Mesopotamian square root approximation by a sequence of rectangles

Abstract

The ancient Mesopotamians were able to estimate $\sqrt{2}$ as $1.41421\overline{296}$, in modern decimal notation. How they calculated this estimate is uncertain. We furnish Neugebauer and Sachs' original explanation with a new geometric description and reconstruct this estimate using contemporary reciprocal approximation.

There are many ways to approach the square root of 2. For example, we might imagine it as the positive solution to $x^2-2=0$, or as an irrational number like Hippasus tragically did according to legend.

The ancient Mesopotamians took a geometric approach and understood $\sqrt{2}$ as the diagonal of a unit square. Two numerical estimates were known at the time. In modern decimal notation, these are $1.41\overline{6}$ and $1.41421\overline{296}$.

Neugebauer and Sachs originally proposed ‘Heron's method’ was used to obtain these estimates. They were able to recover the first estimate, but not the second due to a problematic division.

We provide a new geometric interpretation of this proposal and resolve the division issue using contemporary reciprocal approximation. With these modifications, Neugebauer and Sachs' method does produce the second estimate.

1. The diagonal of a square

Practitioners of mathematics in ancient Mesopotamia were called scribes. Scribes performed their calculations using a sexagesimal (base sixty) number system. We represent their numbers as a sequence of Hindu-Arabic digits separated by colons, similar to the notation used for hours, minutes, and seconds in modern timekeeping. For ease of exposition, we use the semicolon as a radix point to separate the integer and fractional parts of a number. For example, in this notation 1; 24 : 51 : 10 means $1+\frac{24}{60}+\frac{51}{{60}^2}+\frac{10}{{60}^3}$which is $1.41421\overline{296}$ in decimal notation, and a very good approximation to $\sqrt{2}$.

Scribes had no written symbol equivalent to a radix point or zero digit during the Old Babylonian period (1900-1600 BCE). Sometimes the intended position of the radix point is clear, otherwise, its placement involves a degree of interpretation.

Consider the Old Babylonian student exercise YBC 7289 for example. First published by Neugebauer and Sachs (1945, 42–43), this tablet contains a sketch of a square and its diagonals as shown in Figure 1.

Figure 1. A square with diagonals. The side 30 is multiplied by the technical constant 1; 24 : 51 : 10 to obtain the diagonal 42; 25 : 35. Drawing by the author.

We find the number thirty along the side of the square. Due to the aforementioned absence of written symbols equivalent to a radix point or zero, this can be read as the integer 30, or the fraction $0;30=\frac{30}{60}=\frac{1}{2}$, or indeed as anything of the form $30\times {60}^n,n\in \mathbb{Z}$. It is not possible to know which of these values was intended by the scribe, so we leave it simply as an integer.

Next, we find the number 1; 24 : 51 : 10 in the centre of the square. We recognize this as the sexagesimal approximation for $\sqrt{2}$ above and so the position of the radix point is clear. The student is likely to have copied this number from a reference table of technical constants, such as YBC 7243, which describes it as the scaling factor that converts the side of a square into ‘the diagonal of a square’ (Neugebauer and Sachs 1945, 136; Fowler and Robson 1998, 370). Applied to our square, this scaling factor gives a diagonal of $30\times 1;24:51:10=42;25:35.$Indeed, 42; 25 : 35 is the third and final number from YBC 7289, written just below the diagonal.

The accuracy of this scaling factor is intriguing because it is correct to three sexagesimal places (or six decimal places), meaning the three sexagesimal digits after ‘;’ match the true value. There is little mystery about how this technical constant found its way onto a student's exercise because students were expected to copy such numbers from reference tables. The real question is: how was this estimate originally obtained?

2. Estimating the diagonal of a rectangle

The recombination text BM 96957 + VAT 6598 is an Old Babylonian list of thirty problems concerning rectangular prisms and rectangles from Sippar. Of interest here is question 18, which describes the method for approximating the diagonal of a gate (Neugebauer 1935–37, 277–287; Thureau-Dangin 1938, 130; Høyrup 2002, 268–272; Friberg 2007, 304–307). We describe what the tablet says and then offer a new geometric interpretation.

