## TMS IX #3

**19**Surface, length, and width I have heaped, 1 the surface. 3 lengths, 4 widths heaped,

**20**its 17th to the width joined, 30′.

**21**You, 30′ to 17 go: 8°30′ you see.

**22**To 17 widths 4 widths join, 21 you see.

**23**21 as much as of widths posit. 3, of three lengths,

**24**3, as much as of lengths posit. 8°30′, what is its name?

**25**3 lengths and 21 widths heaped.

**26** 8°30′ you see,

**27** 3 lengths and 21 widths heaped.

**28**Since 1 to the length is joined and 1 to the width is joined, make hold:

**29**1 to the heap of surface, length, and width join, 2 you see,

**30**2 the surface. Since the length and the width of 2 the surface,

**31**1°30′, the length, together with 1°20′, the width, are made hold,

**32**1 the joined of the length and 1 the joined of the width,

**33**make hold, ^{¿}1 you see.^{?} 1 and 1, the various (things), heap, 2 you see.

**34**3…, 21…, and 8°30′ heap, 32°30′ you see;

**35**so you ask.

**36**… of widths, to 21, that heap:

**37**… to 3, lengths, raise,

**38**1‵3 you see. 1‵3 to 2, the surface, raise:

**39**2‵6 you see, ^{¿}2‵6 the surface^{?}. 32°30′ the heap break, 16°15′ you
see
.

**40**{…}. 16°15′ the counterpart posit, make hold,

**41** you see. 2‵6 ^{¿}erasure^{?}

**42**from
tear out,
you see.

**43**What is equal? 11°45′ is equal, 11°45′ to 16°15′ join,

**44**28 you see. From the 2nd tear out, 4°30′ you see.

**45**igi 3, of the lengths, detach, 20′ you see. 20′ to 4°30′

**46**{…} raise: 1°30′ you see,

**47**1°30′ the length of 2 the surface. What to 21, the widths, may I posit

**48**which 28 gives me? 1°20′ posit, 1°20′ the width

**49**of 2 the surface. Turn back. 1 from 1°30′ tear out,

**50**30′ you see. 1 from 1°20′ tear out,

**51**20′ you see.

Lines 19 and 20 present a system of two equations about a rectangle, one of the first and one of the second degree. The former is of the same type as the one explained in TMS XVI #1

In agreement with what we have seen elsewhere, the text multiplies the first-degree equation by 17 (using the Akkadian verb “to go,”*as much as*):

This is done in the lines 21–25, while the lines 26 and 27 summarize the result.

Lines 28–30 repeat the trick used in section #2 of the text (see Figure 3.10): the length and the width are prolonged by 1, and the square that is produced when that which the two “joined”^{1} “hold” is “joined” to the “heap”
; out of this comes a “surface 2,” the meaning of which is again explained in lines 30–33.

The lines 34–37 are very damaged, too damaged to be safely reconstructed as far as their words are concerned. However, the numbers suffice to see how the calculations proceed. Let us introduce the magnitudes and . The text refers to them as the length and width “of the surface 2”—in other words, . Further,

In order to facilitate the understanding of what now follows we may further introduce the variables

(but we must remember that the text has no particular names for these—in contrast to
and
which do have names; we now speak *about*, not
the Babylonian author). Lines 36–39 find that

summing up we thus have

We have now come to line 39, and arrived at a problem type which we had not seen so far: A rectangle for which we know the area and the *sum* of the two sides.

Once again, a cut-and-paste

Fig. 4.1: The cut-and-paste method of TMS IX #3.

One part of the new square
(16°15′) is constituted by the gnomon, whose area results from recombination of the original rectangle
; this area is hence 2‵6. We also know the area of the outer square,
(lines 40 and 41). When the gnomon is “torn out” (lines 41 and 42),
remains for the square contained by the gnomon. Its side (that which “is equal”) is 11°45′, which must now be “joined” to one of the pieces 16°15′ (which gives us
) and “torn out” from the other, its “counterpart” (which gives us
). This time, however, it is not *the same* piece that is “joined”

