Ordering Operations in Square Root Extractions, Analyzing Some Early Medieval Sanskrit Mathematical Texts with the Help of Speech Act Theory

Abstract

Procedures for extracting square roots written in Sanskrit in two treatises and their commentaries from the fifth to the twelfth centuries are explored with the help of Textology and Speech Act Theory. An analysis of the number and order of the steps presented in these texts is used to show that their aims were not limited to only describing how to carry out the algorithm. The intentions of authors of these Sanskrit mathematical texts are questioned by taking into account the expressivity of relationships established between the world and the text.¹

Appendices

A Appendix 1: Different Steps in the Algorithm for Extracting Square Roots as Spelled Out in the Corpus

Taking into account all the steps detailed by the authors considered here (with an arbitrary filter—the mesh of the net may at times seem too small and at others too large- that is underlined in paragraph 5.2.3), thirteen steps for extracting a square root can be listed. Step 3, 6 and 12 state common tacit steps. The algorithm may be more efficiently illustrated in Fig. 5.2, although they are not equivalent renderings of the process.⁵⁷

• 1. The number whose square root is to be extracted is noted down in decimal place-value notation. Places are categorized with a grid that enables one to identify square powers of ten. Either positions for square and non-square powers of ten are listed or the series enumerating positions starting with the place with the lowest power of ten is considered. This list categorizes places as even or odd places.

• 2. The highest odd/square place is identified.

• 3. Consider (tacitly) the number made by all the digits to the left of the digit noted down in that place, that digit included.

• 4. Find the largest square contained in the number noted down to the left in the last/highest odd place.

  From here, onwards, one could also start by considering step 8, before turning to step 5 to 7.

• 5. Subtract the square from the number under consideration.

• 6. Replace (tacitly) the minuend by the remainder of the subtraction.

• 7. The root of the subtracted square is the first digit of the square root being extracted.

• 8. The root of this square (Ab family)/The double of the root of this square (PG family) is noted on the same line, to the left of the whole number/ on a line below the line of the number whose root is being extracted. In the PG family then, the doubling of the digit is a separate step in the process. The doubling does not necessarily need to take place immediately, one can note down the digit, and then double it just before it enters the division described in Step 10. This is what the APG recommends.

• 9. Consider the number whose highest digits are the previously noted remainder and which includes the next digit to its right.

• 10. Divide this number by (twice) the partial square root from Step 8.⁵⁸ In the following,

• 11. Replace the dividend with the remainder of the division. Then one should consider the next place on the right, which is a square/uneven place.

• 12. The quotient is the next digit in the partial square-root. It (Ab), or its double (PG), is thus noted down next to the previously found digit, as in Step 8. Its square is what will be subtracted from the number with the next digit as the process is iterated here from Step 5.

• 13. When there is no place on the right, the algorithm is finished. Examples only consider a process that extract a perfect square, consequently, either the square-root, or its double is obtained, according to the procedure followed. If we are in the latter case, the number obtained is halved.

B Appendix 2: Extracting the Square Root of 186 624

This is a numerical example addressed in APG. Footnotes and asterisks indicate non-attested forms.

• 1. The number whose square root is extracted is noted in decimal place-value notation. These decimal places are categorized using a grid: Square (varga), non-square (avarga) powers of ten (Ab), or even (sama, abr. sa) and odd (visama, abr. vi) place ranks - counted starting with the lowest power of ten- (BAB, PG, SYAB, APG).

  $\begin{array}{cccccc} avarga & varga & avarga & varga & avarga & varga \\ \\ sa & vi & sa & vi & sa & vi\\ \\ 10^{5} & 10^{4} & 10^{3} & 10^{2} & 10^{1} & 10^{0} \\ \\ {\bf 1}& {\bf 8} & {\bf 6} &{\bf 6} & {\bf 2} & {\bf 4}\\ \end{array}$

  $$186624= \mathbf{1}.10^{5}+\mathbf{8}.10^{4}+\mathbf{6}.10^{3}+\mathbf{6}.10^{2}+\mathbf{2}. 10^{1}+\mathbf{4}. 10^{0}$$

• 2. Subtract the square from the highest odd place

  The highest “odd” (viṣama) place or “square” (varga) place is 10⁴. The process starts by finding, by trial and error, the highest square number contained in the number noted to the left of this place. In this example, one looks for the highest square that will go into 18. And thus 18 − 4² is the operation carried out.

• 3. Replace the minuend with the reminder, 4, (BAB) or place the reminder, 4, below (APG).

• 4. The root, 4, of the subtracted square (16) is the first digit of the square root being extracted. The root of this square (Ab family-4)/ The double of the root of this square (PG family- 8) is noted down on the same line (BAB)/ or a separate line (PG).