The question asks for the diagonal d of a rectangular gate. The height of the gate is specified as 1/2 nindan 2 kùš, and the breadth as 2 kùš. A ninda is equivalent to about 6 meters, and a kùš is equivalent to 1/12th of a nindan, or about 50 cm (Høyrup 2002, 17). These fractional lengths are first converted into their sexagesimal equivalents. Since 1/2 nindan $=0;30$ nindan and 1 kùš $=0;05$ nindan the gate has a height of $\frac{1}{2\mathrm{nindan}+2\mathrm{k}\stackrel{`}{u}\check{s}=0;30+2\times 0;05\mathrm{nindan}=0;40\mathrm{nindan}}$and breath of 0; 10 nindan.

According to the Pythagorean theorem, the sum of squares built on the sides of the gate equals the square built on the diagonal. In other, more modern, words $0;{10}^2+0;{40}^2=d^2$. The scribe's approach is to build a square with side d by adding the area $0;{10}^2$ onto the square with side 0; 40. The scribe describes this process as¹

You: square 0; 10 the breadth. You will see 0; 01 : 40, the area. Take the reciprocal of 0; 40, the height; multiply by 0; 01 : 40, the area. You will see 0; 02 : 30. Break in half 0; 02 : 30. You will see 0; 01 : 15. Add 0; 01 : 15 to 0; 40, the height. You will see 0; 41 : 15. The diagonal is 0; 41 : 15. The method.

These instructions were originally understood algebraically (Neugebauer 1935–37, 286). Here we follow Høyrup and interpret them as a sequence of geometric steps (Høyrup 2002, 268–272). Take two squares with sides 0; 40 and 0; 10 as shown in Figure 2.

Figure 2. Two squares with total area $A=0;{40}^2+0;{10}^2$.

First the area 0; 01; 40 of the small square is rearranged into a rectangle with sides 0; 40 and $0;01:40/0;40=0;02:30$ as shown in Figure 3.

Figure 3. The squares from Figure 2 are merged into a single rectangle.

Next, the scribe breaks off half this area and spreads it across an adjacent side, making Figure 3 more like a square. If the long and short sides of the rectangle in Figure 3 are $a_0=0;42:30$ and $b_0=0;40$ respectively, then this new shape has one side of length $a_1=b_0+\frac{a_0-b_0}{2}=0;40+0;01:15=0;41:15.$This process is equivalent to averaging the sides of the initial rectangle $\frac{a_1=1}{2(a_0+b_0)=0;41:15.}$This side approximates $\sqrt{A}=d$ because the shape has area A and is approximately a square.

Question 18 from BM 96957 + VAT 6598 finished without ever telling us exactly how half the area was spread across an adjacent side. So we do not know for certain what the final shape was. Some authors, including Høyrup, believe this should be a gnomon (Fowler and Robson 1998, 371–372; Høyrup 2002, 269). Here we entertain the possibility that the final shape was instead a rectangle and we compare the approximations produced by these two geometric interpretations.

2.1. Iteration

Starting from a rectangle with area A and sides $a_0,b_0$, we can apply the above diagonal approximation procedure to iteratively obtain a whole sequence of rectangles with constant area A and sides (1) $\frac{a_n=1}{2(a_{n-1}+b_{n-1})\mathrm{and}{b}_n=A/a_n\mathrm{for}n\in {\mathbb{Z}}^{+}.}$(1) Each rectangle in this sequence is progressively closer to a square, consequently each value of $a_n$ yields a better approximation to $\sqrt{A}$.

This procedure can continue indefinitely in theory, but not in practice. To pinpoint where the difficulty lies for a scribe we explore a hypothetical question: how many iterations could a scribe perform for question 18 from BM 96957 + VAT 6598?

Starting from a rectangle with area $A=0;{40}^2+0;{10}^2$ and sides $a_0=0;42:30$ and $b_0=0;40$, the first iteration produces a rectangle with sides $a_1=\frac{1}{2}(a_0+b_0)=0;41:15$ and $b_1=A/a_1=0;41:12:43:\dots$ as pictured in Figure 4.