## AO 8862 #2

**I**

**30**Length, width. Length and width

**31**I have made hold: A surface I have built.

**32**I turned around (it). The half of the length

**33**and the third of the width

**34**to the inside of my surface

**35**I have joined: 15.

**36**I turned back. Length and width

**37**I have heaped: 7.

**II**

**1**Length and width what?

**2**You, by your proceeding,

**3**2 (as) inscription of the half

**4**and 3 (as) inscription

**5**of the third you inscribe:

**6**igi 2, 30′, you detach:

**7**30′ steps of 7, 3°30′; to 7,

**8**the things heaped, length and width,

**9**I bring:

**10**3°30′ from 15, my things heaped,

**11**cut off:

**12**11°30′ the remainder.

**13**Do not go beyond. 2 and 3 make hold:

**14**3 steps of 2, 6.

**15**igi 6, 10′ it gives you.

**16**10′ from 7, your things heaped,

**17**length and width, I tear out:

**18**6°50′ the remainder.

**19**Its moiety, that of 6°50′, I break:

**20**3°25′ it gives you.

**21**3°25′ until twice

**22**you inscribe; 3°25′ steps of 3°25′,

**23**; from the inside

**24**11°30′ I tear out:

**25** the remainder.
By
, 25′ is equal
.

**26**To the first 3°25′

**27**25′ you join: 3°50′,

**28**and (that) which from the things heaped of

**29**length and width I have torn out

**30**to 3°50′ you join:

**31**4 the length. From the second 3°25′

**32**25′ I tear out: 3 the width.

**33**7 the things heaped.

**34**4, the length; 3, the width; 12, the surface.

The first two words of the first line (I.30) tell us that we are dealing with a figure that is fully characterized by its length and its width, that is, with a rectangle (cf. page *I turned around it* in line I.32 probably means that the surveyor, after having laid out a field, has walked around it; in I.36 he *turned back*).

Before studying the procedure, we may concentrate on certain aspects of the formulation of the text. In line I.31 we see that the operation “to make hold”

The text, almost certainly from Larsa, seems to be from c. 1750 bce and thus to belong to the early phase of the adoption of algebra by the southern scribe school

The topic of the problem is thus a rectangle. Lines I.36–37 tell us that the “heap” of its length and width is 7, while the lines I.32–35 state that “joining” half of the length and one third of the width to the “surface” produces 15:^{2}

The upper part of Figure 4.2 illustrates this situation, with 2 and 3 “inscribed as inscription” of
respectively
of the “projections”^{3} 1 of the length and the width (lines II.2–5); the heavily drawn configuration thus has an area equal to 15.

The solution *could* have followed the pattern of TMS IX #3

However, the present text does not proceed like that—Old Babylonian
algebra was a flexible instrument*physically* to the place where length and width (provided with widths
and
) are to be found. In this way it becomes possible to “cut off” the rectangle
—as long as it was elsewhere that would make no sense. In bottom of Figure (4.2), the area that is eliminated is drawn shaded and black: the rest, in white, will be equal to 11°30′.

Fig. 4.2: The reduction of AO 8862 #2.

In this operation, it is obvious that the (shaded) half of the length that had been “joined” according to the statement has been eliminated. However, more than the (equally shaded) third of the width has disappeared. How much more precisely?

It would be easy to subtract
from
, but that may not have been deemed sufficiently informative.^{4} In any case, the text introduces a detour by the phrase “Do not go beyond!” (the same verb as in the “subtraction by comparison”). A rectangle
(3,2) is constructed (perhaps one should imagine it in the corner where 2 and 3 are “inscribed” in Figure 4.2; in any case Figure 4.3 shows the situation). Without further argument it is seen that the half (three small squares) exceeds the third (two small squares) by one of six small squares, that is, by a sixth—another case of reasoning by “false position.”

Fig. 4.3

We thus know that, in addition to the third of the width, we have eliminated a piece
(*w*,10′) (drawn in black); if
, we therefore have

Fig. 4.4

Once more we therefore have a rectangle of which we know the area and the sum of length and width. The procedure is the same as in the final part of TMS IX #3^{5}

In the end, the final addition of the side of the square precedes the subtraction, as in TMS IX #3. Once more, indeed, it is not the same piece that is involved in the two operations; there is therefore no need to make it available before it is added.

## VAT 7532

Fig. 4.5: The diagram of VAT 7532. The “upper width” is to the left.

