3.2: BM 13901 #2
- Page ID
- 47602
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Obv. I
5 My confrontation inside the surface I have torn out: \(14^{\prime} 30\) is it. 1, the projection,
6 you posit. The moiety of 1 you break, \(30^{\prime}\) and \(30^{\prime}\) you make hold,
7 \(15^{\prime}\) to \(14^{\prime} 30\) you join: by \(14^{\prime} 30^{\circ} 15^{\prime}\), \(29^{\circ} 30^{\prime}\) is equal.
8 \(30^{\prime}\) which you have made hold to \(29^{\circ} 30^{\prime}\) you join: 30 the confrontation.
This problem, on a tablet which contains in total 24 problems of increasing sophistication dealing with one or more squares, follows immediately after the one we have just examined.
From the Old Babylonian point of view as well as ours, it is its “natural” counterpart. Where the preceding one “joins,” this one “tears out.” The basic part of the procedure is identical: the transformation of a rectangle into a gnomon, followed by a quadratic complement.
Initially the problem is stated (line 5): My confrontation inside the surface I have torn out: \(14^{\prime} 30\) is it. Once again the problem thus concerns a square area and side, but this time the “confrontation” \(c\) is “torn out.”
To “tear out” is a concrete subtraction by removal, the inverse of the “joining” operation, used only when that which is “torn out” is part of that magnitude from which it is “torn out.”2 The “confrontation” \(c\) is thus seen as part of (the inside of) the area. Figure 3.3, A shows how this is possible: the “confrontation” c is provided with a width (a “projection”) 1 and thereby changed into a rectangle 
(\(c\),1), located inside the square. This rectangle (shaded in dark grey) must thus be “torn out”; what remains after we have eliminated 
(\(c\),1) from 
(\(c\)) should be \(14^{\prime} 30\). In modern symbols, the problem corresponds to
\(\square(c)-c=14^{\prime} 30\).
Once more, we are left with a rectangle for which we know the area (\(14^{\prime} 30\)) and the difference between the length (\(c\)) and the width \((c-1)\)—and once more, this difference amounts to 1, namely the “projection.
\((c, c-1)\) is composed of a (white) square and a (shaded) "excess" rectangle whose width is the projection 1.\((c, c-1)\), that is, to equal to \(14^{\prime} 30\).Next, the area of the completed square and its side are found: \(15^{\prime}\) to \(14^{\prime} 30\) you join: by \(14^{\prime} 30^{\circ} 15^{\prime}\), \(29^{\circ} 30^{\prime}\) is equal.
Putting back the “moiety” which was moved around, we find the side of the initial square, which turns out to be \(29^{\circ} 30^{\prime}+30^{\prime}=30\): \(30^{\prime}\) which you have made hold to \(29^{\circ} 30^{\prime}\) you join: \(30\) the confrontation.
We notice that this time the “confrontation” of the square is \(30\), not \(30^{\prime}\). The reason is simple and compelling: unless \(c\) is larger than 1, the area will be smaller than the side, and we would have to “tear out” more than is available, which evidently cannot be done. As already explained, the Babylonians were familiar with “subtractive magnitudes,” that is, magnitudes that are predetermined to be “torn out”; but nothing in their mathematical thought corresponded to our negative numbers.
We also notice that the pair (\(14^{\prime} 30^{\circ} 15^{\prime}\), \(29^{\circ} 30^{\prime}\)) does not appear in the table of squares and square roots (see page 23); the problem is thus constructed backwards from a known solution.