Figure 4. A rectangle with area $A=0;{40}^2+0;{10}^2$ and sides $a_1=0;41:15,b_1=A/a_1=0;41:12:43:\dots$. Because the rectangle is approximately square, its sides approximate $\sqrt{A}$.

The second iteration produces a rectangle with side $\frac{a_2=1}{2(a_1+b_1)=0;41:13:51:49:05:\dots .}$We halt the procedure here because calculating $b_2=A/a_2$ is computationally prohibitive without modern tools.

Similarly, a scribe can only follow this procedure up to the point when division becomes intractable. All scribes could follow this procedure up to $a_1=0;41:15$ without difficulty and the evidence discussed below shows that some scribes could continue a little bit further.

3. Reciprocals of regular and irregular numbers

Scribes were not taught a dedicated division algorithm such as long division or Euclidean division. To a scribe, division of a number b by a number a was performed by finding the sexagesimal value of 1/a and multiplying it by b (Sachs 1952, 151). The number 1/a is referred to as the reciprocal of a and is denoted in modern literature by $\overline{a}$. In other words, the reciprocal of a is the number $\overline{a}$ such that $a\times \overline{a}=1.$Numbers whose reciprocals have finite sexagesimal expression are called regular. Other numbers, whose reciprocals have infinite sexagesimal expression, are called irregular. The method to obtain reciprocals was widely understood for numbers of the first kind, but not the second.

Scribes memorized the first few regular numbers and their reciprocals, which appear on the Standard Table of Reciprocals (Table 1). Division by one of these small regular numbers was simply a matter of multiplication by the reciprocal listed on this table, and this was ‘usually sufficient to carry out divisions in actual mathematical problems’ (Sachs 1952, 151). The zeros and radix points in Tables 1, 2 and 3 are not indicated in the original texts and are included for ease of exposition.²

Table 1. The standard table of reciprocals

a	$\overline{a}$	a	$\overline{a}$	a	$\overline{a}$	a	$\overline{a}$	a	$\overline{a}$
0; 02	30	0; 09	6;40	0; 20	3	0; 36	1; 40	1	1
0; 03	20	0; 10	6	0; 24	2; 30	0; 40	1; 30	1; 04	0; 56 : 15
0; 04	15	0; 12	5	0; 25	2; 24	0; 45	1; 20	1; 12${}^{\ast }$	0; 50${}^{\ast }$
0; 05	12	0; 15	4	0; 27	2; 13 : 20	0; 48	1; 15	1; 15${}^{\ast }$	0; 48${}^{\ast }$
0; 06	10	0; 16	3; 45	0; 30	2	0; 50	1; 12	1; 20${}^{\ast }$	0; 45${}^{\ast }$
0; 08	7;30	0; 18	3;20	0; 32	1; 52 : 30	0; 54	1; 06 : 40	1; 21	0; 44 : 26 : 40

Note: Some texts omit duplicate a and $\overline{a}$ pairs marked with ${}^{\ast }$ that appear elsewhere on the table. Zeros and radix points added by the author.

Table 2. M10: An Old Babylonian table of reciprocals for irregular numbers

n	Approximate reciprocal $\tilde{n}$	Description	Calculated to one place beyond text
0; 07	8;34 : 16 : 59	deficit	8;34 : 17 : 08 : 34
0; 07	8;34 : 18	excess	8;34 : 17 : 09
0; 11	5;27 : 16 : 22	deficit	5;27 :16 : 21 : 49
0; 13	4;36 : 55	deficit	4;36 : 55 : 23
0; 14	42; 51 : 25 : 40	deficit	42; 51 : 25 : 42 : 51
0; 17	35;17	deficit	35;17 : 39

Note: Zeros, radix points and final column added by the author.