**Obv. **

**1**A trapezium. I have cut off a reed.

**2**1 sixty (along) the length I have gone. The 6th part

**3**broke off for me: 1‵12 to the length I have made follow.

**4**I turned back. The 3rd part and
kùš broke off for me:

**5**3 sixty (along) the upper width I have gone.

**6**With that which broke off for me I enlarged it:

**7**36 (along) the width I went. 1 bùr

**8**You, by your proceeding, (for) the reed which you do not know,

**9**1 may you posit. Its 6th part make break off, 50′ you leave.

**10**igi 50′ detach, 1°12′ to 1 sixty raise:

**11**1‵12 to
1‵12
join: 2‵24 the false length it gives you.

**12**(For) the reed which you do not know, 1 may you posit. Its 3rd part make break off,

**13**40′ to 3 sixty of the upper width raise:

**14**2‵ it gives you. 2‵ and 36 the lower width heap,

**15**2‵36 to 2‵24 the false length raise,
the false surface.

**16**The surface to 2 repeat,
to
raise

**17** it gives you. And
kùš which broke off

**18**to 3 sixty raise: 5 to 2‵24, the false length,

**19**raise: 12‵.
of 12‵ break, 6‵ make encounter,

**Rev. **

**1** to
join,
it gives you.

**2**By
,
is equal. 6‵ which you have left

**3**to
join,
it gives you. igi
,

**4**the false surface, I do not know. What to

**5**may I posit which
gives me? 25′ posit.

**6**Since the 6th part broke off before,

**7**6 inscribe: 1 make go away

**8**igi 5 detach, 12′ to 25 raise, 5′ it gives you
. 5′ to 25′ join:
nindan, the head of the reed it gives you.

This problem

Everything else, however—that is, that the area of the field is known before it is measured, and also the ways to indicate the measures of the pieces that break off from the reed—shows which
ruses the Old Babylonian school masters had to make use of in order to produce second-degree problems

For once, Figure 4.5 reproduces a diagram that is traced on the tablet itself. In general, as also here, diagrams are only drawn on the tablets

We thus undertake to measure the trapezium by means of a reed of unknown length . We manage to measure 1‵ reed lengths along the length of the trapezium before the reed loses a sixth of its length and is reduced to . What remains of the length turns out to be 1‵12 (lines Obv. 2–3).

Then the reed breaks for the second time. According to lines Obv. 4 and 5, the measure of the “upper width” (to the left)^{6} is 3‵
, where
kùš is the length of the reed after this second reduction.

The piece that broke off last is put back into place, and the “(lower) width” (evidently to the right) is paced out (line Obv. 7) as 36 *r*. Finally we learn that the area of the field is 1 bùr = 30‵ sar (1 sar = 1 nindan^{2}), see page

Lines Obv. 9–11 determine the length in units
by means of a false position

Fig. 4.6: The doubled trapezium of VAT 7532.

The text speaks of 2‵24 as the “false length,” that is, the length expressed in units .

Another false position

Since line Obv. 7 indicates that the false

Yet the text does not calculate this area: *The surface to 2 repeat.* Instead it doubles the trapezium so as to form a rectangle (see the left part of Figure 4.6), and the lines Obv. 14–16 calculate the area of this rectangle (the “false surface”

If the reed had not lost an ulterior piece of
kùš, we might now have found the solution by means of a final false position similar to that of BM 13901 #10^{2} (Obv. 16: *The surface to 2 repeat,
*). However, things are more complicated here. For each of the 3‵ steps made by the twice shortened reed a piece of
kùš is missing from our calculation, in total thus
kùš = 1‵ kùš = 5 nindan (1 kùš =
nindan): *And
kùš which broke off to 3 sixty raise: 5* (Obv. 17–18). Therefore the area of the real field does not correspond to what we see to the left in Figure 4.6 but to that which remains after elimination of the shaded strip to the right. The area of this strip is
: *5 to 2‵24, the false length, raise: 12‵*. The relation

This non-normalized equation is solved in the usual way. First it is multiplied by
: *
to
raise
it gives you *(Obv. 16–17). That leads to the normalized equation

or, with
*r* as unknown,

From here onward, the procedure coincides with that of BM 13901 #2

The area
corresponds to the rectangle of (height)
and breadth
. Half of the excess of the height over the breadth is “broken” and repositioned as seen in the diagram: lightly shaded in the original positions, heavily shaded where it is moved to. The construction of the completing square is described with one of the synonyms

Fig. 4.7

After the usual operations we find that
, and in line Rev. 5 that
. We observe, however, that the “moiety” that was moved around is not put back into its original position, which would have reconstituted
in the vertical direction. Instead, the other “moiety,” originally left in place, is also moved, which allows a horizontal reconstitution
: *6‵ which you have left to 2‶30‵ join, 2‶36‵ it gives you. ^{7}*

