 6–9 sum up.
Section 2 again refers to a rectangle with known dimensions – once
more l = 30 ́, w = 20 ́. Th is time the situation is that both sides are added to
the area, the sum being 1. Th e trick to be applied in the transformation is
identifi ed as the ‘Akkadian method’. Th is time, both length and width are
augmented by 1 (line 11); however, the resulting rectangle ( l + 1, w + 1)
contains more than it should (cf. Figure 11.4 ), namely beyond a quasi-gno-
mon representing the given sum (consisting of the original area ( l , w ),
a rectangle ( l ,1) whose measure is the same as that of l , and a rectangle
(1, w ) = w ), also a quadratic completion (1,1) = 1 (line 12). Th erefore,
the area of the new rectangle should be 1 + 1 = 2 (line 13). And so it is: the
new length will be 1°30 ́, the new width will be 1°20 ́, and the area which
they contain will be 1°30 ́·1°20 ́ = 2 (lines 15–17).
Since extension also occurs in section 1 , the ‘Akkadian method’ is likely
to refer to the quadratic completion (this conclusion is supported by further
arguments which do not belong within the present context).
Aft er these two didactical explanations follows a problem in the proper
sense. In symbolic form it can be expressed as follows:
(l,w) + l + w = 1 ,^1 ⁄ 17 (3l + 4w) + w = 30′
Th e fi rst equation is the one whose transformation into
(λ,ω) = 2
(λ = l + 1, ω = w + 1) was just explained in Section 2. Th e second is multiplied
by 17, thus becoming
3 l + 21w = 8° 30 ′.
and further transformed into
3 λ + 21ω = 32°30,
whereas the area equation is transformed into
(3λ,21ω) = 2′6.