Table 3. YBC 10529: An Old Babylonian table of reciprocals for regular and irregular numbers

n	Reciprocal $\overline{n}$ or $\tilde{n}$	Calculated to one place beyond text
0; 56	…
0; 57	…
0; 58	$1;02:04:\dots$	1; 02 : 04 : 08
0; 59	1; 01 : 01	1; 01 : 01 : 01
1	1
1; 01	0; 59 :00 : 59 : 00	0; 59 :00 : 59 : 00 : 59
1; 02	0; 58 : 03 : 52	0; 58 : 03 : 52 : 15
1; 03	$0;57:08:\mathbf{24}$	0; 57 : 08 : 34 : 17
1; 04	0; 56 : 15
1; 05	$0;55:23:04:\mathbf{30}$	0; 55 : 23 : 04 : 36 : 55
1; 06	$0;54:32:43:\mathbf{30}$	0; 54 : 32 : 43 : 38 : 11
1; 07	0; 53 : 43 : 52	0; 53 : 43 : 52 : 50
1; 08	$0;52:56:\mathbf{53}:14$	0; 52 : 56 : 28 : 14 : 07
1; 09	$0;52:10:\mathbf{28}$	0; 52 : 10 : 26 : 05
1; 10	0; 51 : 25 : 42	0; 51 : 25 : 42 : 51
1; 11	0; 50 : 42 : 15	0; 50 : 42 : 15 : 12
1; 12	0; 50
1; 13	$0;49:18:\mathbf{55}$	0; 49 : 18 : 54 : 14
1; 14	0; 48 : 38 : 55	0; 48 : 38 : 55 : 08
1; 15	0; 48
1; 16	0; 47 : 22 : 06	0; 47 : 22 : 06 : 19
1; 17	0; 46 : 45 : 11	0; 46 : 45 : 11 : 41
1; 18	0; 46 : 09 : 13 : 50	0; 46 : 09 : 13 : 50 : 46
1; 19	$0;45:34:\mathbf{09}:\mathbf{54}$	0; 45 : 34 : 10 : 37 : 58
1; 20	0; 45

Note: Numbers in bold font are inconsistent with the true value. Zeros, radix points and final column added by the author.

For example, consider the rectangle with side 0; 40 and area 0; 01 : 40 from question 18 of BM 96957 + VAT 6598. The other side of this rectangle is determined by division $0;01:40/0;40=0;02:30$, and the scribe explicitly uses a reciprocal to effect this calculation

Take the reciprocal of 0; 40, the height; multiply by 0; 01 : 40, the area. You will see 0; 02 : 30.

Indeed, if we take the reciprocal of 0; 40 from Table 1 and multiply it by the area then we do see $0;01:40\times 1;30=0;02:30.$Reciprocals of larger regular numbers were found by a well-understood process called ‘the Technique’ (Sachs 1947, 222–223). Hence, either directly from the Standard Table of Reciprocals or indirectly by the Technique, scribes could readily obtain the reciprocal of any regular number.

Reciprocals of irregular numbers are more troublesome on account of their endless sexagesimal expression. By far the most common approach was to assert they have igi nu, which translates to ‘no reciprocal’ (Nemet-Nejat 1995, 251; Proust 2012, 390 footnote 19). Scribal students were not expected to find, or use, the reciprocal of an irregular number. This topic lay outside the scope of scribal education, and school texts were deliberately ‘rigged in advance’ to avoid it (Sachs 1952, 151; Høyrup 2002, 29).

Only two tables of reciprocals for irregular numbers are currently known to exist: M10 and YBC 10529, both from the Old Babylonian period. We denote an approximate reciprocal of a by $\tilde{a}$, meaning $\tilde{a}$ is a finite sequence of sexagesimal digits such that $a\times \tilde{a}\approx 1.$M10 is from the John F. Lewis Collection of Cuneiform Tablets in The Free Library of Philadelphia and was published by Sachs (1952). This table fills the initial gaps in the Standard Table of Reciprocals, giving approximate values for the reciprocals of $0;07,0;11,0;13,0;14$ and 0; 17. These approximations are followed by the words SI-NI-ÍB (deficit) or dirig (excess), respectively indicating that the given approximation is too small or too large as seen in Table 2. For comparison, an extra column has been added that gives the approximations to one additional place.

Two reciprocals are given for 0; 07, and according to Sachs this clearly indicates what the scribe was thinking: the initial digits of this reciprocal are 8;34 : 17 because 8;34 : 16 : 59 is too small and 8;34 : 18 too large (Sachs 1952, 153).³ The reciprocals of 0; 14 and 0; 17 are too large by a factor of ten, and it is unclear why (Sachs 1952, 153). Sachs suggests this could be related to reciprocals of another kind found on VAT 3462 (Neugebauer 1935–37, 30) where $a\times \overline{a}=10$ is used to define reciprocals instead of $a\times \overline{a}=1$. The reciprocal of 0; 11 is described as being in deficit when is actually in excess.