In the lines Rev. 6–8, the calculator introduces a third false position

One might believe this problem type to be one of the absolute favorites of the Old Babylonian teachers of sophisticated mathematics. We know four variants of it differing in the choice of numerical parameters. However, they all belong on only two tablets sharing a number of terminological particularities—for instance, the use of the logogram
for the “moiety,” and the habit that results are “given,” not (for example) “seen” or “coming up.” Both tablets are certainly products of the same locality and local tradition (according to the orthography based in Uruk), and probably come from the same school or even the same hand. A simpler variant with a rectangular field, however, is found in an earlier text of northern origin, and also in a text belonging together with the trapezium variants; if not *the* favorite, the broken reed was probably *a* favorite.

## TMS XIII

As TMS VII #2

**1**2 gur 2 pi 5 bán of oil I have bought. From the buying of 1 shekel of silver,

**2**4 silà, each (shekel), of oil I have cut away.

**3** mina of silver as profit I have seen. Corresponding to what

**4**have I bought and corresponding to what have I sold?

**5**You, 4 silà of oil posit and 40, (of the order of the) mina, the profit posit.

**6**igi 40 detach,
you see,
to 4 raise, 6′ you see.

**7**6′ to 12‵50, the oil, raise, 1‵17 you see.

**8** of 4 break, 2 you see, 2 make hold, 4 you see.

**9**4 to 1‵17 join, 1‵21 you see. What is equal? 9 is equal.

**10**9 the counterpart posit.
of 4 which you have cut away break, 2 you see.

**11**2 to the 1st 9 join, 11 you see; from the 2nd tear out,

**12**7 you see. 11 silà each (shekel) you have bought, 7 silà you have sold.

**13**Silver corresponding to what? What to 11 ^{¿}silà^{?} may I posit

**14**which 12‵50 of oil gives me? 1‵10 posit, 1 mina 10 shekel of
silver.

**15**By 7 silà each (shekel) which you sell of oil,

**16**that of 40 of silver corresponding to what? 40 to 7 raise,

**17**4‵40 you see, 4‵40 of oil.

This is another problem*mina* = 40 shekel of silver. For us, familiar with algebraic letter symbolism, it is easy to see that the total purchase price (the investment) must be
, the total sales price
, and the profit in consequence
. Multiplying by
we thus get the equation

and since , the system

This system—of the same type as the one proposed in YBC 6967*igûm-igibûm* problem (page

The cue to their method turns up towards the end of the text. Here the text first finds the total investment and next the profit *in oil* (4‵40 sìla). These calculations do not constitute a proof since these magnitudes are not among the data of the problem. Nor are they asked for, however. They must be of interest because they have played a role in the finding of the solution.

Figure 4.8 shows a possible and in its principles plausible interpretation. The total quantity of oil is represented by a rectangle, whose height corresponds to the total sales price in shekel, and whose breadth is the “sales rate” (sìla per shekel). The total sales price can be divided into profit (40 shekel) and investment (purchase price), and the quantity of oil similarly into the oil profit and the quantity whose sale returns the investment.

The ratio between the latter two quantities must coincide with that into which the quantity bought for one shekel was divided—that is, the ratio between 4 sìla and that which is sold for 1 shekel (thus ).

Modifying the vertical scale

Fig. 4.8: Geometric representation

On the whole, the final part of the procedure follows the model of YBC 6967

In YBC 6967, the *igûm-igibûm* problem (page

## BM 13901 #12

**Obv. II**

**27**The surfaces of my two confrontations I have heaped:
.

**28**My confrontations I have made hold: 10′.

**29**The moiety of
you break:
and
you make hold,

**30**^{8} is it. 10′ and 10′ you make hold,

**31**inside
you tear out: by
,
is equal.

**32** to one
you join: by 15′, 30′ is equal.

**33**30′ the first confrontation.

**34** inside the second
you tear out: by
, 20′ is equal.

**35**20′ 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

Fig. 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

Fig. 4.10: The diagram that corresponds to BM 13901 #8.

However, the author chooses a different method, showing thus the flexibility of the algebraic technique. He takes the two *areas*
and
as *sides* of a rectangle, whose area can be found by making 10′ and 10′ “hold” (see Figure 4.10):

In spite of the geometric character of the operations the Babylonians were thus quite aware that the area of a rectangle whose sides are the squares and coincides with that of a square whose side is the rectangle —which corresponds to our arithmetical rule .