  Bhāskara might have written:⁵⁹

  $$\begin{array}{cccccc} \ast 10^{5}& 10^{4} & 10^{3} & 10^{2} & 10^{1} & 10^{0} \\ & {\bf 4}/2 &6 & 6 & 2 & 4\end{array}$$

  While the APG writes:

  Because $1.10^{5}+ 8.10^{4}=[4.10^{2}]^{2}+2.10^{4}$,

  $186624= [\mathbf{4.10^{2}}]^{2}+2.10^{4}+6. 10^{3}+6.10^{2}+2.10^{1}+4. 10^{0}$.

• 5. Moving one place to the right, one should divide by twice the root.

  In this example, 26 is divided by 8: $26=8\times 3+2$. The quotient is 3, 2 is the remainder.

  This is then set down. Bhāskara’s style

  $$\begin{array}{cccccc} \ast & {\bf 4} &{\bf 3}/2 & 6 & 2 & 4\end{array}$$

  APG style

  $$\begin{array}{ccccc} \ast 2& 6 & 2 & 4 \\ & {\bf 8} & 3 & & \\ \end{array}$$

  In other words, because $2.10^{4}+6.10^{3}=8\times3.10^{3}+2.10^{3}$ $186624= [\mathbf{4.10^{2}}]^{2}+[2\times (4. 10^{2}) (3.10^{1})]+2.10^{3}+6.10^{2}+2.10^{1}+4. 10^{0}$.

• 6. Moving one place to the right, iterate. That is “subtract the square” again. This time the square of the quotient is subtracted. In this example $3^{2}$ is subtracted from 26: $26-9=17$. The remainder is 17. This is noted down again:

  Bhāskara style:

  In other words, writing $26-9=17$ according to the corresponding powers of ten.

  $186624= [\mathbf{4.10^{2}}]^{2}+[2\times (4.10^{2}) (3. 10^{1})] +[\mathbf{3. 10^{1}}]^{2} - [3. 10^{1}]^{2}+2.10^{3}+6.10^{2}+2.10^{1}+4. 10^{0}= [\mathbf{4.10^{2}}]^{2}+[2\times (4.10^{2}) (3. 10^{1})] +[\mathbf{3. 10^{1}}]^{2} +1.10^{3}+7.10^{2}+2.10^{1}+4.10^{0}$

• 7. Moving one place to the right, divide by twice the root. In this example one should divide 172 by $2\times43=86$: $172= 2\times 86$. The quotient is 2 and there is no remainder. This is noted

  Bhāskara style:

  $186624= [\mathbf{4. 10^{2}}]^{2}+[2\times (4. 10^{2}) (3. 10^{1})] +[\mathbf{3.10^{1}}]^{2} +2\times (2. 10^{0})(4.10^{2}+3. 10^{1})+4. 10^{0}= (\mathbf{4.10^{2}+3.10^{1}+2.10^{0}})^{2}$

• 8. The square root is 432. To end the procedure, moving one step to the right, one can “subtract the square of the quotient” ($2^{2}$):

  Bhāskara style: 4 3 2

  APG style:

  8 6 4

  $186624= (432)^{2}$

C Appendix 3: SYAB.2.4

⁶⁰He states (āha) a square root computation (vargamūlānayana) with an ārya⁶¹:

One should divide, repeatedly, the non-square [place] by twice the square root |

When the square has been subtracted from the square [place], the result is a root in a different place ||

In places where numbers are set-down (vinyāsa), the odd places have the technical name (samjñā) “square”. Even places have the technical name “non-square”. In this verse, when a square quantity is chosen (uddiṣṭa), having initially started by marking (cihnayitvā) the square and non-square places, when one is able to subtract (śodhayituṃ śakyate) the square of a special number- among those [squares of the digits] beginning with one and ending with nine- from the last square place, having subtracted (apāsya) that square; this special number is referred to [in the rule] as a root (mūlatvena gṛhṇīyāt). Consequently, here, the result which is a square root (vargamūlaphala) has been mentionned (ucyate). One should divide (bhāgaṃ haret) the next adjacent non-square place by twice that [root]. In this verse, when the square of this quotient has been subtracted (śuddhe) from the next adjacent square place, that quotient from the non-square place, in a different place, in the next square place, that [quotient] becomes (bhavati) the root.⁶² Also, when one has multiplied it (the quotient) by two (dviguṇīkṛtya), dividing (bhāgahāraṇa) in due order both [digits] from its adjacent non-square, as before, the computation of the third root [is accomplished]. Once again with three [digit numbers, the process is carried out]. In this way, one should perform (kuryād) [the process] until no square and non-square [place] remain (bhavanti).⁶³ When the root quantity has been obtained (labdhe), having multiplied it by two (dviguṇī kṛtaṃ), it should be halved (dalayet). Concerning fractions also, having divided (vibhajya) the square root of the numerator by the square root of the denominator the quotient⁶⁴ becomes (bhavanti) the root. One states (āha) in this way:

When the square root of the numerator has been extracted, and the root born from the denominator [also,] the root [is obtained](PG 34) |

In order to obtain the roots of the square which were previously explained (in the commentary of verse 3 which is on squares), setting down: 15 625. The result is the square root 125. Setting down the second: $\frac{4}{9}$. The root of the numerator 2, the root of the denominator 3, having divided (vibhajya) the numerator by that, the result is the square root of the fraction: $\frac{2}{3}$. Thus the fourth rule [has been explained].