YBC 10529 is a much larger table of reciprocals for regular and irregular numbers first published in 1945 by Neugebauer and Sachs (1945, 16–17). The top of this tablet is damaged, with about a third missing. What remains is a list of all numbers between 0; 56 and 1; 20 along with their approximate reciprocals as shown in Table 3. Scholars believe the original list may have started at 0; 40 (Bruins 1973, 184) or 0; 48 (Fowler and Robson 1998, 375 footnote 14). A starting value of 0; 45 would make aesthetic sense because this is the reciprocal of the final value 1; 20, but we will never know for certain. Seven entries are usually seen as errors because they are inconsistent with the true values. These are the approximate reciprocals of: 1; 03 and 1; 08, which appear to be simple copy errors; 1; 19, which appears to be a genuine computational error; and 1; 05, 1; 06, 1; 09 and 1; 13, which differ from the true value in the final digit. We should be hesitant to dismiss these last four as errors because further iterations of the (as of yet unknown) generation procedure might have rectified the issue.⁴ A possible method for generating these reciprocals was proposed by Bruins (1973) however Fowler and Robson find it unconvincing (Fowler and Robson 1998, 375 footnote 14). For now, we simply remark that most of the approximations are accurate to three sexagesimal places.

The topic of reciprocal approximation is absent from Old Babylonian educational texts (Sachs 1952, 151). Høyrup believes M10 and YBC 10529 might have been isolated exercises or were developed for real-life computations ‘where irregular divisors turned up’ (Høyrup 2002, 29–30). While it is possible these texts were exercises, the following discussion explores the latter possibility.

3.1. Iteration with reciprocal approximation

We now return to our hypothetical diagonal approximation example. Having obtained $a_1=0;41:15$ the next step is to compute $b_1=A\times \overline{a_1}$. Most scribes would be forced to abandon the procedure here because the reciprocal of $a_1=3;45\times 0;11$ does not exist, at least not exactly, on account of the irregular factor 0; 11. However, the scribe who authored M10 could approximate this reciprocal as $\begin{array}{rl}\widetilde{a_1}& =\overline{3;45}\times \widetilde{0;11}\\ & =0;16\times 5;27:16:22\\ & =1;27:16:21:52.\end{array}$Which can be used to find $b_1\approx A\times \widetilde{a_1}=0;41:12:43:39:33:20$and $\frac{a_2=1}{2(a_1+b_1)=0;41:13:51:49:46:40.}$Further iterations are infeasible because this value for $a_2$ contains a prime factor five sexagesimal digits long, and the reciprocal of such a large irregular number is well beyond contemporary tables of irregular reciprocals.

4. Comparative analysis of generation procedures

It is natural to expect the approximation 1; 24 : 51 : 10 for the ‘diagonal of a square’ was the result of the standard diagonal approximation procedure applied to a unit square. According to the Pythagorean theorem, the diagonal d satisfies $d^2=1^2+1^2$ and the procedure begins by combining these two squares into a single rectangle with area 2 and sides $a_0=2,b_0=1$. After one iteration we have the approximation $a_1=\frac{1}{2}(a_0+b_0)=1;30$ and further iterations are required for a more accurate result. There are different ways to proceed from here.

4.1. A sequence of rectangles

Neugebauer and Sachs originally suggested that multiple iterations of the diagonal approximation procedure were used to obtain $\sqrt{2}\approx 1;24:51:10$, although they described the method as an alternating sequence of ‘arithmetic and harmonic means’ (Neugebauer and Sachs 1945, 43) instead of geometrically as a sequence of rectangles. This is also known as ‘Heron's method’ because a single iteration occurs in Proposition 8 of Heron's Metrica.