We now have a rectangle for which we know the area and the sum of the two sides, as in the problems TMS IX #3

Fig. 4.11: The procedure used to solve the rectangle problem.

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*genuine algebra* on page

## BM 13901 #23

**Rev. II**

**11**About a surface, the four widths and the surface I have heaped,
.

**12**4, the four widths, you inscribe. igi 4 is 15′.

**13**15′ to
you raise:
you inscribe.

**14**1, the projection, you join: by
, 1°5′ is equal.

**15**1, the projection, which you have joined, you tear out: 5′ to two

**16**you repeat: 10′, nindan, confronts itself.

Whereas the previous problem illustrates the “modern” aspect of Old Babylonian mathematics, the present one seems to illustrate its archaic side—even though they come from the same tablet.

This is no real contradiction. The present problem*intentionally* archaic. In other words, it is *archaizing* and not truly archaic, which explains its appearance together with the “modern” problems of the same collection. The author is not modern and archaic at the same time, he shows his virtuosity by playing with archaisms. In several ways, the formulations that are used here seem to imitate the parlance of Akkadian surveyors*width*^{9} seems to be an abbreviated version of the characteristic formula introducing a mathematical riddle: “if somebody asks you thus about a surface …” (cf. pages ^{10} reflects an interest in what is *really there* and for what is *striking*, an interest that characterizes riddles

The problem can thus be expressed in the following way:

Figure 4.12 makes clear the procedure: 4*c* is represented by 4 rectangles
; the total
thus corresponds to the cross-shaped configuration where a “projection”

Lines 12–13 prescribe to cut out of the cross (demarcated by a dotted line) and the “joining” of a quadratic complement to the gnomon that results. There is no need to “make hold,” the sides of the complement are already there in the right position. But it is worthwhile to notice that it is the “projection” itself that is “joined”: it is hence no mere number but a quadratic configuration identified by its side.

Fig. 4.12: The procedure of BM 13901 #23.

The completion

This method is so different from anything else in the total corpus that Neugebauer

The archaizing aspect, it should be added, does not dominate completely. Line 12, asking first for the “inscription” of 4 and stating afterwards its igi

## TMS VIII #1

**1**The surface 10′. The 4th of the width to the width I have joined, to 3 I have gone … over

**2**the length 5′ went beyond. You, 4, of the fourth, as much as width posit. The fourth of 4 take, 1 you see.

**3**1 to 3 go, 3 you see. 4 fourths of the width to 3 join, 7 you see.

**4**7 as much as length posit. 5′ the going-beyond to the to-be-torn-out of the length posit. 7, of the length, to 4, ^{¿}of the width^{?}, raise,

**5**28 you see. 28, of the surfaces, to 10′ the surface raise, 4°40′ you see.

**6**5′, the to-be-torn-out of the length, to four, of the width, raise, 20′ you see.
break, 10′ you see. 10′ make hold,

**7** you see.
to 4°40′ join,
you see. What is equal? 2°10′ you see.

**8**10′ ^{¿}…^{?} to 2°10′ join, 2°20′ you see. What to 28, of the surfaces, may I posit which 2°20′ gives me?

**9**5′ posit. 5′ to 7 raise, 35′ you see. 5′, the to-be-torn-out of the length, from 35′ tear out,

**10**30′ you see, 30′ the length. 5′ the length to 4 of the width raise, 20′ you see, 20 the length (mistake for width).

In BM 13901 #12

Translated into symbols, the problem is the following;

(“to 3 I have gone” in line 1 means that the “joining” of
in line 1 is repeated thrice). The problem could have been solved in agreement with the methods used in TMS IX #3

Fig. 4.13: The method of TMS VIII #1.

However , once again the calculator shows that he has several
strings on his bow, and that he can choose between them as he finds
convenient*(You, 4, of the fourth, as much as width posit)*, and
the length prolonged by 5′ will be equal to 7, understood as 7*z* *(7 as much as length posit)*. Line 4 finds that the rectangle with sides 7
and 4
—in other words, the initial rectangle prolonged by 5′—consists of
small squares
.^{11} These 28 squares exceed the area 10′ by a certain number of sides (
), the determination of which is postponed until later. As usual, indeed, the non-normalized problem

is transformed into

Line 6 finds
, and from here onward everything follows the routine, as can be seen on Figure 4.14: 28
will be equal to 2°20′, and
hence to 5′.^{12} Therefore, the length
will be
, and the width
.