D Appendix 4: APG

An algorithmic rule (karaṇasūtra) of two āryas for square roots:⁶⁵

PG.25. Having removed the square from the odd term, one should divide the remainder by twice the root that has dropped down to a place [and] insert the quotient on a line||

PG.26. Having subtracted the square of that, having moved the previous result that has been doubled, then, one should divide the remainder. [Finally] one should halve what has been doubled.||

What is the root of a given quantity whose nature is a square? This is the aim of that procedure. One should subtract (tyajet) a possible (saṃ bhavina) square, from the odd (viṣama) of the square quantity, from what is called odd (oja), that is from the first, third, fifth, or seventh etc., ; the places for one, one hundred, ten thousand, or one million, etc.; from the last term (pada), that is from among other places. This should provide (syāt) the root of that square which one should place (sthāpayed) beneath the place of decrease, the place the possible square is subtracted (śodhita) from that, one, a hundred, ten or thousands, etc., the last among the other places. And one should divide (bhāgam apaharet) from above (upariṣṭāt) by twice that, just there. The result should be inserted (viniveśayet) on a line, one should subtract (ś) the square of that from above that, and this should be doubled (dviguṇīkūryāt). If when this is doubled (dviguṇe kṛte) an additional place is created (jāyet), then it should be used (yojayet) as before when it is a result. Both have the quality of being a unique quantity (raśita). This quantity has the name “result”. One should repeat (utsārayet) this, thus one should divide (vibhajet), one should insert (viniveśayet) the result on a line, etc. as before in as much as the serpentine is possible (sarpaṇasaṃbhava)⁶⁶. When finished (samāpta) one should halve (dalayet) the whole result, thus obtaining the square root.

Thus for 186624, for which quantity is this a square?

In due order starting from the first place which consists of four, making (karaṇa) the names: “odd viṣama), even (sama), odd (viṣama), even (sama)”.

$$\begin{array}{*{20}{c}} {{\rm{Setting}}\,{\rm{down}}:}\\ {\begin{array}{*{20}{l}} {{\rm{sa}}}&{{\rm{vi}}}&{{\rm{sa}}}&{{\rm{vi}}}&{{\rm{sa}}}&{{\rm{vi}}}\\ 1&8&6&6&2&4 \end{array}} \end{array}$$

In this case, the odd terms which are the places for the ones, hundreds, and ten thousands, consist of four, six and eight. Therefore the last odd term is the ten thousand place which consists of eight. Then, the first quantity is 18. Having subtracted (apāsya) 16 since it is a possible square for these quantities, the remainder is two. That last quantity is placed separately (vyavatiṣthate) above. Thus, where it is placed (sthite sati) the root of 16, 4, times 2, 8, is to be led (neyaḥ) below the place where the square was subtracted (vargaśuddhi), which consists of six for the place of decrease. And then division (bhāgāpahāraḥ) of twenty six led above (uparitanyā). Setting down:

$$\begin{array}{*{20}{c}}2&6&6&2&4\\ {}&8&{}&{}&{} \end{array}.$$

When 24, which is eight multiplied by 3, is subtracted from below (that is from 26), above two remains. Below, the quotient which is three should be inserted (niveśya) on a line, they (e.g., these three units) should be placed (sthāpya) under consisting of six. Its square is nine. Having subtracted (śodhayitvā) this from above, these (1724) , three is multiplied by two, six is to be made (kartavya). Below, eighty six is produced. This quantity slithers (sarpati) on a line. Below two, there is [six], below seven, eight. Setting down:

$$\begin{array}{*{20}{c}} 1&7&2&4\\ {}&8&6&{}\end{array}$$

Division above of a 100 increased by 72 by that 86. Decreasing from above the dividend without remainder by two, the result is two, having inserted (niveśya) those two on a line, having placed (sthāpyau) both below four, its square is four; having subtracted (śodhayitvā) from above, those two multiplied by two should be made (kartavyā) four, therefore 800 increased by 64 is produced (jayite).

Since above the quantity subtracted has no remainder, there is no sliding like a snake etc. process, it remains just the halving of the quantity obtained which should be carried out. Thus, when that is done (kṛta), the result is 432. Its square is 186624.