The first few iterations of this procedure are given in Table 4 and are obtained without difficulty up to $a_2$. The value for $a_2=1;25$, or $1.41\overline{6}$ in decimal, is itself significant because this is ‘the coefficient of the diagonal of a square’ from the table of technical constants TMS 3 (Bruins and Rutten 1961; Friberg 2007, 450; Fowler and Robson 1998, 374) (see also AO 6484 Neugebauer 1935–37, 104). So this method immediately yields a contemporary approximation for $\sqrt{2}$, just not the one we were looking for.

Table 4. The first three iterations of ‘Heron's method’

n	$a_n$	$b_n$
0	2	1
1	1; 30	1; 20
2	1; 25	1; 24 : 42 : 20
3	1; 24 : 51 : 10

Note: An approximate reciprocal was used to compute $b_2$.

Further iteration requires the scribe to calculate $b_2=2/a_2$, which is difficult because $a_2=1;25$ is irregular and most scribes would be incapable of finding the required reciprocal. Neugebauer and Sachs did not attempt to explain how the calculations were made past this point. The idea that ‘some approximation was known’ for the reciprocal of 1; 25 is briefly mentioned by Fowler and Robson before being dismissed as it ‘would take us too far from our topic here’ (Fowler and Robson 1998, 374–375). Consider, however, that 1; 25 is close to the numbers seen on YBC 10529 and so it is reasonable to conjecture an approximate reciprocal was used, similar to the three-digit approximations found on that tablet: $\widetilde{a_2}=0;42:21:10.$This would enable a scribe to approximate $b_2=2/a_2$ as $b_2\approx 2\times \widetilde{a_2}=1;24:42:20$and then compute $\frac{a_3=1}{2(a_2+b_2)=1;24:51:10.}$In support of our conjecture, we note this is precisely the approximation that appears on the table of technical constants YBC 7243 and student exercise YBC 7289. Since $a_2$ and $a_3$ are contemporary approximations for $\sqrt{2}$ generated by successive iterations of the same procedure, they do indeed appear to be ‘links of the same chain’ (Neugebauer and Sachs 1945, 43). This is strong evidence that Neugebauer and Sachs were right all along. We have reinforced their hypothesis with a geometric description and a reciprocal approximation consistent with contemporary tables.

4.2. A sequence of gnomons

Fowler and Robson give a geometric interpretation based on a different understanding of question 18 from BM 96957 + VAT 6598. They suggest the rectangle in Figure 4 should be a gnomon, similar to contemporary completing-the-square style problems such as BM 13901 # 14 (Høyrup 2002, 73–77, 269) and YBC 6967 (Neugebauer and Sachs 1945, 129–130).

Their approach begins with a square whose side $c_0$ is chosen to be a regular number such that $c_0^2\approx A$. For $n\in {\mathbb{Z}}^{+}$, each subsequent gnomon is constructed from a square with side $c_{n-1}$. If $A>c_{n-1}^2$ then this square is enlarged into a gnomon, and otherwise reduced. Each gnomon has area A and a side length of (2) $c_n=\{\begin{array}{cc}\frac{c_{n-1}+1}{2(A-c_{n-1}^2)\overline{c_{n-1}}}& \mathrm{if}A>c_{n-1}^2\\ \frac{c_{n-1}-1}{2(c_{n-1}^2-A)\overline{c_{n-1}}}& \mathrm{otherwise}.\end{array}$(2) If negative numbers are allowed then these cases can be condensed into a single expression $\frac{c_n=c_{n-1}+1}{2(A-c_{n-1}^2)\overline{c_{n-1}}.}$However, Mesopotamian scribes were not willing to work with negative numbers and so we must keep these cases separate.

Each gnomon in this sequence is progressively closer to a square with area A, consequently each value of $c_n$ yields a better approximation to $\sqrt{A}$. As noted by Fowler and Robson, methods (1(1) $\frac{a_n=1}{2(a_{n-1}+b_{n-1})\mathrm{and}{b}_n=A/a_n\mathrm{for}n\in {\mathbb{Z}}^{+}.}$(1) ) and (2(2) $c_n=\{\begin{array}{cc}\frac{c_{n-1}+1}{2(A-c_{n-1}^2)\overline{c_{n-1}}}& \mathrm{if}A>c_{n-1}^2\\ \frac{c_{n-1}-1}{2(c_{n-1}^2-A)\overline{c_{n-1}}}& \mathrm{otherwise}.\end{array}$(2) ) are algebraically equivalent (Fowler and Robson 1998, 376). These methods initially yield identical results when used to find the diagonal of a unit square: $a_1=c_1=1;30$ and $a_2=c_2=1;25$. However, these methods are not numerically equivalent. This is apparent if we investigate what happens in the third iteration when reciprocal approximation is used.