Fig. 4.14: Resolution of the normalized equation of TMS VIII #1.

## YBC 6504 #4

**Rev.**

**11**So much as length over width goes beyond, made encounter, from inside the surface I have torn out:

**12**. 20′ the width, its length what?

**13**20′ made encounter:
you posit.

**14** to
you join: 15′ you posit.

**15**By 15′, 30′ is equal. 30′, the length, you posit.

So far, everything we have looked at was mathematically correct,
apart from a few calculational and copying errors

The present text offers an example. Translated into symbols, the problem is the following:

Astonishingly, the length is found as that which “is equal by”

—that is, after a transformation and expressed in symbols, as
.

Fig. 4.15: The cut-and-paste operations of YBC 6504 #4.

The mistake seems difficult to explain, but inspection of the geometry of the argument reveals its origin (see Figure 4.15). On top the procedure is presented in distorted proportions; we see that the “joining” of
presupposes that the mutilated rectangle be cut along the dotted line and opened up as a pseudo-gnomon. It is clear that what results from the completion*not*
but instead—if one counts well—
. Below we see the same thing, but now in the proportions of the actual problem, and now the mistake is no longer glaring. Here,
and
, and therefore
. In consequence the mutilated rectangle opens up as a true gnomon, and the completed figure corresponds to
—but only because
.

This mistake illustrates an important aspect of the “naive”*given magnitudes* of a problem from what more they knew about it.

## Footnotes

As the “to-be-joined” of page *wuṣubbûm*) is derived from the verb “to join.”

We should observe that the half that appears here is treated as any other fraction, on an equal footing with the subsequent third. It is not a “moiety,” and the text finds it through multiplication by 30′, not by “breaking.”

Let us also take note that the half of the length and the third of the width are “joined” to the “surface,” not “heaped” together with it. A few other early texts share this characteristic. It seems that the surveyors thought in terms of “broad lines,” strips possessing a tacitly understood breadth of 1 length unit;
this practice is known from many pre-Modern surveying traditions, and agrees well with the Babylonian understanding of areas as “thick,” provided with an implicit height of 1 kùš (as inherent in the metrology of volumes, which coincides with that for areas—see page

The absence of this notion from the text should not prevent it from using it as a technical term of general validity.

Alternatively, the trick used by the text could be a leftover from the ways of surveyors not too familiar with the place-value system; or (a third possibility) the floating-point character of this system might make it preferable to avoid it in contexts where normal procedures for keeping track of orders of magnitude (whatever these normal procedures were) were not at hand.

It is not quite to be excluded that the text does not directly describe the construction but refers to the inscription twice of 3°25′ on a tablet for rough work, followed by the numerical product—cf. above, note 11, page

The position of the “upper” width to the left is a consequence of the new orientation of the cuneiform script (a counterclockwise rotation of 90°) mentioned in the box “Cuneiform writing.” On tablets, this rotation took place well before the Old Babylonian epoch, as a consequence of which one then wrote from left to right. But Old Babylonian scribes knew perfectly well that the *true* direction was vertically downwards—solemn inscriptions on stone (for example Hammurabi’s law) were still written in that way. For reading, scribes may well have turned their tablets 90° clockwise.

This distinction between two halves of which one is “left” is worth noticing as another proof of the geometric interpretation—it makes absolutely no sense unless understood spatially.

By error, line 30 of the text has instead of ; a partial product 25 has been inserted an extra time, which shows that the computation was made on a separate device where partial products would disappear from view once they had been inserted. This excludes writing on a clay surface and suggests instead some kind of reckoning board.

The error is carried over in the following steps, but when the square root is taken it disappears. The root was thus known in advance.

In the original, the word is “surface” marked by a phonetic complement indicating the accusative. An accusative in this position is without parallel, and seems to allow no interpretation but the one given here.

For once, the determinate article corresponds to the Akkadian, namely to an expression which is only used to speak about an inseparable plurality (such as “the four quarters of the world” or “the seven mortal sins”).

The use of a “raising” multiplication shows that the calculator does not construct a new rectangle but bases his procedure on a subdivision of what is already at hand—see the discussion and dismissal of a possible alternative interpretation of the procedure of BM 13901 #10 in note 5, page

Line 10 speaks of this as *5′ the length*—namely the side of the small square. Some other texts from Susa also speak of the side of a square as its “length.”