With the approximation $\widetilde{1;25}=0;42:21:10$, the rectangle method yields $a_3=1;25:51:10$ but the gnomon method yields $\begin{array}{rl}{c}_3& \frac{\approx 1;25-1}{2(1;{25}^2-2)\times 0;42:21:10}\\ & =1;24:51:10:35:25.\end{array}$This is different to $a_3$ because the error introduced by the approximation affects different parts of Equations (1(1) $\frac{a_n=1}{2(a_{n-1}+b_{n-1})\mathrm{and}{b}_n=A/a_n\mathrm{for}n\in {\mathbb{Z}}^{+}.}$(1) ) and (2(2) $c_n=\{\begin{array}{cc}\frac{c_{n-1}+1}{2(A-c_{n-1}^2)\overline{c_{n-1}}}& \mathrm{if}A>c_{n-1}^2\\ \frac{c_{n-1}-1}{2(c_{n-1}^2-A)\overline{c_{n-1}}}& \mathrm{otherwise}.\end{array}$(2) ), which changes the result.

The gnomon method can be made to yield $c_3=1;24:51:10$ if we use the approximation $\widetilde{1;25}=0;42:24$. However, this approximation is only accurate in the first one and a half digits, which is worse than the worst approximation from M10 and so seems inconsistent with contemporary reciprocal approximation.

Fowler and Robson's preferred approach is to replace 1; 25 with a nearby regular number, such as 1; 25 : 20 (Fowler and Robson 1998, 375). This strategy is a ‘good guess’ according to Høyrup (2002, 262). However, the best approximation achieved by their approach is (3) $\begin{array}{rl}{c}_3& \approx 1;25:20-\frac{1}{2}(1;25:{20}^2-2)\times \overline{1;25:20}\\ & =1;24:51:15.\end{array}$(3) We can improve the accuracy of this approach by replacing only the offending reciprocal with $\overline{1;25:20}$ while keeping other occurrences of 1; 25 intact. However, this is still insufficient $\begin{array}{rl}{c}_3& \approx 1;25-\frac{1}{2}(1;{25}^2-2)\times \overline{1;25:20}\\ & =1;24:51:12:39:22:30.\end{array}$Both methods (1(1) $\frac{a_n=1}{2(a_{n-1}+b_{n-1})\mathrm{and}{b}_n=A/a_n\mathrm{for}n\in {\mathbb{Z}}^{+}.}$(1) ) and (2(2) $c_n=\{\begin{array}{cc}\frac{c_{n-1}+1}{2(A-c_{n-1}^2)\overline{c_{n-1}}}& \mathrm{if}A>c_{n-1}^2\\ \frac{c_{n-1}-1}{2(c_{n-1}^2-A)\overline{c_{n-1}}}& \mathrm{otherwise}.\end{array}$(2) ) can produce the desired estimate for $\sqrt{2}$ by approximating the reciprocal of 1; 25. The former uses an approximation consistent with contemporary tables, while the latter does not. So it appears the former is more likely to be correct. Reciprocal approximation has allowed us to reconstruct the desired estimate $\sqrt{2}\approx 1;24:51:10$, and distinguish between algebraically equivalent interpretations of the diagonal approximation procedure.

4.3. Buckle's approximation

Following Fowler and Robson, Buckle (2023) proposes 1; 25 was replaced by the regular number 1; 24 : 22 : 30. Using this value he produces the same result as Equation (3(3) $\begin{array}{rl}{c}_3& \approx 1;25:20-\frac{1}{2}(1;25:{20}^2-2)\times \overline{1;25:20}\\ & =1;24:51:15.\end{array}$(3) ) $\frac{1}{2(1;24:22:30+2\times \overline{1;24:22:30})=1;24:51:15.}$Buckle then uses additional steps and observations to modify this result into 1; 24 : 51 : 10. He argues that if the scribe were to check the accuracy of their result by squaring they would notice $(1;24:51:15{)}^2=2;00:00:13:46:33:45.$While computing this square, the scribe may also have noticed $(1;24:51{)}^2=1;59:59:31:21.$From this information, it is possible to deduce $1;24:51<\sqrt{2}<1;24:51:15.$Hence the fourth digit of this approximation must be between 0 and 15, so only a handful of possible values need to be checked to find the optimal four-digit approximation 1; 24 : 51 : 10.

All of the above proposals agree that the scribe was faced with finding the reciprocal of 1; 25. Buckle, Fowler and Robson believe 1; 25 was discarded in favour of a nearby regular number. Whereas we propose the reciprocal of 1; 25 was approximated directly, as per M10 and YBC 10529.

4.4. Side and diagonal numbers

Høyrup offers an alternative hypothesis and suggests the procedure might have been based on the algorithm of ‘side-and-diagonal’ numbers known from Theon of Smyrna (Høyrup 2002, 263). Starting with an initial approximation of $s_1=1,d_1=1$ the algorithm generates new side $s_{n+1}=s_n+d_n$ and diagonal $d_{n+1}=s_n+2d_n$ numbers. After three iterations we have $s_4=12$ and $d_4=17$ which is in the ratio $17/12=1;25$. After another four iterations, the sides are (in decimal) $s_8=408,d_8=577$ which is in the ratio $577/408\approx 1;24:51:10:35:\dots$ and can be truncated to 1; 24 : 51 : 10. Even if this method was known at the time, the decimal number 408 is irregular in base sixty and so we still encounter the issue of irregular division.

5. Conclusion

Back in 1946, Sachs wrote that ‘we have as yet no mathematical problem-text which explicitly uses approximate values for reciprocals of irregular numbers’ (Sachs 1946, 213) and the truth of this statement remains unchallenged. Students were not expected to approximate the reciprocals of irregular numbers, but nor were they expected to generate technical constants. Generation of technical constants would have the domain of senior scribes.

To be precise, the technical geometric constant known as the ‘diagonal of a square’ seems to have been generated by the standard diagonal approximation procedure applied to a unit square. The first iteration produces $\sqrt{2}\approx 1;30$. The second iteration produces $\sqrt{2}\approx 1;25$ as found on the table of technical constants TMS 3. One further iteration produces $\sqrt{2}\approx 1;24:51:10$ as found on the table of technical constants YBC 7243 and repeated on the student exercise YBC 7289. This final iteration seems to have used a reciprocal approximation technique that was known at the time, but not widely.

The most impressive part of this approximation is not its accuracy but the mathematical ability of the scribe who found it. In particular, all scribes knew how to find the reciprocal of a regular number. Tables such as M10 and YBC 10529 show that some scribes had begun to extend this idea to irregular numbers. It now seems this advanced understanding was more than an isolated incident or exercise. Instead, it was used for real purposes such as improving the accuracy of technical constants.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Notes

1 This translation is based on Fowler and Robson (1998, 371–372). The exception is our translation of qà-qá-ri/qà-qá-ra. This has been variously translated as ‘Fläche/area’ (Neugebauer 1935–37, 282), ‘ground’ (Friberg 2007, 304; Høyrup 2002, 268) and ‘base’ (Fowler and Robson 1998, 371–372). The first translation seems to fit best.

2 The radix point can be placed in many different positions for reciprocals. This is because if a and $\overline{a}$ are reciprocals then so are $a\times {60}^n$ and $\overline{a}\times {60}^{-n}$ for any integer $n\in \mathbb{Z}$.

3 This logic assumes the scribe knows enough to dismiss any number that begins with 8;34 : 16 : 59, such as 8;34 : 16 : 59 : 30. So it seems the scribe knows more than is stated on M10.

4 The approximate reciprocal of 1; 16 given by Neugebauer and Sachs is incorrect in the final digit (Neugebauer and Sachs 1945, 16) and this error was copied by Bruins (1973, 178). This just goes to show that sexagesimal arithmetic is difficult, and if such great minds are capable of computational and copy errors then we should not judge ancient scribes harshly when they do the same.